Introduction to Mathematical Physics Method & Concept

Introduction to Mathematical Physics

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Introduction to Mathematical Physics Methods and Concepts Second Edition

Chun Wa Wong Department of Physics and Astronomy University of California Los Angeles

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Great Clarendon Street, Oxford, OX2 6DP, United Kingdom Oxford University Press is a department of the University of Oxford. It furthers the University’s objective of excellence in research, scholarship, and education by publishing worldwide. Oxford is a registered trade mark of Oxford University Press in the UK and in certain other countries c Chun Wa Wong 2013  The moral rights of the author have been asserted First Edition published in 1991 Second Edition published in 2013 Impression: 1 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, by licence or under terms agreed with the appropriate reprographics rights organization. Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above You must not circulate this work in any other form and you must impose this same condition on any acquirer British Library Cataloguing in Publication Data Data available ISBN 978–0–19–964139–0 Printed and bound by CPI Group (UK) Ltd, Croydon, CR0 4YY

Preface to the second edition

In this second edition of a book for undergraduate students of physics, two long chapters of more advanced topics have been added. One of these chapters deals with the unfolding of invariant Lorentz spacetime scalars into 4D spacetime vectors in relativistic square-root spaces. It tells the story of how Hamilton’s 3D generalization of the complex number, Pauli’s spin matrices, Dirac’s relativistic wave equation, and other actors, including spinor wave functions and tensor operators, play on Einstein’s spacetime stage where physical events take place. The other chapter is concerned with nonlinear phenomena in physics. Nonlinearity has two strangely contradictory personalities. It can cause irregularities, instabilities, turbulence and chaos. Under other circumstances it can give rise instead to unusual coherence and collectivity. We shall explain how these seemingly contradictory manifestations can be understood conceptually by using sometimes analytic and sometimes numeric tools. These added topics are of interest to advanced undergraduate and beginning graduate students. Seven short tutorials on some basic topics of college mathematics have been appended for review or reference. With the all-pervasive personal computer in many academic toolboxes, it is now possible to undertake rather extensive numerical calculations and even formal mathematical manipulations using computer algebra systems such as Mathematica. Two appendices on computer algebra systems in general and Mathematica in particular have been included to introduce the reader to this important class of tools in mathematical physics. A section has been added to describe Internet resources on mathematical physics at this introductory level. The old book list and bibliography of the first edition have also been expanded. I have taken this opportunity to improve and add to the text of the original chapters and to correct the many misprints present in the first edition. Hints have been added to many of the problems. An Instructor’s Solutions Manual is available to instructors who have chosen the book for course adoption. I appeal to their good will to refrain from posting this solutions manual on the Web, so that the book can serve its educational purposes. Several readers and students have taken the trouble to call my attention to misprints in the first edition. I thank them for their kindness. One of them deserves a special mention. L. C. Jones sent me a particularly long list of misprints covering the first two chapters when the first edition first came out. The drawings of Newton and Einstein shown in the endpapers are by the noted Argentine artist Iutta Waloschek. I thank her for permission to use them in the book. Los Angeles February 2012

C. W. W.

Preface to the first edition

This book is based on lecture notes for two undergraduate courses on mathematical methods of physics that I have given at UCLA during the past twenty years. Most of the introductory topics in each chapter have been used at one time or another in these courses. The first of these courses was intended to be a beginning junior course to be taken before the physics core courses, while the second was an elective course for seniors. These courses have evolved over the years in response to the perceived needs of a changing student population. Our junior course is now a prerequisite for junior courses on electricity and magnetism and on quantum mechanics, but not for those on analytic mechanics. Our elective senior course is now concerned solely with functions of a complex variable. In this format I am able to cover in one quarter most of the easier sections of the first three chapters in our junior course, and the entire Chapter 6 in the senior course. Most of our students are physics majors, but many are from engineering and chemistry, especially in the senior course. The latter course is also recommended to our own first-year graduate students who have not been exposed to complex analysis before. The idea of teaching mathematical physics as a subject separate from the physics core courses is to help the students to appreciate the mathematical basis of physical theories and to acquire the expected level of competence in mathematical manipulations. I believe that our courses, like similar courses in other universities, have been quite successful. This is not to say that these courses are easy to teach. My experience has been that our junior course on mathematical methods of physics is one of the most difficult undergraduate courses to teach. Several factors combine to make it so challenging, including the diversity in the background and abilities of our students and the large number of topics that one would like to cover. Not the least of these factors is an adequate list of available textbooks at the right level and written in the right style. It is hoped that this book will address this need. Each chapter in this book deals with a single subject. Chapter 1 on vectors and fields in space is concerned with the vector calculus needed for a study of electricity and magnetism. Chapter 2 on transformations, matrices and operators contains a number of topics in algebra of special importance to the study of both classical and quantum mechanics. It is also concerned with the general mathematical structure of laws of physics. Chapter 3 on Fourier series and Fourier transforms prepares the students for their study of quantum mechanics. The treatment of differential equations in Physics in Chapter 4 covers most of the basic mathematical concepts and analytic techniques needed to solve many equations of motion or equations of state in physics. Special functions are covered in Chapter 5 with emphasis on special techniques with which the properties of these functions can be extracted. Finally, Chapter 6 on functions of a complex variable gives a detailed introduction to complex analysis, which is so basic to the understanding of functions and their manipulations. This chapter provides a firmer mathematical foundation to students

Preface to the first edition

vii

who intend to go on to graduate studies in the physical sciences. Many topics have been left out in order to have a book of manageable size. These include infinite series, tensor analysis, probability theory, the calculus of variations, numerical analysis and computer mathematics. The style of writing and the level of difficulty differ in different chapters, and even in different sections of the same chapter. As a rule, the pace is more leisurely and the derivations are more detailed in the more basic sections that have been used more heavily. My experience has been that even the most detailed derivation is not adequate for some of the students. There is really no substitute for a patient and perceptive instructor. On the other hand, the more advanced sections have been written in a rather concise style on the assumption that the readers who might want to read them can be expected to be more experienced and fearless; they might be able to bridge or tolerate most of the missing steps without too much anguish. Conciseness might also be a virtue if the book is to be used as a reference. It is hoped that, after the book has been used as a textbook, it will remain on the bookshelf as a reference book. I am grateful to the many students who have taken my courses and used previous versions of my lecture notes. Their criticism and suggestions have helped me improve the text in numerous places. I want to thank the many teaching assistants who have been associated with these courses for their help with problems. I want to thank editors, reviewers and colleagues who have read or used this book for their advice and suggestions. Two persons in particular must not remain unnamed. Mrs. Beatrice Blonsky typed successive versions of this book over a period of many years; but for her enthusiasm, this book would not have taken shape. Mr. Ron Bohm worked wonders at the daunting task of entering the manuscript into the computer. To each of these persons, named as well as unnamed, I want to express my deep and sincere appreciation. Los Angeles July 1990

C. W. W

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Contents

1 Vectors and fields in space 1.1 Concepts of space 1.2 Vectors in space 1.3 Permutation symbols 1.4 Vector differentiation of a scalar field 1.5 Vector differentiation of a vector field 1.6 Path-dependent scalar and vector integrations 1.7 Flux, divergence and Gauss’s theorem 1.8 Circulation, curl and Stokes’s theorem 1.9 Helmholtz’s theorem 1.10 Orthogonal curvilinear coordinate systems 1.11 Vector differential operators in orthogonal curvilinear coordinate systems Appendix 1 Tables of mathematical formulas

1 1 4 14 20 25 31 42 48 53 56 65 72

2 Transformations, matrices and operators 2.1 Transformations and the laws of physics 2.2 Rotations in space: Matrices 2.3 Determinant and matrix inversion 2.4 Homogeneous equations 2.5 The matrix eigenvalue problem 2.6 Generalized matrix eigenvalue problems 2.7 Eigenvalues and eigenvectors of Hermitian matrices 2.8 The wave equation 2.9 Displacement in time and translation in space: Infinitesimal generators 2.10 Rotation operators 2.11 Matrix groups Appendix 2 Tables of mathematical formulas

117 125 129 135

3 Relativistic square-root spaces∗ 3.1 Introduction 3.2 Special relativity and Lorentz transformations 3.3 Relativistic kinematics and the mass–energy equivalence 3.4 Quaternions 3.5 Dirac equation, spinors and matrices 3.6 Symmetries of the Dirac equation∗

138 138 139 150 159 165 172

∗ Marks

an advanced topic in Contents, or a long or difficult problem in the chapters.

76 76 77 87 93 97 104 108 114

x

Contents

3.7 Weyl and Majorana spinors, symmetry violations∗ 3.8 Lorentz group 3.9 Cartan spinors and spin transformations in square-root space 3.10 Dyadics 3.11 Cartesian tensors 3.12 Tensor analysis Appendix 3 Tables of mathematical formulas

179 188 195 200 206 217 232

4 Fourier series and Fourier transforms 4.1 Wave–particle duality: Quantum mechanics 4.2 Fourier series 4.3 Fourier coefficients and Fourier-series representation 4.4 Complex Fourier series and the Dirac δ function 4.5 Fourier transform 4.6 Green function and convolution 4.7 Heisenberg’s uncertainty principle 4.8 Conjugate variables and operators in wave mechanics 4.9 Generalized Fourier series and Legendre polynomials 4.10 Orthogonal functions and orthogonal polynomials 4.11 Mean-square error and mean-square convergence 4.12 Convergence of Fourier series 4.13 Maxwell equations in Fourier spaces 4.14 3D Fourier transforms: Helmholtz decomposition theorem Appendix 4A Short table of Fourier cosine series Appendix 4B Short table of Fourier sine series Appendix 4C Short table of Fourier transforms Appendix 4D Short table of 3D and 4D Fourier transforms Appendix 4E Tables of mathematical formulas

244 244 247 250 258 265 269 273 276 280 287 292 295 299 305 313 313 314 314 315

5 Differential equations in physics 5.1 Introduction 5.2 Linear differential equations 5.3 First-order differential equations 5.4 Second-order linear differential equations 5.5 The second homogeneous solution and an inhomogeneous solution 5.6 Green functions 5.7 Series solution of the homogeneous second-order linear differential equation 5.8 Differential eigenvalue equations and orthogonal functions 5.9 Partial differential equations of physics 5.10 Separation of variables and eigenfunction expansions 5.11 Boundary and initial conditions 5.12 Separation of variables for the Laplacian 5.13 Green functions for partial differential equations Appendix 5 Tables of mathematical formulas

319 319 321 324 328 332 337 342 347 350 351 354 359 364 368

Contents

xi

6 Nonlinear systems∗ 6.1 Introduction 6.2 Nonlinear instabilities 6.3 Logistic map and chaos 6.4 Strange attractor 6.5 Driven dissipative linear pendula 6.6 Chaos in parametrically driven dissipative nonlinear pendula 6.7 Solitons 6.8 Traveling kinks 6.9 Nonlinear superposition of solitons 6.10 More general methods for multi-solitons∗ Appendix 6 Tables of mathematical formulas

373 373 374 382 392 400 407 415 424 432 440 451

7 Special functions 7.1 Introduction 7.2 Generating function for Legendre polynomials 7.3 Hermite polynomials and the quantum oscillator 7.4 Orthogonal polynomials 7.5 Classical orthogonal polynomials∗ 7.6 Associated Legendre polynomials and spherical harmonics 7.7 Bessel functions 7.8 Sturm-Liouville equation and eigenfunction expansions Appendix 7 Tables of mathematical formulas

458 458 458 464 469 476 481 487 495 498

8 Functions of a complex variable 8.1 Introduction 8.2 Functions of a complex variable 8.3 Multivalued functions and Riemann surfaces 8.4 Complex differentiation: Analytic functions and singularities 8.5 Complex integration: Cauchy integral theorem and integral formula 8.6 Harmonic functions in the plane 8.7 Taylor series and analytic continuation 8.8 Laurent series 8.9 Residues 8.10 Complex integration: Calculus of residues 8.11 Poles on the contour and Green functions 8.12 Laplace transform 8.13 Inverse Laplace transform 8.14 Construction of functions and dispersion relations 8.15 Asymptotic expansions∗ Appendix 8 Tables of mathematical formulas

502 502 502 509 521 524 528 533 540 546 550 561 571 577 584 590 614

xii

Contents

Appendix A Tutorials A.1 Complex algebra A.2 Vectors A.3 Simple and partial differentiations A.4 Simple and multiple integrals A.5 Matrices and determinants A.6 Infinite series A.7 Exponential functions

620 620 627 630 636 643 650 662

Appendix B Mathematica and other computer algebra systems

670

Appendix C Computer algebra (CA) with Mathematica C.1 Introduction to CA C.2 Equation solvers C.3 Drawing figures and graphs C.4 Number-intensive calculations

677 677 679 683 684

Resources for students

688

Bibliography

694

Name index

699

Subject index

702

1 Vectors and fields in space 1.1

Concepts of space

Physics is concerned with the objective description of recurrent physical phenomena in space. Our intuitive appreciation of space arose in primordial times from the territorial imperative (i.e., the need to secure our immediate environment for our own safety and benefit). In time, it found a more objective realization in response to the search for rigor in reasoning in science and philosophy, and to the practical needs of mensuration in irrigation, construction and navigation. Most of the great mathematicians of antiquity were geometers; we still remember the Elements of Euclid, the geometrical inventiveness of Archimedes, and perhaps also the conic sections of Apollonius. The Greek mathematicians were followed by the Arabs, who made contributions in arithmetic and algebra. The introduction of Arabic textbooks on mathematics into Europe during the Renaissance stimulated the study of algebra. This in turn led to two important developments in the seventeenth century that laid the foundation of modem mathematics—the invention of analytic geometry and calculus. Both are crucial to the description of physical events in space. In his book G´eom´etrie (1637), Descartes brought the power of algebra to bear on geometrical problems. The book had a significant impact on the gradual evolution of calculus, which was established later in the century by Newton and by Leibniz. Newton is the pre-eminent genius of the ages. He made important discoveries in geometry, algebra and calculus. We remember him even more for his contributions in physics. Physics before Newton’s time was composed of many empirical facts concerning physical phenomena both on earth and in the sky. Two major developments occurred in the century before Newton: Galileo had studied inertia, acceleration and falling bodies, and Kepler had abstracted three empirical laws of planetary motion from the astronomical observations of Tycho Brahe. In his famous book Principia Mathematica Philosophiae Naturalis, published in 1687, Newton presented a unified theory of these diverse phenomena in the form of three laws of motion. The theory stated that, given the initial state of motion of a system in space and the forces acting on it, the state of motion at all other times could be computed exactly. He demonstrated the validity of his theory by showing that Kepler’s empirical laws of planetary motion were consequences of a law of forces between massive objects— the law of universal gravitation. This deterministic or mechanical view of dynamics

2

Vectors and fields in space

was highly successful; it was to provide the theoretical basis for the understanding of physical phenomena up to the end of the nineteenth century. In its simplest form, Newtonian mechanics describes the motion of a point mass, say at position r(t). The mass may undergo acceleration as the result of the imposition of an external, action-at-a-distance force, provided, for example, by a second mass at a different point r in space. In other words, there are two gravitating masses. The rest may be considered empty space. The Newtonian concept of space proved quite adequate until the nineteenth century. Then a self-educated genius materialized in the person of Faraday, to make fundamental experimental discoveries on electromagnetic phenomena. Lacking the mathematical tools to handle the complicated interactions among magnets, charges and currents, he made progress by visualizing them instead as lines of force emanating from these objects and forming webs or fields all over space, including the location at which the interaction took place. Faraday’s intuitive but successful picture of lines of force was given respectability by the mathematical physicist Maxwell. Stimulated by the work of Faraday and others, he proposed in 1864 that electromagnetic phenomena could be described by four mathematical equations, the Maxwell equations, satisfied by an electric field and a magnetic field. These fields were supposed to reside in a medium called the ether, which pervaded all space, including regions of vacuum. It was supposed to be capable of receiving and storing energy and of being set into vibrations. The speed of the resulting electromagnetic disturbance could be computed and was found to be close to that of light. Maxwell suggested that light itself was an electromagnetic wave, because its polarization was known to be affected by the electromagnetic properties of a medium. It was not long before nonoptical electromagnetic waves themselves were generated in the laboratory. This was achieved by Hertz, who found that they propagated with the expected speed. In this way, the reality of the electromagnetic fields in all space was convincingly demonstrated. Henceforth, physical space assumes a character rather different from the old mechanistic one, in which the only relevant points were those at which forces acted on masses. A fundamental question was soon raised (by Maxwell in 1879). What was the motion of the earth relative to the ether? A definitive answer was given in 1887 by Michelson and Morley, who could find no relative motion. The question of relative or absolute motion is an ancient one. The sun, the moon, the planets, and the stars march regularly across the sky once a day. The rational men of ancient Greece realized that it could be the earth rather than the heavens that rotated. The sun, the moon, and the planets also move regularly with respect to the stars, the sun’s motion being an annual one. This was interpreted by some to mean that the sun moved around the earth—a view that found a quantitative expression in Ptolemy’s model of planetary motion in the second century. By the sixteenth century the accumulated inaccuracies of the model forced Copernicus to quantify the alternative Greek view that the earth and the planets moved around the sun instead. After Newton, the dynamics of planetary motion was finally understood. It was found to be one of many manifestations of a universal dynamical theory of physical

Concepts of space

3

phenomena Forces caused “absolute” motion, which was the same in all inertial frames moving relative to one another with constant velocities. At the same time, Newton still used the concept of “immovable” space, even though it had no dynamical significance. The concept of time to him was of course entirely independent of that of space. In 1905 a bright star burst upon physics. An unknown patent examiner named Einstein published three fundamental papers on physics. Of these, the work on the electrodynamics of moving bodies was perhaps the most revolutionary. In this paper, he pointed out that Maxwell’s highly successful theory of electrodynamics, like Newton’s dynamics, depended only on relative motion and not on the idea of absolute rest. He then generalized this observation to the principle of relativity by stating that the laws of physics are the same in all inertial frames in which the equations of mechanics hold good and that the idea of absolute rest is superfluous. He further postulated that the speed of light in vacuum is independent of the motion of the source or of the observer. These postulates led him to a startling conclusion. Not only is there no absolute space, there is also no absolute time. That is, on going from a stationary frame to a moving frame of reference, one should find both space coordinates and time changed. In this theory, which Einstein called the Special Theory of Relativity, physics deals with physical events not in space, but in spacetime. The consequences of special relativity, many of which were worked out by Einstein himself, have all been verified experimentally to great precision. One of these, the equivalence between inertia and energy (the famous E = mc2 relation), was to cast its influence far beyond the problems of physics by ushering in the age of nuclear weapons. Remarkable as the union of space and time was, Einstein immediately objected to the preference given to inertial frames. The trouble is that the concept of an inertial frame is ill defined. In an inertial frame, a mass moves uniformly if not acted on by an external force, but we know that there is no external force only when the mass is moving uniformly. This is a classic example of circular reasoning. A solution of the problem was obtained by Einstein in his General Theory of Relativity (1916). He started by noting that an observer falling freely in a gravitational field while inside a closed freely falling elevator would believe that his world was in an inertial frame. Indeed, Einstein saw no need to abandon this local inertial frame (or free-falling frame) in favor of an external (i.e., nonaccelerating), inertial frame in the formulation of physical laws. He was particularly impressed by the effect of gravity, because it had been known since Galileo that the acceleration of free fall was the same for all massive objects. (This comes about because the gravitational mass is always proportional to the inertial mass and may be taken to be equal to it.) This universal effect of gravity made it possible for Einstein to represent the curvature of the path of a gravitating mass as due to the inherent curvature of space itself. Thus space (or more precisely, spacetime) itself took on an important dynamical attribute that was totally unsuspected previously. Einstein spent many years of his life trying to show that electromagnetism was also a structure of space. In this he was unsuccessful. The modern successes in unifying electromagnetism with the weak interaction between subatomic particles

4

Vectors and fields in space

are based instead on the spacetime properties of the phase angles of fields describing physical properties in space. In other words, these interactions are the internal properties of fields residing in space. It is not yet clear whether gravity itself admits a similar description. In this chapter, we make a modest beginning in studying the mathematics used to describe our increasingly subtle appreciation of the concept of physical space. The topics include vectors and fields in space and the use of curvilinear coordinates.

1.2

Vectors in space

Physics deals with physical events in spacetime. In Newtonian mechanics, time is completely independent of space. It is characterized by a single number such as 5 in the statement, “It is now 5 minutes past the hour.” Such single numbers are called scalars. In contrast, a set of three numbers, called a vector, is needed to characterize a point in three-dimensional (3D) space. We use the equivalent notation r = xi + yj + zk = (x, y, z) = xe x + yey + zez = x1 e1 + x2 e2 + x3 e3 = (x1 , x2 , x3 )

(1.1)

to characterize a point whose rectangular (or Cartesian) coordinates as measured from an arbitrarily chosen origin are x, y, z, or equivalently x1 , x2 , x3 . We call r the position vector of a point in space. The numbers (x1 , x2 , x3 ) making up a vector r are called its components. Unlike a scalar, a vector r has both a length and a direction. We know from geometry that the length in a position vector r is r = |r| = (x2 + y2 + z2 )1/2 .

(1.2)

Its direction is then given by the unit vector er = e(r) = r/r = (xi + yj + zk)/r

(1.3)

along the direction of r. It is obviously a vector of unit length parallel to r. Thus r = re(r).

(1.4)

In particular, the vectors i = e x = e1 ,

j = ey = e2 ,

k = ez = e3

(1.5)

in Eq. (1.1) are unit vectors along the arbitrarily chosen x, y, and z axes, respectively. These axes are chosen to be perpendicular to each other so that  0, i  j ei · ej = δij = (1.6) 1, i = j.

Vectors in space

5

The symbol δij is called a Kronecker delta symbol. (They also form a right-handed coordinate system; that is, rotation from e1 to e2 causes a right-handed screw to advance along the e3 direction.) An arbitrary vector A can thus be written in terms of either its rectangular components (A x , Ay , Az ) or its length A and direction e(A) A = A x i + Ay j + Az k =

3 

Ai ei

i=1

= Ae(A). 1.2.1

(1.7)

Algebra of vectors

In the above discussion we used the following two basic algebraic operations that define an algebra of vectors: vector addition: C = A + B = (A x + Bx , Ay + By , Az + Bz ),

(1.8)

and scalar multiplication: C = λA = (λA x , λAy , λAz ).

(1.9)

By an algebra we mean that the results of these operations are objects (i.e., vectors) similar to the original objects, which are vectors. A zero or null vector 0 is defined: 0 = 0i + 0j + 0k = (0, 0, 0).

(1.10)

As a consequence, the negative −A of a vector A is also defined: (−A) + A = 0,

(1.11)

−A = −A x i − Ay j − Az k = −Ae(A).

(1.12)

that is,

Since Eq. (1.2) shows that −A has the same length A as A, we find e(−A) = −e(A). That is, −A points in a direction opposite that of A. Example 1.2.1 The sum of the vectors A=i+j B = j + 3k is C = A + B = i + 2j + 3k.

(1.13)

6

Vectors and fields in space

This vector sum has a length C = (C 2x + Cy2 + Cz2 )1/2 = (1 + 4 + 9)1/2 = (14)1/2 , and a direction e(C) =

C 1 = √ (i + 2j + 3k).  C 14

The black square marks the end of an example. 1.2.2

Geometry of space

The description of a vector A by its length A and its direction e(A) is basically geometrical, since geometry deals with sizes and shapes, that is, with properties in space. In particular, the concept of length is a special case of the concept of scalar product between two vectors A · B = A x B x + Ay By + Az Bz .

(1.14)

It involves the scalar product A · A = A2x + A2y + A2z = A2 .

(1.15)

The concept of direction is most easily understood in 2D space. We decompose e(A) into its x, y components: e(A) = cos θAx e x + sin θAx ey = cos θAx e x + cos θAy ey ,

(1.16)

where θAi is the direction angle between e(A) and the ith axis. These components of e(A) are called its direction cosines (Fig. 1.1). Each of these direction cosines can be isolated by using the scalar product operation. For example, e(A) · ex = (cos θAx e x + cos θAy ey ) · e x = cos θAx .

(1.17)

It is now easy to generalize the direction-cosine decomposition of a unit vector e(A) to 3D space by induction. The result is e(A) = cos θAx e x + cos θAy ey + cos θAz ez .

(1.18)

Substitution of Eq. (1.17) into Eq. (1.18) gives the result e(A) = [e(A) · e x ]e x + [e(A) · ey ]ey + [e(A) · ez ]ez = e(A)(· e x e x + · ey ey + · ez ez ).

(1.19)

Vectors in space

7

y e(A)

cos θAy

θAy

θAx cos θAx

x

Fig. 1.1 Direction cosines.

Since this equation holds for any vector e(A), or any vector A, we have obtained the formal identity 1 = · e x e x + · ey ey + · ez ez ,

(1.20)

called a completeness relation. It expresses symbolically the well-known result that a 3D vector has three components. It further states that its x component, for example, can be obtained by the projection of A on e x , that is, by the scalar product A · ex = A x = Ae(A) · e x = A cos θAx .

(1.21)

Finally, we note that the scalar product, Eq. (1.14), between two vectors can be written in the familiar cosine form A · B = ABe(A) · e(B) = ABcos θAB ,

(1.22)

where θAB is the angle between the vectors A and B. Example 1.2.2 Find the projection of A = i + 2j + 3k on B = j + 2k. The √ projection of A on B is the same as the projection of A on eB = B/B, where B = 5. It is √ A · eB = A · B/B = (i + 2j + 3k) · (j + 2k)/ 5 √ = 8/ 5.  Example 1.2.3 Obtain the unit vector e in space that makes an equal angle with each of the three coordinate axes e1 , e2 , and e3 . What is this angle? We are given the information that θ1 = θ2 = θ3 = θ, where θi is the angle between e and ei . Hence cos θ1 = cos θ2 = cos θ3 = cos θ = a,

8

Vectors and fields in space y B

β

x

α

A

Fig. 1.2 Sum of angles.

where a is some constant. That is, e = ae1 + ae2 + ae3 has equal components along all three axes. Since e has unit length, we must have √ 1 = (a2 + a2 + a2 )1/2 = 3a; √ a = 1/ 3 = cosθ, or θ = 55◦ .  Example 1.2.4 Use the scalar product to show that cos(α + β) = cos α cos β − sin α sin β. For the 2D vectors of Fig. 1.2, we see that θAx = α, θAy

θBx = β, θAB = α + β, π π = + α, θBy = − β. 2 2

Since we are interested in cos(α + β), we should examine the scalar product A · B = AB cos(α + β) = A x B x + Ay By . Hence

A  B  A  B  y y x x + cos(α + β) = A B A B = cos θAx cos θBx + cos θAy cos θBy = cos α cos β + (− sin α)(sin β). 

Vectors in space

1.2.3

9

Vector product

One additional vector operation is important for vectors in 3D space: the vector product B × C = i(ByCz − BzCy ) + j(BzC x − BxCz ) + k(BxCy − ByC x )      i j k   e1 e2 e3  =  Bx By Bz  =  B1 B2 B3  = −(C × B).     C x C y Cz   C1 C2 C3 

(1.23)

In this equation, the six terms of the second expression have been represented by a 3 × 3 array called a determinant. A 3 × 3 determinant can also be written in terms of 2 × 2 determinants          i j k   B B B  = i  By Bz  + j  Bz Bx  + k  Bx By  ,   C C   C C   x y z  Cy Cz  z x x y  C x Cy C z  where    a b  = ad − bc. c d The reader should think of the six-term expression shown in Eq. (1.23) whenever a 3 × 3 determinant appears in this chapter. For example, the antisymmetric property shown in the last expression of this equation comes from a change of sign of the six-term expression when the symbols B and C are interchanged. A more detailed discussion of determinants can be found in Section 2.3. Example 1.2.5    e1 e2 e3  e1 × e1 =  1 0 0  = 0,  1 0 0  because each of the six terms is zero, while    e1 e2 e3  e1 × e2 =  1 0 0  = e3 , e2 × e3 = e1 ,  1 0 0 

e3 × e1 = e2 . 

(1.24)

If in B × C we put e(B) = e1 ,

e(C) = cos α e1 + sin α e2 ,

(1.25)

10

Vectors and fields in space z

B×C

A y β C α

x B

A × (B × C)

Fig. 1.3 The vector product, the triple scalar product, and the triple vector product.

where α = θBC , then B × C = BCe1 × (cos αe1 + sin αe2 ) = BC sin αe3 .

(1.26)

Thus B × C has a length BC sin α and a direction e3 perpendicular to the plane containing B and C. Its length BC sin α is equal to the area of the parallelogram, with sides B and C shown by the shaded area in Fig. 1.3. The triple scalar product A · (B × C) = A|B × C| cos β = ABC sin α cos β

(1.27)

may be interpreted as the volume of the parallelipiped shown in the figure with the vectors A, B, and C on three of its sides. The vector D = e1 D1 + e2 D2 + e3 D3 and the scalar product A · D = A1 D1 + A2 D2 + A3 D3 look very similar in structure. They differ only by the substitutions ei ↔ Ai . The same substitution shows that A · (B × C) can be written directly from Eq. (1.23) as    A1 A2 A3  A · (B × C) =  B1 B2 B3 .  C C C  1

2

3

Since a determinant changes sign when two of its rows are interchanged, we find A · (B × C) = −A · (C × B) = −C · (B × A) = B · (C × A) = C · (A × B).

Vectors in space

11

B s2 C s1 c b

s3

a

A

Fig. 1.4 Area of the triangle ABC.

Since the order of a scalar product is unimportant, we have A · (B × C) = (B × C) · A = (A × B) · C = (C × A) · B. Example 1.2.6 If a, b and c are the position vectors of the points A, B and C in space, what is the area of the triangle ABC? From Fig. 1.4 we see that the sides of triangle ABC may be described by the vectors s1 = a − b s2 = b − c s3 = c − a. These vectors are coplanar because s1 + s2 + s3 = (a − b) + (b − c) + (c − a) = 0, or s3 = −(s1 + s2 ). The area of the triangle is one-half of the magnitude of the vector product s1 × s2 = (a − b) × (b − c) = a × b + b × c + c × a. We also note that the right-hand side of this expression is unchanged under the cyclic permutation abc → bca (i.e., under the substitutions a → b, b → c, c → a) or abc → cab (i.e., under the substitutions a → c, b → a, c → b). As a result, the lefthand side is also unchanged under the cyclic permutation 123 → 231 or 123 → 312. That is, s1 × s2 = s2 × s3 = s3 × s1 .

12

Vectors and fields in space

These equations can also be proved directly. For example, s1 × s2 = −s1 × (s1 + s3 ) = −s1 × s3 = s3 × s1 .  There is also a triple vector product A × (B × C). Since B × C is perpendicular to the BC plane containing B and C, A × (B × C) must lie on the BC plane in a direction perpendicular to A. This direction is shown in Fig. 1.3. A more precise statement of this result is given by the BAC rule A × (B × C) = B(A · C) − C(A · B).

(1.28)

This rule can be proved easily with the help of Eq. (1.26)   e3   e1 e2 A × (B × C) =  A1 A2 A3  = BC sin α(A2 e1 − A1 e2 )  0 0 BC sin α  = BCA2 sin α e1 − BCA1 (sin α e2 ). Since Eq. (1.25) gives e1 = e(B),

sin α e2 = e(C) − cos α e1 ,

we find that A × (B × C) = B(CA2 sin α + CA1 cos α) − CBA1 = B(A2C2 + A1C1 ) − C(A · B) = B(A · C) − C(A · B). Other triple vector products can be deduced from the BAC rule. For example (A × B) × C = −C × (A × B) = C × (B × A), where the different forms are equal to one another by virtue of Eq. (1.23). The BAC rule can now be applied by simply interchanging the symbols A and C in every term of Eq. (1.28): C × (B × A) = B(C · A) − A(C · B).

(1.29)

These results show that in general (A × B) × C  A × (B × C).

(1.30)

These expressions are not the same in general because A × (B × C) lies on the BC plane, according to Eq. (1.28), while (A × B) × C lies on the AB plane. Only when B is perpendicular to both A and C will the two expressions be equal, for then the second term in both Eqs. (1.28) and (1.29) vanishes.

Vectors in space

13

Example 1.2.7 (Helmholtz theorem) If A is an arbitrary vector and e is an arbitrary unit vector, show that A = e(A · e) + e × (A × e) = A + A⊥ , where A and A⊥ are the component vectors parallel and perpendicular, respectively, to the unit vector e. This follows directly from the BAC rule e × (A × e) = A(e · e) − e(A · e). For example, if we take e = e x , then A = A x e x . The remaining term must be A⊥ = Ay ey + Az ez (i.e., that part of A that lies on the plane perpendicular to e). In particular, if A is on the xy plane, Az = 0. Then A × e = (A x e x + Ay ey ) × e x = −Ay ez , and e × (A × e) = −Ay e x × ez = Ay ey , as expected.  Problems 1.2.1 If A = (1,2,3) and B = (3,1,1), calculate A + B, A − B, A · B, the projections of A on B and of B on A, A × B, |A × B|, and e(A × B). 1.2.2 By using suitable vectors, prove the trigonometric identities (a) cos(α − β) = cos α cos β + sin α sin β, (b) sin(α − β) = sin α cos β − cos α sin β. 1.2.3 Prove that the diagonals of a parallelogram bisect each other. 1.2.4 Take one corner of the unit cube as the origin, and its three adjacent sides as the x,y,z axes. From the origin, four diagonals can be drawn across the cube: three on three faces and one across the body to the opposite corner. Calculate all the angles between pairs of these diagonals. 1.2.5 If a, b and c are the position vectors of the points A, B and C in space, the area of the triangle ABC is a × b + b × c + c × a (Example 1.2.6). Show that the perpendicular distance from the origin to the plane containing this triangle is a · (b × c)/|a × b + b × c + c × a|. 1.2.6 Let r be the position vector of any point on a plane in space not containing the origin. Let a be the position vector of that point on the plane nearest the origin. Show that a · r = a2 = constant. 1.2.7 If a is the position vector of a fixed point A in space, what is the nature of the surface defined by the arbitrary position vector r satisfying the equation |r − a| = constant = b?

14

Vectors and fields in space

1.2.8 Use the BAC rule to prove the Jacobi identity a × (b × c) + b × (c × a) + c × (a × b) = 0. 1.2.9 If an unknown vector X satisfies the relations X · b = β, X × b = c, express X in terms of β, b, and c. Hint: See Example 1.2.7. 1.2.10 If D is a linear combination of three arbitrary noncoplanar vectors A, B, C: D = aA + bB + cC, show that a = D · (B × C)/A · (B × C). Obtain corresponding expressions for b and c. 1.2.11 Describe and explain a test for coplanarity of three arbitrary vectors A, B and C. 1.2.12 Show that (A × B) × (C × D) lies on the line of intersection of the plane containing A and B and the plane containing C and D. 1.2.13 If r and v = (d/dt)r are both functions of time, show that d [r × (v × r)] = r2 a + (r · v)v − (v2 + r · a)r, dt where a is the acceleration.

1.3

Permutation symbols

Our intuitive (i.e., geometrical) understanding of physical space is of great antiquity. Archimedes—perhaps the greatest mathematician of ancient Greece—was known for his geometrical work, including formulas for areas of figures and volumes of solid bodies. The well-known principle in hydrostatics bearing his name demonstrates his geometric insight. In contrast, the algebraic description of space, which we used in the preceding section, is of relatively recent origin. It evolved from the algebra of Arabic mathematicians of the ninth through the eleventh centuries. (The word algebra itself arose from the Arabic word al-jabr, meaning the reunion of broken parts, which had appeared in the title of a well-known Arabic book of the ninth century on algebra.) The word vector was coined by Hamilton in the nineteenth century. It is easy enough to see why the algebraic description of space, being so symbolic and abstract, has taken so long to develop. It is perhaps more difficult to appreciate its power. In this section we further illustrate the the power of algebraic recombinations by reconsidering the vector product of vectors in space. Let us first recall that the determinantal form, Eq. (1.23), of the vector product can become very cumbersome when one deals with several successive vector products. (For example, try to write down a determinantal expression for A × [B × (C × D)].) The structure of the vector product is very simple, however. It is a vector in space and must have three components. For example, one of the equations in Eq. (1.24) can be written in the form e1 × e2 = ε121 e1 + ε122 e2 + ε123 e3 , where the components of the vector product have been denoted ε12k . The first two indices (12) refer to unit vectors on the left-hand side (LHS), while the last index

Permutation symbols

15

k is that of the vector component on the right-hand side (RHS). Since e1 × e2 = e3 (a result referred to as the right-hand rule for vector products), we must have ε121 = 0, ε122 = 0, ε123 = 1. There are 3 × 3 = 9 vector products ei × ej , and a total of 9 × 3 = 27 εijk components, Eq. (1.24) shows that these components are zero except the six in which i  j  k, namely ε123 = ε231 = ε312 = 1,

ε213 = ε321 = ε132 = −1.

(1.31)

We shall see that the six sets of indices appearing in Eq. (1.31) are called the permutations, or rearrangements, of the three objects 1, 2 and 3. For this reason the εijk are called permutation (or Levi-Civita) symbols. 1.3.1

Permutations

Let us examine the six sets of indices in Eq. (1.31) more carefully. The first set, 123, is an ordered sequence of three objects 1, 2 and 3. The remaining five sets are also ordered sequences of these objects, but they differ from 123 in the ordering. They are said to be permutations (or rearrangements) of the original ordering 123. The ordering 123 is itself a permutation of each of the other five sets. In other words, all six sets are permutations of one another. We now show that there are only six permutations of three distinct objects such as 1, 2 and 3. In ordering these three objects we can put any one of the three in the first position, and any one of the remaining two objects in the second position. The last remaining object must be put in the last position. Thus the number of permutations equals 3 × 2 × l = 6. Permutations from an original ordered sequence such as 123 can be achieved by successive transpositions, or interchanges, of neighboring objects. A permutation is said to be even (or odd) if an even (or odd) number of transpositions is required for the rearrangement. For example, 213 is an odd permutation of 123 because one transposition 12 → 21 will do the job. On the other hand, 231 is an even permutation and requires the transpositions 12 → 21 and then 13 → 31. Eq. (1.31) shows that the permutation symbols of three indices have the value 1 for even permutations of 123, and the value −1 for odd permutations of 123. We can also start with the ordering ijk instead. An even permutation of ijk will not change the value of the resulting permutation symbol, while an odd permutation gives a permutation symbol of the opposite sign. That is, εijk = εjki = εkij = −εjik = −εkji = −εikj .

(1.32)

Permutation symbols such as ε112 and ε111 with two or more identical indices cannot be permutations of 123. They have the value 0, but they also satisfy Eq. (1.32).

16

1.3.2

Vectors and fields in space

Vector products with permutation symbols

It will be useful to write the nine equations implied by Eq. (1.24) in the abstract form ei × ej =

3 

εijk ek ,

i, j, k = 1, 2, or 3.

(1.33)

k=1

Using this notation we can write the vector product B × C as ⎞ ⎛ 3 ⎞ ⎛ 3 ⎟⎟⎟ ⎜⎜⎜ ⎟⎟⎟ ⎜⎜⎜ Bm em ⎟⎟⎠⎟ × ⎜⎜⎝⎜ Cn en ⎟⎟⎠⎟ B × C = ⎜⎜⎝⎜ m=1

=



n=1

BmCn (em × en )

m, n

=



(BmCn εmnj )ej .

m, n, j

This shows that B × C has components   BmCn εmnj = εjmn BmCn , (B × C)j = m, n

(1.34)

m, n

because jmn is an even permutation of mnj. For example,  (B × C)1 = ε1mn BmCn = ε123 B2C3 + ε132 B3C2 m, n

= B2C3 − B3C2 , in agreement with the usual geometrical result. There are many different ways of writing B × C in this notation, since the different permutations of mnj are in a sense equivalent. The following expressions can be obtained with the help of Eq. (1.32) or by relabeling. ⎧ε ⎧B C e ⎪ ⎪ mnj m n j ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ε B ⎪ ⎪ njm n Cj em ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪   ⎪ εjmn ⎪ BjCm en ⎨ ⎨ B×C= BmCn ej ⎪ = εmnj ⎪ , (1.35) ⎪ ⎪ ⎪ ⎪ (−)εnmj (−)BnCm ej ⎪ ⎪ ⎪ ⎪ m, n, j m,n, j ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (−)εjnm (−)BjCn em ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ (−)εmjn (−)BmCj en where the second set of relations involves the permutation of indices in the BCe factor rather than in ε. Why should we use such an abstract and obscure notation to perform an operation for which we already have a nice intuitive geometrical understanding? The answer has

Permutation symbols

17

to do with successive vector products such as A × [B × (C × D)]. Eq. (1.35) shows that each vector product gives rise to an ε symbol, so that an expression with n vector products is a sum of terms each involving n of these symbols. Such sums of products can often be simplified with the help of the following three reduction formulas: 3 

εmnk εijk = εmn1 εij1 + εmn2 εij2 + εmn3 εij3

k=1

 j,k

= δmi δnj − δmj δni ,

(1.36a)

εmjk εnjk = 2δmn , 

(1.36b)

ε2ijk = 6.

(1.36c)

i, j,k

The δ symbols on the right are Kronecker δ symbols. These reduction formulas can be proved in the following way. Eq. (1.36c) is just a sum of six contributions of 1 from each of the six permutations of 123. Eq. (1.36b) gives zero if m  n, because then each term in the sum contains a permutation symbol with repeated indices. If m = n (= 1 say), there are only two nonzero terms ( jk = 23 or 32), each with a contribution of 1. Eq. (1.36a) gives zero if ij is not a permutation of mn, because then each term in the sum contains a permutation symbol with repeated indices. If ij is a permutation of mn, the permutation is even if ij = mn, and odd if ij = nm: These account for the two terms on the RHS. These arguments are somewhat abstract. The reader who is not sure what they mean should write out some or all of these sums explicitly. This will be left as an exercise (Problem 1.3.5). Thus, under appropriate conditions, the ε symbols can be made to disappear two by two from the expression. Every time this occurs, the reduced expression contains two vector products fewer than before. In this way, a highly complicated expression can be simplified algebraically (i.e., painlessly). This reduction is illustrated by the following example. Example 1.3.1 Prove the BAC rule for A × (B × C). ⎞ ⎛ ⎞ ⎛⎜ ⎟⎟⎟ ⎜⎜⎜ ⎟⎟⎟ ⎜⎜⎜ using Eq. (1.34) A × (B × C) = ⎜⎜⎝ Ai ei ⎟⎟⎠ × ⎜⎜⎝⎜ (B × C)j ej ⎟⎟⎟⎠⎟, i

=

 i, j,k

=

 i, j,k

=−

j

Ai (B × C) j εi jk ek ,

using Eq. (1.33)

⎛ ⎞ ⎜⎜⎜ ⎟⎟⎟ Ai ⎜⎜⎜⎝ BmCn εmn j ⎟⎟⎟⎠ εijk ek ,

using Eq. (1.34)

m,n

 i,k m,n

Ai BmCn ek (δmi δnk − δmk δni ),

using Eq. (1.36a).

18

Vectors and fields in space

The overall negative sign in the last expression comes from writing εijk as −εikj . The sum over i and k (or m and n) can next be performed readily with the help of the expressions 

Ai δmi = A1 δm1 + A2 δm2 + A3 δm3

i



= Am , ek δnk = en ,

k

since δmi vanishes unless m = i. This means that the summation simply picks up the term Am or en . Hence A × (B × C) = −



Ai BmCn (δmi en − em δni )

i,m,n

=−



BmCn (Am en − em An )

m,n

= −(A · B)



C n en + B

n



C n An

n

= B(A · C) − C(A · B),

(1.37)

where the final steps are made by rearranging a sum of nine terms into a product of sums of three terms each: ⎛ ⎞⎛ ⎞  ⎜⎜⎜ ⎟⎟⎟ ⎜⎜⎜ ⎟⎟ Am BmCn en = − ⎜⎜⎝ Cn en ⎟⎟⎠ ⎜⎜⎝ Am Bm ⎟⎟⎟⎠ − m,n

n

m

= −C(A · B). (If you are not sure what this means, you should write out all nine terms explicitly by letting both m and n go through the values of 1, 2 and 3.)  The purely formal manipulations used in this algebraic method are not without disadvantages. Problems can arise because the calculation is too abstract or mechanical. As a result, the geometrical significance of the intermediate steps or of the final results is easily lost. It is usually a good policy to use the elementary geometrical method of Section 1.2 whenever possible, and to call upon the abstract algebraic method only to handle the really complicated expressions. When the expression is very complicated, the algebraic method is usually the only decent method of calculation.

Permutation symbols

19

Problems 1.3.1 Write the nine numbers ε1jk as a 3 × 3 table (or matrix), with j labeling the rows and k labeling the columns. Do this for ε2jk and ε3jk . 1.3.2 (a) Show that there are n! permutations of n distinct objects. (b) Write out the 4! permutations of the four distinct objects a, b, c, and d. 1.3.3 The word permutation is also used to denote a different arrangement of objects when they are not all distinct. (a) Let n objects be made up of nl indistinguishable objects of one kind, n2 indistinguishable objects of a second kind, . . ., and nk indistinguishable objects of the k-th kind, with n = n1 + n2 + . . . + nk . Show that the number of permutations is n!/n1 !n2 ! . . . nk !. (b) Write out explicitly the 4!/2! permutations of the four objects a, a, c and d. 1.3.4 Show that    δim δin δil    εi jk εmnl =  δjm δjn δjl     δkm δkn δkl  by noting the following: (a) Both sides vanish if either permutation symbol has two or more indices in common. (b) If both permutation symbols involve distinct indices, mnl can only be one of the six permutations of ijk. 1.3.5 Show that  εmjk εnjk = 2δmn , j,k



εmnl εijl = δmi δnj − δmj δni ,

l

by using (a) the result of Problem 1.3.4 and (b) the arguments given in the text but expressed in greater detail.  For method (a) you will need the identity l δln δjl = δnj . 1.3.6 Show that (A × B) · (C × D) = (A · C)(B · D) − (A · D)(B · C). 1.3.7 Show that (A × B) × (C × D) = (A · B × D)C − (A · B × C)D = (A · C × D) B − (B · C × D)A by using (a) the BAC rule; and (b) permutation symbols.

20

Vectors and fields in space

1.4

Vector differentiation of a scalar field

Newton was the first scientist to appreciate the relationships between infinitely small changes in related physical properties, as well as the cumulative effects of an infinitely large number of such small changes. Indeed, his method of fluxions or generalized velocities (discovered in 1665–66, but published only in 1736, nine years after his death) represents the first invention of calculus. Using this method, he established the law of universal gravitation and obtained from it Kepler’s empirical laws on planetary motion. However, it was the mathematician Leibniz, the co-inventor of calculus and the master of mathematical notations, who first published in 1684-86 an account of this new mathematical theory. Leibniz introduced the symbols we now use in calculus and in many other branches of mathematics, as well as the names differential and integral calculus. The first text book on calculus, published in 1696 by l’Hospital, already contained much of what is now undergraduate calculus. It is now easy for us to take a scalar function of a single variable, such as the distance of fall of an apple as a function of time, and differentiate it successively to determine its velocity, acceleration, etc. (A scalar function is one whose functional value is specified by a scalar, i.e., a single number. A scalar function may still be multivalued; the important feature is that each of these multiple values is a scalar.) It is almost as easy to deal with a scalar function of several variables. For example, the total differential of a smooth (i.e., differentiable) scalar function Φ(s, t, u) of three independent variables s,t,u dΦ(s, t, u) =

∂Φ ∂Φ ∂Φ ds + dt + du ∂s ∂t ∂u

is expressible in terms of the partial derivatives. 1.4.1

Scalar field

A scalar function Φ(x, y, z) of the position coordinates (x,y,z) is called a scalar field. That is, it is a rule for associating a scalar Φ(x,y,z) with each point r = (x,y,z) in space. For this reason, it is also called a scalar point function. Scalar fields are of obvious importance in physics, which deals with events in space. By a smooth scalar field we mean a scalar field that is differentiable with respect to the position coordinates (x,y,z). The total differential corresponding to an infinitesimal change dr = (dx,dy,dz) in position is then dΦ(x, y, z) =

∂Φ ∂Φ ∂Φ dx + dy + dz. ∂x ∂y ∂z

(1.38)

It is useful to express dΦ as a scalar product of two vectors: dΦ(x, y, z) = [∇Φ(x, y, z)] · dr,

(1.39)

Vector differentiation of a scalar field

21

where ∇Φ(x, y, z) =

∂Φ ∂Φ ∂Φ i+ j+ k ≡ ∇Φ(r) ∂x ∂y ∂z

(1.40)

is a vector field, or vector point function. By this we mean that to each point r in space we associated a vector ∇Φ(x, y, z) as specified by its three components (∂Φ/∂x, ∂Φ/∂y, ∂Φ/∂z). 1.4.2

The gradient operator ∇

The operation that changes a scalar field to a vector field in Eq. (1.40) is denoted by the symbol ∇=i

∂ ∂ ∂ +j +k , ∂x ∂y ∂z

(1.41)

called a gradient operator. The vector field ∇Φ(r) itself is called the gradient of the scalar field Φ(r). The operator ∇ contains both partial differential operators and a direction, and is known as a vector differential operator. Both features are important in the generated vector field ∇Φ(r), which like other vectors has a length |∇Φ(r)| and a direction e(∇Φ): ⎡ 2  2  2 ⎤1/2 ⎢⎢ ∂Φ ∂Φ ∂Φ ⎥⎥⎥⎥ ∇Φ(r) . (1.42) |∇Φ(r)| = ⎢⎢⎢⎣ + + ⎥⎦ , e(∇Φ) = ∂x ∂y ∂z |∇Φ(r)| All calculations in the following examples are performed in rectangular coordinates. Example 1.4.1 Φ(r) = Φ(r) = r2 = x2 + y2 + z2 : ∂ 2 r = 2x, ∂x

∂ 2 r = 2y, ∂y

∂ 2 r = 2z, ∂z

∴ ∇r2 = 2xi + 2yj + 2zk = 2r. Thus |∇r2 | = 2r,

e(∇r2 ) = er . 

Example 1.4.2 Φ(r) = r : ∂ 2 ∂r d 2 ∂r r = r = 2r , (chain rule) ∂x ∂x dr ∂x 1 ∴ ∇r = ∇r2 = er . 2r Thus ∇r is a unit vector everywhere.



(1.43)

22

Vectors and fields in space

Example 1.4.3 Φ(r) = f (r): df (r) (∇r) dr df = er .  dr

∇ f (r) =

(1.44)

Example 1.4.4 Φ(r) = exp(−ar2 ): ∇[exp(−ar2 )] =

d exp(−ar 2 )(∇r) dr

= er (−2ar) exp(−ar2 ).  Example 1.4.5 Φ(x, y) = xy in a 2D space. ∇Φ(x, y) = iy + jx. Hence |∇Φ| = (y2 + x2 )1/2 = r,

y x e(∇Φ) = i + j .  r r

It is clear from Eq. (1.40) that ∇Φ(r) at the point r is a vector. So is dr. Hence dΦ(r) = [∇Φ(r)] · dr = |∇Φ|dr[e(∇Φ) · e(dr)] = |∇Φ|dr cos θ,

(1.45)

where dr = |dr|. If the displacement dr in 3D space lies on a surface on which Φ(r) is a constant (say C), then dΦ(r) = Φ(r + dr) − Φ(r) = C − C = 0. Since both |dr| and |∇Φ| are not necessarily zero, we must conclude that in general cos θ = 0,

or

θ = π/2,

when dΦ = 0.

Thus, e(∇Φ) must be perpendicular to any dr on a surface of constant Φ, that is, e(∇Φ) is normal to this surface. Some of the most common scalar fields in physics are potential fields, which give potential energies of systems in space. As a result, the surfaces of constant Φ in a 3D space are often called equipotential surfaces. In 2D spaces, the equipotentials are lines. A good example of the latter is a contour line on a topological map, giving the elevation (or gravitational potential) of a point on the earth’s surface. Example 1.4.6 If Φ(x, y) = axy gives the elevation (in m) on a topological map, the “equipotential” condition is Φ = axy, or y = Φ/ax, where a is a contant (in units of m−1 . The contour lines are the hyperbolas shown in Fig. 1.5, where a = 10−4 m−1 is used. The arrows point to the valleys on both sides of the mountain pass. 

Vector differentiation of a scalar field

23

y (km) 2 Φ = –200 m –100 0 100 200

–2

0

x (km)

2

200 m 100 0 –100 –200

–2

Fig. 1.5 The contour lines of a mountain pass or “saddle”.

We have seen that ∇Φ(r) is everywhere perpendicular to the equipotential passing through r. If Φ(r) is a gravitational potential, F(r) = −m∇Φ(r) gives the gravitational force on a point mass m. Hence forces are also perpendicular to equipotentials. A mass will be accelerated along e(F(r)) at the point r. A slowly moving mass will therefore “flow” along a flow curve, or a line of force, whose tangent at r gives the direction e(F(r)) of the force acting on it. Flow curves are shown as rivers on a map. The change dΦ shown in Eq. (1.45) depends also on the direction e(dr) of the displacement. If dr is along a flow curve in the “uphill” direction e(∇Φ), we find e(dr) = e(∇Φ), and cos θ = 1. The change dΦ then takes on its maximal value (dΦ)max = |∇Φ|dr. This means that |∇Φ| =

(dΦ)max . dr

Thus at the point r, the scalar field Φ(r) changes most rapidly along e(∇Φ) with the maximal slope of |∇Φ|. This direction is perpendicular to the equipotential and is antiparallel to the flow curve, or force line, passing through r. Example 1.4.7 Calculate the directional derivative of the scalar field Φ(r) = x2 + y2 + z2 along the direction of the vector A = i + 2j + 3k at the point r = (2, 0, 0). By a directional derivative we mean dΦ = ∇Φ · e(A), ds

24

Vectors and fields in space

that is, the rate of change of Φ along the direction e(A). Hence √ dΦ = (2xi + 2yj + 2zk) · (i + 2j + 3k)/ 14 ds √ = (2x + 4y + 6z)/ 14. √ This gives a value of 4/ 14 at r = (2, 0, 0). By the way, the quipotentials of Φ are spherical surfaces.  1.4.3

The operator ∇2

The scalar product ∇2 = ∇ · ∇ =

∂2 ∂2 ∂2 + + ∂x2 ∂y2 ∂z2

(1.46)

is a scalar differential operator called the Laplacian, named after Laplace, a French mathematician of the eighteenth century. Being a scalar, it will not change the vectorial character of the field on which it operates. Thus ∇2 Φ(r) is a scalar field if Φ(r) is a scalar field, while ∇2 [∇Φ(r)] is a vector field because ∇Φ(r) is a vector field. The Laplacian is also unchanged when the sign of one or more of the coordinates x,y,z are changed. A function Φ(r) that is even (or odd) under one of these space reflection operations (x ↔ −x, etc.) will remain even (or odd) after the ∇2 operation. The evenness or oddness of a function or operator under a sign change of its variable is called its parity. Hence the ∇2 operator is said to have even parity. ∇2 is the simplest differential operator that will not change the vectorial and parity properties of a field on which it operates. These are some of the reasons why ∇2 appears so frequently in the equations of motion or the equations of state of physical systems. Example 1.4.8



 ∂2 ∂2 ∂2 2 + + r = 6, ∇ r = ∂x2 ∂y2 ∂z2 2 2

since (∂2 /∂x2 )r2 = 2.



Problems 1.4.1 Show that the vector field A(r) = r × ∇Φ(r) is orthogonal to both r and ∇Φ(r): A(r) · r = 0, 1.4.2 If r 2 = x2 + y2 + z2 , verify that

A(r) · ∇Φ(r) = 0.

Vector differentiation of a vector field

25

(a) ∇(ln r) = r/r2 , (b) ∇(3z2 − r2 ) = −2xi − 2yj + 4zk, (c) ∇2 (ln r) = 1/r2 , (d) ∇2 (3z2 − r2 ) = 0, and (e) ∇2 (1/r) = 0, for r  0. 1.4.3 If φ(r) and ψ(r) are two scalar fields such that ∇φ(r) × ∇ψ(r) = 0 over all space, how are their equipotential surfaces and lines of force related? 1.4.4 Consider the scalar potential field Φ(r) = q+ Φ+ (r) + q− Φ− (r),  1/2 r± = (x ∓ 1)2 + y2 + z2 ,

Φ± =

1 , r±

due to the charges q± located at x = ±1, y = 0, z = 0. (a) For q+ = 1, q− = 0, find the equipotential surfaces Φ(r) = const, and the force field F(r) = −∇Φ(r). (b) For q+ = 1, q− = −1, show that the x = 0 plane is an equipotential surface and that on this surface F(r) = −∇Φ(r) = −(2/r3 )i. Show that equipotential surfaces with Φ > 0 are located entirely in the x > 0 half space, and that as r → (1, 0, 0), they are spheres centered at r = (1, 0, 0) where Φ→∞ 1.4.5 Show that the equation x + y + z = 3 is satisfied by any point r = (x, y, z) on a plane tangent to a sphere √ at the point (1,1,1) on the surface of the sphere, if the sphere has radius 3 and is centered at the origin. (Hint: consider the scalar field Φ(x, y, z) = x +√y + z.) 1.4.6 Show that the point (0, 1, 2) lies on the curve of intersection of the surfaces defined by the equations x2 + y2 + z2 = 3 2x + 2y + z2 = 4. Calculate the angle between the normals to these surfaces at this point. 1.4.7 (a) Show that the gravitational potential of a thin uniform spherical shell centered at the origin, of radius a and total mass m, at an external point r = (0, 0, z), z > a, is unchanged when the mass of the shell is concentrated at the origin. (b) Show that the result of part (a) also holds if the spherical shell is replaced by a uniform solid sphere of radius a < z and total mass M.

1.5

Vector differentiation of a vector field

A vector function F(t) of a single variable t is made up of three components F x (t), Fy (t), Fz (t), each of which is a scalar function of t. similarly, a vector function

26

Vectors and fields in space

of several variables is made up of components, each of which is a scalar function of these variables. Of particular interste in physics are vector functions of the position vector r = (x, y, z) or vector fields. A vector field V(r) = (Vx (x, y, z), Vy (x, y, z), Vz (x, y, z)) is simply a rule for assigning the values Vi (x, y, z), i = x, y, z, to the components of the vector at r. A vector field is differentiable, if all the partial derivatives of its components exist. Examples of vector fields are ∇Φ(r), where Φ(r) is a scalar field, and a force field F(r), which defines a force F at every point r in space. Vector differential operations on vector fields are more complicated because of the vectorial nature of both the operator and the field on which it operates. Since there are two types of products involving two vectors, namely, the scalar product and the vector (i.e., cross) product, vector differential operations on vector fields can also be separated into two types called the divergence and the curl. The divergence of a vector field V(r) is defined by the scalar product   ∂ ∂ ∂ ∇ · V(r) = i + j + k · (iVx + jVy + kVz ) ∂x ∂y ∂z =

∂Vx ∂Vy ∂Vz + + . ∂x ∂y ∂z

(1.47)

The result is a scalar field. It differs from the scalar differential operator V(r) · ∇ = Vx

∂ ∂ ∂ + Vy + Vz , ∂x ∂y ∂z

(1.48)

as one can see in the following example. Example 1.5.1 

 ∂x ∂y ∂z r(∇ · r) = r = 3r, + + ∂x ∂y ∂z   ∂ ∂ ∂ (r · ∇)r = x + y + z (ix + jy + kz) = r ∂x ∂y ∂z  r(∇ · r).



The curl (or rotation) of a vector field V(r) is defined by the vector product    i j k   ∂ ∂ ∂  ∇ × V(r) =  ∂x  ∂y ∂z   Vx Vy Vz 

Vector differentiation of a vector field

27



     ∂Vy ∂Vx ∂Vz ∂Vy ∂Vx ∂Vz =i +j +k − − − ∂y ∂z ∂z ∂x ∂x ∂y  ∂ εmnl em Vl . = ∂xn m,n,l

(1.49)

The result is a vector field, not the vector differential operator  ∂ εmnl em Vn . V(r) × ∇ = ∂x l m,n,l Example 1.5.2 e1 × (∇ × r) = 0, because    i j k  ∂ ∂ ∂  = 0. ∇ × r =  ∂x  ∂y ∂z  x y z However,      i j k  ∂ ∂   1 0 0   (e1 × ∇) × r =  × r = −j + k ×r ∂z ∂y  ∂ ∂ ∂  ∂x ∂y ∂z = −2i  e1 × (∇ × r) = 0. The differences derive from the fact that both differential operations and vector products depend on the positions of the objects in the expression.  Some of the most interesting vector fields in physics satisfy the property that the results of such vector differential operations vanish everywhere in space. Such special fields are given special names: If ∇ · V(r) = 0, V is said to be solenoidal (or divergence-free). If ∇ × V(r) = 0, V is said to be irrotational. Example 1.5.3 Show that the gradient of any scalar field Φ(r) is irrotational and that the curl of any vector field V(r) is solenoidal.    i j k   ∂ ∂ ∂ ∇ × (∇Φ) =  ∂x ∂y ∂z  Φ(x, y, z) = 0 (1.50)  ∂ ∂ ∂   ∂x ∂y ∂z  because there are two identical rows in the determinant. Alternatively,  ∂ ∂ ∇ × (∇Φ) = εijk ei Φ(x, y, z) = 0, ∂x ∂x j k i, j,k

28

Vectors and fields in space

because εijk is antisymmetric in j, k, while ∂2 /∂xj ∂xk is symmetric. Thus each term in the sum is always cancelled by another term: εijk

∂ ∂ ∂ ∂ + εikj = 0.  ∂xj ∂xk ∂xk ∂xj

Similarly,  ∂ (∇ × V)i ∂xi i ⎛ ⎞ ⎟⎟⎟  ∂ ⎜⎜⎜ ∂ ⎜⎜⎜ ε ⎟⎟⎟ = 0, = u ijk k ⎜⎝ ⎟⎠ ∂x ∂x i j i j,k

∇ · (∇ × V) =

(1.51)

because εijk is antisymmetric in i, j. The physical interpretation of these results will be given in the next two sections. Finally, both the scalar operator ∇2 and the vector operator ∇ can operate on a vector field V(r): ∇2 V(r) = i(∇2 Vx ) + j(∇2 Vy ) + k(∇2 Vz )

(1.52)

is obviously a vector field, while ∇V(r) = i(∇Vx ) + j(∇Vy ) + k(∇Vz )   ∂Vx ∂Vx ∂Vx +j +k =i i ∂x ∂y ∂z   ∂Vy ∂Vy ∂Vy +j i +j +k ∂x ∂y ∂z   ∂Vz ∂Vz ∂Vz +k i +j +k ∂x ∂y ∂z

(1.53)

is a more complicated mathematical object known as a dyadic field. (A dyad or dyadic is a bilinear combination of vectors. In 3D space it has nine components associated with the nine unit dyads ei ej .) 1.5.1

Using vector differential operators

In dealing with vector differential operators (VDO), it is important to remember that they are both differential operators and vectors. For example, the ordering of factors is important in differentiation:     ∂ ∂h ∂ ∂g f (gh) = fh + fg  (fgh). (1.54) ∂x ∂x ∂x ∂x

Vector differentiation of a vector field

29

The bookkeeping of vector components could be quite unpleasant if one used the cumbersome determinantal notation for vector products. In all but the simplest cases, the use of permutation symbols is far more convenient. Example 1.5.4 If Φ(r) is a scalar field and V(r) is a vector field, show that ∇ × (ΦV) = Φ(∇ × V) + (∇Φ) × V. We first note that both sides of the equation are vector fields. Method 1       j k   i  i j k   i j k    ∂ ∂ ∂   ∂Φ ∂Φ ∂Φ  ∂ ∂ ∂  ∇ × (ΦV) =  ∂x ∂y ∂z  = Φ  ∂x ∂y ∂z  +  ∂x ∂y ∂z ,       ΦVx ΦVy ΦVz  Vx Vy Vz   Vx Vy Vz  where the last step has been made with the help of identities of the form   ∂Vy ∂ ∂Φ Vy . (ΦVy ) = Φ + ∂x ∂x ∂x Method 2 ∇ × (ΦV) =



εijk ei

i, j,k

∂ (ΦVk ) ∂xj

   ∂ ∂Φ =Φ εijk ei Vk + εijk ei Vk ∂xj ∂xj i, j,k i, j,k 

= Φ(∇ × V) + (∇Φ) × V. 

(1.55)

Other useful relations involving vector differential operations on products of fields are ∇ · (ΦV) = Φ(∇ · V) + (∇Φ) · V, ∇ · (A × B) = B · (∇ × A) − A · (∇ × B), ∇ × (∇ × A) = ∇(∇ · A) − ∇2 A. Example 1.5.5   df (r) ∇ f (r) = ∇ · ∇ f (r) = ∇ · er dr   df (r) df (r) + = er · ∇ ∇ · er . dr dr 2

(1.56) (1.57) (1.58)

30

Vectors and fields in space

Now according to Eq. (1.44)  d2 f df = 2 er . ∇ dr dr 

Therefore the first term is d2 f /dr2 . The second term requires the calculation of   r 1 1 ∇ · er = ∇ · = ∇·r+r· ∇ r r r   1 2 3 = + r · er − 2 = . r r r Hence ∇2 f (r) =

d2 dr

2

f (r) +

2 d f (r).  r dr

(1.59)

Example 1.5.6 Show that if B(r) and C(r) are vector fields, then ∇ × (B × C) = B(∇ · C) + (C · ∇)B − [C(∇ · B) + (B · ∇)C].

(1.60)

Method 1 

∂ (B × C)k , using Eq. (1.34) ∂xj i, j,k ⎛ ⎞  ⎟⎟ ∂ ⎜⎜⎜⎜ ⎜⎝⎜ εkmn BmCn ⎟⎟⎟⎟⎠ = εijk ei ∂xj m,n i, j,k ⎛ ⎞  ⎜⎜⎜ ⎟ ⎜⎜⎜ εijk εkmn ⎟⎟⎟⎟⎟ ei ∂ (BmCn ) = ⎝ ⎠ ∂xj i, j,m,n k

∇ × (B × C) =

=

εijk ei



(δim δjn − δin δjm )ei

i, j,m,n

=

 i, j

ei

∂ (BmCn ), ∂xj

using Eqs. (1.32) and (1.36a)

∂ (BiCj − BjCi ) ∂xj

  ∂Cj ∂Bj ∂Bi ∂Ci = + Cj ei − e i C i − Bj ei . ei Bi ∂xj ∂xj ∂xj ∂xj i, j

(1.61)

These four terms correspond exactly to the four terms of Eq. (1.60). We note that when permutation symbols are used, the proof proceeds in a very straightforward and orderly manner.

Path-dependent scalar and vector integrations

31

Method 2 Many of the steps used to obtain Eq. (1.61) are already contained in the BAC rule of Eq. (1.28). The only novel feature is that A is now the vector differential operator ∇ here placed to the left of the vector fields B(r) and C(r). The chain rule of differential calculus, Eq. (1.54), now requires the replacements B(A · C) = (C · A)B → B(∇ · C) + (C · ∇)B −C(A · B) = −(B · A)C → −C(∇ · B) − (B · ∇)C, with two terms appearing for each term of the BAC rule. Thus Eq. (1.60) obtains. We shall refer to this procedure as the operator form of the BAC rule.  Problems 1.5.1 Derive the following identities: (a) ∇ · (A × B) = B · (∇ × A) − A · (∇ × B); (b) ∇ × (∇ × A) = ∇(∇ · A) − ∇2 A; (c) ∇(A · B) = (B · ∇)A + (A · ∇)B + B × (∇ × A) + A × (∇ × B). 1.5.2 If A(r) is irrotational, show that A × r is solenoidal. 1.5.3 A rigid body is rotating with constant angular velocity ω about an axis passing through the origin. Show that the velocity v of any point r in the rigid body is given by the equation V = ω × r. Show that (a) v is solenoidal; (b) ∇ × v = 2ω; and (c) ∇(ω · r) = ω. 1.5.4 Calculate the divergence of the inverse-power force Fn = ker /r n . Show that for n = 2 the force is solenoidal, except at r = 0, where no conclusion can be drawn. Show that all these forces are irrotational. Note: We shall find in Section 5.13 that for n = 2, ∇ · F2 (r) = 4πkδ(r)  0 at r = 0. 1.5.5 If B(r) is both irrotational and solenoidal, show that for a constant vector m ∇ × (B × m) = ∇(B · m). 1.5.6 Show that ∇ · er = 2/r and ∇ × er = 0.

1.6

Path-dependent scalar and vector integrations

Integrals are easy to visualize when the number of integration variables equals the number of variables in the integrand function. For example, the integrals

32

Vectors and fields in space





b

 Φ(x, y, z) dx dy dz,

f (x) dx, a

volume Ω

V(x, y, z) dx dy dz

volume Ω

give a scalar constant, a scalar constant, and a vector constant, respectively. Integrals can also be defined when the number of integration variables is less than the number of variables in the integrand, but the situation is more complicated. For example, the integral  (x2 ,y2 ) I= Φ(x, y)dx (x1 ,y1 )

is not yet completely defined, because we do not know the value of y in Φ(x, y). What is needed in addition is a statement such as y = yc (x)

(1.62)

that specifies y for each value of x. The integrand then reduces to Φ(x, yc (x)) = fc (x), so that the integral becomes well defined:  x2  fc (x)dx = Φ(x, y)dx, x1

(1.63a)

c

where c refers to the constraint in Eq. (1.62). The constraint (Eq. (1.62)) specifies a path c on the xy plane connecting the starting point (x1 , y1 ) to the ending point (x2 , y2 ). The x integration in Eq. (1.63a) is carried out along this path. For this reason we call Ic a path-dependent integral, or simply path integral. It is clear from their definition that both fc (x) and Ic can be expected to change when the path is changed, even though the endpoints on the xy plane remain the same.  (1,1) Example 1.6.1 Integrate (0,0) (x2 + y2 )dx along the straight line (path 1) and the circular arc of radius 1 (path 2) shown in Fig. 1.6. Path 1: y = x:  1 2 2x2 dx = . I1 = 3 0 Path 2: (x − 1)2 + y2 = 1:



I2 = 

1

{x2 + [1 − (x − 1)2 ]} dx

0 1

= 0

2x dx = 1  I1 .



Path-dependent scalar and vector integrations

33

y

Pa t

h

1

Pa t

h

2

(1,1)

x

Fig. 1.6 Different paths for a path integral.

In a similar way, the line integral  Ic =

(x2 ,y2 ,z2 )

(x1 ,y1 ,z1 )

Φ(x, y, z)dx

(1.63b)

of a scalar field in space is a 1D integral over a path c in 3D space. It gives a single number since, along the path c, y = yc (x), z = zc (x); hence Φ(x, yc (x), zc (x)) = fc (x). Example 1.6.2 The line integral  Ic =

(1,1,1)

(x2 + y2 + z2 )dx

(0,0,0)

gives different results along the following paths: 1. Along the straight line connecting (0,0,0) and (1,1,1), we find y = x and z = x. Hence  1 3x2 dx = 1. Ic = 0

2. Along the following three edges of the cube (α) (0,0,0) to (1,0,0), (β) (1,0,0) to (1,1,0), and (γ) (1,1,0) to (1,1,1), we find Ic = Iα + Iβ + Iγ = 13 , because 1 Iα = 0 x2 dx = 13 ,  Iβ =  Iγ =

(1,1,0)

(x2 + y2 + z2 )dx = 0,

(1,0,0) (1,1,1) (1,1,0)

(x2 + y2 + z2 )dx = 0. 

34

Vectors and fields in space

The above discussion also makes clear that the 2D integral   I= Φ(x, y, z)dx dy of a scalar field in space cannot lead to a unique number unless a surface S has been defined on which z = z s (x, y) is known. In that case, the integrand becomes Φ(x, y, z) = Φ(x, y, zS (x, y)) = gS (x, y). The resulting integral then yields a single number, and is called a surface integral. Example 1.6.3 the surface integral  (x2 + y2 + z2 )dx dy IS = S

gives different results on the following surfaces: 1. Inside the circle x2 + y2 = 1 on the z = 0 plane:   1 π 2 2 IS = (x + y )dx dy = 2π ρ3 dρ = . 2 0 circle

2. On the surface of the hemisphere x2 + y2 + z2 = 1 with z  0:    1 IS = dx dy = dx dy = 2π ρ dρ = π. hemisphere

circle



0

We thus see that the specification of a path (in the case of a line integral) or a surface (in the case of a surface integral) gives us a rule for the elimination of undesirable variables (or variable) in favor of the integration variable (or variables). Such an elimination or substitution is conceptually very simple and elementary; there is really nothing interesting in the procedure itself. It is the scalar field involved that turns out to be interesting, because certain integrals of related scalar fields can be shown to be equal to each other. We call these formal mathematical relations integral theorems. They describe important mathematical properties of scalar and vector fields in space. Before studying these integral theorems we shall first show how more complicated path-dependent integrals can be constructed from the basic line and surface integrals discussed previously. 1.6.1

Vector line integrals

By a (vector) line integral is meant an integral in which the integration variable is the vector line element dr = i dx + j dy + k dz

Path-dependent scalar and vector integrations

in space. Three types of line integrals can be distinguished     Φ(r)dr = i Φdx + j Φdy + k Φdz = a, 

c

c



V(r) · dr =

c



Vx dx +

c



Vz dz = w,

c

(1.64)

c

Vy dy +

c

35

(1.65)

c

    i j k    V(r) × dr =  Vx Vy Vz  = b, c c  dx dy dz 



(1.66)

 i.e., b x = c (Vy dz − Vz dy), etc. All these integrals are path-dependent in general, and give for each chosen path a scalar or a vector constant. 1.6.2

Vector surface integrals

In describing the flow of fluids across a closed surface it is important to distinguish a flow out of an enclosed volume from a flow into the volume. This distinction can be made conveniently by giving a surface element dσ a direction n also, so that dσ = n dσ

(1.67)

is a vector differential surface element. By convention, n (or en ) is chosen to be the normal or perpendicular direction to the surface element, with the sign chosen so that n points outward away from the interior if dσ is part of a closed surface. Since dσ is a vector, we may write in rectangular coordinates dσ = i dσ x + j dσy + k dσz ,

(1.68)

where dσx = ±dy dz,

dσy = ±dz dx,

dσz = ±dx dy,

with the signs chosen to match the given dσ on the surface. We shall show how this is achieved in Example 1.6.4. Three types of surface integral can be defined:     Φ(r)dσ = i Φdσ x + j Φdσy + k Φdσz = A, (1.69) 

S

S



V(r) · dσ = S

S



Vx dσ x + S

S



Vy dσy + S

    i j k   V(r) × dσ =  Vx Vy Vz  = B, S S  dσ x dσy dσz 

Vz dσz = Φ,

(1.70)

S



(1.71)

 with Bx = s (Vy dσz − Vz dσy ), etc. Each of these integrals requires the specification of a surface S over which the integrand becomes a function of the two integration

36

Vectors and fields in space z n k

y

x

Fig. 1.7 A surface integral over a hemisphere.

variables only. Also, we have simplified the notation by using only one integration symbol for a multivariable integral. Example 1.6.4 Show that

 r · dσ = 2πa3 ,

(1.72)

S

where S is the curved surface of a hemisphere of radius a. On the hemispherical surface the longitudinal displacement is adθ, while the latitudinal displacement is a sinθ dφ. The resulting directed surface element is dσ = er a2 d(cos θ)dφ. Hence

 I=a

1

3

 d cos θ

0

2π

dφ = 2πa3 .

0

It is also instructive to do this problem in rectangular coordinates    I= x dσ x + y dσy + z dσz S

S

= I x + Iy + Iz ,

S

(1.73)

where the surface S is defined by the equation x2 + y2 + z2 = a2 .

(1.74)

Suppose the hemisphere is the Northern Hemisphere, with the z axis passing through the North Pole, as shown in Fig. 1.7. Now dσz = dx dy, in the Northern Hemisphere because r · k is always positive. Hence

Path-dependent scalar and vector integrations

37

 Iz =

(a2 − x2 − y2 )1/2 dx dy 

= 2π

a

(a2 − ρ2 )1/2 ρ dρ

0

= 2πa /3 3

after the integration over the equatorial circle of radius a in the xy plane. The integral Ix is more complicated, because the dy dz integration is performed over a semicircle in the yz plane. The hemispherical surface now separates into two equal pieces, one half lying in front of the yz plane in the positive x direction. Here r · i is positive, and hence dσx = dy dz. The remaining half lies behind the yz plane and has dσx = −dy dz. Thus I x = I x+ + I x−   = x dy dz − front

x dy dz.

(1.75)

back

Actually, both halves contribute the same amount of πa3 /3, so that I x = 2πa3 /3. Finally, Iy can be calculated in the same way as I x give another to 2πa3 /3. Thus the total surface integral is I = Ix + Iy + Iz = 2πa3 , in agreement with Eq. (1.72). A third method for doing this problem is perhaps the most general. It turns out to be possible to integrate over just the equatorial circle in the xy plane. This possibility arises because dσ is related to its projection on the xy plane through the equation dx dy = |dσ|n · k,

(1.76)

where n is the normal direction to the spherical surface, as shown in Fig. 1.7. Hence  I = r · n|dσ|  dx dy = r·n . k·n The normal n must be calculated from the gradient of the LHS of Eq. (1.74) ∇r 2 = 2r, or n = er . Hence r · n = a, k · n = z/a, and  a I=a dx dy. z

38

Vectors and fields in space y y + Δy

3

D

C

4 y

2

A

0

B

1

x

x + Δx

x

Fig. 1.8 A small rectangular closed loop.

When expressed in circular coordinates ρ = (x2 + y2 )1/2 and φ in the xy plane, this simplifies further to  a −1/2 2 (a2 − ρ2 ) ρ dρ I = 2πa 0

= 2πa . 3



Surface integrals tend to be difficult to calculate or to visualize. We shall see in the following two sections how it is sometimes possible to express certain special types of surface integrals in terms of simpler volume or line integrals. In the rest of this section we show a few common surface integrals that can be evaluated by using simple geometrical considerations alone. Example 1.6.5 Show that



 r × dr = 2

c

dσ

(1.77)

S

where c is an arbitrary closed curve (its closure being denoted by the small circle superposed on the integration sign), and S is any surface bounded by c. A simple way to show this is to note that r × dr gives twice the vectorial area dσ of the triangle with r and dr as two of its sides. As the boundary c is traversed, we just add up the vectorial areas swept by the radius r. The result holds for any choice of the origin from which r is measured. It is instructive to demonstrate the result in a more convoluted way in order to illustrate a way of thinking that will be useful in more complicated situations. Consider first a small closed rectangular loop c on the plane z = z0 = constant, as shown by the arrows A → B → C → D → A in Fig. 1.8. For this closed path      i j k    r × dr =  x y z0  Δc Δc  dx dy 0     = i(−z0 ) dy + jz0 dx + k (x dy − y dx). Δc

Δc

Δc

Path-dependent scalar and vector integrations

39

Δci

Fig. 1.9 Decomposition of a closed curve into small rectangular closed loops.

The first two closed-path integrals vanish identically because  dy = y(final) − y(initial) = 0; similarly,  dx = 0. We are left with 

 Δc

r × dr = k

(x dy − y dx)

= k{(−y Δx)1 + [(x + Δ x)Δy]2 + [−(y + Δy)(−Δx)]3 + [x(−Δy)]4 }, where a subscript denotes the contribution of the side marked in Fig. 1.8. In this way We find  r × dr = k(2Δx Δy) = 2k Δσz . Δc

Since the planar curve can be tilted in space, we have actually proved the more general result  r × dr = 2n Δσ = 2 Δσ, Δc

where n is normal to the plane containing Δc. We now go back to the original closed curve c, and break it up into small rectangular closed loops Δci , as shown in Fig. 1.9. Adjacent sides of neighboring loops always involve integrations in opposite directions, with zero net contribution. As a result, only the outer exposed sides of closed loops will contribute. These exposed sides add up to just the original closed curve c. Hence

40

Vectors and fields in space

(a)

(b)

Fig. 1.10 Two different surfaces enclosed by the same boundary curve.

 r × dr = c

lim

Δci →0

i

=



 

Δci



lim 2 Δσi = 2

Δσi →0

i

r × dr dσ. S

It is important to realize that each small closed loop Δci can be made planar by making it infinitesimally small. However, the finite closed curve c does not have to be planar. Furthermore, the enclosed surface S is not uniquely defined by the boundary curve c. For example, if c is the circle shown in Fig. 1.10, the surface S can be either the flat surface inside the circle shown in Fig. 1.10(a), or part of the spherical surface shown  in Fig. 1.10(b). That the surface integral dσ gives the same result on either surface can be seen from the following examples.  Example 1.6.6 Show that

 dσ = 0

(1.78)

S

over a closed surface. (The closure of the surface is denoted by the small circle superposed on the integration sign.) Let us first consider a small brick-like volume of sides Δx, Δy, Δz. Over the six sides of its closed surface ΔS , we find  dσ = i(Δy Δz − Δy Δz) + j(Δz Δx − Δz Δx) + k(Δx Δy − Δx Δy) ΔS

= 0, where the contribution from a front surface of the brick is exactly cancelled by that from its back surface if the normal directions always point to the outside of the volume. If the volume enclosed by the original closed surface is separated into such small brick-like volumes with closed surfaces ΔSi , we find that    dσ = dσ = 0 = 0.  S

i

ΔSi

i

Path-dependent scalar and vector integrations

41

 Example 1.6.7 Show that the surface integral dσ over a surface bounded by a closed curve c is the same for any surface bounded by c. Take two distinct surfaces S1 and S2 bounded by c, as shown, for example, in Fig. 1.10. These two surfaces together form a closed surface S, if the normal to one of the surfaces (say S1 ) is reversed to make it always point away from the volume. We then find    dσ = 0 = − dσ + dσ.  S

S1

S2

Example 1.6.8 Show that  r · dσ = 3V,

(1.79)

S

where V is the volume enclosed by the closed surface S. We start by noting that   r · dσ = (x dσ x + y dσy + x dσz ) = 3 Δx Δy Δz = 3ΔV ΔS

ΔS

for a brick-like volume ΔV = Δx Δy Δz. The final result then follows from the usual rule of integral calculus.  Of these integrals of vector fields in space, the two involving scalar products, namely Eqs. (1.65) and (1.70), are so important as to deserve special names. The first  (1.80) w = V(r) · dr c

is called the generalized work done against the vector field V(r) along the path c. The reason for this name is that w is the mechanical work done when V is the mechanical force F(r). It is well known that if the mechanical system is a conservative one, that is, if F(r) = −∇Φ(r) is derivable from a potential field Φ(r), then    w = F · dr = − ∇Φ(r) · dr = − dΦ(r) c

c

c

= −[Φ(r2 ) − Φ(r1 )], where use has been made of Eq. (1.39). Under these circumstances, w depends only on the endpoints r1 and r2 , but not on the path of integration. This is an example of the kind of mathematical results of interest in the study of integral properties of fields in space.

42

Vectors and fields in space

The closed-path integral

 C=

V(r) · dr

(1.81)

c

is called the circulation of the vector field V(r) along the closed path c. For a conservative mechanical system, all circulations of the force field vanish. Conversely, if all circulations of a vector field vanish, the vector field must be proportional to the gradient of a scalar potential field. According to Eq. (1.50), such a vector field must be irrotational. In fluid mechanics, a fluid flow is said to be a potential flow or an irrotational flow in a region of space in which all circulations of its velocity field vanish. The second important integral is the scalar Φ defined in Eq. (1.70). It is called the flux of the vector field V(r) across the surface S. Its physical significance is described in the next section. Problems

 1.6.1 Show the closed-path line integral r · dr = 0. 1.6.2 (a) Show that  1 V(r) · dr = x02 y0 − y30 3 c if V(r) = 2xyi + (x2 − y2 )j, and the path c is made up of two parts: c1 : (0, 0) → (x0 , 0) on the x axis, c2 : (x0 , 0) → (x0 , y0 ) parallel to the y axis. (b) Show that the integral along the straight line connecting (0,0) to (x0 ,y0 ) gives the same result.  1.6.3 Evaluate the surface integral s x2 y2 z2 dσ, where dσ always points away from the volume and S is (a) the curved-cylindrical surface of the cylinder x2 + y2 = 1 between z = 0 and 1; (b) the curved spherical surface of the unit sphere for positive x, y, z, that is, in the first octant. Obtain the result in each of the remaining seven octants.

1.7

Flux, divergence and Gauss’s theorem

The flux (or flow) ΦS of a vector field j(r) across a surface S is defined by the scalar integral  j(r) · dσ ΦS = S

Flux, divergence and Gauss’s theorem

43

v Δz Δy Δx

Fig. 1.11 Liquid flow in a rectangular pipe.

It can be visualized readily by considering a liquid of density ρ(r) flowing in a pipe of rectangular cross section Δy Δz with velocity v(r) = ivx (r), as shown in Fig. 1.11. The vector field j(r) = ρ(r)v(r)

(1.82)

is called a mass flux density. It has the dimension of mass per second per unit area. The flux of j(r) across the shaded surface in Fig. 1.11 is the mass flux or flow (mass per second)  Φ2 = j(r) · dσ  Δy Δz ρ(r2 )vx (r2 ), ΔyΔz

where r2 is a representative point on the surface. It gives the total mass crossing the surface per second along the direction of its normal (n = i). Similarly Φ1  −Δy Δz ρ(r1 )vx (r1 ) is the mass flux on the back surface, the negative sign appearing because n = −i is so chosen that a flow into a closed volume is negative. Finally, we note that the total mass flux out of the volume is Φ x = Φ2 + Φ1  Δy Δz Δ j x = Δτ

Δ jx , Δx

(1.83)

where Δτ = Δx Δy Δz, 1.7.1

Δ j x = ρ(r2 )vx (r2 ) − ρ(r1 )vx (r1 ).

Divergence

We can also consider a mass flux density for a general velocity field v(r) = iv x (r) + jvy (r) + kvz (r) in the body of the liquid itself, rather than in the rectangular pipe of Fig. 1.11. A rectangular volume of space in the fluid, of sides Δx, Δy and Δz, can still be imagined.

44

Vectors and fields in space

The only difference is that in general there will be mass fluxes across the remaining four sides. Hence the total mass flux out of the volume per unit volume, in the case of infinitesimal volumes, is    1 Δ j x Δ jy Δ jz j · dσ = lim = ∇ · j(r), (1.84) + + lim Δ→0 Δτ ΔS Δx,Δy,Δz→0 Δx Δy Δz where the integration symbol denotes integration over a closed surface. The divergence of a vector field at a point r in space gives its total flux (or net outflow) per unit volume “coming out of the point r”, that is, passing through an infinitesimally small closed surface surrounding the point r. 1.7.2

Gauss’s theorem

A finite closed surface S can be subdivided into many smaller closed surfaces. This is done by dividing the enclosed volume into many smaller volumes. The closed surfaces of the small volumes are then the small closed surfaces in question. By applying Eq. (1.84) to each of these small closed surfaces, we find that     1 dσ · j(r) = dσ · j(r) lim Δτi →0 Δτi ΔSi S i  dτ∇ · j(r). (1.85) = Ω

This integral relation is known as Gauss’s theorem. It states that the net outflow across a closed surface S is equal to the total divergence in the volume Ω inside S. As a result, we may say that the enclosed divergence “causes” the outflow; that is, the enclosed divergence is a “source” of the outflow. The most familiar example of the application of Gauss’s theorem is in electrostatics, where the electric field intensity E(r) is known to be related to the charge density ρ(r) as follows: ∇ · E(r) = ρ(r)/ε0 , where ε0 is the permittivity of free space. As a result,   ρ Q dτ = = dσ · E. ε0 ε0 Ω S

(1.86)

(1.87)

This Gauss’s law of electrostatics states that the total flux of the electric field intensity coming out across a closed surface is proportional to the total charge Q enclosed by the surface. For this reason, we consider Q, or more specifically ρ(r), to be the scalar source of E(r). In this language, a solenoidal (divergence-free) vector field is said to exist without the help of a scalar source, because by definition a solenoidal field is divergence-free. Its source density is not a scalar field, but a vector field, as we shall see in Section 1.9.

Flux, divergence and Gauss’s theorem

45

Example 1.7.1 Show that the electrostatic field intensity E(r) of a point charge Q at the origin has an inverse-square dependence on r. We first note that the angular components of E must vanish, because with the point charge at the origin there is no preferred angular direction. Hence E(r) = E r (r)er . Consider next a spherical surface of radius r surrounding Q. On this surface dσ = r2 d2 Ωer ,

(1.88)

d2 Ω = d cos θdφ,

(1.89)

where

Ω being called a solid angle. Gauss’s law then gives  Q = (r2 d 2 Ωer ) · [Er (r)er ] ε0  2 = r Er (r) d 2 Ω = 4πr2 E r (r), since the full solid angle of the sphere is   1  2 d Ω= d cos θ −1

2π

dφ = (2)2π = 4π.

0

Hence E(r) =

1.7.3

Q er . 4πε0 r2



(1.90)

Continuity equation

If a fluid is in a region of space where there is neither a source nor a sink, the total mass flow out of a volume Ω expresses itself as a rate of decrease of the total mass in the volume. Mass conservation requires an exact balance between these effects:   ∂ ρ(r)dτ = ρ(r)v(r) · dσ − ∂t Ω S  = ∇ · (ρv)dτ Ω

on applying Gauss’s theorem. The first and the last expressions show that mass conservation requires that the equation ∂ρ(r) + ∇ · [ρ(r)v(r)] = 0 ∂t

(1.91)

46

Vectors and fields in space

must be satisfied everywhere in the volume. This is called the continuity equation for a conserved current, the current being j(r) = ρ(r)v(r). 1.7.4

Operator identity

Gauss’s theorem, Eq. (1.85), can be applied to the vector field V(r) = ei Φ(r) to give   ∂ dσi Φ(r) = dτ Φ(r). ∂xi S Ω Since this is true for any i, it follows that    dσΦ(r) = ei dσi Φ(r) S

S

i

 =

Ω

dτ∇Φ(r).

This result generalizes Gauss’s theorem to the operator identity   dσ = dτ∇ Ω

S

(1.92)

(1.93)

for operations on any field in space. It gives rise to other integral theorems such as the following:   dσ × A(r) = dτ∇ × A(r), (1.94) Ω

S





dσ · (u∇v − v∇u) = S

 =

Ω

Ω

dτ∇ · (u∇v − v∇u) dτ(u∇2 v − v∇2 u).

The last identity is called Green’s theorem. The application of these formulas is illustrated in the following examples: Example 1.7.2 Show that

 dσ ·

r = r2

 dτ

1 . r2

This is equivalent to showing that ∇·

r 1 = 2, 2 r r

a result that can be demonstrated readily in rectangular coordinates. 

(1.95)

Flux, divergence and Gauss’s theorem

Example 1.7.3 Calculate

 s

47

dσ × r over a closed surface. 

 dσ × r =

dτ∇ × r = 0

S

over any closed surface.

 

Example 1.7.4 Show that s dσ · r is three times the volume V enclosed by the closed surface S.    dσ · r = dτ∇ · r = 3 dτ = 3V.  S

R

Problems

 1.7.1 Show that s dσ = 0 over a closed surface S.  1.7.2 Calculate I = dσ · V(r) over the unit sphere for (a) V(r) = xy2 i + yz2 j + zx2 k, (b) V(r) = A(r) × r, where A(r) is irrotational. (Answers: I = 4π/5, 0.) 1.7.3 Calculate the total flux (or net outflow) over a spherical surface of radius R about the origin for the vector field F(r) =

r−a , |r − a|3

a = ai,

when R < a and when R > a. (Answers: Φ = 0, 4π.) 1.7.4 Verify Gauss’s theorem by showing separately that over a sphere of  radius R the surface integral dσ · (r/r n+1 ) = 0 and the volume integral  dτ∇ · (r/rn+1 ) are both equal to 4πR2−n . Explain why this is true even for n = 2 when ∇ · (r/rn+1 ) = 0 for all finite r. 1.7.5 (Green’s theorem in the plane) If V(x, y) = u(x, y)e x + ν(x, y)ey , use Gauss’s theorem to show that     ∂u ∂v dx dy = (u dy − v dx), + ∂y S ∂x c where c is an arbitrary continuous, closed curve forming the boundary of the surface S . This is just Gauss’s theorem in 2D. Hint: Change the theorem into a standard Gauss’s theorem in 3D by Z multiplying both sides by the constant Z = 0 dz. The volume of the theorem is that of the right cylinder based on the surface S in the xy plane. There is no outflow from either flat ends of the cylinder. To find the outflow from the

48

Vectors and fields in space

curved cylinder surface, denote the differential line element along c in the xy plane by dρ = dx ex + dy ey . Then the surface element for the curved surface dσ = dρ × ez dz lies in the xy plane. In the convention where the positive direction of c is counter-clockwise, the normal n = dσ/dσ to the closed curve c points to the right away from the enclosed area S . 1.7.6 If B = ∇ × A is the magnetic induction and D = ε0 E is the electric displacement satisfying the equation ∇ · D = ρ, evaluate the net outflow integrals over a closed surface S   B · dσ and D · dσ. S

S

Use these results to show that the normal component of B is continuous across the boundary between two media, while that of D has a discontinuity given by the surface charge density (per unit area) at the boundary. 1.7.7 Show that in electrostatics (where E = −∇Φ, D = ε0 E, ∇ · D = ρ)   ρΦdτ = D · E dτ if the electrostatic potential Φ(r) vanishes sufficiently rapidly for large r. If Φ(r) ∝ r−p , what inequality should p satisfy for the above identity to hold? (Answer: p > 1/2) 1.7.8 If T (r) is the temperature field in space, the heat flux (or flow) at a point r is q(r) = K∇T (r), where K is the thermal conductivity (in calories per second per meter squared for a thickness of 1 m and a temperature difference of 1K). When heat flows out of the region, its temperature decreases because its internal energy per unit mass e decreases by an amount Δe = CΔT , where C is the specific heat. From the energy conservation requirement   ∂e ρ dτ = − q · dσ, Ω ∂t S where ρ is the mass density, obtain a continuity equation known as the 2 diffusion equation for heat conduction ρC ∂T ∂t = K∇ T.

1.8

Circulation, curl and Stokes’s theorem

The circulation Γc of a vector field F(r) along a closed curve c is defined by the line integral  Γc ≡ (F x dx + Fy dy + F z dz). (1.96) c

Circulation, curl and Stokes’s theorem

49

The path c contains a direction of integration, the counter-clockwise direction being taken to be positive. The integral changes sign when its direction is reversed:   =− . (1.97) −c

c

Example 1.8.1 A helical pipe of helical radius a carries water at a constant speed v. The velocity circulation after n turns of flow in a counter-clockwise direction is   v · dr = (v x dx + vy dy). Γn = n turns

n turns

If x = a cos θ,

y = a sin θ,

v x = −v sin θ, vy = v cos θ,  Γn = av (sin2 θ + cos2 θ)dθ = n(2πav).  n turns

Another way to visualize the circulation is to consider a small rectangular closed loop on the plane z = constant, such as the one shown in Fig. 1.8. We find  (v x dx + vy dy) Γ= Δc

 Δx[v x (x1 , y, z) − v x (x3 , y + Δy, z)] + Δy[vy (x + Δx, y2 , z) − vy (x, y4 , z)], where xi , yi are suitable points on side i. Hence     Δy v x Δ x vy ∂v x ∂vy Γ = ΔxΔy −  Δx Δy − , + + Δy Δx ∂y ∂x

(1.98)

where the partials are calculated at a suitable point (x, y, z) on or inside the rectangular path. 1.8.1

Curl

The right-hand side of Eq. (1.98) involves (∇ × v)z . Hence Γ  (∇ × v)z (Δσ)z , = (∇ × v) · Δσ = (∇ × v) · n Δσ,

where

(Δσ)z = ΔxΔy

50

Vectors and fields in space

where n(= k here) is the normal to the surface Δσ enclosed by the loop. Since a scalar product does not depend on the absolute orientations of the vectors, but only on their relative orientation (i.e., cos θ), it is clear that the resulting circulation  v · dr  (∇ × v) · n Δσ (1.99) Γ= Δc

is actually independent of the choice of the coordinate system. As a consequence, the same result is obtained whether or not the small loop lies on the xy plane. If we now make the loop infinitesimally small, we obtain the identity  1 lim v · dr = [∇ × v(r)] · n. (1.100) Δσ→0 Δσ Thus the curl of a vector field v(r) at a point r in space has the simple interpretation that its component along any direction n is given by the circulation per unit enclosed area of v around an infinitesimally small closed loop surrounding r on a plane perpendicular to n. This circulation is to be calculated along the positive (i.e., righthanded) direction relative to n. 1.8.2

Stokes’s theorem

By “subdividing” a given closed loop into small closed loops ci , we find      1 v · dr = Δσi lim v · dr Δσi →0 Δσi ci c i  dσ · (∇ × v). =

(1.101)

S

This integral relation is known as Stokes’s theorem. It states that the circulation around a closed path is equal to the flow of the curl across the enclosed area. That is, the circulation is “caused” by a “flow of curl”, and vice versa. Example 1.8.2 Show that the gradient of a scalar field is irrotational: ∇ × ∇Φ(r) = 0. This can be done by using Stokes’s theorem to change a surface integral to a line integral:    dσ · [∇ × ∇Φ(r)] = dr · ∇Φ(r) = dΦ(r) S

c

c

= Φ(r0 ) − Φ(r0 ) = 0, where r0 is any point on c at which the closed curve can both begin and end. Since this result holds for any surface S, ∇Φ(r) must be irrotational. 

Circulation, curl and Stokes’s theorem

51

Thus a conservative force, which is derivable from a potential Φ(r), is irrotational. The converse of this is also true. Wherever ∇ × F = 0, F is conservative and is derivable from a potential. In fluid dynamics, the curl of the velocity field v(r) (which gives the velocity of the fluid at the point r) is called its vorticity—vortices being the whirls one creates in a cup of coffee on stirring it. If the velocity field is derivable from a potential v(r) = −∇Φ(r), it must be irrotational, according to Eq. (1.50). For this reason, an irrotational flow is also called a potential flow. Its circulation vanishes. Being thus free of vortices and eddies, it describes a steady flow of the fluid. One of Maxwell’s equations in electromagnetism is Ampere’s law: ∇ × B = μ0 J,

(1.102)

where B is the magnetic induction, J is the current density (per unit area), and μ0 is the permeability of free space. As a result, current densities may be visualized as “vortices” of B. An application of Stokes’s theorem to Eq. (1.102) gives Ampere’s circuital law: 

 B · dr = μ0

J · dσ = μ0 I.

(1.103)

c

This states that the circulation of the magnetic induction is proportional to the total current passing through the surface enclosed by c. It is worth pointing out that Stokes’s theorem in Eq. (1.101) is valid whether or not the closed curve c lies in a plane. This means that in general the surface S is not a planar surface. Indeed, S does not have to be planar even when c is planar, because the small closed loops ci can be taken out of the plane. Thus Stokes’s theorem holds for any surface bounded by c.  Example 1.8.3 Calculate s v(r) · dσ for the vector field v(r) = ω × r, where ω is a constant vector, over the surface S enclosed by a unit circle in the xy plane centered at the origin. The result is unchanged if the surface bounded by c is deformed into a hemispherical surface. Then dσ is proportional to r, and v · dσ vanishes everywhere on the hemisphere because ω × r · r = 0. Alternatively, ω × r can be integrated over the original planar surface enclosed by the unit circle c. The integral vanishes because contributions from equal areas on opposite sides of the origin always cancel because ω × r is odd in r. 

52

1.8.3

Vectors and fields in space

Operator identity

Stokes’s theorem, Eq. (1.101), can be applied to the vector field v(r) = ei Φ(r) to give   dxi Φ(r) = (dσ × ∇)i Φ(r). c

S

Since this is true for any i, it follows that    dr Φ(r) = ei dxi Φ(r) c

c

i



(dσ × ∇)Φ(r).

=

(1.104)

S

This result generalizes Stokes’s theorem to the operator identity   dr = (dσ × ∇) c

(1.105)

S

for operations on any field in space. It gives rise to other integral theorems such as the following:   (dσ × ∇) × V(r), (1.106) dr × V(r) = S



 dr · (u∇v) =

(dσ × ∇) · (u∇v) S

 =

dσ · [∇ × (u∇v)] S

 =

dσ · [(∇u) × (∇v)].

(1.107)

S

Example 1.8.4 If v(r) = φ(r)ex − ψ(r)ey , Stokes’s theorem for this vector field reads   v(r) · dr = [φ(r)dx − ψ(r)dy] c



c



S

[∇ × v(r)] · dσ

= =

[(∇ × v) x dy dz + (∇ × v)y dz dx + (∇ × v)z dx dy S

 =

S

 !  ∂ψ ∂φ ∂φ ∂ψ dx dy . dy dz + dz dx − + ∂z ∂z ∂x ∂y

Helmholtz’s theorem n

53

z Medium 1

y Δz

x

Medium 2

Fig. 1.12 Interface between two media.

If φ(r) and ψ(r) are both independent of z, the result simplifies to     ∂ψ ∂φ [φ(x, y)dx − ψ(x, y)dy] = − dx dy. + ∂y c S ∂x This is known as Green’s theorem in the plane.



Problems

 1.8.1 Use Stokes’s theorem to calculate r · dr. 1.8.2 Use Stokes’s theorem to show that Δc r × dr = 2 Δσ, where Δσ is the area enclosed by the small closed curve Δc. 1.8.3 Use Stokes’s theorem to identify the conservative forces in the following: (a) F(r) = xn i + yn j + zn k, (b) F(r) = zn i + xn j + yn k, (c) F(r) = e x sin y i + e x cos y j, (d) F(r) = rn e(r), n = any integer. 1.8.4 Use Stokes’s theorem to calculate the circulation of the following vector fields along the closed unit circle in the yz plane centered at the origin. (a) v(r) = yzi + zxj + xyk, (b) v(r) = ω × r, where ω is a constant vector. 1.8.5 A time-independent magnetic field H satisfies the equation ∇ × H = J, where J is the current density (per unit area). By considering the narrow loop perpendicular to the interface between two media 1 and 2 shown in Fig. 1.12, show that (H1 − H2 ) x = Jy Δz. Generalize this result to read n × (H1 − H2 ) = K, where K is the surface current density (per unit length) at the interface.

1.9

Helmholtz’s theorem

It is not accidental that the divergence and curl of a vector field play such important roles. Their significance is made clear by Helmholtz’s theorem, which we now state without proof: A vector field is uniquely determined by its divergence and curl in a region of space, and its normal component over the boundary of the region. In particular,

54

Vectors and fields in space

if both divergence and curl are specified everywhere and if they both disappear at infinity sufficiently rapidly, then the vector field can be written as a unique sum of an irrotational part and a solenoidal part. In other words, we may write V(r) = −∇Φ(r) + ∇ × A(r),

(1.108)

where −∇Φ is the irrotational part and ∇ × A is the solenoidal part. (This can be seen with the help of Eqs. (1.50) and (1.51), which show that both ∇ × (∇Φ) and ∇ · (∇ × A) vanish identically). The fields Φ(r) and A(r) are called the scalar and vector potential, respectively, of V(r). The Helmholtz theorem (1.108) will be derived in Section 4.14 using the method of 3D Fourier transform. It turns out to be related to the decomposition of a 3D vector into components parallel and perpendicular to a certain direction ek in a certain Fourier or reciprocal space, as illustrated in Example 1.2.7. It is useful to express the divergence and curl of V(r) in terms of the scalar and vector potentials. From Eq. (1.108), we obtain directly the results ∇ · V = −∇2 Φ,

∇ × V = ∇(∇ × A) = ∇(∇ · A) − ∇2 A,

(1.109)

where the last step is made with the help of the operator form of the BAC rule discussed in Section 1.5. It turns out that the term ∇(∇ · A) vanishes under certain conditions (e.g., when ∇ × V is either bounded in space or vanishes more rapidly than 1/r for large r). Then ∇ × V = −∇2 A.

(1.110)

In this way we can see that ∇ · V is related to the scalar potential Φ while ∇ × V is related to the vector potential A. It is possible to make these relations quite explicit, although we are not in a position to derive the final results until section 4.14. We begin by pointing out that Eq. (1.109) has the form of a Poisson equation ∇2 Φ(r) = −s(r),

s(r) = ∇ · V(r),

(1.111)

where the charge density ρ(r) of electrostatics is now denoted s(r), the scalar source density of Φ(r). From electrostatics, we know that the electrostatic potential due to a charge distribution is  s(r2 ) 1 dτ2 , r12 = |r − r2 |. (1.112) Φ(r) = 4π r12 Hence the scalar potential in Eq. (1.108) can be written explicitly as  1 ∇ · V(r2 ) Φ(r) = dτ2 . 4π r12

(1.113)

Helmholtz’s theorem

55

Similarly, Eq. (1.110) shows that each component Ai (r) of the vector potential also satisfies a Poisson equation. Hence A(r) has the explicit solution  1 ∇ × V(r2 ) dτ2 . (1.114) A(r) = 4π r12 Thus the divergence and curl (or vorticity) of a vector field can be interpreted as the source density of its scalar and vector potential, respectively. Once the structure of a vector field is understood, it is easy to form a physical picture of a vector field used in physics. As an example, let us consider Maxwell’s equations in vacuum in SI units: ∇·E=

ρ , ε0

∇ · B = 0,

∂B , ∂t ∂E + μ0 J. ∇ × B = μ0 ε0 ∂t ∇×E=−

(1.115)

We see by inspection that these equations summarize the following experimental facts concerning the two vector fields E and B of electromagnetism: 1. The source of the electroscalar potential (i.e., the scalar potential of the electric field, or the electrostatic potential) is proportional to the charge density ρ(r). The source of the electrovector potential is Faraday’s induction term (∂/∂t)B(r). 2. There is no magnetoscalar potential, because the magnetic monopole density is found experimentally to vanish everywhere. The magnetovector potential (i.e., the vector potential A) originates from either an actual current density (Ampere’s law) or Maxwell’s displacement current density ε0 ∂E/∂t. In other words, Maxwell’s equations describe the physical nature of the four source densities of the electromagnetic fields. Eqs. (1.115) are known as Maxwell’s equations in honor of his discovery (in 1865) of the displacement current. In the absence of this displacement current term, the fourth equation (Ampere’s law) gives a result ∇ · (∇ × B) = 0 = μ0 ∇ · J, that violates charge conservation as expressed by the continuity equation, Eq. (1.91): ∇·J+

∂ρ =0 ∂t

when the charge density ρ changes in time. The missing term ∂ρ/∂t when expressed in terms of ε0 E with the help of Coulomb’s law, the first of Eqs. (1.115), gives rise to Maxwell’s displacement current.

56

Vectors and fields in space

Problems 1.9.1 With the help of Helmholtz’s theorem, show that a velocity field v(r) free of vortices, ∇ × v = 0, is completely determined by a scalar potential. 1.9.2 The Maxwell equations (1.115) are four coupled first-order partial differential equations (PDEs) for the vector fields E and B. (PDEs are DEs in more than one variable. Four variables, x, y, z, t, are needed to describe events in spacetime.) Show by eliminating the coupling between E and B that two uncoupled PDEs of second order in the spacetime variables can be obtained, one for E and one for B:   1 1 2 2 ∇ − 2 ∂t E = ∇ρ + μ0 ∂t J, ε0 c   1 ∇2 − 2 ∂2t B = −μ0 ∇ × J, c where ∂t = ∂/∂t. These are the inhomogeneous wave equations with which Maxwell √ showed that the speed c = 1/ μ0 ε0 of electromagnetic waves they described should be close to that of light.

1.10

Orthogonal curvilinear coordinate systems

In discussing vectors and fields in space, we have so far used only the rectangular or Cartesian coordinate system. This system has great intuitive appeal, since it makes use of the straight lines and perpendicular directions of the flat space in which we live. The simplicity of vector differential and integral operators in rectangular coordinates reinforces our fondness for them. However, many physical systems are not naturally rectangular. A sphere is a good example. The rectangular coordinates of a point on its surface are changing from point to point, but in spherical coordinates the spherical surface is specified simply as a surface of constant radius r. Thus the choice of coordinate systems can be important in the description of physical systems. A good choice may lead to greater simplification and insight in the description of physical properties. There is, of course, a price to pay for this improvement. Coordinate systems other than the rectangular are less intuitive and harder to visualize. Integral and differential operators have more complicated forms, and these are harder to memorize. It is the purpose of this and the following sections to show that the task is quite tractable, perhaps even enjoyable, when approached from a certain point of view. This turns out to be just a matter of changing directions and changing scales, as we shall see. 1.10.1

Generalized coordinates

Let us begin by noting that any three independent variables (u1 , u2 , u3 ) can be used to form a coordinate system if they specify uniquely the position of a point in space.

Orthogonal curvilinear coordinate systems

57

z r

θ y

x

Fig. 1.13 Spherical coordinates as related to rectangular coordinates.

It is convenient to start with the familiar rectangular coordinates (x, y, z) = (x1 , x2 , x3 ) and specify the new, generalized coordinates in terms of these: ui = ui (x, y, z).

(1.116a)

For the transformations between these two coordinate systems to be well defined and unique, it is necessary that the inverse relations xi = xi (u1 , u2 , u3 ),

i = 1, 2, 3,

(1.116b)

also exist, and that all these relations are single-valued functions. For example, the spherical coordinates may be defined in terms of the rectangular coordinates by either of the following sets of equations (see Fig. 1.13): r = (x2 + y2 + z2 )1/2 θ = arctan[(x2 + y2 )/z2 ]1/2

(1.117a)

φ = arctan(y/x), or x = r sin θ cos φ, y = r sin θ sin φ,

(1.117b)

z = r cos θ. A coordinate axis now becomes a coordinate curve, along which only of the coordinates is changing. The coordinate curves in spherical dinates are radii (for variable r), longitudes (for variable co-latitude angle θ), and latitudes (for variable longitudinal angle φ).

58

Vectors and fields in space

It is easy to find algebraic expressions describing these coordinate curves, because by definition only one of the coordinates changes along such a curve, while all the others remain unchanged. For example, r = rer in spherical coordinates. Then the partial derivatives ∂ r = er , ∂r

∂ ∂er r=r , ∂θ ∂θ

∂ ∂er r=r ∂φ ∂φ

describe the vectorial changes along these coordinate curves. Each of these derivatives is a vector in space; it has a length hi and a direction ei , i = r, θ or φ. Their explicit forms can be obtained readily with the help of rectangular coordinates: ∂ ∂r = (xe x + yey + zez ) ∂r ∂r = sin θ cos φ e x + sin θ sin φ ey + cos θ ez = hr er , ∂r = r cos θ cos φ e x + r cos θ sin φ ey − r sin θ ez = hθ eθ , ∂θ ∂r = −r sin θ sin φ e x + r sin θ cos φ ey = hφ eφ . ∂φ

(1.118)

A simple calculation then yields hr = 1,

hθ = r,

hφ = r sin θ,

(1.119)

and therefore dr =

∂r ∂r ∂r dr + dθ + dφ ∂r ∂θ ∂φ

= hr er dr + hθ eθ dθ + hθ eφ dφ = dr er + r dθ eθ + r sin θ dφ eφ . For arbitrary generalized coordiriates (u1 , u2 , u3 ), coordinate curves can be determined in a similar way by examining the displacement dr =

∂r ∂r ∂r du1 + du2 + du3 . ∂u1 ∂u2 ∂u3

Each partial derivative ∂r ∂x ∂y ∂z = ex + ey + ez . ∂ui ∂ui ∂ui ∂ui

(1.120a)

is a vector (the tangent vector) with a length and a direction: ∂r = hi (r)ei (r), ∂ui

(1.120b)

Orthogonal curvilinear coordinate systems

59

where ⎡ 2  2  2 ⎤1/2 ⎢⎢ ∂x ∂y ∂z ⎥⎥⎥⎥ hi (r) = ⎢⎢⎢⎣ + + ⎥ . ∂ui ∂ui ∂ui ⎦ In terms of these quantities, the original displacement may be written in the form   dr = hi (r)ei (r)dui = ei (r)dsi (r). (1.121) i

i

Thus ei (r) defines a coordinate curve, since it gives the unit vector tangent to the curve at r. The infinitesimal scalar displacement dsi (r) = hi (r)dui gives the displacement along this coordinate curve. The function hi (r) is called a scale factor, since it ensures that the displacement has the dimension of a length irrespective of the nature and dimension of the generalized coordinate ui itself. Other geometrical quantities can be calculated readily in terms of these scale factors and unit tangent vectors. Thus the infinitesimal scalar displacement ds along a path in space is  (ds)2 = dr · dr = gij dui duj , (1.122) i, j

where gij = hi hj (ei · ej ) are called the metric coefficients of the generalized coordinate system. The differential elements of surface and volume can be written down with the help of Eqs. (1.26) and (1.27) of Section 1.2:     ∂r ∂r dui × duj = hi hj (ei × ej )dui duj dσij = ∂ui ∂uj = dsi dsj (ei × ej ),

(1.123)

and dτ = ds1 ds2 ds3 e1 · (e2 × e3 )

(1.124a)

= du1 du2 du3 J, where J=

dx dy dz = h1 h2 h3 e1 · (e2 × e3 ) du1 du2 du3

(1.124b)

60

Vectors and fields in space

is called a Jacobian. It is the 3D generalization of J = dx/du in the 1D change of variable dx = (dx/du)du = Jdu. 1.10.2

Orthogonal curvilinear coordinates

If at every point r the three unit tangets ei (r) are orthogonal to one another, that is, if ei (r) · ej (r) = δij ,

(1.125)

or equivalently if ∂r ∂r · ∝ δij , ∂ui ∂uj then the generalized coordinates ui are said to form an orthogonal curvilinear coordinate (OCC) system. In this system, the unit tangents e1 (r), e2 (r) and e3 (r) form a Cartesian coordinate system at every point r. The only complication is that their orientations change from point to point in space. In an OCC system, the metric coefficients  2  2  2 ∂x ∂y ∂z 2 2 gij = hi δij , hi = + + , (1.126) ∂ui ∂ui ∂ui are diagonal, and the squared length (ds) = 2

3 

h2i (dui )2 =



(dsi )2

(1.127)

i

i=1

does not contain cross terms. (The quantity dsi always has the dimension of length.) The differential elements of surface and volume also simplify to  dσij = dsi dsj εijk ek , (1.128) k

and dτ = ds1 ds2 ds3 .

(1.129)

Thus the dsi are very much like the rectangular coordinate dxi . However, the tangent directions ei (r) do change from point to point, except in the special case of rectangular coordinates for which they are constant unit vectors. For spherical coordinates, direct calculations with the tangent vectors of Eq. (1.118) show that er · eθ = 0 = eθ · eφ = eφ · er ,

Orthogonal curvilinear coordinate systems

61

Table 1.1 Two common orthogonal curvilinear coordinate systems

Coordinates

Cylindrical

Spherical

ρ, φ, z

r, θ, φ

Definitions

x = ρ cos φ y = ρ sin φ z=z

x = r sin θ cos φ y = r sin θ sin φ z = r cos θ

Scale factors

hρ = 1 hφ = ρ hz = 1

hr = 1 hθ = r hφ = r sin θ

and er × eθ = eφ ,

eθ × eφ = er ,

eφ × er = eθ .

These orthogonality properties are easily visualized geometrically because er points radially outward from the origin, eθ points south along a longitude, while eφ points east along a latitude. Also with dsr = dr,

dsθ = rdθ,

dsφ = r sin θdφ,

we find (ds)2 = (dr)2 + (rdθ)2 + (r sin θdφ)2 , dσrθ = rdr dθ eφ = −dσθr , dσθφ = r 2 sin θdθ dφ er = −dσφθ , dσφr = r sin θ dφ dr eθ = −dσrφ , dτ = r2 sin θdr dθ dφ. We can also define coordinate surfaces on which one of the coordinates is constant, while the remaining two coordinates change. The coordinate surfaces for spherical coordinates are spherical surfaces of constant r, conical surfaces of constant θ, and half-planes of constant φ. For OCC systems, coordinate surfaces are perpendicular to coordinate curves along which the third coordinate changes value. For example, spherical surfaces of constant r are perpendicular to the coordinate curve e r = er . We list in Table 1.1 the definitions of ui , and their scale factors, in two common OCC systems. The coordinate surfaces for cylindrical coordinates are the cylindrical surfaces of constant ρ, the planes of constant z, and the half-planes of constant φ.

62

Vectors and fields in space

Finally we should note that the coordinate r(t) of a physical event is a function of time. This can be differentiated with respect to time to give successively its velocity, acceleration, etc. If curvilinear coordinates are used, there will be contributions from the time variations of both the tangent vectors and the curvilinear coordinates themselves. More specifically, we find by dividing Eq. (1.121) by dt that v=

dr  u˙ i hi ei , = dt i

u˙ i =

dui . dt

(1.130a)

A direct differentiation then gives

a=

#  dv  " u¨ i hi + u˙ i h˙ i ei + u˙ i hi e˙ i . = dt i

(1.130b)

In spherical coordinates, for example, Eq. (1.130) reads ˙ θ + φr ˙ sin θeφ . v = r˙er + θre

(1.131)

Since it is also true that v=

d (rer ) = r˙er + r˙er , dt

we see that ˙ θ + φ˙ sin θeφ , e˙ r = θe

(1.132a)

a result that can also be obtained by a direct differentiation of er from Eq. (1.118). Similarly, we find by a direct differentiation of eθ ˙ sin θ cos φ e x − sin θ sin φ ey − cos θ ez ) e˙ θ = θ(− ˙ cos θ sin φ e x + cos θ cos φ ey ) + φ(− ˙ r + φ cos θ eφ , = −θe

(1.132b)

˙ cos φ e x − sin φ ey ) e˙ φ = φ(− ˙ = −φ(sin θ er + cos θ eθ ).

(1.132c)

These are the basic relations that can be used to simplify the time rate of change of any vector field expressed in terms of spherical coordinates. Eleven other OCC systems in frequent use are listed in Table 1.2.

Table 1.2 Orthogonal curvilinear coordinate systems.

Coordinates (a) Elliptic cylindrical

x

y

z

a cosh u cos v

a sinh u sin v

z

− ξ2)

z

(b) Parabolic cylindrical

ξη

(c) Bipolar (cylindrical)

a sinh η cosh η − cos ξ

a sin ξ cosh η − cos ξ

z

(d) Prolate spheroidal

a sinh u sin v cos φ

a sinh u sin v sin φ

a cosh u cos v

(e) Oblate spheroidal

a cosh u cos v cos φ

a cosh u cos v sin φ

a sinh u cos v

ξη cos φ

ξη sin φ

a sinh η cos φ cosh η − cos ξ a sin ξ cos φ cosh η − cos ξ

a sinh η sin φ cosh η − cos ξ a sin ξ sin φ cosh η − cos ξ

1 2 (η − ξ 2 ) 2 a sin ξ cosh η − cos ξ a sinh η cosh η − cos ξ

(f) Parabolic (g) Toroidal (h) Bispherical  (i) Confocal ellipsoidal

1 ξ2 ξ3

1/2



(a2 −ξ1 )(a2 −ξ2 )(a2 −ξ3 ) (b2 −a2 )

(b2 −ξ1 )(b2 −ξ2 )(b2 −ξ3 ) (b2 −a2 )(b2 −c2 )



#1/2

bc

 (k) Confocal parabolic

(a2 −ξ1 )(a2 −ξ2 )(a2 −ξ3 ) (a2 −b2 )(a2 −c2 )

"ξ

(j) Conical

1 2 2 (η

1/2



ξ12 (b2 −ξ22 )(b2 −ξ32 ) b2 (b2 −c2 )

1/2

1/2

(b2 −ξ1 )(b2 −ξ2 )(b2 −ξ3 ) (a2 −b2 )

1/2



(c2 −ξ1 )(c2 −ξ2 )(c2 −ξ3 ) (c2 −a2 )(c2 −b2 )



ξ12 (c2 −ξ22 )(c2 −ξ32 ) c2 (c2 −b2 )

1 2 2 (a

1/2

1/2

+ b2 − ξ1 − ξ2 − ξ3 )

64

Vectors and fields in space

Problems 1.10.1 Obtain the unit tangent vectors ei (r) for the cylindrical coordinate system and show that they are orthogonal to each other. 1.10.2∗ Obtain the scale factors and the unit tangent vectors, demonstrate the orthogonality, and describe the coordinate curves and surfaces of the generalized coordinate systems (a) and (b) defined in Table 1.2. Plots of some of these curves and surfaces can be found in Morse and Feshbach, Methods of Theoretical Physics, pp. 657–66, and in Arfken, Mathematical Methods for Physicists, 2nd ed., pp. 95–120. The latter reference contains the answers to this question. The parameters a, b and c in the table are constants having the dimension of length. They are not the generalized coordinates. Hints: (a) (Elliptic cylinder) The parametric equations for ellipses and hyperbolas are, respectively: x = A cos v,

y = B sin v;

x = C cosh u,

y = D sinh u.

Also

h2u = h2v = a2 (sinh2 u + sin2 v). (b) (Parabolic cylinder) The equation of a parabola along the y axis with vertex at y0 is x2 = 2p(y − y0 ). h2ξ

=

h2η

Also

=ξ +η . 2

2

1.10.3 Calculate the differential elements dsi , (ds)2 , dσij , and dτ for the curvilinear coordinate systems of Problem 1.10.2. Hint: Use the scale factors and unit tangent vectors obtained in Problem 1.10.2. 1.10.4 Express v = dr/dt and a = dv/dt in cylindrical coordinates. 1.10.5 Express a = d 2 r/dt2 in spherical coordinates. 1.10.6 Express v = dr/dt and a = dv/dt in the parabolic cylinder coordinate system. Partial answer: e˙ ξ =

˙ ηξ ˙ − ξη eη , 2 ξ + η2

e˙ η =

˙ − ηξ ξη ˙ eη . 2 ξ + η2

1.10.7 (2D Jacobian) The quantities ∂x/∂ui , dσij , etc., defined by Eqs. (1.120, 1.123) are valid even for 2D/3D curvilinear coordinates ui (x, y, z), i = 1, 2, 3 that are not orthogonal. Consider such a general nonorthogona1 coordinate system in the xy plane defined by the functions ui (x, y), i = 1, 2, while u3 = z.

Vector differential operators

65

(a) Show from Eq. (1.120a) that ∂x/∂ui = hi cix , where cix = ei · e x . (b) Show that e1 × e2 = (c1x c2y − c1y c2x )ez . (c) Show from Eq. (1.123) that the vector differential area is   x, y du1 du2 ez , dσ12 ≡ dxdyez = J u1 , u2    ∂x ∂x    x, y 1 ∂u2  J =  ∂u ∂y ∂y , u 1 , u2  ∂u1 ∂u2  where J is called a Jacobian for the change of variable. 1.10.8 (3D Jacobian) Eq. (1.124) for the differential volume dτ is valid even for the 3D curvilinear coordinates ui (x, y, z), i = 1, 2, 3, that are not orthogonal. Use the notation of Problem 1.10.7 to show that    c1x c1y c1z  (a) e1 · (e2 × e3 ) =  c2x c2y c2z ;  c c c  3x 3y 3z   x, y, z (b) dτ ≡ dx dy dz = J du1 du2 du3 , u1 , u2 , u3  ∂x ∂x ∂x    ∂u1 ∂u2 ∂u3    ∂y ∂y ∂y  x, y, z . =  ∂u J u1 , u2 , u3  1 ∂u2 ∂u3  ∂z ∂z ∂z   ∂u ∂u ∂u 1

1.11

2

3

Vector differential operators in orthogonal curvilinear coordinate systems

Since the generalized coordinates r = (u1 , u2 , u3 ) define a point in space, scalar and vector fields in space can be written as Φ(u1 , u2 , u3 ) or V(u1 , u2 , u3 ). In performing vector differential operations on such fields, it is of course possible to change variables back to the rectangular coordinates x, y and z before applying the VDO in its familiar rectangular form. This procedure is often unwise, because the calculation may become rather complicated, and the physical insights to the problem that might have motivated the use of a special system of generalized coordinates might be lost this way. The purpose of this section is to show how such calculations can be made directly by using the generalized coordinates themselves. The procedures are not very complicated, and are well worth mastering because they give insight to our understanding of generalized coordinates. We shall restrict ourselves to orthogonal curvilinear coordinates.

66

Vectors and fields in space

Let us start with ∇Φ(u1 , u2 , u3 ). At every point r, it describes a vector that can be decomposed into components along the local unit tangents ei (r): ∇Φ(r) =

3 

ei (r)(∇Φ)i ,

i=1

where (∇Φ)i = (∇Φ) · ei (r)   ∂Φ ∂Φ ∂Φ = ex + ey + ez · ei (r) ∂x ∂y ∂z The direction cosines involved turn out to be just the partial derivatives e x · ei = e x · =

∂r ∂ = ex · (xe x + yey + zez ) ∂si ∂si

∂x . ∂si

Hence (∇Φ)i = =

∂Φ ∂x ∂Φ ∂y ∂Φ ∂z + + ∂x ∂si ∂y ∂si ∂z ∂si ∂Φ , ∂si

and ∇Φ(u1 , u2 , u3 ) =

3  i=1

ei (r)

∂Φ . ∂si

(1.133)

That is, ui in ∇Φ(u1 , u2 , u3 ) can be treated as if they were rectangular coordinates if the local tangents ei (r) are used together with the dimensioned displacements dsi = hi dui . Example 1.11.1 Calculate ∇(r2 sin θ). According to Eq. (1.119)   ∂ 1 ∂ 1 ∂ Φ(r, θ, φ). + eφ ∇Φ(r, θ, φ) = er + eθ ∂r r ∂θ r sin θ ∂φ Hence ∇(r2 sin θ) = er 2r sin θ + eθ r cos θ. 

Vector differential operators

67

Perhaps the most interesting gradients in a curvilinear coordinate system are those constructed from the curvilinear coordinates uj (r) themselves: ∇uj (r) =



ei (r)

i

1 ∂uj ej (r) . = hi ∂ui hj (r)

(1.134)

This result expresses the fact that ∇uj (r) changes most rapidly along the uj coordinate curve. It also shows that ej is normal to the coordinate surface defined by the equation uj (r) = const. Now the gradient of a scalar field is always irrotational; hence ∇ × (ej /hj ) = 0.

(1.135)

Since −1 ∇ × (ej /hj ) = (∇h−1 j ) × ej + hj (∇ × ej ),

we find that a generalized coordinate curve has a natural curliness of ∇ × ej = −hj (∇h−1 j ) × ej . One consequence of Eq. (1.136) is that the curl of a vector field  ej (r)Vj (r) V(r) =

(1.136)

(1.137)

j

has contributions from both ej and Vj . Both contributions can be combined into a simple expression with the help of Eqs. (1.55) and (1.135):    ej ∇×V= ∇× hj Vi hj j  !   ej ej = ∇× hj Vj + ∇(hj Vj ) × hj hj j =



∇(hj Vj ) ×

j

=

 i, j,k

εijk

ej hj

1 ∂ (hj Vj )ek . hj ∂si

(1.138a)

The result shows that, besides the usual “Cartesian” contribution proportional to ∂Vj /∂si , there is an additional effect proportional to (Vj /hj )(∂hj /∂si ) that arises from the curliness of a coordinate curve.

68

Vectors and fields in space

Another useful way of writing Eq. (1.138a) is ∇×V=



εijk

i, j,k

1 ∂ (hj Vj )ek hi hj ∂ui

   h1 e1 h2 e2 h3 e3  1  ∂  ∂u1 ∂u∂2 ∂u∂3  . = h1 h2 h3   h1 V1 h2 V2 h3 V3 

(1.138b)

To derive a similarly compact and useful expression for ∇ · V, we start with the curl of um ∇un : ∇ × (um ∇un ) = (∇um ) × (∇un ) + um ∇ × (∇un )     em en = × , using Eq. (1.134) hm hn  ek = εmnk . hm hn k

(1.139)

Since this vanishes for m = n for which the permutation symbol has repeated indices, the nonzero cases involve m  n. For example, ∇ × (u1 ∇u2 ) =

e3 = −∇ × (u2 ∇u1 ). h1 h2

Now, according to Eq. (1.51), the curl of any vector field is always solenoidal and free of divergence. This holds also for the curl of the last equation. That is, ∇ · (ei /pi ) = 0,

if

pi = h1 h2 h3 /hi .

(1.140)

Since −1 ∇ · (ei /pi ) = (∇p−1 i ) · ei + pi ∇ · ei ,

this means that ∇ · ei =

−pi (∇p−1 i )

  ∂ 1 · ei = −pi ∂si pi

does not in general vanish. In other words, the curliness of the coordinate curve contributes to the divergence. We are now in a position to put these results together to determine the divergence of a vector field written in generalized coordinates

Vector differential operators

69



 ! ei ∇·V= ∇ · (ei Vi ) = ∇· p i Vi pi i i  !   ei ei = ∇· pi Vi + ∇(pi Vi ) · pi pi i 

=



1 ∂ (pi Vi ). pi ∂si i

(1.141)

We have thus reduced ∇ × V and ∇ · V in the OCC system to something involving derivatives of scalar fields that are similar in structure to those in rectangular coordinates. All one has to do is to put in an appropriate scale function before a differentiation and to divide it out afterwards, as shown in both Eqs. (1.138) and (1.141). It is also not accidental that the formula for ∇ × V comes from ∇ui , while that for ∇ · V is related to um ∇un . The reason is that the surface integral of ∇ × V is related to a line integral (i.e., the circulation of v) according to Stokes’s theorem. Hence it depends on the linear changes ∇ui . For ∇ · V, Gauss’s theorem relates a volume integral to a surface integral in which the scale factor hi appears quadratically. The derivation of Eqs. (1.138) and (1.141) directly from these integral theorems will be left as exercises (Problems 1.11.7 and 1.11.8). Example 1.11.2 Calculate ∇ · [(er + eθ + eφ )/r2 ]. The scale functions for ∇ · V in spherical coordinates are pr = hr hθ hφ /hr = r2 sin θ, pθ = r sin θ, pφ = r. Hence ∇ · (er Vr + eθ Vθ + eφ Vφ ) = =

=

1 ∂ 1 ∂ 1 ∂ (pr Vr ) + (pθ Vθ ) + (pφ Vφ ) pr ∂sr pθ ∂sθ pφ ∂sφ ∂ 2 (r sin θVr ) ∂r 1 1 ∂ (r sin θVθ ) + r sin θ r ∂θ 1 1 ∂ + (rVφ ) r r sin θ ∂φ 1

r2 sin θ

1 ∂ 1 ∂ 2 (r Vr ) + (sin θVθ ) r sin θ ∂θ r2 ∂r 1 ∂ + Vφ . r sin θ ∂φ

70

Vectors and fields in space

Therefore

  er + e θ + e φ cos θ 1 ∂ sin θ = 3 . = r sin θ ∂θ r2 r2 r

∇· 1.11.1



Repeated operations

To avoid possible confusion and error in calculating repeated VDOs such as ∇2 = ∇ · ∇, it is best to proceed one step at a time, because the pj (r) functions may differ in different operators. This stepwise procedure is illustrated by the following examples. Example 1.11.3 Obtain ∇2 = ∇ · ∇ in an OCC system. For definiteness we may consider ∇2 Φ(u1 , u2 , u3 ) = ∇ · V(u1 , u2 , u3 ), where V = ∇Φ.  (1) V = ∇Φ, Vi = ∂s∂ i Φ = hi ∂∂ui Φ, V = i Vi ei . (2) ∇ · V →

∂ ∂s1 V1

+

∂ ∂s2 V2

∇ · V(u1 , u2 , u3 ) =

+

∂ ∂s3 V3

if Cartesian. Hence

1 ∂ 1 ∂ 1 ∂ (p1 V1 ) + (p2 V2 ) + (p3 V3 ), p1 ∂s1 p2 ∂s2 p3 ∂s3

where pi = h1 h2 h3 /h1 ,

dsi = hi dui .

(1.142)



Example 1.11.4 Calculate ∇2 Φ in spherical coordinates. We note that For gradient hr = 1

(∇Φ)r =

hθ = r

(∇Φ)θ =

hφ = r sin θ

(∇Φ)φ =

∂Φ ∂r 1 ∂Φ r ∂θ 1 ∂Φ r sin θ ∂φ

For divergence pr = hθ hφ = r2 sin θ pθ = hφ hr = r sin θ pφ = hr hθ = r

Hence 3  1 ∂ (pi Vi ) ∇ Φ= p ∂si i=1 i      ∂ 1 1 ∂ ∂ 2 ∂ r sin θ + r sin θ = 2 ∂r r sin θ r∂θ r∂θ r sin θ ∂r   ! 1 ∂ ∂ + r Φ(r, θ, φ) r r sin θ∂φ r sin θ∂φ   1 ∂ 2∂ ∂2 1 ∂ 1 ∂ = 2 r Φ(r, θ, φ). + sin θ + ∂θ r2 sin2 θ ∂φ2 r ∂r ∂r r2 sin θ ∂θ 2



Vector differential operators

71

In both examples, the additional factors between differential operators come from the additional scale functions appearing in OCC systems.

Problems 1.11.1 Show that ∂/∂x, ∂/∂y, ∂/∂z expressed in terms of spherical coordinates are: ∂ 1 ∂ sin φ ∂ ∂ = e x · ∇ = sin θ cos φ + cos θ cos φ − , ∂x ∂r r ∂θ r sin θ ∂φ ∂ ∂ 1 ∂ cos φ ∂ = sin θ sin φ + cos θ sin φ + , ∂y ∂r r ∂θ r sin θ ∂φ ∂ 1 ∂ ∂ = cos θ − sin θ . ∂z ∂r r ∂θ 1.11.2 Calculate ∇2 Φ( ρ, φ, z) in cylindrical coordinates. 1.11.3 Calculate ∇Φ( ρ, φ, z) and ∇2 Φ in cylindrical coordinates for (a) Φ = ρ, (b) Φ = ρ2 + z2 , (c) Φ = ρ2 tan φ + ρz tan2 φ. 1.11.4 Calculate ∇Φ(r, θ, φ) and ∇2 Φ in spherical coordinates for (a) Φ = Φ(r), (b) Φ = r2 (sin θ + cos φ), (c) Φ = Ar cos θ cos φ + Br3 cos3 θ sin θ, A and B being constants. 1.11.5 Calculate ∇ · V( ρ, θ, z) and ∇ × V in cylindrical coordinates for (a) V = ρeρ + zez , (b) V = eρ , (c) V = eφ , (d) V = ln ρ ez , (e) V = ln ρ eφ . 1.11.6 Calculate ∇ · V(r, θ, φ) and ∇ × V in spherical coordinates for (a) V = rer , (b) V = f (r)er , (c) V = eθ , (d) V = −(cot θ/r)eφ . 1.11.7 Use Gauss’s theorem in the form of Eq. (1.84) 1 ∇ · V = lim Δτ→0 Δτ

 ΔS

V · dσ

to derive the formula for ∇ · V in an OCC system.

72

Vectors and fields in space

1.11.8 Use Stokes’s theorem in the form of Eq. (1.100)  1 (∇ × V)i = lim V · dr Δσi →0 Δσi ci to derive the formula for ∇ × V in an OCC system. 1.11.9 Obtain a general expression for each of the following VDO in OCC systems: (a) ∇(∇ · V); (b) ∇ · (∇ × V); (c) ∇ × (∇ × V); (d) ∇ × (∇Φ);  where V = i Vi (u1 , u2 , u3 )ei , Φ = Φ(u1 , u2 , u3 ) are understood to be given functions of the generalized coordinates.

Appendix 1 Tables of mathematical formulas 1 Vectors A · B = A x Bx + Ay By + Az Bz = B · A    i j k   A × B =  A x Ay Az  = εijk ei Aj Bk   i, j,k Bx By Bz A × (B × C) = B(A · C) − C(A · B)

   A x Ay Az  A · (B × C) = B · (C × A) = C · (A × B) =  B x By Bz   C C C  x y z (A × B) · (C × D) = (A · C)(B · D) − (A · D)(B · C) (A × B) × (C × D) = −A[B · (C × D)] + B[A · (C × D)] = C[(A × B) · D] − D[(A × B) · C].

2 Vector differential operators ∇(φψ) = φ∇ψ + ψ∇φ ∇ · (φA) = φ∇ · A + (∇φ) · A ∇ × (φA) = φ∇ × A + (∇φ) × A ∇ · (A × B) = B · (∇ × A) − A · (∇ × B) ∇ × (A × B) = (B · ∇)A − B(∇ · A) − (A · ∇)B + A∇ · B ∇(A · B) = (B · ∇)A + (A · ∇)B + B × (∇ × A) + A × (∇ × B) ∇ × ∇φ = 0 ∇ · (∇ × A) = 0

Tables of mathematical formulas

∇ × (∇ × A) = ∇(∇ · A) − ∇2 A ∇·r=3 ∇×r=0 ∇r2 = 2r ∇2 r2 = 6. 3 Orthogonal curvilinear coordinates (u1 , u2 , u3 ) ∂r = hi (r)ei (r) ∂ui ⎡ 2  2  2 ⎤1/2 ⎢⎢ ∂x ∂y ∂z ⎥⎥⎥⎥ hi (r) = ⎢⎢⎢⎣ + + ⎥ ∂ui ∂ui ∂ui ⎦ ei (r) · ej (r) = δij   dr = hi (r)ei (r)dui = dsi (r)ei (r) i

i

dsi = hi dui dσij = dsi dsj (ei × ej ) dτ = ds1 ds2 ds3 = h1 h2 h3 du1 du2 du3 ∇ui (r) = ei (r)/hi (r), ∇ × (ei /hi ) = 0  εijk ek , ∇ · (∇ × ui ∇uj ) = 0 ∇ × (ui ∇uj ) = hi hj k ∇φ(u1 , u2 , u3 ) =

 ∂φ ei ∂si i

∇ · (V1 e1 + V2 e2 + V3 e3 ) =

1 ∂ (pi Vi ), pi ∂si i

∇ × (V1 e1 + V2 e2 + V3 e3 ) = 

∇2 φ(u1 , u2 , u3 ) =



εijk ei

i, j,k

1 ∂ (hk Vk ) hk ∂sj

 1 ∂ ∂ pi φ. p ∂s ∂s i i i i

4 Cylindrical coordinates ( ρ, φ, z) x = ρ cos φ, hρ = 1,

y = ρ sin φ,

hφ = ρ,

eρ = cos φi + sin φj

hz = 1,

pi =

z=z dτ = ρdρdφdz

h1 h2 h3 hi

73

74

Vectors and fields in space

eφ = − sin φi + cos φj ez = k eρ × eφ = ez ,

eφ × ez = eρ ,

ez × eρ = eφ

r = ρeρ + zez d ˙ φ eρ = φe dt ˙ ρ e˙ φ = −φe e˙ ρ =

e˙ z = 0 ˙ φ + z˙ez v = r˙ = ρe ˙ ρ + ρφe 1 ∂ 1 ∂ ∂ (ρAρ ) + A φ + Az ρ ∂ρ ρ ∂φ ∂z    eρ ρeφ ez  1 ∂ ∂ ∂   ∇ × (Aρ eρ + Aφ eφ + Az ez ) =  ∂ρ ∂φ ∂z  ρ  Aρ ρAφ Az  ∇ · (Aρ eρ + Aφ eφ + Az ez ) =

∇2 u(ρ, φ, z) =

1 ∂ ∂u 1 ∂2 u ∂2 u ρ + 2 2 + 2. ρ ∂ρ ∂ρ ρ ∂φ ∂z

5 Spherical coordinates (r, θ, φ) x = r sin θ cos φ,

y = r sin θ sin φ,

hr = 1,

hφ = r sin θ,

hθ = r,

z = r cos θ

dτ = r 2 sin θdr dθ dφ

er = sin θ cos φ i + sin θ sin φ j + cos θ k eθ = cos θ cos φ i + cos θ sin φ j − sin θ k eφ = − sin φ i + cos φ j er × eθ = eφ ,

eθ × eφ = er ,

eφ × er = eθ

r = rer ˙ θ + φ˙ sin θeφ e˙ r = θe ˙ r + φ˙ cos θeφ e˙ θ = −θe ˙ e˙ φ = −φ(sin θer + cos θeθ ) ˙ θ + φr ˙ sin θeφ v = r˙ = r˙er + θre ∇ · (Ar er + Aθ eθ + Aφ eφ ) =

1 ∂ 2 1 ∂ (r Ar ) + (sin θAθ ) r sin θ ∂θ r2 ∂r 1 ∂ + Aφ r sin θ ∂φ

Tables of mathematical formulas

   er reθ r sin θeφ   1  ∂ ∂ ∂   ∇ × (Ar er + Aθ eθ + Aφ eφ ) = 2 r sin θ  ∂r ∂θ ∂φ   Ar rAφ r sin θAφ      1 ∂ 2 ∂u 1 ∂u 1 ∂ ∂2 u 2 ∇ u(r, θ, φ) = 2 . r + 2 sin θ + ∂r ∂θ r ∂r r sin θ ∂θ r2 sin2 θ ∂φ2 6 Jacobians dx = J1 du,

J1 =

dx du

 ∂x  ∂u J2 =  ∂y1  ∂u

∂x ∂u2 ∂y 1 ∂u2

dx dy = J2 du1 du2 ,

 ∂x  ∂u  ∂y1 J3 =  ∂u  ∂z1 

∂x ∂u2 ∂y ∂u2 ∂z ∂u1 ∂u2

dx dy dz = J3 du1 du2 du3 ,

7 Integral theorems



 dσ =

Gauss’s : 

   

S



Ω

dr =

Stokes’s : c

dτ∇

(dσ × ∇). S

∂x ∂u3 ∂y ∂u3 ∂z ∂u3

    .  

75

2 Transformations, matrices and operators 2.1

Transformations and the laws of physics

Physics deals with physical events in spacetime. Given a physical system at a given point in spacetime, we would like to know how it looked in the past, and how it will evolve in the future. In addition, we wonder if it might be related to other relevant events in the past, in the future, and at other points and orientations in space. To answer these questions we need spacetime machines to move us back and forth in spacetime so that we can examine the situation ourselves. These spacetime machines are called spacetime transformation operators. In this chapter, we show that the transformation operators in spacetime can appear as square matrices and as differential operators. Square matrices can be multiplied into column vectors to give other column vectors; functions can be differentiated to yield other functions. If we represent the state of a physical system by a column vector or by a function (of position and/or time), its evolution in spacetime can then be symbolized mathematically as the consequence of an operation by a square matrix or by a differentia1 operator. Transformation operators also appear in statements of physical laws. A law of physics codifies the behavior of a class of physical events. Since the evolution of physical events can be symbolized by transformation operators, a physical law is essentially a statement concerning one or more of these operators. For example, Newton’s equation of motion in its simplest form m

d2 x(t) = F(x, t) dt2

describes how the position of a mass is changed, or transformed, in response to a driving force. An even simpler physical law is contained in the (1D) wave equation 

 ∂2 1 ∂2 − u(x, t) = 0. ∂x2 v2 ∂t2

This states that wave motion will appear if the transformation operator ∂2 /∂x2 in space induces a change with the same sign as, and proportional to, that caused by the

Rotations in space

77

transformation operator ∂2 /∂t2 in time. We thus see that a wave is a manifestation of a certain correlation between disturbances in space and those in time. Besides simple spacetime properties, a physical system may also have additional “internal” attributes that may or may not be describable directly in the spacetime language. Their transformations are often represented in matrix form. Since successive transformations add up to a total transformation, matrices, like spacetime transformations, form mathematical structures called groups. Modern physics is often a detective story in which an unknown group structure behind relevant experimental observations is to be identified. In this way, new physical laws can be established even before we can describe the internal attribute in spacetime terms. To help the reader in his future adventures in group theory, this chapter closes with a short section describing briefly a number of common matrix groups.

2.2

Rotations in space: Matrices

Besides scalars and vectors, which are linear arrays of numbers, we also use rectangular arrays of numbers called matrices. Matrices are particularly useful in describing rotations in space. By a rotation in space we mean either the rotation of a point in space about the origin of the coordinate system, the coordinate axes being fixed, or the rotation of the coordinate axes about the origin, the point in space being fixed. It is obvious that these two rotations are simply related to each other. In the present context it is more convenient to consider the second type of rotation—the rotation of coordinate axes. Let us therefore suppose that we are interested in the position r of a fixed point in space in two coordinate systems related by a rotation: r = x1 e1 + x2 e2 + x3 e3 = x1 e1 + x2 e2 + x3 e3 .

(2.1)

We ask: What is the relation between the “new” coordinates (x1 , x2 , x3 ) and the old coordinates (x1 , x2 , x3 )? The answer can be obtained readily by writing the old unit vectors ei in terms of the new vectors e1 with the help of the completeness relation Eq. (1.20): ej = e1 (e1 · ej ) + e2 (e2 · ej ) + e3 (e3 · ej ) =

3 

ei λij ,

(2.2)

i=1

where λij = ei · ej = enew · eold j = cos θij i is a direction cosine. Thus r = x1 (e1 λ11 + e2 λ21 + e3 λ31 ) + x2 (e1 λ12 + e2 λ22 + e3 λ32 ) + x3 (e1 λ13 + e2 λ23 + e3 λ33 ).

(2.3)

78

Transformations, matrices and operators

e2

e2

e1 θ

e1

Fig. 2.1 A 2D rotation.

That is, xi

= λi1 x1 + λi2 x2 + λi3 x3 =

3 

λij xj ,

(2.4)

j=1

with the indices j for the old coordinate system appearing together side by side in λ and x. The summations in Eqs. (2.2) and (2.4) are the same as those for matrix multiplications, as we shall see shortly. For this reason, Eq. (2.4) can be written conveniently in the compact “matrix” form ⎛ ⎞ ⎛ ⎜⎜⎜ x1 ⎟⎟⎟ ⎜⎜ λ11 ⎜⎜⎜⎜ x2 ⎟⎟⎟⎟ = ⎜⎜⎜⎜ λ21 ⎝ ⎠ ⎝ λ31 x3

λ12 λ22 λ32

⎞⎛ ⎞ λ13 ⎟⎟ ⎜⎜ x1 ⎟⎟ ⎟⎜ ⎟ λ23 ⎟⎟⎟⎠ ⎜⎜⎜⎝ x2 ⎟⎟⎟⎠, λ33 x3

(2.5)

or x  = λx,

(2.6)

where x and x  are the column matrices (also called column vectors). The 3 × 3 matrix λ is a tabular array containing the nine direction cosines between the two sets of coordinate axes. It is called a transformation matrix, or more specifically, a rotation matrix. Example 2.2.1 The xy (or 12) axes are rotated by an angle θ in a positive (i.e., counter-clockwise) direction, as shown in Fig. 2.1. What is the resulting rotation matrix for the transformation of the coordinates of a fixed point in space? We note that λ11 = e1 · e1 = cos θ,

λ12 = e1 · e2 = sin θ,

λ13 = e1 · e3 = 0,

etc.

Rotations in space

79

Table 2.1 Matrix operations.

Operation

Matrix

Element

Equality Addition Zero Multiplication by a scalar Matrix multiplication

C=A C = A+B A=0 C = αA C = AB

Cij Cij Aij Cij

 BA

Cij =

Therefore

⎛ ⎜⎜⎜ cos θ λ(θ) = ⎜⎜⎜⎝ − sin θ 0

= Aij = Aij + Bij =0 = αAij n 

Aik Bkj

k=1

sin θ cos θ 0

A

B

C

m×n m×n m×n m×n

— m×n — —

m×n m×n — m×n

m×n

n×l

m×l

⎞ 0 ⎟⎟ ⎟ 0 ⎟⎟⎟⎠ 1

(2.7)

is the rotation matrix describing the coordinate changes under a rotation.  2.2.1

Matrices

It is worthwhile to review the algebra of matrices at this point. A rectangular array of numbers (called elements or matrix elements) Aij ⎞ ⎛ ⎜⎜⎜ A11 . . . A1n ⎟⎟⎟ ⎟⎟⎟ ⎜⎜⎜ . A = ⎜⎜⎜ .. ⎟⎟⎟ = (Aij ) ⎠ ⎝ Am1 . . . Amn containing m rows and n columns is called an m × n matrix if it satisfies the matrix operations of Table 2.1. The definition of a zero m × n matrix also implies the existence of a negative matrix (−A)ij = −Aij . Thus m × n matrices are “closed under addition,” meaning that the result of an addition is also a matrix. Table 2.2 shows the operations that can be performed on matrices. A matrix that remains unchanged under one of these operations is given a special name. For an operation involving a transposition (under which rows become columns), only square matrices (with m = n) can be invariant under the operation. A square matrix is symmetric if Aji = Aij ; it is antisymmetric if Aji = −Aij . The number of rows or columns of a square matrix is called its order or degree. A square matrix in which only diagonal matrix elements are nonzero is a diagonal matrix. A diagonal matrix with ones along the diagonal is the identity, or unit, matrix I. Its matrix elements are

80

Transformations, matrices and operators Table 2.2 Operations on matrices.

Operation

Matrix

Element

A

C

if C = A

Transposition Complex conjugationb Hermitian conjugationc

C = AT C = A∗ C = A† = AT∗

Cij = Aji Cij = A∗ij Cij = A∗ji

m×n m×n m×n

n×m m×n n×m

Symmetrica Real Hermitiana

a

For square matrices only. √ = −i, i = −1. c Or adjoint. b i∗

Iij = δij .

(2.8)

The square matrix A−1 that satisfies the properties A−1 A = AA−1 = I is called the inverse of the square matrix A. A square matrix A that does not have an inverse is a singular matrix. The sum of the diagonal matrix elements of a square matrix is called its trace: n  Tr A = Aii . (2.9) i=I

An orthogonal matrix is one whose transpose is its inverse. Thus (OT O)ij = δij = (OOT )ij , that is,



Oki Okj = δij =

k



Oik Ojk .

(2.10)

k

These are called orthogonality relations. A unitary matrix is one whose Hermitian conjugate or adjoint is its inverse. Thus (U † U)ij = δij = (UU † )ij , that is,

 k

Uki∗ Ukj = δij =



Uik Ujk∗ .

(2.11)

k

These are called unitarity relations. Finally, we note that the order of the matrix multiplication is important. For example, if AB is defined, as shown in the last entry of Table 2.1, the product BA does not even exist, unless m = l, For square matrices (m = n = l), BA is always defined, but may not be equal to AB.

Rotations in space

81

Example 2.2.2 

 1 2 A= is symmetric, Tr A = 5. 2 4   0 2 is antisymmetric. B= −2 0   0 i C= is Hermitian because −i 0     0 −i 0 −i T ∗ , and C = = CT. C = i 0 i 0   −2 1 AC = i , while −4 2   2 4 CA = i  AC.  −1 −2 The simplest matrices are 2 × 2 matrices, each of which has four matrix elements. It is convenient to express them in terms of the four linearly independent 2 × 2 Hermitian matrices     1 0 0 1 I= , σ1 = , 0 1 1 0     0 −i 1 0 σ2 = , σ3 = (2.12) i 0 0 −1 so that a1 σ1 + a2 σ2 + a3 σ3 + bI = a · σ + bI     A11 A12 b + a3 a1 − ia2 = = A, = a1 + ia2 b − a3 A21 A22

(2.13)

where a = a1 e1 + a2 e2 + a3 e3 , σ = σ1 e1 + σ2 e2 + σ3 e3 . The matrices σi were used by Pauli to describe the three components of the spin (a vector) of the electron. They are called Pauli spin matrices. One can verify by direct calculation that they satisfy the relation  σi σj = δij I + i εijk σk , (2.14a) k

82

Transformations, matrices and operators

where εijk is a permutation symbol. Eq. (2.14a) is equivalent to the relations σ2i = I,

σi σj = iσk ,

(2.14b)

where (i, j, k) is a cyclic permutation of (1,2,3). It is also useful to think of ±I, ±σ1 , ±σ2 , ±σ3 as eight of the square roots of the 2 × 2 matrix I. Unlike simple numbers, these matrices do not commute, and are said to make up a noncommuting algebra. The expression (2.13) of 2 × 2 matrices in terms of Pauli matrices may appear very complicated, but it is actually no worse than the representation of vectors by their components. Vector components are extracted from a given vector by taking scalar products. The components ai and b in Eq. (2.13) can be calculated from a given matrix A by taking traces. This is done by first noting that Tr I = 2,

Tr σi = 0.

As a consequence Tr A = a1 Tr σ1 + a2 Tr σ2 + a3 Tr σ3 + b Tr I = 2b Tr(Aσ1 ) = Tr(a1 I + a2 σ2 σ1 + a3 σ3 σ1 + bσ1 ) = Tr[a1 I + a2 (−iσ3 ) + a3 (iσ2 ) + bσ1 ] = a1 Tr I = 2a1 , Hence A=



etc.

σi 21 Tr(Aσi ) +

"

1 2 Tr

# A I.

i

2.2.2

Transformation matrices

Let us return to matrices for coordinate transformations. Since coordinate transformations are reversible by interchanging old and new indices, we must have new T = enew · eold (λ−1 )ij = eold i · ej i = λji = (λ )ij . j

(2.15)

Hence transformation matrices are orthogonal matrices. Example 2.2.3 The rotation matrix of Eq. (2.7) is orthogonal because ⎛ ⎞ ⎞⎛ ⎜⎜⎜ cos θ − sin θ 0 ⎟⎟⎟ ⎜⎜⎜ cos θ sin θ 0 ⎟⎟⎟ T λ (θ)λ(θ) = ⎜⎜⎝⎜ sin θ cos θ 0 ⎟⎟⎠⎟ ⎜⎜⎝⎜ − sin θ cos θ 0 ⎟⎟⎠⎟ = I.  0 0 1 0 0 1 A rotation matrix such as that given in Eq. (2.7) is a continuous function of its argument θ. As a result, its determinant is also a continuous function. In fact, it is equal to 1 for any θ. In addition, there are matrices of coordinate transformations with a determinant of −1. These transformations change the handedness of the coordinate system. Examples of such parity transformations are

Rotations in space z P1

z Right-handed

Left-handed x

=

y x

83

y

z x

P3

Right-handed

= y y

Left-handed

x z

Fig. 2.2 Parity transformations of the coordinate system.

⎛ ⎜⎜⎜−1 P1 = ⎜⎜⎜⎝ 0 0

0 1 0

⎛ 0 ⎜⎜⎜−1 P3 = ⎜⎜⎜⎝ 0 −1 0 0

⎞ 0 ⎟⎟ ⎟ 0 ⎟⎟⎟⎠, 1

⎞ 0 ⎟⎟ ⎟ 0 ⎟⎟⎟⎠, −1

P2i = I.

They change the signs of an odd number of coordinates of a fixed point r in space, and therefore must involve a change in handedness of the coordinate system, as shown in Fig. 2.2. A change of handedness, unlike a simple rotation, cannot be made in a continuous manner. We note finally that a rotation matrix such as that given in Eq. (2.7) can often be obtained by a geometrical method, but it is easier to use the simple algebraic (i.e., abstract) method described here. All we need here is the ability to calculate the nine direction cosines of Eq. (2.3). 2.2.3

Successive transformations

One advantage of the matrix method for coordinate transformations is that successive transformations 1, 2, . . . , m of the coordinate axes about the origin are described by successive matrix multiplications as far as their effects on the coordinates of a fixed point are concerned. This is because, if x(1) = λ1 x, x(2) = λ2 x(1) , . . ., then x(m) = λm x(m−1) = (λm λm−1 . . . λ1 )x = Lx, where L = λm λm−1 . . . λ1

(2.16)

is the net transformation matrix for the m successive transformations taken in the specified manner. Note that λk acts on the kth coordinate system obtained after k − 1 rotations.

84

Transformations, matrices and operators

Example 2.2.4 Consider a rotation of the xy axes about the z axis by an angle θ. If this rotation is followed by a back-rotation of the same angle θ in the opposite direction, that is, by −θ, we recover the original coordinate system. Thus ⎛ ⎞ ⎜⎜⎜ 1 0 0 ⎟⎟⎟ R(−θ)R(θ) = I = ⎜⎜⎜⎝ 0 1 0 ⎟⎟⎟⎠ = R−1 (θ)R(θ). 0 0 1 Hence ⎛ ⎜⎜⎜ cos θ R (θ) = R(−θ) = ⎜⎜⎜⎝ sin θ 0 −1

− sin θ cos θ 0

⎞ 0 ⎟⎟ ⎟ 0 ⎟⎟⎟⎠ = RT (θ). 1



(2.17)

Example 2.2.5 Use rotation matrices to show that sin(θ1 + θ2 ) = sin θ1 cos θ2 + cos θ1 sin θ2 . We make two successive 2D rotations of angles θ1 , and θ1 , R(θ2 )R(θ1 ) = R(θ1 + θ2 ), that is



cos θ2 − sin θ2

sin θ2 cos θ2



cos θ1 − sin θ1

   sin θ1 cos (θ1 + θ2 ) sin (θ1 + θ2 ) = − sin (θ1 + θ2 ) cos (θ1 + θ2 ) . cos θ1

A direct matrix multiplication of the LHS gives   sin θ1 cos θ2 + cos θ1 sin θ2 cos θ1 cos θ2 − sin θ1 sin θ2 , − sin θ1 cos θ2 − cos θ1 sin θ2 − sin θ1 sin θ2 + cos θ1 cos θ2 where the 12 matrix element is the desired result.  2D rotations (i.e., successive rotations about the same axis ei ) are Abelian, that is, Ri (θ2 )Ri (θ1 ) = Ri (θ1 )Ri (θ2 ). However, rotations about different axes do not generally commute. Thus ⎛ ⎞ ⎞⎛ ⎜⎜⎜ cos α sin α 0 ⎟⎟⎟ ⎜⎜⎜ cos β 0 − sin β ⎟⎟⎟ ⎜ ⎜ ⎟ 1 0 ⎟⎟⎠⎟ Rz (α)Ry (β) = ⎜⎝⎜ − sin α cos α 0 ⎟⎠⎟ ⎜⎝⎜ 0 0 0 1 sin β 0 cos β ⎞ ⎛ ⎜⎜⎜ cos α cos β sin α − cos α sin β ⎟⎟⎟ ⎟ ⎜ = ⎜⎜⎜⎜ − sin α cos β cos α sin α sin β ⎟⎟⎟⎟, ⎠ ⎝ sin β 0 cos β

Rotations in space H

Ry

z

π H 2

Rx

π R π H 2 y 2

Rx

π H 2

Ry

85

π R π H 2 x 2

x

Fig. 2.3 Rotations of the right hand about its center of mass.

whereas

⎛ ⎜⎜⎜ cos α cos β Ry (β)Rz (α) = ⎜⎜⎜⎝ − sin α cos α sin β

⎞ sin α cos β − sin β ⎟⎟ ⎟ cos α 0 ⎟⎟⎟⎠ sin α sin β cos β

 Rz (α)Ry (β). This distinction is illustrated in Fig. 2.3. Example 2.2.6 The orientation of a rigid body about its center of mass is specified by three angles, not two. The reason is that while an axis rigidly attached to the body can be specified by only two angles (e.g., the longitude and latitude on a sphere), a third angle is needed to specify the orientation of the rigid body about this axis. A convenient choice of the three orientational angles is given by the Euler angles (α, β, γ) defined by first rotating the body about a z axis by an angle α, then about the new x axis by an angle β, and finally about the latest z axis by an angle γ: R(α, β, γ) = Rz (γ)R x (β)Rz (α).

(2.18)

We should note the following: (1) An earlier rotation appears further to the right in the matrix multiplication. (2) The rotational axes of two successive rotations cannot be chosen the same, otherwise the rotation angles are not independent.  Problems 2.2.1 (a) Show that any square matrix B can be written in the form B = S + A, where S is symmetric and A is antisymmetric, (b) Obtain the symmetric and antisymmetric parts of ⎛ ⎞ ⎜⎜⎜ 1 2 3i ⎟⎟⎟ B = ⎜⎜⎜⎝ 4 5i 6 ⎟⎟⎟⎠. 7i 8 9 2.2.2 Use rotation matrices to obtain explicit expressions for sin(θ1 − θ2 ) and cos(θ1 − θ2 ). 2.2.3 Show explicitly that Rn (θ) = R(nθ) for 2D rotations.

86

Transformations, matrices and operators

2.2.4 There is a constant vector V = (Vx , Vy ) = (2, 1) at the point (x, y) = (5, 10) in a 2D space. If the coordinate axes are rotated by 30◦ in the positive sense in the xy plane (i.e., about the z axis), calculate the components of V in the new (rotated) coordinate system. 2.2.5 Obtain explicitly the rotation matrices Rx (β) and R(α, β, γ) of Eq. (2.18). 2.2.6 For any square matrix A, show that A + A† is Hermitian, while A − A† is skewHermitian or anti-Hermitian, changing sign under the adjoint operation. 2.2.7 Show that Tr(AB) = Tr(BA). 2.2.8 Show that two n × n square matrices do not necessarily commute even when one is diagonal. 2.2.9 (a) Show that (AB)† = B† A† . (b) Show that the commutator [H1 , H2 ] = H1 H2 − H2 H1 of two Hermitian matrices is skew-Hermitian or anti-Hermitian. (A square matrix A is skew-Hermitian if A† = −A). 2.2.10 Verify that the 2 × 2 Pauli matrices defined in Eq. (2.12) satisfy the multiplicative relations shown in Eqs. (2.14a) and (2.14b). 2.2.11 With the help of Eq. (2.14), show that Tr[(σ · a)(σ · b)] = 2a · b, Tr[(σ · a)(σ · b)(σ · c)] = 2ia · (b × c). 2.2.12 With the help of Eq. (2.14), show that (σ · a)(σ · b) = a · b + iσ · (a × b). 2.2.13 A rectangular coordinate system for the earth has its origin at the center of the earth, its z axis passing through the North Pole, and its x axis passing through the point 0◦ N 0◦ E. Its z axis is then rotated from the North Pole to Los Angeles ( 30◦ N 120◦ W) along a longitude, so that the final z axis goes from the center of the earth to Los Angeles. Show that the resulting transformation matrix can be expressed in terms of three successive rotations Rz (−α)Rx (β)Rz (α), where Rz (α) rotates the x axis along the equator to the point 0◦ N 30◦ W. (Hint: The first rotation Rz (α) transforms to a certain coordinate system, while the last rotation Rz (−α) transforms back from that coordinate system.) 2.2.14 Find the transformation matrix that rotates a rectangular coordinate system through an angle of 120◦ about an axis through the origin, making equal angles with the original three coordinate axes. When viewed along the axis of rotation, the coordinate axes look like Fig. 2.4.

Determinant and matrix inversion

87

z

Direction of rotation

x

y

Fig. 2.4 Looking down the rotation axis in problem 2.2.14.

2.3

Determinant and matrix inversion

To facilitate the use of matrices, we need two additional mathematical objects that can be calculated from a square matrix: its determinant and its inverse matrix. 2.3.1

Determinant

The determinants of small square matrices are well known:     A A12 A12  A , det 11 = A11 A22 − A21 A12 =  11 A21 A22 A21 A22  ⎛ ⎞     ⎜⎜⎜ A11 A12 A13 ⎟⎟⎟ A23  A13  A A   22 12 ⎜ ⎟  − A21   det ⎜⎜⎝ A21 A22 A23 ⎟⎟⎠ = A11  A32 A33  A32 A33  A31 A32 A33   A12 A13   . + A31  A22 A23 

(2.19)

(2.20)

More generally, we define the determinant of an n × n matrix to be    A11 · · · A1n     A21  = εij···l Ai1 Aj2 · · · Aln , det A =  ..  ij···l  .  An1 · · · Ann 

(2.21)

where εij···l , is a permutation (or Levi-Civita) symbol of order n (i.e., with n indices). It is defined as follows: ⎧ ⎪ +1, if ij . . . l is an even permutation of 12 . . . n, ⎪ ⎪ ⎨ −1, if ij . . . l is an odd permutation of 12 . . . n, (2.22) εij···l = ⎪ ⎪ ⎪ ⎩ 0, if any index is repeated.

88

Transformations, matrices and operators

This defines nn permutation symbols. Of these only n! symbols are nonzero, since there are exactly n! permutations of the n distinct characters 1, 2, . . . , n. Thus Eq. (2.21) contains n! terms. The order n of the matrix is called the order of the determinant. Using the definition given by Eq. (2.21), we write directly   A11 A12 det = ε11 A11 A12 + ε12 A11 A22 + ε21 A21 A12 + ε22 A21 A22 A21 A22 = A11 A22 − A12 A21 . Similarly, ⎛ ⎜⎜⎜A11 det ⎜⎜⎜⎝A21 A31

A12 A22 A32

⎞ A13 ⎟⎟ ⎟ A23 ⎟⎟⎟⎠ = A11 A22 A33 − A11 A32 A23 + A21 A32 A13 A33 − A21 A12 A33 + A31 A12 A23 − A31 A22 A13 ,

(2.23)

where the six terms correspond to the six permutations (three even, three odd) of the indices 123. These expressions are equal to those shown in Eqs. (2.19) and (2.20), respectively. Properties of determinants that can be proved directly from the definition in Eq. (2.21) are 1. det A = det A. ⎛ ⎜⎜⎜kA11 A12 ⎜⎜⎜kA21 A22 2. det ⎜⎜⎜⎜⎜ .. .. ⎜⎜⎝ . . kAn1 An2 T

(2.24) ...

⎞ Aln ⎟⎟ ⎟⎟⎟ ⎟⎟ .. ⎟⎟⎟⎟ = k det A, . ⎟⎟⎠ Ann

det (kA) = kn det A.

(2.25)

3. If A and B differ only in the exchange of two rows or two columns, then det B = − det A.

(2.26)

In particular, if A contains two identical rows or columns, then det A = 0. 4. det (AB) = (det A)(det B) = det (BA).

(2.27) (2.28)

The proofs of these results will be left as exercises. The expression of a determinant as a sum of determinants of a lower order, as shown for example in Eq. (2.20), is called a Laplace development. To obtain a Laplace development from Eq. (2.21), we factor out any one matrix element, say Akq , so that  AkqCkq , (2.29) det A = k

Determinant and matrix inversion

89

where the cofactor Ckq =





εij···l (Ai1· · · Aln ) 

(2.30)

ij···l (k)

of the element Akq does not contain Akq , while the sum does not involve k. These restrictions are denoted by the primes in Eq. (2.30). Consider next the “reduced” determinant of order n − 1    Mkq = εij···l (Ai1· · · Aln )  (2.31) ij···l (k)

obtained from the matrix in which the kth row and the qth column are simply removed. It is identical to Ckq except that the permutation symbol εij···l of n indices is replaced by the reduced permutation symbol εij···l (k) of n − 1 indices obtained from it by dropping q from the standard arrangement 12 . . . n, and dropping k from the permutation ij . . . l. If q = 1 and k = 1 are both in the first position, they can both be dropped from C kq to give Mkq . If q is in the qth position, it takes q − 1 transpositions to move it to the first position. Similarly it takes k − 1 transpositions to move k from the kth position to the first. Each transposition changes the “parity” of the permutation (i.e., even  odd permutation). Hence C kq = (−1)k+q Mkq ,

(2.32)

and det A =

n 

Akq (−1)k+q Mkq .

(2.33)

k=1

Going back to the expression in Eq. (2.31), it is easy to see that Mkq is the determinant of the reduced matrix constructed from A by removing the kth row and qth column. We call Mkq the minor of the element Akq , and the expansion in Eq. (2.33) the Laplace development of the qth column, with 1 ≤ q ≤ n. Eq. (2.20) is thus a Laplace development of the first column. Since rows and columns of a matrix can be interchanged without changing the value of the determinant, as stated in Eq. (2.24), we can also obtain the Laplace development of the kth row: det A =

n 

Akq (−1)k+q Mkq .

q=1

The Laplace development of the second row, for a 3 × 3 matrix, is       A12 A13  A11 A13  A11 A12      + A22   − A23  . det A = −A21  A32 A33  A31 A33  A31 A32 

(2.34)

90

Transformations, matrices and operators

All six Laplace developments of a third-order determinant (three by column and three by row) are contained in Eq. (2.23). They are just different regroupings of the same six terms. 2.3.2

Matrix inversion

An important application of the theory of determinants is concerned with the calculation of the inverse of a square matrix. Let us first recall that a 2 × 2 matrix   A11 A12 A= A21 A22 has the inverse  A22 −1 A = −A21

 $ −A12 A22 (A11 A22 − A12 A21 ) = A11 −A21

$ −A12 det A. A11

(2.35)

This result can be verified directly by showing that the inversion relations A−1 A = I = AA−1 are indeed satisfied. Eq. (2.35) shows clearly that the inverse exists only if det A  0. The inverse matrix of a general n × n matrix A can be constructed readily by using a slightly more general form of Eq. (2.34) for the Laplace development of the ith row δik det A =

n 

AiqCkq = (ACT )ik = (CAT )ki ,

(2.36a)

q=1

while a Laplace development of the ith column gives δki det A = (C T A)ki .

(2.36b)

(The proof of these results is left as an exercise.) Since δik = Iik = Iki , we find that I=

ACT CT A = . det A det A

Thus A−1 = C T / det A,

(2.37)

where C is the n × n cofactor matrix whose matrix elements are the cofactors Ckj . Since the cofactors [see Eq. (2.30)] always exist, it follows that A−1 exists if det A  0. If det A  0, or if A−1 exists, the matrix A is said to be nonsingular or invertible. Example 2.3.1 Invert the matrix ⎛ ⎜⎜⎜ 1 A = ⎜⎜⎜⎝ 1 1

1 2 1

⎞ 1 ⎟⎟ ⎟ 1 ⎟⎟⎟⎠ . 3

Determinant and matrix inversion

91

First calculate det A to determine if A is invertible. Suppose we do a Laplace development of the first row. The required minors are     2 1  1 1   = 5,  = 2, M12 =  M11 =  1 3 1 3   1 2    = −1. M13 =  1 1 Therefore det A = A11 M11 − A12 M12 + A13 M13 = 2, and A is invertible. The cofactor matrix is constructed from all the signed minors, that is, ⎛ ⎞ ⎛ ⎞ M13 ⎟⎟ ⎜⎜ 5 −2 −1⎟⎟ ⎜⎜⎜ M11 −M12 ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ M22 −M23 ⎟⎠⎟ = ⎜⎝⎜−2 2 0⎟⎟⎠⎟ C = ⎜⎝⎜−M21 −1 0 1 M31 −M32 M33 after calculating remaining minors. Hence ⎛ ⎜⎜⎜ 5 1 A−1 = ⎜⎜⎜⎝−2 2 −1

−2 2 0

⎞ −1⎟⎟ ⎟ 0⎟⎟⎠⎟. 1

Finally, it is necessary to check that A−1 A (or AA−1 ) = I to ensure against numerical mistakes.  2.3.3

Simultaneous equations

Determinants are also involved in the solution of simultaneous algebraic equations. Suppose we have n algebraic equations in n unknowns x1 , . . . , xn : A11 x1 + A12 x2 + . . . + A1n xn = c1 .. . An1 x1 + An2 x2 + . . . + Ann xn = cn . These equations can be written very compactly in matrix form as Ax = c,

(2.38)

where A is an n × n matrix, and x = (x1 , x2 , . . . , xn ) and c = (c1 , . . . , c2 ) are nD column vectors. Left multiplication with A−1 then yields the unique solution x = A−1 c,

(2.39)

92

Transformations, matrices and operators

if A−1 exists. This requires that det A  0, that is, that A be nonsingular. Eq. (2.39) gives xi = (A−1 c)i =

 (C T )ij cj j

det A

=

 cjCji . det A j

(2.40)

The numerator of this expression looks like a Laplace development of the ith column of a determinant. This is indeed the case. The determinant is that of A modified by replacing its ith column by c. Eq. (2.40) is called Cramer’s rule. Example 2.3.2 Solve the matrix equation ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎜⎜⎜ 1 1 1 ⎟⎟⎟ ⎜⎜⎜ x1 ⎟⎟⎟ ⎜⎜⎜ 1 ⎟⎟⎟ ⎜⎜⎜ 1 2 1 ⎟⎟⎟ ⎜⎜⎜ x2 ⎟⎟⎟ = ⎜⎜⎜ 2 ⎟⎟⎟. ⎝ ⎠⎝ ⎠ ⎝ ⎠ 1 1 3 x3 3 Since det A = 2, we have by Cramer’s rule   1 1 1   1 x1 =  2 2 1  = −1 2  3 1 3    1 1 1   1 x2 =  1 2 1  = 1 2  1 3 3    1 1 1   1 x3 =  1 2 2  = 1. 2  1 1 3  That these are the solutions can easily be verified by doing the matrix multiplication Ax.  Problems 2.3.1 Verify Eqs. (2.24)–(2.27). 2.3.2 Show that det(A−1 ) = (det A)−1 if A is nonsingular.    1 2 3  2.3.3 Calculate  4 5 6  by Laplace developments of  7 8 9  (a) the second row, and (b) the third column. 2.3.4 Verify Eqs. (2.36a) and (2.36b). 2.3.5 Verify that (a) (A−1 )−1 = A, and (b) (AB)−1 = B−1 A−1 .

Homogeneous equations

⎛ ⎜⎜⎜ 1 2.3.6 Invert the matrix ⎜⎜⎜⎝ 4 7

2 5 8

93

⎞ 3 ⎟⎟ ⎟ 6 ⎟⎟⎟⎠. 1

2.3.7 Prove the relation det (AB) = (det A)(det B) for (a) 2 × 2 matrices, and (b) n × n matrices, n ≥ 3. 2.3.8 Use Cramer’s rule to solve the matrix equation ⎛ ⎜⎜⎜ 1 ⎜⎜⎜⎝ 4 7

2.4

2 5 8

⎞⎛ ⎞ ⎛ ⎞ 3 ⎟⎟ ⎜⎜ x1 ⎟⎟ ⎜⎜ 1 ⎟⎟ ⎟⎜ 6 ⎟⎟⎟⎠ ⎜⎜⎜⎝ x2 ⎟⎟⎟⎟⎠ = ⎜⎜⎜⎜⎝ 2 ⎟⎟⎟⎟⎠. 1 x3 3

Homogeneous equations

It is also possible to solve a set of n homogeneous algebraic equations written in matrix form as Ax = 0,

(2.41)

that is, Eq. (2.38) with c = 0. It is obvious that there is a unique solution x = A−1 c = 0 if A is nonsingular. We call this a trivial case. A nontrivial solution is one with x  0. This can only occur if det A = 0, for then we have the quotient of two zeros, which can be finite. To see what we are getting into, let us first consider the simple case of 2 × 2 matrices. Then Ax = 0 means the two simultaneous algebraic equations A11 x1 + A12 x2 = 0 A21 x1 + A22 x2 = 0. These equations state that x = (x1 , x2 ) is perpendicular to the row vectors a1 = (A11 , A12 ),

a2 = (A21 , A22 )

contained in A. That is, x is a vector, any vector, perpendicular simultaneously to both a1 , and a2 . Its length is of no consequence in the homogeneous equation; only its direction e(x) or e(−x) can be determined. If the two row vectors are nonparallel, det A does not vanish. There is then no nontrivial solution. The reason is that a1 and a2 already span the entire 2D plane of the problem. An x perpendicular to this plane cannot also lie in it. There is thus no solution x in the plane. This is the reason why det A = 0 is required. Det A = 0 can be obtained by making a1 and a2 parallel. They then occupy only one of the two dimensions of the problem, leaving the other dimension for x. The direction of x is then uniquely determined up to a sign.

94

Transformations, matrices and operators

For 3 × 3 matrices, Ax = 0 means that x must be perpendicular to all three row vectors of A: ai · x = 0.

(2.42)

Suppose two of these row vectors are not parallel to each other. To obtain det A = 0, the third row vector, say a3 , must be made up of a linear combination of these nonparallel vectors a3 = c1 a1 + c2 a2 .

(2.43)

Equation (2.43) states that a3 lies in the plane containing a1 and a2 . Thus the three row vectors span only a 2D plane. Eq. (2.42) now shows that x is the normal to this plane through the origin. (All vectors are measured from the origin.) Again, only its direction is determined, not its sense or length. Det A = 0 is also realized when all three row vectors are parallel to one another. These vectors then span only the 1D subspace specified by the common direction e(a1 ). Any vector in the plane normal to e(a1 ) that contains the origin is an acceptable solution. That is, the solutions x span this 2D plane perpendicular to e(a1 ). In both cases, the solutions span the subspace not spanned by the row vectors of A. These considerations can immediately be generalized to n × n matrices. The row vectors ai of A span an r-dimensional subspace Sa , with r < n if det A = 0 is to be realized. The solutions x of the homogeneous matrix equation A = 0 span the remaining (n − r)-dimensional subspace S x . The subspace Sx is called the orthogonal complement of the subspace Sa . These subspaces are orthogonal to each other. Together they make up the entire nD space. If r = n − 1, n − r = 1. Then Ax = 0 determines one direction only. More generally Ax = 0 determines the n − r linearly independent directions that specify the subspace Sx . Example 2.4.1 The homogeneous equation ⎛ ⎞⎛ ⎞ ⎜⎜⎜ 1 1 1 ⎟⎟⎟ ⎜⎜⎜ x1 ⎟⎟⎟ ⎜⎜⎝⎜ 1 1 1 ⎟⎟⎠⎟ ⎜⎜⎝⎜ x2 ⎟⎟⎠⎟ = 0 1 1 1 x3 has solutions which satisfy the equation x1 + x2 + x3 = 0. They are vectors lying in the plane perpendicular to the row vectors a1 = a2 = a3 = (1, 1, 1). Example 2.4.2 Solve the homogeneous equation ⎛ ⎞⎛ ⎞ ⎜⎜⎜−1 1 1 ⎟⎟⎟ ⎜⎜⎜ x1 ⎟⎟⎟ ⎜⎜⎝⎜ 1 1 1 ⎟⎟⎠⎟ ⎜⎜⎝⎜ x2 ⎟⎟⎠⎟ = 0. 0 2 2 x3



Homogeneous equations

95

First verify that det A = −0 − 0 + 0 = 0 by a Laplace development of the first column. Then note that a3 = a1 + a2 so that the three vectors are linearly dependent. Finally we throw out a1 to find the two independent equations x1 + x2 + x3 = 0 2(x2 + x3 ) = 0. The solutions are x1 = 0,

x2 + x3 = 0,

along the line x = x2 (0, 1, −1) in the 23 plane. The solution can be obtained even more readily by using the result that x = (xi , x2 , x3 ) is perpendicular to both a1 and a2 . Hence x = const(a1 × a2 ) = const(0, 2, −2). 2.4.1



Linear independence of vectors and the rank of a matrix

The solution of the problem of homogeneous matrix equations given so far is adequate but quite sketchy for the general case. The inquisitive reader who must know how things work out in detail is urged to read on. Suppose only the first r row vectors ai of A are linearly independent. The remaining n − r vectors ak are linearly dependent on the former in the sense that they can be expressed as the linear combinations ak =

r 

ci ai .

(2.44)

i=1

This means that if a vector x is orthogonal to the ai s, ai · x = 0,

i = 1, . . . , r,

(2.45)

it is automatically orthogonal to the ak s as well. Eq. (2.45) can be written more compactly in the matrix form Bx = 0,

(2.46)

where B is an r × n nonsquare submatrix of A containing the r linearly independent row vectors ai , and x is an n × 1 column matrix.

96

Transformations, matrices and operators

The r × n nonsquare matrix B contains n column vectors bj . Not all of these are linearly independent of one another, because an arbitrary rD vector has at most r independent components. Consequently, no more than r of these column vectors are linearly independent. A column vector, say bn , linearly dependent on the other column vectors bn =

n−1 

d j bj ,

j=1

can be dropped from the matrix B without affecting the linear independence of the remaining (n − 1)D row vectors. This is because bn does not contain any information not already contained in the remaining column vectors bj . There are exactly n − r of these linearly dependent column vectors that can be discarded for the time being. We now have an r × r square submatrix R of B made up of the remaining columns of B. If we try to discard any more columns from R, we will end up with a nonsquare matrix in which the row vectors have fewer than r components each. Should this happen, not all the r row vectors still in the matrix can be linearly independent. Thus by construction, we have obtained an r × r square matrix R containing r linearly independent column vectors. Its determinant does not vanish. Hence R is a nonsingular invertible matrix. The order r of this submatrix R is called the rank of the matrix A or B. Problems 2.4.1 Solve the homogeneous equation ⎛ ⎜⎜⎜ 1 ⎜⎜⎜ 4 ⎝ 5

2 5 7

⎞⎛ ⎞ 3 ⎟⎟ ⎜⎜ x ⎟⎟ ⎟⎜ ⎟ 6 ⎟⎟⎟⎠ ⎜⎜⎜⎝ y ⎟⎟⎟⎠ = 0. 9 x

2.4.2 By expressing the last row vector in terms of the others, determine the rank of the following matrices   1 2 3 4 (a) ; 5 6 7 8 ⎞ ⎛ ⎜⎜⎜ 1 2 3 ⎟⎟⎟ ⎜ (b) ⎜⎝⎜ 4 5 6 ⎠⎟⎟⎟; 7 8 9 ⎞ ⎛ 2 3 4 ⎟⎟ ⎜⎜⎜ 1 ⎟ 6 7 8 ⎟⎟⎟⎠. (c) ⎜⎜⎜⎝ 5 9 10 11 12

The matrix eigenvalue problem

97

e(ω) = ez

L=r×p m θ

r

m

Fig. 2.5 A rotating dumbbell.

2.5

The matrix eigenvalue problem

It is an experimental fact that the linear momentum of a massive object p = mv is always parallel to its linear velocity v. In contrast, the angular momentum L of a rigid body is not always parallel to its angular velocity ω. A classic illustration of this observation is the rotating dumbbell shown in Fig. 2.5. It is made up of two point masses m connected by a massless rigid rod of length 2r. It is rotated about its midpoint 0 about the z axis with angular speed ω. The angular momentum L = r × p is always perpendicular to the dumbbell axis (if θ  0). Consequently, it is not always parallel to ω. However, both L and ω are 3D column vectors when expressed in matrix notation. Hence there exists a 3 × 3 square matrix I that can connect them:  Iij ωj . (2.47) L = Iω, or Li = j

I is called the inertial matrix. For the special case of a point mass, we have with the help of the BAC rule, L = mr × v = mr × (ω × r) = m[r2 ω − r(r · ω)]. Thus I = m(r2 − rr·) in vector notation. To write the relation in matrix notation, we note that  #  " Li = m r2 ωi − xi xj ωj = [m(r2 δij − xi xj )]ωj . j

j

98

Transformations, matrices and operators

Comparison with Eq. (2.47) gives Iij = m(r2 δij − xi xj ) in matrix notation. Having established that L is in general not parallel to ω, we may ask next if occasionally they might be. It is easy to see that for the rotating dumbbell there are indeed three distinct directions along which this occurs. One of these ωs is parallel to the dumbbell axis (i.e., θ = 0), while the remaining two are perpendicular to the dumbbell axis (i.e., θ = π/2). These directions are called the principal axes of inertia. The corresponding inertial parameters (the proportionality constants λ = L/ω are called the principal moments of inertia. The existence of three principal axes of inertia turns out to be a universal property of rigid bodies, as we shall now show. The parallelism between L and ω along these special directions is expressed by the condition Iω = λω,

(2.48)

where the scalar proportionality constant λ is called an eigenvalue (or characteristic value) of I. (Eigen is the German word for “own, proper, characteristic, special.”) Eq. (2.48) itself is called an eigenvalue equation. It is an interesting mathematical observation that this equation can be solved for these special directions ω and for the eigenvalue λ associated with each of these directions. 2.5.1

Solution of the eigenvalue problem

We shall write the general matrix eigenvalue equation as the homogeneous equation (B − λI)u = 0.

(2.49)

It is clear that the matrix B should be square, otherwise Bu will not have the same dimension as u and cannot possibly satisfy the equation. It also follows from the results of the last section that nontrivial solutions with u  0 can appear only if the determinant   B12 . . .   B11 − λ  B22 − λ  φ(λ) = det(B − λI) =  B12 (2.50) .. ..   . . vanishes. φ(λ) is called a characteristic (or secular) determinant. If matrix B is n × n, φ(λ) is a polynomial of degree n in λ. The determinantal condition φ(λ) = 0,

(2.51)

called the characteristic (or secular) equation, therefore has n roots λ1 , λ2 , . . . , λn , which are not necessarily different. When λ = λi , det (B − λI) vanishes; Eq. (2.49) can then have a nontrivial solution ui . Thus the n roots of Eq. (2.51) are the only possible eigenvalues of Eq. (2.49) when the matrix B is n × n.

The matrix eigenvalue problem

99

To illustrate the calculation of eigenvalues, let us consider a general 2 × 2 matrix   a c B= . d b The associated secular equation is quadratic in λ φ(λ) = det(B − λI) = λ2 − (a + b)λ + ab − cd = 0. It has two roots λ1,2 = 12 [(a + b) ±

% (a − b)2 + 4cd]

(2.52a)

that satisfy the useful identity λ1 + λ2 = a + b = Tr B. This result turns out to be a special case of the general theorem that states that the sum of the eigenvalues of a matrix of any order is equal to its trace  λi = Tr B. (2.53) i

The proof of this general theorem is left as an exercise (Problem 2.5.5). As we have discussed in the last section, only the directions ei = e(ui ) of the nontrivial solutions ui of the eigenvalue equation (2.49) are determined, not their lengths. Thus the matrix eigenvalue equation may be written more precisely as (B − λi I)ei = 0,

or

Bei = λi ei .

(2.54)

We call ei the eigenvector “belonging” to the eigenvalue λi . Since we already know how to solve homogeneous matrix equations, the eigenvalue equation (2.54) can be solved to obtain ei . The procedure is illustrated in the following examples. Example 2.5.1 A uniform solid sphere has an inertial matrix (in suitable units) of ⎛ ⎞ ⎜⎜⎜ 1 0 0 ⎟⎟⎟ ⎜⎜⎜⎝ 0 1 0 ⎟⎟⎟⎠. 0 0 1 The characteristic equation is φ(λ) = (1 − λ)3 so that the three roots are λ1 = λ2 = λ3 = 1. Identical eigenvalues like these are said to be degenerate. The eigenvectors are to be calculated from the homogeneous equation Aω = 0, where ⎛ ⎞ ⎜⎜⎜ 0 0 0 ⎟⎟⎟ A = I − λI = ⎜⎜⎝⎜ 0 0 0 ⎟⎟⎠⎟. 0 0 0

100

Transformations, matrices and operators

This shows that ω can be any vector in space. Since there are three linearly independent vectors in space, we may choose ω1 = e x , ω2 = ey , ω3 = ez .  Example 2.5.2 Obtain the eigenvalues and eigenvectors of the real, symmetric 2 × 2 matrix   a c S = , a  b. c b According to Eq. (2.52a) λ1,2 = 12 (a + b) ∓

% 1 2

(b − a)2 + 4c2 .

(2.52b)

That is, λ1 is below the mean (a + b)/2 of the diagonal matrix elements by an amount % Δ = 12 (b − a)2 + 4c2 , and below a itself by δ = Δ + 12 (a − b), while λ2 is above the mean by an amount Δ and above b by an amount δ. Any normalized 2D vector can be written in the form   cos θ , e= sin θ since cos2 θ + sin2 θ = 1 gives its normalization. The eigenvector ei satisfies the homogeneous equation (S − λi I)ei = 0. That is,



a − λi c

c b − λi



 cos θi = 0; sin θi

or tan θi =

λi − a c = . c λi − b

More specifically, λ1 − a δ =− , c c c c = = − cot θ1 . tan θ2 = λ2 − b δ tan θ1 =

Hence if θ = θ1 , then cos θ2 = − sin θ,

sin θ2 = cos θ.

The matrix eigenvalue problem

This shows that

  − sin θ e2 = cos θ

is perpendicular to e1 .

101

(2.55)



Example 2.5.3 Obtain the eigenvalues and eigenvectors of the Hermitian matrix   γ iβγ B= −iβγ γ with real constants β and γ. (This is a transformation matrix for the Lorentz transformation in 2D Minkowski spacetime (x, ict).) From Eq. (2.52a), the eigenvalues are found to be λ1 = γ(1 − β),

λ2 = γ(1 + β).

The first eigenvector e1 is obtained from the equation    γ − λ1 iβγ q1 = 0. −iβγ γ − λ1 q2

(2.56)

That is, (γ − λ1 )q1 + iβγq2 = 0, or q2 γ − λ1 = = i, q1 −iβγ

and

  1 . e 1 = c1 i

The question now arises as to how the normalization constant c1 is to be chosen. The usual scalar product   1 T 2 =0 e1 · e1 = c1 (1 i) i seems to suggest that e1 might have no length. This cannot be correct, since it has nonzero components. We recall that the length (or absolute value) of a complex number z = x + iy is not the square root of z2 = x2 − y2 + 2ixy, but that of the real non-negative number z∗ z = x2 + y2 = |z|2 . For the same reason, the length of e1 should be calculated from the Hermitian scalar product (usually referred to as just the scalar product)   1 † T∗ 2 e1 · e1 = e1 · e1 = |c1 | (1 − i) = 2|c1 |2 . (2.57a) i

102

Transformations, matrices and operators

This gives 1 |c1 |2 = , 2

1 |c1 | = √ . 2 It is clear that c1 itself can be a complex number of phase φ1 . Hence   1 iφ1 1 . e1 = √ e i 2 or

(2.58)

Any choice of φ1 is acceptable, since Eq. (2.57) is satisfied for any choice of φ1 . We refer to this undetermined degree of freedom as a phase, or gauge, transformation of the eigenvector. (In advanced physics, gauge degrees of freedom can be used to describe certain internal properties of the system. For our purposes here, however, this gauge transformation is as yet undetermined, and therefore unimportant.) In a similar way, we can determine the eigenvector belonging to λ2 :   1 iφ2 1 e2 = √ e . (2.59) −i 2 The orthogonality between e1 and e2 can be checked by examining the scalar product   1 † 1 i(φ1 −φ2 ) (1 i) =0 e2 · e1 = 2 e i = e†1 · e2 .

(2.57b)

Because this scalar product vanishes, these complex vectors are said to be orthogonal to each other.  The eigenvalues of a 3 × 3 matrix are the roots of a cubic equation in λ. They can also be expressed in closed form with the help of Cardan’s formula for these roots. Unfortunately, the formula is rather long, and it involves cube roots as well as square roots. It is more profitable to restrict ourselves to simple special cases. Example 2.5.4 The special matrix of any order n where all matrix elements are real and equal (say Bij = 1) is of considerable interest in physics. It describes a system of n degrees of freedom with equal interactions among them. The eigenvalues and eigenvectors of B can be found readily, and are well worth knowing. √ To begin with, the matrix B for n = 2 has eigenvectors e2,1 = (1, ±1)/ 2. e2 is an eigenvector because e2 ∝ a1 = a2 = a = (1, 1), the two identical row vectors of B. Hence its eigenvalue is λ2 = n = 2. e1 ⊥ e2 is also an eigenvector because of Eq. (2.55). Its eigenvalue λ1 = 0 can be found by direct multiplication, but more exhausted the trace, leaving λ1 = 0. interestingly by noting that λ2 = Tr B has √ For n = 3, one eigenvector is e3 = a/ n, where a = (1, 1, 1) is the common row vector of B. Its eigenvalue is λ3 = n = 3. By an extension of Eq. (2.55), one can show that the remaining two eigenvectors lie on the plane perpendicular to e3 . (A general proof will be given in Section 2.7.) B does not show any preferred direction on this

The matrix eigenvalue problem

103

plane. So the remaining eigenvalues λi , i  n, are the same, or degenerate. Furthermore, λn has already exhausted Tr B. So λi = 0, i  n. Any two orthogonal vectors√on this plane can be used √ as the eigenvectors. It is convenient to use e1 = (1, −1, 0)/ 2 and e2 = (1, 1, −2)/ 6. Then the ei s form a right-handed coordinate system. √ The result for any n is now obvious. en = a/ n, where a is the common row vector of B, is the eigenvector belonging to the collective eigenvalue λn = n that exhausts Tr B. The remaining eigenvectors all have the same eigenvalue λi = 0, i  n. They span an (n − 1)D subspace orthogonal to en . They can be chosen to be the orthonormal basis vectors: √ e1 = (1, −1, 0, . . .)/ 2, √ e2 = (1, 1, −2, 0, . . .)/ 6, & em = (1, 1, . . . , −m, 0, . . .)/ m2 + m, m < n. Note that em contains m components of 1 and one component −m. Any remaining components are 0. The simple matrices ±B lie at the heart of a number of important collective physical phenomena of quantum origin. −B gives a simple model of many low-lying (or low-temperature) collective states including those describing superconductivity and collective rotations and vibrations of quantum systems. The collective eigenvalue λn = −n of −B gives the collective energy gap that enhances the stability of the collective mode en . The high-lying collective mode en of B also appears in physical systems, but it tends to be unstable because of coupling to other degrees of freedom.  Some of the eigenvalues in the following problems can be obtained by simple considerations without using general formulas. The eigenvalues and eigenvectors of more complicated matrices are usually calculated by using a computer. Problems 2.5.1 Calculate the eigenvalues and eigenvectors of the following matrices:   0 1 (a) σ1 = ; 1 0   0 −i (b) σ2 = ; i 0 ⎛ ⎞ ⎜⎜⎜ 0 0 1 ⎟⎟⎟ ⎜ (c) ⎜⎝⎜ 0 0 1 ⎟⎟⎠⎟; 1 1 0 ⎞ ⎛ ⎜⎜⎜ 0 0 1 1 ⎟⎟⎟     ⎜⎜⎜⎜ 0 0 1 1 ⎟⎟⎟⎟ 0 B 1 1 (d) ⎜⎜⎜ ⎟= B 0 , B= 1 1 . ⎜⎝ 1 1 0 0 ⎟⎟⎟⎠ 1 1 0 0 (Hint: The eigenvectors are V = (v, ±v), where v = v1 = (1, −1) or v2 = (1, 1).)

104

Transformations, matrices and operators

k1

k12

k2

q1

q2

Fig. 2.6 Two masses connected by three springs.

2.5.2 Obtain the eigenvalues and eigenvectors of an n × n diagonal matrix. 2.5.3 Show that if B is singular it has one or more zero eigenvalues. 2.5.4 By equating the coefficients of the λn−1 terms in the equation |B − λI| = (λ1 − λ)(λ2 − λ) . . . (λn − λ) for an arbitrary n × n matrix B, show that λi . . . λn =

n '

λi = det B,

i=1 n 

λi = Tr B.

i=1

Hint: You may want to do this separately for n = 2 and n ≥ 3.

2.6

Generalized matrix eigenvalue problems

Fig. 2.6 shows a system of two masses m1 and m2 connected by three springs with spring constants k1 , k12 , and k2 . Their motion is described by Newton’s equations of motion: m1 q¨ 1 = −k1 q1 − k12 (q1 − q2 ), m2 q¨ 2 = −k2 q2 − k12 (q2 − q1 ),

(2.60)

where qi is the displacement of mass mi from its position of equilibrium. This system of two coupled differential equations can be written in matrix form ¨ = −Kq(t), Mq(t) where

  q1 (t) , q(t) = q2 (t)

 m M= 1 0

 0 , m2

(2.61a) 

k + k12 K= 1 −k12

 −k12 . k2 + k12

Generalized matrix eigenvalue problems

105

The equation can be solved by first asking the following question: Is it at all possible that Kq is parallel not to q itself, but to Mq Kq = λMq,

(2.62a)

where λ is a scalar proportionality constant? It turns out not to be important (in principle) whether M is diagonal. It can be a general square matrix with nonzero offdiagonal matrix elements. Eq. (2.62a) will be referred to as the generalized matrix eigenvalue problem. The special values of λ for which Eq. (2.62a) is satisfied are called its eigenvalues. The corresponding vector q is its eigenvector. Writing Eq. (2.62a) as an homogeneous matrix equation (K − λM)q = 0,

(2.62b)

we see that nontrivial solutions q exist only if det(K − λM) = φ(λ) = 0.

(2.63)

With 2 × 2 square matrices K and M, the determinant is a quadratic function of λ. Hence the secular equation (2.63) has two roots, the eigenvalues λ1 and λ2 , for which Eq. (2.62a) can be satisfied. The actual vector q(r) along which Eq. (2.62a) is realized (K − λr M)q(r) = 0

(2.62c)

is just the eigenvector “belonging” to λr . We thus see that the mathematics of this generalized matrix eigenvalue problem is essentially the same as that of the simple eigenvalue problem of the last section. We should add that in the problem of coupled oscillators, the mass matrix M is diagonal if q contains actual coordinates. M can be off diagonal if q contains generalized coordinates such as angles. Example 2.6.1 Solve Eq. (2.62a) when     2 −1 1 1 . , K=k M=m −1 2 1 6    2ω20 − λ −ω20 − λ   det(K − λM) = m  2 −ω0 − λ 2ω20 − 6λ  = m(5λ2 − 16ω20 λ + 3ω40 ) = 0, where ω20 = k/m. The roots are λ1 = ω20 /5,

λ2 = 3ω20 .

The corresponding eigenvectors satisfy the homogeneous equation (K − λr M)q(r) = 0.

106

Transformations, matrices and operators

They can be shown to be   2 (1) q = const × = η1 e1 , 3



(2)

q

 4 = const × = η2 e2 , −1

(2.64a)

where er is the unit vector along q(r) . These vectors are not orthogonal to each other, but neither are they parallel to each; they are linearly independent of each other.  The reason why we look for these special directions is that along each of these directions Eq. (2.61a) has the simple form M q¨ (r) (t) = −λr Mq(r) (t).

(2.61b)

Each q(r) has a time-independent direction represented by er . Therefore its time dependence resides solely in its length ηr (t): q(r) (t) = ηr (t)er .

(2.64b)

As a result, we may write Eq. (2.61b) as [¨ηr (t) + λr ηr (t)]Mer = 0.

(2.61c)

Thus the length ηr (t), called a normal or natural coordinate, satisfies the differential equation for simple harmonic oscillations: η¨ r (t) + λr ηr (t) = 0.

(2.61d)

The solutions are known to be ηr (t) = Cr eiωr t ,

or

Dr e−iωr t ,

(2.65)

a result that can be verified by substitution into Eq. (2.61d). They describe a normal vibration of normal frequency ωr = λ1/2 r .

(2.66)

If λr > 0, ωr is real. The amplitude of the vibration, |ηr (t)| as defined in Eq. (2.65), remains constant in time, and the vibration is said to be stable. If λr < 0, ωr is purely imaginary. |ηr (t)| then changes with time, and the vibration is unstable. If q(1) (t) is a solution of Eq. (2.61a) and q(2) (t) is another solution, we can see by direct substitution that their sum q(t) =

2 

q(r) (t)

(2.67)

r=1

is also a solution. This is referred to as a superposition principle. The generalization of these results to n × n matrices is straightforward, and will be left to the reader.

Generalized matrix eigenvalue problems

107

It turns out that the generalized matrix eigenvalue equation (2.62) is mathematically equivalent to the simple matrix eigenvalue equation (2.49) only when the eigenvalues of both K and M are positive definite. If M itself has m zero eigenvalues, the number of solutions λ will be only n − m, where n is the dimension of the matrices (Problem 2.6.2). The interested reader will find a number of other interesting possibilities in Problem 2.6.3. Problems 2.6.1 Calculate the eigenvalues and eigenvectors of the generalized eigenvalue problem Eq. (2.62a) for the following matrices:     1 0 3 1 (a) M = , K= , 0 3 1 1     1 1 3 1 (b) M = , K= , 1 3 1 1 Partial answers: (a) λ1,2 = 3.12, 0.21. (b) λ1,2 = 3.73, 0.27. 2.6.2 Show that if the matrix M itself has m zero eigenvalues, then Eq. (2.62a) for n × n matrices has at most n − m eigenvalues. (Hint: Assume that M is diagonal and has m zeros on the diagonal.) 2.6.3 Calculate the eigenvalues and eigenvectors of Eq. (2.62a) for the following matrices where a is real:     1 0 1 1 (a) M = , K= ; 0 0 1 6     1 1 a 2 (b) M = , K= ; 1 1 2 1     1 0 1 1 (c) M = , K= , for a = −3, 1, 2. 0 −1 1 a Answers: (a) λ = 5/6, q = (6, −1). (b) No solution if a = 3. One solution if a  3 : λ = (a − 4)/(a − 3), q = (1, 2 − a). (c) If a = −3: one solution λ = 2, q = (1, 1). If a = 1: one solution λ = 0, q = (1, −1). √ If a = 2: two solutions λ1,2 = (1 ± 5)/2, q(1) = (1, −2.62), q(2) = (−2.62, 1).

108

Transformations, matrices and operators

2.7

Eigenvalues and eigenvectors of Hermitian matrices

Many matrices appearing in eigenvalue problems in physics are Hermitian matrices for which H † = H ∗T = H.

(2.68)

Hermitian matrices can be real or complex. If real, they are symmetric (S T = S ). Hermitian matrices appear frequently in physics because of the following properties of their eigenvalues and eigenvectors: 1. Their eigenvalues are real. 2. Eigenvectors belonging to distinct eigenvalues are orthogonal. 3. Eigenvectors belonging to degenerate eigenvalues (e.g., λ1 = λ2 ) can be orthogonalized. Property (1) is appropriate to physics because all measurable quantities are real. Properties (2) and (3) show that the eigenvectors of a Hermitian matrix define a Cartesian coordinate system in the space of nD vectors. This is a feature of considerable convenience in describing the physics of a system in which the original matrix H plays a significant role. To derive these properties we first note that a Hermitian matrix is in general a complex matrix. Its eigenvectors ei in the simple eigenvalue equation Hei = λi ei

(2.69)

can be complex; that is, they can have complex components. The scalar product e†i ei = eT∗ i ei is always non-negative. Two complex vectors ei , ej are said to be orthogonal if † e†i ej = eT∗ i ej = δij = ej ei .

(2.70)

Coming back to our eigenvalue problem in Eq. (2.69), we see that e†i Hej = λi e†j ei .

(2.71)

Since H is Hermitian, the left-hand side can also be written as (e†j Hei )†† = (e†i H † ej )† = (e†i Hej )† = (λj e†i ej )† = λ∗j e†j ei .

(2.72)

The difference between the last two equations is O = (λi − λ∗j )e†j ei .

(2.73)

Thus if j = i, e†i ej > 0; hence λ∗i = λi = real. If λj  λi , then e†j ei = 0 is required to satisfy Eq. (2.73). Hence the corresponding eigenvectors are orthogonal.

Eigenvalues and eigenvectors of Hermitian matrices

109

Eq. (2.73) does not exclude the possibility that two or more linearly independent eigenvectors may belong to the same eigenvalue. Since any linear combination of these eigenvectors also belongs to the same eigenvalue, we are free to choose linear combinations that lead to orthogonal eigenvectors. That there are indeed distinct orthogonal eigenvectors belonging to degenerate eigenvalues can be seen as follows. Suppose the Hermitian matrix H0 has two degenerate eigenvalues λ1 = λ2 , while the eigenvalues of the Hermitian matrix H0 + H1 are all distinct. We now consider the Hermitian matrix H(ε) = H0 + εH1 , and decrease ε from 1 to 0. The eigenvalues λ1 (ε) and λ2 (ε) are distinct for finite ε, but coincide in the limit ε → 0. When they are distinct, the corresponding eigenvectors are orthogonal. Since there is no abrupt change in the mathematical structure of the problem as ε → 0, the orthogonal eigenvectors remain orthogonal as ε → 0. 2.7.1

Matrix diagonalization

Consider the square matrix ⎛ ⎞ ⎜⎜⎜ e11 e21 · · · en1 ⎟⎟⎟ ⎟⎟⎟ ⎜⎜⎜ e12 ⎟⎟⎟ U = ⎜⎜⎜⎜⎜ .. ⎟⎟⎟ ⎜⎜⎝ . ⎟⎠ · · · enn e1n

(2.74)

made up of the orthonormalized eigenvectors ej = (ej1 , ej2 , . . . , ejn ) of a Hermitian matrix H stored columnwise. The matrix element (U † U)i j involves the row vector e†i making up the ith row of U † and the column vector ej making up the jth column of U. It is just the scalar product (U † U)ij = e†i ej = δij .

(2.75)

Hence U † U = I, and U must be a unitary matrix. Consider next the matrix element  (U † )ik Hkl Ulj . Dij = (U † HU)ij = k,l

This also involves the ith row of U † and the jth column of U. Hence Dij = e†i Hej = λj e†i ej = λj δij ,

(2.76)

and D is a diagonal matrix containing the eigenvalues on its diagonal. The matrix U † HU is called a unitary transformation of H. We see that H is diagonalized if the unitary matrix U of the transformation contains the orthonormal eigenvectors of H stored columnwise.

110

Transformations, matrices and operators

The eigenvalue equation itself also changes under U: (H − λα )eα = 0 = U † (H − λα )eα = U † (H − λα )UU † eα = (D − λα )eα . Since D in the transformed eigenvalue equation is diagonal, the new eigenvectors eα = U † eα must be unit vectors along the new coordinate axes. This result can be deduced directly from the last equation itself, for the components of the new eigenvector are (eα )β = (U † eα )β = e†β eα = δαβ .

(2.77)

From the perspective of the coordinate transformation of Section 2.2, the eigenvalue system has not changed. Rather it is the coordinate system that has rotated. The transformation U † has rotated the coordinate system so that the eigenvectors have become the new coordinate axes. In physics, the eigenvectors are often called the principal axes of H. The unitary transformation U is then referred to as a principalaxes transformation.   1 1 Example 2.7.1 Diagonalize the matrices H = , H 2 , H n and eH . 1 1 We have shown in Example 2.5.4 that the eigenvalues of H are λ1 = 0 and λ2 = 2. The corresponding eigenvectors are     1 1 1 1 , e2 = √ . e1 = √ 2 −1 2 1   11 The unitary matrix U = √1 containing these normalized eigenvectors colum2 −1 1 nwise will therefore diagonalize H through the unitary transformation       1 1 −1 1 0 2 0 0 † U HU = √ = = D. 1 √2 0 2 0 2 2 1 The diagonalized matrix D contains the eigenvalues 0 and 2 along its diagonal. The same unitary transformation applied to H 2 yields U † H 2 U = U † HU † UHU = D2 . Since the product of two diagonal matrices AB is a diagonal matrix whose diagonal matrix elements are the products of corresponding matrix elements, that is, (AB)ij = Aii Bii δij , we have

Eigenvalues and eigenvectors of Hermitian matrices

111



 0 0 D = . 0 4 2

Similarly, †



U H U=D = n

n

 0 . 2n

0 0

We come finally to the matrix function eH of a matrix H. It may be defined by the usual power series for the exponential function e = H

∞ 

H n /n!,

n=0

now applied to the matrix H. Since each term on the right-hand side is a 2 × 2 matrix, eH itself must also be a 2 × 2 matrix. Its diagonalization can now be achieved as follows: ⎞ ⎛∞ ⎜⎜ H n ⎟⎟⎟ † H †⎜ ⎟⎟⎠⎟ U U e U = U ⎜⎜⎝⎜ n! n=0 =

∞  Dn n=0

n!

= eD .

Since the product of diagonal matrices is diagonal, eD is diagonal. Its matrix elements are ∞  Dnii (eD )ii = = eDii . n! n=0 Hence

 1 e = 0 D

 0 . e2



Exercise 2.7.1 Show explicitly that the matrix   1 1 1 U= √ 2 i −i  containing the eigenvectors of Eq. (2.56) columnwise will also diagonalize the matrix Is U unitary?

 0 i . −i 0

112

2.7.2

Transformations, matrices and operators

The generalized problem

Let us next turn our attention to the generalized matrix eigenvalue problem of Eq. (2.62a): Kei = λi Mei . This turns out to be equivalent to Eq. (2.69) if M is nonsingular. To see this, we first note that the matrix inverse square root M −1/2 , which is itself a square matrix, exists if M is nonsingular. If it is left multiplied into Eq. (2.62a), we obtain M −1/2 Kei = M −1/2 K M −1/2 (M 1/2 ei ) = λi (M 1/2 ei ),

(2.78a)

or K  bi = λi bi , where K  = M −1/2 K M −1/2 , bi = M 1/2 ei .

(2.79)

As usual, the eigenvalue Eq. (2.78a) defines only the direction ei of bi K  ei = λi ei .

(2.78a)

It is clear that ei differs in general from ei (the direction of q(i) ) unless M 1/2 is a scalar. In that case, Eq. (2.62a) is identical to the simpler matrix eigenvalue problem of Eq. (2.49). Let us examine the matrix K  when both M and K are general Hermitian matrices. Since the inverse and the square root of Hermitian matrices are also Hermitian, K  is Hermitian. The situation is therefore the same as in Eq. (2.69). Eigenvalues are real, and eigenvectors are, or can be made, orthogonal. The only difference from Eq. (2.69) is that, while the ei are orthogonal, the ei are not, although they are necessarily linearly independent of one another. (A proof of this statement is requested in Problem 2.7.6.) An example of such nonorthogonal eigenvectors has been given in Eq. (2.64a) of the preceding section. We note finally that if M is Hermitian, it can be diagonalized. Hence the weight matrix M may be taken to be diagonal without any loss of generality. The orthogonality relation then has the explicit form  b†i bj = e†i Mej = δij |eik |2 Mkk , (2.80) k

where eik is the kth component of ei .

Eigenvalues and eigenvectors of Hermitian matrices

2.7.3

113

Matrix transformations

We shall conclude this section by introducing some mathematical terms describing matrix transformations. Two square matrices A and B are said to be equivalent to each other if there exist nonsingular matrices P and Q such that B = PAQ.

(2.81)

(A useful result: every matrix is equivalent to some diagonal matrix.) The transformation PAQ of A is called an equivalent transformation. The important special cases are: 1. If P = Q−1 , then Q−1 AQ is a similarity transformation of A. 2. If P = QT , then QT AQ is a congruent transformation of A. 3. If P = Q† , then Q† AQ is a conjunctive transformation of A. 4. If P = Q−l = QT , it is an orthogonal transformation. 5. If P = Q−1 = Q† , it is a unitary transformation. Finally, we mention three useful mathematical results: 1. The trace of a matrix is unchanged under a similarity transformation: Tr (Q−1 AQ) = Tr (QQ−1 A) = Tr A. 2. The trace of a Hermitian matrix is real and is equal to the sum of its eigenvalues:  Tr H = Tr (U † HU) = Tr D = λi . i

3. A matrix A is normal if [A, A† ] = 0. A normal matrix can be diagonalized by a unitary transformation. Hermitian, anti-Hermitian, unitary, and antiunitary matrices are examples of normal matrices. Problems 2.7.1 Show that if two Hermitian matrices A and B have the same eigenvectors, A and B commute. 2.7.2 Show that two commuting Hermitian matrices A and B can be diagonalized by the same unitary transformation if the eigenvalues λi of A are nondegenerate, i.e., λi  λj if i  j. 2.7.3 Show that two commuting Hermitian matrices A and B can be diagonalized by the same unitary transformation if the eigenvalue λ1 of A are m-fold degenerate, with λ1 = λ2 = . . . = λm , while the remaining eigenvalues are distinct from λ1 and from one another. Hint: The unitary transformation U that diagonalizes A will also diagonalize B except in the mD subspace S m spanned by the m orthonormalized eigenvectors ei , i = 1, . . . , m, of A belonging to the m-fold degenerate eigenvalue λ1 .

114

Transformations, matrices and operators

Let Bm be the m × m Hermitian submatrix of B in S m . Show that in this A representation, the off-diagonal matrix elements of B connecting the subspace S m to the remaining subspace S n−m are all zero. Diagonalizing Bm in S m will provide simultaneous eigenvectors for both A and B in S m and therefore in the original nD space S . 2.7.4 Show that the eigenvalues of an anti-Hermitian matrix are purely imaginary. 2.7.5 Show that the eigenvalues of a unitary matrix are complex numbers of unit amplitude. 2.7.6 If H is a Hermitian matrix, show that (a) Tr H = Tr D, where D = U † HU is its diagonalized form; (b) det H = det D. 2.7.7 Given n orthonormal unit vectors ei , show that the n vectors Qei are linearly independent if Q is a nonsingular matrix. 2.7.8 Discuss the solution of the generalized matrix eigenvalue problem described by Eq. (2.62a) if one of the eigenvalues of M is zero. (You may assume that both M and K are Hermitian.) 2.7.9 (a) A function F(x) = f0 + f1 x + f2 x2 + . . . is defined by its Taylor series. Show that this also defines a matrix function f(B) of an n × n matrix B and that f(B) is also an n × n matrix. (b) If matrix B has the eigenvalues b1 , b2 , . . . , bn , show that F(B) has the eigenvalues f (b1 ), f (b2 ), . . . , f (bn ). (c) If H is a Hermitian matrix, show that det(eH ) = eTrH .

2.8

The wave equation

One of the most important equations of motion in classical physics is the wave equation. To facilitate subsequent discussions, it is profitable to describe briefly a simple physical system whose motion satisfies this equation. It is hoped that this will provide a meaningful physical context in which the often abstract considerations of several subsequent sections can be embedded. The reader who already understands the wave equation can move immediately to the next section. The simple system in question is that of n identical masses m connected by identical massless springs each of length d and spring constant k. The displacement qi (t) of the ith mass from its position of equilibrium satisfies the Newtonian equation of the form of Eq. (2.60) m q¨ i = −(qi − qi−1 ) − (qi − qi+1 ) = qi+1 − 2qi + qi−1 , k i = 1, . . . , n.

(2.82)

The wave equation

115

Here we have chosen q0 = qn+1 = 0 as the (zero) displacements of the fixed ends. We would like to go to the limit n, k → ∞, m, d → 0 in such a way that M = nm,

L = (n + 1)d,

τ = kd

are all finite. In this “continuum” limit, it is more convenient to replace the position index i of the ith mass along the length of the springs by the finite variable xi = id: qi (t) = q(xi , t). Dividing Eq. (2.82) by d2 and taking this limit of n → ∞, we find lim

m kd

2

= lim

mn knd

2

=

M 1 ≡ 2 = finite, τL v

and lim

1 ∂2 [q(x , t) − 2q(x, t) + q(x , t)] = q(x, t), i+1 i−1 d2 ∂x2

where the permissible values (or spectrum) of the variable x have changed in character from discrete to continuous. Eq. (2.82) now reads   ∂2 1 ∂2 − q(x, t) = 0, (2.83) v2 ∂t2 ∂x2 where the parameter v has the dimension of a speed (i.e., distance divided by time). This is called a 1D wave equation. It describes the longitudinal vibrations of a continuous, uniform, massive spring. (The mass comes from the point masses, not from the originally massless springs.) The solution of Eq. (2.83) can be written in the form q(x, t) = f (x − vt) + g(x + vt),

(2.84a)

where f and g are completely arbitrary functions of their arguments. To see this, first set g = 0, and denote x − vt by s. Then ∂ d f (s) = f (s), ∂x ds

1∂ d f (s) = − f (s). v ∂t ds

The left-hand side of Eq. (2.83) is now d2 d2 f (s) − f (s), ds2 ds2 which must necessarily be zero. Hence any function f (x − vt) is a solution. In a similar way one can show that any function g(x + vt) is also a solution. Finally their sum, Eq. (2.84a), is also a solution, since each term contributes nothing on the lefthand side of Eq. (2.83). The fact that a sum of solutions is also a solution is referred to as a superposition principle.

116

Transformations, matrices and operators

Although f (s) does not change when s is constant, the wave does move in general in space and in time because s itself is a function of x and t. In particular, ds = dx − vdt. Hence when ds = 0, we must have dx = v. dt Thus the point of constant s, and constant f (s), must move along the +x axis with speed v. We call this point of constant s a wave front of constant phase (which refers to s itself). The speed v is called the phase velocity of the wave. In a similar way, one can show that in g(x + vt) the wave front of constant phase x + vt moves with a velocity dx = −v dt along the −x axis. The relative negative sign between the two partial differential operators in Eq. (2.83) is of importance in wave propagation. It is quite conceivable that there exist physical phenomena that can be described by the differential equation   1 ∂2 ∂2 Π(s, t) = 0, (2.85) + v2 ∂t2 ∂x2 with the opposite (i.e., positive) relative sign. If so, its solution must be expressible in the form Π(x, t) = f (x − ivt) + g(x + ivt) with a purely imaginary speed of propagation iv. All measurable physical quantities are real. An imaginary speed is not a speed at all. It simply gives a position divided by time scale over which the physical phenomenon in question is appearing and disappearing. In fact, we do not know of any important physical phenomenon that is described by Eq. (2.85). Let us now return to the wave equation, Eq. (2.83). It could also describe a wave propagating along the ±x axis, but in a 3D space. (Obviously, we are no longer considering the problem of the massive linear spring.) Such a wave is called a plane wave, since the wave front is the entire yz plane at x. If we had chosen the coordinate axes along different directions, the same plane wave propagating along the same direction would have involved disturbances along the new y and z coordinates as well. Thus the equation of motion should contain ∂2 /∂y2 and ∂2 /∂z2 in a way symmetrical with ∂2 /∂x2 . This consideration leads to the 3D wave equation   1 ∂2 2 (2.86) − ∇ u(r, t) = 0. v2 ∂t2

Displacement in time and translation in space

117

For this equation, the general solution is of the form u(r, t) = f (r − vt) + g(r + vt).

2.9

(2.84b)

Displacement in time and translation in space: Infinitesimal generators

The wave equation (2.83) describes a certain relation between the acceleration of masses and the restoring force responsible for it. More specifically, it states that wave propagation is the consequence of a simple correlation between the curvature (i.e., the second derivative) of the wave function q(x, t) in time to its curvature in space. Like other laws of nature, it makes a nontrivial statement relating these apparently distinct physical properties. In particular, the relationship is that between a differential operation in time and those in space, again like many other laws of nature. One might well wonder why so many laws of nature can be expressed as differential equations in space and in time. The answer, which we would like to discuss in more detail in this section, is that the differential operators in space and in time are related to time displacements and space translations. Physics is concerned not with a dead and static universe, but with one that is dynamic and changing both in space and in time. Physical laws relate properties at different points of space and at different times. Let us consider time translation first, because it is simpler. We are familiar with the idea that, under an infinitesimal time displacement dt, a differentiable function f (t) of time t changes to   d df f (t). (2.87) f (t + dt) = f (t) + dt = 1 + dt dt dt After another time displacement dt, we have    2 d d f (t + 2dt) = 1 + dt f (t). f (t + dt) = 1 + dt dt dt Since a finite time displacement τ can be constructed from an infinite number of successive infinitesimal time displacements, we have   n  d τd f (t) = exp τ f (t) f (t + τ) = lim 1 + n→∞ n dt dt = T t (τ) f (t),

(2.88)

where T t (τ) = exp(−iτW),

W=i

d . dt

(2.89)

118

Transformations, matrices and operators

Thus the increase by τ of the time argument t in f(t) can be expressed formally as the result of a left multiplication of f(t) by a time displacement operator T t (τ). The operator W itself is called the infinitesimal generator of time displacement. To displace the time of an arbitrary function f(t) by the amount τ, it might be necessary to compute the operator algebraic series   d f (t) f (t + τ) = exp τ dt ⎡ ⎤  2 ⎢⎢⎢ ⎥⎥ d d 1 = ⎢⎢⎣1 + τ + (2.90) τ + . . .⎥⎥⎥⎦ f (t) dt 2! dt term by term. This is of course just the Taylor expansion of f (t + τ) about f(t), as one might suspect. Example 2.9.1 Calculate sin ω(t + τ) ⎡ ⎤  2  3 ⎢⎢⎢ ⎥⎥ d d d 1 1 sin ω(t + τ) = ⎢⎢⎣1 + τ + τ τ + + . . .⎥⎥⎥⎦ sin ωt dt 2! dt 3! dt = sin ωt + ωτ cos ωt +

1 (ωτ)2 (− sin ωt) 2!

1 (ωτ)3 (− cos ωt) + . . . 3!   1 2 = sin ωt 1 − (ωτ) + . . . 2!   1 3 + cos ωt ωτ − (ωτ) + . . . 3! +

= sin ωt cos ωτ + cos ωt sin ωτ.  There are certain advantages to using the exponential time displacement operator of Eq. (2.89) to represent a Taylor expansion of a function of time. Two successive time displacements T t (τ1 )T t (τ2 ) = exp[−i(τ1 + τ2 )W] = T t (τ1 + τ2 ) is equivalent to a single composite displacement in the same way that two rotation matrices can be multiplied into a single rotation matrix. Perhaps more important, the time displacement operator takes on a very simple form for those functions that satisfy the equation Wf (t) = ω f (t),

(2.91)

Displacement in time and translation in space

119

with a constant ω. This is called a differential eigenvalue equation, in which f(t) is the eigenfunction belonging to the eigenvalue ω. For such a function, Eq. (2.90) simplifies to   1 2 f (t + τ) = 1 + τ(−iω) + (−iωτ) + . . . f (t) 2! = exp(−iωτ) f (t).

(2.92)

In particular, if at t = 0, f (t = 0) = 1, we find f (τ) = exp(−iωτ),

(2.93)

that is, f (t) = exp(−iωt), which is obviously a solution of Eq. (2.91). This eigenfunction exp(−iωt) = cos ωt − i sin ωt

(2.94)

should be familiar to us as the function describing the time dependence of a system oscillating with frequency ω. This suggests that time displacement is related to the frequencies ω with which the system can oscillate. We may therefore call W a frequency operator. Another interesting feature of W arises from the fact that it is a differential operator: d d [tf (t)] = t f (t) + f (t). dt dt That is, it satisfies the commutation relation [W, t] = Wt − tW = i.

(2.95)

This is a direct consequence of the fact that time is changed by the time displacement operation. In a similar way, an infinitesimal translation dr in space results in a change of a scalar field (or a component of a vector field) Φ(r) to Φ(r + dr) = Φ(r) + dr · ∇Φ(r) = (1 + dr · ∇)Φ(r). A finite translation ρ of the position r can be constructed from an infinite number of successive infinitesimal space translations. The resulting scalar field is  n ρ Φ(r + ρ) = lim 1 + · ∇ Φ(r) = exp(ρ · ∇)Φ(r) n→∞ n = T r (ρ)Φ(r), (2.96) where T r (ρ) = exp(iρ · K),

K=

1 ∇. i

(2.97)

120

Transformations, matrices and operators

Thus the space coordinate in Φ(r) is increased by ρ as the result of a left multiplication of Φ(r) by the space translation operator T r (ρ), which is an exponential function of the generator K of space translation. Its eigenfunction belonging to the eigenvalue k is proportional to exp(ik · r). The generator K satisfies the commutation relations [Ki , xj ] = −iδij ,

(2.98)

where xj is the jth component of r. Example 2.9.2 Calculate the electrostatic potential of a point charge at r s = (0, 0, d) const = exp(−ir s · K)Φ(r) = exp(−idK z )Φ(r) |r − r s | ⎡ ⎤    2 ⎢⎢⎢ ⎥⎥ ∂ 1 ∂ = ⎢⎣⎢1 + −d + −d + . . .⎥⎥⎥⎦ Φ(r). ∂z 2! ∂z

Φ(r − r s ) =

These derivatives can be calculated with the help of the identity ∂ 1 1 ∂ mz = 2z 2 2 m/2 = − m+2 m ∂z r ∂r (r ) r =−

m cos θ . rm+1

Thus ∂ 1 cos θ z =− 3 =− 2 , ∂z r r r ∂2 1 3 cos2 θ − 1 . = ∂z2 r r3 Hence 

 1 d cos θ d 2 3 cos2 θ − 1 + + ... . + Φ(r − r s ) = const × r 2 r2 r3 These terms are referred to as the monopole, dipole and quadrupole terms, respectively. The expansion is called a multipole expansion.  2.9.1

Transformation groups

The exponentiated form of these spacetime transformation operators show that they satisfy the following multiplicative properties:

Displacement in time and translation in space

121

1. Closure under multiplication: Tt (τ2 )Tt (τ1 ) = Tt (τ2 + τ1 ). Tr (ρ2 )Tr (ρ1 ) = Tr (ρ2 + ρ1 ).

(2.99)

That is, the product of operators is itself an operator of the same type. 2. Associative property: Tt (τ3 )[Tt (τ2 )Tt (τ1 )] = [Tt (τ3 )Tt (τ2 )]Tt (τ1 ).

(2.100)

3. Identity operator: Tt (0) = 1,

Tr (0) = 1.

(2.101)

4. Inverse operators: Tt−1 (τ)Tt (τ) = 1 = Tt (τ)Tt−1 (τ), i.e.,

Tt−1 (τ) = Tt (−τ).

(2.102)

A collection or set of objects (called elements) satisfying the algebraic properties (1)–(4) is said to form a group with respect to the specified multiplication rule between group elements. For this reason, we talk about the group of time displacements and the group of space translations. In addition, our group multiplications are commutative: Tt (τ1 )Tt (τ2 ) = Tt (τ2 )Tt (τ1 ), Tr (ρ1 )Tr (ρ2 ) = Tr (ρ2 )Tr (ρ1 ).

(2.103)

We call groups with commutative products Abelian groups, after Abel, a mathematician of the nineteenth century. Rotation matrices also form a group, but they do not in general commute. Hence they form a non-Abelian group. 2.9.2

Infinitesimal generators, invariance principles and physical laws

A study of the powerful mathematics of group theory is outside the scope of this book; so let us return to the spacetime transformation operators themselves. Their importance arises from the basic postulate of physics that physical laws should remain essentially unchanged in all space and for all time. If this were not true, physics would be just a part of history, which involves a study of the past, and never an objective prediction of the future. Unlike historical events, there is every reason to believe that the fall of an apple here tomorrow can be predicted with the same precision using Newton’s laws as that of an apple Newton saw in England, or of the balls Galileo dropped from the Leaning Tower in Pisa at a time when Newton’s laws of motion and gravitation had not yet been discovered. These nice invariant properties of physical laws can readily be ascertained with the help of the space and time displacement operators. Let us suppose that the state

122

Transformations, matrices and operators

of motion of a system here and now is described by a function f (r, t) that satisfies the equation of state D f (r, t) = 0.

(2.104)

where D is an appropriate operator. Under a time displacement it is changed into 0 = T t (τ)[D f (r, t)] = T t (τ)D T t−1 [T t (τ) f (r, t)] = Dt (τ) f (r, t + τ). Since the choice of the starting time t = 0 is arbitrary, we see that the invariance of the equation of state (2.104) under time displacement requires that Dt (τ) = T t (τ)D T t−1 (τ) = D .

(2.105)

That is, the operator must commute with Tt (τ), or equivalently with the frequency operator W. In a similar way, invariance under space translation requires that D commutes with the space translation operator T r (ρ), or equivalently with its generator K. One operator that will commute with K is K itself. It changes sign when the coordinate r changes sign. Its vector components also change when the coordinate system is rotated. If the physical description is to be independent of these changes, only scalar products such as K 2 = K · K should appear. The simplest choice satisfying these invariance principles is D = K 2 − k2 .

(2.106)

Its use in Eq. (2.104) gives the 3D Helmholtz equation (K 2 − k2 )u(r) = 0,

(2.107)

which describes the spatial part of wave motion. The Helmholtz equation can be solved as follows. The infinitesimal generator K of space translation has eigenvalues ±k and eigenfunctions exp(±ik · r), the latter being the spatial version of Eq. (2.93): K exp(±ik · r) = ±k exp(±ik · r).

(2.108)

Applying K twice in the form K · K exp(±ik · r) = ±k · K exp(±ik · r) = k2 exp(±ik · r), we get just the Helmholtz equation. Thus the two eigenfunctions of K belonging to the eigenvalues ±k are also the linearly independent eigenfunctions of K 2 belonging to the ”doubly degenerate” eigenvalue k2 . We thus see that the eigenfunctions themselves appear in the description of physical disturbances for the simple reason that they describe states of the system in which these physical disturbances appear in their simplest form.

Displacement in time and translation in space

123

In a similar way, the time part of u(r, t) at fixed r will satisfy a differential operator that is even in the frequency operator W: (W 2 − ω2 )u(r = fixed, t) = 0. This shows that the complete time-dependent wave function u(r, t) satisfies the partial differential equation (i.e., a differential equation in both r and t)     W2 ω2 K 2 − 2 u(r, t) = k2 − 2 u(r, t) = 0, (2.109) v v if k = ±ω/v. This is just the wave equation, Eq. (2.86),   1 ∂2 2 −∇ + 2 2 u(r, t) = 0, v ∂t where v is the wave speed. If space (or r) and time are originally independent of each other, the eigenfunctions u(r,t) must be products (actually linear combinations of products) of the eigenfunctions for K (for the spatial part) and W (for the time part). There are four independent products with the same eigenvalues k2 for K 2 and ω2 for W 2 : u1 (r, t) = exp i(k · r − ωt),

u2 (r, t) = exp i(−k · r − ωt),

u3 (r, t) = exp i(k · r + ωt),

u4 (r, t) = exp i(−k · r − ωt),

(2.110)

Two of these solutions, u1 and u4 , describe waves moving with speed v along the direction k, while the remaining two solutions, u2 and u3 , describe waves propagating along −k, as the reader is asked to confirm in Problem 2.9.2. It is the wave equation itself that sets up the specific correlation between changes in position and changes in time, which leads to the propagation of a wave along k or −k. For this reason, K may be called a propagation operator. 2.9.3

Wave equation in quantum mechanics

One of the greatest discoveries of physics in the twentieth century is that matter in motion shows wave properties, an observation that is of utmost importance in describing the physics of atomic phenomena. The mathematical description of matter waves, like that of any other wave motion, requires a wave equation. This is achieved in quantum mechanics with the help of the following “quantization rules” deduced from experimental facts. The concepts of momentum and energy for a material particle must be generalized to the momentum operator and the Hamiltonian operator, respectively, involving Planck’s constant : pop = K,

and

H = W.

(2.111)

The classical energy–momentum relation, such as the E 2 = p2 c2 + m2 c4

(2.112)

124

Transformations, matrices and operators

relation of Einstein, should next be generalized to its operator equivalent, H 2 = p2op c2 + m2 c4 in our example. But since H and pop are differential operators of independent variables r and t, respectively, this operator equivalent cannot be taken literally as an operator equation. It is conceivable, however, that H 2 operating on a special class of functions u(r, t) gives a result exactly equal to p2op c2 + m2 c4 operating on the same function. That is, H 2 u(r, t) = (p2op c2 + m2 c4 )u(r, t).

(2.113)

When the differential forms of H and pop are used, this equation can be simplified to the form of a Schr¨odinger-like wave equation   m2 c2 1 ∂2 2 −∇ + 2 + 2 2 u(r, t) = 0, (2.114)  c ∂t known as the Klein–Gordon equation. Problems 2.9.1 Calculate the electrostatic potential at large distances of (a) An electric dipole with charges Q at s/2 and −Q at −s/2; (b) A linear electric quadrupole with charges Q at ±s/2 and −2Q at the origin. 2.9.2 Determine the direction of propagation of the wave described by each of the four wave functions shown in Eq. (2.110). 2.9.3 Show that all m × n matrices form a group under matrix addition as the group multiplication. What is the identity element of the group? What is the inverse operation? 2.9.4 Show that all n × n square nonsingular matrices form a group under matrix multiplication as the group multiplication. 2.9.5 Show that all 3 × 3 rotation matrices form a group under matrix multiplication. 2.9.6 Show that the four 2 × 2 matrices     1 0 0 i I= , A= , 0 1 i 0     −1 0 0 −i B= , C= 0 −1 −i 0 are closed under matrix multiplication. (That is, their products with each other can be expressed in terms of the matrices themselves.) Show that these matrices form a group. Show that this group has the same mathematical structure (i.e., group multiplicative properties) as any one of the following groups of four elements: (a) I = 1, A = i, B = −1, C = −i;

Rotation operators

125

(b) I = R(0), A = R(π/2), B = R(π), C = R(3π/2), of 2D rotation matrices R(θ), θ = 0, π/2, π, 3π/2.

2.10

Rotation operators

Let us now return to rotation matrices, because our earlier discussion is somewhat incomplete. In Section 2.2 we obtained the matrices for finite rotations. According to the ideas discussed in Section 2.9, it should be possible to generate finite rotations from infinitesimal rotations. The associated infinitesimal generators are in a sense more interesting than the finite rotation matrices themselves, since the former may appear as operators in terms of which the states of rotating systems can be described. In this section we concentrate on these generators of infinitesimal rotations. 2.10.1

Infinitesimal rotations

The rotation matrix for an infinitesimally small rotation of amount dθ3 about the 3 axis is from Eq. (2.7) ⎛ ⎞ dθ3 0⎟⎟ ⎜⎜⎜ 1 ⎟ 0⎟⎟⎟⎠ = 1 + idθ3 J3 , (2.115) R(dθ) = ⎜⎜⎜⎝−dθ3 1 0 0 1 where

⎛ ⎜⎜⎜0 J3 = i ⎜⎜⎜⎝1 0

−1 0 0

⎞ 0⎟⎟ ⎟ 0⎟⎟⎟⎠ . 0

(2.116a)

Similarly, for rotations about the 1 and 2 axes separately, 1 + idθ2 J2 , with ⎛ ⎛ ⎞ 0⎟⎟ ⎜⎜⎜0 0 ⎜⎜⎜ 0 0 ⎟ J1 = i ⎜⎜⎜⎝0 0 −1⎟⎟⎟⎠ , J2 = i ⎜⎜⎜⎝ 0 0 0 1 0 −1 0

we have 1 + idθ1 J1 , and ⎞ 1⎟⎟ ⎟ 0⎟⎟⎟⎠ . 0

(2.116b)

The matrices Ji , i = 1, 2, 3, are the generators of infinitesimal rotations. By direct manipulation, one can show that these matrices satisfy the following properties: 1. They are Hermitian matrices ⎛ ⎛ ⎞ ⎜⎜⎜0 0 0⎟⎟⎟ ⎜⎜⎜1 2 2 2. J1 = ⎜⎜⎝⎜0 1 0⎟⎟⎠⎟ , J2 = ⎜⎜⎝⎜0 0 0 1 0

(2.117) 0 0 0

⎞ 0⎟⎟ ⎟ 0⎟⎟⎟⎠ , 1

J32

⎛ ⎜⎜⎜1 = ⎜⎜⎜⎝0 0

0 1 0

⎞ 0⎟⎟ ⎟ 0⎟⎟⎟⎠ ; 0

(2.118)

hence J 2 = J · J = J12 + J22 + J32 = 2I.

(2.119)

126

Transformations, matrices and operators

3. They do not commute with one another. More specifically, their commutators are  εijk Jk . (2.120) [Ji , Jj ] ≡ Ji Jj − Jj Ji = i k

2.10.2

Finite rotations

Let us next rotate our coordinate system about an arbitrary but fixed direction eθ by an infinitesimal amount dθ. The appropriate rotation matrix is R(dθ) = 1 + idθ · J,

dθ = dθeθ .

(2.121)

This can also be written more generally as R(θ + dθ) = R(dθ)R(θ) = (1 + idθ · J)R(θ). That is, dR(θ) = R(θ + dθ) − R(θ) = idθ · JR(θ).

(2.122)

For repeated rotations about the same direction eθ , Eq. (2.122) can be integrated to give θ = θeθ .

R(θ) = exp(iθ · J),

(2.123)

This is called a rotation operator. The rotation operator, being a function of the 3 × 3 matrix operator J, is also a 3 × 3 matrix operator. This can be seen readily in the power-series expansion R(θ) = exp(iθ · J) =

∞  (iθ · J)n n=0

n!

.

(2.124)

The series expansion is also useful in understanding the properties of R(θ). For example, one can show in this way that R3 (θ) = exp(iθJ3 ) is identical to the rotation matrix of Eq. (2.7). (See Problem 2.10.2.) The form of Eq. (2.124) is the more flexible because the direction e(θ) can be oriented along any direction in space. It is also more compact. Various matrix operations can be performed on functions of matrices. For example, R† (θ) = exp(−iθ · J† ) = exp(−iθ · J) = R−1 (θ).

(2.125)

This shows that the orthogonal matrix R(θ) is also unitary. R (θ) is orthogonal by virtue of Eq. (2.15).

Rotation operators

2.10.3

127

Vector algebra in matrix form

There is a useful connection between the generators Ji of infinitesimal rotations and the cross product of vector algebra that can be brought out by writing vector algebra in a matrix form. We first point out that the scalar product may be written as the matrix product of a row vector and a column vector ⎛ ⎞ ⎜⎜⎜ B1 ⎟⎟⎟ A · B = (A1 A2 A3 ) ⎜⎜⎝⎜ B2 ⎟⎟⎠⎟ = AT B. (2.126) B3 It is unchanged by a coordinate transformation, because in the new coordinate system we have A · B  = (λA)T λB = AT (λT λ)B = A · B.

(2.127)

This invariant property can also be seen in the familiar cosine form A · B = AB cos θAB of the scalar product, since neither A, B, nor θAB depends on the choice of the coordinate axes. We refer to this invariance property by calling scalar products rotational scalars. The matrix form of the vector product is ⎛ ⎞⎛ ⎞ A2 ⎟⎟ ⎜⎜ Bl ⎟⎟ −A3 ⎜⎜⎜ 0 ⎟⎜ ⎟ 0 −A1 ⎟⎟⎠⎟ ⎜⎜⎝⎜ B2 ⎟⎟⎠⎟ = −i(A · J)B, A × B = ⎜⎜⎝⎜ A3 (2.128) −A2 A1 0 B3 where A · J = A1 J1 + A2 J2 + A3 J3 = J · A

(2.129)

is an antisymmetric matrix. In this way, we can see explicitly that the cross product is related to rotation. The result shown in Eq. (2.128) has some useful applications. For example, the change in the coordinates of a constant vector r under an infinitesimal rotation of the coordinate axes is from Eq. (2.122). dr = r  − r = idθ · Jr = −dθ × r.

(2.130)

As a result v=

dr = −ω × r, dt

ω=

dθ . dt

(2.131)

If the coordinate axes are fixed, but the point is rotated, its linear velocity vfixed in the fixed coordinate system is the reverse of this vfixed = −v = ω × r. This result is useful in describing the kinematics of rotating objects.

(2.132)

128

Transformations, matrices and operators

The purpose of the above discussion is not the derivation of simple results by a complicated procedure. It is rather to show how pervasive the generators Ji are in the description of rotations, although we may not be aware of them. Problems 2.10.1 Obtain the matrix that generates infinitesimal rotations about an axis making equal angles with the three coordinate axes. 2.10.2 Use Eq. (2.122) or (2.123) to obtain the identity

R3 (θ) = eiθJ3 2.10.3 Verify Eq. (2.120): [Ji , Jj ] = i

 k

⎛ ⎜⎜⎜ cos θ = ⎜⎜⎜⎝ − sin θ 0

sin θ cos θ 0

⎞ 0⎟⎟ ⎟ 0⎟⎟⎟⎠. 1

εijk Jk .

2.10.4 Show that (a) [exp(A)]† = exp(A† ); (b) If H is Hermitian, exp(iaH) is unitary if a is real. 2.10.5 If the commutator [B, A] = λI of n × n square matrices A and B is proportional to the identity matrix I, show that (a) [B, An ] = nλAn−1 ; ∂ (b) [B, f (A)] = λI ∂A f (A); (c) e x(A+B) = exA exB e(1/2)x λI (by showing that both sides have the same d/dx derivative), and hence that eA+B = eA eB e(1/2)[B,A] ; (d) eA eB = eB eA e−[B,A] 2

2.10.6 In deriving Eq. (2.123), why is it all right not to worry about the fact that the three components of J do not commute? 2.10.7 A point mass m is moving with velocity vrot relative to a rotating coordinate system that is rotating with constant angular velocity ω relative to a fixed Newtonian inertial frame. Show that (a) Its velocity in the fixed frame is  vfixed = 

d dt

 r fixed

 d = r + ω×r dt rot = vrot + ω × r;

Matrix groups

129

(b) Its acceleration in the fixed frame is   d afixed = vfixed + ω × vfixed dt rot = arot + 2ω × vrot + ω × (ω × r). Thus the effective force on the point mass in a rotating frame differs from the Newtonian force mafixed by ΔF = m(arot − afixed ) = m[ω × (ω × r) + 2ω × vrot ]. The first term is called the centrifugal force, while the second term is called the Coriolis force. 2.10.8 Prove the Jacobi identity for the commutators of matrices [A, [B,C]] + [B, [C, A]] + [C, [A, B]] = 0.

2.11

Matrix groups

Matrices are also important in physics because they are the simplest mathematical objects that form groups under noncommutative multiplications. The definition of a group has been given in Section 2.9 in connection with the study of transformations. The use of group properties of matrices is indeed related to the transformation and classification of physical properties. Although it is not our intention to study group theory in this book, it is nevertheless useful to recognize the more common matrix groups. All invertible (i.e., nonsingular) complex n × n matrices form a group under matrix multiplication. This group is called the complex general linear group of order n and is denoted by the symbol GL(n,C). It is a very large group, and it contains many subgroups, each of which satisfies all group properties. Interesting subgroups of GL(n,C) of complex matrices are (l) the special linear group SL(n,C) of unimodular matrices (i.e., matrices with unit determinant, (2) the unitary group U(n) of unitary matrices, and (3) the special unitary group SU(n) of unitary unimodular matrices. Interesting subgroups of GL(n,C) with real matrices are (1) the linear group GL(n) of all real invertible n × n matrices, (2) the special linear group SL(n) of unimodular matrices, and (3) the orthogonal group O(n) of orthogonal matrices. Some of these groups are of particular interest in physics. We shall discuss a few of these. 2.11.1

The group O(n)

The group O(n) is the group of coordinate transformation matrices λ in nD space. Under the O(n) transformation, the scalar product A · B  = (λA)T λB = AT (λT λ)B = A · B

130

Transformations, matrices and operators

is invariant. Matrices of O(n) are characterized by 12 n(n − 1) real parameters, because there are n2 real matrix elements that satisfy 12 n(n + 1) orthogonality relations λT λ = I. (See Problem 2.11.1.) The group O(3) of rotation matrices in space is characterized by three parameters. These may be taken to be the Euler angles α, β, γ defined in Eq. (2.18) in Section 2.2, R(α, β, γ) = Rz (γ)R x (β)Rz (α), where Ri (θ) is the matrix for rotation about the axis ei by an angle θ. Alternatively, we may use Eq. (2.123) R(θ) = exp(iθ · J), where θ = θ1 e1 + θ2 e2 + θ3 e3 contains three real parameters. 3D rotation matrices R actually have det R = 1. Such matrices form a subgroup of O(3) called the special orthogonal group SO(3). The remaining matrices of O(3) have the form −R, with det (−R) = −1. The matrix −R changes the sign of all three coordinates in addition to the rotation. 2.11.2

The groups U(n) and SU(n)

The group U(n) contains all complex unitary matrices of order n. The n2 complex matrix elements are equivalent to 2n2 real parameters, but the unitarity relation U † U = I imposes n2 constraints. (Prove this statement.) Hence the matrices of U(n) are characterized by n2 real parameters. A unitary matrix of order n can always be written in the form U = exp(iH),

(2.133)

where H is an n × n Hermitian matrix. One can show directly that n × n Hermitian matrices are characterized by n2 real parameters. The determinant of a unitary matrix can be calculated readily using the method described in Example 2.7.1. One finds det U = det[exp(iH)] = exp(iTrH) = exp(iα). where α=



(2.134)

λi

i

is the sum of the eigenvalues λi of H. Since λi and therefore α are real, we see that det U is a complex number of unit magnitude. It is also clear that the group U(1) is made up of the complex numbers exp(iβ), where β is a real number. That is, U(1) is the group of complex phase factors. Unitary matrices with det U = 1 (i.e., α = 0) form the group SU(n). If these matrices are denoted U0 , then U 0 = exp(iH0 ),

Tr H0 = 0,

det U0 = 1.

(2.135)

Matrix groups

131

If U = exp(iH) is a matrix of U(n), it can be associated with a matrix U 0 = exp(iH 0 ) of SU(n) as follows:  α  α α H = H0 + I, U = exp i U 0 = U0 exp i . (2.136) n n n Since the phase factor belongs to U(1), we may represent these relations by the expression U(n) = S U(n) ⊗ U(1),

(2.137)

meaning that a matrix of U(n) can be expressed as a direct product of a matrix in SU(n) and a matrix (the phase factor) in U(1). Matrices of SU(n) are characterized by n2 − 1 real parameters. For example, a general matrix of SU(2) has the form   sin β + i sin α cos β cos α cos βeiγ U0 (α, β, γ) = (2.138) − sin β + i sin α cos β cos α cos βe−iγ containing three real parameters α, β and γ. It is easy to verify that U0 is both unitary and unimodular. 2.11.3

Continuous groups and infinitesimal generators

A matrix group is called a continuous group if the matrices are continuous functions of the group parameters. Since the identity matrix I belongs to all the matrix groups, there are matrices of a continuous group G infinitesimally close to I. Suppose the matrices of G in a small neighborhood of I can be specified by a small number of matrices S of G. Then following the idea of integral calculus, we see that all matrices of G can be reached by making an infinite number of successive infinitesimal transformations of the type shown in Eq. (2.121), each involving only the generators S . Indeed, from what we know for rotation operators, we see that matrices of G are exponential functions of the generators and that the number of generators is equal to the number of real parameters characterizing the group matrices. We have met such group generators before. The rotation matrix exp(iθ · J) of Eq. (2.123) is expressible in terms of the three generators Ji of Eqs. (2.116). Other generators we have used are the frequency operator W of Eq. (2.89), which generates the time displacement operators exp(−iτW) of the time displacement group, and the propagation operator K of Eq. (2.97), which generates the space translation operator exp(iρ · K) of the space translation group. It is possible to obtain the generators of a matrix group when we are given the general form of the matrices of the group. For example, in the case of SU(2), one obtains directly from Eq. (2.138) the matrices for infinitesimal changes from I:   1 + i dγ dβ + i dα U0 (dα, dβ, dγ) = −dβ + i dα 1 − i dγ = I + i(σ1 dα + σ2 dβ + σ3 dγ),

(2.139)

132

Transformations, matrices and operators

where

 0 σ1 = 1

 1 , 0



0 σ2 i

 −i , 0



1 σ3 = 0

0 −1

 (2.140)

are the Hermitian Pauli matrices of Eq. (2.12). If φ = αi + βj + γk,

σ = σ1 i + σ2 j + σ3 k,

U0 (dφ) = I + iσ · dφ.

(2.141)

This matrix for an infinitesimal change from the identity matrix I can be integrated (in a way similar to the treatment of the rotation operator in Section 2.10) to give the general form of matrices in SU(2): U0 (φ) = exp(iσ · φ).

(2.142)

It is clear from the above derivation that this U0 (φ) is exactly equal to U0 (α, β, γ) of Eq. (2.138). The generators iσk , k = 1, 2, 3, of SU(2) can also be obtained directly from Eq. (2.135) if one knows that all 2 × 2 Hermitian matrices (which are characterized by four parameters) can be expressed in terms of the Pauli matrices: H =σ·φ+

α I, 2

H0 = σ · φ,

Tr H0 = 0.

(2.143)

Pauli matrices satisfy the useful multiplicative relation described by Eq. (2.14a)  σi σj = δij I + i εijk σk . (2.144a) k

2.11.4

Lie algebra

It was recognized by Lie, a Norwegian mathematician of the nineteenth century, that the important property of the generators of continuous groups is not so much that they can be identified with numerical matrices, as in Eq. (2.140) or (2.116). It is rather that they satisfy certain commutation relations. Thus for the three generators Jk of S O(3), we have Eq. (2.120),  [Ji , Jj ] = i εijk Jk , k

while for the three Pauli spin matrices of SU(2) we have  εijk σk [σi , σj ] = 2i k

(2.144b)

Matrix groups

133

or 

1 1 2 σi , 2 σj



= j



εijk 21 σk .

(2.144c)

k

Thus commutators of generators are linear combinations of the generators themselves. Such a mathematical structure is called a Lie algebra. The linear coefficients ii jk are called its structure constants. What we have shown is that the three matrices 12 σk (but not the σk themselves) constitute the same Lie algebra as the three generators Jk for rotations. Therefore, if exp(iθ · J) is a rotation matrix for rotation of a vector angle θ, exp(iφ · 12 σ) must be a rotation matrix for rotation of a vector angle φ. But wait! exp(iθ · J) is a 3 × 3 matrix that operates on 3D position vectors, but exp(iφ · 12 σ) is a 2 × 2 matrix. What does it operate on? The answer, obtained by Pauli from studies of atomic spectra, is that it operates on an internal attribute of atomic particles that can have one of only two possible values. Since we are concerned with rotation, the property is called an intrinsic spin, which can either point up or point down. (These directions may be chosen arbitrarily to be the +z or −z direction.) An intrinsic-spin state can thus be represented by a 2D state vector   a u= b

(2.145)

of unit length (a2 + b2 = 1) on which the rotation matrices act. There is in addition an extra factor 12 in the rotation matrix for intrinsic spins. As a result, a rotation of φ = 2θ is necessary to produce the same effect in the spin vector u as a rotation of θ produces in the position vector r. This is readily seen by considering successive rotations of angle π about the y axis. The relevant rotation matrices are ⎞ ⎛ ⎜⎜⎜cos θ2 0 − sin θ2 ⎟⎟⎟ 1 0 ⎟⎟⎟⎠ , Ry (θ2 ) = eiθ2 J2 = ⎜⎜⎜⎝ 0 sin θ2 0 cos θ2 ⎛ ⎞ 0⎟⎟ ⎜⎜⎜−1 0 ⎟ 0⎟⎟⎟⎠ ; Ry (π) = ⎜⎜⎜⎝ 0 1 0 0 −1 ⎛ ⎞ 1 1 ⎜ ⎟⎟⎟ φ sin φ cos ⎜ 2 2 2 2 ⎟⎟⎠ , Uy (φ2 ) = eiφ2 (1/2)σ2 = ⎜⎜⎜⎝ − sin 12 φ2 cos 12 φ2   0 1 = iσy . Uy (π) = −1 0

(2.146)

(2.147)

134

Transformations, matrices and operators

Therefore ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜⎜⎜ x⎟⎟⎟ ⎜⎜⎜−x⎟⎟⎟ ⎜⎜⎜ x⎟⎟⎟ ⎜⎜⎜−x⎟⎟⎟ ⎜⎜⎜ x⎟⎟⎟ Ry (π) ⎜⎜⎜⎝y⎟⎟⎟⎠ = ⎜⎜⎜⎝ y⎟⎟⎟⎠ , Ry (2π) ⎜⎜⎜⎝y⎟⎟⎟⎠ = Ry (π) ⎜⎜⎜⎝ y⎟⎟⎟⎠ = ⎜⎜⎜⎝y ⎟⎟⎟⎠ ; −z z z z −z         b −a a a , Uy (2π) , = = Uy (π) −a −b b b         a −b a a Uy (3π) = , Uy (4π) = . b a b b

(2.148)

(2.149)

We see that it takes two full turns, not just one, to restore the spin vector u to its original form. Mathematical objects like u that change their sign after one full turn around the rotation axis and resume their original value only after two full turns are called spinors. Problems 2.11.1 Show that for square matrices λ of order n, the orthogonality relation λT λ = I = λλT gives 12 n(n + 1) conditions on its matrix elements. Hint: The unit matrix I contains n diagonal matrix elements of 1 and n(n − 1) off-diagonal matrix elements of 0. 2.11.2 Show that for square matrices U of order n, the unitarity relation U † U = I = UU † gives n2 conditions on its matrix elements. 2.11.3 (Closure properties) (a) Show that the matrix property of orthogonality [unitarity] is closed under group multiplication for O(n) [U(n)]. (b) Show that the determinantal property is closed under group multiplication for SO(n) and SU(n). (c) Show that the O(n) [U(n)] matrices with determinant −1 do not form a group. 2.11.4 Show that a general SU(2) matrix has the form   a b S = , −b∗ a∗ where a, b are complex numbers with aa∗ + bb∗ = 1. 2.11.5 [U(2) and SU(2)] Show that the 2D rotation matrix complexified by additional phase factors   cos φeiζ sin φeiη R= − sin φeiγ cos φeiδ

Tables of mathematical formulas

135

is unitary if ζ + δ = η + γ, and has unit determinant if ζ + δ = η + γ = 0. Given the closure properties described in Problem 2.11.3, these matrices form the U(2) group and the SU(2) group, respectively.

Appendix 2 Tables of mathematical formulas 1 Coordinate transformations xnew = λxold ,

λij = enew · eold j i

λT = λ−1 ⎛ ⎞ ⎜⎜⎜ cos θ sin θ 0 ⎟⎟⎟ Rz (θ) = ⎜⎜⎜⎝ − sin θ cos θ 0 ⎟⎟⎟⎠, 0 0 1 ⎛ ⎞ 0 0 ⎟⎟ ⎜⎜⎜1 ⎟ R x (θ) = ⎜⎜⎜⎝0 cos θ sin θ ⎟⎟⎟⎠ 0 − sin θ cos θ

⎛ ⎜⎜⎜cos θ Ry (θ) = ⎜⎜⎜⎝ 0 sin θ

⎞ 0 − sin θ⎟⎟ ⎟ 1 0 ⎟⎟⎟⎠, 0 cos θ

R(α, β, γ) = Rz (γ)R x (β)Rz (α), for the Euler angles α, β,γ. 2 Determinant det A =



εij···l Ai1 Aj2 · · · Aln

ij···l

=



AkqCkq =



k

=

 q

δik det A =

Akq (−1)k+q Mkq

k

AkqCkq 

(Laplace development of the qth column)

(Laplace development of the kth row)

AijCkj

j

A−1 = C T / det A det (AT ) = det A det (kA) = kn det A,

if k is a scalar

det (AB) = (det A)(det B) = det (BA). 3 Matrix equations Ax = c has solutions xi = Σj cjCji /det A (Cramer’s rule); Ax = 0 has nontrivial solutions if det A = 0. The solutions x are orthogonal to the row vectors ai of A. 4 Matrix eigenvalue problem (K − λM)q = 0 requires det(K − λM) = 0. If H in Hei = λi ei is Hermitian, the eigenvalues λi are real and the eigenvectors ei , are orthogonal or can be

136

Transformations, matrices and operators

orthogonalized. If the square matrix U contains the orthonormahzed eigenvectors ei columnwise, then U † U = I,

U † HU = D = (λi δi j ).

Tr A = Σi λi , where λi are the eigenvalues of A. 5 Infinitesimal generators of transformations   d f (t + τ) = exp τ f (t) dt φ(r + ρ) = exp(ρ · ∇)φ(r) R(θ) = exp(iθ · J) ⎛ ⎛ ⎞ ⎜⎜⎜ 0 0 0 ⎟⎟⎟ ⎜⎜⎜ 0 ⎜ ⎟ J1 = ⎜⎝⎜ 0 0 −i ⎟⎠⎟, J2 = ⎜⎜⎝⎜ 0 0 i 0 −i ⎛ ⎞ ⎜⎜⎜0 −i 0⎟⎟⎟ 0 0⎟⎟⎟⎠ J3 = ⎜⎜⎜⎝i 0 0 0  [Ji , Jj ] = i εijk Jk

⎞ 0 i ⎟⎟ ⎟ 0 0⎟⎟⎟⎠, 0 0

k

U0 (φ) = exp(iφ · 12 σ)     0 1 0 −i , σ2 = , σ1 = 1 0 i 0  [ 12 σi , 12 σj ] = i εijk 21 σk

 1 σ3 = 0

 0 −1

k

σi σj = δij + i



εijk σk .

k

6 Differential eigenvalue equations Helmholtz equation: K 2 u(r) = k2 u(r),

K = −i∇ :

u(r) = e±ik · r ;

Klein–Gordon equation: (p2op c2 − H 2 )u(r, t) = −m2 c4 u(r, t), u(r, t) = e±ik · r e±iωt ,

H = i

∂ : ∂t

(mc2 /)2 = ω2 − k2 c2 .

Tables of mathematical formulas

137

7 Matrix relations Tr(AB) = Tr(BA). P−1 AP is an orthogonal (unitary) transformation of A if P is an orthogonal (unitary) matrix. OTO = I = OOT if O is orthogonal (OT = O−1 ). U † U = I = UU † if U is unitary (U † = U −1 ). (σ · A)(σ · B) = A · B + iσ · A × B, where A and B are vectors. det(eA ) = eTrA if A is a normal matrix, i.e., one that can be diagonalized by a unitary transformation.

3 Relativistic square-root spaces∗ 3.1

Introduction

In Chapter 1, physical space is described mathematically and therefore quantitatively and objectively. In Chapter 2, spacetime events are formalized as transformations involving matrices and differential operators. The purpose of this chapter is to show that these two chapters, simple as they are, provides the mathematical framework in which the story of modern physics, the physics of the last century, can be told. Modern physics is dominated by two conceptual revolutions in the early twentieth century: relativity and quantum mechanics. We shall begin with Einstein’s derivation of the Lorentz transformation matrix in spacetime, including his important conclusion that time is not absolute, but changes with motion. Einstein’s transparent formulation of relativistic kinematics should be contrasted with the failure of Poincar´e, who could not take the relativity of time at face value. Einstein’s achievement can be appreciated if one remembers that Poincar´e was the leading mathematical physicist when modern physics emerged from discoveries of radioactivity and certain strange properties of light propagation during the two decades when the nineteenth century ended and the twentieth century began. While we celebrate successes, we also remember failures and try to learn from them what lessons we can. A good example is the mathematical system of quaternions, invented by the mathematical physicist Hamilton. Once considered the key that could unlock unknown mysteries in physics, they fall short because they could not naturally accommodate the quadratic invariance of spacetime scalar products. They suffer the indignity of having their most useful results split off as vector analysis. We shall show that they are still good for describing 3D rotations. The relativistic kinematics of Lorentz spacetime and momentum-energy vectors is readily applied to practical problems by using their scalar products that are Lorentzinvariant, the same in every Lorentz frame. The converse process, the unfolding of quadratic spacetime invariant scalars into their original spacetime vectors, may appear trivial. At the hand of Dirac, who at the dawn of quantum mechanics made it his business to reconcile quantum mechanics with relativity and classical mechanics, it produces the linearized Dirac wave equation. This equation is surely one of the most beautiful equations in physics, an equation that describes a point particle that nevertheless possesses such unusual properties as intrinsic spin, intrinsic parity, antimatter and other magical properties of square-root spaces. We shall cover the symmetries of the Dirac equation, and how these symmetries can be broken, all from the simple perspective of elementary mathematics.

Special relativity and Lorentz transformations

139

Not everything can be made elementary. The more advanced parts of the mathematics of square-root spaces requires the use of group theory and complex square-root coordinates. Here our objective is to give the reader an introduction to these ideas so that they will not be total strangers the next time we meet them. Mathematics is built up of simple parts: √ number theory from whole numbers, functions of complex variables by using i = −1. In fact, i provides the simplest realization of a nontrivial square-root degree of freedom that permits a simple description of wave phenomena as the coherence and interference between two coupled degrees of freedom, the reals and the imaginaries. A surprising amount of theoretical and experimental physics involves this little unintrusive symbol i and the phenomena it can describe. Physics too can build up imposing structures with simple building blocks. Systems with spins can be built from elementary objects of spin 1/2. The Lorentz group is the direct product of two SU(2) groups. Vectors can be placed together as multivectors. We shall introduce the reader to some of the multivectors in common use: multispinors, dyadics, Cartesian tensors and finally general tensors suitable for the description of curved space. Tensor analysis, the differential geometry of curved space, is what Einstein needed when he looked for a mathematical language to express his thoughts on the dynamical origin of a curved spacetime. The topics covered in this chapter are more advanced than those of the last two chapters. The problems too are correspondingly more difficult. The serious reader should do the problems. Many of the sections are understandable if one’s interest is more casual.

3.2

Special relativity and Lorentz transformations

Einstein’s special theory of relativity is based on two fundamental postulates: 1. Principle of relativity: Physical laws are the same in all inertial frames. 2. Constancy of light speed: Light propagates in empty space with the same universal speed c in all inertial frames independently of the motion of its source. These postulates appear contradictory at first sight. The contradiction is easily seen by a passenger on a moving train viewing a car traveling with the same velocity on a nearby highway. The moving car appears stationary to the train passenger. Thus velocities do not seem to have the same values in different inertial frames. Yet the constancy of light speed in all inertial frames is needed to explain a variety of experimental observations on light from distant stars. Light speed also appears as a universal constant in Maxwell’s equations of electromagnetism that are solidly based on many electomagnetic observations. Some of these observations actually involve moving sources or moving observers. To resolve the paradox, Einstein made the bold suggestion that time is not absolute but changes in different inertial frames, in a way already suggested tentatively by Lorentz that is now known as

140

Relativistic square-root spaces

the Lorentz transformation (LT). The different inertial frames so connected are now called Lorentz frames. Before proceeding further, Einstein first defines the times at different locations of the same inertial frame as determined by identical local clocks. These clocks must first be synchronized with one another—a task that must be done even in Newtonian physics. In the latter, old-fashioned way, one can imagine an infinite number of identical clocks all set at the same time at the same location, as one does in real life. These synchronized clocks are then taken and placed at different locations of the coordinate system. If motion affects time, however, this intuitive way will no longer work precisely. To synchronize clocks at different locations of the same inertial frame, Einstein notes that the time it takes light to travel from any point, say A to any other point, say B, must be the same as the time it takes light to travel from B to A. Suppose a light signal is emitted at A at time tA = 0, is reflected back to A by a mirror at B, and arrives back at A at time tA = 2t. The clock at B is properly synchronized with the clock at A if the reflection occurs at B at time tB = t. The next task is to synchronize clocks on different Lorentz frames in relative motion, for without such synchronization the times on different frames are totally unrelated. Since the times on each frame are already synchronized, it is only necessary to synchronize one clock in one frame with one clock in the second frame. This synchronization is quite easy to do because at every instant of time, a point in the stationary frame S coincides with a point in the moving frame M. We only need to agree to set the times of the two local clocks at a single chosen location, one clock from each frame, to the same time value. This synchronization should be done only once, for each resetting of the clocks from different frames will destroy any evidence of previous time differences caused by their prior relative motion. We are now ready to analyze the motional effect on time. Take for simplicity a one-dimensional (1D) space where a light wave is emitted from a source placed at the origin of the stationary coordinate system S at time t = 0. The positions of the two light fronts at time t after emission are given by three equivalent expressions x = ±ct,

x2 = (ct)2 ,

x2 + (ict)2 = X 2 = 0.

(3.1)

The last expression contains a sum of two squares, like the squared distance in 2D space, but with the non-Pythagorean result that the sum X 2 is zero. The unusual mathematical space where time is represented as a purely imaginary number ict in terms of the distance ct light travels during the time t was introduced by Minkowski to give an intuitive geometrical interpretation to the motional effect on time caused by LTs. This spatial treatment of time gives rise to the concept of spacetime as a unified geometrical concept. We shall use this idea right from the beginning. It is a bit of a detour that will turn out to be a shortcut instead. An event X = (x, ict) in a 2D Minkowski spacetime is a 2D vector. Its square distance X 2 from the origin of coordinates (0, i0) is zero on the light front, as shown in Eq. (3.1). More generally, if X 2 < 0, the event is called time-like. If X 2 > 0, the

Special relativity and Lorentz transformations O'

141

v

M

x O

S

0

x

vt

Fig. 3.1 Lorentz transformation from the stationary frame S to the moving frame M moving with velocity v in the stationary frame. The two coordinate origins are at the same location initially (at time t = 0 = t ).

event is space-like, and more Pythagorean. The same adjectives can be used for the event separation X1 − X2 = (x1 − x2 , ict1 − ict2 ). Consider now two inertial frames S and M with M moving relative to S in such a way that its spatial origin x = 0 has the spatial position x = vt when described in S , as shown in Fig. 3.1. Let their times t and t be synchronized to t = t = 0 when their spatial origins coincide, x = x = 0. That is, in the 2D Minkowski spacetime these two inertial frames have the same spacetime origin. Their coordinate axes can differ only by a rotation: 



X =

x ict



 =

cos θ sin θ − sin θ cos θ



x ict



= R(θ)X.

(3.2)

The 2 × 2 transformation matrix R(θ) is an orthogonal matrix because RT (θ) = R(−θ) = R−1 (θ),

(3.3)

as one could directly verify by calculating the matrix product R(θ)RT (θ). As a result, the squared distance X T X  = [R(θ)X]T R(θ)X = X T X

(3.4)

is invariant under rotation. That is, it is an invariant scalar. This invariance property is one of the most important properties of orthogonal transformations. In Minkowski space, the time coordinate ict is purely imaginary. The rotation angle θ = iα must also be purely imaginary, otherwise it will give an imaginary contribution to the position coordinate. Hence cos θ = cosh α,

sin θ = i sinh α,

sin θ + cos θ = cosh2 α − sinh2 α = 1. 2

2

(3.5)

142

Relativistic square-root spaces

The LT for the spacetime coordinates between two different Lorentz inertial frames then has the general form       x cosh α i sinh α x  = X = −i sinh α cosh α ict ict = λX,

(3.6)

where the LT matrix is now denoted λ. The coordinate rotation angle θ = iα, or equivalently the Lorentz boost or rapidity α can be found by using in Eq. (3.6) the information that events at x = 0 in M occur at different Minkowski coordinates X = (vt, ict) in S . The spatial part of the Minkowski rotation in that matrix equation then reads x = 0 = x cosh α − ct sinh α, v tanh α = ≡ β. c

or (3.7)

The geometrical relation (3.5) gives next the matrix elements 1

cosh α ≡ γ = &

1 − tanh2 α sinh α = tanh α cosh α = βγ

of the LT matrix. Hence

 λ(β) =

= &

1 1 − β2 (3.8)

 γ iβγ . −iβγ γ

(3.9)

Using the LT matrix in Eq. (3.6), the time part of the matrix equation can be found to be ict = −iβγ(vt) + ictγ = t = τ =

t . γ

ict , γ

or (3.10)

Two events, at t = t = 0 and at t = τ = t/γ, are thus seen in the frame M as occurring at the same spatial location, here x = 0. This happens only in one inertial frame, the frame M now called the proper frame for these events. The times t = τ in the proper frame for events occurring at the same location are called proper times. Eq. (3.10) then states that proper times are smaller than the times for these events in any other frame such as S where the events do not occur at the same location. We shall see in a following subsection that this result has a simple intuitive geometrical interpretation.

Special relativity and Lorentz transformations

3.2.1

143

Velocity addition

Einstein’s velocity addition formulas come from the elementary trigonometric identities tan θ1 + tan θ2 , 1 − tan θ1 tan θ2 cos(θ1 + θ2 ) = cos θ1 cos θ2 (1 − tan θ1 tan θ2 ). tan(θ1 + θ2 ) =

(3.11)

For purely imaginary angles in the notation of Eqs. (3.7) and (3.8), these expressions read β12 =

β1 + β 2 , 1 + β1 β2

γ12 = γ1 γ2 (1 + β1 β2 ).

(3.12)

The result for β12 shows that if one or both βi = 1 (i.e., if vi = c), then β12 = 1 follows for the composite transformation resulting from two successive LTs. Hence light speed is c in all inertial frames. This desired outcome is actually put into the solution by hand by insisting that the time coordinate ict involves the same universal light speed c in all inertial frames. The velocity addition formula has an interesting implication. If a particle becomes superluminal (faster than light), the velocity addition (3.12) becomes ill-behaved. To see this, let β2 > 1 describes the velocity of this superluminal particle in a moving frame 1, and β1 be the velocity of this moving frame relative to the stationary frame S . The velocity β12 of the particle in S then contains a denominator 1 + β1 β2 that vanishes when the velocity of the moving frame β1 = −1/β2 is subluminal. This is a mathematical pathology called a singularity, where the particle velocity becomes infinite and is therefore ill-behaved. One can show in a similar way that the description of a subluminal particle in a moving frame 1 is singular in a stationary frame S if frame 1 is moving superluminally relative to S . No such embarrassment occurs if particle and frame are both subluminal or both superluminal. We know from experience that subluminal motion including particles at rest are possible. Hence superluminal motion is inconsistent with the Lorentz transformation, or the spacetime structure it implies. In other words, if spacetime is Einsteinian, no superluminal particle could coexist in a world with massive particles that are subluminal. 3.2.2

Lorentz vectors and scalars

The Minkowskian notation has the advantage of being intuitive, and the disadvantage of involving factors of i here and there. One can also use the so-called Euclidean notation with a real time variable by eliminating all factors of i by suitable redefinitions. The problem there is that there are two different notations in common use that differ in signs. For this reason, we shall use only the Minkowskian notation, where the appearance of i can be turned to an advantage as a reminder of all the nonEuclidean aspects of the problem as they appear in calculations. The final results are independent of the notation chosen. We shall henceforth refer to Minkowski vectors

144

Relativistic square-root spaces

by their generic name of Lorentz vectors unless we want to emphasize their nonEuclidean features. The scalar product of two Lorentz vectors, say X  with Y  , is # " (3.13) X T Y  = (λX)T λY = X T λT λ Y = X T Y by virtue of the orthogonality of the LT matrix λ, namely λT λ = I, the identity matrix. Thus a Lorentz scalar is invariant under a LT, the same in all Lorentz frames. The squared separation in spacetime between two events X1 = (x1 , ict1 ) and X2 = (x2 , ict2 ), namely 2 2 2 2 2 X12 = (X1 − X2 )2 = x12 − c2 t12 = x12 − c2 t12 ,

(3.14)

2 is zero for events on the light front where x = ct in is likewise a Lorentz scalar. X12 i i any Lorentz frame. Such Lorentz vectors on the light front are called null vectors. If 2 > 0, the event separation is space-like. It is time-like if X 2 < 0. X12 12 The frame-independent Lorentz scalar (3.14) makes it possible to explain both length contraction and time dilation from a simple Lorentz-invariant perspective.

3.2.3

Length contraction and time dilation

According to the Lorentz invariant expression (3.14), the spatial separation between two events is shortest in that frame (say the moving frame M) where they occur at the  = 0: same time, i.e., t12 2 2 2 2 = x12 − c2 t12 < x12 . x12

(3.15)

The measured length between two spatial points therefore contracts to its smallest value in that unique frame M where the measurements are made simultaneously, using properly synchronized clocks. In a similar way, the time interval between two events is smallest in that frame (using M again) where they occur at the same location r12 = 0 (and therefore measured by the same local clock): 2 2 t12 = t12 −

1 2 2 r < t12 . c2 12

(3.16)

M is the proper frame for these events. The proper time interval τ between these  |, is thus smaller than the time interval |t | in any Lorentz frame events, here |t12 12 where they do not occur at the same location. Thus the proper frame is unique. Any other Lorentz frame cannot be proper, because the two events cannot occur at the same location there. From the perspective of all these other Lorentz frames, the proper frame is a special moving frame. It is the inertial frame of an observer of events whose watch gives the proper elapsed time between events because the observed events always occur at the watch location. That

Special relativity and Lorentz transformations

145

is, a moving watch runs slow compared to the time intervals shown by clocks in all the other Lorentz frames—a phenomenon called time dilation. Time dilation is probably a confusing name. Dilation means expansion, here the expansion of the time unit between two ticks of the watch, not the increase of the time value, but rather its decrease. It is as if by “thermal length expansion” we mean the expansion of a heated ruler with which we measure an unheated rod. In real life, we mean rather the length expansion of a heated rod measured by an unheated ruler. But “time dilation” means that the traveling watch runs slow, whether you like the term or not. The real age of an animal is its proper age because its biological activities always occur at the same locations inside the moving animal. Returning to physics, we note that the squared proper time is a Lorentz invariant 2 : just like X12 τ2 = −

2 X12

c2

.

(3.17)

The fact that τ2 is the smallest of the time separations is a quintessential nonEuclidean or Minkowskian attribute. The infinitesimal squared proper time separation dr2 − c2 dt2 = dt2 (1 − β2 ) c2  2 dt . = γ

dτ2 = −

(3.18)

is also the smallest time separation. Its Lorentz invariance is an important property because if dτ and not dt is used to define velocity and acceleration, then the Lorentz transformation character of the changing Lorentz vector will not be changed by these calculus operations, as we shall see in the next section. Time dilation can be related to length contraction. A person can determine the length of an object passing her with speed v by reading off the passage time τ of the object on her watch. She then deduces a length vτ. The result is minimal, according to Eq. (3.17). Hence this deduced length must be the Lorentz contracted length. The real length of the object as measured by a meter stick in the object’s rest frame is the uncontracted length |x12 | = γvτ. In contrast, the unique elapsed time between two events is the proper time. As a special case of Eq. (3.16), consider two events that occur simultaneously in the proper frame where they occur at the same location, i.e., τ = 0 and X12 = 0. Then in all other Lorentz frames moving relative to the proper frame, we shall find 2  0 and hence t2  0. This means that the events are no longer simultaneous that r12 12 in these moving frames. As a result, “we cannot attach any absolute significance to the concept of simultaneity”, using Einstein’s own words in his famous 1905 paper on special relativity. In a similar way, the same two events cannot occur at the same spatial location in all Lorentz frames. So the concept of spatial localization is also not frame-independent when it involves events at different times.

146

3.2.4

Relativistic square-root spaces

Twin puzzle

After deriving the LT and finding Eq. (3.10) showing that the proper time between two events is smaller than their time duration in any other inertial frame where they do not occur at the same location, Einstein adds succinctly in a short paragraph of his 1905 paper: “It is at once apparent that this result still holds good if the clock moves from A to B in any polygon line, and also when the points A and B coincide.” That is, when the moving clock returns to its starting point in 3D space, it will give a smaller elapsed time τ = tγ than the reading t of an identical clock left at A. Most of us who do not have Einstein’s clarity of thought wonder if a paradox might be involved if motion is all relative and yet the traveling clock runs slower than the non-traveling clock. When applied to the asymmetric aging of two identical twins, the puzzle is called the twin paradox or puzzle. To illustrate the puzzle, consider the following story of two twins: At age 18, two identical twins are separated after their birthday celebration. The sister leaves Earth in an outgoing rocket moving with constant velocity v (and γ = 30) towards a distant space station. The traveler arrives at the space station on her 19th birthday. She then jumps into a return rocket traveling with velocity −v and returns to Earth on her 20th birthday. On her return she finds her twin brother just celebrating his 78th birthday. These Einsteinian ages must be correct because on the outgoing leg of the journey, both the departure and the transfer occur at the same location for both sister and the outgoing rocket. So their times are the proper times. These two events are seen on the Earth frame as occurring at different locations, however, the departure from the Earth and the transfer at the space station being almost one light year apart. So the time between these two events for the non-traveling brother on earth must be t = γτ = 30 years. On the return leg of the journey, the transfer and the arrival jump back to Earth a year later occur at the same location for both the traveler and the return rocket. Hence their elapsed traveling time of one year is also a proper time. In the Earth frame, these two events occur at locations separated by almost one light year. The elapsed time is then another 30 years. There is thus a clear asymmetry between the experiences of the traveling sister and the stay-at-home brother. The reader who is still unpersuaded should examine and verify the time readings summarized in Table 3.1 for the traveler’s watch and some of the synchronized clocks of the three Lorentz frames involved. Note that at the time of transfer at the space station where the sister jumps “rocket”, the clock readings of the outgoing rocket and the return rocket are not the same. The difference is an artifact of the clock setting or synchronization used on line 1 of the table, and has absolutely no significance in the mathematics or physics of the problem. It is just an example of a common experience that arriving passengers at airports often have to reset their watches to the local time if they want to talk time with the locals. Of course, physics treats the twins impartially. After the sister’s departure from Earth, the twins’s motions are symmetrical to each other. The brother too can break the symmetry if on his 19th birthday he jumps into an express rocket that moves with velocity v relative to the outgoing rocket in which his sister is traveling. On his

Special relativity and Lorentz transformations

147

Table 3.1 Clock readings in different locations in different Lorentz frames in the twin problem. The clocks in each row are all located at the same point in space, but their position readings may be different in different Lorentz frames, except at start when all coordinate origins are on Earth and all clocks and watches are synchronized to time zero. γE = γ2 (1 + β2 ) = 2γ2 − 1 is from Eq. (3.12). Dashed entries are inapplicable. Unspecified clock readings are not involved in the puzzle.

Location (1) On Earth at start (2) Sister jumps rocket (3) Brother leaves (4) Sister returns (5) Brother jumps rocket

Sister’s watch

Outgoing rocket

Earth

0

0

0

τ

τ

– 2τ –

Return rocket

Brother’s watch

Express rocket

0

0

0

γτ

2γ2 τ − τ

–

γτ

τ

γτ

τ

2γ2 τ

2γτ

2γ2 τ

– 2τ

γE τ

γE τ + τ

20th birthday he will catch up with his sister, assuming that she has not de-rocketed at the space station. When he jumps over to the sister’s outgoing rocket, he will find his sister celebrating her 78th birthday. Such is the strange but true life in the ultrarelativistic land. For more information, see Problem 3.2.3.

Problems 3.2.1 (a) Explain why λ−1 (β) = λ(−β) if the Lorentz transformation matrix λ is considered a function of β = v/c. Verify by direct matrix multiplication that λ−1 (β)λ(−β) = I, the 2 × 2 unit matrix. (b) Explain why λ(α2 )λ(α1 ) = λ(α2 + α1 ) if the Lorentz transformation matrix λ is considered a function of the Lorentz boost α = −iθ. 3.2.2 (Twin puzzle) (a) Verify all entries of Earth and outgoing rocket clocks shown in Table 3.1 by using suitable Lorentz transformations between the Earth and outgoing rocket frames. (b) Verify all the other entries of the table from the results of part (a) by using arguments based on symmetry, analogy, dead reckoning or any other way. Use LTs only if you have no other choice. 3.2.3 (Return rocket time) Use a single Lorentz transformation between the earth frame S and the return rocket frame M  to verify that the return rocket clock does indeed read t = (2γ2 − 1)τ at the moment the traveler transfers to it at the space station, as stated in Table 3.1. Show that one obtains the same

148

Relativistic square-root spaces

result if the Lorentz transformation is between the outgoing rocket and the return rocket. 3.2.4 (Touring photographers) Alice and Bob are seated Δx = xA − xB = 10 m. apart, front and back, on a bus touring the countryside at relativistic speed with γ = 4. Each person agrees to take a photograph of a famous landmark, Alice at time tA and Bob at time tB . (a) Find all possible values of Δt = tA − tB when the squared event separation (ΔX  )2 = Δx2 − c2 Δt2 of the two camera shots (i) has the largest possible value, (ii) is space-like, (iii) is light-like, (iv) is time-like. (b) Discuss the relationship between the spatial and time separations Δx and Δt of these events in the Lorentz frame of the countryside under each of these four scenarios. Show that Δx = γΔx in scenario (i), and Δx = γ(1 ± β)Δx or Δx = Δx /γ(1 ∓ β) in scenario (iii). Find the corresponding Δt. (c) Show that the curve Δx/Δx versus cΔt is a straight line. Mark the regions/points on a plot of this function of cΔt , where the event separation is space-like, light-like, time-like, and where Δt = 0 in the countryside frame. 3.2.5 (The photon rocket) In the photon rocket, fuel is first converted into photons before being emitted as the propellant. A photon rocket has mass M0 when it is at rest in an inertial frame when it starts firing its engine. Show that when it reaches the velocity v = βc in that inertial frame, its mass has decreased to M=

M0 . γ(1 + β)

Hint: Use Lorentz momentum-energy conservation: d(Procket + Pγ ) = 0. 3.2.6 (Lorentz acceleration vector in 2D Minkowski space) Consider a 2D Minkowski space, where the Lorentz position and velocity vectors of a particle are X = (x, ict), U ≡ dX/dτ, τ being the proper time. (a) Show that its Lorentz acceleration vector is A≡

dU = aγ4 (1, iβ), dτ

A2 = a2 γ6 .

and (3.19)

Here β = dx/d(ct), a = d2 x/dt2 . (b) The instantaneous rest frame (of interest in general relativity) is the inertial frame where v = dx /dt = 0 at time t . Show that in this frame, a2 = (d 2 x /dt2 )2 is larger than the corresponding a2 in any other Lorentz frame. Show that A2 is either 0 or nonzero in all Lorentz frames.

Special relativity and Lorentz transformations

149

Show that if the acceleration a is independent of time in one frame, it is time-independent in all Lorentz frames. ∗ 3.2.7 (Hyperbolic motion under constant acceleration) The engine of a rocket gives it constant acceleration a = d2 x /dt2 in the rocket’s instantaneous rest frame (the frame where the rocket is momentarily at rest, or v = dx /dt = 0). (a) Show that it can outrun a beam of light if given sufficient headstart x0 > c2 /a at the starting time t = 0 (of the inertial frame of an outside observer) and an inexhaustible supply of fuel (whose mass is to be neglected for the purpose of this problem). Sketch the position x(ct) in the observer frame for both rocket and light beam. Hints: You may use the result of a previous problem that the squared Lorentz acceleration A2 = a2 γ6 is a Lorentz scalar. Then show that the time and position of the rocket in the observer frame are t=

c sinh α, a

x=

c2 cosh α. a

(b) A rocket leaves Earth for two space stations in the same direction with the second twice as far away as the first. At takeoff the rocket engine is fired to give the rocket a constant acceleration a relative to the rocket’s instantaneous rest frame in the direction of the space stations. The rocket’s rest frame is not an inertial frame. As the rocket passes the first space station, its engine is turned around to give the rocket a constant deceleration (or acceleration −a ) in its own rest frame. The rocket will come to rest just as it reaches the second space station. With the engine still firing, the rocket begins to return to the Earth. As it passes the first space station, the rocket engine is turned around once more to give an acceleration a so that the rocket will just come to a stop when it returns to Earth. If the rocket returns after t years according to the Earth’s clock, show that the distance x of the second space station from the Earth and the duration tp of the roundtrip for the rocket passenger are ⎡( ⎤ ⎥⎥⎥ ⎢⎢⎢   2 2  2c ⎢ at ⎥ x =  ⎢⎢⎢⎢ 1 + − 1⎥⎥⎥⎥ , ⎦ a ⎣ 4c tp

   at 4c . =  arcsinh a 4c

Hint: For the rocket passenger, dt = (dt /dt)dt. (c) Let the traveling twin sister of the twin puzzle travels instead on the rocket of the present problem to the second space station 1 light-year from Earth. Show that when she returns to the Earth to find her twin

150

Relativistic square-root spaces

brother celebrating his 78th birthday 60 years after her departure, she is younger by only 16 days. Hint: You may use the series expansion arcsinh z = z − z3 /6 + 3z5 / 40 − . . . instead of an arcsinh table.

3.3

Relativistic kinematics and the mass–energy equivalence

Relativistic kinematics in 4D spacetime starts with the four-velocity U=

dX d = γ (r, ict) = γ(u, ic) dτ dt

(3.20)

that is a four-vector like the four-position X = (r, ict) = (x, y, z, ict). It satisfies the same LT as the four-position X. This result follows from the fact that the proper time τ used is a Lorentz scalar. The name Lorentz scalar or vector is a more inclusive name that does not depend on the number of spatial dimensions involved. The scalar product of the four-velocity with itself U ·U =

u 2 − c2 = −c2 2 1 − (u/c)

(3.21)

has a universal or invariant value in all Lorentz frames. The value is also the same for all particles, massive or massless. For a particle of finite inertial mass m, the product mU is a four-vector called its four-momentum  E P = mU = mγ(u, ic) = p, i . c

(3.22)

The four scalar P2 = p2 −

 E 2 c

= −(mc)2

(3.23)

is Lorentz invariant, having the same value in all inertial frames. Its value in its own rest frame (where u = 0) is given in terms of the inertial mass alone, as shown in the last expression in the equation. The result can be written more transparently as E 2 = p2 c2 + m2 c4 .

(3.24)

Two applications of this energy-momentum relation are worthy of note. If p = 0, we get the famous statement of mass–energy equivalence E = mc2 .

(3.25)

Relativistic kinematics and the mass–energy equivalence

151

(a) M S

v

γ

(p, ip)

E < E (b)

M

v γ

E > E

S (–p, ip)

Fig. 3.2 In the relativistic Doppler effect, the photon energy E = pc appears to an observer to be (a) E  < E when the observer is receding from the source, and (b) E  > E when the observer is approaching the source.

This equation is probably the best known equation in physics. Equation (3.24) is actually valid also for massless particles such as photons. It then gives E = pc,

(3.26)

thus showing that the four-momentum of a photon is (p, ip). The Lorentz scalar P2 = 0 vanishes for all massless particles. That is, it is a light-front or null Lorentz scalar. 3.3.1

Relativistic Doppler effect

Suppose p(1, i) is the Lorentz momentum of a photon in the rest frame S of its source in a 2D Minkowski spacetime. Its energy E  as seen by an observer in the moving frame M receding from S with speed v [Fig. 3.2(a)] is readily found from the LT Eq. (3.9) to be      γ iβγ p p = . (3.27) −iβγ γ ip ip That is, the photon energy is ( 

E = Eγ(1 − β) = E

1−β . 1+β

(3.28)

This result is called the relativistic Doppler effect, as we shall explain further in Example 3.2.4. If the observer is approaching the same stationary light source in S from the left as shown in Fig. 3.2(b), the backward moving photon has negative momentum −p and the higher observed energy of

152

Relativistic square-root spaces

( E  = Eγ(1 + β) = E

1+β . 1−β

(3.29)

This is called a blueshift, in contrast to the redshift described by Eq. (3.28) for a receding source or observer. 3.3.2

Applications of relativistic kinematics

In the following examples we shall use the natural system of units where c = 1. By thus omitting all c factors, all formulas become shorter and easier to read. Of course, the reader should know how to translate all results back to the traditional system of units when needed. Since the four-momentum contains both three-momentum and energy, kinematical calculations for a system of particles become greatly simplified by using two simple equations: (1) The momentum of the center of mass (c.m.) of a system of n particles is its total momentum: n  Pcm = Pi = (Pcm , iEcm ), where i=1

Pcm =



pi ,

Ecm =



i

Ei ,

(3.30)

i

where Pi = (pi , iEi ) is the four-momentum of ith particle. The coordinate frame where Pcm = 0 is called the c.m. frame. (2) The squared four-momentum P2cm = PTcm Pcm is a Lorentz scalar that has the same value in all Lorentz frames: 2 P2 cm = Pcm ,

P2 cm

−

2 E cm

=

P2cm

or

2 − Ecm .

(3.31)

From this simple equation, many kinematical quantities in the c.m. frame can be obtained readily from those in the laboratory frame, or vice versa, as we shall illustrate in the following examples. Example 3.3.1 Center-of-mass momentum of two particles A particle of mass m1 and three-momentum p1 = plab is incident on another particle of mass m2 at rest in the laboratory. (a) What are the velocity and the (rest) mass of the center-of-mass (c.m.) in the lab frame? (b) What are the momenta and energies of the particles in the center-of-mass frame? (a) In the laboratory, the c.m. behaves like a point particle with mass denoted M = Ecm and velocity U = βcm . Momentum conservation requires that the momentum of the c.m. is just the total momentum. Therefore the Lorentz momentum vector of the c.m. in the lab frame is, by Eq. (3.30),

Relativistic kinematics and the mass–energy equivalence

Pcm = (p1 , i[E 1 + m2 ]) = γcm E cm (βcm , i).

153

(3.32)

This expression can be used to obtain βcm = %

p1 , E 1 + m2

M = Ecm =

E1 + m2 , γcm

(3.33)

where γcm = 1/ 1 − β2cm . An expression similar to Eq. (3.32) can be written for the Lorentz momentum in the c.m. frame if one used βcm = 0 instead. One can verify that M = Ecm is indeed the total energy in the c.m. frame. (b) The c.m. frame is that frame in which the total momentum vanishes: pcm = p1 + p2 = 0. This means that p1 = |p1 | = p2 ≡ p . To answer the second question, we note that the velocity in the c.m. frame of particle 2, the resting particle in the lab, is equal and opposite to that of the c.m. in the laboratory frame. Hence p = m2 β2 γ2 = m2 βcm γcm m2 p1 = . E cm

(3.34)

So p1 = p p1 /p1 , p2 = −p1 . With the particle momenta known, their energies in the c.m. frame are just % %   2 2 (3.35) E1 = m1 + p , E2 = m22 + p2 . For a more direct, but also more laborious and less elegant method, see Problem 3.3.2.  Example 3.3.2 Antiproton creation A proton of mass m and energy E collides with a proton target at rest in the laboratory. Find the minimum lab energy E(min) needed to create an antiproton in the nuclear reaction pp → 3p + p. ¯ Note that a third proton has to be created at the same time because of the experimental law of the conservation of fermion number that requires fermions to be created in fermion–antifermion pairs only. A nucleon is a fermion, a particle of spin 1/2. The initial state has a nucleon number of 2, while the antiproton in the final state has nucleon number –1. This means that its production requires the creation of another nucleon, namely the third proton. There are thus four particles in the final state, with the same mass m and four-momenta P(i), where i = 1, . . . , 4, with 1 denoting the incident proton. The minimum energy in the c.m. frame is realized when all four particles are at rest. 2 (min) = (4m)2 = −P2 in the c.m. frame, where P =  P (i) is also the Then Ecm i cm cm total four-momentum of the system. The Lorentz scalar P2 cm is a frame-independent Lorentz invariant quantity that is conserved during the reaction producing the extra proton–antiproton pair. It can be calculated in the lab frame directly in terms of the properties defined there

154

Relativistic square-root spaces 2 Ecm (min) = 16m2 = −P2 cm

= −P2tot = (E + m)2 − p21 = 2mE + 2m2 .

(3.36)

We have used here the given information E1 = E, p2 = 0 in the lab. Hence E ≡ E(min) = 7m is the minimum lab energy of the incident proton needed for the production.  Example 3.3.3 Decay length of charged muons A negative muon (μ− ) has mass mμ = 105.7 MeV (or MeV/c2 if c  1). It decays into an electron (plus two other particles) via the reaction μ− → e + νμ + ν¯ e with a mean lifetime of τ = 2.20 × 10−6 s in its own rest frame. What is the decay length L of a negative muon of energy E = 1 GeV in the laboratory?  A population of negative muons decays in time as e−t /τ in the proper frame where the muons are at rest. In the laboratory this population decays in time as e−t/tlab , where t = γt and tlab = γτ. If the population is moving with velocity v in space, the population decays as e−x/L , where x = vt = βγt and L = vtlab = βγτ. L is called the mean decay length of the population. In writing these expressions, we have made use of the fact that (x, it) are the Minkowskian spacetime coordinates of a muon moving with speed v = β in the laboratory. The Lorentz factor of a 1-GeV muon is γ = E/m = 9.46 = cosh α, so & βγ = sinh α = cosh2 α − 1 = 9.41. (3.37) Hence L = βγτ = 2.07 × 10−5 s in natural units, or 2.07 × 10−5 c = 6.2 km in normal units.  Example 3.3.4 Relativistic Doppler effect An observer is receding with speed v = β (or v = βc in a normal system of units) along the line of sight from a stationary light source emitting a continuous light wave of wavelength λ and frequency ν = 1/λ (or ν = c/λ in normal units), as shown in Fig. 3.3(a). What is the frequency λ seen by the observer? The spacetime separation between two successive crests 1 and 2 of the light wave along the line of sight is the light-front vector ΔX = X2 − X1 = (−λ, iλ) in frame S of the light source, where x1 > x2 , but t1 < t2 . This spacetime separation seen by the receding observer traveling with velocity v = β > 0 in the S frame is given by the standard LT that in matrix form reads      γ iβγ −λ −λ = . (3.38) −iβγ γ iλ iλ Hence

( λ = γ(1 + β)λ =

1+β λ. 1−β

(3.39)

Relativistic kinematics and the mass–energy equivalence

155

(a)

v

M

2 1

S

λ

λ > λ

(b)

v

M

1 2

S

λ

λ < λ

Fig. 3.3 In the relativistic Doppler effect, the wavelength λ of a light wave appears to an observer to be (a) λ > λ when the observer is receding from the source, and (b) λ < λ when the observer is approaching the source.

The observed frequency is ( 1 ν =  = λ

1−β ν 1+β

(3.40)

is smaller. The effect is called a redshift in optics, especially in connection with the light from stars. The name Doppler effect refers specifically to this change in the observed frequency of the wave. The relativistic Doppler effect was first described in coordinate space to first order in β by W. Voigt in 1887. The effect was named after Christian Doppler, who first pointed out in 1842 that the color of a spectral line from a distant star changes with the star’s motion. The Doppler effect was confirmed observationally by W. Huggins in 1868. The crest-to-crest time t = λ (λ /c in normal units) is just the period of one light oscillation, the time unit used to measure time values in the observer frame. It too increases in a redshift situation, meaning that the time unit has lengthened or dilated. So the clock for the moving observer is running slow, as we would say colloquially. Note that Eq. (3.40) has the same structure as Eq.(3.28) for the change in value of the corresponding photon momentum or energy. This shows that E = hν

(3.41)

holds for any light wave. The proportionality constant turns out to be the Planck’s constant of quantum mechanics. The fact that a light wave can be treated as a particle is a consequence of quantum mechanics.

156

Relativistic square-root spaces

For an observer approaching a stationary light source, the Doppler shift in the observed frequency ν can be obtained from Eq. (3.40) by changing the sign of β, as the sign of β is changed in the LT matrix. The result is then a blueshift. The same result can be obtained without changing the LT matrix by having the observer approach the stationary light source from the left, as shown in Fig. 3.3(b). Then the crest-to-crest light-front vector is ΔX = (λ, iλ), with a positive sign in the spatial separation.

Problems 3.3.1 (Relativistic kinetic energy) In 1D space, the kinetic energy T of mass m increases when an external force F does work on it:  2 T2 − T1 = Fdx. 1

Here F = dp/dt, p = γmu, and u is the instantaneous velocity dx/dt. Evaluate this integral to find the result T = mc2 (γ − 1),

or

E = T + mc2 = γmc2 .

3.3.2 (Particle momentum in the c.m. frame) Derive Eq. (3.34) for the particle momentum p = |p1 | = |p2 | = m2 p1 /Ecm in the c.m. frame of a two-particle system by using the fact that the squared total momentum P2 = (P1 + P2 )2 is a frame-independent Lorentz invariant quantity that has the same value in both the lab frame and the c.m. frame. Hints: (a) In the lab frame, the total momentum is Ptot = (p1 , i[E 1 + m2 ]). In the c.m. frame, Pcm = (0, iEcm ), where Ecm = E1 + E2 , and the particle ener% gies are Ei = m2i + p2 i . (b) You need to square certain expressions and extract a quadratic equation for p after canceling p4 terms. 3.3.3 (Two-body decay in the c.m. frame) A particle of mass M decays at rest into two particles of masses m1 and m2 . Verify that in the c.m. frame p2 = |p1 |2 = |p2 |2  1  2 2 2 2 2 2 = − m − m ) − 4m m (M ; 1 2 1 2 4M 2 1 (M 2 + m21 − m22 ). E1 = 2M Note: This problem is very similar to the previous problem.

(3.42)

Relativistic kinematics and the mass–energy equivalence

157

3.3.4 (Electron–positron annihilation) (a) Show that the annihilation of an electron and a positron can produce a single massive particle (say, X), but not a single photon. (b) A positron beam of energy E can be made to annihilate against electrons by hitting electrons at rest in a fixed-target machine or by hitting electrons moving in the opposite direction with the same energy E in an electron– positron collider (colliding-beam accelerator). Show that the minimum energy Emin of a positron beam needed to produce neutral particles X of mass M  m (the electron mass) is much greater in a fixed-target machine than in a collider. 3.3.5 (Pion production) A proton of mass m = 940 MeV (in c = 1 units) and lab energy E collides with a proton target at rest in the laboratory to produce a neutral pion π0 of mass 135 MeV by the production reaction p + p → p + p + π0 . (a) Show that the minimum proton laboratory energy needed for the reaction is Emin = m + 2mπ +

m2π . 2m

Show that the minimum proton kinetic energy needed is T min = Emin = 280 MeV. (b) If E > E min , show that the minimum and maximum three momenta of the π0 in the c.m. frame are pπ min = 0, √ √ pπ max = ( s − 2m − mπ )( s − 2m + mπ ), s = 2m(m + E). 3.3.6 (Doppler redshift) How fast must you drive away from a green traffic light for it to appear red when you look back at it? Do you see the same color if you look at the rear-view mirror instead? Explain. The wavelength is λg = 530 nm for green light and λr = 660 nm for red light. 3.3.7 (Gravitational redshift) When a photon of energy E = hν is emitted, the inertial mass of the emitter decreases by m = E/c2 according to Einstein’s mass–energy equivalence relation. There is an older equivalence principle that states that the gravitational mass mg of an object is equal to its inertial mass m. Extending this principle to photons has the consequence that a photon also carries an effective or pseudo-gravitational mass μγ = Eγ /c2 . It therefore also interacts with the gravitational field. A photon traveling upwards away from the center of the Earth will lose energy to the gravitational attraction of the Earth in the same way as a stone thrown skyward. The photon energy therefore decreases. If h = E/ν is a universal constant independent of the gravitational field, then the frequency

158

Relativistic square-root spaces

of the photon will also decrease proportionally. The color of a photon rising skyward is therefore redshifted by the conservation of energy in a gravitational field. This is the gravitational redshift. (a) A γ-ray is an electromagnetic radiation like light but with higher energy or frequency, E = hν. A γ-ray is emitted by a source at the top of a tower of height Δz > 0. It is detected by an observer at the base of the tower. Show that the detected energy of the γ-ray is E  = E + ΔE, with ΔE Δν gΔz = = 2 E ν c ( 1+β = − 1 ≈ β, 1−β assuming that β  1. Here gΔz is the gravitational energy the photon has absorbed by falling to the base of the tower. The photon frequency is thus blueshifted. (b) In the 1960 experiment of Pound and Rebka, Δz = 74 ft = 22.6 m. The measurement was done twice, the second time after interchanging the positions of source and detector. The total two-way fractional energy change is thus 2β. The measured result is (5.13 ± 0.51) × 10−15 . Calculate the result predicted by Einstein to decide if his prediction has been proved or disproved. The physical constants are g = 9.807 m s−2 , c = 2.998 × 108 m s−1 . (c) Sunlight received here on Earth has actually been redshifted by the solar gravitational field. The Earth is so far away from the Sun that the solar gravitational field strength here on Earth is practically zero. Show that the fractional solar redshift of sunlight is Δν GM  = −2.12 × 10−6 . =− ν R The solar mass and radius are M = 1.97 × 1030 kg, and R = 6.95 × 108 m, respectively. The gravitational constant is G = 6.67 × 10−11 m3 kg−1 s−2 . Note: The period τ = 1/ν of the electromagnetic radiation emitted by the same nuclear or atomic transition at a lower elevation z1 as seen by an observer at a higher elevation z2 = z1 + Δz is inversely proportional to the detected frequency. Hence the higher observer detects a fractional change Δτ Δν gΔz =− = 2 τ ν c in the period of the radiation emitted by the lower source. If the lower clock is calibrated by the period of the electromagnetic radiation emitted there,

Quaternions

159

the time between two ticks of the lower clock is larger that that of the higher clock. So the lower clock runs more slowly. This effect is called the gravitation time dilation.

3.4

Quaternions

The √ real number system is incomplete because it does not contain the solutions ±i = ± −1 of the algebraic equation x2 + 1 = 0. Imaginary numbers such as ib with real bs are specified in units of i. Complex numbers have the form a + ib, where a, b are real. Other kinds of number systems have also been used. For example, the three 2 × 2 Pauli matrices σj , j = x, y, z, or 1, 2, 3, introduced in Section 2.2 are also square roots of the 2 × 2 identity matrix I = diag(1, 1), like I and −I themselves. They too can be used as units for advanced number systems. What is intriguing about the Pauli spin matrices is that there are just three of them, the same as the number of spatial dimensions. As a result, we can construct a Pauli spin vector σ = σ x e x + σy ey + σz ez =

3 

σi ei ,

(3.43)

i=1

where the x, y, z components are also denoted 1, 2, 3 to match  up more easily with the summation notation. Its scalar product with the vector a = i ai ei gives a scalar expression σ·a=

3 

σi ai

(3.44)

i=1

that forms a 3D number system in units of the Pauli matrices. Together with the diagonal matrix a0 I where a0 can play the role of a time variable, one has a 4D real number system (a0 I, σ · a) expressed as 2 × 2 Hermitian matrices. A rather similar number system called quaternions was already discovered in the mid-nineteenth century. Quaternions were discovered in 1843 by William Hamilton after a long search for a multidimensional generalization of the complex numbers t + ix. Recall that these complex numbers themselves generalize the real number system t in the 1D space of scalars t. Hamilton found that the linear combination or quaternion (from quaternio, Latin for a quartet) Q = t + xi + yj + zk

(3.45)

containing four independent components t, x, y, z satisfies a closed algebra of the usual kind except that the unit basis quaternions i, j and k do not commute. They are defined to satisfy instead the noncommutative multiplication algebra

160

Relativistic square-root spaces

ij = k = −ji, jk = i = −kj, ki = j = −ik :

(3.46)

i2 = j2 = k2 = −1 = ijk.

(3.47)

The algebra is closed in the sense that products of quaternions remain quaternions. Quaternions can thus be considered both an extension and a complexification of the complex numbers: Q ≡ (t, x, y, z) = (t + xi) + (y + zi)j ≡ (t + xi, y + zi).

(3.48)

Quaternions form a division algebra that contains the multiplicative inverses Q−1 except for the zero quaternion Q = 0, as we shall see shortly. These inverses are defined by the multiplication rule Q−1 Q = QQ−1 = 1.

(3.49)

Q−1 = Q∗ (QQ∗ )−1

(3.50)

They have the formal structure

in common with complex numbers. Here the quaternion complex conjugate is Q∗ = t + xi∗ + yj∗ + zk∗ = t − xi − yj − zk,

(3.51)

with all three unit basis quaternions i, j and k changing sign under complex conjugation. The squared absolute value is also similar to that for complex numbers |Q|2 = Q2 ≡ Q∗ Q = t2 + x2 + y2 + z2 = QQ∗ .

(3.52)

Just like complex numbers, Q = 0 only when all its components vanish. Quaternions differ from complex numbers in one important way. The quaternion complex conjugate Q∗ is an algebraic function of Q itself 1 Q∗ = − (Q + iQi + jQj + kQk), 2

(3.53)

as one can easily verify by writing the expression on the right in terms of the quaternion components t, x, y, z. In contrast, the corresponding expression for the complex variable C = t + xi is C + iCi = 0 always. Furthermore (t + ix)∗ = t − ix is independent of t + ix itself, and cannot be obtained from the latter by multiplications alone. This difference may appear minor, but it is not. Functions of a quaternion variable is severely constrained

Quaternions

161

by Eq. (3.53). They do not have the rich structure seen in functions of a complex variable. Nature seems to prefer the richness of complex numbers over the relative simplicity of quaternions. According to the Frobenius theorem, the only finite dimensional division algebras over the real number system are the real numbers, the complex numbers, and the quaternions. In spite of this favorable property, quaternions are not as useful in the description of the physical world as many mathematicians and physicists expected. The basic problem with quaternions is that their squared norms Q2 are not Lorentz scalars. 3.4.1

Rotation quaternions

The popularity of quaternions declined drastically after its most practical parts were split off and given new notation as vector analysis in the 1880s. In the last few decades they have enjoyed a partial revival due to their ability to describe spatial rotations of rigid bodies more economically than matrices containing Euler angles. Quaternions are now the preferred tools for describing the orientation of rigid bodies for most software engineers working in aircraft and spacecraft control systems, in computer vision and robotics, and in computer graphics and animation. It is therefore worthwhile to give some specific examples of quaternion functions and calculations. Quaternions can be represented by matrices, the simplest matrix form being 2 × 2. A convenient choice for the unit basis quaternions are the Pauli spin matrices i = iσTx ,

j = iσTy ,

k = iσTz .

(3.54)

Since the Pauli matrices are Hermitian, and both σ x and σz are real and symmetric, the transposition changes the sign of the purely imaginary and antisymmetric matrix σy . We shall keep the transposition symbol T on the Pauli spin vector to distinguish it from the untransposed spin vector. The commutation relation (2.144a)  σi σj = δij + i εijk σk (3.55) k

then gives a similar commutation relation for the transposed spin matrices  σTi σTj = δij − i εijk σTk .

(3.56)

k

One can verify explicitly that the 2 × 2 matrix representation (3.54) reproduces the stated noncommutative multiplication rules for quaternions. Note that the second term of Eq. (3.56) carries a negative sign. This means that the permutation ordering ijk in Eq. (3.56) is a left-handed one. A quaternion in any representation can be denoted by its quartet of real numbers Q = (q0 , q1 , q2 , q3 ) = (q0 , q). One can call q0 the real scalar part of Q and q its

162

Relativistic square-root spaces

imaginary vector part. The quaternion Q = (q0 , 0) is a scalar quaternion, while Q = (0, q) is a vector quaternion. The 2 × 2 Pauli matrix representation of a quaternion is Q = q0 + iσT · q.

(3.57)

Some authors refer to a vector quaternion simply as a vector. It is that, but it also contains the Pauli scalar σT · q. The commutation relation (3.56) involved in quaternion multiplications can be expressed in a more transparent and useful form as the Pauli scalar identity (σT · a)(σT · b) = a · b − iσT · (a × b).

(3.58)

The quaternion product of P = p0 + i σT · p and Q is then easily evaluated by longhand, PQ = p0 q0 − p · q + iσT · (p0 q + q0 p + p × q).

(3.59)

In the shorthand notation Q = (q0 , q), the result is PQ = (p0 q0 − p · q, p0 q + q0 p + p × q)  QP.

(3.60)

The noncommutativity resides in the last term containing the vector product q × p. For any quaternion, Q = q0 + iσT · q = q0 + iqσTq ,

where

σTq = σT · eq , eq =

q = (c1 , c2 , c3 ), q

ci = eq · ei .

(3.61)

Here eq is the unit vector along q, ci are its direction cosines, and (σTq )2 = 1. As a result, Q has the Euler representation Q = Qeiψq σq = Q(cos ψq + iσTq sin ψq ), % q Q = q20 + q2 , sin ψq = . Q T

where (3.62)

The quaternion inverse is then just the matrix inverse Q−1 =

Q† , Q2

where use has been made of the fact that σT is Hermitian.

(3.63)

Quaternions

163

Quaternion multiplication is just the matrix analog of complex number multiplication: PQ = PQeiψp σp eiψq σq T

T

 QP = QPeiψq σq eiψ p σ p . T

T

(3.64)

The quaternion product remains noncommutative in general because σTp and σTq do not in general commute. The absolute value Q of Q is called its tensor (“stretcher” in Latin). Its exponent iψq σTq is called its versor (“turner” in Latin). These terms describe the multiplicative effect of Q on another quaternion. Unit quaternions are those with Q = 1. They do not stretch another quaternion on multiplication, only turn or rotate them. As matrices, unit quaternions are unitary matrices with unit determinant. They form an SU(2) matrix group. They can be used as rotation operators, but not the usual kind associated with rotations of actual vectors in ordinary space like those described in Section 2.2. Indeed, the matrix representation of quaternions has certain formal advantages over the abstract form of quaternions by displaying explicitly why quaternions differ in significant ways from ordinary vectors in space. First, quaternion multiplications do not commute because they are matrix multiplications, as we have already noted. Second, each quaternion, being a matrix, has two indices, unlike vectors in space that have only one index each specifying its spatial components. So a rotational transformation of a quaternion must transform both of its indices. That is, the operation is a “two-sided” operation with two operators, one on each side of the quaternion. Each unit quaternion Ea (ψ/2) = eiψσa /2 = c + iσTa s, T

c = cos (ψ/2),

s = sin (ψ/2).

(3.65)

is a rotor that takes care of half of the rotation angle ψ about the axis ea . Hence the rotation of a vector quaternion X = (0, r) = iσT · r takes the form X = Ea (ψ/2)XE†a (ψ/2).

(3.66)

It is always possible to choose ea = e1 when there is only one rotation. We then T find by using the quaternion Euler formula E1 (ψ/2) = eiψσ1 /2 and after a straightforward calculation that X = ix + jy + kz = E1 (ψ/2)XE†1 (ψ/2) = ix + j(y cos ψ − z sin ψ) + k(y sin ψ + z cos ψ).

(3.67)

164

Relativistic square-root spaces

The result agrees with that obtained for the rotation of an object about the 1-axis in ordinary 3D space. It differs from the rotation of the coordinate axes treated in Section 2.2 by the sign change ψ → −ψ. For successive rotations, it will be necessary to use the general expression for a rotation about an arbitrary axis ea = (α, β, γ) defined by its direction cosines α, β, γ. The quaternion calculations involved are more easily done using the abstract ijk basis for the rotation quaternion (3.65) Ea (ψ/2) = c + s(iα + j β + kγ). The result can be written as a standard 3 × 3 rotation matrix ⎛ ⎞ ⎛ ⎞ ⎜ x⎟ ⎜⎜⎜ x ⎟⎟⎟ ⎜⎜⎜ y ⎟⎟⎟ = R(ψ) ⎜⎜⎜⎜⎜ y ⎟⎟⎟⎟⎟ , where ⎝ ⎠ ⎝ ⎠ z z ⎛ 2 ⎜⎜⎜ α (1 − C) + C αβ(1 − C) − γS αγ(1 − C) + βS R(ψ) = ⎜⎜⎜⎜⎝ βα(1 − C) + γS β2 (1 − C) + C βγ(1 − C) − αS γα(1 − C) − βS γβ(1 − C) + αS γ2 (1 − C) + C

(3.68)

⎞ ⎟⎟⎟ ⎟⎟⎟ . ⎟⎠ (3.69)

Here C = cos ψ, S = sin ψ. Problem 3.4.4 gives some hints on how this expression can be derived. The most general rotation of a general quaternion Q is Q = Ea (ψ)QE†b (χ).

(3.70)

When the rotation angles ψ, χ are both imaginary, the transformation turns out to be a Lorentz transformation, but we shall examine the situation from a different perspective in the next section.

Problems 3.4.1 Verify Eq. (3.53) for Q∗ . 3.4.2 Verify Eqs. (3.46) and (3.47) for the Pauli matrix representation (3.54) of quaternions. 3.4.3 Verify Eq. (3.67) by using the Euler formula E1 (ψ/2) = c + is, where c = cos (ψ/2), s = sin (ψ/2). Hint: The calculation is a little easier if done with the abstract ijk form of the rotation quaternions. 3.4.4 Verify the 3 × 3 quaternion rotation matrix R(ψ) of Eq. (3.69) in two steps: (a) Calculate just the first column of R(ψ). Hint: The calculation is a little easier if done using the abstract ijk form of the rotation quaternion. (b) Write down the second and third columns of R(ψ) by cyclic permutations of the first column.

Dirac equation, spinors and matrices

3.5

165

Dirac equation, spinors and matrices

Many functions describing physical systems are single-value functions in 3D space that resume their original value after one full turn around any axis. They embody the commonsense expectation that an observer will see the same world after a full turn on the merry-go-round. Nature is more subtle, however. We have seen in Section 2.11 how the Pauli matrices σi are the mathematical generators of rotations of the 2D spin vectors or spinors describing the spin states of spin 1/2 particles like electrons. In particular, we have seen how after one full rotation in 3D space, these 2D spinors change sign, unlike the world seen from a merry-go-round. It will take two full turns to restore them to their original values. This two-turn periodicity is a quintessential characteristic of spinors. ´ Spinors were discovered in 1913 by Elie Cartan in his study of null vectors  Z = i xi ei in a complexified 3D space whose coordinates x1 , x2 , x3 are complex numbers. Null vectors are vectors with zero scalar products (Z · Z = 0 without any complex conjugation). He found that these null vectors span a 2D surface describable by two complex coordinates that are square roots of linear combinations of the original real coordinates xi . It is the square root functional dependence that gives rise to the two-turn spinorial periodicity. The use of spinors in physics began in a totally different way with Pauli’s discovery of 2 × 2 spin matrices. Because σ2i = 1, these Pauli matrices are the simplest nontrivial square roots of 1 after ±1. Hence Pauli matrices provide access to a certain 3D square-root space. In the preceding section we have already shown that the old idea of quaternions is actually based on the access to the same square-root space when Pauli matrices are used to represent unit quaternions. The trouble with quaternions is that their squared absolute values are not Lorentz-invariant. In this section we shall consider square-root spaces with well-defined Lorentz properties. Such square-root spaces are called spin spaces. As before, the 3D square-root space can be reached by using the Pauli scalar identity derivable from Eq. (3.55): (σ · a)(σ · b) = a · b + iσ · (a × b),

(3.71)

where a and b are 3D vectors. One can then write down by inspection the Pauli expressions for the Minkowski spacetime 4-scalars (in units of c = 1): τ2 = t2 − r2 = t2 − (σ · r)(σ · r),

(3.72)

m2 = E 2 − p2 = E 2 − (σ · p)(σ · p).

(3.73)

In this way, one can establish an intimate connection between the squared quantities and their square roots in square-root space, as we shall see presently. Before doing that, let us first note that in quantum mechanics, the energymomentum relation (3.73) is first rewritten as an operator expression for a squaredmass operator m ˆ 2 = Hˆ 2 − pˆ 2

(3.74)

166

Relativistic square-root spaces

by the replacement or quantization m → m, ˆ d d E → Hˆ = i = i , dt dt 1 p → pˆ = ∇, i

(3.75)

where quantum operators wear hats and  = 1. The energy operator is traditionally ˆ It is named after Hamilton, the inventor of called the Hamiltonian and denoted H. quaternions, for his work on classical mechanics. The differential eigenvalue equation (m ˆ 2 − m2 )ΨKG (r, t) = 0

(3.76)

then gives the Klein–Gordon quantum wave equation for the wave function 1 ΨKG (r, t) = √ ei(p · r−E± t) , V % where E± = ± m2 + p2 ,    Ψ 2 d 3 r = 1 d3 r = 1 KG V

(3.77)

for a free particle of mass m and spin 0 in a spatial volume V. To access the square-root space directly, Dirac in 1928 looked for and found the linearized Dirac Hamiltonian for a spin 1/2 particle in free space Hˆ D = α · pˆ + βm,

(3.78)

where the four quantities β, αi , i = 1, 2, 3 must be chosen to satisfy Eq. (3.74) in the ˆ 2 + pˆ 2 . That is, they must have unit squares and anticommute in pairs: form Hˆ D2 = m α2i = β2 = 1, {αi , αj } = αi αj + αj αi = δij , αi β + βαi = 0.

(3.79)

Since these objects anticommute, they cannot be numbers, but they can be matrices. In common with other quantum mechanical operators describing physical properties, they are Hermitian matrices with real eigenvalues. With unit squares, they are just the anticommuting matrix square roots of the unit matrix I. However, the matrix dimension must be greater than 2, because there are only three independent Hermitian matrices of dimension 2 that are the anticommuting square roots of I, namely the Pauli matrices. So the eigenvalues of ±1 must appear more than once. The minimal number of their appearances is therefore two. So these objects are 4 × 4 Hermitian matrices. The resulting Dirac equation

Dirac equation, spinors and matrices

(Hˆ D − E)ψD (r, t) = 0

167

(3.80)

defines Dirac wave functions ψD of a relativistic spin 1/2 particle that are 4 × 1 column vectors called Dirac spinors. Two of the matrix dimensions describe the two spin up/down degrees of freedom for a spin 1/2 particle, as dictated by the appearance of 2 × 2 Pauli matrices in Eq. (3.73). The remaining two degrees of freedom come √ from the fact that the energy eigenvalues appear with ± signs in the form E = ± E 2 of positive and negative energy states. In deriving Dirac’s matrices, it is instructive to show directly how this doubling of the matrix dimension comes from a direct access to the square root space. We begin by factoring the quadratic Lorentz invariant ˆ Hˆ + σ · p). ˆ m2 = (Hˆ − σ · p)(

(3.81)

The resulting differential eigenvalue equation ˆ Hˆ + σ · p)φ ˆ L (r, t) = m2 φL (r, t) (Hˆ − σ · p)(

(3.82)

defines the two-component spin wave functions       φ1 1 0 φL = + φ2 = φ1 0 1 φ2 = φ1 | ↑ + φ2 | ↓.

(3.83)

that are 2 × 1 column vectors matching the matrix dimension of the Pauli matrices. The vector (3.83) providing a simple match up for the Pauli matrices as generators of the SU(2) group is said to be the fundamental representation of the group. The three generators σi themselves are said to make up an adjoint representation of the group. To continue, let 1 ˆ ˆ L, (H + σ · p)φ m 1 ˆ R φL = (Hˆ − σ · p)φ m

φR = then

(3.84)

define two sets of 2 × 1 Pauli spinors. The subscripts L, R denote a specific choice of these spin states to be made in the next section. The results in the rest of this section are valid for any choice. The sum and difference of these two Pauli spinors are Hˆ σ · pˆ φ+ − φ− , m m Hˆ σ · pˆ φ− = φR − φL = φ+ − φ− . m m φ+ = φR + φL =

(3.85)

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Relativistic square-root spaces

This can be written compactly as a single 4 × 4 matrix equation      φ+ −Hˆ σ · pˆ φ+ = −m , ˆ −σ · pˆ H φ− φ−

(3.86)

ˆ ±σ · pˆ are themselves 2 × 2 matrices embedded in where the matrix elements ±H, another 2 × 2 structure shown explicitly in the equation. The outer matrix skeleton can be written compactly in terms of another set of Pauli matrices ρi having the same numerical value as the σi matrices that act instead to connect the physical space to its square-root partner, in the sense of Eq. (3.86). We shall refer to this square-root space structure as the ρ matrix structure. Before getting more deeply into the ρ matrix structure, let us first point out that the Dirac equation should look the same in every Lorentz frame. This means that the operators it contains must be Lorentz scalars. The mass m is already a Lorentz scalar, and the matrix operator on the left must also be expressible as a Lorentz scalar product. To make the rest of this discussion more useful, we now change our Minkowski notation to a more modern Euclidean notation in general use in relativistic quantum mechanics and particle physics. In this Euclidean notation, the Minkowski vectors such as the spacetime coordinate vector (r, it) can be written in two different but related forms, with an upper or contravariant index and a lower or covariant index, respectively: xμ = (x0 , r),

xμ = (x0 , −r);

xμ = gμν xν ,

xμ = gμν xν ,

gμν = gμν ,

g = diag(1, −1, −1, −1).

(3.87)

A Lorentz scalar product between two Lorentz vectors (a0 , a) and (b0 , b) is then a0 b0 − a · b = aμ bμ = aμ bμ = aμ gμν bμ = aμ gμν bμ ,

(3.88)

where each matrix multiplication involves an upper and a lower index taken in any order, using the Einstein notation that repeated indices in the same term implies summation. All the vector components are now real or Euclidean. The distinction between spatial and time components is made only in the diagonal metric tensor or matrix g. The mathematical origin of this tensor index notation will be given in Section 3.12. The metric g is chosen so that the squared energy-momentum Lorentz scalar E 2 − p2 is nonnegative and has the Lorentz invariant value of m2 in any Lorentz frame. This Euclidean notation is easier to use in complicated relativistic kinematical calculations than the equivalent Minkowskian expression involving the imaginary unit i. In general relativity where the emphasis is in the spatial components, the metric is chosen to be the negative of that used here.

Dirac equation, spinors and matrices

169

Returning now to Eq. (3.86), we note that the Lorentz scalar expression on the left can be written in the Euclidean notation as " # γμ pˆ μ − m ψ = # " (3.89) iγμ ∂μ − m ψ = 0. Here   ∂ ∂ ∂μ = μ = ,∇ , ∂x ∂t

 ∂ ∂ ∂ , ∇= , , ∂x ∂y ∂z 

(3.90)

and ψ = (φ+ , φ− )T is the four-component Dirac spinor. With γμ = (γ0 , γ) :

ˆ γμ pˆ μ = γ0 Hˆ − γ · p,

the γ matrices can be read off directly from Eq. (3.86):  γ = 0

 γ = k

I 0 0 −I

 = ρ3 I,

 0 1 σk = iρ2 σk = −γk , −1 0

k = 1, 2, 3,

(3.91)

where the Pauli matrix ρi refers to the skeletal 2 × 2 matrix structure shown explicitly in Eq. (3.86). We do not use ρi or σk. As I and the σi are themselves 2 × 2 matrices, γμ are 4 × 4 matrices called Dirac matrices in the Euclidean Dirac–Pauli (DP) representation or basis. These matrices can be used to show that the 4 × 4 matrices in Dirac’s linearized Hamiltonian (3.78) are (Problem 3.5.1) α = ρ1 σ,

β = ρ3 .

(3.92)

We shall presently describe two other choices of these Euclidean gamma matrices as used (a) by Weyl, and (b) by Majorana. The operator pˆ μ or its eigenvalue pμ for a free particle is a (covariant) Lorentz vector. The Lorentz invariance of the Dirac equation is ensured if γμ pˆ μ is a Lorentz scalar, the same in all frames. This requires that γμ transforms as a (contravariant) Lorentz vector, as explained in detail in section 3.12. The Dirac equation of a massive particle is simplest in the rest frame where p = 0. Eq. (3.86) then shows that the spinors φ± are decoupled from each other, with φ+ having the positive energy of E + = m and φ− having the negative energy E − = −m. Hence ρ3 can be interpreted as the energy signature matrix of a massive particle at rest.

170

3.5.1

Relativistic square-root spaces

More about Dirac matrices

The gamma matrices are related to four of the sixteen 4 × 4 matrices E made up of a direct product of two independent sets of 2 × 2 Pauli matrices: E = ρ ⊗ σ,

or Eμν = ρμ σν ,

μ, ν = 0, 1, 2, 3;

ρ = (ρ0 = I, ρ1 , ρ2 , ρ3 ) = (I, ρ), σ = (σ0 = I, σ1 , σ2 , σ3 ) = (I, σ).

(3.93)

Of the sixteen Eμν matrices, E44 = I has a trace of 4. The remaining fifteen matrices are traceless generators of the direct product SU(2) ⊗ SU(2) group. The gamma matrices satisfy an interesting algebra. With γ = iρ2 σ, γ0 = ρ3 I, the spatial components satisfy the Pauli spin relations γi γj = −ρ22 σi σj = −δij − iεijk σk , γ0 γi = ρ1 σi = −γi γ0 .

(3.94)

Hence they satisfy the anticommutation relations {γμ , γν } ≡ γμ γν + γν γμ = 2gμν

(3.95)

that include those for the Pauli spin matrices. These anticommutators define an algebra called a Clifford algebra. These sixteen Eμν matrices can be separated into five sets with different spacetime properties. Besides the single I that is a spacetime scalar and the quartet γμ that is a spacetime vector, we have i σμν = {γμ , γν } = iγμ γν , 2

μ  ν;

γ5 = iγ0 γ1 γ2 γ3 = ρ1 = γ5 ; γμ γ 5 ,

(3.96)

forming sets of six, one and four elements, respectively. Table 3.2 shows explicitly how they are related to the group elements Eμν = ρμ σν of SU(2) ⊗ SU(2) which are Hermitian and have unit squares. Seven of the ΓA s have an extra factor of i. So these are antiHermitian and have squares of –1. This slightly awkward situation has no significance; it is a small price to pay for the convenience of the Euclidean notation. The tensor set of six refers to the antisymmetric bivectors σμν . (Clifford algebra is the algebra of multivectors or direct products of vectors including these antisymmetric bivectors. Called wedge products, antisymmetric bivectors generalize the antisymmetric vector products in space to the case where one of the coordinates is timelike.) The remaining five matrices are conveniently expressed in terms of a γ5 matrix. It has a unit square and anticommutes with γμ . As ρ1 , it is off-diagonal in the φ± states.

Dirac equation, spinors and matrices

171

Table 3.2 Five sets of Euclidean gamma matrices and their bilinear products of different spacetime properties expressed in terms of the sixteen group elements of S U(2) ⊗ S U(2).

Γm

N

Space

Time

I γμ σμν γμ γ5 γ5

1 4 6 4 1

iρ σ  2 i k εijk σk ρ 3 σi

ρ3 σ0i = iρ1 σi iρ2

ψ¯ Γm ψ Scalar Vector Tensor Pseudovector Pseudoscalar

¯ m ψ is called a density Note: μ, ν are spacetime labels, i, j are space labels, and γ5 = ρ1 . ψΓ ¯ μψ a of the type shown in the table. ψ¯ = ψ† γ0 contains an additional γ0 needed to make ψγ conserved vector current (Problem 3.5.3).

The sixteen matrices E μν are closed under matrix multiplication. This property simplifies the study of the symmetries of the Dirac equation.

Problems 3.5.1 Show that the Klein–Gordon wave function (3.77) solves the Klein–Gordon wave equation (3.76). 3.5.2 Verify Eq. (3.92) for the Dirac α and β matrices of Eq. (3.78) expressed in terms of the ρ and σ matrices. 3.5.3 (Conserved 4-current) A Dirac 4-current density μ ¯ jμ (x) = ψ(x)γ ψ(x)

can be defined in terms of the Dirac spinor ψ(x) and an adjoint or conjugate ¯ spinor ψ(x) = ψ† (x)X, where X is an unknown 4 × 4 matrix. Show that jμ (x) ¯ satisfies the adjoint equation is conserved, meaning ∂μ jμ = 0, if ψ(x) ¯ ¯ ∂μ ψ(x) = imψ(x). Show that this adjoint equation is satisfied if X = γ0 . 3.5.4 (Free-particle spinors) The Dirac equation for a massive spin 1/2 particle in free space has plane-wave solutions of definite momentum p and energy E of the form: 1 ψr (r, t) = √ ur (p)ei(p · r−E± t) , V

r = 1 − 4,

(3.97)

where ur (p) is a 4 × 1 spinor called a free-particle spinor. (a) Show that there are four linearly independent free-particle spinors, and that in the DP basis they can be chosen to be

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Relativistic square-root spaces

⎛ ⎞ ⎜⎜⎜ 1 ⎟⎟⎟ ⎜⎜⎜ 0 ⎟⎟⎟ u1 (p) = N ⎜⎜⎜⎜ p3 ⎟⎟⎟⎟ , ⎜⎜⎝ E+m ⎟⎟⎠ p+

⎛ ⎞ ⎜⎜⎜ 0 ⎟⎟⎟ ⎜⎜⎜ 1 ⎟⎟⎟ u2 (p) = N ⎜⎜⎜⎜⎜ p− ⎟⎟⎟⎟⎟ ; ⎜⎜⎝ E+m ⎟⎟⎠ −p3

E+m

E+m

⎛ −p3 ⎞ ⎜⎜⎜ |E|+m ⎟⎟⎟ ⎜⎜⎜ −p ⎟⎟⎟ ⎜ + ⎟ u3 (p) = N ⎜⎜⎜⎜⎜ |E|+m ⎟⎟⎟⎟⎟ , ⎜⎜⎜ 1 ⎟⎟⎟ ⎝ ⎠ 0

⎛ −p− ⎞ ⎜⎜⎜ |E|+m ⎟⎟⎟ ⎜⎜⎜ p3 ⎟⎟⎟ ⎟ ⎜ (3.98) u4 (p) = N ⎜⎜⎜⎜ |E|+m ⎟⎟⎟⎟ . ⎜⎜⎜ 0 ⎟⎟⎟ ⎠ ⎝ 1 & 2 2 Here p± = p1 ± ip2 , u1,2 have positive & energy E+ = m + p = E, and u3,4 have negative energy E− = − m2 + p2 = −|E|. (b) Show that these free-particle spinors are orthogonal: ur (p)† u s (p) = 0,

r  s.

(c) Show that the normalization constant should be N =

√

(|E| + m)/2m if

ur (p)† ur (p) = |E|/m. Note: The DP basis of the γ matrices used here is particularly convenient for slow particles with p/m  1. In fact, when p/m = 0, the four spinors in this problem are just the spin up/down states of positive/negative energies. 3.5.5∗ Show that the m = 1–16 matrices E μν = ρμ σν , μ = 0–3, ν = 0–3, of SU(2) ⊗ SU(2) satisfy the following properties: (a) Em En = I, only if m = n; † −1 ; (b) Em = Em = Em (c) det Em =1; 4, if μν = 00 ; (d) Tr E m = 0, otherwise  (e) n an E n = 0 only if all an = 0. Note: Property (e) states that the sixteen matrices are linearly independent of one  another. This property is obtained by first showing that 4am = Tr (Em n an En ).

3.6

Symmetries of the Dirac equation∗

The Dirac equation has symmetry S if its form is unchanged under a similarity transformation S " # 0 = S γμ pˆ μ − m S −1 S ψ   = (γμ ) ( pˆ μ ) − m ψ , where

(γμ ) = S γμ S −1 , ψ = S ψ.

( pˆ μ ) = S pˆ μ S −1 , (3.99)

Symmetries of the Dirac equation

173

This means that the transformation operator S is expressible at least partially in terms of one or more of the sixteen ΓA group elements. If the transformation does not change any known physical attribute of the system, but (γμ )  γμ , it is said to change the mathematical basis of the gamma matrices to another basis. If the transformation changes an important physical attribute, the transformed Dirac equation is said to violate such a physical symmetry. Of the physical symmetries, one distinguishes between spacetime symmetries involving properties of spacetime variables such as t, r, E and p, and internal symmetries involving intrinsic properties not expressible in terms of spacetime variables. 3.6.1

Parity

There are only two spacetime symmetries. One is parity or space inversion S = P under which r → r = P r P−1 = −r, p → p = P p P−1 = −p,

(3.100)

while t and E remain unchanged. The Dirac equation transforms under parity as   0 = P γμ (P−1 P) pˆ μ P−1 − m Pψ(r, t)   = (γμ ) ( pˆ μ ) − m ψ (r , t), where (γμ ) = Pγμ P−1 ,

( pˆ μ ) = P pˆ μ P−1 ,

ψ (r , t) = Pψ(r, t) = Pψ(−r, t).

(3.101)

Here P is a 4 × 4 matrix containing that part of P that operates on the 4 × 4 γs. The Dirac equation is invariant under parity if the transformed equation is the same equation. This happens if (γμ ) ( pˆ μ ) = (γμ )( pˆ μ ). Since the spatial part pˆ  = −pˆ has changed sign, this sign change has to be canceled by a compensating sign change in γ = −γ. Thus P commutes with γ0 but anticommutes with the spatial γk matrices. So P ∝ γ0 = ρ3 , shorthand for ρ3 ⊗ I. The 4 × 4 matrix ρ3 has eigenvalues 1, 1, −1, −1 in the Dirac basis states |+ ↑ = (1, 0, 0, 0)T , |+ ↓ = (0, 1, 0, 0)T , |− ↑, |− ↓, where ± refer to the two distinct eigenvalues ±1 of ρ3 . The normalization of these basis states is unchanged under parity if P = ηP γ0 , where ηP = eiαP is a global phase factor containing a real phase αP that is independent of the spacetime variable xμ . The phase αP cannot be determined experimentally; so ηP can be taken to be 1. Then the 4 × 4 matrix P = γ0

(3.102)

is not only unitary but also orthogonal. Thus the energy signature matrix ρ3 for a massive particle at rest serves double duty as the parity operator for the matrix part of Dirac spinors. A positive (negative) energy eigenstate, of energy m (−m), is simultaneously a state of even (odd) parity.

174

Relativistic square-root spaces

This shows the Dirac equation admits the concept of intrinsic parity that is welldefined even for a particle of momentum p = 0 and for a wave function (3.97) that is independent of the position r. We should add that the sign of the energy considered here is physically well defined as it is determined by the mass m that cannot be chosen arbitrarily. Our energies are therefore different from the relative energies of classical mechanics or nonrelativistic quantum mechanics that can be chosen relative to any convenient energy “landmark”. Under P, a more general Dirac spinor changes as     +ψ+ (−r, t) ψ+ (r, t) = , (3.103) P ψ− (r, t) −ψ− (−r, t) where the wave functions ψ± (r, t) depends on the position r that is odd under parity. (It can also depend on other parity-odd quantities such as the momentum p that changes sign under parity. Note the absence of a final P−1 operator on the right because the wave function is in some sense a column vector.) As a result, spatial wave functions can be constructed that are even/odd when r changes sign. These wave functions are then said to have even/odd orbital parities in addition to the intrinsic parity. Parity deals with sign changes that come from different factors of a wave function. So the total parity of a spinor is the product of intrinsic and orbital parities. All symmetry transformations deal with signs; they all have similar multiplicative eigenvalues. 3.6.2

Time reversal

The second spacetime symmetry is time reversal (TR) S = T . Under it, t → −t and p → −p, while r and for a conservative system E too are left unchanged. The TR operator T is unfortunately rather complicated and requires a long discussion. ˆ −1 = H, ˆ the timeIn quantum mechanics with a TR invariant Hamiltonian T HT dependent wave function satisfies the Schr¨odinger equation:   ∂ ˆ ψ(r, t) = Hψ(r, t) (3.104) i ∂t should be TR invariant. The situation is in marked contrast to all first-order differential equations in time with real coefficients such as the diffusion (thermal conduction) equation ∂ ∂2 Φ(x, t) = κ 2 Φ. ∂t ∂x

(3.105)

for the temperature Φ of an object of thermal conductivity κ. A video of a puddle of water solidifying into a block of ice on a hot summer day must have been the video of the melting of the ice block run backwards. It cannot occur naturally, because

Symmetries of the Dirac equation

175

the diffusion equation describing the heat transfer needed to melt the ice is not TRinvariant. Under TR, the left side of the equation changes sign when t → −t. Wigner explained in 1932 that the first-order Schr¨odinger equation is TR invariant because the factor i that goes with ∂/∂t is transformed into −i under TR: ! ∂ −1 ˆ −1 T )ψ, or T i(T T ) (T −1 T )ψ(r, t) = T H(T ∂t ! ∂ ˆ ψ), (T ψ) = H(T (3.106) −i ∂(−t) ˆ −1 = H. ˆ So time reversal must involve a complex conjuif T −1 i T −1 = −i and T HT gation operation K under which KiK = −i. Wigner thus concluded that T = UK is antiunitary, meaning that a K appears to the right of the unitary operator U. What exactly is the TR transformed wave function T ψ(r, t; p, . . .) = Uψ∗ (r, −t; −p, . . .)?

(3.107)

First note that T changes the sign of all quantities in the wave function that are odd under TR. Besides t, they can include the momentum p and other TR odd variables. For the Klein–Gordon wave function (3.77) of a spinless particle, the unitary matrix is just a phase factor U = ηT = eiαT , the simplest unitary transformation that will not change the probability density |ψ|2 = ψ† ψ of the wave function used to define its normalization. Let us choose U = ηT = 1 for simplicity. For example, for the (unnormalized) plane-wave wave function of momentum p ψ(r, t; p) = ei(p · r−Et) : T ψ(r, t; p) = e−i[(−p) · r−E(−t)] = ψ(r, t; p)

(3.108)

turns out to be just the original wave function when expressed in terms of the original variables. This result is not surprising, because if the wave equation is unchanged under TR, then the wave function too is unchanged. The example thus confirms or illustrates Wigner’s conclusion that the imaginary unit i is odd under TR. The fact that the wave function in the original variables is unchanged under TR is also not surprising intuitively. If we take a video of a stone dropped from a balcony and run the video forwards, we shall see just a stone falling from the balcony. To see the TR motion in which the stone appears to have been thrown skyward, we need to run the video backwards, with t = −t as the operating variable. Then we see the same mathematical object 



T ψ(r, t; p) = e−i(−p · r−Et ) = ei(p · r+Et )

(3.109)

from a different time perspective, with an apparent skyward velocity dz/dt = −E/p < 0, where z = r · ep still increases in the downward direction.

176

Relativistic square-root spaces

The TR operators T for particles with spin are more complicated, because of the nontrivial unitary transformations U. To find them, we first note that the ˆ and more generally any component Jˆi of a spin operator, angular momentum rˆ × p, change sign under time reversal. This is true for Jˆz in particular: T |JM ∝ |J − M

(3.110)

in the eigenstate notation of quantum mechanics, showing that M, the eigenvalue of Jˆz , changes sign under TR. This flipping of Jˆz can be undone by rotating the coordinate axes by an angle π about an axis perpendicular to the z axis. The rotation axis is usually taken to be the y axis. For a spin 1/2 particle, for example, spin flipping can be realized by using U = ηT σ2 , containing an arbitrary phase factor ηT = eiαT :       1 0 1 = iηT , =U T 0 1 0     0 0 = −iηT . T (3.111) 1 1 Hence T 2 = UKUK = ηT σ2 (η∗T σ∗2 )KK = −1.

(3.112)

We have thus derived a special case (for J = 1/2) of the general formula T 2 = (−1)2J

(3.113)

The general formula describes the undoing of the action of T 2 by a 2π rotation of about a perpendicular axis, usually taken to be the y axis, for any spin state |JM. For integer J or a boson state, such a rotation returns the wave function to its original value. For half-integer J or a fermion state, the rotated wave function is the negative of the original wave function. It will take two complete turns to restore the original value. Under TR, the transformed Dirac equation (3.99) reads   0 = (γμ ) ( pˆ μ ) − m ψ , (3.114) where (γμ ) ( pˆ μ ) = γμ pˆ μ , ˆ −pˆ  = p) ˆ = pˆ μ , ( pˆ μ ) = (H, (γμ ) = U(γμ )∗ U −1 = (γ0 , −γ) = γμ , ψ (r, t = −t) = T ψ(r, t) = Uψ∗ (r, t = −t), U = ηT σ2 .

(3.115)

The verification that the chosen U generates (γμ ) will be left to Problem 3.6.1.

Symmetries of the Dirac equation

3.6.3

177

Charge conjugation

We turn finally to an internal symmetry called charge conjugation that changes the sign of the electric charge of a charged particle such as the electron. The negative-energy states of the Dirac equation for an electron are hard to interpret, as their energies go all the way to −∞. In 1930 Dirac had the ingenious idea that all the trouble will disappear if in the vacuum or space free of anything, these negative-energy states are completely filled with electrons, one to each distinct spacespin state because of the Pauli exclusion principle. Thus occupied, these negativeenergy states form an inert background called the Dirac sea. However, if an electron is excited from a filled negative-energy state to a positive energy state higher in energy by ΔE ≥ 2m, an electron/hole pair would be created. Since the original vacuum is nothingness, the hole would have physical attributes that are opposite in sign to those of the associated electron: positive charge, negative energy, negative momentum (compared to the electron momentum) and negative spin. Only a product or quotient of two such negatives can remain a positive. This was how the antielectron or positron was first proposed before Anderson discovered it unexpectedly in a cloud chamber in 1932. If the positron can have the same physical attributes as the electron except for the sign of its charge, it deserves its own Dirac equation. One must of course add the electromagnetic (e.m.) interaction 4-momentum pˆ μ → pˆ μ + eAμ so that one can tell the sign of its electric charge. How, then, are the charge-dependent negative-energy (hole) solutions of the electron Dirac equation related to the positive-energy solutions of the positron Dirac equation that treats a positron as a real particle? To describe a positron as a particle, we need a charge conjugation operator C|q, psE+  = | − q, psE+ ,

(3.116)

that is chosen to change the electron charge q = e = −|e| into the positron charge, without changing its mass, momentum p, spin s and positive energy E+ . To find C, we first complex conjugate the Dirac equation of an electron in an e.m. field  *∗ ) 0 = iγμ (∂μ − ieAμ ) − m ψ   = i(−γμ )∗ (∂μ + ieAμ ) − m ψ∗ . (3.117) The complex conjugation changes the electron charge in the e.m. momentum term to a positron charge |e|. It also changes the negative signs of the electron hole energy and momentum appearing in the (Klein–Gordon) spatial wave function contained in the Dirac wave function to the positive energy and momentum of a positron particle. One complex conjugation will take care of all three sign changes. Before the resulting equation qualifies as a Dirac equation for a positron, however, the (−γμ )∗ factor needs to be changed back to γμ by a similarity matrix transformation: C(−γμ )∗C −1 = γμ .

(3.118)

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Relativistic square-root spaces

Eq. (3.118) can be solved for C. The solution can be extracted more easily if the transformation is transferred to the better known γμ :  μ if μ = 2 γ , −1 μ μ ∗ . (3.119) C γ C = −(γ ) = −γμ , otherwise With γ2 purely imaginary while the remaining 3 γμ s are real, we see that C commutes with γ2 and anticommutes with the remaining γμ . So C = γ2 .

(3.120)

Most authors use a TR matrix that in our notation is Cγ0 . When written out in terms of the γ, all expressions should agree. Note that it is the ρ2 in C that performs the interchange between negative and positive energy states. The reason for the standard notation using Cγ0 is that it is the adjoint spinor ¯ mψ ψ¯ = ψ† γ0 , not ψ† itself, that appears in the physical Lorentz covariant densities ψΓ ¯ of Table 3.2 and Problem 3.5.3. The additional factor of γ0 will cancel the γ0 in ψ. Eq. (3.119) shows that Eq. (3.117) can be written as a Dirac equation for the positron as a normal particle:   0 = C −1 iγμ (∂μ + ieAμ ) − m Cψ∗ , or   0 = iγμ (∂μ + ieAμ ) − m ψc , where ψc = Cψ = ηc γ2 ψ∗ .

(3.121)

An arbitrary phase factor ηc has been added to C. The arbitrariness of ηc can be demonstrated explicitly (Problem 3.6.3(a)). It is the complex conjugation K that changes the sign of the charge between a particle and its antiparticle. K also appears in time reversal. For this reason, in Feynman’s diagrammatic description of processes in quantum field theory, an antiparticle line is shown as a particle line that points backwards in time.

Problems 3.6.1 (Quantum commutator) (a) Derive the 1D commutator relation [ p, ˆ x] = 1/i where pˆ = (1/i)d/dx. (b) Use the fundamental quantum commutator obtained in (a) to show that the imaginary unit i is unchanged under parity, but changes sign under time reversal. 3.6.2 Show that (γμ ) = T (γμ )T −1 obtained from the time-reversal operator T = ηT σ2 K agree with the results found in Eq. (3.115). 3.6.3 (Charge conjugation) (a) (Dirac wave function) Use C = ηc γ2 K to show explicitly that (ψc )c = ψ for any Dirac wave function.

Weyl and Majorana spinors, symmetry violations

179

(b) (KG wave function) Write down the Klein–Gordon (KG) wave equation for an electron in an electrostatic potential Aμ = (A0 = Φ, −A = 0). If its solution is ψKG (r, t), show that ψKG (r, t)∗ is the solution of a KG equation for a positron in the same Aμ . (c) (Free-particle Dirac wave function) Show that in the notation of Problem 3.5.4 and in the Dirac–Pauli basis:   C u3 (−p)ei(−p · r−E− t) = iu2 (p)ei(p · r−E+ t) ,   C u4 (−p)ei(−p · r−E− t) = −iu1 (p)ei(p · r−E+ t) , & where E± = ± m2 + p2 . You may use without proof the results of Problem 3.5.4.

3.7

Weyl and Majorana spinors, symmetry violations∗

Weyl spinors are used in connection with the concept of chirality used to describe the interactions between elementary particles, the basic building blocks of nature. Majorana spinors are introduced to describe a charge neutral, spin 1/2 particle that is identical to its own antiparticle. No Majorana fermions have yet been discovered. Their discovery or continued absence has important implications in our understanding of the physical world. In this section we present a brief introduction to these spinors. 3.7.1

Weyl spinors

The DP basis for the Dirac equation is particularly useful for slowly moving massive particles with p  m. In fact, a particle at rest is described by the 2-component spinor φ± for energy E±, respectively. The σ · p term for a slow, nonrelativistic particle can be treated as a perturbation. For fast, relativistic particles with p  m, however, both spinors appear with comparable amplitudes in a Dirac spinor in the DP basis, resulting in a complication that is avoidable. This is because the Dirac equation itself actually simplifies to just γμ pμ ψ = 0 when m = 0. Note that γ5 does not appear explicitly in the Dirac equation, and anticommutes with γμ . So if ψ is a solution, so is γ5 ψ. It is then convenient to use instead the linear superpositions called chiral projections 1 φL,R = (1 ∓ γ5 )ψ = PL,R ψ, 2

(3.122)

because they are simultaneous eigenstates of γ5 with eigenvalues ∓1, respectively: 1 γ5 PL,R = (γ5 ∓ 1) = ∓PL,R , 2 P2L = PL ,

P2R = PR ,

PL PR = PR PL = 0,

(3.123)

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Relativistic square-root spaces

where (γ5 )2 = 1 has been used. Because of this chiral projection, each wave function φL,R occupies only one position or dimension in ρ space, but it is actually a 2-component Pauli spinor when its spin structure is made explicit. φL,R are called the left- and right-chiral Weyl spinors, γ5 being the chirality (meaning handedness) operator. These spinors have chirality ∓1, respectively. The description given so far of the Weyl spinors is actually basis-independent. In the DP basis, γ5 = ρ1 . So its normalized eigenvectors are     1 1 1 1 , uR = √ , uL = √ 2 −1 2 1   1 1 1 U= √ 2 −1 1 1 = (I + iρ2 ) 2

while

(3.124)

5 = −ρ . This new is the unitary matrix that will diagonalize ρ1 into a new diagonal γW 3 basis is called the Weyl basis. A general Dirac spinor in the new Weyl basis is



φ ψW = L φR



   1 1 −1 φ+ =U ψ= √ , φ− 2 1 1   1 φ+ − φ− . = √ 2 φ+ + φ− †

(3.125)

√ The inverse relation φ± = (φR ± φL )/ 2 is the same, apart from an unimportant normalization constant, as that given in Eq. (3.85). The arbitrary spinors φL,R used there are finally shown here to be Weyl spinors. Returning to the general Dirac equation (3.84) for a massive particle in the Weyl basis: 

−m Hˆ − σ · pˆ ˆ H + σ · pˆ −m



 φL = 0, φR

(3.126)

we see that when written in the Lorentz covariant form " # μ γW pˆ μ − m ψW = 0,

(3.127)

Weyl and Majorana spinors, symmetry violations

181

the γW matrices are 0 = ρ1 , γW

k γW = iρ2 σk = γk ;

5 0 1 2 3 γW = iγW γW γW γW = −ρ3 .

(3.128)

α = U −1 γα U, α = They can also be calculated from the unitary transformation γW 0–3, 5 (Problem 3.7.1). Specializing now to massless particles, the Dirac equation can be written in a 0 Hamiltonian form by left multiplication with γW

ˆ W = γ5 Σ · pψW , Hψ W

(3.129)

where a capitalized symbol Σ = I ⊗ σ is used to emphasize its 4 × 4 matrix structure. 5 appears here because γ0 γ = ρ (iρ σ) = γ5 σ. γW 1 2 W This massless Hamiltonian equation has a very interesting property. For a free particle, the Lorentz 4-vector operator pˆ μ in the Dirac equation has the eigenvalue pμ = (E, −p) after operating on the Klein–Gordon wave function ψKG = ei(p·r−Et)

(3.130)

contained in any ψ. Under Lorentz transformations, the spatial momentum p of massless particles behaves very differently from that of massive particles. A massive particle moving forward in one Lorentz frame becomes stationary in its own rest frame. It appears to move backwards in all Lorentz frames moving forward at faster speeds. In contrast, a massless particle in free space always moves with light speed in any Lorentz frame. It has no rest frame. Its momentum direction is Lorentz-invariant. For this reason, the spin matrix called the helicity operator Σp = Σ · ep

(3.131)

simplifies to Σ3 when the 3-axis is taken to be the Lorentz invariant direction ep . Its four 4 × 1 helicity eigenspinors are the special case of the DP free-particle energy spinor (3.98) when p1,2 = p± = 0, p = |p| = p3 . It is convenient to show the results more compactly as:     1 1 E = |p| : u1 ∝ χ1 , u2 ∝ χ−1 ; 1 −1     −1 1 E = −|p| : u3 ∝ χ1 , u 4 ∝ χ−1 . (3.132) 1 1 They have the helicities of h = 1, −1, 1, −1, respectively, because χ±1 = (1, 0)T , (0, 1)T are the Pauli eigenspinors of σ3 with eigenvalues ±1, respectively. A particle of helicity 1 is said to be right-spinning (or right-handed in spin), while that of helicity –1 is left-spinning (or left-handed in spin).

182

Relativistic square-root spaces

In the Weyl basis, the chiral spinor factor simplifies further under the unitary transformation U of Eq. (3.124)     1 0 χ1 , uW2 ∝ χ−1 ; uW1 ∝ 0 1     0 1 uW3 ∝ χ1 , uW4 ∝ χ−1 . (3.133) 1 0 5 = −ρ of chiralDenoting by ξ−1 = (1, 0)T , ξ1 = (0, 1)T the chiral eigenspinors of γW 3 ity κ = ∓1, respectively. The Hamiltonian Eq. (3.129) simplifies to

ˆ κ χh ψKG ) = κh|p|(ξκ χh ψKG ). H(ξ

(3.134)

Since the energy eigenvalue E can be found directly from ψKG , we have thus derived the nontrivial relation E = κh|p|,

(3.135)

where each eigenvalue comes from a different eigen function/vector in the composite wave function. Some authors use a different sign notation where the same physical situation is described by a κ of the opposite sign. In our notation, positive energy states have h and κ of the same sign, while these quantities have opposite signs in negative energy states. Weyl’s 1929 theory of massless particles of spin 1/2 allows only left-chiral parti5 in Eq. (3.129) replaced by its eigenvalue κ = −1, we have finally found cles. With γW the Weyl equation ˆ L = −σ · pφL , Hφ

(3.136)

where the small σ is used to emphasize the fact that φL is a two-component spinor. The right side of the equation involves a parity-odd operator σ · p, while a parityeven operator Hˆ appears on the left. The Weyl equation thus violates parity. It does so maximally in the sense that the parity violation on the right is undiluted by a parity-conserving term. 3.7.2

P and CP violations

Both at and long after Weyl’s time, parity conservation was considered an inviolate law of nature. So Weyl’s equation was viewed as a mere mathematical curiosity until Lee and Yang in 1957 pointed out that many puzzling features of the weak interaction between elementary particles can be understood if parity is violated. The flood of experimental studies that followed their suggestion proved that they were right. Indeed, a fast neutrino was found experimentally to have negative helicity, spinning left-handedly in a direction antiparallel to its motion, while fast antineutrinos are right-spinning. The reason for the parity violation is soon pinned down: the weak interaction involves only left-chiral fermions whether they are massless or massive.

Weyl and Majorana spinors, symmetry violations

183

It seems appropriate that it was Weyl the mathematician who first wrote down the equation, and that it took the physicists Lee and Yang to start the process that shows its place in the physical world. Dirac, the physicist with a mathematical taste in the class of Einstein’s, never believed in the general validity of the parity invariance, or time reversal invariance either. The violation of the combined CP invariance (equivalent to time reversal invariance because CPT together is a good symmetry under very general conditions) was found in 1964 by Cronin and Fitch in the decay through the left-chiral weak interaction of certain elementary particles called neutral kaons when they are transformed into their own antiparticles. More specifically, physically prepared neutral kaons of definite mass have small admixtures with the “wrong” CP eigenvalue. Kobayashi and Maskawa (KM) explained in 1973 how this situation can occur mathematically, but only if another group of elementary particles not known at that times exists in nature. To understand their argument we need a short lesson on particle physics. In 1972, all known elementary particles were made up from two generations i = 1, 2 of certain unobservable charged, spin 1/2 particles called quarks:       ! {ui } t u c . (3.137) ; , = {di } d s b Each generation contains a doublet, shown by the first two columns on the right of the expression: (up, down)T and (charm, strange)T for the two known generations. KM showed that CP violation cannot occur with only two generations, but it can occur if a third generation of quarks, shown by the column (top, bottom)T , exists. These quarks are usually referred by the first letters of their names. The upper members u, c, t have charge (2/3)|e|, while the lower members have charge −(1/3)|e|. Generations are named by the upper members u, c, t. Quarks are massive. They can be left- or right-chiral. The weak interaction operates only between left-chiral fermions and right-chiral antifermions. Kaons are quark–antiquark pairs of two different generations. Their “weak” decay can be described in terms of the mixing of left-chiral quarks of different generations. With generations named by their upper members, these upper members can be considered unmixed, as a matter of convention. Generational mixing then means that the physically observable mass states d , s , b are related to the natural states d, s, b by a unitary matrix V: ⎛ ⎞ ⎛ ⎞ ⎜⎜⎜ d ⎟⎟⎟ ⎜d⎟ ⎜⎜⎜ s ⎟⎟⎟ = V ⎜⎜⎜⎜⎜ s ⎟⎟⎟⎟⎟, where ⎝ ⎠ ⎝ ⎠ b b ⎛ ⎞ ⎜⎜⎜ Vud Vus Vub ⎟⎟⎟ V = ⎜⎜⎜⎝ Vcd Vcs Vcb ⎟⎟⎟⎠. (3.138) Vtd Vts Vtb Note that the physical states are labeled by the generation names u, c or t in V. The important observation of KM is that when the matrix dimension of V is N ≥ 3, a

184

Relativistic square-root spaces

general unitary matrix V cannot always be reduced to a form where all matrix elements are real. CP violation appears when an irreducibly complex matrix V changes value under the complex conjugation K contained in the CP operator. To understand why N ≥ 3 generations are needed, consider the scalar u† Vd = u† (ηΦu )∗ (ηΦu )V(ηΦd )∗ (ηΦd )d, = u† VKM d , where

d = (ηΦd )d,

u = (ηΦu )u,

VKM = (ηΦu )V(ηΦd )∗ , Φu = diag (1, α2 , . . . , αN ), Φd = diag (β1 , β2 , . . . , βN ).

(3.139)

Here u, d are arbitrary column vectors, η is the overall complex phase factor with |η| = 1, and Φu,d are diagonal square matrices containing a total 2N − 1 phase factors αi , βi relative to η. The purpose of the equivalent transformation of V by these relative phase matrices is to make VKM as real as possible. Suppose VKM is real. Then it reduces to an SO(N) matrix, an element of a subgroup SO(N) of SU(N). Recall from Section 2.11 that the real matrices of the orthogonal group SO(N) contains N 2 real numbers. An orthogonal matrix satisfies N(N + 1)/2 orthogonality relations, leaving only N(N − 1)/2 real numbers that serve as rotation angles. The SO(1) group has only two elements ±1 (the only real numbers that are its own inverses) and no rotation angle. Rotation angles appear only for N ≥ 2. SU(N) is a much larger group. An SU(N) matrix contains in general N 2 complex numbers, or 2N 2 real numbers. These numbers have to satisfy N 2 unitarity relations, leaving N 2 independent real numbers. N(N − 1)/2 of these independent real numbers are the rotation angles of its SO(N) substructure. This leaves N(N + 1)/2 real parameters that can be used to define N(N + 1)/2 phase factors that complexify the rotations into complex rotations of a special kind. The overall phase factor η of U(1) does nothing to V because it always cancels its partner η∗ . So it does not count. Only the 2N − 1 relative phases contained in the two phase matrices Φu,d can be taken from outside V to the inside, or vice versa. When taken outside V, these relative phases can be merged into the arbitrary phases of the wave functions residing in the u, d column vectors. This leaves a reduced unitary matrix VKM containing 1 N(N + 1) − (2N − 1) 2 1 = (N − 1)(N − 2) 2

NKM =

(3.140)

irreducible CP-violating KM phases that are physically meaningful. Now the number of relative phases increases linearly with N, while the number of orphaned real numbers to be found, namely N(N + 1)/2, increases quadratically

Weyl and Majorana spinors, symmetry violations

185

as N 2 . So the possibility that NKM = 0 can be realized only for the very special choices of N = 1, 2. This result shows that for N ≥ 3, VKM cannot in general be real. It describes in general a complexified rotation in ND space with (N − 1)(N − 2)/2 complexified 2D rotations that can be changed under complex conjugation. Hence CP violation is the rule rather than the exception when three or more generations of left-chiral quarks are mixed, no matter how the third and higher generations appear. The CP conserving solutions are already familiar to the particle physicists. For N = 1, there is only the arbitrary phase factor η that can be taken to be 1. For N = 2, VKM reproduces the known two-generation mixing model of Cabbibo: 

 cC sC VC = . −sC cC

(3.141)

Here cC = cos θC , sC = sin θC of the Cabbibo quark mixing angle θC between the u and c generations. It was not known before KM that CP violation is impossible when there are only two families of quarks. For N = 3, V contains one irreducible complex phase factor ζ = eiδ , where δ is called the CP-violating KM phase: ⎛ ⎜⎜⎜ c2 VKM = R1 ⎜⎜⎜⎝ 0 −s2 ζ ⎛ ⎜⎜⎜ 1 0 R1 = ⎜⎜⎜⎝ 0 c1 0 −s1

⎞ 0 s2 ζ ∗ ⎟⎟ ⎟ 1 0 ⎟⎟⎟⎠ R3 , where 0 c2 ⎞ ⎛ ⎞ 0 ⎟⎟ ⎜⎜⎜ c3 s3 0 ⎟⎟⎟ ⎟⎟⎟ s1 ⎟⎠, R3 = ⎜⎜⎜⎝ −s3 c3 0 ⎟⎟⎟⎠. c1 0 0 1

(3.142)

Here ci = cos θi , si = sin θi . In the literature, the quark mixing angle θi is written as θjk , ijk = cyclic. For N = 4 generations, there are six quark mixing angles and three irreducible CP-violating KM phase angles. The weak interaction between quarks also involves certain charge neutral, spin 1/2 elementary particles called neutrinos. Three types or flavor of neutrinos have been found—electron, muon and tau neutrinos, denoted νe , νμ , ντ respectively—one for each generation. Experimental data show definitively that there are only three neutrino flavors. Experimental data to date on CP violation are consistent with the appearance of only one KM phase angle. However, other sources of CP violation different from the KM mechanism are possible. The KM description of CP violation shows that mathematically the observed effects can indeed appear. It does not explain the size of the KM phase δ, or its physical origin. Such an explanation will involve an understanding of (a) why there are generations, (b) why three generations, and (c) why the KM phase does not vanish. That is, the first two conditions are necessary but not sufficient. Indeed, in phenomena involving the so-called strong interaction among the same three families of quarks, now with both right- and left-chiral quarks interacting, CP violation has not been seen. The reason for strong CP conservation is also not understood.

186

3.7.3

Relativistic square-root spaces

Majorana spinors

Neutrinos have no known electromagnetic or strong interactions. Long considered massless, at least two of them have been found to be massive, like neutrons, in 1998. The neutrino mass eigenstates have mixed flavors. Neutrino flavors control how neutrinos interaction with elementary particles containing different generations of quarks. It is the flavor mixing that makes it possible to measure neutrino mass differences. These mass differences are now known to be of the order of 10 meV— about 108 times smaller than the mass of the electron. The mixing of three flavors involves a mixing matrix similar to the quark mixing matrix. Flavor mixing also generates CP violation. The effect resides in one irreducible phase angle if the neutrinos are Dirac particles, and three irreducible phase angles if they are Majorana neutrinos, the subject of this subsection. Massive neutrinos have rest frames where their helicity disappears. So their helicity is no longer a Lorentz invariant. Like neutrons, they can appear in both leftand right-chiral forms. The right-chiral forms do not interact even “weakly” with other elementary particles if the neutrinos are Dirac neutrinos with only left-chiral interactions. They may have gravitational interaction, but this is so very weak that it does not count for all practical purposes. So right-chiral Dirac neutrinos and leftchiral Dirac antineutrinos are said to be sterile. Let us now examine the mass term of the Dirac equation more critically. Dirac’s √ square-root space involves the Lorentz momentum operator pˆ μ , but leaves m = m2 a nonnegative scalar parameter. Majorana pointed out in 1937 that m2 too can have its own square root space. Suppose there exists a special Majorana mass operator mSM , with S2M = 1. With m → mSM , the Dirac equation (3.89) is changed into a pair of Majorana equations γμ pˆ μ ψ = mSM ψ = mψM , γμ pˆ μ ψM =

1 μ 2 (γ pˆ μ ) ψ = mψ. m

(3.143)

With S2M = 1, SM has eigenvalues S M = ±1, which can be called an S M parity. The corresponding eigenstates are just the SM projected states 1 Ψ (±) = (1 ± SM )ψ. 2

(3.144)

These are also the mass eigenstates of mass eigenvalues ±m. Physical mass states are states of positive mass, here Ψ (+) . Majorana actually used SM = C = γ2 K, the charge conjugation operator. So the ± mass eigenstates are also states of good charge-conjugation C = ±1 parity. Now if ψ is the left-chiral Weyl spinor φL of Eq. (3.122) for a spin 1/2 particle, then φLc ≡ CφL = γ2 KφL

(3.145)

is a right-chiral Weyl spinor for the antiparticle, because C anticommutes with γ5 . It exists in an antiparticle space similar to but distinct from the space φL of left-chiral

Weyl and Majorana spinors, symmetry violations

187

Weyl particles. The projected states of definite C-parity, being a linear superposition of particle and antiparticle states, is physically meaningful only for charge-neutral particles such as a neutrino (a fermion) and neutral mesons (bosons). Only the neutrino and other spin 1/2 fermions satisfy the Dirac equation. The states of C-parity C = ±1 are also ± mass eigenstates, respectively, of the Majorana mass operator mC, namely the (projected) Majorana spinors 1 Φ(±) = (1 ± C)φL ≡ P± φL : L 2   Φ(±) = CΦ(±) = ±Φ(±) . L L L c

(3.146)

They describe Majorana fermions that are their own antiparticles. For them, the distinction between matter and antimatter vanishes. They are equally at home in both worlds and antiworlds. has left-chiral particle The positive mass, even C parity Majorana spinor Φ(+) L component and a right-chiral antiparticle component, both of which can interact weakly if they have the same dynamics as the corresponding Dirac neutrinos. In the same way, the Majorana spinor # 1 1 " (+) (+) (3.147) Φ(+) = = + φ φ (1 + C)φ R Rc R 2 2 R made up of a right-chiral particle component and a left-chiral antiparticle component would be sterile. On the other hand, Majorana neutrinos may have totally different dynamics from Dirac neutrinos. Then all bets are off. Also a neutrino can have both Majorana and Dirac masses that are different from each other in value as well as in dynamics. The importance of Majorana neutrinos comes from the fact that the neutrinos νe , νμ , ντ presently known are Dirac neutrinos that come in distinct neutrino and antineutrino forms with interaction properties that are different with matter and antimatter. For this reason, a neurino number can be assigned to them, 1 for a neutrino and –1 for an antineutrino. In their interactions, the neutrino number is known to be conserved. This means that matter and antimatter are distinct even for neutrinos that do not carry any charge. In the case of an electron, one can well imagine that the conservation of electron number (1 for an electron and –1 for a positron that is also a hole in the occupied electron “sea”) is another face of charge conservation. The conservation of neutrino number shows that this is not the case. One may well ask if there is an exception to this rule of neutrino number conservation. The Majorana neurino, being a 50/50 combination of a neutrino and an antineutrino, describes a maximum violation of neutrino number. A discovery of a Majorana neutrino will overturn the idea that neutrino number conservation is a universal law of nature. We have not found any Majorana neutrino, however, after repeated searches. So we do not know if these neutrinos exist or what they might be like. Neutrinos are very difficult to detect. We have no experimental information on any irreducible CPviolating phase angle in the neutrino flavor mixing matrix either.

188

Relativistic square-root spaces

Problems μ

3.7.1 Show that γW = S (γμ )S T , μ √ = 0, 1, 2, 3, 5, obtained from the orthogonal transformation S = (I − iρ2 )/ 2 connecting the Dirac-Pauli basis to the Weyl basis agree with the results given in Eq. (3.128). 3.7.2 Use the Weyl basis to show that (a) the chiral projection operators PL,R = 12 (1 ∓ γ5 ) anticommute with C and γ0 , (b) φLc is a right-chiral state, and (c) under charge conjugation, a left-chiral particle is changed into a rightchiral antiparticle, and (outside the confines of the Weyl equation) a right-chiral particle is changed into a left-chiral antiparticle.

3.8

Lorentz group

Recall that rotations of the coordinate axes in 3D space about the same coordinate origin are described by orthogonal matrices with unit determinants, i.e., by the SO(3) group. Under 3D rotations, the squared distance x12 + x22 + x32 remains unchanged. More generally, SO(N) matrices for rotations in ND space contain N(N + 1)/2 rotation angles. When the time variable x0 is also added to form a 4D spacetime, the possible 4D rotations leave the squared spacetime separation τ2 = x02 − x12 − x22 − x32 invariant. To use these real, Euclidean coordinates, different signatures are needed in the terms in τ2 . Other than this complication, 4D spacetime rotations are mathematically similar to rotations in ordinary 4D Euclidean space, with N(N + 1)/2 = 6 generalized rotation angles. We have seen in Section 3.1 that the three additional rotation angles involving the time variable are purely imaginary when the Minkowski notation is used. They are concerned with Lorentz transformations along the three coordinate directions. Lorentz transformations together with spatial rotations form a group called the Lorentz or the SO(1, 3) group, where the argument gives the numbers of positive and negative signatures appearing in τ2 . With six generalized rotations, SO(1, 3) contains six generators of infinitesimally small rotations. The generators for the three spatial rotations are just the angular momentum operators Ji , i = 1, 2, 3, of SO(3), now embedded in 4 × 4 matrices. For a rotation R3 (dθ) = I + iJ3 dθ about the spatial 3-axis, for example,   cos dθ sin dθ in 12 space, R3 (dθ) = − sin dθ cos dθ ⎞ ⎛ ⎜⎜⎜ 0 0 0 0 ⎟⎟⎟ ⎜⎜ 0 0 −1 0 ⎟⎟⎟ ⎟⎟. J3 = i ⎜⎜⎜⎜⎜ (3.148) ⎜⎝ 0 1 0 0 ⎟⎟⎟⎠ 0 0 0 0 We have chosen here not to use 3-axis as a quantization axis along which J3 would have been a diagonal matrix. Instead we choose J3 to be the generator of a 2D

Lorentz group

189

rotation about the 3-axis. When the remaining two components J1 , J2 are defined in an analogous way, these operators are related to one another by cyclic permutations of indices, thereby making explicit all the symmetries in the problem. The sign of Ji is also chosen to satisfy the standard Lie algebra for angular momentum matrices in a right-handed coordinate system: [Ji , Jj ] = iεijk Jk .

(3.149)

In the Euclidean notation where upper component indices are used for Lorentz vectors, the Lorentz transformation λ(β) in (3.9) due to the motion of the 1-axis is  0     x γ βγ x0 = . (3.150) βγ γ x1 x1 Hence an infinitesimal change dβ induces a transformation matrix λ(dβ) = 1 + dβK1 containing the boost operator ⎞ ⎛ ⎜⎜⎜ 0 1 0 0 ⎟⎟⎟ ⎜⎜ 1 0 0 0 ⎟⎟⎟⎟ (3.151) K1 = ⎜⎜⎜⎜⎜ ⎟. ⎜⎝ 0 0 0 0 ⎟⎟⎟⎠ 0 0 0 0 For a boost along the 3-axis, one finds similarly ⎛ ⎜⎜⎜ 0 0 0 ⎜⎜ 0 0 0 K3 = ⎜⎜⎜⎜⎜ ⎜⎝ 0 0 0 1 0 0

1 0 0 0

⎞ ⎟⎟⎟ ⎟⎟⎟ ⎟⎟⎟. ⎟⎟⎠

(3.152)

These matrix operators show that K3 commutes with J3 as they are in different matrix dimensions. However, J3 involves the same spatial dimension x1 as K1 . So these two matrices interfere with each other and do not commute. A direct calculation gives [J3 , K1 ] = iK2 . For product of Ki among themselves, the time dimension is always involved. So they do not commute either. A direct calculation gives [K1 , K2 ] = −iJ3 , showing that the Ki do not form a closed algebra among themselves. Using cyclic permutations, one can verify that the Lie algebra of these six generators of SO(1, 3) is given by the SU(2) Lie algebra (3.149) and the following relations: [Ji , Kj ] = iεijk Kk , [Ki , Kj ] = iεijk Jk .

(3.153)

The purpose of these infinitesimal generators is to generate an arbitrary finite transformation by exponentiation A = eG ,

G = iθ · J − η · K.

(3.154)

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Relativistic square-root spaces

The resulting 4 × 4 matrix operates on the 4 × 1 column coordinate vector x = (x0 , x1 , x2 , x3 )T , x = Ax,

(3.155)

to give a new vector x in a spatially rotated and Lorentz-boosted coordinate frame. The six generators of A have one particularly simple feature. Each involves only two matrix dimensions in which a J is just a Pauli matrix σ2 in two spatial dimensions, while a K is a Pauli matrix σ1 where one matrix dimension is that of time. Hence all even powers of the same generator is just the identity matrix in these two matrix dimensions, while all odd powers reduce to the generator itself. This means that for pure rotation about the 3-axis, A = R3 (θ) = eiθJ3 = (I − I(12) ) + I(12) (cos θ + J3 sin θ),

(3.156)

when the exponential matrix function is expanded in a power series of the matrix exponent. Here I(12) = diag(0, 1, 1, 0) is the unit matrix in the spatial 1, 2 matrix dimensions. We then recover our starting point for a simple 2D rotation. For a pure Lorentz boost involving K1 , a similar power-series expansion gives A = B1 (η) = e−ηK1 = (I − I(01) ) + I(01) (cosh η − K1 sinh η) ⎛ ⎞ ⎜⎜⎜ cosh η − sinh η 0 0 ⎟⎟⎟ ⎜⎜ sinh η cosh η 0 0 ⎟⎟⎟⎟ . = ⎜⎜⎜⎜⎜ 0 1 0 ⎟⎟⎟⎟⎠ ⎜⎝ 0 0 0 0 1

(3.157)

After the expansion, the power series sums to hyperbolic functions because the exponent here does not contain a separate factor of i. On comparing the result to the starting point of Eq. (3.150), we conclude that in general η = eβ arctanh β.

(3.158)

The general pure Lorentz transformation matrix B(η) along an arbitrary direction eβ is given in Problem 3.8.3. The commutators (3.149) and (3.153) of the Lorentz group appear rather intricate at first sight. However, if one uses instead the six complex combinations N±i = (Ji ± Ki )/2, the new commutators simplify to [N+i , N+ j ] = iεijk N+k , [N−i , N− j ] = iεijk N−k , [N+i , N− j ] = 0.

(3.159)

Lorentz group

191

The first two new commutators describe two separate SU(2) Lie algebras. The third commutator shows they are independent of each other, a most useful result. So SO(1, 3) has the same structure as (or is isomorphic to) the direct product group SU(2)+ ⊗ SU(2)− . Both matrices Ji and Ki are Hermitian, the Ji are purely imaginary and antisymmetric, while the Ki are real and symmetric. Such a factorized structure is particularly helpful in deriving many results. For example, the arbitrary transformation matrix considered so far can be written as A = A+ A− = A− A+ ,

with

det A = (det A+ )(det A− ) = 1,

(3.160)

because we know that det A± = 1 for each SU(2) group. (The Pauli spin matrices are traceless, Hermitian matrices. Each can be diagonalized by a unitary matrix. In the diagonal form, det A = eTrG = 1.) Our As are the proper Lorentz transformations that have det A = 1. They do not include improper Lorentz transformations with det A = −1. An example of the latter is the space inversion ei → −ei of all three spatial axes that is realized by the transformation matrix AP = diag (1, −1, −1, −1). The (Pauli) spinors χ±α , α = 1, 2 (↑, ↓), of the two SU(2)± groups can be used to construct representations of the proper Lorentz group SO(1, 3) of the type ( j+ , j− ) = (0, 0), (1/2, 0), (0, 1/2), (1/2, 1/2), (1, 0), (0, 1), . . . We shall explain what this cryptic statement means in the next subsection. 3.8.1

SU(2) representations

Roughly speaking, the vectors on which elements of a matrix group act, when properly classified, form a vector space called a representation (rep) of the group. Take, for example, the SU(2) group with its three generators Ji = σi /2 expressed in terms of the 2 × 2 Pauli matrices σi . They operate on 2 × 1 column vectors that can be expressed as linear combinations of the two Pauli spinors χ+ =

  1 = | ↑, 0

χ− =

  0 = | ↓. 1

(3.161)

These Pauli spinors are eigenstates of J3 = σ3 /2 with eigenvalues m = ±1/2, respectively, of a particle of spin j = 1/2. When the dimension of the vector representation space matches that of the matrix generators, here both 2, the representation is said to be a fundamental representation. In the case of Pauli spinors, the fundamental representation is also called a spinor representation. The word “spinor” refers specifically to the fact that the Pauli spinors change sign after one turn around a rotation axis. It takes two complete turns to restore them to their original values. Because of this sign change, the particle is said to be a fermion, as distinct from a boson for which one complete rotation is enough to restore the spin state to its original value. The –1 signature of a Pauli spinor expresses the fact that the spinor resides in a square-root space.

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Relativistic square-root spaces

Strictly speaking, the words “fermion” and “boson” have their traditional meaning describing the nature of a point particle in physics only when the Pauli spin involved is that describing the intrinsic spin of the particle, the degree of freedom originally discovered by Pauli in atomic physics. Their mathematical properties are present in any SU(2) group, however, including the SU(2)± under discussing here. Both these subgroups of the Lorentz group are concerned with the external properties related to the spacetime description of an event. They have nothing to do with Pauli’s intrinsic spin degree of freedom. However, these two words, “fermion” and “boson”, are so descriptive of the mathematical properties they embody that we shall use them frequently in a more general mathematical context in our explanation of group representations. We may call this usage a pseudo-spin language. The essence of group representations can be understood from the quantum theory of angular momentum for which the words “fermion” and “boson” have their traditional meaning: 1. All spin states or vectors can be obtained from multispinors of degree or rank + r made up of the direct product of r Pauli spinors ri=1 χi . Here the column or particle label i contains the spinor signature label ± as well. On one complete rotation around any axis, each Pauli spinor changes sign. So a multispinor of rank r suffers a sign change (−1)r , which is -1 if r is odd, but 1 if r is even. The composite system is thus a fermion if r is odd, and a boson if r is even. 2. Along the quantization 3-axis, the total projected spin value of a multispinor is  the sum of the individual projections: m = i mi . With mi = ±1/2, the eigenvalues of the individual σ3 /2, m is half integer if r is odd, and integer if r is even. The distinct m states of the multispinor differ from one another by an integer, because this is true for a single spinor for which m = ±1/2 only. 3. The possible m values of the quantum spin j are  m = ± j, ±( j − 1), . . . ,

0, integer j . ±1/2, half integer j

(3.162)

m and j are thus both integers, or both half integers. This means that fermions described by multispinors of odd degrees have half integer j, while bosons with their multispinors of even degrees have integer j. This connection between spin and spinor signature is part of the spin-statistics theorem of quantum mechanics. (The theorem states that the wave function of a system of two fermions changes sign when the fermions interchange their positions, while no sign change results when the particles are bosons.) We also see from Eq. (3.162) that the total number of distinct m values for j is d = 2 j + 1. It is even for a half integer spin j, and odd for an integer spin. d is called the dimension of the rep ( j). 4. The total dimension of a multispinor of degree or rank r is the direct product 2r . This product can be reduced to a sum of the dimensions of the possible spin values { j} that can be constructed from the multispinor:

Lorentz group

2r =

 (2 j + 1).

193

(3.163)

{ j}

The actual decomposition is done sequentially, following rigidly the rule that any intermediate multispinor factor of rank r is fermionic (bosonic) if r is odd (even): r=2:

2 ⊗ 2 = 1 ⊕ 3,

r=3:

2 ⊗ (2 ⊗ 2) = 2 ⊗ (1 ⊕ 3), = 2 ⊕ 2 ⊕ 4.

(3.164)

Note that the product on the left, giving the dimension of the product space, is equal to the sum on the right, giving the sum of dimensions of states of definite j. We have thus described how two 1/2 spins can be “added” uniquely to a total spin of j = 0, 1 of dimensions 1, 3, respectively. The expression on the right cannot be decomposed further. For this reason, the j states on the right are called irreducible representations (or irreps). For three 1/2 spins, we find two ( j = 1/2) irreps and one (3/2) irrep. The term spin or angular momentum addition refers to the fact that additions are involved in constructing the irreps of the product of two reps. The basic construction of SU(2) irreps like those shown in Eq. (3.164) can be obtained by using the general expression for the addition of two spins j< , j> , where j< ≤ j> : ( j< )( j> ) =

j< 

( j> + m).

(3.165)

m=− j<

We are now in a position to understand the representations of the Lorentz SO(1, 3) ↔ SU(2)+ ⊗ SU(2)− group. One can label these representations with two independent pseudo-spin labels j+ , j− : ( j+ , j− ) = (0, 0), (1/2, 0), (0, 1/2), (1/2, 1/2), (1, 0), (0, 1), . . .

(3.166)

This means that the vectors in the vector space of the representation are products of two (pseudo-) angular momentum states of the type used in quantum mechanics. The word “state” means a vector in this vector space. In the famous Dirac bracket notation of quantum mechanics, the state vectors are states of the type | j+ m+ ; j− m−  = | j+ m+ | j− m− ,

(3.167)

where the symbol | . . . denotes a column vector of dimension dj = 2 j + 1, j = j± in the vector space, a vector like the column eigenvector of a matrix eigenvalue equation.

194

Relativistic square-root spaces

Problems 3.8.1 (SO(1, 3) generators) Verify the following results: (a)

(b) (c) (d) (e)

3 2 J = 2 diag (0, 1, 1, 1), i=1 2 i i Ki = diag (3, 1, 1, 1); T Ji = −Ji , KiT = Ki ; [Ji , Jj ] = iεijk Jk ; [Ji , Kj ] = iεijk Kk ; [Ki , Kj ] = iεijk Jk ; Eq. (3.159).

3.8.2 (Finite Lorentz boost) Verify Eq. (3.157). 3.8.3 (Finite Lorentz boost) Let β = v/c = βeβ be an arbitrary dimensionless velocity vector along the direction eβ = (c1 , c2 , c3 ) defined by the direction cosines ci = eβ · ei , i = 1, 2, 3, and M = Kβ = eβ · K. Verify the following results: (a) M is a 4 × 4 matrix with only six nonzero matrix elements in its first row and first column: M0i = Mi0 = ci , i = 1, 2, 3. (b) M 2 has only ten nonzero matrix elements where M has zero matrix elements: (M 2 )00 = 1, (M 2 )ij = (M 2 )ji = ci cj ,

i, j = 1–3.

(c) M 3 = M (by direct multiplication). Hence for a pure Lorentz boost along an arbitrary direction, the 4 × 4 Lorentz transformation matrix is B = e−ηM = I − M sinh η + M 2 (cosh η − 1). 3.8.4 (SU(2) representations) Show that the irreps that can appear for a system of n spin 1/2 particles are: n = 4 : ( j = 0)2 , (1)3 , (2); n = 5 : (1/2)5 , (3/2)4 , (5/2). The symbol (1/2)5 means that five distinct angular momentum states of j = 1/2 appear.

Cartan spinors and spin transformations

3.9

195

Cartan spinors and spin transformations in square-root space

´ Spinors were discovered by Elie Cartan in his study of Lie algebras of complex variables. We are concerned here with only two aspects of his spinors: the use of complex numbers in the construction of square roots of the four spacetime coordinates themselves when they are on the light cone, and the use of the complex 2D special linear group SL(2,C) to describe the Lorentz group. 3.9.1

Complex square-root coordinates on the light cone

Light-cone coordinates describe light rays emitted from the origin r = 0 of the coordinate system at time t = 0. These light rays reside on the light cone define by the condition t2 − r2 = t2 − (x2 + y2 + z2 ) = 0.

(3.168)

Because of this constraint, light-cone vectors are often called null vectors because of their zero length. The light cone has two parts: r = t for the future-pointing light cone, and r = −t for the past-pointing light cone. The direction of a light ray on the light cone can be specified by a dimensionless complex variable ζ=

x + iy t+z = . t−z x − iy

(3.169)

ζ is a projective or directional variable because it is unchanged when all spacetime coordinates are scaled by the same nonzero complex constant λ. Functions like this that are invariant under a uniform scale change of all its variables (t → λt, xi → λxi ) are also said to be homogeneous in mathematics. A projective variable is often called an homogeneous variable. It is useful to write ζ as the ratio of two complex numbers: ξ ζ= . η

(3.170)

ζ defines a complex projective line because η = ζ −1 ξ is the equation of a straight line of complex slope ζ −1 in the complex 2D space (ξ, η). The ratios contained in Eqs. (3.169) and (3.170) can be expanded into four linear equations involving the spacetime coordinates: x + iy = λξ = aξη∗ , t − z = λη = aηη∗ , t + z = μξ = aξξ∗ , x − iy = μη = aηξ ∗ .

(3.171)

We have chosen λ = aη∗ , μ = aξ ∗ , where a is a real constant, in order to satisfy Eq. (3.169) and the fact that the spacetime coordinates are real.

196

Relativistic square-root spaces

These linear equations are easily solved for the coordinates: a t = (ξξ ∗ + ηη∗ ), 2 a x = (ξη∗ + ηξ ∗ ), 2

a ∗ (ξξ − ηη∗ ), 2 a y = (ξη∗ − ηξ ∗ ). 2i z=

(3.172)

Neither the binary products ξξ∗ , ξη∗ , . . . nor the complex square-root coordinates ξ, η are projective coordinates independent of scale changes. Both the spacetime coordinates and the spinor coordinates involve four real variables. The mapping (3.172) is a one-to-two or twofold map that is ultimately responsible for the spinorial property that it takes two full turns in 3D space to restore a spinor vector to its original value. The 2D complex spinors   ξ , π† = (ξ∗ η∗ ) π= (3.173) η are the spinors studied by Cartan. They give access to the square-root (or spin) space. π has a spinor norm 2 π = π† π = ξ∗ ξ + η∗ η = t a

(3.174)

that is a real number. The word “norm” is used here in a nontraditional way for the squared quantity because this squared quantity can be negative as well as zero or positive. In contrast, the binary product   ξ ∗ ∗ X ≡ ππ† = (ξ η ) η  ∗  ξξ ξη∗ = ηξ∗ ηη∗   1 t + z x + iy (3.175) = a x − iy t − z is a complex Hermitian 2 × 2 matrix in square-root/spin space giving the 4-position on the light cone. The second to last expression has been obtained here by standard matrix multiplication of non-square matrices. In terms of Pauli matrices, X=

1 (t + σ · r) a

(3.176)

differs from the quaternion Q of Eq. (3.57) in that σ appears instead of iσT . Hence X is not a quaternion, and it does not satisfy quaternion algebra.

Cartan spinors and spin transformations

3.9.2

197

Spinor representation of spacetime coordinates and the Lorentz group

While ππ† is always on the light cone (see Problem 3.4.5), the 4-position matrices   t + z x + iy X≡ , (3.177) x − iy t − z using a = 1 for simplicity, are defined everywhere in spacetime. They satisfy the property that det X = t2 − (x2 + y2 + z2 ) = X

(3.178)

is nonzero outside the light cone. Here X = (t, x, y, z)T

(3.179)

is the usual spacetime vector written as a 4D column vector in a Euclidean notation. (We have changed our notation slightly from that used in preceding sections in order to keep up with the variety of notations in common use.) The vector norm is defined to be X ≡ XT gX,

(3.180)

where the diagonal signature matrix g = diag(1, −1, −1, −1) is the usual one containing the signs of the terms in the scalar product (3.178). On the light cone, det X vanishes. When det X > 0 (< 0), the t2 term dominates (spatial terms dominate). The resulting position matrix is said to be time-like (space-like). Lorentz transformations of the 4-vector can be expressed in terms of 4 × 4 matrices L X = LX.

(3.181)

As explained in the last section, the transformation matrices L (denoted A there) form a subgroup of the 4D orthogonal group O(1, 3), where the arguments 1, 3 refer to the number of positive and negative signs, respectively, appearing in the metric matrix g. Those Ls with unit determinant form a subgroup called the special orthogonal group SO(1, 3). The corresponding transformations are called proper Lorentz transformations. In common with other O(4) groups, matrices of both O(1, 3) and SO(1, 3) are characterized by n(n − 1)/2 = 6 real parameters when n = 4. Three of the parameters are the rotational parameters for the subgroup O(3) or SO(3) for rotations in 3D space. The three remaining parameters describe velocity boosts in three spatial directions. In the 2D spinor space, the situation is somewhat different. The 4-position matrix X has matrix elements Xij with two indices. Each index has to be transformed by a 2 × 2 transformation matrix A or A† . Hence Lorentz transformations of X take the two-sided form

198

Relativistic square-root spaces

X  = AXA† ,

(3.182)

det X  = (det X)(det A)(det A† ) = (det X)| det A|2 .

(3.183)

The most general form of A appearing here is the complex invertible (i.e., det A  0) matrix   α β A= . (3.184) γ δ Such matrices form a group, the 2D complex general linear group GL(2,C). If det A = ±1, then det X = X remains unchanged. The matrices with det A = 1 form a subgroup SL(2,C). Such matrices are called spin-matrices, and their associated transformations in spinor space are called spin transformations. Matrices of GL(2,C) are characterized by four complex, or eight real, parameters. The requirement det A = 1 for complex As contains two real conditions, thus reducing the number of parameters in SL(2,C) matrices to six. These parameter numbers agree with those for the groups O(1, 3) and SO(1, 3), respectively. Furthermore, if an SL(2,C) matrix A maps into an SO(1, 3) matrix L, the matrix −A will also map into the same L. Hence the mapping is two-to-one. SL(2,C) is thus said to give a twofold or double cover of SO(1,3). 3D rotations can be studied by considering 4-position matrices X at the same time t. If t = 1 is chosen, light rays on the light cone satisfy the condition det X = 0 = 1 − (x2 + y2 + z2 )

(3.185)

describing the unit sphere in 3D space (called a 2-sphere in mathematics). The proper Lorentz group SO(1, 3) then reduces to its subgroup SO(3) for 3D rotations, while the double covering group SL(2,C) simplifies to its subgroup SU(2), both with three real parameters. Experience with spinors on the light cone can be gained by doing Problem 3.9.2 where the SU(2) spin analog of the 3D rotation matrix for rotation by the Euler angles is derived.

Problems 3.9.1 (Light-cone vector) Show that any matrix of the form ππ† , where π is any complex 2D column vector, has zero determinant and cannot be used to describe a 4-vector off the light cone. 3.9.2∗ (Long problem on SU(2) ↔ SO(3) mapping) Consider the spin transformations of the light-cone spinors of Eq. (3.173):   ξ  π =  = Aπ, π† = (ξ∗ η∗ ). η

Cartan spinors and spin transformations

199

(a) The nonlinear map π → X defined by Eq. (3.171) generates the SO(3) transformation r = (x , y , z )T = Rr. Use this map (with a = 1 for simplicity) and Eq.(3.172) to verify the following: (i) The two-sided spin (or SU(2)) rotation  iφ/2  e 0 Az (φ) = ± 0 e−iφ/2 maps into the same x , y coordinates of the rotated vector generated by the SO(3) rotation ⎛ ⎞ ⎜⎜⎜cos φ − sin φ 0 ⎟⎟⎟ Rz (φ) = ⎜⎜⎜⎝ sin φ cos φ 0 ⎟⎟⎟⎠. 0 0 1 Note: This Rz (θ) has off-diagonal matrix elements with signs opposite to those of the Rz (θ) of Example 2.2.5 and Section 3.8. The latter rotation matrix is for the rotation of the coordinate axes while the vector remains fixed in space. In this problem, we consider the rotation of a vector while the coordinate frame remains fixed. The purpose of the exercise is to make the reader aware of the sign difference between these two kinds of rotations. (ii) The spin rotation   cos θ/2 − sin θ/2 Ay (θ) = ± sin θ/2 cos θ/2 maps into the same SO(3) rotation Ry (θ) derivable from the related rotation matrix of Example 2.2.5. (iii) (Optional) The spin rotation   cos χ/2 i sin χ/2 A x (χ) = ± i sin χ/2 cos χ/2 maps into the same SO(3) rotation R x (χ). (b) (Successive transformations) Verify that A(φ, θ, ψ) = Az (ψ)Ay (θ)Az (φ)   cos 2θ ei(ψ+φ)/2 − sin 2θ ei(ψ−φ)/2 . =± sin 2θ e−i(ψ−φ)/2 cos 2θ e−i(ψ+φ)/2

200

Relativistic square-root spaces

Show formally that they map into the SO(3) rotation R(φ, θ, ψ) = Rz (ψ)Ry (θ)Rz (φ). (Do not calculate the matrix elements of R(φ, θ, ψ).) Note: (i) SU(2) matrices were used to describe rotations of rigid bodies in classical mechanics before spinors were discovered. The complex matrix elements of A(φ, θ, ψ) are called Cayley–Klein parameters in classical mechanics. (ii) In this problem and in Penrose and Rindler (I, 1984, p.20), the spinor or vector itself is rotated. In Sections 2.2, 3.8 and in Goldstein (I, 1950, p. 116), the coordinate axes are rotated instead.

3.10

Dyadics

Dyadics are bivectors (two vectors placed side by side) that can handle matrix multiplications as well as the vector operations of vector analysis. They can be very useful when these two types of operations are needed together. The basic limitation of dyadics is that they cannot be extended easily to more than two vectors. A dyad is the combination (or direct product) ab of two vectors a and b of the same dimension placed side by side. A dyadic  is a sum of dyads. However, if the vectors are expanded into components a = i ai ei in a coordinate system taken (for simplicity) to be orthogonal ei · ej = δij , then ab =



ai bj ei ej

(3.186)

i,j

is also a dyadic. Even e1 e1 when expanded in another coordinate system is a dyadic. So we can use the term dyadic to cover all cases. If c is another vector with the same dimension, the following scalar and vector products with the dyadic ab can be defined: c · ab = (c · a)b, c × ab = (c × a)b.

(3.187)

The scalar product here is a vector along b, while the vector product is a dyadic. If c is a vector or the differential operator ∇, ab · c = a(b · c), ab × c = a(b × c).

(3.188)

Dyadics

201

If a and b are vector fields, ∇ · ab = (∇ · a)b + (a · ∇)b, ∇ × ab = (∇ × a)b − (a × ∇)b.

(3.189)

In the above operations, we treat the left and right side separately in the manner of vector algebra. There are other operations unique to dyadic such as double scalar and vector products between two dyadics. See Chen (1983). In matrix representation, the dyadic A = ab may be written as a square matrix A: ab = A = A = (Aij ) = (ai bj ) = (ai )(bj )T ,

(3.190)

where (ai ) is a column vector. The two vector products between the vectors in the same dyadic ab are  Aii , a · b = |A| = Tr A = a × b = A =



i

ei εijk Ajk ,

(3.191)

i, j,k

where εijk is the permutation (Levi–Civita) symbol of three indices. |A| is called the expansion factor of the dyadic A, and A its rotation factor. A dyadic can still be defined when the factorized ab form is unavailable:  A= ei Aij ej . (3.192) i, j

The dyadic components Aij are the matrix elements Aij . The dyadic A is thus the matrix A in a dyadic notation. The dyadic (3.192) can be rewritten two other ways:   Aj ej = ei ATi (3.193) A= j

where Aj =



ei Aij ,

i

ATi =



i

Aij ej

(3.194)

j

are column and row vectors, respectively. A transposed or conjugate dyadic is defined as AT = (ab)T = ba.

(3.195)

It can be obtained from A by transposing either the dyads ei ej → ej ei or the dyadic/matrix components Aij → Aji . All matrix-like operations can be performed on the associated matrices with the final expression translated back to the dyadic

202

Relativistic square-root spaces

notation, as we shall see. Matrices are older and more widely used than dyadics. Matrices were invented by Sylvester in 1850, and dyadics by Gibbs in 1884. The unit dyadic is I = e1 e1 + e2 e2 + e3 e3 .

(3.196)

It leaves any vector a unchanged on scalar multiplications: a · I = I · a = a.

(3.197)

Vector multiplications with any vector a (which can always be written as ae1 ) give a dyadic that is orthogonal to a on both left and right: a × I = a(e3 e2 − e2 e3 ) = I × a.

(3.198)

The use of dyadics can give a nice intuitive picture of the dynamics of certain physical problems. Consider, for example, the deformations or strains produced in an elastic body or medium by forces or stresses acting on it. We follow the description given by Morse and Feshbach (1957, pp. 70–2): An arbitrarily oriented differential vector surface element in the medium has the general form dA = e x dydz + ey dzdx + ez dxdy  ei dAi . =

(3.199)

i

Let the stress vectors Ti , i = 1, 2, 3, or x, y, z, be so defined that the force acting on the partial surface element dAi = ei dAi is dFi = Ti dxj dxk , ijk = cyclic. Equivalently, ΔFi . ΔAi →0 ΔAi

Ti ≡ lim

(3.200)

These stress vectors have the dimension of pressure. Note that T i does not have to be parallel to ei . A fluid in which T i is always parallel to ei is said to be a normal fluid. The total force acting on dA is thus   dF = dFi = Ti dAi . (3.201) i

i

It can be written neatly as dF = T · dA,

where

T = T1 e1 + T2 e2 + T3 e3  = Ti ei i

is called the stress dyadic.

(3.202)

Dyadics

203

3 T33

T32

T31

T23

T13 T22 T21

T12

T11

2

O

1 Fig. 3.4 The nine-component stress dyadic or tensor T = {T ij }. Its diagonal elements are normal stresses that generate no torque. Its off-diagonal elements are shear stresses.

The stress component T 12 = T1 · e2 is that component of T1 acting on the vector surface element dA1 = e1 dx2 dx3 that points along e2 . Similarly, T 21 = T2 · e1 acts on the surface element dA2 = e2 dx3 dx1 and points along e1 . These stresses are shown in Fig. 3.4. They generate rotational torques about the coordinate origin O parallel to the 3-axis that is perpendicular to the 12-plane of the stresses T12 and T21 : τ3 = dx1 (T 12 dx2 dx3 ) − dx2 (T 21 dx3 dx1 )

(3.203)

are directed along the 3-axis by the right-hand rule. When dV of the medium is in rotational equilibrium, the total torque τ3 must vanish, giving T 12 = T 21 .

(3.204)

Similar considerations apply to other pairs of stresses. Hence the equilibrium stress matrix or dyadic T is symmetric:  T =TT = ei Ti . (3.205) i

The stress matrix T is also real, and hence Hermitian. It has real eigenvalues Pi , i = 1, 2, 3, called its pressures or principal stresses, and orthonormal eigenvectors aˆ i . Thus the stress dyadic has the diagonal, principal-axis representation T = P1 aˆ 1 aˆ 1 + P2 aˆ 2 aˆ 2 + P3 aˆ 3 aˆ 3  = Pi aˆ i aˆ i . i

(3.206)

204

Relativistic square-root spaces

Different physical situations can then be recognized readily. For example, if P1 = P2 = P3 , the stress is isotropic, just like the pressure in a normal fluid. If P2 = P3 = 0, there is tension along aˆ 1 if P1 > 0 and compression if P1 < 0. The stress P3 = 0 and P2 = −P1  0 is a shearing stress perpendicular to aˆ 3 . The differential volume dV of a rigid body will respond to imposed stresses by undergoing translation and rotation. If the medium is elastic and isotropic, any line element in it will experience the change dr = dr · (I + D),

(3.207)

where the dimensionless deformation dyadic D = A +  can be separated into two parts: (a) an antisymmetric part A that describes an additional rotation of dV caused by the elastic deformation (see Problem 3.10.2), and (b) a symmetric part  called the pure strain, with dyadic elements 1 Aij = (Dij − Dji ), 2

1 ij = (Dij + Dji ). 2

(3.208)

For small deformations of an elastic medium, Hooke’s law states that stress and strain are linearly related. Since both stress and strain are dyadics, their linear relationship is described by a super-matrix called the elasticity tensor  T ij = cijk k , (3.209) k

where c may be considered a 9 × 9 matrix with the super indices ij and k. Like any symmetric dyadic considered as a matrix, pure strain can be expressed as a sum of two parts =

1 (Tr )I + Δ. 3

(3.210)

The first term is a constant dyadic describing the spherical strain. The second term Δ is traceless, and describes non-spherical strain. It is customary to write Hooke’s law for an isotropic elastic medium in the slightly different form T = λ(Tr )I + 2μ,

(3.211)

where λ, μ are real (Lam´e) elastic constants. Hooke’s law (3.211) for an isotropic elastic medium shows that the principal axes of T and  are the same. The eigenvalues e1 , e2 , e3 of the pure strain  = e1 aˆ 1 aˆ 1 + e2 aˆ 2 aˆ 2 + e3 aˆ 3 aˆ 3 ,

(3.212)

are called the principal extensions of the medium. The connection between the Lam´e elastic constants and the usual bulk modulus, Young’s modulus and Poisson ratio are clarified in Problem 3.10.2.

Dyadics

205

Problems 3.10.1∗ (Unit dyadic) (a) Verify Eq. (3.198). Show that (a × I ) · a = a · (a × I ) = 0. (b) Show that more generally a×I =−



εijk ai ej ek .

i, j,k

Use this general expression to show that (a × I ) · (b × I ) = ab − a · bI . 3.10.2∗ (Elastic moduli) An isotropic elastic medium satisfies Hooke’s law (3.211) that can be rewritten as T = K||I + 2GΔ,

|| = Tr .

(a) Show that K = λ + (2/3)μ and G = μ. Show that K is the bulk modulus usually defined as K≡−

P . dV/V

Hint: Under an external isotropic compression T = −PI , with P > 0, the pure strain is also isotropic,  = eI with e < 0. The isotropic stress causes the radius r of a small sphere of the medium to change by Δr = er. The volume of the sphere then changes by ΔV = 3eV. (b) If the stress is the 1D tension T = T e x e x , show that ejk = ejj δjk , (λ + μ)T , μ(3λ + 2μ) λT . ey = ez = − 2μ(3λ + 2μ)

e x ≡ exx =

The fractional stretching along e x is Δx/x = e x is related to Young’s modulus Y, while the ratio of lateral shrinkage to longitudinal stretch is called the Poisson’s ratio ν. Show that Y≡

μ(3λ + 2μ) T = , Δx/x λ+μ

ν≡

|ey | λ = . ex 2(λ + μ)

206

Relativistic square-root spaces

(c) A simple shear is caused by the stress dyadic T = T (e x e x − ey ey ). Use the results of part (b) to show that the resulting principal extensions are e x = −ey =

3.11

T ; 2μ

ez = 0.

Cartesian tensors

Cartesian tensors of dimension N and rank r are multivectors consisting of r ND vectors placed side by side. The word Cartesian refers to the fact that the vector components are defined by a Cartesian coordinate system with the same orthogonal coordinate axes ei , i = 1–N, everywhere in the ND dimensional space. They are thus simpler than more general tensors whose components are defined in a curvilinear coordinate system with coordinate axes that change in direction and scale from point to point in space. These general tensors, needed in Einstein’s theory of general relativity, are the subject of the next section. In this section we are interested in two related aspects of Cartesian tensors that are conveniently described in related but not identical mathematical languages: • tensors as elements of a tensor algebra that deals directly with the multiple vector indices appearing in these multivectors, and • tensors as group representations, both in the wave functions or states in quantum mechanics and in the quantum operators acting on these wave functions. 3.11.1

3D Cartesian tensors

3D tensors are of special interest because physical space is 3D. Consequently, we are almost always dealing with 3D tensors in nonrelativistic physics. It is thus worthwhile to begin with 3D tensors which are among the simplest and most useful of tensors. Let us first recall that a 3D vector V can be expressed as an expansion V=

3 

Vj ej ≡ Vj ej .

(3.213)

j

in an orthonormal basis {ej }, j = 1, 2, 3. The orthonormal basis vectors form a righthanded coordinate system: ei · ej = δij , ei × ej · ek = εijk .

(3.214)

The symbols that appear are the Kronecker δ and the Levi–Civita permutation symbols, respectively, for this basis. In the last expression of Eq. (3.213), we have used the Einstein summation notation where an automatic summation is required when the same index (here j) appears twice in the same term of an expression.

Cartesian tensors

207

In Section 2.2 we showed how the Cartesian components Vi = ei · V of V change when coordinate axes are rotated to the new basis {ei }. The new Cartesian components are Vi = ei · V = ei · ej Vj = λij Vj ,

(3.215)

where the orthogonal transformation matrix λ contains the direction cosines λij = ei · ej .

(3.216)

In fact, the vector V can be defined by the transformation Eq. (3.215) of its components under a rotation of the coordinate axes instead of the expansion given in Eq. (3.213) in terms of basis vectors . In analogy, a Cartesian tensor of rank r containing r vector indices can be defined by an expansion in terms of any set of basis vectors as T = T m...n e em . . . en .

(3.217)

It can also be defined in terms of the transformation property under a rotation of the coordinate axes of the tensor components T m...n = (e em . . . en ) · T :  = (ei ej . . . ek ) · T T ij···k

= λi λjm · · · λln T m···n ,

(3.218)

after using the multiple scalar product (ei ej . . . ek ) · (e em . . . en ) ≡ (ei · e ) . . . (ek · en ).

(3.219)

Tensor algebra

As generalizations of vectors and matrices, tensors can be manipulated in similar ways for tensors of the same dimension N. For example, if c is a scalar, then cT has tensor components cT m···n . Two tensors of the same dimension and rank can be added (or subtracted): Aij···k + Bij···k = Cij···k .

(3.220)

The vector algebra of scalar products is realized through the use of δ symbols. Vector products too take on the familiar form for tensor dimension N = 3 by using the 3D permutation symbols of Eq. (3.214). For N > 3, less familiar ND permutation symbols appear. Tensor algebra goes far beyond vector and matrix algebras in its ability to handle indices and products. One of its powerful tools is the contraction of two identical indices, say j, no matter where they appear in the same term of an expression, whether in the same tensor or in two different tensors multiplied together. Then according to the Einstein summation convention, the two identical indices should be summed over. Summed indices differ from unsummed indices because the former do not contribute

208

Relativistic square-root spaces

to the rank of the tensor/tensors involved. So the rank of the term is reduced by 2 for each contraction. For example, T jj = Tr T is the contraction of a rank-2 tensor T or the trace of its matrix alter ego. Both are scalars of rank 0. If a and b are ND vectors, then ai bj is a rank-2 tensor, while aj bj = a · b is the rank-0 scalar product of the two vectors. The matrix product Aij Bjk is a matrix or rank-2 tensor, but for tensors the repeated index j can appear anywhere. The results are different rank-2 tensors. Of course, the power of tensor, its raison d’ˆetre, is the possibility of having 3 or more indices. If Cij···k is a tensor of rank r with r indices, the once contracted tensor C( .. j.. j . . .) has rank r − 2 no matter where the common indices j appear. The same result applies to a direct product of two tensors of the same dimension but arbitrary ranks r1 and r2 , respectively: Aij···k Bm···n = Cij···km···n .

(3.221)

If all the indices of A and B are independent, C is a tensor or rank r = r1 + r2 . If B contains one of the indices of A anywhere, the repeated index will have to be summed over and removed from the rank count. Hence the contracted product has rank r − 2. Thus contraction generalizes matrix multiplication with respect to both rank of the tensors and the positions of the summed indices. Repeated contractions reduce any tensor of rank r to either a vector or a scalar. The last vector left standing can contract with itself to give its own scalar product. This great power of tensor products is magnified by the fact that the direct product of two tensors of different dimensions is also allowed. In fact, we have used them before in studying the representations of SU(2) in Section 3.8.1. The multispinors studied there are spinorial tensors of rank r and dimension 2. The spin addition rule (3.165) for constructing the irreps of two product reps ( j< ) and ( j> ) of generally different dimensions d = 2 j + 1 describes in broad strokes how the tensor product involving tensors of the same or different dimensions can be separated into independent irreps of different dimensions. Each irrep ( j) is actually a vector of one index i denoting the dj = 2 j + 1 components of the irrep. Thus a direct product of two tensors of arbitrary dimensions and ranks can be reduced into a sum of vectors of different dimensions. In group theory, the reduced irreps contained in a tensor of rank r are sometimes called its tensor representations. We shall discuss the tensor representations of ND tensors (SO(N) multivectors) later in this section. Permutation tensors

Returning to Eq. (3.214), we note that each δ or permutation symbol is a scalar product, but it is also a tensor component. The δ tensor, of rank 2, is just the unit matrix I. The permutation tensor is of rank 3. Both tensors have the same values for their components in any coordinate system that is rotated relative to the original frame used in Eq. (3.214). They are thus isotropic tensors that look the same in any direction. Both tensors are basis tensors that describe the properties of the chosen basis {ei }. They have no other physical content.

Cartesian tensors

209

Recall that the only nonzero elements of the permutation tensor are ε123 = ε231 = ε312 = 1 for all even permutations of 123, and ε132 = ε213 = ε321 = −1 for all odd permutations. An odd permutation is one requiring an odd number of transpositions of any two indices to achieve the required permutation from the standard order 123 of three distinct objects. It then follows that two permutation tensors can be contracted to simpler expressions: εijk εijn = 2δkn , εijk εimn = δjm δkn − δjn δkm .

(3.222)

A simple product of two symbols is nonzero only if both are nonzero: εijk εmn

   δi δim δin  =  δj δjm δjn .  δ δ δ  k km kn

(3.223)

The determinant on the right contains all six permutations of lmn: det A = |A| =



εlmn Ail Ajm Akn .

(3.224)

lmn

The result is easily extended to N D space where each permutation symbol carries N indices. Each symbol has value 0 unless the indices are all distinct, 1 if the indices is an even permutation of a standard ordering of these indices, say 12 . . . N, and −1 for odd permutations. The resulting permutation symbol is a rank-N Cartesian tensor. Then εij...k εlm...n is also expressible as a determinant like Eq. (3.223) but of dimension N. When N = 3, the vector product A × B = εijk ei Aj Bk

(3.225)

is a vector because (a) it contains one unit vector ei , or (b) the tensor εijk Aj Bk is rank 1. Parity transformation and pseudotensors

In a 3D space, it is useful to include among the physically interesting coordinate transformations λ the space inversion of all three coordinate axes ei → ei = −ei , i = 1, 2, 3. Under this parity transformation, a right-handed coordinate system becomes left-handed. The parity transformation matrix is thus λP = −I = λ−1 P , with det λP = −1. As a result, the new components of the same vector fixed in space is Vi = −Vi . That is, V though fixed in space changes sign when decomposed into the new basis. Such vectors are called true or polar vectors.

210

Relativistic square-root spaces

Under the parity transformation, the vector algebraic operations involving three polar vectors A, B and C transform as follows: A · B → A · B = A · B, A × B → A × B = A × B, A × B · C → A × B · C = −A × B · C.

(3.226)

Note in particular that the vector product A × B remains unchanged under the parity transformation. Such a vector does not remain fixed in space like true vectors but actually changes its direction in space so that its coordinates in the new left-handed basis are the same as those of the original vector in the old right-handed basis. Such a behavior can be attributed to an additional attribute called an intrinsic parity that is odd under a parity transformation. It is no longer a polar vector, and is called an axial vector or a pseudovector instead. The odd intrinsic parity resides in the cross product sign ×, that is, the permutation tensor εijk is a pseudotensor. The triple scalar product A × B · C, the scalar product of a pseudovector with a true vector, is a pseudoscalar. This description refers to its sign change under the parity transformation caused by the intrinsic odd parity of the pseudovector. It is worth repeating that the vector product A × B is a vector and not a rank 2 tensor, because it is left with only one set of vector components, not two. Similarly, the triple scalar product A × B · C is a scalar because there is no component index left unsummed. These considerations lead to the definition of a general pseudotensor P of rank r as one satisfying the more general coordinate transformation λ Pi j···k = sgn(det λ)λi λjm . . . λkn Pm···n ,

(3.227)

where det λ = ±1 and sgn(det λ) = det λ/| det λ|. For example, a pseuodscalar is invariant under rotations because det λR = 1, but changes sign under parity because det λP = −1. It is worth pointing out that the correct ND generalization of the vector product is the ND vector product A(A . . . B) ≡ εij···k ei Aj . . . Bk ,

(3.228)

where εij···k is the permutation symbol of N indices and A, . . . , B are the N − 1 vectors in the product. This vector is just the ND determinant that generalizes the 3D vector product A × B written as a 3D determinant. The ND scalar product D · A(A . . . B) ≡ εij···k Di Aj . . . Bk ,

(3.229)

is similarly the ND determinant which generalizes the 3D determinant that is the triple scalar product. These determinants change sign when any two vectors are interchanged. The parity transformation matrix is λP = −I, with det λP = −1 if the space dimension N is odd. To get the same sign for the determinant when N is even, the parity

Cartesian tensors

211

operation is redefined as the spatial reflection involving an odd number of spatial coordinates. This means that when N is even, an odd number of coordinates are left unchanged under a parity transformation. Such partial spatial reflections of an odd number of coordinates are also allowed when N is odd. All such parity transformations are discrete transformations that cannot be reached from the original coordinate system by continuous coordinate transformations. Parity and time-reversal symmetries

The laws of classical physics remain unchanged under parity r → −r in the physical 3D space and under time reversal t → −t. This invariance can be satisfied if the physical quantities appearing in physical laws have well-defined behavior under these symmetry transformations. Quantities like r that change sign under parity are said to have odd parity. Those that remain unchanged have even parity. Quantities that change sign under time reversal, such as v = dr/dt are time-odd. Those that remain unchanged are time-even. A tensor of rank r in 3D space is a true tensor if it transforms under parity with a sign (−1)r . If the sign change is (−1)r+1 , it is a pseudotensor. Tensor fields

The tensors we have described so far are properties associated with a point particle in space (or more generally in spacetime), such as the position of a point mass in space. There are in addition mathematical objects called fields that exist everywhere in space rather than localized at only one point. For example, the gravitational potential field of the Sun is present everywhere in space so that a distant mass can still feel its gravitational potential. This potential field is a scalar field Φ(x, y, z), so defined that a number Φ(x, y, z) is assigned to every point (x, y, z) in space. Such extended fields behave under rotation as tensor of well-defined ranks. A tensor field T (x, y, z) is made up of the totality of tensors one at each location in space. The rank of a tensor field is then the rank of any one of these tensors, say the one located at (x, y, z), as the coordinate system is rotated while both the tensor field and the chosen representative tensor at (x, y, z) remain fixed in space. 3.11.2

Tensor representations

The tensor components T m··· can be written in multivector form as a direct product of vectors in column form. The multispinors of dimension N = 2 introduced in Section 3.8.1 are among the simplest multivectors. Yet the SU(2) irreps discussed there contain all the irreps appearing in tensors of any dimension. This is to be expected, because the state of any spin j can be constructed from a suitable number of spin 1/2 particles by the algebra of spin addition. However, for multivectors of ordinary or bosonic vectors in ordinary ND space (i.e., vectors of the SO(N) group), the correspondence between SU(2) irreps and SO(N) irreps are not exact because they can be fermionic in SU(2) but always bosonic in SO(N). The translation between these two types of irreps is not hard to find, and will be left to a problem.

212

Relativistic square-root spaces

For odd N dimensions, both the SU(2) and SO(N) irreps are bosonic, and have identical properties. The 3D case is so important that it is worthwhile to enumerate its irreps for tensors of low rank r. The fundamental representation of SO(3) is made up of the (vector components associated with the) three Cartesian unit vectors ei , i = 1, 2, 3. It is more convenient to use the spherical or angular-momentum basis instead: ⎛ ⎞ ⎜⎜⎜ 0 ⎟⎟⎟ | j = 1, m = 0 = e3 = ⎜⎜⎝⎜ 1 ⎟⎟⎠⎟, 0 1 | j = 1, m = ±1 = √ (e1 ± ie2 ) 2 ⎛ ⎞ ⎛ ⎞ ⎜⎜⎜ 1 ⎟⎟⎟ ⎜⎜⎜ 0 ⎟⎟⎟ = ⎜⎜⎜⎝ 0 ⎟⎟⎟⎠, ⎜⎜⎜⎝ 0 ⎟⎟⎟⎠. 0 1

(3.230)

These are the three states of the fundamental rep ( j = 1) of dimension N = 2 j + 1 = 3. The basic sequential construction of tensor irreps of SO(3) is then based on the usual reduction of the direct product of two spin irreps ( j< )( j> ) given by Eq. (3.165): ( j< )( j> ) =

j< 

( j> + m) :

m=− j<

r=2:

(1)2 = (0), (1), (2);

r=3:

(1)3 = (1)[(0), (1), (2)] = (0), (1)3 , (2)2 , (3).

(3.231)

Here the subscript in (1)3 denotes the number of times the irrep appears, and commas are used in stead of + signs, all for the sake of notational clarity. One can easily check that the total dimension 3r of the tensor is reproduced by the irreps on the right. Another general property is worth noting. A rank 2 tensor, T ij , is nothing but a matrix. It can be separated into antisymmetric and symmetric parts 1 Aij = (T ij − T ji ), 2

1 S ij = (T ij + T ji ). 2

(3.232)

For N = 3, these matrices have three and six matrix elements, respectively. The symmetric part S can further be separated into isotropic and traceless parts Tr T δij + S ij(2) , 3 Tr T = S ij − δij , 3

S ij = S ij(2)

(3.233)

Cartesian tensors

213

of one and five elements, respectively. An isotropic tensor such as δij is one with components that have the same values in all rotated frames. All three separated parts remain matrices, or rank-2 Cartesian tensors. On the other hand, one can define another tensor rank, called the spherical tensor rank or equivalently the integer spin j, by how the tensor components transform under rotation. For example, the unit matrix has the same value of 1 in any rotated coordinate system. It can therefore be said to be a scalar, or a spherical tensor of rank 0, an object with spin j = 0, or an object that has only one component. The antisymmetric part Aij has three components like a vector or tensor of rank 1. In both cases, the rotational rank is smaller than the Cartesian rank. Such Cartesian tensors are then said to be reducible to spherical tensors of lower ranks. The remaining five components form the irreducible spherical tensor S (2) of rank or j = 2 (shown as a superscript) that has been reduced from the original nine components of the Cartesian tensor T ij or the six components of the symmetric Cartesian tensor S ij . In this way, we have derived the r = 2 irreps of Eq. (3.231) using symmetry arguments instead of spin addition algebra. The symmetry used is the permutation symmetry of two indices. All permutations of N indices or objects form a group, the permutation or symmetric group S N . All group representations have definite permutation symmetries, but the theory of permutation symmetry is much more complicated than the simple reduction formula (3.165) for spin additions. Spherical harmonics

The prototypes of spherical tensors of rank or integer spin  are the spherical harmonics Ym (θ, φ) of degree . They are functions of the angles θ and φ of spherical coordinates. They form a complete basis in terms of which any angular function f (θ, φ) =

 ∞  

fm Ym (θ, φ)

(3.234)

=0 m=−

on the unit sphere can be expanded. They satisfy the following properties: (a) Their degree  = 0, 1, 2, . . . is a nonnegative integer, while the order m is restricted to the 2 + 1 values − ≤ m ≤  for each . The d = 2 + 1 components Ym of the same degree  form an irrep () of dimension or multiplicity d. In quantum mechanics, the d states of () form a multiplet of states or particles. The spherical harmonics are constructed to satisfy the orthonormality condition  π  2π ∗ sin θdθ dφYm (θ, φ)Y m (θ, φ) = δ δmm . (3.235) 0

0

As orthonormal functions, they form an orthonormal basis in the angular function space of dimension 2 + 1 for each degree .

214

Relativistic square-root spaces

Table 3.3 Spherical harmonics of the first three degrees. The normalization constant is C  = √ (2 + 1)/4π, where 2 + 1 is the dimension d of the spherical tensor, and 4π is the total solid angle of the sphere.



m

0 0 1 0 1 ±1 2 0 1 ±1 1 ±2

Ym ( xˆi )/C

Ym (θ, φ)/C

1 xˆ3

cos θ

1

√ ∓( xˆ1 ± i xˆ2 )/ 2

√ ∓ sin θe±iφ / 2

(3ˆz2 − 1)/2 % ∓ 32 xˆ3 ( xˆ1 ± i xˆ2 ) % 3 2 8 ( xˆ1 ± i xˆ2 )

(3 cos2 θ − 1)/2 % ∓ 32 cos θ sin θe±iφ % 3 2 ±2iφ 8 sin θe

(b) They are th degree polynomials of the components  cos φ xˆ3 = cos θ, xˆ1,2 = sin θ sin φ

(3.236)

of the unit vector er = rˆ . A more convenient spherical basis, xˆ0 = xˆ1 ,

xˆ± = xˆ1 ± i xˆ2 = sin θe±iφ ,

(3.237)

is used for calculations in spherical coordinates. (c) They form the angular part of quantum mechanical wave functions of a state of definite angular momentum L = r × p, where p is the momentum of the state. In this state, L2 has the value ( + 1)2 and Lz the value m, where  = h/2π is the reduced Plank constant. The first three spherical harmonics are given in Table 3.3. The  = 1 spherical harmonics of first degree are (up to a normalization) just the fundamental rep (1) = ({ xˆi }) of S O(3). Those of degrees 0 and 2 are both contained in the symmetric rank-2 Cartesian tensor S ij = xˆi xˆj . Its trace is S ii = xˆ1 xˆ1 + xˆ2 xˆ2 + xˆ3 xˆ3 = 1. The remaining five components make up the  = 2 spherical harmonics Y2m of irrep (2). Note that the order m is just the integer appearing in the φ-dependent factor eimφ . Its integer value ensures that the φ dependence is modulo 2π. In Section 5.6 we shall return to spherical harmonics as polynomial solutions of certain differential equations and to the orthonormality condition (3.235).

Problems 3.11.1 (Triple scalar product) (a) Verify Eqs. (3.222) and (3.223).

Cartesian tensors

(b) Use Eq. (3.223) to derive the identity   A · D A · E A · F (A × B · C)(D × E · F) =  B · D B · E B · F  C · D C · E C · F

  . 

215

(3.238)

3.11.2 (δ and permutation tensors) (a) Verify that the δ and permutation symbols defined by the scalar products of Eq. (3.214) are isotropic Cartesian tensors of rank 2 and 3, respectively. Hint: Isotropic tensors have the same components in a rotated frame. Under a rotation of the coordinate axes, ei · ej transforms into ei · ej . (b) Show that Tr δ is a tensor of rank 0. That is, δii = δii . Show that δ − (Tr δ/3)I is a null tensor of rank 2. (c) Write ε1jk , ε2jk , ε3jk explicitly as numerical 3 × 3 matrices. 3.11.3 (SO(2) irreps) Verify that 2D rotations of SO(2) tensors (not the coordinate axes) satisfy the following properties: (a) The rotation matrices are   cos ϕ − sin ϕ R(ϕ) = = e−iϕJ , sin ϕ cos ϕ   0 −i J= , J 2 = 1. (3.239) i 0 These rotations are Abelian: R(ϕ2 )R(ϕ1 ) = R(ϕ1 )R(ϕ2 ). R(ϕ2 ) is modulo 2π. So the irreps are all bosonic. All spins are integers by the spin-statistics theorem. Note: We use e−iϕJ for the rotation of vectors and eiϕJ for the rotation of the coordinate axes so that the same infinitesimal generator J appears in both. (b) The fundamental rep contains just the two eigenvectors of J of eigenvalues m = ±1:     1 1 1 i φ1 = √ , φ−1 = √ . 2 i 2 1 Its rep dimension is d = 2. So SO(2) irreps are identical to SU(2) irreps in structure. This means that SU(2) irreps ( j) can be used in SO(2) if they are relabeled with integer J = 2 j, as shown in the following example: SU(2) : (1/2)3 = (1/2)2 , (3/2) → SO(2) : (1)3 = (1)2 , (3).

216

Relativistic square-root spaces

Table 3.4 Transformation properties of particle and field quantities under rotation, space inversion (parity) and time reversal (TR) of the spacetime coordinate axes. The symmetry under transformations of space inversion r → −r or time reversal t → −t is traditionally described as even (odd) if the quantity remains unchanged (changes sign) under the transformation. A tensor of rank r is a true tensor if its parity is (−1)r and a pseudotensor if its parity is (−1)r+1 . Thus a pseudovector has even parity. After Jackson (1999, p. 271).

Quantity

Rank

Parity

TR

Particle Position Velocity Momentum Ang mom Force Torque Energies

r v = dr/dt p = mv r×p F = dp/dt r×F 2 p /2m, Φ

1 1 1 1 1 1 0

Odd Odd Odd Even Odd Even Even

Even Odd Odd Odd Even Even Even

Field Potential Elec field Vector pot Magn ind

Φ E A B

0 1 1 1

Even Odd Odd Even

Even Even Odd Odd

The important constraint is that the dimension of the SO(2) irrep (J) remains d = 2 j + 1 = J + 1. Find the SO(2) irreps for tensors of rank r = 4. 3.11.4 (SO(3) irreps) Verify Eq. (3.231) and find the irreps for tensors of rank r = 4. Show that these irreps span the same dimension of the tensors. 3.11.5 (3D reducible tensor) Consider the Cartesian tensor T ij = ai bj , where ai , bj are the components of the polar vectors a and b. Show that it can be written in the form T=

a·b I + A + S (2) , 3

where

(i) (a · b/3)I is an isotropic spherical tensor of rank 0, (ii) Aij = (1/2)εijk (a × b)k is a component of a spherical antisymmetic pseudotensor of rank 1, and (iii) S (2) = S − (a · b/3)I a symmetric spherical tensor of rank 2. 3.11.6 (Transformations of particle and field quantities) Verify the tensor rank, parity and time reversal (TR) symmetries of the particle and field quantities shown in Table 3.4. The electric field E and the magnetic induction field B are related to the electromagnetic scalar potential Φ and vector potential A = (A1 , A2 , A3 ) as

Tensor analysis

E = −∇Φ −

217

∂A , ∂t

B = ∇ × A. 3.11.7 (Orthonormality of spherical harmonics) Verify the normalization √ C = (2 + 1)/4π and the orthogonality of the spherical harmonics of degree  = 0, 1.

3.12

Tensor analysis

The Cartesian tensors of the last section are defined with components specified in a Cartesian coordinate system. If the coordinate system used is not only curvilinear but also nonorthogonal in general, a vector needs to be decomposed in two distinct but related “biorthogonal” bases { i } and { i }, as we shall explain in this section. A tensor of rank r that results then takes the more general form j1 jm n T = T ij11···i ··· jm  i1 . . .  in  . . . 

(3.240)

that has n (≤ r) upper or contravariant indices and m = r − n lower   or covariant indices. Such a tensor is said to be of type or valence (n, m) or mn . We shall see that a lower index describes a component that transforms like a rotated coordinate system. An upper index describes a component that transforms like a component of a rotated vector. Simple types of tensors have special names: scalars are type (0,0), vectors are type (1,0), bivectors are type (2,0), covectors are type (0,1), scalar products are of type (1,1), and metric tensors are type (0,2). The objective for this section is twofold: • Show how the biorthogonal bases are defined and used in tensor algebra and how the use of curvilinear coordinates leads naturally to the concept of local coordinate transformations, the forerunner of the transformations of local inertial frames in general relativity. • Show how the curliness of curvilinear coordinates affects vector differential operations on tensor fields. The many basic ideas of tensor analysis, including Christoffel symbols, parallel transport and Riemann tensors, are described in the problems instead of the text, because they require a certain amount of effort for the reader to derive. The casual reader can still find out what these ideas are by reading the problems without answering the questions. 3.12.1

Tensor algebra

We shall first explain the biorthogonal representation of vectors in a nonorthogonal coordinate system, then discuss the idea of local coordinate transformation, and

218

Relativistic square-root spaces

finally summarize the differences between general tensor algebra and the algebra of Cartesian tensors. Nonorthogonal basis in 3D space

A nonorthogonal basis for a 3D space is made up of any three non-coplanar and not necessarily normalized covectors  1 ,  2 ,  3 : dr = dx1  1 + dx2  2 + dx3  3 = dxi  i = dxj   j , i =

∂r , ∂xi

j =

where

∂r . ∂xj

(3.241)

We have given dr in two different choices of coordinates to emphasize the fact that dr is an abstract object that is independent of its representation, i.e., the choice of coordinates. If these basis vectors change from point to point in space, the coordinate system is said to be curvilinear. Note that the formula for the contravariant vector dr = (dxi ) i contains as usual an implicit sum over the repeated indices i. For curvilinear coordinates, Eq. (3.241) is not the same as d(xi  i ) = (dxi ) i + xi d i . To avoid confusion, we must agree not to use the latter expression for dr when the coordinates are curvilinear. The scalar products  i ·  j = gij ,  i ×  j ·  k = eijk

(3.242)

are elements of the metric tensor and the permutation tensor, respectively. The chosen basis ordering 123 is right-handed if e123 > 0. Both these tensors are covariant tensors. The permutation tensor is a pseudotensor that changes sign under the parity transformation involving the inversion  i → − i of an odd number of spatial coordinate axes. The permutation tensor can be defined for any spatial dimension n ≥ 2. They all have the interesting property that they are isotropic when space itself is spherically symmetric. That is, they have values that remain the same under spatial rotation. The permutation tensor is here denoted e. One can show (using Eq. (3.238)) that    gi gim gin  eijk emn =  gj gjm gjn . (3.243)   g g k km gkn  Now the tensor element v = eijk is the volume of the parallelepiped defined by the three edge vectors  m , m = i, j, k, while the differential volume element is dτ = vdx1 dx2 dx3 . In particular (with no summation over repeated indices),    gii gij gik  v2 =  gji gjj gjk   g g g  ki kj kk = |gij | = det g = g.

Tensor analysis

219

So the permutation tensor is simply proportional to the permutation symbol √ (3.244) eijk = g εijk . With a nonorthogonal basis, two kinds of components can be defined. The projective components of V are Vi = V ·  i ,

(3.245)

while its expansion components V j are those appearing in its expansion formula V = V jj.

(3.246)

From their definitions, we find these components are related as Vi =  i ·  j V j = gij V j .

(3.247)

This is an example of the lowering of a contravariant component index to a covariant position by an Einstein sum over the repeated index j—one in an upper position and one in a lower position. A sum like this defines the important operation called a contraction for general tensors. A contraction reduces the rank of the tensor product by two units, from 3 to 1 here. Dual basis

The existence fo two kinds of components shows that there is another basis { j } dual or biorthogonal to { i } so defined that the expansion component of a vector in one basis is its projective component in the other basis, and vice versa: V = Vj  j = V i  i , V j = V ·  j,

Vi = V ·  i .

(3.248)

The relationship between projective and expansion components is summarized by the biorthogonal completeness relations 1 = (· i  i ) = ( i  i ·) = (· i  i ) = ( i  i ·)

(3.249)

at each point in space. The remaining metric tensors are defined by the scalar products j

j

 i ·  j = gi = δ i ,  i ·  j = gij = δij ,  i ·  j = gij , gij  j =  i ,

gij  j =  i .

(3.250)

220

Relativistic square-root spaces

The Kronecker δ symbols appearing here have their usual numerical values, but keep their tensor type designations so that any contraction over repeated indices can be made in the appropriate tensor way. The relations between the two biorthogonal bases give rise to raising and lowering rules for the associated tensor indices. As an example, one finds the identity gij g jk = gki = δki .

(3.251)

This result shows that g jk is the inverse of the metric tensor gij . A raising/lowering operation is also needed before one can sum over two repeated indices that are originally both covariant or both contravariant. An additional g tensor will then appear to give the correct final result in nonorthogonal algebra. To summarize, a contravariant vector V = V j  j is expanded in terms of a covariant basis { j }, while a covariant vector C = Ci  i is expanded in terms of a contravariant basis { i }. Local coordinate transformations

The differential position vector dr of Eq. (3.241) as an abstract object is unchanged under a change of differential coordinates from dxi to dxj . The differential coordinates themselves change as dxj =   j · dr = (  j ·  i )dxi ,   j ·  i = λi = j

where

∂xj ∂xi

(3.252)

are elements of the transformation matrix λ between these two sets of differential coordinates for the same unchanging abstract vector dr. Note the simple mnemonic correlations in our notation. The coordinate xj associated with the left basis vector   j of the scalar product or the first index of λ appears at the top position of the partial derivative. These transformation matrices satisfy the orthonormality relations λ−1 λ = I. When written out explicitly, it reads: δi = (λ−1 )k λki = ( j ·   k )(  k ·  i ) =  j ·  i j

j

=

∂xj ∂xk ∂xj = i, ∂x ∂xk ∂xi

δi = (λ−1 )i λ = ( i ·    )(  ·  j ) =  i ·  j j

j

=



∂xi ∂x ∂xi = , ∂x ∂xj ∂xj

(3.253)

where the second expression is for the dual representations dr = dxi  i = dxj   j .

(3.254)

Tensor analysis

221

Two similar expressions for λλ−1 = I will be left to a problem. In following Ricci by writing the coordinate transformation Eq. (3.252) as the functional derivative ∂xj /∂xi , we are assuming that the functions xi = xi (x1 , x2 , . . . , xN ), xj = xj (x1 , x2 , . . . , xN ),

(3.255)

are both well-defined, with all necessary derivatives. Furthermore ∂xi /∂xj = (∂xj /∂xi )−1 exists. This means that the Jacobian determinant det (∂xi /∂xj )  0. The Ricci notation is also very intuitive, with the completeness relations (3.253) appearing simply and naturally as chain rules in partial differential calculus. Tensor analysis became a recognizable branch of mathematics after Ricci’s comprehensive study of differential invariants in 1892. A simple example of a differential invariant can be constructed from a scalar function Φ(x1 , x2 , . . . , xN ) of the N coordinates xi of an ND space. Its total differential dΦ =

∂Φ(xi ) i dx ∂xi

(3.256)

gives the difference in Φ between two points in space infinitesimally close to each other. In another coordinate system xj , we find instead dΦ =

∂Φ (xj ) j dx . ∂xj

(3.257)

However, dΦ = dΦ must hold since they are the same mathematical object in different “dresses”. That is, they have the same value everywhere even though they have different functional dependences on their coordinates. The equality dΦ = dΦ defines a differential invariant. An equivalent definition is that dΦ is a differential invariant if  i ∂Φ ∂x ∂Φ = , and j ∂x ∂xj ∂xi  i 1 ∂x 1 = (3.258) j ∂xj dxi dx transform the same way under coordinate transformations. Even though these transformation properties define the tensorial character of a mathematical object, it will prove very convenient to have an abstract, coordinate-free notation. This is achieved by writing the differential scalar invariant under discussion in an abstract form as dΦ = dr · (∇Φ),

(3.259)

where every symbol denotes an object in the abstract without the need to specify a coordinate system. If a basis { j } is nevertheless used, the vector differential operator appearing here has the specific coordinate representation

222

Relativistic square-root spaces

∇ =  j ∇j

(3.260)

in terms of its expansion components ∇j in the chosen basis. We have further refined our notation by replacing ∂ with the more general symbol ∇ called a covariant derivative used to operate on objects in the abstract. The old symbol ∂j is now reserved for the specific differentiation of a function of the coordinates xi such as the transformation function xj (xi ) or a component Cj (xi ) of a covector field C. Tensors as abstract objects

We can now define a tensor of rank r = n + m and type (n, m) as j1 jm n T = T ij11···i ··· jm  i1 . . .  in  . . . 

(3.261)

in any choice of two sets of basis vectors ( i1 . . .  j1 . . .). This is possible only if its components in another bases ( k1 . . .  1 . . .) are related to those of the first bases as T  k···1 ··· = ( k1 . . .  1 . . .) · T = λk1 i1 . . . λ1 j1 T ij11··· ··· , where 1

λki =  k ·  i = λj =   ·  j =

∂xk , ∂xi ∂x ∂xj

(3.262)

are the transformation matrix elements between the two bases. Why is it necessary to define covariant, contravariant and various mixed tensors when they are different versions of the same abstract tensor expanded in contravariant, covariant or mixed bases that are dual to one another? This has to do with common usage in differential geometry where the vector differential operator ∇ is considered a covector and the position vector dr a contravariant vector. Of course, indices can be raised or lowered, and tensor types can be changed accordingly to an appropriate dual tensor. One just has to use expressions the same way other people do, in order to communicate with them without misunderstanding. Differential invariants are particularly useful in displaying tensor types. For example, the representations dr = dxi  i = dxj   j can be used to construct different versions of the scalar invariant ds2 = dr · dr = dxi ( i ·  j )dx j = dxi gij dx j = dxi dxi = dxj dx j = dxj g jk dxk .

(3.263)

So the metric tensor gij is a (0,2)-tensor, while its inverse gk is a (2,0)-tensor. New features in tensor algebra

Tensor algebra of Cartesian tensors has to be modified to handle the two sets of indices in general tensors. Tensor additions involve tensors of the same type (n, m).

Tensor analysis

223

The direct or tensor product two tensors T and S, of types (n, m) and (n , m ), respectively, is the tensor T ⊗ S with merged indices of type (n + n , m + m ). For example, the direct product of a vector V i with a covector Cj is the mixed (1,1)-tensor (V ⊗ C)ij = V iCj . A contraction of two repeated and therefore summed indices in a term in a tensor expression must involve one upper (contravariant) index and one lower (covariant) index, and never two indices in the same type. A contraction reduces the rank of the tensor product (the number of unsummed indices) by 2, from (n, m) to (n − 1, m − 1). In particular, the trace T ii of a mixed (1,1) tensor is a scalar. Implicit summations over repeated indices are also used to expand a tensor in terms of both its components and its basis vectors such as the expression or display formula for the contravariant vector V = V i  i . These summations do not change the transformation properties of the symbols that appear. That is, they are not contractions. Contractions occur only when both repeated indices appear in the tensor element/elements. The reduction of a tensor product of two tensors into a sum of irreps that are themselves vectors of various dimensions can be handled by the elementary method described in preceding sections only if each of the tensors in the product is separable into a factor involving all upper indices and a factor involving all lower indices. These irrep constructions are the concerns of group theory rather than of tensor analysis. 3.12.2

Vector differential operations on tensor fields

We finally turn to the differential calculus that is at the heart of differential geometry. We shall find here the results of a number of vector differential operations on tensor fields. The covector operator ∇ =  i ∇i can act on a scalar field Φ in the abstract to generate the covariant vector field: ∇Φ =  k ∇k Φ =  k ∂k Φ(xi ),

(3.264)

where ∂k = ∂/∂xk is used when Φ(xi ) is considered a function of the coordinates xi . As we have noted when discussing curvilinear coordinates in Section 1.11, there is nothing unusual about ∇Φ. This observation remains valid when the coordinate system is nonorthogonal as well as curvilinear. Something unusual appears in the direct product ∇V, the covariant derivative of the contravariant vector field V = V i  i in the abstract. The expression ∇V =  k ∇k V =  k  i (∇V)ik , (∇V)ik = ( k  i ) · (∇V),

where (3.265)

has the index structure of a (1, 1) tensor. (The scalar product used in the last expression is a multiple scalar product involving both unit vectors separately and taken in the

224

Relativistic square-root spaces

order shown.) The verification that it is indeed a (1, 1)-tensor is left to a problem. Its component (∇V)ik = ( k  i ) · (∇V) =  i · (∇k V) =  i · ∇k (  V  ) = ∂k V i + V   i · (∂k   )

(3.266)

can be separated into the two terms shown in the final expression. Neither of these terms considered separately is the component of a tensor even though it has the correct summation indices, a necessary but not sufficient condition. The reason is that each has the wrong transformation property. To see the problem, examine the first term in another local frame:  k ∂ ∂x  j ∂ V = [(  j ·  i )V i ]  ∂x ∂xk  2 j !  k   j  ∂x ∂x i i ∂ x V + V ∂ . (3.267) = k ∂xi ∂x ∂xk ∂xi This object does not transform like the element of a (1,1)-tensor because of the presence of a nonzero second or curvilinear term in the final expression, a term that arises when ∂/∂xk differentiates on the transformation matrix element ∂xj /∂xi that is position-dependent. If the local coordinate system is nonorthogonal but not curvilinear, this extra term would vanish because ∂xj /∂xi would then be positionindependent. The extra curvilinear term in both Eqs. (3.266) and (3.267) is proportional to Γik ≡  i · (∂k   ) = Γik

(3.268)

called a Christoffel symbol of the second kind or an affine connection. It is a connection in the sense of an addition or correction due to the curvature of the curvilinear coordinate axes. It is affine because it is related to a special type of projective transformations called affine transformations where parallel lines remain parallel but their lengths and absolute orientations in space are changed, as happens when one moves along a curvilinear coordinate axis. Its k ↔  symmetry is the consequence of the identity ∂k   = ∂k ∂ r = ∂ ∂k r = ∂  k .

(3.269)

The component of the covariant derivative ∇V of a contravariant vector field V is thus (∇V)ik = ∂k V i + Γik V 

(3.270)

Tensor analysis

225

in our notation. Other common notations for it are ∇k V i and V i;k , while ∂k V i is also written as V i k . For a covector field C = Cj  j , the covariant derivative is ∇C =  k  j (∇C)kj , (∇C)kj = ∂k Cj −

where

ΓkjC .

(3.271)

The derivation of this expression is very similar to that for ∇V and will be left to Problem 3.5.2. The formula for the covariant derivative of an (n, m)-tensor is thus ∇T =  k ( i1 . . .  j1 . . .)(∇T )ik1j······ , 1

where

i1 ··· (∇T )ki1j······ = ∂k T ij1 ··· ··· + Γk T j1 ··· + . . . 1

1

i1 ··· − Γk j1 T ··· − ... .

(3.272)

The tensor divergences are simply expressed as a contraction. In the case of rank-1 fields, the results are (∇V)kk = V k, k or gkj (∇C)jk are both scalar fields. The expression (∇C)kk is not a tensor because it does not behave correctly under rotation. The great advantage of the tensor notation is that the repeated index k can appear anywhere in the label of a tensor field of rank r > 1. Tensor analysis is the indispensable tool of general relativity which is concerned with the curvature of spacetime. It is also used, but to a much lesser extent, in electromagnetic theory and in special relativity. Some of the special tensors used in these applications are introduced in the problems. General relativity differs from tensor analysis in that spacetime itself can be intrinsically curved without reference to any possible embedding in a higher-dimensional space. The mathematical language must then be refined further to handle this additional degree of abstraction. For an easy introduction to general relativity, we suggest Landau and Lifshitz (1951, 1975), an oldie but goodie.

Problems 3.12.1 (∇V) Show that the components j

(∇V) =

∂xk ∂xj (∇V)ik i  ∂x ∂x

of ∇V transform as a (1,1)-tensor. 3.12.2 Show that the orthonormality relations corresponding to the transformation matrix equation λλ−1 = I are the chain rules ∂x ∂x ∂xj = , ∂xj ∂xk ∂xk

∂x

∂x ∂xj = . ∂xk ∂xj ∂xk

3.12.3 (Affine connection) Show that the affine connection does not transform as a (1,2)-tensor in curvilinear coordinates. That is,

226

Relativistic square-root spaces j

Γnm =   j · (∂m   n )  !  j  ∂2 xi ∂x ∂xk i ∂x Γ + = ∂xn ∂xm ∂xi ∂xn ∂xm k contains an inhomogeneous curvilinear term that ruins its tensorial property. 3.12.4 (∇C) Derive Eq. (3.271). 3.12.5∗ (Spherical coordinates) The spherical coordinates dxi = (dr, dθ, dφ) form a curvilinear but orthogonal coordinate system with the orthonormal basis vectors ei , i = r, θ, φ, as described in more details in Section 1.10. In the following formulas, repeated indices are not summed. (a) From the line element dr = drer + rdθeθ + r sin θ eφ , show that ∂r = hi ei : ∂xi  r = er ,  θ = reθ , i =

hr = er ·  r = 1, grr = h2r = 1, grr = h−2 r = 1, e i : i = hi  r = er ,

θ =

 φ = r sin θeφ ;

hθ = r,

gθθ = r2 ,

hφ = r sin θ; gφφ = r2 sin2 θ;

gθθ = r−2 ,

eθ , r

φ =

gφφ = (r sin θ)−2 ;

eφ ; r sin θ

 r ·  r = 1 =  θ ·  θ =  φ ·  φ; ds2 = dr2 + r2 dθ2 + r2 sin2 θ dφ2 . (b) For the surface and volume elements, show that dσθφ = r2 sin θ dθdφ er , dσφr = r sin θ dφdr eθ , dσrθ = r drdθ eφ , √  r ×  θ ·  φ = g, g = det (gij ) = r4 sin2 θ √ dτ = g drdθdφ = r2 sin θ drdθdφ.

Tensor analysis

227

(c∗ ) (Optional) The affine connections can be calculated readily by using the Christoffel symbols of the first kind (Problem 3.12.6(b)). However, you can practice your mathematical skill in curvilinear coordinates by verifying the following results for spherical coordinates directly from the basis vectors  i ,  i , i = r, θ, φ, expressed in terms of the unit vectors ei in spherical and Cartesian coordinates: ∂θ  θ = −r er , ∂φ  φ = −r sin θ (cos φ e x + sin φ ey ), ∂r  θ = eθ = ∂θ  r , ∂r  φ = sin θ eφ = ∂φ  r , ∂θ  φ = r cos θ eφ = ∂φ  θ ; φ

φ

φ

Γθrθ = Γθθr = Γrφ = Γφr =

Γrθθ = −r,

1 , r

φ

Γθφ = Γφθ = cot θ, Γrφφ = −r sin2 θ,

Γθφφ = − sin θ cos θ.

All other members of these two groups of symbols are zero. 3.12.6 (Christoffel symbols) Christoffel symbols are made up of standard tensor objects. So their indices can also be lowered or raised. The Christoffel symbol with its superscript lowered is called a Christoffel symbol of the first kind and is denoted j

[k, i] ≡  i · (∂k   ) = gij Γk . (a) Show that [k, i] = [k, i] : ∂k gij = [ki, j] + [kj, i] = [ik, j] + [jk, i], 1 [ij, k] = (∂i gkj + ∂j gki − ∂k gij ). 2 (b) For spherical coordinates, show that ∂r gθθ = 2r,

∂r gφφ = 2r sin2 θ,

∂θ gφφ = 2r2 sin θ cos θ; [θθ, r] = −r,

[rθ, θ] = r,

[rφ, φ] = r sin θ, 2

[φφ, r] = −r sin2 θ,

[θφ, φ] = r2 sin θ cos θ, [φφ, θ] = −r2 sin θ cos θ.

(3.273)

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Relativistic square-root spaces

Show that the affine connections calculated from these symbols agree with the values given in Problem 3.12.5(c). 3.12.7 (Compact form for (∇V)kk ) (a) The contractions appearing in the divergence (∇V)kk permits a further simplification. Show that 1 kj g ∂ (gkj ) 2 1 √ 1 = ∂ g = √ ∂ g; 2g g

Γkk = gkj  j · (∂  k ) =

1 √ (∇V)kk = ∂k V k + V  √ ∂ g g 1 √ = √ ∂k ( gV k ). g (Hint: Use the results of Section 2.3 that the matrix G = (gkj ) is symmetric, and its determinant has the cofactor expansion detG = gkjCkj /3, because the double sum contains three copies of detG. The cofactor matrix is related to the inverse matrix G−1 = C T / detG.) (b) For a spherical-coordinate system where V = V i  i , i = r, θ, φ, show explicitly that the expression (∇V)ii from part (a) agrees with the formula for ∇ · V given in Chapter 1. In the notation of Chapter 1, i e , where V i = V i h , h being the scale the vector field is V = Vsph i i i sph functions defined there. 3.12.8 (Riemann curvature tensor) A function f (x) of a single variable x has nonzero curvature wherever its second derivative d2 f /dx2  0. In higher dimensions, one uses the partial derivatives ∂k ∂ = ∂ ∂k . However, the covariant derivatives ∇k ∇ do not commute. The commutator (∇k ∇ − ∇ ∇k )V that quantifies the path-dependence of the operations is thus nonzero. If V is a contravariant vector field, the commutator is a component of a (1,2)-tensor field that can be written as a contraction of a (1,3)-tensor R with V. R is the Riemann–Christoffel curvature tensor. The Riemann–Christoffel (RC) tensor R depends entirely on the affine connections and their derivatives. Since the affine connections vanish for Cartesian coordinates, R vanishes too for them. So R measures the intrinsic curvature of the curvilinear coordinates themselves when they are in ordinary space. For example, functions on a sphere embedded in ordinary 3D space have nonzero R. In general relativity, however, a RC tensor describes the possible intrinsic curvature of spacetime itself even if spacetime is not embedded in a higher-dimensional space.

Tensor analysis

229

Show that  i · (∇k ∇ V) = ∂k (∂ V i + Γij V j ) + Γimk (∂ V m + Γmj V j ) i − Γm k (∇m V ),

 i · [(∇k ∇ − ∇ ∇k )V] = Ri jk V j ,

where

Ri jk = ∂k Γij − ∂ Γijk + Γimk Γmj − Γim Γm jk , Ri jk + Ri jk + Ri kj = 0.

(3.274)

Hint: In the last equation, the three sets of subscripts are the three cyclic permutations of the indices. Once the twelve terms are written out explicitly, their pairwise concellation will become obvious. 3.12.9 (Parallel transport) It is easy to visualize a 2D vector lying entirely in a 2D plane being transported from point p to point q along any curve in the plane in such a way that it will keep its original length and direction. In contrast, parallelism for vectors in two separate locations on a curved surface is a non-intuitive concept that has to be defined carefully. Consider the 2D surface of a sphere embedded in 3D space. First define vectors at a point p on the sphere to be those lying in the tangent plane at p. This tangent plane is just the local cartographic map where angles can be measured from the local cartographic East direction that coincides with the east-pointing latitudinal direction. In fact, the two orthonormal vectors eθ,φ of spherical coordinates point east and south, respectively, on the local cartographic map. In the embedding 3D space, the orientation of this local cartographic map, or the tangent plane it describes, changes from point to point on the sphere. Two vectors are said to be parallel at two different points p and q on the sphere if they have the same orientation angle relative to the local coordinate axes eθ,φ . Parallelly transported vectors have the same length as well in every cartographic map, as the rotation does not change their lengths. They are the closest things to constant vectors at different locations on the sphere. In the absolute 3D space where the 2D sphere is embedded, a parallelly transported vector can differ from the original vector by only a rotation. There are two points on the sphere where the spherical coordinates are ill-defined. These points are the North and South Poles where the latitude circles have shrunk to a point and therefore the latitudinal east/west directions are not defined, or the north direction either. If you are walking north along the Greenwich longitude or meridian, your right arm raised

230

Relativistic square-root spaces

sideways at a right angle perpendicular to your body will point along the latitudinal east direction. This east direction remains well defined until you reach the North Pole. Then you will suddenly find that your raised right arm is pointing south along the meridian 90◦ east. At the North Pole, all directions points south, while at the South Pole, all directions point north. The local cartographic maps at the North/South Pole have no east, west, or north/south directions. This is the reason why the poles are said to be the singular points of the spherical coordinate system. Now the problem: (a) A unit vector pointing east at map coordinates (latitude, longitude) = (0◦ , 0◦ ) (Greenwich meridian at the equator) is parallelly transported first westward to the equator at the Los Angeles meridian (−120◦ ), and then northward along the meridian to Los Angeles at the map coordinates ≈ (34◦ , −120◦ ). Sketch the parallelly transported unit vector in the local tangent plane or cartographic map as it arrives (i) at the equator at (0◦ , −120◦ ), and (ii) at Los Angeles. Repeat the sketches when the parallel transportation is (iii) first northward to the North Pole, and then (iv) southward along the Los Angeles meridian to Los Angeles. (b) A geometric interpretation can be given to the covariant derivative. To this end, consider two points p and q on the sphere infinitesimally close together such that xqk = xkp + dxk . The total change in a (contravariant) vector field V between q to p is dV = dxk ∂k V ≡ V(q) − V(p), where V(p), V(q) are the vector field at p and q, respectively. Part of this change, d V = V (p → q) − V(p) =  i dxk Γ˜ ik V  (p), comes from the parallel transport from p to q of V(p). Parallel transport thus contains two effects both described by the still undetermined generator Γ˜ that is a (1,2) tensor. There is first a translation dx = x(q) − x(p) of the field position involving a lower index k of the generator. The remaining i,  indices refers to the rotation of component V  (p) of the transported vector to the ith direction. All indices span only the dimensions of the tangent plane. On the surface of the sphere, these are just the angular directions ei , i = θ, φ that are independent of the radius of the sphere. The rotation of the orientation of the tangent plane in the embedding 3D space is intuitively obvious as one moves from one point to another on the sphere. This rotation is a property of the curvilinear angular coordinate axes

Tensor analysis

dV

V (p

231

q) DV

V(p)

dV V(q)

dr p

q

Fig. 3.5 The change dV = V(q) − V(p) = DV + d V in a vector field V(p) from point p to point q expressed as a sum of a covariant change DV and a change d V = V (p → q) − V(p) due to the parallel transport of V(p) to the position q.

themselves. It vanishes if the orientation of the tangent plane in the embedding space never changes. The rest of dV is DV = dxk ∇k V = dV − d V, " # ∇k V =  i ∂k V i + Γik V  .

where

DV is the intrinsic change in the vector from p to q relative to the parallelly transported vector at q. It is called a covariant change. The relation between these changes is shown schematically in Fig. 3.5. Show that the generator of parallel transport is the negative of the affine connection: Γ˜ ik = −Γik . What is the significance of the negative sign in the last equation? (c) Show explicitly that ∇k g = 0, where g = gij  i  j is the metric tensor. Hence ∂k gij contain only parallel transports. Hint: Show that (∇k g)ij = ∂k gij − Γik g j − Γjk gi = 0. Note that since g = gji  i  j = gij  i  j can also be expressed in other choices of basis, similar results hold for these other bases as well. 3.12.10 (Covariant Riemann tensors) (a) The sole upper index of the (1,3)-Riemann tensor Rijk of Eq. (3.274) can be lowered to give a (0,4)-tensor. Show that its component can be

232

Relativistic square-root spaces

written as " # Rijk ≡ gin ∂k Γnj − ∂ Γnjk + Γnmk Γmj − Γnm Γm jk 1 = (∂j ∂k gi + ∂i ∂ gjk − ∂j ∂ gik − ∂i ∂k g j ) 2 " # n m + gnm Γni Γm jk − Γik Γ j . Hint: Use Eq. (3.273) to show that the four second derivative terms come from ∂k (gin Γnj ) − ∂ (gin Γnjk ). (b) Verify the following symmetry properties of Rijk : (i) symmetric under the exchange i j ↔ k; (ii) antisymmetric under the exchange k ↔  from the defining commutator origin of Riemann tensors, and under the exchange i ↔ j; (iii) vanishes on cyclic summation of any three indices, as in the result Rijk + Rijk + Rik j = 0 following from a similar identity for the Riemann (1,3)-tensor given in Eq. (3.274). (c) (Ricci tensor) Show that the Ricci tensor R j = gik Rijk satisfies the symmetry R j = Rj and the antisymmetric properties Rj = −gik Rjik = −gik Rijk = gik Rjik , gij Rijk = gk Rijk = 0. Note: There is essentially only one way to contract Rijk to a scalar curvature because R = gik g j Rijk = −gik g j Rjik , 0 = gij gk Rijk .

Appendix 3 Tables of mathematical formulas 1 Special relativity and Lorentz transformations Lorentz transformation and velocity addition:

Tables of mathematical formulas



233

    x γ iβγ x = , ict −iβγ γ ict β12 =

β1 + β2 , 1 + β1 β2

γ12 = γ1 γ2 (1 + β1 β2 ). Length contraction, time dilation and proper time: 2 2 2 2 = x12 − c2 t12 < x12 , x12

2 if t12 = 0,

1 2 2 r < t12 , c2 12  2 dt dX 2 2 dτ = − 2 = . γ c

2 if x12 = 0,

2 2 t12 = t12 −

The traveling twin who returns is younger than the twin who stays at home. 2 Relativistic kinematics and the mass–energy equivalence Four-momentum of a particle of mass m and its square are: P = mU = m

d (r, ict) = (p, iE/c), dτ

P2 = p2 − E 2 /c2 = P2 . The Lorentz scalar P2 has the same value in any Lorentz frame. Relativistic Doppler effect: The photon is massless: m = 0. Its energy is E = pc. Its Lorentz momentum in a 2D spacetime transforms as (      1−β p γ iβγ p  . = , p =p ip −iβγ γ ip 1+β Relativistic kinematics: Four-momentum of the center of mass (cm) of a system of n particles of momenta Pi = (pi , iEi ) when c = 1 is: Pcm =

n 

Pi = (Pcm , iEcm ),

i=1

Pcm =



pi ,

i 2 P2 cm = Pcm .

Ecm =

 i

Ei ;

where

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Relativistic square-root spaces

3 Quaternions Q = t + xi + yj + zk = t + iσT · r : ij = k = −ji,

jk = i = −kj,

ki = j = −ik :

i2 = j2 = k2 = −1 = ijk, i∗ = −i = −iσTi ∗ = −iσ†i ,

also for

j, k,

Q∗ Q = t2 + x2 + y2 + z2 = QQ∗ , Q−1 = Q∗ (QQ∗ )−1 , 1 Q∗ = − (Q + iQi + jQj + kQk). 2 Pauli matrix representation: Q = (q0 , q) = q0 + iσTq q,

σTq = σT · q,

(σT · a)(σT · b) = a · b − iσT · (a × b), Q = Qeiψq σq = Q(cos ψq + iσTq sin ψq ), % q Q = q20 + q2 , sin ψq = , Q T

Q−1 = Q−1 e−iψq σq = Q† /Q2 ; T

PQ  QP,

where

(3.275)

because [σTq , σTp ]  0.

A rotor is the unit quaternion Ea (ψ/2) = eiψσa /2 , T

that rotates a vector quaternion X = (0, r) to X = (0, r ) = Ea (ψ/2)XE†a (ψ/2) by an angle ψ about the axis ea after a two-sided multiplicative operation. 4 Dirac equation, spinors, matrices and symmetries Light-cone variables in spin space: τ2 = t2 − r2 = t2 − (σ · r)(σ · r), m2 = E 2 − p2 = E 2 − (σ · p)(σ · p). Dirac equation: 

−Hˆ σ · pˆ −σ · pˆ Hˆ



   φ+ φ+ = −m ; φ− φ−

Tables of mathematical formulas

# " # " γμ pˆ μ − m ψ = iγμ ∂μ − m ψ = 0 : ˆ p), ˆ pˆ μ = (H, ∂ ∂μ = μ = ∂x



 ∂ ,∇ , ∂t

γμ = (γ0 , γ) = (ρ3 I, iρ2 σ), {γμ , γν } = γμ γν + γν γμ = 2gμν , γ5 = iγ0 γ1 γ2 γ3 = ρ1 = γ5 . Symmetries of the Dirac equation: Parity P :

r → r = PrP−1 = −r, p → p = PpP−1 = −p; Pψ(r, t) = Pψ(−r, t) : P = γ 0 = ρ3 .

Time Reversal T = UK : T ψ(r, t; p, . . .) = Uψ∗ (r, −t; −p, . . .), U = σ2 , T 2 = (−1)2J . Charge conjugation C : C|q, psE+  = | − q, psE+ , ψc = Cψ = γ2 ψ∗ . 5 Weyl and Majorana spinors, symmetry violations Weyl spinor: 1 (1 ∓ γ5 )ψ = PL,R ψ, 2 1 = (γ5 ∓ 1) = ∓PL,R , 2

φL,R = γ5 PL,R

P2L = PL ,

P2R = PR ,

PL PR = PR PL = 0,

235

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Relativistic square-root spaces

Weyl representation : " # μ γW pˆ μ − m ψW = 0, 0 = ρ1 , γW

k γW = iρ2 σk = γk ;

5 0 1 2 3 γW = iγW γW γW γW = −ρ3 .

Massless equation : ˆ W = γ5 (Σ · p)ψW , Hψ

Σ= I⊗σ

E = κh|p|,

(eigenvalues) Weyl equation :

only left-chiral fermions ˆ L = −σ · pφL , Hφ ⇒ maximal parity violation.

CP violation: Kobayashi and Maskawa characterize CP-violation in terms of an irreducible complex phase factor ζ = eiδ in a complexified 2D rotation appearing in the mixing of N = 3 quark generations d, s, b: ⎛ ⎞ ⎛ ⎞ ⎜d⎟ ⎜⎜⎜ d ⎟⎟⎟ ⎜⎜⎜ s ⎟⎟⎟ = V ⎜⎜⎜⎜⎜ s ⎟⎟⎟⎟⎟, ⎝ ⎠ ⎝ ⎠ b b ⎛ ⎜⎜⎜ Vud Vus Vub V = ⎜⎜⎜⎝ Vcd Vcs Vcb Vtd Vts Vtb u† Vd = u† VKM d , 

d = (ηΦd )d,

⎞ ⎟⎟⎟ ⎟⎟⎟ : ⎠

where 

u = (ηΦu )u,

VKM = (ηΦu )V(ηΦd )∗ , Φu = diag (1, α2 , . . . , αN ), Φd = diag (β1 , β2 , . . . , βN ), ⎛ ⎞ ⎜⎜⎜ c2 0 s2 ζ ∗ ⎟⎟⎟ ⎜ VKM = R1 ⎜⎝⎜ 0 1 0 ⎟⎟⎠⎟ R3 , −s2 ζ 0 c2

⎛ ⎞ ⎜⎜⎜ c3 s3 0 ⎟⎟⎟ ⎜ R3 = ⎜⎜⎝ −s3 c3 0 ⎟⎟⎟⎠. 0 0 1

Here u, d are arbitrary column vectors, αi , βi are reducible complex phases factors, Ri , i = 1, 3, are the real 2D rotation matrices about the i-axis, and ci = cos θi , si = sin θi , i = 1, 2, 3, where θi is the real rotation angle about the i-axis. The number of irreducible CP-violating phases when N quark generations are mixed is

Tables of mathematical formulas

237

NKM = (N − 1)(N − 2)/2. Majorana equation: m2 = (mSM )2 : γμ pˆ μ ψ = mSM ψ = mψM , γμ pˆ μ ψM =

1 μ 2 (γ pˆ μ ) ψ = mψ. m

Majorana spinors: SM = C: 1 = (1 + C)φL,R ≡ P+ φL,R . Φ(+) L,R 2 6 Lorentz group Lorentz group SO(1,3): Its infinitesimal generators and finite rotation matrices based on the invariant squared spacetime separation τ2 = x02 − x12 − x22 − x32 are   cos dθ sin dθ in 12 space, R3 (dθ) = − sin dθ cos dθ ⎞ ⎛ ⎜⎜⎜ 0 0 0 0 ⎟⎟⎟ ⎜⎜ 0 0 −1 0 ⎟⎟⎟⎟ J3 = i ⎜⎜⎜⎜⎜ ⎟; ⎜⎝ 0 1 0 0 ⎟⎟⎟⎠ 0 0 0 0 A = R3 (θ) = eiθJ3 = (I − I(12) ) + I(12) (cos θ + J3 sin θ), I(1,2) = diag (0, 1, 1, 0).  0     x γ βγ x0 = in 01 space, βγ γ x1 x1 ⎞ ⎛ ⎜⎜⎜ 0 1 0 0 ⎟⎟⎟ ⎜⎜ 1 0 0 0 ⎟⎟⎟⎟ K1 = ⎜⎜⎜⎜⎜ ⎟; ⎜⎝ 0 0 0 0 ⎟⎟⎟⎠ 0 0 0 0 A = B1 (η) = e−ηK1 = (I − I(01) ) + I(01) (cosh η − K1 sinh η).

238

Relativistic square-root spaces

The generators satisfy the Lie algebra [Ji , Jj ] = iεijk Jk , [Ji , Kj ] = iεijk Kk , [Ki , Kj ] = iεijk Jk . SO(1, 3) ↔ SU(2) ⊗ SU(2): N±i = (Ji ± Ki )/2 :

[N+i , N+ j ] = iεijk N+k , [N−i , N− j ] = iεijk N−k , [N+i , N− j ] = 0.

SU(2) and SU(2) ⊗ SU(2) irreps: Quantum spin j is integer or half integer, and has dj = 2 j + 1 possible values for its m = j3 component:  0, integer j . m = ± j, ±( j − 1), . . . , ±1/2, half integer j These states make up a dj dimension column vector or representation (rep) denoted ( j). Half integer spins have even-dimensional reps called spinors that change sign when rotated one complete turn about any axis in space. It takes two complete turns to restore them to their original values. They describe particles called fermions. Integer spins describe boson reps of odd dimensions that return to their original values after only one complete turn about any axis in space. SU(2) multispinors of rank r are direct products of r of the d = 2 column spinors of ( j = 1/2). They can be reduced to dj dimensional column vectors called irreps by repeated applications of the spin addition rule on products of reps: ( j< )( j> ) =

j< 

( j> + m) :

m=− j<

r=1:

(1/2) = fundamental rep

r=2:

(1/2)2 = (0), (1);

r=3:

(1/2)3 = (1/2)2 , (3/2);

r=4:

(1/2)4 = (0)2 , (1)3 , (2).

SU(2) ⊗ S U(2) irreps: ( j+ , j− ) = (0, 0), (1/2, 0), (0, 1/2), (1/2, 1/2), (1, 0), (0, 1), . . .

Tables of mathematical formulas

7 Cartan spinors in square-root space Complex projective coordinates ξ, η on the light cone: x + iy t+z ξ = = : t−z x − iy η a a ∗ t = (ξξ + ηη∗ ), z = (ξξ ∗ − ηη∗ ), 2 2 a ∗ a x = (ξη + ηξ ∗ ), y = (ξη∗ − ηξ ∗ ). 2 2i

ζ=

Complex spinor X on the light cone:   ξ , π† = (ξ∗ η∗ ): π= η 2 π† π = ξ∗ ξ + η∗ η = t, a   ξ ∗ ∗ X ≡ ππ† = (ξ η ) η  ∗    1 t + z x + iy ξξ ξη∗ = = ηξ∗ ηη∗ a x − iy t − z =t+σ·r

if

a = 1.

Complex spinor X in Minkowski space and its Lorentz transformations:   t + z x + iy X≡ = t + σ · r, x − iy t − z det X = t2 − (x2 + y2 + z2 ), X  = AXA† , det X  = (det X)(det A)(det A† ) = (det X)|det A|2 . 8 Dyadics Dyadics are bivectors (direct product of two vectors):  ai bj ei ej : ab = i, j

∇ · ab = (∇ · a)b + (a · ∇)b,

239

240

Relativistic square-root spaces

∇ × ab = (∇ × a)b − (a × ∇)b. I = e1 e1 + e2 e2 + e3 e3 : a · I = I · a,

a × I = I × a.

Stress tensor describes the force acting on a surface element dA: F = T · dA,

where

T = T x e x + Ty ey + Tz ez = =





Ti e i ,

i

Pi aˆ i aˆ i

aˆ i = principal axis.

if

i

Deformation dyadic D describes the change in the line element under stress dr − dr = dr · D, D = A + , 1 1 Aij = (Dij − Dji ), ij = (Dij + Dji ). 2 2 Hooke’s law for small deformations: The linear stress–strain relation  ci jk k , T ij = k

T = λTr()I + 2μ,

simplifies to

where

 = e1 aˆ 1 aˆ 1 + e2 aˆ 2 aˆ 2 + e3 aˆ 3 aˆ 3 for an isotropic elastic medium. 9 Cartesian tensors Cartesian tensors are higher-rank generalizations of vectors and dyadics (using the Einstein summation notation): V = Vj ej , T = T m···n e em . . . en , T ij···k = λi λjm . . . λkn T m···n , λi = ei · e . Pseudotensors have different transformation properties from tensors when det λ = −1: Pi j···k = sgn(det λ)λi λjm . . . λkn Pm···n , sgn (det λ) = det λ/| det λ|.

Tables of mathematical formulas

241

det λ = −1 holds for a general parity transformation, one where ei = −ei for an odd number of coordinate axes. For space inversion r → −r in the 3D physical space, a tensor of rank r is a true tensor if its sign changes as (−1)r and a pseudotensor if its sign change is (−1)r+1 . 3D Cartesian tensors are direct products of SO(3) column vectors of degree or rank r. Their irreps are a subset of SU(2) irreps: r=1:

(1) = fundamental rep;

r=2:

(1)2 = (0), (1), (2);

r=3:

(1)3 = (0), (1)3 , (2)2 , (3).

All group irreps have well-defined permutation symmetries. A 3D matrix or Cartesian tensor T ij of rank 2 can be reduced into three spherical tensors of ranks 0, 1, and 2, with the permutation symmetries of symmetric and traceless, antisymmetric and symmetric, respectively: T ij = Aij + S ij : 1 Aij = (T ij − T ji ), 2

S ij =

Tr T δij + S ij(2) . 3

Spherical harmonics: Ym (θ, φ) of rank  is a polynomial of xˆ1 = sin θ cos φ, xˆ2 = sin θ sin φ, xˆ3 = cos θ. of degree  and order or dimension d = 2 + 1. 10 Tensor analysis Nonorthogonal basis in 3D space: dr = dxi  i = dxj   j , ∂r , ∂xi  i ·  j = gij ,   gi eijk emn =  g j  g k i =

∂r ; ∂xj  i ×  j ·  k = eijk :  gim gin  gjm gjn . gkm gkn  j =

242

Relativistic square-root spaces

Expansion and projective components, and dual basis: V = V j j :

Vi = V ·  i = gij V j ,

V = Vj  j :

V i = V ·  i = gij Vj ,

j

j

 i ·  j = gi = δ i , gij  j =  i ,

 i ·  j = gij = δij ,

 i ·  j = gij = δij ,

gij  j =  i ,

1 = (· i  i ) = ( i  i ·) = (· i  i ) = ( i  i ·). Coordinate transformations: dxj =   j · dr = (  j ·  i )dxi ,   j ·  i = λi = j

∂xj , ∂xi

δi = (λ−1 )k λki = ( j ·   k )(  k ·  i ) =  j ·  i j

j

=

∂xj ∂xk ∂xj = i. ∂x ∂xk ∂xi

Differential invariant of a scalar field Φ: dΦ = dr · (∇Φ)    j   ∂Φ(xi ) ∂Φ (x ) i = dx = dxj = dΦ, ∂xi ∂xj ∇ =  j ∇j . Tensors as abstract objects: A tensor of rank r = n + m and type (n, m) is j1 jm n T = T ij11···i ··· jm  i1 . . .  in  . . .  ,

T  k···1 ··· = (  k1 . . .  l1 . . .) · T = λki11 . . . λ 1 T ij11··· ··· , j

1

1

λki =   k ·  i =

∂xk , ∂xi

λ =   ·  j = j

∂x ∂xj

where

.

Covariant derivatives: The covariant derivative ∇V of a contravariant vector field V =  i V i is the (1,1)tensor ∇V =  k ∇k V =   k ∇k V =  k  i (∇V)ik , (∇V)ik = ( k  i ) · (∇V) = ∂k V i + V   i · (∂k   ) Γik ≡  i · (∂k   ) = Γik .

Tables of mathematical formulas

243

If (∇V)ik = 0, ∂k V i = −V  Γik contains only parallel transport. Other covariant derivatives are ∇C =  k  j (∇C)kj ,

(∇C)kj = ∂kCj − ΓkjC ,

∇T =  k ( i1 . . .  j1 . . .)(∇T )ki1j······ , 1

(∇T

)ki1j······ 1

=

∂k T ij11··· ···

+

Γik1 T ··· j1 ···

+ ...

i1 ··· − Γk j1 T ··· − ... .

Parallel transport of a vector field from point p to point q separated by the distance dxk = xk (q) − xk (p): V (p → q) = V(p) −  i dxk Γik V  (p). Christoffel symbols: [i j, k] ≡  k · (∂i  j ) = gk Γij , 1 = (∂i gkj + ∂j gki − ∂k gij ), 2 √ 1 1 √ Γkk = √ ∂ g, ∇ · V = √ ∂ ( gV  ). g g Riemann–Christoffel curvature tensor: ∇k V i ≡ (∇V)ik :

(∇k ∇ − ∇ ∇k )V i = R i jk V j ,

i m i m Γj − Γm Γjk , R i jk = ∂k Γji − ∂ Γjki + Γmk

0 = R i jk + R i jk + R i kj ; 1 Rijk = (∂j ∂k gi + ∂i ∂ gjk − ∂j ∂ gik − ∂i ∂k gj ) 2 " # + gnm Γin Γjkm − Γikn Γjm , Rj = gik Rijk = −gik Rjik = −gik Rijk = gik Rjik , gij Rijk = gk Rijk = 0, R = gik gj Rijk = −gik gj Rjik .

4 Fourier series and Fourier transforms 4.1

Wave–particle duality: Quantum mechanics

In a famous lecture delivered on 14 December 1900, Planck proposed an explanation for the observed energy density per unit frequency of the radiation emitted by a black body in thermal equilibrium. His formula contained as special cases known theoretical expressions that worked well only for high frequencies (Wien’s radiation formula) or for very low frequencies (Rayleigh-Jeans law). However, he had to introduce a totally new idea to achieve this—the assumption that the radiation energy is emitted or absorbed by mechanical oscillators only in discrete amounts ε proportional to its frequency ν ε = hν.

(4.1)

The universal proportionality constant h, Planck’s constant, can readily be deduced from the measured radiation density. It characterizes the discontinuous character of Planck’s quantum oscillators, in marked contrast to the continuous values of the energies or amplitudes of classical oscillators. Planck’s hypothesis of the energy quantum marked the beginning of quantum physics. In technical terms, Planck’s quantum hypothesis is concerned with the partition of energy in statistical mechanics. A similar observation on energy partition led Einstein in 1905 to the “heuristic point of view” that light may appear as particles. He pointed out that Wien’s radiation formula could be obtained by imaging that the black-body radiation itself was a gas of particles, the light quanta, of energies hν. He immediately applied this theory to several physical phenomena, including the photoelectric effect (which describes the observation that the energies of electrons emitted from a metal irradiated by ultraviolet light depend only on its frequency and not on its intensity). It was Niels Bohr who first saw the far-reaching implications of the quantum postulate in understanding the structure of atoms. Bohr was able to explain the distinctive colors of light emitted by excited atoms by making a number of revolutionary postulates. He supposed that an atom consisted of Z electrons bound electrostatically to a very small nucleus with a positive charge Ze, e being the magnitude of the electronic charge. He next supposed that the electron of an excited atomic system existed in a stationary state of discrete energy E i , and that, when the electron made a transition to a final atomic state of energy E f , the excess energy was emitted as light of frequency ν: hν = E i − E f = ω,

(4.2)

Wave–particle duality

245

where ω = 2πν,  = h/2π. By assuming that the orbital angular momentum of the electron moving around the positively charged atomic nucleus could exist only in integral multiples of , he was able to derive expressions for the energies of these stationary states that agreed with values deduced from experiment. Bohr’s work stimulated the search for quantum conditions or quantization rules. In this, much progress was made by following Bohr’s correspondence principle. This states that the quantum theory of matter should approach classical mechanics in the domain (the so-called limit of large quantum numbers) where classical mechanics is known to be valid. Successes of this “old” quantum theory in accounting for the structure of simple atomic systems were followed by significant difficulties with more complicated systems. The quantization rules also appeared empirical and ad hoc. This unsatisfactory state of affairs led in 1925 to two different but equivalent formulations that gave quantum mechanics its present logical foundation—the matrix mechanics of Heisenberg, Born, and Jordan, and the wave mechanics of de Broglie and Schr¨odinger. According to Heisenberg, the difficulties of the old quantum theory arose because of its reliance on physically unobservable concepts such as electronic orbits, that is, the position x(t) of classical mechanics. The reason is that in a quantum system it is not possible to determine a position without introducing disturbances that radically change the state of motion of the system. One should use only physical observables instead. The obvious question now is how these physical observables are to be constructed. Heisenberg’s solution is as follows. The position of an atomic state n should be expanded in the Fourier series xn (t) =

∞ 

x(n, m) exp[iω(n, m)t],

m=−∞

where x(n, m) has the physical interpretation of a transition amplitude between the states n and m, and ω(n, m) = (E m − En )/

(4.3)

is the quantum frequency of the transition as given by the Bohr–Einstein relation of Eq. (4.2). The physical observables are x(n, m) and ω(n, m), not xn (t) itself. The distinction between these two types of quantities might appear trivial, but the situation is quite subtle. The reason is that in order to preserve consistency in the theoretical description, the amplitudes x(n, m) should satisfy a certain recombination rule that Heisenberg worked out. This turned out to be the rule of matrix multiplication. The resulting theory was soon formalized by Born, Jordan and Heisenberg into the matrix formulation of quantum mechanics. In the same year (1925), de Broglie noted that the symmetrical treatment of energy and momentum in relativity implied that momentum should also have a quantum character, namely p = k.

(4.4)

246

Fourier series and Fourier transforms

For light p = E/c = ω/c = k; hence k is just the wave number of its wave motion. de Broglie then suggested that since the energy–momentum relation also held for material particles, they should also have wave properties. In this way, the concept of wave–particle duality was now applied to massive particles as well as to light–waves. This was a remarkable suggestion. The wave–particle nature of light had been a subject of speculation since ancient times. Newton himself preferred a corpuscular theory, while his respected contemporary Huygens advocated the wave theory. Newton’s preference remained unquestioned for over a century until Young demonstrated the diffraction of light by two slits in 1803. This is a wave property that cannot be explained in the corpuscular picture. Gradually, with the mathematical analyses and experimental observations on interference phenomena by Fresnel and others, the wave theory gained the upper hand. We have seen how, a century later, Einstein finally rescued the particle theory of light from disrepute. However, it had never been seriously suspected before that massive particles had wave properties. The wave nature of particles must be verified by observing interference phenomena for particles. This was provided a few years later by Davisson and Germer, and by G. P. Thomson and others who successfully observed the diffraction of electrons by crystals. The significance of the concept of matter waves in quantum mechanics was realized by Schr¨odinger, who applied it in 1926 to the archetypical conservative mechanical system in nonrelativistic mechanics described by the expression p2 + V(r) = E, 2m

(4.5)

where V(r) is the potential energy. We recall from Section 2.7 that the frequency ω = E/ is the eigenvalue of the time displacement operator i ∂/∂t. Similarly the wave vector k in Eq. (4.4) is the eigenvalue of the space translation operator −i∇. Hence the terms in Eq. (4.5) may be interpreted instead as operators operating on what is now called the time-dependent Schr¨odinger wave function Ψ (r, t): ! 1 ∂ 2 (−i∇) + V(r) Ψ (r, t) = i Ψ (r, t). 2m ∂t

(4.6)

For a system in a stationary state of energy ω, the time dependence in Ψ (r, t) has to be an eigenfunction, say exp(−iωt), of the Hamiltonian operator H = i ∂/∂t belonging to the eigenvalue E = ω. Hence Ψ (r, t) = ψ(r) exp(−iωt).

(4.7)

Fourier series

247

This allows a differential equation for the time-independent wave function ψ(r) to be separated from Eq. (4.6): ! 2 2 − ∇ + V(r) − E ψ(r) = 0. (4.8) 2m The result is an eigenvalue problem in which the eigenvalue E and the eigenfunction ψ(r) are to be determined. This can be done by requiring that ψ(r) satisfies certain boundary conditions at r = 0 and ∞. Thus at one stroke, the mathematics of differential equations, which had been developed since the time of Newton and Leibniz, was brought to bear on the problems of quantum mechanics. Schr¨odinger himself demonstrated the equivalence between his wave mechanics and the matrix mechanics of Heisenberg, Born and Jordan. The matrix element F f i in the latter formulation describing the transition amplitude from the initial stationary state i to the final state f caused by the operator F is just the integral  (4.9) F f i = ψ∗f (r)F(r)ψi (r)d3 r involving the wave functions of these states. The operator F is here “represented” by an appropriate function or operator F(r) of r. This turns out to be part of the mathematics of Fourier analysis of operators and functions describing ordinary wave phenomena. This familiar language gives wave mechanics a very intuitive appeal. It enables the physical basis of quantum mechanics to be described in concrete physical terms. In contrast, matrix mechanics is much more compact and abstract, and hence more elegant. In this chapter we study the representation of functions by Fourier series and integrals. Certain important wave properties of quantum mechanics, including the uncertainty principle (1927) of Heisenberg, are derived. Finally, Fourier analysis is used to show that the Maxwell equations of classical electrodynamics make simple statements in Fourier spaces on the physical origins of the components of the electromagnetic fields E and B parallel and transverse to the wave vector k.

4.2

Fourier series

The world is full of vibrations. The sound we hear is an acoustic wave; the sight we see is an electromagnetic wave; surfers in Malibu, California, ride on the gravity waves of the ocean. The simplest wave motion in one dimension is described by the 1D wave equation   2 1 ∂2 ∂ − u(x, t) = 0. (4.10) ∂x2 v2 ∂t2 which we derived in Section 2.8. If the time dependence is exp(±iωt), the wave motion has a definite frequency ω. We then obtain the Helmholtz equation

248

Fourier series and Fourier transforms



d2 2

dx

 + k X(x) = 0, 2

k=

ω . v

(4.11)

It can be shown by direct substitution that a solution of this equation is the function X(x) = a cos kx + b sin kx,

(4.12)

where a and b are constants. (The theory of differential equations is discussed in Chapter 5). In particular, the shape of a vibrating string of length π fixed at its ends is given by Xn (x) = sin nx,

(4.13)

where n is a positive integer. This shape is called the nth normal mode of vibration of the string. Historically, the wave equation (4.10) was first studied around the middle of the eighteenth century. Around 1742, D. Bernoulli stated that vibrations of different modes could coexist in the same string. In 1753, stimulated by the work of D’Alembert and of Euler, he further proposed that all possible shapes of a vibrating string even when its ends were not fixed were representable as f (x) =

∞ 

bn sin nx.

(4.14)

n=1

His argument was that the infinity of coefficients could be used to reproduce any function at an infinity of points. The possibility of adding a cosine series to take care of even functions of x was considered in the controversy that followed this claim. However, all the other eighteenth-century mathematicians, while agreeing that the series in Eq. (4.14) solved Eq. (4.11), disputed the possibility that all solutions could be so represented. In 1807, Fourier submitted a paper on heat conduction to the Academy of Sciences in Paris in which he claimed that every function in the closed interval [−π, π] (i.e., −π ≤ x ≤ π) could be represented in the form ∞  1 S = a0 + (an cos nx + bn sin nx). 2 n=1

(4.15)

His integral formulas for the coefficients an , bn , were not new, as they had been obtained by Clairaut in 1757 and by Euler in 1777. However, Fourier broke new ground by pointing out that these integral formulas were well defined even for very arbitrary functions, and that the resulting coefficients were identical for different functions that agreed inside the interval, but not outside it. This paper of Fourier was rejected by Lagrange, Laplace and Legendre on behalf of the Academy on the grounds that it lacked mathematical rigor. Although a revised version won the Academy’s grand prize in 1812, it too was not published in the academy’s M´emoires until 1824, when Fourier himself became the secretary of the Academy. This occurred two years after he published his book, Theorique Analytique

Fourier series

249

de la Chaleur (The Analytic Theory of Heat), in which arguments were advanced that an arbitrary function could be represented by the series in Eq. (4.15). This book had great impact on the development of mathematical physics in the nineteenth century and on the concept of functions in modern analysis. Let us suppose that we are interested in those series (4.15) that converge uniformly to functions f (x) in the closed interval [−π, π]. By uniform convergence in the specified interval, we mean convergence for any value of x in the interval. (Uniform convergence for a series of terms fk (x) is assured if | fk (x)| ≤ Mk and the series for Mk is convergent.) A uniformly convergent infinite series is known to be integrable and differentiable term by term. Given a uniformly convergent series f (x) of the above type, the series ∞  1 f (x) cos mx = a0 cos mx + (an cos nx cos mx 2 n=1

+ bn sin nx cos mx)

(4.16)

also converges uniformly, because | cos mx| ≤ 1. It can therefore be integrated term by term, with the result that  π  π 1 f (x) cos mx dx = a0 cos mx dx 2 −π −π   π ∞   π  + cos nx cos mx dx + bn sin nx cos mx dx . (4.17) an −π

m=1

−π

The right-hand side can be evaluated with the help of the formulas  π sin mx dx = 0; −π  π



−π π

−π  π



cos mx dx =

sin nx cos mx dx = 0, 

cos nx cos mx dx =

−π  π

−π

 sin nx sin mx dx =

2π, 0,

m=0 m  0; for all m, n;

(4.18a) (4.18b) (4.18c)

0, m  n π, m = n;

(4.18d)

0, m  n π, m = n,

(4.18e)

for integers m, n. These formulas can be derived by using trigonometric identities such as sin nx sin mx =

1 1 cos(n − m)x − cos(n + m)x. 2 2

(4.19)

250

Fourier series and Fourier transforms

One can then show that



π

−π  π −π

f (x) cos mx dx = am π,

(4.20a)

f (x) sin mx dx = bm π,

(4.20b)



π

−π

f (x) dx = a0 π.

(4.20c)

If the function f (x) in Eq. (4.20) is replaced by an arbitrary function F(x) in the closed interval [−π, π], the resulting coefficients a0 , an , and bn are called its Fourier coefficients, and the trigonometric series (4.15), namely, ∞  1 f (x) = a0 + (an cos nx + bn sin nx), 2 n=1

(4.21)

is called its Fourier representation or Fourier series. Note that two distinct steps are involved here. The Fourier coefficients are first calculated by using the original function F(r). They are then used to reconstruct the original function with the help of the trigonometric series (4.21). In numerical calculations, the infinite series is truncated after a sufficiently large number of terms. A number of questions immediately arise. Do these Fourier coefficients exist at all? Is the Fourier series convergent? Does it converge to the original function F(x)? These are among the questions that will be discussed in this chapter.

4.3

Fourier coefficients and Fourier-series representation

Whether Fourier coefficients an =

1 π

bn =

1 π



π

−π



F(x) cos nx dx,

(4.22)

π −π

n ≥ 0,

F(x) sin nx dx,

n > 0,

exist depends not only on the nature of the function F(x), but also on the precise meaning of an integral. The definite integral of a continuous function, defined in elementary calculus as the limit of a sum, is due to Cauchy. Riemann generalized the concept to certain bounded functions that may have an infinite number of discontinuities in the interval of integration. In 1902, Lebesgue put forward a new theory of integrals in which more functions are integrable. As a result, more functions can be represented by Fourier series. In this book we restrict ourselves to ordinary, that is, Riemann, integrals. For these, it is known that a function is integrable in the finite closed interval [a, b] if

Fourier coefficients and Fourier-series representation

251

it is continuous, or bounded and continuous except for a finite number of points. If it is bounded, the function is integrable even when it has an infinite number of discontinuities, provided that they can be enclosed in a finite or infinite number of intervals with a total length that can be made as small as one pleases. In particular, this bounded function is integrable if its points of discontinuity are “countable” like the integers 1, 2, 3, . . . . Besides the integrability of F(x) itself, there are two other useful concepts of integrability that will be used in our discussion. A function F(x) is absolutely integrable if its absolute value |F(x)| is integrable. If F(x) is absolutely integrable, it is certainly integrable, but an integrable function may not be absolutely integrable. A function is square integrable if both F(x) and F 2 (x) are integrable. Every bounded integrable function is also square integrable, but an unbounded integrable function may not be square integrable. A simple example of the latter is x−1/2 in the interval [0, b] since the square integral (ln x) is infinite at x = 0. In elementary applications we shall deal only with functions that are both absolutely and square integrable. The relevant integrals are often tabulated in mathematical books and tables (such as the Handbook of Chemistry and Physics) as finite sine and cosine transforms. They can also be obtained from computer algebra programs. Let us now concentrate on the construction and use of Fourier series. 4.3.1

Calculation of Fourier coefficients

Example 4.3.1 Obtain the Fourier coefficients for F(x) = x, −π ≤ x ≤ π. Since F(x) is odd, all an vanish. For bn we have 

π −π

 x sin nx dx = 2

x sin nx dx 0

      π d 1 d cos kx dx = 2 − sin kπ =2 − dx 0 dk k   1 π cos kπ = 2 2 sin kπ − k k k=n =

2 ∴ bn = (−1)n+1 . n

π

2π (−1)n+1 n



Example 4.3.2  G(x) =

1, 0≤x≤π −1, −π ≤ x ≤ 0.

252

Fourier series and Fourier transforms 2.0 1.8 1.6 1.4 1.2 GM(x) 1.0 0.8 0.6 0.4 0.2 0.0 0.00

.314

.628

.942

Fig. 4.1 The partial sums G M (x) = (4/π) positive x.

Again an = 0, while

1.25

1.57 x

1.88

2.19

2.51

M

n=1 sin(2n + 1)x/(2n + 1)

2.82

3.14

for M = 1, 10, 100 and for

 2 π 2 sin nx dx = − cos nx|π0 bn = π 0 πn  0, n even = 4 πn , n odd.

Three partial sums of the Fourier series for G(x) are shown in Fig. 4.1. Example 4.3.3 Obtain the Fourier-series representation for the function F(x) = x2 ,

−π ≤ x ≤ π,

and use it to show that ∞  1 , (a) π2 = 6 n2 (b)

π2

n=1 ∞ 

= 12

n=1

(−1)n+1 . n2

1 a0 = π



π

−π π

x2 dx =

2 2 π , 3

 2  π  d 2 − x cos nx dx = cos kx dx π dk2 0 −π k=n  ! 1 2 d π − sin kπ + cos kπ =− π dk k2 k k=n

1 an = π



2

Fourier coefficients and Fourier-series representation

253

 ! 2 π2 2 2π = − − 2 cos kπ + sin kπ 3 − π k k k=n k 4 (−1)n , n2 bn = 0. ∞  cos nx π2 ∴ x2 = + 4 (−1)n 2 . 3 n n=1 =

(a) At x = π: π2 =

∞  1 π2 , +4 3 n2 n=1

i.e.,

π2 = 6

∞  1 . n2 n=1

(b) At x = 0: ∞  π2 (−1)n , = −4 3 n2 n=1

i.e.,

π2 = 12

∞  (−1)n+1 n=1

n2

.

(Questions: Which series would you rather use for a numerical calculation of π2 ? Why?).  4.3.2

Modified Fourier series

We see in these examples that, when a given function is odd in x, only the sine terms will appear. Similarly, the Fourier series representation of a function even in x contains only cosine terms. More generally, any function F(x) can be separated into an even and an odd part: F(x) = Feven (x) + Fodd (x), where 1 Feven (x) = [F(x) + F(−x)] = G1 (x), 2 1 Fodd (x) = [F(x) − F(−x)] = G2 (x). 2 Therefore it follows that

 1 Feven (x) = a0 + an cos nx, 2 n  Fodd (x) = bn sin nx. n

This separation is illustrated in Fig. 4.2.

(4.23a) (4.23b)

254

Fourier series and Fourier transforms (b)

(a) 1.0

–π

π

0

x

(c)

–π

π

0

x

(d) 1.0 0.5

–π

0

π

x

π –π

x

Fig. 4.2 The even and odd parts of a function: (a) F(x); (b) F(−x); (c) Feven (x); (d) Fodd (x).

If the function G1 (x) is originally defined only in the positive half-interval [0, π], we can expand it in a cosine series by first making an even extension to the negative half-interval [−π, 0]. The extended function is just Feven (x), and the cosine series in question is just Eq. (4.23a). In a similar way, a function G2 (x) in the half-interval [0, π] has an odd extension Fodd (x) to [−π, 0], that is the sine series of Eq. (4.23b). Other pieces of F(x) can be isolated in a similar way. For example, if Feven,even (x) = cosine terms with even n only,

(4.24a)

it must also be even about the points x = ±π/2 for each even half of F even (x), since this is the property of cos nx with even n. Similarly, Feven,odd (x) = cosine terms with odd n only

(4.24b)

has two even halves, each of which is odd about its own middle point. Fig. 4.3 shows these Fourier series constructed from the functions of Fig. 4.2, together with the corresponding forms for the sine terms: Fodd,even (x) = sine terms with odd n only

(4.24c)

Fodd,odd (x) = sine terms with even n only.

(4.24d)

Note that the subscripts on F refer to the symmetry of the function, not the evenness or oddness of n.

Fourier coefficients and Fourier-series representation (a)

255

(b) 0.75

0.25 –π

π

0

(d)

(c)

0.25

0.25

Fig. 4.3 Additional Fourier series that can be constructed from the function F(x) of Fig. 4.2(a). (a) F even,even (x): cosine terms with even n only; (b) Feven,odd (x): cosine terms with odd n only; (c) Fodd,even (x): sine terms with odd n only; (d) Fodd,odd (x); sine terms with even n only.

Interesting results can also be obtained by adding Fourier series together. For example, the sum of ∞  cos nx n=1

n

 x = − ln 2 sin , 2

(0, 2π),

and ∞   cos nx x (−1)n+1 = ln 2 cos , n 2 n=1

(−π, π)

is 2

∞  cos(2n + 1)x n=0

2n + 1

 x = − ln tan . 2

This sum is valid only in the open interval (0, π) (i.e., 0 < x < π) common to both the original intervals. (The original intervals are chosen differently in order to avoid the infinity caused by ln(0).) 4.3.3

Periodic extension of F(x) by its Fourier representation

While F(x) is defined only in the interval [−π, π], its Fourier representation f (x), assuming that it does exist in this interval, is also defined for all other values of x. In particular,

256

Fourier series and Fourier transforms

–5π

0 π

–π

–3π

3π

5π

Fig. 4.4 Periodic extension of Fig.4.2(a) outside the interval [−π, π]. ∞  1 f (x + 2mπ) = ao + {an cos[n(x + 2mπ)] + bn sin[n(x + 2mπ)]} 2 n=1

= f (x),

if m = any integer.

Therefore, f (x) for |x| > π is a periodic repetition of the basic function defined in the original interval, with a period of 2π, the length of the original interval. For example, Fig. 4.4 shows the periodic extension of Fig. 4.2(a). 4.3.4

Fourier series for an arbitrary interval

If f (x) is defined for the interval [−L, L] of length 2L, a simple change of variables to y=

π x L

will modify the interval to the standard form [−π, π], now for the variable y. We then have ∞   πx πx  1 an cos n + bn sin n f (x) = a0 + , (4.25) 2 L L n=1 where



 L  F x = y dy π −π  L 1 = F(x) dx, L −L

1 a0 = π

π

and similarly an = bn =



 πx  dx, F(x) cos n L −L   πx  1 L dx. F(x) sin n L −L L

1 L

L

(4.26)

Fourier coefficients and Fourier-series representation

257

For example, if L = 1, the trigonometric functions used are cos nπx and sin nπx. An asymmetric interval A ≤ x ≤ B is sometimes used. The necessary formulas can be obtained readily by changing the variable to y=

π (x − x0 ), L

1 L = (B − A), 2

where

1 x0 = (B + A). 2

The resulting development will be left as an exercise. A change of variable can also be made in a given Fourier series to generate another. For example, given the Fourier series ∞  sin(2n + 1)x

2n + 1

n=0

=

π 4

for the open interval (0, π) (or 0 < x < π), a change of variables to 1 t = x− π 2 yields ∞  n=0

(−1)n

cos(2n + 1)t π = . 2n + 1 4

Integrations and usually differentiations of a given Fourier series will also generate others. Care should be taken with differentiation, since the resulting infinite series is always less convergent than the original one.

Problem 4.3.1 Find the Fourier-series representations of the following functions in the interval −π ≤ x ≤ π: (a) F(x) = 0 if x < 0; F(x) = x, if x > 0; (b) F(x) = 0, if x < 0; F(x) = 1, if x > 0; (c) F(x) = sin ax, where a is an arbitrary constant; (d) F(x) = cos ax. 4.3.2 Find the Fourier-series representation of the following functions in the interval −L ≤ x ≤ L: (a) F(x) = e x ; (b) F(x) = |x|; (c) F(x) = |x|, if |x| ≤ L/2; F(x) = L − |x|, if |x| > L/2; (d) F(x) = −1, if x < −L/3; F(x) = 1, if x > L/3; F(x) = 0, if −L/3 < x < L/3; (e) F(x) = x, if |x| < L/2, F(x) = (L − |x|)x/|x|, if |x| > L/2. 4.3.3 Find the Fourier cosine series in the interval −L ≤ x ≤ L, which will reproduce the functions of Problem 4.3.2 in the half-interval 0 ≤ x ≤ L.

258

Fourier series and Fourier transforms

4.3.4 Find the Fourier sine series in the interval −L ≤ x ≤ L, which will reproduce the functions of Problem 4.3.2 in the half-interval 0 ≤ x ≤ L. 4.3.5 Obtain the Fourier series for an asymmetric interval A ≤ x ≤ B. 4.3.6 Find the Fourier-series representations of the functions of Problem 4.3.2 in the interval 0 ≤ x ≤ L with period L. 4.3.7 Find the function F(x) in [−π, π] whose Fourier-series representation is 1 1 cos 3x + cos 5x + · · · . 9 25

f (x) = cos x +

Hint: This is related to the Fourier series for x2 . 4.3.8 Describe the functions  π ∞  sin nx F(t) sin nt dt, (a) g1 (x) = π1 n=1 −π  π ∞  (b) g2 (x) = π1 sin(2n − 1)x F(t) sin(2n − 1)t dt, n=1 −π  π ∞  (c) g3 (x) = π2 sin nx F(t) sin nt dt, n=1 0  π ∞ 2  (d) g4 (x) = π sin(2n − 1)x F(t) sin(2n − 1)t dt n=1

0

for −2π ≤ x ≤ 2π in terms of a given arbitrary function F(t) defined for −π ≤ t ≤ π. 4.3.9∗ Show that ∞  1 (a) 12 − x = π1 for 0 < x < 1; n sin 2nπx = u(x), (b)

1 4

=

1 π

∞  n=1

n=1 sin (2n−1)πx (2n−1)

= v(x),

for

0 < x < 1;

(c) n(x) = x + u(x) − 8v2 (x) is an integer and that it gives the largest integer in any positive number x. Hint: u(x) is a Fourier representation of 12 − x in [0,1] with a period of 1, while v(x) is an odd extension of 14 in [0,1] to the interval [−1, 1] in a Fourier sine series with a period of 2. 4.3.10 By using a Fourier cosine series given in Appendix 4A, show that ∞  n=0

4.4

(−1)

 1)x π  π  = F(x) = − x − 4 2  (2n + 1)2

n sin(2n +

 π   . 2

Complex Fourier series and the Dirac δ function

The Fourier representation [Eq. (4.25)] of a function F(x) in the interval [−L, L] can be written in another useful form involving complex coefficients:

Complex Fourier series and the Dirac δ function

259

∞   exp[i(nπ/L)x] + exp(−i(nπ/L)x] 1 f (x) = a0 + an 2 2 n=1  exp[i(nπ/L)x] − exp[−i(nπ/L)x] + bn 2i ∞ 

=

cn exp[i(nπ/L)x],

(4.27)

n=−∞

if 1 c0 = a0 , 2

1 cn>0 = (an − ibn ), 2

1 cn 0. This is just what is obtained if δ(ax) = δ(x)/a. If a < 0, the integration over y goes from a positive value −aL to a negative value aL. To bring the direction of integration back to the normal order of lower limit to upper limit, we must write  aL  −aL  −aL dy dy dy =− = . a |a| −aL a aL aL The additional factor |a|−1 can be interpreted as the change in area under the rectangular function Dε (ax) when it is plotted as a function of x rather than as a function of ax.

262

Fourier series and Fourier transforms

The demonstration of Eq. (4.34d) requires a number of steps. An integration over a δ function represents a function lookup at the zero of the argument, here x2 − b2 . As a function of the integration variable x, there are actually two zeros: x = b and −b. Since x2 − b2 = (x − b)(x + b), it behaves like (x − b)2b near x = b, and like −2b(x + b) near x = −b. Hence δ(x2 − b2 ) = δ(2b(x − b)) + δ(−2b(x + b)), where we have used the fact that an integration requires an addition of contributions. Each of the δ functions can be evaluated with the help of Eq. (4.34c) to give the stated result. An even more complicated relation is the identity  ∞ f (x0 ) f (x)δ(g(x))dx =  , (4.35a) |g (x0 )| −∞ if g(x) has a single root or zero at x0 . Here the slope g (x) =

d g(x) dx

at x0 must be nonzero. Thus near x0 the function g(x) behaves like (x − x0 )g (x0 ). Such zeros are called simple zeros. In a similar way, if g(x) has N “simple” zeros at x = xi where g(x)  g (xi )(x − xi ) with nonzero slopes, then N  δ(x − xi )

δ(g(x)) =

|g (xi )|

i=1

.

(4.35b)

The demonstration of these relations will be left as exercises. (See Problem 4.4.5.) Example 4.4.1  ∞ −∞



−x

e δ(x − a )dx = 2

2

∞

e−x

−∞ −a

δ(x − a) + δ(x + a) dx 2a

+ ea )/2a,

= (e

if

a > 0.

Example 4.4.2 

∞

−∞

−x2

e

 δ(sin x)dx = =

∞

−∞

e

∞   n=−∞

=

−x2

∞  n=−∞

⎞ ⎛ ∞ ⎜⎜⎜  δ(x − nπ) ⎟⎟⎟ ⎟⎟⎠ dx ⎜⎜⎝ | cos nπ| n=−∞

∞ −∞

e−x δ(x − nπ)dx 2

e−(nπ) .  2



Complex Fourier series and the Dirac δ function

263

Example 4.4.3 

5

sin x δ((x − 2)(x − 4)(x − 6))dx =

0

sin 4 sin 2 + |g (2)| |g (4)|

on applying Eq. (4.35b). There is no contribution for the δ function at x = 6 because it is outside the limits of integration. To evaluate the derivatives, we note that g (x) = (x − 4)(x − 6) + (x − 2)(x − 6) + (x − 2)(x − 4). Hence g (2) = 8, g (4) = −4, and the integral is 18 sin 2 + 14 sin 4.  Even derivatives of the δ function can be defined and used. For example, if δ (x − x ) ≡

d δ(x − x ), dx

(4.36)

then an integral involving it can be calculated by an integration by parts:  L F(x ) δ (x − x )dx = F(L) δ(x − L) − F(−L) δ(x + L) −L

 −

L

−L

F  (x ) δ(x − x )dx .

(4.37)

An advanced but noteworthy feature involving the δ function is connected with Eq. (4.34b). This states that xδ(x) is identically zero. As a result, we may add any finite multiple of this zero to one side of an equation A(x) = B(x) = B(x) + cxδ(x), where c is an arbitrary finite constant. However, if we should divide both sides by x, the addition cδ(x) is not longer zero at x = 0. Then A(x)/x = B(x)/x + c δ(x)

(4.38)

is not necessarily true for arbitrary values of c. For example, x

d ln x = 1 = 1 + cx δ(x) dx

is valid for any finite constant c. However, d 1 ln x = + c δ(x) dx x

(4.39)

is true only for special values of c. To determine these values, we must examine the behavior of these functions in the neighborhood of x = 0 by integrating both sides of the expression from −ε to ε:

264

Fourier series and Fourier transforms

 I=

ε

d ln x dx = ln ε − ln(−ε) = ln(−1), −ε dx

while on the right-hand side

 I=

ε

dx + c = c, −ε x

because l/x is an odd function of x. Thus c = ln(−1) = ln(ei(2n+1)π ) = i(2n + 1)π,

(4.40)

where n is any integer. The need for the additive constant c in Eq. (4.39) will become very clear in Chapter 8, where we shall find that d ln x/dx is a multivalued function, while x−1 is a single-valued function. The different values of c are thus needed to match the former function’s many values. The δ-function term in Eq. (4.39) is not just a mathematical curiosity. It plays an important role in quantum mechanics.

Problems 4.4.1 Verify that the Dirac δ function for the real Fourier series in the interval [-L, L] is   ∞  nπx  1 nπx 1  cos cos + δ(x − x ) = 2L L n=1 L L  nπx   nπx ! sin . + sin L L 4.4.2 Obtain the Dirac δ function δ(x − x ) for (a) The Fourier cosine series; and (b) the Fourier sine series, in the half interval [0,L]. Why are these two answers not the same? 4.4.3 By integrating the Dirac δ function in Problem 4.4.1 or 4.4.2, obtain the Fourier series for the function  1, x>0 G(x) = −1, x < 0. 4.4.4 Given a function

⎧ −1 1  ⎪ ⎪ ⎨ε , |x − x | < 2 ε Dε (x − x ) = ⎪ ⎪ ⎩0, |x − x | > 12 ε 

in the interval, (−L, L). (a) Obtain its Fourier-series representation.

Fourier transform

265

(b) Show that in the limit ε → O this Fourier-series representation agrees with that for the Dirac δ function obtained in Problem 4.4.l or 4.4.2. 4.4.5 Derive Eqs. (4.35a) and (4.35b). 4.4.6 If F(x) is discontinuous at x = 0, show that ∞ (a) 0 F(x) δ(x) dx = 12 limε→0 F(ε), 0 (b) −∞ F(x) δ(x) dx = 12 limε→0 F(−ε).

4.5

Fourier transform

The complex Fourier series has an important limiting form when L → ∞. Suppose that as L → ∞ 1. nπ/L ≡ k remains large (ranging in fact from −∞ to ∞), and. 2. cn → 0 (because it is proportional to L−1 ), but  ∞ L 1 g(k) ≡ lim cn = f (x)e−ikx dx = finite. L→∞ π 2π −∞ c →0

(4.41)

n

Then f (x) =

∞ 

cn eikx ,

n=−∞

k=

nπ L

∞  π = lim g(k)eikx . L→∞ L c →0 n=−∞ n

The sum over n in steps of Δn = 1 can be written as a sum over k, which is proportional to n. Hence ⎞ ⎛  π ⎜⎜⎜ L ⎟⎟⎟ π  ⎟⎟⎟⎠ = . = ⎜⎜⎜⎝ L π L n Δk Δk However, Δk = (π/L)Δn becomes infinitesimally small when L becomes large. The sum over Δk then becomes an integral π  = dk, (4.42) lim L→∞ L c →0 n n

and

 f (x) =

∞

−∞

dk g(k)eikx .

(4.43)

We call g(k) of Eq. (4.41) the Fourier transform of f (x), and Eq. (4.43) the Fourier inversion formula.

266

Fourier series and Fourier transforms

To obtain the Dirac δ function for this transformation, we substitute Eq. (4.41) into Eq. (4.43) to obtain  ∞  ∞  ikx 1 f (x) = dk e dx e−ikx f (x ) 2π −∞ −∞  ∞ dx f (x )δ(x − x ). = −∞

Hence 1 δ(x − x ) = 2π 

4.5.1



∞

−∞



dkeik(x−x ) .

(4.44)

Conjugate symmetry

The Fourier transform and its inverse are similar in structure. This symmetry can be made explicit by using the symmetrical definitions:  ∞ 1 e−ikx f (x)dx, (4.45a) g(k) = √ 2π −∞  ∞ 1 f (x) = √ eikx f (k)dk. (4.45b) 2π −∞ Thus the roles of f (x) and g(k) can be interchanged under the substitutions x ↔ k and i → −i. We call this substitution a conjugate transformation. The variables x and k are said to form a conjugate pair of variables. The conjugate pair x and k are joined together by the fact that either f (x) or g(k) contains all the information about the function. This is an important feature of Fourier transforms and of wave functions to which we shall return in a later section. 4.5.2

Properties and applications of Fourier transforms

Fourier transforms involve complex integrations, which are usually studied as part of the theory of functions of complex variables. This theory is described in Chapter 8. For the time being, we shall be content with the transforms of very simple functions, or with the use of tables of Fourier transforms in mathematical handbooks. A short table is appended to this chapter for the reader’s convenience. In referring to Fourier transforms, it is convenient to use the notation  ∞ 1 F { f (x)} ≡ √ e−ikx f (x)dx = g(k). (4.46) 2π −∞ Example 4.5.1 The Fourier transform of the box function  1, |x| ≤ α b(x) = 0, |x| ≥ α

(4.47)

Fourier transform

267

Table 4.1 Properties of Fourier transforms.

Property

If f (x) is

Its Fourier transform is

Complex conjugation

Real Real and even Real and odd

g∗ (k) = g(−k) g∗ (k) = g(−k) = g(k) g∗ (k) = g(−k) = −g(k)

Translation Attenuation

f (x − a) f (x)eax

e−ika g(k) g(k + ai)

Derivatives

d dx f (x) dn f (x) dxn

ikg(k) =

f (n) (x)

(ik)n g(k)

is α  α 1 1 e−ikα  −ikx F {b(x)} = √ e dx = √  2π −α 2π −ik −α 1 2 = √ sin kα.  2π k

(4.48)

Example 4.5.2 

 ,   ∞ d 1 −ikx d F e f (x) = √ f (x) dx dx dx 2π −∞ ∞  1 (−ik) ∞ −ikx −ikx  = √ f (x)e  − √ e f (x)dx −∞ 2π 2π −∞ after an integration by parts. If f (x) → 0 as |x| → ∞, only the second term survives. Therefore  , d F f (x) = ikF { f (x)}.  dx This and other simple properties of the Fourier transform are summarized in Table 4.1. Their derivations are left as exercises.

4.5.3

Calculation of Fourier transforms

The formulas for derivatives are particularly useful because they reduce differential expressions to algebraic expressions, as the following example shows.

268

Fourier series and Fourier transforms

Example 4.5.3 Solve the inhomogeneous differential equation   2 d d + p + q f (x) = R(x), −∞ ≤ x ≤ ∞, dx dx2 where p and q are constants. We transform both sides  2 , df d f F + p + qf = [(ik)2 + p(ik) + q] f˜(k) dx dx2 ˜ = F {R(x)} = R(k), where a tilde is used to denote a Fourier transform. Hence  ∞ 1 f (x) = √ eikx g(k)dk 2π −∞  ∞ ˜ 1 R(k) dk.  eikx 2 = √ −k + ipk + q 2π −∞

(4.49)

(4.50)

(4.51)

The formal solution is called an integral representation of the solution. Of course, it will not do us any good if we do not know how to evaluate this complex integral. Fortunately, this is a simple problem in the theory of functions of complex variables, which will be discussed in Chapter 8. Hence we have gained by obtaining an integral representation. Example 4.5.4 Use the translation property of Table 4.1 to obtain the Fourier transform of the translated box function b(x − β) of Eq. (4.47). 1 2 F {b(x − β)} = e−ikβ F {b(x)} = e−ikβ √ sin kα, 2π k where F {b(x)} is from Eq. (4.48). 

Problems 4.5.1 Obtain the Fourier transform of  0, f (x) = −ax e sin bx,

x 0.

4.5.2 Find the 3D Fourier transform of the wave function of a 1s electron in the hydrogen atom: ψ1s (r) =

1 (πa20 )1/2

where a0 is the radius of the orbit.

exp(−r/a0 ),

Green function and convolution

4.5.3 Use the Fourier transform  , sinh ax 1 cos a F = √ , sinh πx 2π cosh k + cos a to find

269

|a| < π,



, 1 − exp(−bx) −cx F e , 1 − exp(−2πx)

where b > 0 and c > 0. Should the value of b be restricted? 4.5.4 If g(k) = F { f (x)}, show that   1 k−b ibx , a > 0; (a) F { f (ax)e } = g a a    ! 1 k−b k+b g +g , a > 0; (b) F { f (ax) cos bx} = 2a a a    ! k−b k+b 1 g −g , a > 0. (c) F { f (ax) sin bx} = 2ai a a What are the results if a < 0? 4.5.5 Use Eq. (4.44) to show that  ∞  ∞  ∞ 2 ∗ | f (x)| dx = f (x)dx δ(x − x ) f (x )dx −∞

 =

−∞ ∞ −∞

−∞

|g(k)|2 dk.

4.5.6 Use Entry 2 of the table of Fourier transforms given in Appendix 4C to obtain the following results:  , √ 1 (a) F = −i 2πe−ka−ikb Θ(k), x − b + ia ,  √ 1 = i 2πeka−ikb Θ(−k), if a > 0. F x − b − ia Here Θ(x) is 0 if x < 0, and 1 if x > 0. ,  & 1 (b) F 2 = π/2 exp(−|k|). x +1

4.6

Green function and convolution

If the inhomogeneity function in Eq. (4.49) is R(x) = δ(x),

270

Fourier series and Fourier transforms

˜ is just (2π)−1/2 . Equation (4.51) then states that the solution its Fourier transform R(k) of the differential equation 

 d + p + q G(x) = δ(x), dx dx2 d2

(4.52)

which is now denoted by G(x), has a Fourier transform that is (2π)−1/2 times the reciprocal of the Fourier transform of the differential operator. Being essentially the inverse of a differential operator, the function G(x), called a Green function, is potentially a very useful idea. To appreciate its power, let us suppose that the solution of any inhomogeneous linear differential equation L (x) f (x) = R(x)

(4.53)

involving an arbitrary differential operator L (x) can be written in the integral form  f (x) =

∞

−∞

G(x − x )R(x )dx .

(4.54)

Substitution of Eq. (4.54) into Eq. (4.53) gives the result 

b

[L (x)G(x − x )]R(x )dx = R(x).

a

Thus the original differential equation is satisfied if L (x)G(x − x ) = δ(x − x ),

(4.55)

that is, if G(x − x ) is the Green function for the differential operator L (x). An integral of the form (4.54) is called a convolution, and may be denoted by the symbol 1 (G ∗ R) x ≡ √ 2π



∞ −∞

G(x − x )R(x )dx .

(4.56)

˜ R(k) ˜ of transforms. This Its Fourier transform turns out to be always the product G(k) convolution theorem can be derived directly as follows: 1 F {(G ∗ R) x } ≡ √ 2π

 e

−ikx



1 dx √ (2π)







G(x − x )R(x )dx





Green function and convolution

271



  1 −ik(x−x )   = √ G(x − x )d(x − x ) e 2π    1 −ikx   R(x )dx e × √ 2π = F {G}F {R}.

(4.57)

This shows, perhaps more clearly than Eq. (4.54), that the Green function may be interpreted as the inverse of the differential operator L (x). If L (x) involves functions ˜ of x rather than just constant coefficients, it is not so easy to calculate G(k), but that is another story. (See Chapter 5 for a general method for calculating Green functions.) In engineering, the inhomogeneity function R(x) in Eq. (4.53) is called an input to, and the solution f (x) is called an output from, the system, while the Green function in Eq. (4.54) is called a response function, since it describes how the system responds to the input. Newton’s equation of motion in the presence of a driving force m

d2 dt2

r(t) = F(t)

is another example of Eq. (4.53). A δ-function force proportional to δ(t) is called an impulsive force. Thus the Green function also describes the response of a mechanical system to an impulsive driving force. Example 4.6.1 Obtain a solution to the equation of a driven harmonic oscillator x¨(t) + 2β x˙(t) + ω20 x(t) = R(t),

(4.58)

where β and ω0 are positive real constants. According to Eq. (4.54), x(t) has the form  ∞ x(t) = G(t − t )R(t )dt ,

(4.59)

−∞

where the Green function G(t − t ) is the solution of the differential equation with an impulsive driving force:  2  d d 2 + 2β + ω0 G(t − t ) = δ(t − t ). dt dt2 Let 1 ˜ F {G(t − t )} = G(ω) ≡ √ 2π



∞

−∞



e−iω(t−t )G(t − t )d(t − t ).

Proceeding in the same way as in Eq. (4.50), we obtain 1 1 ˜ . G(ω) = √ 2 2 2π (ω0 − ω ) + i2βω

272

Fourier series and Fourier transforms

The Green function 1 G(t) = √ 2π



∞ −∞

˜ eiωtG(ω)dω

can now be computed with the help of Entry 2 of the table of Fourier transforms given in Appendix 4C. The result (to be worked out in Problem 4.6.2) is G(t) =

1 −βt e sin ω1 t Θ(t), ω1

where

 Θ(t) =

is the unit step function and ω1 =

(4.60)

0 if t < 0 1, t > 0,

% ω20 − β2 .

The use of Eq. (4.60) in Eq. (4.59) gives the explicit solution  t 1  x(t) = e−β(t−t ) sin ω1 (t − t )R(t )dt . ω1 −∞

(4.61a)

We should note the interesting feature that the integral over t takes into account the effects of all driving forces occurring in the past (t < t). It contains no effect due to driving forces in the future (t > t), because these forces have not yet occurred. Hence the result is explicitly consistent with the physical requirement of causality. To check that the Green function in Eq. (4.60) is correct, we can apply it to obtain the response to the simple driving force R(t ) = eiΩt . Eq. (4.61a) then gives (to be worked out in Problem 4.6.2) x(t) =

eiΩt . (ω20 − Ω2 ) + i2βΩ

(4.61b)

It can be verified by direct substitution that this is the solution of Eq. (4.58) for the given driving force. 

Problems 4.6.1 Suppose the input (as a function of time t) to a system  0, t 0,

Heisenberg’s uncertainty principle

273

gives rise to the output  f (t) =

0, t 0,

where α and β are positive constants. (a) Find the Fourier transform 1 ˜ G(ω) = √ 2π



∞ −∞

e−iωt G(t)dt

˜ of the response function G(t). [G(ω) is also called a response function.] (b) Obtain the response of the system to the input R(t) = Aδ(t). 4.6.2 (a) Use Entry 2 of the table of Fourier transforms given in Appendix 4C to verify Eq. (4.60). (b) Verify Eq. (4.61b).

4.7

Heisenberg’s uncertainty principle

The Fourier transform of the Gaussian function 1 f (x) = N exp(− cx2 ) 2 is of considerable interest: 1 g(k) = √ 2π



∞

−∞

1

(4.62)

e−ikx Ne− 2 cx dx. 2

It can be calculated by completing the square in the exponent  2 1 2 c k2 ik 1 − cx − ikx = − x + − = − cy2 − k2 /2c, 2 2 c 2c 2 so that N 2 g(k) = √ e−k /2c cπ



∞

−∞

−(1/2)cy2

e

&



2 N c/2dy = √ e−k /2c . c

(4.63)

We see that a Gaussian transforms into a Gaussian but with an inverted falloff constant c−1 that is inversely proportional to the old falloff constant c in f (x). Thus a narrow f (x) gives rise to a broad g(k), and vice versa. This is illustrated in Fig. 4.5. The reciprocal width relations shown in Fig. 4.5 turn out to be a general property of Fourier transforms. It can be characterized more precisely as follows. Let us denote integrals by a “bracket” symbol called a scalar, or inner, product.  ∞ ( f1 , f2 ) ≡ f1∗ (x) f2 (x)dx. (4.64) −∞

274

Fourier series and Fourier transforms g(k)

f(x) Large c:

Narrow

Broad

x

k g(k)

f(x)

Narrow

Broad

Small c:

k

x

Fig. 4.5 Reciprocal width relations in Fourier transforms.

If gi (k) is the Fourier transform of fi (x), then  ( f 1 , f2 ) =

∞

−∞ ∞

 =

−∞

 = =

f1∗ (x) f2 (x)dx 1 √ 2π

∞ ∞

−∞  ∞ −∞

−∞



∞

−∞

∗ 

ikx

e g1 (k)dk

g∗1 (k)g2 (k )



1 √ 2π



1 √ 2π

∞ −∞

dxe



∞ −∞

ik x

e

−i(k−k )x

g∗1 (k)g2 (k)dk = (g1 , g2 ).









g2 (k )dk dx

dkdk (4.65a)

This shows that the inner product can be calculated by using either f or g. This is to be expected since g contains the same information as f ; they describe the same mathematical system, which may be denoted by the symbol φ such that φ = f (x) or g(k) depending on whether the integration variable is x or k. We call f (x) or g(k) the x, or k, representation of φ. In this notation, we may write Eq. (4.65a) as (φ1 , φ2 ) = ( f1 , f2 ) = (g1 , g2 ).

(4.65b)

More general integrals can also be represented readily in this new notation. For example, 

∞

−∞

f1∗ (x)A(x) f2 (x)dx = ( f1 , A(x) f2 ) = (φ1 , A(x)φ2 ),

(4.66)

Heisenberg’s uncertainty principle

while



∞ −∞

275

g∗1 (k)B(k)g2 (k)dx = (g1 , B(k)g2 ) = (φ1 , B(k)φ2 ).

(4.67)

Here we refer explicitly to the x dependence of A(x) or the k dependence of B(k) to remind ourselves in which representation we know these functions. To illustrate these expressions, let us calculate a few inner products for the Gaussian function (4.62):  ∞ & 2 e−cx dx = N 2 π/c, (φ, φ) = ( f, f ) = N 2  (φ, xφ) = N 2 (φ, x2 φ) = N 2

−∞

∞

−∞  ∞ −∞

e

−cx2

xdx = 0,

e−cx x2 dx = 2

N2 & π/c. 2c

The average value of x in the system φ is called its expectation value x ≡ (φ, xφ)/(φ, φ) = x¯,

(4.68)

while its uncertainty or dispersion Δx is defined by (Δx)2 = (φ, (x − x¯)2 φ)/(φ, φ) = (φ, x2 φ)/(φ, φ) − x¯2 .

(4.69)

Hence the Gaussian function (4.62) has x¯ = 0,

(Δx)2 =

1 . 2c

(4.70)

Similarly, it is easily to see that for the same Gaussian function k¯ = 0,

c (Δk)2 = , 2

(4.71)

since the Gaussian falloff constant in g(k) is c−1 . Thus for the Gaussian function (4.62), the product of uncertainties has a unique value Δx Δk =

1 , 2

(4.72)

independent of c. We have thus found that the conjugate variables x and k in φ cannot simultaneously be known with infinite precision. If one is known better, knowledge of the other

276

Fourier series and Fourier transforms

must unavoidably be reduced proportionally. A complete knowledge of one, say k, is possible only when there is complete ignorance of the other. To see in physical terms why x and y cannot be known simultaneously with infinite precision, let us recall that, if x is the position of a wave, then k is its wave number. A wave with a unique value of k is infinitely long, for otherwise it will have a beginning and an end where the wave does not oscillate in the same manner as over its length. An infinitely long wave like this does not have a definite position, since x can be anywhere along its length. Hence the position uncertainty Δx must be infinite in order for Δk to be zero. For Gaussian functions, the uncertainty product Δx Δk is a constant. It turns out that for other functions the uncertainty product can be larger, but never smaller. That is, Δx Δk ≥

1 2

(4.73)

is generally true, as we shall prove in the next section. For matter waves, k is the particle momentum according to Eq. (4.4). Hence the uncertainty relation can be written in the form Δx Δp ≥ /2,

(4.74)

called the Heisenberg uncertainty principle (1927). This states that the position and the momentum of a massive particle cannot be known simultaneously with infinite precision, in dramatic contrast to the situation assumed in Newtonian mechanics. The uncertainty principle describes an important feature of quantum-mechanical systems. It has interesting epistemological implications as well, since it imposes a limit on the precision of our knowledge of physical systems.

Problems 4.7.1 Obtain the Fourier transform of the sequence of functions n δn (x) = √ exp(−n2 x2 ) π and show that lim(n → ∞) δn (x) = δ(x). 4.7.2 For the Gaussian functions (4.62) and (4.63) show directly that (a) (φ, k2 φ) = −(φ, (d/dx)2 φ) (b) (φ, eika φ) = (φ, ea(d/dx) φ). Hint: Use the displacement property discussed in Section 2.9.

4.8

Conjugate variables and operators in wave mechanics

Given mathematical systems φi that have x representations fi (x) and k representations gi (k), it is clear that the integrals (φi , A(x)φj ) involving a given function A(x) is

Conjugate variables and operators in wave mechanics

277

conveniently calculated by an x integration, while the integral (φi , B(k)φj ) is readily calculated by a k integration, It is also possible to calculate ( fi , A(x) fj ) in the k representation, however. To do this, we note that, according to Eq. (4.65b), an inner product is unchanged under Fourier transformation (φi , φj ) = ( fi , fj ) = (F { fi (x)}, F { fj (x)}) = (gi , gj ).

(4.65c)

Hence (φi , A(x)φj ) = ( fi , A(x) fj ) = (F { fi (x)}, F {A(x) fj (x)}). F { fi (x)} is just gi (k), while F {A(x) fj (x)} can be evaluated readily with the help of the convolution theorem    ∞  ∞ 1 1 −ikx  ik x ˜  F {A(x) fj (x)} = √ dxe dk e A(k ) √ 2π −∞ 2π −∞    ∞ 1  ik x  dk e gj (k ) × √ 2π −∞   1 ˜  = δ(k + k − k)dk dk √ A(k )gj (k ) 2π  1 ˜ − k )gj (k ). = dk A(k 2π

According to Eq. (4.56), the final expression is a convolution F {A(x) fj (x)} = A˜ ∗ gj ,

(4.75)

a result that actually can be read off directly from Eq. (4.57). Hence (φi , A(x)φj ) = ( fi , A(x) fj ) = (gi , A˜ ∗ gj )

(4.76)

describes how the integral can be calculated in the k representation. There is another way of representing this change of representation that is even more suggestive and useful. According to Table 4.1, ,  d F f (x) = ikF { f (x)} = ikg(k), dx and  F

d dx

n

, f (x) = (ik)n g(k),

(4.77a)

278

Fourier series and Fourier transforms

when certain boundary terms vanish, as shown explicitly in Example 4.5.2. Similarly, ,  n d F g(k) = (−ix)n f (x), (4.77b) dk where we have used the conjugate transformation x ↔ k, i ↔ −i. Hence a change of integration variable can be presented symbolically as follows:     d gj , (4.78a) ( fi , A(x) fj ) = (φi , A(x)φj ) = gi , A i dk     1 d (gi , B(k)gj ) = (φi , B(k)φj ) = fi , B (4.78b) fj . i dx Indeed, this is the more flexible procedure, because it can accommodate an expression appearing between the φs in which both conjugate variables appear simultaneously:     1 d fj (φi ,C(x, k)φj ) = fi ,C x, i dx     d = gi ,C i , k gj . (4.79) dk When this occurs, we must remember that one variable is expressible as a differential operator of the other, so that they do not commute:     d d [k, x] = kx − xk = −i x − x −i = −i dx dx     d d =k i − i k = −i. (4.80) dk dk Thus when both conjugate variables x and k appear in C, they are no longer simple variables but noncommuting operators whose orderings in the expression must be carefully preserved. We are now in a position to prove Heisenberg’s uncertainty principle 1 (Δx)(Δk) ≥ , 2 where Δu is the uncertainty of the expectation value of u as defined by Eq. (4.69). This inequality turns out to be related to the so-called Schwarz inequality in vector algebra, which states that for two real vectors A and B A2 B2 ≥ (A · B)2 = (ABcos θ)2 ,

(4.81)

Conjugate variables and operators in wave mechanics

279

¯ then where A2 = A · A. If we now take A = (x − x¯)φ, B = (k − k)φ, A2 = ((x − x¯)φ, (x − x¯)φ) = (φ, (x − x¯)2 φ) = (Δx)2 ,

(4.82)

where we have used the fact that the expression appearing to the left of the comma is transposed. Therefore, A2 B2 = (Δx)2 (Δk)2 ≥ (A · B)2 ¯ 2 = |(φ, (x − x¯)(k − k)φ)| ¯ 2 = |(x − x¯)(k − k)| = |αβ|2 , where (Problem 4.8.1) α = x − x¯,

β = k − k¯

(4.83)

are Hermitian operators. If we work in terms of symmetric and antisymmetric products, we can express the last expression as 1 1 |αβ|2 = | αβ − βα + αβ + βα|2 2 2 1 1 1 = |αβ − βα|2 + |αβ + βα|2 + Reαβ − βααβ + βα. 4 4 2 The third term on the right vanishes because αβ − βααβ + βα is purely imaginary. (This is because αβ + βα is real and αβ − βα is purely imaginary. See Problem 4.8.2.) The second term 14 |αβ + βα|2 is non-negative. Hence 1 (Δx)2 (Δk)2 ≥ |αβ|2 ≥ |αβ − βα|2 . 4 The remaining term involves the commutator ¯ = [x, k] = i, [α, β] = [x − x¯, k − k] where use has been made of Eq. (4.80). Hence |αβ − βα|2 = 1 and 1 (Δx)(Δk) ≥ . 2

Problems 4.8.1 An operator H is said to be Hermitian if Hij = (φi , Hφj ) = Hji∗ = (φj , Hφi )∗ .

(4.84)

280

Fourier series and Fourier transforms

(a) Show that the position operator xˆ is Hermitian if it has only real values in the x representation. You may assume, for simplicity, that the possible position eigenvalues are all discrete and that the eigenvectors are orthonormal. d is a Hermitian operator in the k representation, that is (b) Show that x = i dk 

! ,∗ d gi (k) dk dk −∞   ∞ ! ,∗ d ∗ = −i gj (k) gi (k)dk + boundary terms −∞ dk !  ∞ d ∗ = gi (k) i gj (k) dk, dk −∞ ∞

g∗j (k) i

only if gj (k) and gi (k) satisfy suitable boundary conditions at k = ±∞. ¯ show thatαβ + βα 4.8.2 If x, k are Hermitian operators and α = x − x¯, β = k − k, is real, while αβ − βα is purely imaginary. 4.8.3 Show that an eigenvector of a matrix is simultaneously an eigenvector of all other matrices that commute with it.

4.9

Generalized Fourier series and Legendre polynomials

The developments of the preceding sections suggest that there is considerable similarity between the expansion of functions by Fourier series and transforms and the decomposition of simple vectors into components. It is now useful to develop this similarity in more detail in order to gain additional understanding into the nature of the Fourier-series expansion, or equivalently into the mathematics of wave motion. Let us start by reminding ourselves that a vector A in 3D space has three Cartesian components  A = A1 e1 + A2 e2 + A3 e3 = Ai ei . (4.85) i

Given A, its component Ai along ei can be calculated by means of the scalar product ei · A = ei · (A1 e1 + A2 e2 + A3 e3 ) = Ai .

(4.86)

The Fourier series in Eq. (4.27) [or Eq. (4.21)] has a similar structure. The Fourier series f (x), now called F(x) if we believe it to be as good as the original function F(x), F(x) =

∞  n=−∞

cn ψn (x)

(4.87)

Generalized Fourier series and Legendre polynomials

281

is also a sum of terms, each of which is made up of a Fourier coefficient cn (the analog of a vector component) and a unique function  nπ  ψn (x) = exp i x (4.88) L which plays the role of ei . The functions ψn (x) satisfy the integral relation  L  L  πx  ψ∗m (x)ψn (x)dx = exp i (m − n) dx L −L −L  2L, m=n = 2L sin(m − n)π, m  n (m−n)π

= 2Lδmn .

(4.89)

If we left multiply Eq. (4.87) by ψ∗m (x) and integrate over x, we get 

L

−L

ψ∗m (x)F(x)dx =



 cn

n

=



L −L

ψ∗m (x)ψn (x)dx

cn 2Lδmn = 2Lcm .

(4.90)

n

In this way, the formula (4.28) for Fourier coefficients can be derived directly. In the language of vector algebra, the extraction of a Fourier coefficient has involved a scalar product, as in Eq. (4.86). Hence the integrals over x in both Eq. (4.89) and Eq. (4.90) may also be called scalar or inner products. Scalar products of vectors describe their orthonormalization properties. In a similar way, inner products of functions describe the orthonormalization of functions. In particular, Eq. (4.89) states that the functions ψm (x) are orthogonal to each other and that they can be normalized into the unit functions 1 en (x) = √ exp(inπs/L) 2L satisfying the orthonormality relations  L e∗m (x)en (x)dx = δmn , (em , en ) ≡ −L

(4.91)

(4.92)

where we have used the inner-product notation of Section 4.7. If we now write the Fourier series in the vectorial form F(x) =

∞  n=−∞

F n en (x),

(4.93)

282

Fourier series and Fourier transforms

reminiscent of Eq. (4.85), the components F n of F(x) must necessarily be given by the inner product  L F(x) = (en , F) ≡ e∗n (x)F(x)dx. (4.94) −L

Even the Dirac δ function (4.29) has the simple form ∞ 



δ(x − x ) =

en (x)e∗n (x )

(4.95)

n=−∞

in this notation. The number of distinct unit functions en (x) appearing in these expressions is infinite. Hence the space involved is said to be infinite dimensional. A particularly important feature of this space is that inner products are defined, so that the concepts of length and orthogonality become meaningful. In recognition of this fact, the space is called an inner-product space. It is well known that any choice of three perpendicular axes can be used in the expansion (4.85) of vectors in space. In the case of functions, a different choice of coordinate axes means the use of different unit functions en (x). To illustrate this idea, let us consider the complex Fourier series (4.87) for the interval [−1, 1], that is, for L = 1. The original functions in Eq. (4.88) can be expanded into a convergent infinite series in powers of x: ψn (x) = exp(inπx) = 1 + inπx +

1 (inπx)2 + . . . . 2!

Since powers of x are even simpler than exponential functions, we ask if we cannot use these powers directly in the expansion instead of ψn (x). To study this question, let us take the first two basis functions to be P0 (x) = 1,

P1 (x) = x.

These satisfy the orthogonality relations  1 (P0 , P0 ) = dx = 2, −1

(4.97a) 

(P0 , P1 ) = (P1 , P0 ) =  (P1 , P1 ) =

1

−1

x2 dx =

1 −1

dx = 0,

2 . 3

The next power x2 is orthogonal to P1 (x)  (P1 , x ) = (x , P1 ) = 2

2

(4.96a)

1 −1

x3 dx = 0,

(4.97b) (4.97c)

Generalized Fourier series and Legendre polynomials

283

but not to P0 (x)  (P0 , x ) = (x , P0 ) = 2

2

1 −1

x2 dx =

2 . 3

(4.98)

Hence it cannot be one of a set of mutually orthogonal functions that can be used for expanding functions in what might be called generalized Fourier series. If x2 is not orthogonal to P0 , part of it must be parallel to it. Indeed, Eq. (4.97a) shows that the part parallel to P0 in x2 must be 13 P0 , since it gives the same inner product 23 with P0 , according to Eqs. (4.97a) and (4.98). In other words, x2 − 13 P0 must be orthogonal to P0 . This can be checked explicitly 1 1 2 1 (P0 , x2 − P0 ) = (P0 , x2 ) − (P0 , P0 ) = − (2) = 0. 3 3 3 3 It has become customary to use not x2 − 13 P0 but the function 3 1 P2 (x) = (x2 − ) = (3x2 − 1)/2. 2 3

(4.96b)

proportional to it chosen such that P2 (x = 1) = 1. The next power is x3 , which is orthogonal to all even powers of x, but not to the odd powers. The combination P3 (x) = (5x3 − 3x)/2.

(4.96c)

can easily be shown to be orthogonal to P1 (x) and also satisfies the (arbitrary) convention P3 (x = 1) = 1. Similarly, P4 (x) = (35x4 − 30x2 + 3)/8,

(4.96d)

P5 (x) = (63x5 − 70x2 + 15x)/8

(4.96e)

are orthogonal to polynomials of degree m < 4 and 5, respectively. Proceeding in this manner, we can construct the polynomials Pn (x), n = 0, 1, 2, . . . , ∞, such that Pn (x = 1) = 1 and Pn (x) is orthogonal to all Pm (x) with m < n. These polynomials are called Legendre polynomials. They can be shown to satisfy the orthogonality relation  1 2 Pm (x)Pn (x)dx = (4.99) δmn . 2n + 1 −1 a result we shall derive in Chapter 7. Given a function F(x) defined in [−1, 1], we can expand it in the Legendre series F(x) =

∞  n=0

cn Pn (x),

(4.100)

284

Fourier series and Fourier transforms 1 P0 P1

Pn(x)

P2 1 x

–1

0 P3

–1

Fig. 4.6 Pn (x) for n = 0, 1, 2, and 3.

where the Legendre coefficient cn can be obtained by multiplying Eq. (4.100) by Pm (x) and then integrating over x: 

1 −1

Pm (x)F(x)dx =

 n

= cm

 cn

1

−1

Pm (x)Pn (x)dx

2 . 2m + 1

That is, 2n + 1 cn = 2



1 −1

Pm (x)F(x)dx.

(4.101)

The Legendre polynomials are sketched in Fig. 4.6. The expansion of functions in Legendre series is illustrated by two examples. Example 4.9.1 Expand the function  G(x) =

1, −1,

0≤x≤1 −1 ≤ x ≤ 0,

(4.102)

in a Legendre series. Since G(x) is odd in x, the Legendre coefficients of even n vanish. The odd-n coefficients are

Generalized Fourier series and Legendre polynomials

3 c1 = 2 c3 = = c5 =

7 2 7 2





1

−1



0

1

G(x)

−1



0

11 2

1



1

G(x)xdx = 3 5x3

285

3 xdx = , 2

− 3x dx 2

7 (5x3 − 3x)dx = − , 8 1

0

63x5 − 70x3 + 15x 11 dx = , 4 16

etc.

Hence G(x) =

7 11 3 P1 (x) − P3 (x) + P5 (x) − . . . . 2 8 16

(4.103)

The first three partial sums of this Legendre series are shown in Fig. 4.7. They should be compared with those for the Fourier expansion of the same function given in Example 4.3.2.  Example 4.9.2 Expand F(x) = cos πx in [−1, 1] in a Legendre series. Since F(x) is even, only the coefficients with even n can be nonzero. To calculate these we need the integrals  2  1  d x cos πx dx = 2 − 2 cos kx dx k=π dk −1 0   2 d sin k   = −2 2 k k=π dk



1

2

=−

4 , π2

    d2 sin k  8 6  x cos πx dx = 2 2 = 1− 2 , k k=π π2 π dk −1



1

4

etc.

Hence the Legendre coefficients are 1 c0 = 2 5 c2 = 2 =−

 

1 −1

cos πx dx = 0,

   3 2 1 15 1 2 x cos πx dx x − cos πx dx = 2 4 −1 −1 2 1

15 , π2

286

Fourier series and Fourier transforms

1

1

0

–1

1

x

1

1

0

1

x

0

–1

1

x

–1

–1

1

1

1

0

–1

x

–1

–1

–1

1

0

–1

x

–1

0

1

x

–1

–1 (b)

(a)

Fig. 4.7 The first few partial sums of (a) the Legendre series [Eq. (4.103)] and (b) the Fourier series (Example 4.3.2) for the same function (4.102).

c4 =

91 28



1

−1

(35x4 − 30x2 − 3) cos πx dx

   ! 9 6 15 8 = 35 2 1 − 2 − 30 − 2 16 π π π   9 730 1680 = − 4 , etc.  16 π2 π If there are different sets of functions in terms of which a given function in a given interval can be expanded, it is often necessary to decide which set is the best to use. The choice is not always obvious, and may depend on the structure of a problem as well as the efficacy of the expansion for the specific problem.

Orthogonal functions and orthogonal polynomials

287

Problems 4.9.1 Show that the Legendre polynomials P3 (x) and P4 (x) are orthogonal to polynomials of lower degrees. 4.9.2 Obtain the first three coefficients of the Legendre expansions of the following functions in the interval [−1, 1]: (a) eikx ; (b) e−γx ; (c) e−1/n , interger n > 1, 4.9.3 Give the orthonormal version Eq. (4.93) for the Legendre expansion. Obtain an explicit expression for the Dirac δ function for the Legendre expansion.

4.10

Orthogonal functions and orthogonal polynomials

Other functions of interest to mathematical physics can be introduced in a rather natural way by generalizing the discussion of the previous section on Legendre polynomials. To do this, let us consider the expansion of a function F(x) in an internal [a, b] in terms of a suitable set of functions ψn in [a, b]: F(x) =



cn ψn (x),

n

as we have done for the Fourier series (4.21) or (4.27). The calculation of the expansion coefficient cn becomes simple if these ψn (x) are orthogonal functions satisfying an orthogonality relation of the general form  a

ψ∗m (x)ψn (x)w(x)dx = hn δmn .

(4.104)

It can then be extracted by a simple integration: 1 cn = hn



b a

ψ∗n (x)F(x)w(x)dx.

(4.105)

Equation (4.104) differs from the orthogonality relations used earlier in this chapter in the appearance of an additional weight function w(x). The separation of w(x) from ψn (x) is just a matter of convenience, but it might also be desirable on physical grounds. For example, it might give the load on a loaded string (related to the mass matrix M of Sections 2.6 and 2.7) which we might want to distinguish from the shape of its vibration. In particular, Eq. (4.104) has the same structure as the corresponding expression Eq. (2.80) for matrices.

288

Fourier series and Fourier transforms

The simplest functions one can use for ψn (x) are the powers of x. However, the integral  b 1 (bm+n+1 − am+n+1 ) xm xn dx = (4.106) m + n + 1 a are not orthogonal; they may not even be finite if |a| or |b| becomes infinite. The weight function w(x) now comes to the rescue. It can be chosen to fall off sufficiently rapidly with increasing |x| so as to give finite integrals. The chosen powers must next be orthogonalized. One systematic way of doing this is called Gram-Schmidt orthogonalization, which we have already used in the previous section. To formalize this procedure, let us note that, given a set of n linearly independent but nonorthogonal vectors φi in an nD linear vector space, we can construct from them a set of orthogonal vectors ψi , in the following way: We take ψ1 = φ1 = ψ1 e1 ,

e1 = e(ψ1 ).

(4.107)

For ψ2 we can take that part of φ2 that is perpendicular to ψ1 or e1 . That is, ψ2 = φ2 − e1 (e1 · φ2 ),

(4.108)

since e1 · φ2 is the component along e1 . The operator e1 e1 · = P1 = ·e1 e1

(4.109)

is called the projection operator on e1 . It can be used to simplify Eq. (4.108) to ψ2 = (1 − P1 )φ2 = ψ2 e2 .

(4.110)

We now define ψ3 to be that part of φ3 perpendicular to both ψ1 and ψ2 : ψ3 = (1 − P1 − P2 )φ3 = ψ3 e3 ,

(4.111)

and more generally ⎞ ⎛ n−1  ⎟⎟⎟ ⎜⎜⎜ ψn = ⎜⎜⎜⎝1 − Pi ⎟⎟⎟⎠ φn = ψn en .

(4.112)

i=1

The linear independence of the original φi ensures that none of the ψn will be zero. This is exactly what we have done in the previous section in constructing the Legendre polynomials from the powers xn , n = 0, 1, 2, . . . , ∞. We shall further illustrate the procedure by constructing orthogonal polynomials in the interval [a, b] = [−∞, ∞]. For this infinite interval, the integrals (4.106) are infinite. Hence a weight function w(x) is needed to ensure convergence. The interval is symmetric about x = 0; hence we look for an even function of x. One possibility is the Gaussian function exp(−x2 ) for which

Orthogonal functions and orthogonal polynomials



∞ −∞

289

e−x dx = f (1), 2

where  f (a) =

∞ −∞

e−ax dx = 2

&

π/a,

while 

∞

−∞

xm e−x dx = 0, 2

if m is odd,

(4.113a)

 m  d dx = − f (a) da a=1

(4.113b)

and 

∞ −∞

2m −x2

x e

are all finite. We begin by noting that even and odd powers are orthogonal to each other by virtue of Eq. (4.113a). That is,  (x , x ) = m

n

∞

−∞

e−x xm+n dx = 0, 2

if m + n = odd integer.

Hence we may use for the first two polynomials H0 (x) = 1,

and

H1 (x) = 2x.

(4.114)

Here we have used the convention that the term in Hn (x) with the highest power of x is (2x)n . The functions (4.114) may next be normalized by computing their lengths  (H0 , H0 ) = (H1 , H1 ) = 4

∞

e−x dx = 2

−∞  ∞

√

π,

  √ 2 d e−x x2 dx = 4 − f (a) = 2 π. da −∞ a=1

Hence the orthonormal polynomials are e0 (x) = π−1/4 ,

e1 (x) = 2x(4π)−1/4 .

290

Fourier series and Fourier transforms

The next orthogonal polynomial H2 (x) is just that part of (2x)2 orthogonal to e0 (x): H2 (x) = (1 − P0 )4x2 = 4x2 − e0 (x)(e0 , 4x2 )   2 −1/4 −1/4 π = 4x − π

∞ −∞

e

−x2

 2

4x dx

= 4x2 − 2.

(4.115)

Similarly, H3 (x) is just that part of (2x)3 orthogonal to e1 (x): H3 (x) = (1 − P1 )8x3 = 8x3 − e1 (x)(e1 , 8x3 )  ∞  2x 3 −x2 2x 3 = 8x − e 8x dx . (4π)1/4 −∞ (4π)1/4 Since the integral involved is [from Eq. (4.113b)] ⎡ ⎤   ∞ ⎢⎢⎢ d 2 & ⎥⎥⎥ 3√ −x2 4 ⎢ e x dx = ⎢⎣ − π/a⎥⎥⎦ = π, da 4 −∞ a=1 we get H3 (x) = 8x3 − 12x.

(4.116)

In this way, we find ⎞ ⎛ n−1  ⎟⎟⎟ ⎜⎜⎜ ⎜ Pi ⎟⎟⎟⎠ (2x)n Hn (x) = ⎜⎜⎝1 − i=0

= (2x)n −

n−1 

ei (x)(ei , 2n xn ).

(4.117)

i=0

These orthogonalized polynomials are called Hermite polynomials. A number of such orthogonal polynomial systems appear frequently in physics and engineering. They may be specified in terms of the interval [a, b], the weight function w(x), and the convention commonly agreed on for the normalization constant hn in Eq. (4.104). A few common examples are defined in Table 4.2. The first few polynomials themselves are given in Table 4.3. We note that the Chebyshev polynomials T n (x) are defined over the same interval as the Legendre polynomials, but involve a different weight function, namely (1 − x2 )−1/2 rather than 1. This awkward-looking weight function becomes easily recognizable when x = cos θ, for then

Orthogonal functions and orthogonal polynomials

291

Table 4.2 Orthogonality relations of certain orthogonal polynomials.

ψn (x)

Name

a

b

w(x)

hn

Pn (z) Ln (x) Hn (x)

Legendre Laguerre Hermite

–1 0 −∞

1 ∞ ∞

1 e−x 2 e−x

T n (x)

Chebyshev of the first kind

–1

1

(1 − x2 )−1/2

2/(2n+1) 1 √ n ⎧ π 2 n! ⎪ ⎪ ⎨π/2, n  0 ⎪ ⎪ ⎩π, n=0

(1 − x2 )−1/2 dx = −dθ. When the weight function is changed, the meaning of orthogonality changes, too. The polynomials P0 (x) = T 0 (x) = 1 and P1 (x) = T 1 (x) = x are naturally orthogonal because of their different parities. However, the part [ 23 P2 (x)] of x2 orthogonal to P0 (x) = 1 is not the same as the part [ 12 T 2 (x)] of the same x2 orthogonal to the same T 0 (x) = 1. Here everything is the same except for the meaning of orthogonality, that is, the choice of the weight function. One family of orthogonal polynomials defined over the same interval with a different weight function might be better than another in the expansion of a certain class of functions. Indeed, in the expansion of functions, the weight function itself can often be chosen to improve the quality of the expansion.

Problems 4.10.1∗ Use Gram-Schmidt orthogonalization to obtain the four polynomials shown in Table 4.3 for the Laguerre and Chebyshev polynomials. Table 4.3 Special cases of certain orthogonal polynomials.

ψn (x)

ψ0 (x)

ψ1 (x)

ψ2 (x)

ψ3 (x)

Pn (z)

1

x

(3x2 − 1)/2

(5x3 − 2x)/2

Ln (x)

1

−x + 1

(x2 − 4x + 2)/2!

(−x3 + 9x2 − 18x + 6)/3!

Hn (x)

1

2x

4x2 − 2

8x3 − 12x

T n (x)

1

x

2x2 − 1

4x3 − 3x

292

4.11

Fourier series and Fourier transforms

Mean-square error and mean-square convergence

Now that we have seen a variety of functions expanded in terms of different sets of orthogonal functions, it is time to ask if these expansions are good representations of the original functions. To be more precise, let a function F(x) in the interval [a, b] be expanded in terms of a set of orthonormal functions en (x), n = −∞, . . . , ∞, for which  b e∗m (x)en (x)dx = δmn . (4.118) a

In terms of the functions of Eq. (4.104), these orthonormal functions are en (x) = [w(x)/hn ]1/2 ψn (x). Suppose the generalized Fourier coefficients  b fn = e∗n (x)F(x)dx

(4.119)

a

exist and the generalized Fourier series f (x) =

∞ 

fn en (x)

(4.120)

n=−∞

is uniformly convergent. Is f (x) exactly identical to the original function F(x)? The answer is “not necessarily so”. The reason is that it is very difficult to reproduce a function exactly over the uncountably infinite number of points in the interval [a, b] when there are only a countably infinite number of expansion coefficients to adjust. However, it is possible to analyze the situation a little further, in an illuminating way. To begin, we must quantify the discrepancy between an infinite series, say  g(x) = gn en (x) (4.121) n

and the given function. One simple (but not unique) way of doing this is through the mean square deviation or error  b D= |F(x) − g(x)|2 dx, (4.122) a

which has has the virtue of being a single number, thus giving an average measure of the discrepancy. The integrand of D is non-negative; hence D itself is non-negative. It is obvious that there is no upper limit to how large this error can be. The interesting question is how small one can reduce the error. This question is answered by the following theorem: The Fourier series f (x) of Eq. (4.120) gives the best infinite-series representation

Mean-square error and mean-square convergence

293

of F(x) in the sense that the error D is minimized when g(x) = f (x). Any other choice of g(x) will result in a larger error. This theorem can be easily proved in the following way. For simplicity of notation, let us assume that F(x) and en (x) are real. Then using Eq. (4.121), we have  b D= [F 2 (x) − 2g(x)F(x) + g2 (x)]dx a



b

=

F (x)dx − 2 2

a

+





b

=

b

gn

en (x)F(x)dx a

n=−∞ b

gn gn

en (x)en (x)dx a

n,n







F 2 (x)dx − 2

a



gn fn +



n

g2n .

(4.123)

n

This shows that for any n ∂D = −2 fn + 2gn . ∂gn

(4.124)

Hence the choice gn = fn gives ∂D = 0, ∂gn

∂2 D = 2 > 0, ∂g2n

(4.125)

so that the Fourier series (4.120) minimizes the mean-square error D. The minimum error is  b Dmin = |F(x) − f (x)|2 dx 

a b

= a

F 2 (x)dx −



fn2 ≥ 0.

(4.126)

n

This is non-negative because the integrand |F(x) − f (x)|2 is non-negative. We have thus derived the Bessel inequality  b  F 2 (x)dx ≥ fn2 . (4.127) a

4.11.1

n=−∞

Mean-square convergence

The Fourier series f (x) of the function F(x) is said to converge in the mean to F(x) if Dmin = 0.

294

Fourier series and Fourier transforms

When this happens, the Bessel inequality becomes the Parseval equation  b ∞  2 F (x)dx = fn2 . (4.128) a

n=−∞

The set of functions {en (x)} is said to be complete with respect to all functions F(x) satisfying Parseval’s equation. Parseval’s equation itself is often called a completeness relation. What type of functions satisfy Parseval’s equation for the trigonometric functions of the Fourier series (4.21)? This question is answered by Parseval’s theorem (1893): The set of functions {en (x), n = −∞ to ∞} is complete with respect to piecewise continuous functions in the interval [−π, π]. A function is piecewise continuous if it is continuous in the interval except at a finite number of points. For a proof of this theorem, see Carslaw (1950, p. 284). The following example illustrates the usefulness of Parseval’s equation, which applies also to inner products of two different functions: 2 Example 4.11.1 Obtain the result π2 = 8Σ∞ odd n 1/n with the help of the following two Fourier series in [−π, π]

F(x) = x = f (x) =

∞  2

n n=1

G(x) = x/|x| = g(x) =

(−1)n+1 sin nx,

 4 sin nx, nπ odd n

(4.129) (4.130)

obtained in Examples 4.3.1 and 4.3.2. According to Parseval’s equation (4.128)  π x dx = π2 (F, G) = 2 0

 π ∞   2 m+1 4 = 2 (−1) sin mx sin nx dx. m nπ 0 m=1 odd n =8

 1 . n2 odd n



In technical terms, we have been considering an infinite-dimensional inner product space. If this space is complete (roughly in the sense of the Parseval equation), it is called a Hilbert space. Sometimes it is also required that the dimension is a countable infinity, as in the case in Eq. (4.120). The space is then said to be separable. To further illustrate the idea of completeness for the interested reader, we note that the space of all continuous functions F(x) in the interval [−π, π] is not considered to be complete because there are smooth functions such as

Convergence of Fourier series

 0, F n (x) = 1/n x ,

x 0,

295

(4.131)

which becomes discontinuous in the limit n → ∞. Thus discontinuous functions must also be included in order to have a Hilbert space. The space of wave functions of the simple wave equation (2.83) contains discontinuous functions, and is a Hilbert space. Finally we note that the converse of Parseval’s theorem is also true: A series of the form (4.120) for which the sum Σn fn2 converges is the Fourier series of a function that is square integrable. This result is known as the Fischer-Riesz theorem.

Problems 4.11.1 Use Parseval’s equations and the Fourier series of Appendix 4A or 4B to evaluate the following infinite sums: ∞  1 (a) , n4 (b)

n=1 ∞  n=1

(−1)n+1 . n6

4.11.2 Use Parseval’s equation and the Fourier series of Appendix 4A and 4B to evaluate the integral  π-  x .2 ln 2 cos dx. 2 0

4.12

Convergence of Fourier series

Fourier’s claim for his series representation of functions has given rise to many important questions. Their study has led to a deeper understanding of the nature of functions. However, our study of Fourier series has been motivated by our interest in quantum-mechanical applications, rather than by a desire to understand its subtle structure. To compensate for this neglect, we offer in this section a few qualitative remarks of a more mathematical nature. A basic question is whether a Fourier series like Eq. (4.21) converges at all. This convergence problem does not appear to have been solved. Certain properties are known to be necessary but not sufficient; other properties are known to be sufficient but not necessary. The property that is both necessary and sufficient is unknown. In this connection, it is useful to recall that a sufficient condition describes a special example of a general property, while a necessary condition describes a general property characteristic of this example, and possibly of other examples as well. The statement “if p, then q” or “p implies q” can be read as “p is a sufficient condition for q,” or as “q is a necessary condition for p.” For example, “chicken implies bird” means “being a chicken is a sufficient condition for being a bird,” or “a chicken is an

296

Fourier series and Fourier transforms

example of a bird.” It also means “being a bird is a necessary condition for being a chicken”, that is, “a chicken is necessarily a bird”. However, to be a chicken it is not sufficient to be a bird. Thus a necessary condition may not be sufficiently restrictive. In a similar way, to be a bird it is not necessary to be a chicken. That is, a sufficient condition may not be necessary when there are other possibilities. A condition that is both sufficient and necessary satisfies also the converse statement “q implies p”. In this case, p and q are said to be equivalent since one implies the other. The statement is equivalent to the statement that “p is true if, and only if, q is true”, or “q is true, if and only if, p is true”. For example, the statement “equilateral triangles are equivalent to equiangular triangles” means that each is a sufficient and necessary condition for the other. It is clear from this discussion that we have not yet isolated the exact property of a function that makes its Fourier series convergent. Several sufficient conditions for the convergence of Fourier series are known. For example, if F(x) is an absolutely integrable function in [−π, π], then its Fourier series f (x) converges to the value F(x) at every continuity point (i.e., a point where the function is continuous) where the left-hand and right-hand derivatives exist. This includes the special case where F(x) has a unique derivative. At a point of discontinuity where F(x) has left and right derivatives, the Fourier series converges to the arithmetical mean value: 1 f (x) = [F(x + 0) + F(x − 0)]. 2

(4.132)

The same result also holds for a piecewise smooth function, that is, a function that is smooth except for a finite number of jump discontinuities where F(x) is discontinuous or kink discontinuities (where its derivative is discontinuous). Since the Fourier series is made up of continuous trigonometric functions cos nx and sin nx, it might be supposed that a convergent Fourier series must itself be continuous, at least in the basic interval. In 1826 Abel pointed out that the Fourier representation f (x) of x in [−π, π] (Example 4.3.1) is actually discontinuous at the end points ±π and other similar points x = (2n + 1)π, n being any integer, when f (x) is considered a periodic function of x. Furthermore, when a Fourier series f (x) tries to represent a finite discontinuity at x = a of a function F(x) that is piecewise continuous, it overshoots by a calculable amount on both sides of a. This overshoot can already be seen in Fig. 4.1. Indeed, one can show that 1 f (a − 0) = F(a − 0) − pD, 2

1 f (a + 0) = F(a + 0) + pD, 2

where D = F(a + 0) − F(a − 0) is the size of the discontinuity, and 2 p=− π

 π

∞

sin x dx = 0.179, x

Convergence of Fourier series

297

so that d = f (a + 0) − f (a − 0) = (1 + p)D,

(4.133)

rather than the original discontinuity D. This overshoot is known as the Gibbs phenomenon. A Fourier representation of a function is useful even when its convergence is not known. If the function F(x) is absolutely integrable in [−π, π], its integral in this interval can be found by term-by-term integrations:  d ∞  a0 F(x)dx = (d − c) + [an (sin nd − sin nc)/n 2 c n=1 − bn (cos nd − cos nc)/n].

(4.134)

Conversely, if F(x) has an absolutely integrable derivative, a term-by-term differentiation of F(x) gives the Fourier series for F  (x). That is, 



F (x) ∼ f (x) =

∞ 

n(−an sin nx + bn cos nx).

(4.135)

n=1

Because of the additional factor n, the differentiated series may not be convergent even when f (x) is convergent. Another interesting problem concerns the reconstruction, or approximation, of the function F(x) whose Fourier series is known to be convergent. One systematic procedure is to expand the Fourier coefficients in inverse powers of n and to replace each Fourier series in inverse powers so obtained by the appropriate function from tables of Fourier series such as those shown in Appendix 4B and 4C. This procedure is illustrated by an example. Example 4.12.1 Approximate the function giving rise to the Fourier series f (x) =

∞  cos nx n=1

n+s

.

Since 1 s2 1 s = − 2 + 3 + ... , n+s n n n we find ∞ 



1 s cos nx − 2 + . . . f (x) = n n n=1



 x 3x2 − 6πx + 2π2 = − ln 2 sin − s + .... 2 12

298

Fourier series and Fourier transforms

Note that each succeeding term in this expansion has a more rapid convergence than those preceding it.  It might appear that the operation of finding the original function F(x), referred to technically as the summation of the given Fourier series f (x), makes sense only if f (x) is convergent. This is really a trivial case, since we already know that a sum does exist. The interesting case turns out to be the summation of a divergent series! The heart of the matter is a definition for the sum of a divergent series. Many different definitions are possible that all yield the same answer for a convergent series, as they should. However, the summability of a divergent series may vary with the definition. If summable, the sum may also have different values in different definitions! To give the reader a taste of this fascinating subject, let us discuss the summation by the method of arithmetic means (AM). The AM σn of the first n partial sums s0 = f0 ,

s1 = f0 + f1 , . . . ,

(4.136)

s n = f0 + f 1 + . . . + fn of an infinite series of terms fk is defined to be σn = (s0 + s1 + . . . + sn )/n.

(4.137)

The sum by AM is then defined as the limit σ = lim σn .

(4.138)

n→∞

For example, the infinite series of terms 1, −1, 1, −1, . . . has the alternating partial sums s0 = 1,

s1 = 0,

s3 = 0,

... ,

s2 = 1,

Hence the series is divergent, but its AM  1 0, n even σn = + 1/2n, n odd 2 has the limit σ = 12 . Hence the series is summable by the method of AM to a value of 12 . The relevance of this discussion is that the Fourier series of a continuous function is summable by this method of arithmetic means.

Problems 4.12.1 From the Fourier series for x2 , obtain the Fourier series for x. 4.12.2 From the Fourier series for x, obtain the Fourier series for x2 .

Maxwell equations in Fourier spaces

299

4.12.3 Derive the following formula for the Fourier series of the integral  x ∞ ∞  bn  sin nx   F(x )dx = (1 − cos nx) + [an + (−1)n+1 a0 ] n n=1 n 0 n=1 expressed in terms of the Fourier coefficients for F(x) itself. 4.12.4 Obtain approximate functions for the Fourier series ∞  n2 (a) sin nx; n3 + 1 n=1 (b)

∞  n=1

4.13

n3

n cos nx. +2

Maxwell equations in Fourier spaces

Physical laws are usually stated as partial differential equations (PDEs) in spacetime, with three space coordinates r = (x, y, z) and one time variable t. A familiar example is the Maxwell equations in vacuum we have examined in Section 1.9: ∇ · E = ρ/0 , ∇ × E = −∂t B, ∇ · B = 0, ∇ × B = μ0 0 ∂t E + μ0 J.

(4.139)

Here the electric field E(r, t) and the magnetic induction B(r, t) are 3D vector fields in spacetime. The shorthand symbol ∂t =

∂ ∂t

(4.140)

has been used. The constant ε0 (μ0 ) describes the electric (magnetic) property of the vacuum. It is called the permittivity (permeability) of free space. These PDEs are relatively simple because the terms containing PD operators ∇ and ∂t all carry constant coefficients. Complications come mostly from the two inhomogeneous source terms, the scalar (electric) charge density ρ(r, t) and the 3D vector current density J(r, t) = v(r, t)ρ(r, t).

(4.141)

Here v = dr/dt is the velocity at the field point r(t). We would like to show in this section that the use of Fourier transforms, either in 3D space or in 4D spacetime, will also transform these seemingly complicated coupled PDEs into much simpler coupled algebraic equations. Furthermore, these algebraic equations have a simple geometrical interpretation in Fourier spaces, the

300

Fourier series and Fourier transforms

spaces of the Fourier variables (k, ω) that are conjugate to and replace the spacetime variables (r, t). 4.13.1

Algebraic Maxwell equations

Suppose the electric field in spacetime has the 4D Fourier representation   ∞ dω d3 k ik · r−iωt e E(k, ω). E(r, t) = (2π)3 −∞ 2π

(4.142)

Suppose further that the representation remains valid when differentiated twice. For example,  ∞  dω d3 k ik · r−iωt ∇ · E(r, t) = e ik · E(k, ω) (4.143) (2π)3 −∞ 2π is assumed to exist. The same assumptions are made for the magnetic induction B. Note that on the right of Eq. (4.143), the differential operator ∇ has acted directly on the Fourier basis function ψk,ω (r, t) = eik · r−iωt .

(4.144)

The method of Fourier transform has been introduced in Section 4.5. In this method, E(k, ω) may be considered to be the Fourier transform F of E(r, t), namely E(k, ω) = F {E}  ∞  dt d3 re−ik · r+iωt E(r, t). ≡ −∞

(4.145)

The representation (4.142) is then its inverse Fourier transform. The downside of using Fourier spaces is that it is necessary to perform this Fourier inversion if one wants the result in spacetime. So at an introductory level, Fourier spaces are most useful for the qualitative understanding of electromagnetic phenomena for which the complete Fourier inversion back to spacetime is not essential. The 2π normalization convention used in this section differs slightly from that adopted √ in Section 4.5. There both forward and inverse Fourier transforms carry the same 1/ 2π normalization factor for every integration variable dk as well as dx. Here the dx integration carries no normalization factor at all, while dki /2π in the Fourier variable ki carries the combined normalization factor of 1/2π. This asymmetric notation is the one usually used in physics. This normalization convention, placing all 2π normalizations in the Fourier space integrations alone, applies to all formulas, not just the forward and inverse Fourier transforms shown here. It is also important to note that the Fourier basis function (4.144) has been chosen to be a function of the Lorentz scalar k · r − ωt = k · r + (iω)(it)

(4.146)

Maxwell equations in Fourier spaces

301

in the 4D Minkowski spaces (r, it) and (k, iω) discussed in section 3.1. For this reason, the basis functions are Lorentz-invariant, the same in all Lorentz frames. In the Fourier space (k, ω), the Maxwell equations simplify to the algebraic equations ik · E = ρ/ ˜ 0, ik × E = iωB,

(4.147)

ik · B = 0, ik × B = −iωμ0 0 E + μ0 J .

(4.148)

Here ρ˜ ≡ F {ρ} and J = vρ. ˜ 4.13.2

Longitudinal/transverse decomposition of the electromagnetic fields

We have seen in Example 1.2.7 that any vector E can be decomposed into component vectors parallel and transverse (or perpendicular) to the direction ek = k/k: E = E ek + E⊥ ,

(4.149)

where E = ek · E. The transverse component vector E⊥ can be found by applying the BAC rule of vector algebra to the left expression of ek × (E × ek ) = E(ek · ek ) − ek (ek · E).

(4.150)

So the left expression is in fact E⊥ . Furthermore, k × E = k × E⊥

(4.151)

has no contribution from E . Using these results, the algebraic Maxwell equations can be simplified to ρ˜ , ik0 k × E⊥ = ωB, E =

(4.152)

B = 0, k × B⊥ = −ωμ0 0 E − iμ0 J .

(4.153)

The first equation in Eq. (4.152) is called the Gauss law, and the second equation the Faraday law of induction. The first equation in Eq. (4.153) states that no magnetic charge (usually called monopole) is present. Finally, the last equation contains Maxwell’s displacement current −iω0 E. These equations have been written in SI (or the International System of) units. The second equation of each pair has an interesting structure. On the left, the expression is purely transverse, meaning perpendicular to k. Hence the longitudinal or parallel component of the expression on the right must vanish too. So we see by inspection that

302

Fourier series and Fourier transforms

B = 0,

(4.154)

J − iω0 E = 0.

(4.155)

The first result, B = 0, states that no monopole is present. It repeats the information already contained in another Maxwell equation (4.153). This no-monopole condition is hardwired into the third Maxwell Eq. (4.139). No monopole (magnetic charge) can appear in the system. The second result does not appear elsewhere. It also describes a conserved quantity, here the electric charge, but this interpretation is not at all obvious. To display this hidden conservation law, we use Eq. (4.152) to eliminate E in Eq. (4.155) to get the relation kJ = k · J = ωρ. ˜

(4.156)

This algebraic equation can be transformed back to spacetime with the help of the identity (4.143). The result is the PDE ∂ ρ + ∇ · (vρ) = 0. ∂t

(4.157)

The left side of the equation turns out to be the total time rate of change dρ/dt, according to Problem 4.13.2. Hence dρ(r, t)/dt = 0. This result states that a charge can move around in space (because ∂ρ/∂t  0), but it cannot disappear from the system. Note in particular that charge conservation does not involve J ⊥ . The remaining terms of the second equation of each pair involve transverse field components only:

4.13.3

k × E⊥ = ωB⊥ ,

(4.158)

k × B⊥ = −ωμ0 0 E⊥ − iμ0 J ⊥ .

(4.159)

Electromagnetic waves in free space

In 1861 Maxwell published his famous equations that unified electricity and magnetism. A few years later, in 1864, he showed that his equations had wave solutions where oscillating transverse electric and magnetic fields produced wave disturbances traveling along ek . Their wave speed turned out to be the same as that of light. Electromagnetic waves with the predicted properties were first produced and detected by Hertz in 1887. The key to Maxwell’s prediction is to untangle the coupled equations (4.158, 4.159) by multiplying them with k × on the left, using the BAC identity k × ( k × E⊥ ) = −k2 E⊥ .

(4.160)

Maxwell equations in Fourier spaces

The resulting equations can be solved for the ⊥ fields to give   ω2 k2 − 2 E⊥ = iμ0 ωJ ⊥ , c   ω2 2 k − 2 B⊥ = iμ0 k × J ⊥ , c

303

(4.161)

where c= √

1 μ0 0

(4.162)

has the dimension of a velocity. These algebraic equations are inhomogeneous wave equations in the Fourier space (k, ω). c, the velocity or speed of electromagnetic waves in free space, turns out to be the speed of light. To see that these equations describe waves, we go back to the associated PDEs in spacetime with the help of identities like Eq. (4.143):   1 ∇2 − 2 ∂2t E⊥ (r, t) = μ0 ∂t J⊥ (r, t), c   1 2 2 ∇ − 2 ∂t B⊥ (r, t) = −μ0 ∇ × J⊥ (r, t). (4.163) c These PDEs are called wave equations. Their solutions, here the transverse fields, may be called wave functions. In free space far from the source J⊥ , the inhomogeneous driving terms can be dropped. The resulting homogeneous PDEs describe electromagnetic waves in free space. The Fourier basis functions (4.144) are in fact the electromagnetic wave functions in free space, provided that ω = ±ck.

(4.164)

Note that electromagnetic waves are produced not by the charge density ρ, but by its motion or J⊥ = v⊥ ρ transverse to the wave direction. What about J ? According to the continuity Eq. (4.156), its Fourier component J is related to ρ. ˜ ρ˜ itself produces E , as described by the Gauss law (4.152) in the Fourier space k. Their relationship is purely spatial, as one can see more clearly in Eq. (4.139), the original Gauss law in space. The same time t appears on both sides. That is, the response E produced by ρ appears instantaneously without any time lag. This is an example of the celebrated action at a distance of classical physics.

Problems 4.13.1 (Convolution theorem) In the notation of Section 4.6, the convolution of two functions G(x) and R(x) in 1D space is defined as (G ∗ R) x ≡

304

Fourier series and Fourier transforms

F −1 {F {G}F {R}}. Show that for the 2π normalization used in this section, the result is  ∞ G(x − x )R(x )dx . (G ∗ R) x = −∞

4.13.2 (Charge conservation) (a) Show that the PDE (4.157) is transformed into the algebraic Eq. (4.156) in the Fourier space (k, ω). (b) Consider a function f (x(t), t) in 1D space, where the coordinate x(t) changes in time. Show that the total differential change df of the function contains two contributions df ∂f ∂f = + v(t), dt ∂t ∂x in a more familiar notation, with v(t) = dx/dt. The second term is a motional contribution that comes from the motional change in time of x(t). (c) Show that in 3D space, the charge density ρ(r, t) has a total time rate of change of dρ ∂ρ ∂ρ = + v · ∇ρ = + ∇ · J, dt ∂t ∂t where J = vρ. 4.13.3 (Wave equations in Fourier and real spaces) (a) Verify Eq. (4.161). (b) The verification of Eq. (4.163) is much harder, because of the need to project out the transverse component in real space. It can actually be done using the Helmholtz theorem which we shall revisit in the next section. However, the derivation without using the Helmholtz theorem is much easier and more instructive if it is separated into two steps. The first step is the problem to be solved here. Starting from the Maxwell equations (4.139) in real spacetime, show that the electric and magnetic fields satisfy the wave equations " 1 1 # ∇2 − 2 ∂2t E = ∇ρ + μ0 ∂t J, 0 c # " 1 ∇2 − 2 ∂2t B = −μ0 ∇ × J. c

(4.165)

(c) In the second step, the longitudinal/transverse separation is done in Fourier space. Show that in the Fourier space (k, ω), the wave equations (4.165) can be separated into four equations. The two for the transverse components are just Eq. (4.161). The remaining two equations involving the longitudinal components are

3D Fourier transforms

  ω2 kρ˜ 2 k − 2 E = −i + iμ0 ωJ 0 c   ω2 k2 − 2 B = 0. c

305

(4.166)

Actually B = 0, because no monopoles are present. (d) (Spurious causality) The wave equation for E shown in part (c) seems to state that E is causal, meaning that it depends on the light speed c. Show that this conclusion is incorrect by finding that the right-hand side of this wave equation can be written as   ω2 ρ˜ 2 . (4.167) k − 2 c ik0 There are thus two canceling nonzero factors of k2 − ω2 /c2 in the wave equations. Show that after the cancellation, the “instantaneous” result given in Eq. (4.152) obtains. Note: The factor k2 − ω2 /c2 in Eq. (4.167) is nonzero because it comes from the inhomogeneous equation (4.165).

4.14

3D Fourier transforms: Helmholtz decomposition theorem

In the last section, 3D Fourier transforms with asymmetric normalizations are used to study the structure of the Maxwell equations in Fourier space. In this section we show how 3D Fourier transforms are calculated and inverted. They are also used to derive the Helmholtz theorem introduced in Section 1.9. 4.14.1

3D Fourier transforms

In many application in science and engineering, the 3D Fourier transform is defined to be  F { f (r)} = g(k) = d3 re−ik · r f (r). (4.168) Its Fourier inversion formula is F −1 {g(k)} = f (r) =



d 3 k ik · r e g(k). (2π)3

Inserting Eq. (4.168) into Eq. (4.169) yields the identity  f (r) = d 3 r δ(r − r ) f (r ),

(4.169)

(4.170)

306

Fourier series and Fourier transforms

where δ(r − r ) =



d3 k ik · (r−r ) e (2π)3

(4.171)

is a 3D Dirac δ-function. Many of the 3D integrals appearing in 3D Fourier transforms can be calculated analytically. We shall show how this can be done for spherically symmetric functions in the following example. Example 4.14.1 If f (r) = f (r) is a spherical symmetric function, show that its 3D Fourier transform g(k) can be reduced to a 1D integral. In spherical coordinates, d3 r = r2 dr d cos θdφ. Since f (r) is spherically symmetric, independent of the angles θ and φ, the angular integrals can be determined readily: 



∞

F { f (r)} = 2π

r dr 0

2π = ik

2



∞

1

−1

d cos θe−ikr cos θ f (r)

" # rdr eikr − e−ikr f (r).

(4.172)

0

The final 1D definite integral can often be integrated analytically, found in a table of definite integrals, or in a last resort, done numerically in a computer.  The method is nicely illustrated by the direct integration to yield the transform  −μr , e 4π F . (4.173) = 2 r k + μ2 The detail is left as an exercise, in Problem 4.14.1. A short table of 3D Fourier transforms can be found at the end of the chapter. The Fourier transform and its inverse are basically symmetric with respect to each other. A spherically symmetric function f (r) in real space will transform into a spherically symmetric function g(k) in Fourier space. More generally, if f (r) has a certain angular symmetry in real space, then g(k) will also have the same angular symmetry in Fourier space. Such a symmetry invariance can be demonstrated explicitly using angular functions called spherical harmonics, introduced in Section 7.6. The inverse transform is a Fourier transform in its own right, except for the difference in the 2π normalization. The 3D integrals involved are usually easier to do in one direction than in the other direction. Many Fourier transforms are defined with symmetrical normalizations, so that the tables can be used in either direction without modification. In the following example we show how the change in normalization is taken care of when our asymmetric table is used in the backward direction. Example 4.14.2 Given the Fourier transform (4.173), find F {1/(r2 + R2 )}, where R is a constant.

3D Fourier transforms

Our inverse transform carries an extra factor (2π)−3 . So we must have  , 1 4π F 2 = (2π)3 e−kR . 2 k r +R

307

(4.174)

The answer is 

, 1 (2π)3 1 −kR F 2 = e .  4π k r + R2

(4.175)

The use of Fourier transforms in actual problems in physics is illustrated in the following example and in other examples in the next subsection. Example 4.14.3 (Coulomb field of a resting point charge ) A stationary charge q is located at the origin. Its charge density is ρ(r) = qδ(r). Find E (r). We begin with F {ρ} = ρ˜ = q of the charge. The magnitude of its transverse electric field in Fourier space is given by Eq. (4.152). Hence  , q −1 1 E (r) = F i0 k =

q 4π 1 . i0 (2π)3 r2

(4.176)

The result (4.175) has been used in the final step. This is just proportional to the magnitude of the inverse-square electric field E(r) surrounding the point charge q in space. It would be nice to include the direction er too. To do this, we need to start with the complete vector field in Fourier space with E (k) = k

1 q . k2 i0

(4.177)

Having eliminated ek = k/k, we find a k factor that can be inverted first, giving just −i∇ in ordinary space. This purely imaginary result also explains why Eq. (4.176) is also purely imaginary. Continuing the calculation, we find  , q −1 1 E (r) = − ∇F 0 k2   q 1 = −∇ , (4.178) 0 4πr using the Fourier transform (4.173). We recognize that the expression within the parentheses is just the electrostatic potential.

308

Fourier series and Fourier transforms

To finish, we need to calculate the negative gradient with the help of the general formula   df 1 2 ∇r ∇ f (r) = dr 2r = er

df . dr

(4.179)

The final result is just the inverse-square Coulomb field of a resting point charge q E (r) = 4.14.2

q er . 4π0 r2



(4.180)

Convolution and Helmholtz theorems

The convolution or folding of two functions G(r) and R(r) is defined to be the integral over all space  (G ∗ R)r ≡ G(r − r )R(r )d3 r = (R ∗ G)r .

(4.181)

The convolution is unchanged if the two functions appearing in either the integral or the convolution interchange their positions. By direct substitution (Problem 4.14.3), convolutions can be shown to satisfy the convolution theorem F {(G ∗ R)r } = F {G}F {R}.

(4.182)

This theorem can be written alternatively as ˜ R(k)}, ˜ (G ∗ R)r = F −1 {G(k)

(4.183)

˜ ˜ = F {R(r)}. where G(k) = F {G(r)} and R(k) Many PDEs of physics become algebraic equations in Fourier spaces. If the ˜ R(k), ˜ solutions there can be written as a product G(k) then the solutions in spacetime are just the convolutions (4.181). This simple result finds many uses in science and engineering. It is next illustrated by a number of common examples. Example 4.14.4 (Poisson equation) The electric field can be expressed as the negative gradient of a scalar potential Φ(r) E(r) = −∇Φ(r).

(4.184)

The Gauss law in Eq. (4.139) can then be rewritten as a Poisson equation for the scalar potential ∇2 Φ = −ρ/0 . Find the solution Φ(r) as a convolution integral.

(4.185)

3D Fourier transforms

309

In the Fourier space k, the Poisson equation simplifies to ˜ = ρ/ ˜ 0. k2 Φ

(4.186)

Hence the scalar potential in real space is Φ(r) = (G ∗ ρ)r /0 ,

(4.187)

where the Green function is G(r) = F according to Eq. (4.173). Hence  Φ(r) =

−1



, 1 1 = , 2 4πr k

1 ρ(r ) 3  d r.  4π|r − r | 0

(4.188)

(4.189)

Example 4.14.5 (Helmholtz decomposition theorem) We are now in a position to derive the Helmholtz theorem that is so important in classical electrodynamics. Recall that in the Fourier space k, the electric field E can be decomposed into longitudinal and transverse parts relative to ek using the BAC rule: ek × (E × ek ) = E − ek (ek · E).

(4.190)

E = E + E⊥ .

(4.191)

Hence in real space

Here , 1 E (r) = F k 2 (k · E) k  , −1 1 = −∇F (ik · E) k2 −1



= −∇Φ(r)

(4.192)

is expressed in terms of a scalar potential Φ(r). The scalar potential is just Eq. (4.189) expressed in terms of the scalar source density ρ(k)/ ˜ 0 = ik · E, or equivalently in terms of ∇ · E(r) = ρ(r)/0 as  1 Φ(r) = ∇ · E(r ) d3 r . (4.193) 4π|r − r |

310

Fourier series and Fourier transforms

The transverse component can be found in a similar way. The result is ,  −1 1 (ik × E) E⊥ (r) = ∇ × F k2 = ∇ × AE (r), where the Helmholtz electric vector potential is  1 ∇ × E(r ) d3 r . AE (r) = 4π|r − r |

(4.194)

(4.195)

The longitudinal/transverse separation of a vector field given in Eqs. (4.191–4.195) is called the Helmholtz decomposition theorem. The magnetic induction field B has only a transverse part B = B⊥ (r) = ∇ × A(r),  1 ∇ × B(r ) d3 r , A(r) = 4π|r − r |

(4.196)

where the magnetic vector potential is the vector potential. Faraday’s induction law ∇ × E⊥ = −∂t B relates the electric vector potential AE to the magnetic vector potential E⊥ = −∂t A = ∇ × AE .

(4.197)

So it is not necessary to use the electric vector potential explicitly.  Example 4.14.6 (Green functions for the wave equation) Consider the wave equation in 4D spacetime:   1 2 2 (4.198) ∇ − 2 ∂t B(r, t) = −μ0 ∇ × J(r, t). c Let the solution for the inhomogeneous wave equation   1 2 2 ∇ − 2 ∂t G(r, t) = −δ(r)δ(t), c

(4.199)

be denoted G(r, t). The convolution theorem then states that the solution of Eq. (4.198) can be expressed in the integral form or representation   / 0 (4.200) B(r, t) = G(r − r , t − t ) μ0 ∇ × J(r , t ) d3 r  dt .

3D Fourier transforms

311

(The result can be verified by direct substitution into the wave equation (4.198).) Find G(r, t), the Green function for the 3D wave equation. For greater clarity, we do the problem in several steps. The details are left as exercises in Problem 4.14.3. (a) In the 4D Fourier space (k, ω), Eq. (4.199) simplifies to 

 ω2 ˜ k − 2 G(k, ω) = 1, c 2

(4.201)

according to the 4D Fourier representation (4.142). (b) The Fourier transform (4.173) in the modified form shown as entry 2 of Table 4D can be used to get two possible inverse transforms from k space to r space: G˜ ± (r, ω± ) ≡ F =

−1



1 2 k − ω2± /c2

,

e±iω± r/c . 4πr

(4.202)

Here r = |r|, and ω± = ω ± i contains  → 0+ (positive zero) that generates a convergence factor to ensure that the Fourier transform contains no contribution from the upper limit of integration (as r → ∞). Note that the same symbol G˜ has been used for both full and partial Fourier transforms. We rely on the arguments of the function to tell us in which Fourier space the function resides. (c) The final Fourier inversion is elementary:  G± (r, t) = =

∞ ±iωr/c e −∞

4πr

e−iωt

dω 2π

1 c r = δ t∓ δ(r ∓ ct). 4πr c 4πr 

(4.203)

The negative sign appearing on the right of Eq. (4.199) is the common sign convention used in physics. However, we have used the opposite sign convention in Section 4.6. The reader should always check the sign convention used before reading anything on Green functions.  Example 4.14.7 (Retarded and advanced Green functions) It is useful to give a physical interpretation of the two mathematical solutions G± . For G+ , the δ-function in Eq. (4.203) requires that r = ct.

(4.204)

312

Fourier series and Fourier transforms

This equation states that the radial distance from the origin increases linearly in time. Hence the solution G + describes an outgoing spherical wave. Used in the convolution integral (4.200), expression (4.204) is changed into R = |r − r | = c(t − t ).

(4.205)

The physical picture is changed too. A spherical outgoing wave leaves the (vector) source density μ0 ∇ × J(r , t ) at source position r and time t . The wave reaches the field position r at a time t that is R/c after its emission from the source. The Green function G+ is therefore said to be retarded or delayed when observed. Take a movie of this retarded event and run it backwards. You will see an ingoing spherical wave described by the equation r = −ct.

(4.206)

satisfied by the solution G − . For the convolution integral (4.200), the backwardrunning movie shows a spherical wave collapsing onto the source position r at a time t that is R/c after it has passed the observer position r. So the observation of the wave at the observer position occurs before the wave finds its source. G − is called the advanced Green function, very much like the advance announcement of a future event. Common sense requires that a wave must have started somewhere before it can be observed somewhere else. Such a requirement is called the principle of causality. The retarded Green function satisfies this requirement. It is therefore called the physical solution. The advanced Green function violates causality and is unphysical. 

Problems 4.14.1 (3D Fourier transform in spherical coordinates) Verify entry 1 of the table in Appendix 4D for (a) n = 0, and (b) n > 0. 4.14.2 (3D Fourier transform in rectangular coordinates) Verify entry 4 of the table in Appendix 4D as follows: (a) Show that g(k) = g1 (k x )g1 (ky )g1 (kz ), where g1 (k x ) = A exp (k2x /4a). Hint: Complete a square. (b) Find the proportionality constant A by evaluating the integral  ∞ ∞ 2 2 π A2 = dx dy e−a(x +y ) = . a −∞ −∞ 4.14.3 Complete Example 4.14.6 by filling in all missing steps.

Short table of Fourier sine series

Appendix 4A Short table of Fourier cosine series ∞ 

F(x) = an

an cos nx

n=1

1.

1 n

F(x) " # − ln 2 sin 2x

2.

1 n2

1 2 12 (3x

3.

(−1)n+1 /n

# " ln 2 cos 2x

4.

(−1)n+1 /n2

1 2 12 (π

5.

1 n,

− 12 ln tan |x| 2

6.

1 , n2

odd n odd n

"

π π 4 2

− 6πx + 2π2 )

− 3x2 )

# − |x|

Interval (0, 2π) [0, 2π] (−π, π) [−π, π] (−π, 0), (0, π) [−π, π]

Reference: G. P. Tolstov, Fourier Series (Dover, New York, 1976), p. 148–9.

Appendix 4B Short table of Fourier sine series F(x) =

∞ 

bn sin nx

n=1

bn

F(x)

Interval

1.

n−1

(π − x)/2

(0, 2π)

2.

n−3

(x3 − 3πx2 + 2π2 x)/12

[0, 2π]

3.

(−1)n+1 /n

x/2

(−π, π)

4.

(−1)n+1 /n3

(π2 x − x3 )/12

[−π, π]

5.

1 n,

π x 4 |x|

(−π, 0), (0, π)

6.

1 , n3

odd n odd n

πx 8 (π −

|x|)

[−π, π]

Reference: G. P. Tolstov, Fourier Series (Dover, New York, 1976), p. 148–9.

313

314

Fourier series and Fourier transforms

Appendix 4C Short table of Fourier transforms g(k) =

f (x) 1.

2.

δ(x) ⎧ ⎪ ⎪ ⎨0, ⎪ ⎪ ⎩e−ax ,

x0

1 1 √ a + ik 2π

exp(− 12 cx2 )

√1 c

4.

1 1 + x2

√

5.

[(a2 + x2 )(b + ix) p ]−1

6.

[(a2 + x2 )(b − ix) p ]−1 ⎧ ⎪ ⎪ ⎨Pn (x), |x| < 1 ⎪ ⎪ ⎩0, |x| > 1

8.

e−λk a + e−x

Conditions

√1 2π

3.

7.

∞ √1 e−ikx f (x)dx 2π −∞

√ √

" 2# k exp − 2c

π/2 exp(−|k|) π/2

e−ak a(a + b) p

π/2 (b − a) p eak /a

√ (−i)n Jn+1/2 (k)/ k √

Re a > 0

π/2 aλ−1+ik csc(πλ + iπk)

Re p > −1, Re a > 0, Re b > 0 Re p > −1, Re a > 0, Re b > 0, a  b

0 < Re λ < 1, −π < arg a < π

Reference: A Erd´elyi et al., Table of Integral Transforms (Bateman Manuscript Project) (McGraw-Hill, New York, 1954), Vol. I, Chap. III.

Appendix 4D Short table of 3D and 4D Fourier transforms g(k, ω) = F { f (r, t)}  ∞  dt d3 re−ik·r+iωt f (r, t), = −∞

f (r, t) = F −1 {g(k, ω)}   ∞ dω d3 k ik·r−iωt e g(k, ω). = (2π)3 −∞ 2π

Tables of mathematical formulas

f (r, t) 3D: 1.

rn−1 e−μr

2.

1 ±ik± r re

g(k, ω)

Conditions

"

No t or ω. n > 0, μ > 0.

# d n 4π − dμ k2 +μ2 4π k2 −k±2

k± = k0 ± i; k0 > 0,  → 0+ .

3.

1 r2 +R2

4.

e−ar

4D: 5.

1 r δ(t

(2π)3

1 −kR 4π k e " #3/2 2 π e−k /4a a

2

∓ r/c)

4π k2 −(ω2± /c2 )

ω± = ω ± i;  → 0+ .

Appendix 4E Tables of mathematical formulas 1 Fourier series ∞  1 (an cos nx + bn sin nx), F(x)  f (x) = a0 + 2 n=1  1 π an = F(x) cos nx dx π −π  1 π bn = F(x) sin nx dx π −π  fi ei (x), a ≤ x ≤ b : F(x)  f (x) = i



b

(ei , ej ) = 

a b

fi =  a

a b

|F(x)|2 dx ≥

 i

e∗i (x)ej (x)dx = δij e∗i (x)F(x)dx = (ei , F)

| fi |2 ,

Bessel inequality.

−π ≤ x ≤ π :

315

316

Fourier series and Fourier transforms

2 Dirac δ function



a

F(x) =

F(x )δ(x − x )dx

b

δ(−x) = δ(x) δ(ax) = δ(x)/|a| δ(x − a2 ) = [δ(x − a) + δ(x + a)]/2|a|  δ(g(x)) = δ(x − xi )/|g (xi )|, where 2

 b

g(xi ) = 0.

i a

F(x )

d δ(x − x )dx = −F  (x) under certain conditions dx  en (x)e∗n (x ). δ(x − x ) = n

3 Fourier transform

 ∞ 1 g(k) = √ e−ikx f (x)dx = F { f (x)} 2π −∞  ∞ 1 f (x) = √ eikx g(k) dk. 2π −∞

4 Conjugate variables and operators (φi , φj ) = ( fi , fj ) = (gi , gj ),

gi = F { fi (x)}

x¯ = (φ, xφ)/(φ, φ) (Δx)2 = (φ, (x − x¯)2 φ)/(φ, φ) 1 (Δx)(Δk) ≥ , Heisenberg’s uncertainty principle 2   d (φi , xφj ) = i φi , φj dk   1 d (φi , kφj ) = φi , φj i dx [k, x] = kx − xk = −i     1 d (φi ,C(x, k)φj ) = fi ,C x, fj i dx     d = gi ,C i , k gj . dk

Tables of mathematical formulas

317

5 Maxwell equations in Fourier spaces Maxwell equations (in SI units): ∇ · E = ρ/0 , ∇ × E = −∂t B; ∇ · B = 0, ∇ × B = μ0 0 ∂t E + μ0 J; simplify to algebraic equations in the 4D Fourier representation   ∞ dω d3 k ik·r−iωt e E(k, ω). E(r, t) = (2π)3 −∞ 2π In the Fourier (k, ω) space, vector fields are readily decomposed into longitudinal and transverse parts: E⊥ = ek × (E × ek ) = E(ek · ek ) − ek (ek · E) = E − E . Maxwell equations become algebraic ρ˜ , ik0 k × E⊥ = ωB; E =

B = 0, k × B⊥ = −ωμ0 0 E⊥ − iμ0 J ⊥ , J = iω0 E . The continuity equation ∂t ρ + ∇ · (vρ) = 0 and the wave equations for the transverse field components also become algebraic kJ = ωρ; ˜   ω2 k2 − 2 E⊥ = iμ0 ωJ ⊥ , c   ω2 2 k − 2 B⊥ = iμ0 k × J ⊥ . c √ The speed of electromagnetic waves is c = 1/ μ0 0 , the speed of light.

318

Fourier series and Fourier transforms

6 3D Fourier transforms: Helmholtz decomposition theorem E = E + E⊥ , = −∇Φ(r) + ∇ × AE (r),  1 ∇ · E(r ) d3 r , Φ(r) = 4π|r − r |  1 AE (r) = ∇ × E(r ) d3 r ; 4π|r − r | B = B⊥ = ∇ × A(r),  1 ∇ × B(r ) d3 r ; A(r) = 4π|r − r | E⊥ = −∂t A = ∇ × AE .

5 Differential equations in physics 5.1

Introduction

We are familiar with the fact that, given a smooth function p(t), its derivative can be calculated: d p(t) = f(t). dt

(5.1)

If on the other hand the derivative f(t) is given, but p(t) is as yet unknown, the equation is called a differential equation (DE). A DE is solved when a function p(t) is obtained that satisfies the stated functional relationship. The most familiar of such DE in physics is probably the Newton’s force law shown in Eq. (5.1). This states that an external force f(t) causes a time rate of change of the unknown momentum p(t) of a particle. Several other DEs of interest in physics can be derived from Newton’s force law. For example, we have seen in Section 2.8 that its application to the 1D motion of a system of identical masses connected by identical springs yields the one-dimensional wave equation   1 ∂2 ∂2 − u(x, t) = 0 (5.2) ν2 ∂t2 ∂x2 in the continuum limit. In this equation, the curvature term involving the spatial derivative gives the net force acting on the mass located at x by virtue of its displacement from equilibrium. The mass responds by accelerating, so that its kinetic energy is changed. When it returns to an equilibrium position, it does not immediately come to a stop, but overshoots that position because of its “inertia”. The springs connected to it eventually bring it to rest. This happens when its kinetic energy is completely converted into the potential energy stored in the springs. The springs next push or pull masses around so that the potential energy reverts to the kinetic form. This completes one of the countless cycles of the transmutation of energies. The mathematical solution of the wave equation tells us that the motion of each mass does not necessarily repeat itself in exactly the same way. Rather, the wave disturbance, where the energy is located, tends to move away from the source (with velocities ±v) to distances beyond the range of motion of the original mass. It is the wave disturbance that is of primary importance here, like the story of the Norman Conquest depicted on the Bayeux tapestry. The motion of the masses themselves, like the makeup of that linen roll, is only of secondary interest.

320

Differential equations in physics

The wave equation can be modified to describe a number of other interesting physical situations. We may ask for the equilibrium (i.e., time-independent) configuration of masses when they are subject to a given external force. The answer is given by a solution of the one-dimensional Poisson equation −

∂2 φ(x) = f (x). ∂x2

(5.3)

We may even ask for force-free configurations consistent with boundary constraints. These are provided by solutions to the Laplace equation ∂2 φ(x) = 0. ∂x2

(5.4)

Returning to the time-dependent wave equation, we might add a frictional term for a resistive medium. If friction is proportional to the instantaneous velocity, we find   1 ∂2 1 ∂ ∂2 + − u(x, t) = 0. (5.5) ν2 ∂t2 κ ∂t ∂x2 The added frictional term will cause a damping of the wave function u(x, t) as the wave propagates in spacetime. Other physical attributes might behave like the mechanical system described by Eq. (5.5). The special case in which their effective mass vanishes is of unusual interest. In this situation, the first, or inertial, term in Eq. (5.5) is absent, leading to the equation   1∂ ∂2 (5.6) − 2 u(x, t) = 0. κ ∂t ∂x In the absence of inertia, the displacement from equilibrium cannot overshoot the equilibrium position. Instead it disperses in a way rather similar to the relaxation of a heavily damped mechanical system. The rate of dispersal is controlled by the frictional term, where the constant κ is called a diffusion constant in the diffusion of gas molecules or a thermal conductivity in the transmission of heat. Equation (5.6) itself is called a diffusion equation. In addition to differential equations, we also use integral equations in physics. These have the form  (5.7) K(t, t )y(t )dt + M(t)y(t) = R(t). where K(t, t ), M(t), R(t) are given functions, while y(t) is the unknown function to be determined. Integral equations are far less common in physics than differential equations. There are good reasons for the popularity of differential equations. They have simpler mathematical structures, so that they are easier to solve and to use. Certain

Linear differential equations

321

important invariance principles (such as the invariance under space translations and time displacements) can be included simply and explicitly in differential equations. These are important technical advantages, but the basic superiority of differential equations is probably this. They correctly reflect the fact that our experimental knowledge of physical events and physical systems tends to be rather limited both in details and in spacetime extensions. As a result, the coherence and regularities that connect physical properties, and the disturbances and responses that describe sequences of physical events, tend to be the properties of our immediate neighborhood. This limitation and localization of our knowledge favor a differential approach. It is also historical fact that in science we have been successful in relating only properties at essentially the same location and only a few properties at a time. Thus a complete knowledge of many physical attributes all over the universe is not needed before objective predictions of the outcomes of certain events can be made. This is indeed fortunate, for otherwise science might not have emerged. The possibility of doing science in relative ignorance of the actual nature of the universe has its obvious limitations. The regularities that we can discover and quantify are usually among the simplest possible relations. The physical properties so related tend to exist in spacetime in a reasonably smooth manner. Their relations as a rule do not involve derivatives of very high orders, or very complicated functions of their low-ordered derivatives. As a result, known physical laws are often simple differential equations. In this chapter we shall concentrate on linear differential equations. Nonlinear systems will be described in Chapter 6.

5.2

Linear differential equations

The order and degree of a differential equation (DE) refer to those of the derivative of the highest order after the DE has been rationalized. Thus the DE d 4 y(x) dx

4



dy(x) + dx

1/2 + x2 y(x) = R(x)

(5.8)

is of fourth order and second degree, since after rationalization it contains [d4 y(x)/dx4 ]2 . The DE is said to be inhomogeneous if there is a term R(x) that is independent of the unknown y(x). The DE is an ordinary differential equation if there is only one variable, as is the case of Eq. (5.8). If there are two or more variables, it is a partial differential equation. A DE L [y(x)] = R(x)

(5.9)

is linear if the differential operator L is a linear operator satisfying the linearity property L [a1 y1 (x) + a2 y2 (x)] = a1 L [y1 (x)] + a2 L (y2 (x)].

(5.10)

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Differential equations in physics

where the ai are constants. For example, 

1/2 d D [y(x)] ≡ y(x) dx is nonlinear because 1/2 1/2 1/2   √ d d d ay(x) y(x) y(x) D [ay(x)] = = a a . dx dx dx 

In a similar way, the second-order DE d2 dt

2

θ(t) +

g sin θ(t) = 0 l

(5.11)

describing the large-amplitude motion of a pendulum is also nonlinear. This is because sin(λθ1 + μθ2 )  λ sin θ1 + μ sin θ2 . However, if the angular displacement is so small that sin θ  θ is a good approximation, Eq. (5.11) can be linearized to d2

g θ(t) + θ(t) = 0. l dt

(5.12)

2

From this we can see that physical systems not far from states of equilibrium are basically linear systems. Linear differential equations (LDE) are simpler than nonlinear DE. Their relative simplicity is a consequence of the following two superposition principles that their solutions satisfy: 1. If y1 (x) and y2 (x) are any two solutions of a homogeneous LDE, L y(x) = 0,

then yh (x) = c1 y1 (x) + c2 y2 (x),

ci = constant,

(5.13)

is also a solution. This is becuase L (c1 y1 + c2 y2 ) = c1 L y1 + c2 L y2 = 0.

2. If yh (x) is a (i.e., any) solution of the homogeneous LDE L y = 0, and yp (x) is a particular (i.e., any) solution of the inhomogeneous LDE L [yp (x)] = R(x),

Linear differential equations

323

then the linear combination y(x) = ayp (x) + byh (x),

a, b = constant,

(5.14)

is a solution of the inhomogeneous LDE L [y(x)] = aR(x).

This is because L (ayp + byh ) = aL yp + bL yh = aR + 0 = aR.

Examples of linear differential equations are the simple, or hyperbolic, wave equation discussed earlier φtt − ν2 φxx = 0,

where φtt =

∂2 φ, ∂t2

etc.,

and the linearized Korteweg-deVries equation φt + c0 φ x + vφxxx = 0, where c0 and v are are constants. Examples of nonlinear differential equations are the Korteweg-deVries equation (describing long waves in shallow water) φt + (c0 + c1 φ)φ x + νφxxx = 0, and the sine-Gordon equation φtt − φxx + sin φ = 0 (describing certain persistent (called solitary) waves in many physical situations). Problems 5.2.1 Identify the linear differential equations in the following d (a) x2 y + xyy = 0, y = y(x), y = dx y(x); 2  2 (b) x y + y = 0; (c) xy2 + x2 = 0; x(1+y2 )1/2 ; y(1+x2 )1/2  x y =e ; y = ey ; a2 y2 = (1 + y2 )3 .

(d) y = (e) (f) (g)

324

Differential equations in physics

5.3

First-order differential equations

First-order DEs can be solved by direct integration if the equation is separable: dy P(x) = . dx Q(y) Then



 Q(y)dy =

P(x)dx.

Example 5.3.1 % dy − x 1 − y2 = 0, dx dy = x dx. & 1 − y2  dy 1 ∴ = sin−1 y = x2 + c, & 2 1 − y2 or



 1 2 y = sin x + c . 2



The procedure works for both linear and nonlinear DEs, but it is not always easy to separate the variables x and y. A change of variables will sometimes do it, as illustrated by the following example Example 5.3.2 Solve dy = (x − y + 3)2 . dx Let us first change variables to z = x − y + 3. Then dz dy =1− = 1 − z2 , dx dx   dz = dx, 1 − z2 tanh−1 z = x + c, z = tanh(x + c). Therefore y = x + 3 − tanh(x + c).



First-order differential equations

325

The next best thing one can do for a nonlinear DE is to change it into a linear DE by a change of variables. This is because, LDEs can always be solved in closed form, as we shall see. The following example shows how a nonlinear DE can be changed into a linear DE by a change of variables. Example 5.3.3 The nonlinear DE dy + f (x)y = g(x)y , dx

n  1,

can be written as 1 dy + f (x)y1−n = g(x), yn dx or dv + (1 − n) f (x)v = (1 − n)g(x), dx where v(x) = y1−n (x).



We finally turn to linear DEs. They have the general form   d + p(x) y(x) = R(x). dx

(5.15)

The solution to the homogeneous equations, with R(x) = 0, can be obtained readily by separating out the dependence on x and y: 1 dy(x) = −p(x)dx. y(x) Hence



x





x

d ln y(x ) = −

a

p(x )dx .

a

That is,   y(x) = y(a) exp −

x







p(x )dx ,

(5.16)

a

where the multiplicative constant has been so chosen that y(x = a) is equal to a predetermined value y(a). The number y(a) is called the boundary condition at the point x = a.

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Differential equations in physics

Example 5.3.4 

has the solution

 d + x y(x) = 0 dx

  y(x) = y(0) exp −

x

0

  1 2 x dx = y(0) exp − x . 2 





If the boundary condition is y(0) = 1, then   1 2 y(x) = exp − x . 2



Given a solution yh (x) of the homogeneous equation, a particular solution of the inhomogeneous equation can be obtained by first writing yp (x) = c(x)yh (x).

(5.17)

Direct substitution into Eq. (5.15) yields   d R(x) = L yp (x) = + p(x) c(x)yh (x) dx   d d = yh (x) c(x) + c(x) + p(x) yh (x). dx dx Since the last term vanishes, we have R(x) d c(x) = , dx yh (x) so that c(x) = c(a) +

 x a

Thus yp (x) =

 x a

 R(x ) dx . yh (x )

 ! R(x ) dx yh (x), yh (x )

(5.18)

where we have dropped the term involving c(a), since according to Eq. (5.14), it can be made to appear in the second (or homogeneous) term of the general solution y(x) = yp (x) + byh (x). The solution satisfying the boundary condition y(x = a) = y(a) is obtained by choosing the factor b such that y(a) = yp (a) + byh (a).

First-order differential equations

327

That is, y(x) = yp (x) + yh (x)[y(a) − yp (a)]/yh (a).

(5.19)

This method is called a variation of constants, since yp (x) is “proportional” to yh (x) but the proportionality “constant” is itself a function of x. Example 5.3.5 Solve the first-order LDE   d + s y(x) = e−tx . dx a. Get a homogeneous solution: yh (x) = e−sx b. Get the proportional function 

x

c(x) = 0



e−tx  1 (e−(t−s)x − 1).  dx = − −sx t−s e

c. Thus 1 (e−(t−s)x )e−sx t−s 1 −tx = e , s−t

yp (x) = −

where we have ignored a constant term in c(x). d. Check:     d 1 −tx −t + s  −tx = e = e−tx .  +s e dx s−t s−t 5.3.1

Uniqueness of the solution

The solution y(x) has been obtained, for both homogeneous and inhomogeneous equations, by specifying one boundary condition y(a) at x = a. We now show that this is enough to determine the solution uniquely for both linear and nonlinear firstorder DEs. Consider the Taylor expansion of y(x) about the boundary point x = a: y(x) = y(a) + (x − a)y (a) +

1 (x − a)2 y (a) + . . . . 2!

Given y(a), the general DE of first order y = f (x, y(x))

(5.20)

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Differential equations in physics

gives y = f (a, y(a)), y (a) =

d f (x, y(x))| x=a , dx

and all, higher derivatives by direct differentiation. Hence, if the Taylor series converges, the specification of y(a) alone is enough to determine the solution y(x) uniquely. Problems 5.3.1 Suppose R radioactive atoms are produced per second in a sample by neutron irradiation. If each atom so produced decays with the probability λ per second, show that the number N(T ) of radioactive atoms in the sample satisfies the differential equation dN(t) = R − λN(t) dt while under neutron irradiation. Calculate N(t) if N(0) = 0. 5.3.2 A raindrop initially at rest falls down from a height h. If air resistance gives rise to a retarding force −λv(t) proportional to the instantaneous velocity v(t), calculate the speed of the raindrop (a) as a function of time; and (b) when it hits the ground.

5.4

Second-order linear differential equations

Many DEs in physics are second order in the differential operator with the general form  2  d d + P(x) + Q(x) y(x) = R(x). (5.21) dx dx2 Examples are

2. The 3. The 5.4.1

d2 r(t) dt2

F(t) m . " 2 # d + ω2 x(t) = simple harmonic oscillator: 2 " 2 dt # d + k2 ψ(x) = 0. 1D Helmholtz equation: dx2

1. Newton’s force law:

=

0.

Boundary conditions

Consider the Taylor expansion (5.20) about the boundary x = a:

Second-order linear differential equations

y(x) = y(a) + (x − a)y (a) +

329

1 (x − a)2 y (a) + . . . . 2!

For the second-order DE, the numbers y(a) and y (a) are freely adjustable, but not the higher derivatives y (a), y (a), etc. This is because y (a) = −[P(a)y (a) + Q(a)y(a)] + R(a), y (a) = −(P y + Py + Q y + Qy ) x=a + R (a),

etc.,

are uniquely determined once y(a) and y (a) are chosen. Thus these two numbers control the solution y(x) uniquely, assuming again that the Taylor series converges. The discussion can be generalized readily to an nth-order LDE. Its solution is uniquely determined by n boundary conditions that may be taken to be y(a), y (a), . . . , y(n−1) (a) at any point x = a. 5.4.2

The homogeneous differential equation

Suppose y1 (x) and y2 (x) are two solutions of the homogeneous equation satisfying the boundary conditions y1 (a) = 1,

y1 (a) = 0;

(5.22a)

y2 (a) = 0,

y2 (a)

(5.22b)

= 1.

It is clear that y1 (x) and y2 (x) can never be proportional to each other, since they already differ at the boundary a. Thus they are linearly independent. They are also the only linearly independent solutions because second-order DEs have only two independent boundary conditions, such as y(a) and y (a). Indeed, a general solution satisfying the boundary conditions y(a) = c1 and y (a) = c2 is just the linear combination y(x) = c1 y1 (x) + c2 y2 (x). 5.4.3

Test of linear independence of solutions: Wronskian

The converse of the above situation occurs when we have to determine the linear independence of two given homogeneous solutions y3 (x) and y4 (x). This can be done by inspection since two functions are linearly independent if they are not simply proportional to each other. This simple procedure cannot be used for n given homogeneous solutions of an nth-order LDE when n > 2. Hence a more general procedure, applicable to a DE of any order, is desirable. We now describe this general procedure, but, for the sake of notational simplicity, we shall specialize to second-order DEs. It turns out that the result is useful even for second-order equations. If the functions y1 (x) and yz (x) are linearly independent, any solution y(x) of a second-order LDE and its slope y (x) can be expressed as

330

Differential equations in physics

c1 y1 (x) + c2 y2 (x) = y(x) c1 y1 (x) + c2 y2 (x) = y (x),

(5.23a)

with unique linear coefficients c1 and c2 . This is because, if the functions were linearly dependent, with say y2 (x) = ay1 (x), only the combination y = (c1 + ac2 )y can be determined. Then y(x) is also proportional to y1 (x) and therefore cannot be a general solution of a second-order LDE. If Eq. (5.23a) is written in the matrix form 

y1 (x) y2 (x) y1 (x) y2 (x)



   y(x) c1 =  , y (x) c2

(5.23b)

we see that a unique solution requires that the determinant  y (x) y (x) W(x) =  y1 (x) y2 (x) 1 2

  

(5.24)

does not vanish. W(x) is called a Wronskian, named after the nineteenth century mathematician Wronski. Suppose W(x = a)  0. Then the coefficients c1 and c2 can be determined at x = a. Since Eqs. (5.23) are satisfied for other values of x, it follows that W(x)  0 is guaranteed everywhere. A more rigorous demonstration will be given below. The Wronskian for n functions contains these functions and their first n − 1 derivatives. It can be used to test for the linear independence of n solutions of an LDE of order n. The Wronskian satisfies a number of interesting properties. For a second-order homogeneous LDE, W(x) satisfies a first-order LDE: d d    W(x) = (y1 y2 − y2 y1 ) = y1 y2 + y1 y 2 − y2 y1 − y2 y1 dx dx = −y1 (Py2 + Qy2 ) + y2 (Py1 + Qy1 ) = −P(x)W(x).

(5.25)

Therefore W(x) has the explicit form   x    W(x) = W(a) exp − P(x )dx .

(5.26)

a

This solution shows clearly that (1) if W(a)  0, then W(x)  0 everywhere, and (2) if W(a) = 0, then W(x) = 0 everywhere. Thus it is not necessary to test for the linear independence of two solutions at more than one point. Finally, Eq. (5.26) shows that the Wronskian can be calculated before the LDE is solved.

Second-order linear differential equations

Example 5.4.1 Two solutions of the DE   2 d 2 + k y(x) = 0 dx2 are easily seen to be cos kx and (1/k) sin kx. Thus   cos kx 1 sin kx k W(x) =   −k sin kx cos kx

331

(5.27)

   = 1.

If we had written the second solution as sin kx, the Wronskian would be k, which vanishes as k → 0. In this limit, sin kx = 0 is indeed not linearly independent of cos kx = 1, while 1 lim sin kx = x k

k→0

is linear independent of 1.  Problems 5.4.1 Determine if the functions in each of the following collections are linearly independent: (a) eikx , e−ikx , sin kx; (b) eikx , e−ikx , tan kx; (c) x2 − 1, x2 + 1, x. State any additional condition under which your answer is valid. Note: Wronskians of arbitrary functions can behave differently from Wronskians of the solutions of a DE. 5.4.2 Show that the solution of a third-order, homogeneous linear differential equation is determined uniquely by specifying three boundary conditions. 5.4.3 Calculate the displacement x(t) of a damped linear oscillator that satisfies the differential equation m x¨ + b x˙ + kx = 0, with (a) x(t = 0) = 1, x˙(t = 0) = 0; (b) x(t = 0), x˙(t = 0) = 1. 5.4.4 Calculate the Wronskian W(x) of the differential equation in Problem 5.4.3. 5.4.5 Calculate the Wronskian of the following differential equations (a) Associated Legendre equation:   m2 2   y = 0; (1 − x )y − 2xy + l(l + 1) − 1 − x (b) Bessel equation: x2 y + xy + (x2 − m2 )y = 0;

332

Differential equations in physics

(c) Spherical Bessel equation: x2 y + 2xy + [x2 − l(l + 1)]y = 0; (d) Mathieu equation:

! 1 y + c − d2 cos(2x) y = 0; 2

(e) Confluent hypergeometric equation: xy + (c − x)y − ay = 0 5.4.6 Show that the specification of y(a) and y (a) at the boundary x = a will also define a unique solution for a nonlinear differential equation of second order.

5.5

The second homogeneous solution and an inhomogeneous solution

If one solution y1 (x) of a second-order homogeneous LDE is known, a second solution y2 (x), linearly independent of the first, can be obtained with the help of the Wronskian:   y1 y2 − y2 y1 W(x) d y2 = 2 . = dx y1 y21 y1 (x) A simple integration gives y2 (x) y2 (b) − = y1 (x) y1 (b) Hence

 y2 (x) = g(x)y1 (x),

g(x) =

x

b

 b

x

W(t)dt = g(x). y21 (t)

  t $ !  2 exp − P(t )dt y1 (t) dt,

(5.28)

b

where we have dropped a term proportional to y1 (x). Example 5.5.1 If one solution of d2 dx2

ψ(x) = 0

(5.29)

is 1, show that the linearly independent second solution is x. y1 (x) = 1 y2 (x) = g(x)y1 (x) = g(x)  x = dt = x − b = x, b

if

b = 0,

The second homogeneous solution

333

where we have used the result P(x) = 0. Conversely, given y2 (x) = x, we find y1 (x) = cg(x)x, where  x  x 1 1 1 1 dt = − . g(x) = dt = 2 2 b x b y2 (t) b t We may take b = ∞, c = −1 to get y1 (x) = 1.  5.5.1

Inhomogeneous solutions

The solution of the inhomogeneous second-order LDE (5.21) L y(x) = R(x)

has the general form y(x) = yp (x) + [c1 y1 (x) + c2 y2 (x)],

(5.30)

where y p (x) is a solution (i.e., any solution) of the inhomogeneous equation and y1 (x), y2 (x) are the two linearly independent homogeneous solutions. We call y p (x) a particular solution (or a particular integral) and [c1 y1 (x) + c2 y2 (x)] a complementary function. Why do we bother with y(x) if we already have a y p (x)? The reason is that y p (x) satisfies only the boundary conditions y p (a), yp (a) at x = a. Suppose we want instead a solution satisfying the boundary conditions y(a) = α, y (a) = β. We do not want to look for the particular solution with just the right boundary conditions. This is because particular solutions are hard to obtain; we are lucky if we manage to find one, with or without the correct boundary conditions. The complementary function now comes to the rescue, because it can change boundary conditions without contributing anything to the inhomogeneity of the DE. The reader can easily verify that the required complementary function is c1 y1 (x) + c2 y2 (x) with the coefficients chosen to satisfy the correct boundary conditions at x = a: c1 y1 (a) + c2 y2 (a) = α − yp (a), c1 y1 (a) + a2 y2 (a) = β − yp (a). Furthermore, the existence of the linear coefficients  −1     α − yp (a) y1 (a) y2 (a) c1 = y (a) y (a) β − yp (a) c2 1 2

(5.31)

(5.32)

is guaranteed by the linear independence of the homogeneous solutions y1 , and y2 , since then the Wronskian W(a)  0 and therefore the inverse matrix in Eq. (5.32) exists. 5.5.1.1

A particular solution: Method of variation of constants.

We still need one (any one) particular solution y p (x) of the inhomogeneous DE. To obtain this, we first observe that the function y p (x) contaius two degrees of freedom in the sense that at a point x = x1 its value y p (x1 ) and its slope yp (x1 )

334

Differential equations in physics

can be chosen arbitrarily. These two arbitrary numbers may be expressed in terms of the values and slopes of the two linearly independent homogeneous solutions yi (x), i = 1, 2: yp (x) = v1 y1 (x) + v2 y2 (x),

(5.33a)

yp (x)

(5.33b)

=

ν1 y1 (x)

+

ν2 y2 (x),

because the right-hand sides also describe a system with two degrees of freedom, as represented by the two linear coefficients v1 and v2 . Indeed, these linear coefficients are uniquely determined by Eq. (5.33) at x = x1 because the Wronskian W(x1 ) does not vanish if the yi (x) are linearly independent. However, these linear coefficients v1 and v2 , cannot be constants independent of x, for then y p (x) solves the homogeneous LDE, not the inhomogeneous equation. We therefore conclude that vi = vi (x),

i = 1, 2,

are functions of x. It is to this necessary dependence on x that the name “variation of constants” refers. Direct differentiation of Eqs. (5.33a) does not yield Eqs. (5.33b) unless υ1 (x)y1 (x) + υ2 (x)y2 (x) = 0.

(5.34a)

This single requirement is insufficient to determine the two unknowns υi . We need another relation. This can be obtained from the original DE (5.21) with the help of Eqs. (5.33): υ1 (x)y1 (x) + υ2 (x)y2 (x) = R(x).

(5.34b)

The two Eqs. (5.34) can now be used to solve for υ1 and υ2 . The results are υ1 (x) = −y2 (x)R(x)/W(x), υ2 (x) = y1 (x)R(x)/W(x),

W(x)  0.

(5.35)

(The verification of Eqs. (5.34b, 5.35) is left as an exercise.) These are first-order DEs whose solutions are  x  x y2 (t)R(t) y1 (t)R(t) υ1 (x) = − dt, υ2 (x) = dt, (5.36) W(t) W(t) a a where the integration constants have been chosen arbitrarily to give υ1 (a) = υ2 (a) = 0. With this choice, the particular solution satisfies the boundary conditions y p (a) = 0 and yp (a) = 0, according to Eqs. (5.33). Example 5.5.2 Obtain a particular solution of the DE  2  d 2 + k y(x) = A sin qx. dx2

(5.37)

The second homogeneous solution

335

From the earlier discussion on the homogeneous equation (5.27), we know that its two linearly independent solutions are y1 (x) = cos kx,

y2 (x) =

1 sin kx, k

and that the Wronskian is W(x) = 1. Hence Eq. (5.36) gives  x A v1 (x) = − sin kx sin qx dx −L k   A sin(k − q)x sin(k + q)x + C1 , − =− k 2(k − q) 2(k + q) where the constant C1 comes from the lower limit of integration and may be dropped. Similarly  x v2 (x) = A cos kx sin qx dx −L

⎧ " cos(k−q)x ⎪ ⎪ ⎨ A 2(k−q) − =⎪ ⎪ ⎩ A sin2 kx,

cos(k+q)x 2(k+q)

#

+ C2 , k  q;

2k

k = q.

(5.38)

Thus a particular solution from Eq. (5.33a) is, for q  k, A [sin(k − q)x cos kx − cos(k − q)x sin kx] 2k(k − q) A [sin(k + q)x cos kx − cos(k + q)x sin kx] + 2k(k + q)

yp (x) = −

= A sin qx/(k2 − q2 ).

(5.39a)

For q = k, the answer from Eq. (5.38) is   A sin 2kx 1 A yp (x) = − x cos kx + 2 sin3 kx k 4k 2 2k =−

A A x cos kx + 2 sin kx, 2k 2k

where we may drop the sin kx term.

(5.39b)



Eq. (5.37) appears in classical mechanics as the equation of motion (if x = t, the time variable) of a driven harmonic oscillator with no damping. The possible zero in the denominator k2 − q2 of the solution (5.38) suggests the possible appearance of an amplitude resonance at k = q where the amplitude y(t) of the oscillation is maximal. In the special case considered here, there is no frictional damping. As a result, the amplitude at the k = q resonance grows without limit, as the resonance term −(A/2k)t cos kt in Eq. (5.39b) shows. The linear dependence on A of the solutions

336

Differential equations in physics

shows that the amplitude of oscillation is twice as large when the system is driven twice as hard. This arises because the system is linear; that is, it is a consequence of the superposition principle shown in Eq. (5.14). As another example, let us use the Dirac δ function R(x) = δ(x − x ) in Eq. (5.21). Eq. (5.36) now gives  0, v1 (x) = y2 (x )/W(x ),

x < x x > x

= −[y2 (x )/W(x )]Θ(x − x ),

(5.40a)

where Θ(t) is a step function. Similarly v2 (x) = [y1 (x )/W(x )]Θ(x − x ).

(5.40b)

The solution of an inhomogeneous DE with a δ-function inhomogeneity is called a Green function. From Eqs. (5.30) and (5.33), we find that it has the general form G(x, x ) = Gp (x, x ) + c1 y1 (x) + c2 y2 (x),

(5.41)

Gp (x, x ) = {[−y1 (x)y2 (x ) + y1 (x )y2 (x)]/W(x )}Θ(x − x ).

(5.42)

where

As usual, there are two arbitrary constants c1 , c2 that can be chosen to fit two boundary conditions, either at the same point or at two different points. Problems 5.5.1 Verify Eqs. (5.34b) and (5.35). 5.5.2 Obtain the solution of the homogeneous LDE   2 d 2 + k y(x) = 0 dx2 satisfying the boundary conditions 1 y(a) = , 2

y (a) =

1 k 4

at

x=a=

π . 2k

5.5.3 A frictionless simple harmonic oscillator driven by a force F(t) satisfies the inhomogeneous LDE  2  d 2 + ω0 x(t) = F(t). dt2

Green functions

337

If the driving force is F(t) = A cos ωt, real A, show that a particular solution is xp (t) = A cos ωt/(ω20 − ω2 ). Describe the motion of this driven oscillation when the initial conditions at t = 0 are x(0) = 1, x˙(0) = 12 ω0 . Note: x p (t) is much easier to find if one first solves the inhomogeneous DE with the complex driving force Aeiωt . This is called the complex method. If possible, find x p (t) using first the complex method, and then again by using the real method, i.e., by solving the original DE directly. 5.5.4 Consider the differential equation x2 y + (1 − 2a)xy + a2 y = xb ,

a  b.

(a) Show that one solution of the homogeneous equation is y1 (x) = xa . (b) Obtain the linearly independent second solution of the homogeneous equation. (c) Obtain the solution of the inhomogeneous equation satisfying the boundary conditions y(1) = c,

y (1) = d

5.5.5 Obtain a particular solution of the equation for the driven harmonic oscillator with damping  2  d d 2 + 2β + ω0 y(t) = A δ(t − t ) dt dt2 by the method of variation of constants.

5.6

Green functions

We have seen in Section 4.6 that given a solution of the linear DE in x at fixed x L (x)G(x, x ) = δ(x − x )

(5.43)

satisfying specified boundary conditions, the solution of the general inhomogeneous DE L (x)y(x) = R(x)

satisfying the same boundary conditions can be written in the integral form  b y(x) = G(x, x )R(x )dx , a

(5.44)

(5.45)

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Differential equations in physics

where [a, b] is the interval for the variable x. The Green function G(x, x ) is independent of the inhomogeneity function R(x), so that every solution of Eq. (5.44) satisfying the same boundary conditions is given by Eq. (5.45) involving the same Green function even though R(x ) itself might change. We would like to show that G(x, x ) is made up of suitable homogeneous solutions joined together at x = x in a well-defined way. If x  x , the δ function is identically zero. Equation (5.43) is then an homogeneous equation in x. Therefore for fixed x  G< (x, x ), x < x  G(x, x ) = (5.46) G> (x, x ), x > x are just solutions of the homogeneous equations. Suppose a full set of n boundary conditions (for an nth-order DE) has been specified, say at the lower limit a of the interval. G< (x, x ) is now uniquely defined up to a multiplicative constant until we reach x = x . On passing this point, we must switch over to another homogeneous solution G> (x, x ) before continuing to the upper limit x = b. This second homogeneous solution must satisfy a full set of n boundary conditions, say at x = b. This does not completely determine G, however, because we have not yet determined if the normalizations of G< and G> can be chosen arbitrarily. The answer is “no”. If G is a solution of Eq. (5.43), 2G is not; it is a solution of an equation with the inhomogeneity 2 δ(x − x ). Thus the overall normalization is not arbitrary. This is because Eq. (5.43) is an inhomogeneous equation, not an homogeneous one. We now show that the relative normalization between G< and G> is also not arbitrary. Let the nth-order inhomogeneous DE (5.43) for y(x) = G(x, x ) take the form y(n) (x) − f (x, y(x), . . . , y(n−1) ) = δ(x − x ), y(m) (x) =

dm y(x). dxm

(5.47)

The δ-function that appears can be considered the derivative of the unit step function Θ(x − x ) in x at fixed x , because  x +ε d Θ(x − x )dx = Θ(ε) − Θ(−ε) = 1. x −ε dx Hence Eq. (5.47) will be satisfied if y(n−1) (x) has a unit step discontinuity at x = x , while y(n−2) (x) is continuous but has a kink there. The δ function cannot be associated with y(n−1) (x) or one of the lower derivatives, for otherwise on differentiations in Eq. (5.47) it will generate derivatives of the δ function that are not present on the right-hand side. This shows that the only solution is one for which y(n−1) (x) has a unit step or jump discontinuity at x = x . That is, G(n−1) (x , x ) = G(n−1) (x , x ) + 1, > <

(5.48a)

Green functions

339

while (m)     G(m) > (x , x ) = G < (x , x ),

for

m < n − 1.

(5.48b)

To summarize, all derivatives G(m) (x, x ) are smooth for m ≤ n except at x = x , where G(n−2) has a kink, G(n−1) has a jump discontinuity, and G (n) has a δ-function discontinuity. The n boundary conditions needed to specify a Green function for an nth-order DE do not have to be specified all at one point. They may be specified at two or more points in the interval [a, b]. For example, the Green function for a second-order DE may be specified by one boundary condition at a and another at b. Example 5.6.1 Obtain the Green function for the DE  2  d 2 + k y(x) = δ(x − x ) dx2

(5.49)

describing the motion of a driven string of length L with fixed ends at x = 0 and L. Solutions of the homogeneous DE are sine and cosine functions. Since the string ends are fixed [i.e., y(0) = y(L) = 0], we must have G< (x, x ) = a sin kx,

G> (x, x ) = b sin k(x − L).

(5.50)

The unknown coefficients are now determined from Eqs. (5.48) with n = 2: a sin kx = b sin k(x − L), ak cos kx + 1 = bk cos k(x − L). These simultaneous equations have the unique solutions    −1   0 a sin kx − sin k(x − L) = −1 b k cos kx −k cos k(x − L)   sin k(x − L) = (k sin kL)−1 . sin kx

(5.51)

From these results we can see that the Green function in Eq. (5.50) can be written more compactly in the form G(x, x ) = sin kx< sin k(x> − L)/(k sin kL).

(5.52)

This possibility is a consequence of the type of boundary conditions it satisfies.  Example 5.6.2 Obtain a particular integral of the driven damped oscillator equation  2  d d 2 + 2β + ω0 y(t) = eγt , β > 0, (5.53) dt dt2

340

Differential equations in physics

giving its motion as a function of the time t. We shall look for a solution of the form  yp (t) =

∞ −∞



G(t, t )eγt dt ,

(5.54)

where the Green function satisfies the DE 

 d2 d 2 + 2β + ω0 G(t, t ) = δ(t − t ). dt dt2

(5.55)

Since the boundary conditions (here called initial conditions) are not of interest, we may take the trivial solution G < (t, t ) = 0,

for

t < t .

(5.56)

This is actually a physically useful choice because we expect that nothing special happens for t < t before the impulsive driving force δ(t − t ) acts on the system. To obtain g> (t, t ), we first calculate the two homogeneous solutions. These turn out to be exp(−αt) with two different values of α α1,2 = β ± (β2 − ω20 )1/2 .

(5.57)

G> (t, t ) = a1 e−α1 t + a2 e−α2 t .

(5.58)

Hence

It should satisfy the initial conditions at t = t obtainable from Eq. (5.48b): G> (t , t ) = 0,

G(1) > =

d G> (t , t ) = 1. dt

The linear coefficients ai can now be calculated. They are readily found to be 

a1 =

e α1 t , α2 − α 1



a2 =

e α2 t . α1 − α 2

Thus 



G> (t, t ) = (e−α1 (t−t ) − e−α2 (t−t ) )/(α2 − α1 ),

t > t .

(5.59)

Green functions

341

Finally y p is calculated from Eq. (5.54)  yp (t) =

t



−∞

G> (t, t )eγt dt

  1 1 1 eγt − = α2 − α1 α1 + γ α2 + γ =

eγt eγt . = (α1 + γ)(α2 + γ) γ2 + 2βγ + ω20

(5.60)

One can check by inspection that this is indeed a solution of Eq. (5.53).  For DEs with constant coefficients such as Eq. (5.53), y p itself can be obtained more readily by other means. This is not the case with the more general Eq. (5.21), for which the present approach is very useful. The reader can check that when this is done, one recovers the same result as Eq. (5.42), obtained in the previous section by the method of separation of constants. (See Problem 5.6.4). Problems 5.6.1 Calculate the Green function for the following differential equations in the interval (−∞, ∞):  (a) (b)∗

 d2 2 − λ y(x) = R(x), dx2

d2 y(x) = R(x), dx2

with y(−∞) = 0, y(∞) = 0.

with y(0) = 0, y(1) (0) = 0;

5.6.2 Calculate the Green function for the differential equation 

 d 2 d l(l + 1) + y(x) = R(x) − dx2 x dx x2

in the interval (0, ∞) satisfying the boundary conditions y(0) = y(∞) = 0. 5.6.3∗ Solve the driven damped oscillator equation 

 d2 d 2 + 2β + ω0 y(t) = F(t) dt dt2

satisfying the initial conditions y(0) = 0, y˙ (0) = v. 5.6.4 Obtain Eq. (5.42) by using the method of this section.

342

Differential equations in physics

5.7

Series solution of the homogeneous second-order linear differential equation

We have seen in Section 5.5 that, given one solution of an homogeneous secondorder LDE, a second solution linearly independent of the first can be obtained by integration. If we are given both solutions, a particular solution of the inhomogeneous LDE can be calculated by integration. It remains for us to obtain at least one solution of the homogeneous equation. It is often possible and useful to obtain a solution y(x) of the homogeneous LDE L y(x) = 0 in the form of a power series in powers of x. According to Frobenius, the power series should have the general form y(x) = x s (a0 + a1 x + a2 x2 + . . .) =

∞ 

aλ xλ+s ,

a0  0,

(5.61)

λ=0

where the power s is not necessarily zero. (If s = 0, the series is a Taylor series.) By direct differentiation, we get dy(x) = x s−1 [sa0 + (s + 1)a1 x + (s + 2)a2 x2 + . . .] dx ∞   = aλ (λ + s)xλ +s−1 ,

(5.62)

λ

d 2 y(x) = x s−2 [s(s − 1)a0 + (s + 1)sa1 x + (s + 2)(s + 1)a2 x2 + . . .] dx2 ∞   = aλ (λ + s)(λ + s − 1)xλ +s−2 .

(5.63)

λ =0

Substitution of these into the LDE shows that the DE can be satisfied if the coefficient of each power of x vanishes. This set of conditions is usually sufficient to define a solution y(x), as the following two examples will demonstrate. Example 5.7.1 Obtain a Frobenius-series solution for Eq. (5.27) describing harmonic oscillations   2 d 2 + k y(x) = 0. (5.27) dx2 Direct substitution of Eqs. (5.62) and (5.63) gives 0 = x s−2 [s(s − 1)a0 + (s + 1)sa1 x + (s + 2)(s + 1)a2 x2 + . . . + k2 x2 (a0 + a1 x + a2 x2 + . . .)] = a0 s(s − 1)x s−2 + a1 (s + 1)sx s−1 + [a0 k2 + (s + 2)(s + 1)a2 ]x s +...

Series solution of the linear differential equation

343

+ [aλ k2 + (s + λ + 2)(s + λ + 1)aλ+2 ]x s+λ + ....

(5.64)

The coefficient of the lowest power x s−2 must vanish; that is, a0 s(s − 1) = 0.

(5.65)

Since a0  0, this requires that s = 0 or 1. Equation (5.65), which determines the power s of the Frobenius series (5.61) is called an indicial equation. The coefficient of the next power x s−1 must also vanish: a1 (s + 1)s = 0.

(5.66)

This expression shows that if s = 0, a1 can be nonzero, while a1 = 0 is required if s = 1. In a similar way, the coefficient of a higher power x s+λ vanishes if aλ+2 =

k2 aλ . (s + λ + 2)(s + λ + 1)

(5.67)

In this way, the coefficients aλ of the Frobenius series can all be determined step by step. Equation (5.67) is called a recurrence relation. When Eq. (5.67) is used to generate unknown coefficients aλ with increasing index λ, the recurrence relation is said to be used in the forward direction. The recurrence relation (5.67) contains the special feature that it steps up λ by 2. As a result, the coefficients aλ are separated into two disjoint groups—one with even λ and another with odd λ. For the even chain, we find a. for s = 0 :

a2 = −

k2 a0 , 2

for s = 1 :

a2 = −

k2 a0 , (3)(2)

b.

a4 = −

k2 k4 a2 = a0 , . . . , (4)(3) 4!

a4 = −

k2 k4 a2 = a0 , . . . (5.68) (5)(4) 5!

The resulting solutions are, respectively, a. b.

k2 k4 y(x) = yeven (x) = 1 − x2 + x4 − . . . , 2! 4!   2 k 2 k4 4 y(x) = yodd (x) = x 1 − x + x − . . . , 3! 5!

(5.69)

where we have used a0 = 1.  What about the odd chain of coefficients? There is no solution for s = 1, since a1 = 0 and therefore all higher odd coefficients vanish by virtue of Eq. (5.67). For s = 0, a1  0; hence from Eq. (5.67)

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Differential equations in physics

a3 = −

k2 a1 , (3)(2)

a5 = −

k2 k4 a3 = a1 , . . . . (5)(4) 5!

The resulting solution for a1 = 1 is just yodd (x). Thus this solution is not new. This is not unexpected, because Eq. (5.27) has only two linearly independent solutions, and they are already given in Eq. (5.69). The formal solutions in Eq. (5.69) are meaningful only if these infinite series converge. To determine their convergence, we examine the ratio of successive terms, and find x2 aλ+2 2 x = −k2 −→ 0 aλ (s + λ + 2)(s + λ + 1) λ→∞ for any x. Hence both solutions converge to well-behaved functions of x for all finite values of x. Indeed, Eqs. (5.69) are just the Taylor expansions about x = 0 of the functions yeven (x) = cos kx,

yodd (x) =

1 sin kx, k

which may be seen by inspection to be the two linearly independent solutions of Eq. (5.27). 5.7.1

Parity property

According to Eq. (5.69) yeven (−x) = yeven (x),

yodd (−x) = −yodd (x)

(5.70)

are, respectively, an even and an odd function of x. The function yeven (yodd ) then said to have even (odd) parity. For a solution of a LDE to have a definite parity, the linear operator L must be even (or invariant) under the parity operation x → −x; that is, L (−x) = L (x).

(5.71)

To see this, we first note that if y1 (x) is a solution of the DE, then y1 (−x) must be a solution of the DE L (−x)y(−x) = 0. However, if Eq. (5.71) is also satisfied, we have 0 = L (−x)y1 (−x) = L (x)y1 (−x), so that y2 (x) = y1 (−x) is also a solution of the DE L (x)y(x) = 0. Furthermore, if y1 (−x) is linearly independent of y1 (x), then the following two solutions of definite parity can be constructed: 1 yeven (x) = [y1 (x) + y1 (−x)] 2 1 yodd (x) = [y1 (x) − y1 (−x)]. 2

(5.72)

Series solution of the linear differential equation

345

Thus it is the parity invariance of the differential operator in Eq. (5.27) that permits the extraction of solutions of definite parities shown in Eq. (5.69). Physical fields (describing physical properties) in space are also said to have even (odd) parity if they are even (odd) functions of x. If they satisfy an LDE L (x)y(x) = 0, then the linear operator involved must be parity even. An LDE whose solution is a physical field is called a field equation or an equation of state, in the sense that its solution describes a state of existence of the physical property in space. Physical systems whose field equations contain only parity-even operators are said to be parity conserving because states (i.e., solutions) of definite parities can be constructed. On the other hand, there are states of physical systems with unavoidably mixed parities. This feature is a consequence of the lack of parity invariance in the differential operators appearing in their field equations. Such systems are said to be parity nonconserving or parity violating. An example of a DE whose solutions cannot have a definite parity is the equation (for a classical oscillator with damping if x = t)  2  d d 2 + 2β + ω0 y(x) = 0. dx dx2 The recurrence relation for a parity-nonconserving equation such as this does not break up into two disjoint chains of different parities. That the use of recurrence relations may require some ingenuity is illustrated by the following example. Example 5.7.2 Solve the Bessel equation   2 d 2 d 2 2 x + x − (x − μ ) y(x) = 0. dx dx2

(5.73)

This differential operator is parity invariant, so that there are separate even and odd chains of coefficients. The DE itself may be written with the help of Eqs. (5.61)– (5.63) in the form ∞ 

bλ xλ+s = 0,

(5.74)

λ=λmin

where λmin = 0. The conditions for a solution are therefore bλ = aλ [(λ + s)2 − μ2 ] + aλ−2 = 0,

λ  0,

(5.75)

where aλ = 0 if λ < 0. (The derivation of the expression for bλ is left as an exercise in problem 5.7.1.) The coefficient of the lowest power (λ = 0) gives the indicia1 equation a0 (s2 − μ2 ) = 0,

s = ±μ.

(5.76)

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Differential equations in physics

The coefficient of the next power (λ = 1) determines the coefficient a1 of the odd chain: a1 [(s + 1)2 − μ2 ] = a1 (2s + 1) = 0,

(5.77)

where use has been made of Eq. (5.76). Equation (5.77) is satisfied 1.

1 for s  − , 2

if a1 = 0;

2.

for a1  0,

1 if s = − . 2

(5.78)

For λ  2, Eqs. (5.75) and (5.76) can be used to simplify the recursion formula to aλ−2 . (5.79) aλ = − λ(λ + 2s) However, trouble might develop when λ + 2s vanishes in the denominator. This occurs when 3 λ (5.80) s = − = −1, − , −2, . . . , for λ ≥ 2. 2 2 Two situations can be distinguished: (1) If s = − 32 , − 52 , . . . is a negative half-integer, trouble might occur at λ = 3, 5, . . ., that is, in the odd chain. (2) If s = −1, −2, . . ., is a negative integer, trouble might develop at λ = 2, 4, . . ., that is, in the even chain. To see how we can get out of these problems, we first consider the trouble in the odd chain. Suppose 2s = −m, m = 3, 5, . . .. Since s  − 12 , a1 = 0 must hold. The next higher coefficients also vanish one by one until we reach λ = m, when am = −

0 am−2 = = anything. m(m − m) 0

If we now take am  0 and go on to higher coefficients with Eq. (5.79), we find the Frobenius series y s=−m/2 (x) = x−m/2 (a1 + . . . + am xm + am+2 xm+2 + . . .)

(5.81)

actually contains zeros before the term with am . Removing these “ghosts”, we find a series with an effective index seff = 12 m: y s=−m/2 (x) = xm/2 (am + am+2 x2 + . . .). Furthermore, the recursion formula for the coefficients beyond am may be written as am+λ−2 am+λ = − . (5.82) (m + λ)λ This is actually identical to the recursion formula for the Frobenius solution with s = 12 m. The only difference is in the subscripts of the coefficients due to the counting

Differential eigenvalue equations

347

of the ghosts in Eq. (5.81). We therefore conclude that the Frobenius solution for s = − 12 m “collapses” into the solution for s = 12 m, and is not linearly independent of the latter. The same result can also be obtained for the even series when s is a negative integer. (The demonstration is left as an exercise.) In both cases we conclude that the Frobenius method does not necessarily give two linearly independent solutions even when s has two distinct solutions. This is acceptable, because the linearly independent second solution can always be obtained from the first by the method of Section 5.5.  Problems 5.7.1 Verify Eq. (5.75). 5.7.2 Show that the Frobenius-series solution of the Bessel equation (5.73) for s = −m, m any positive integer, is proportional to the solution for s = m. If we denote the solution of Eq. (5.73) as J s (x), then we may write J−m (x) = (−1)m Jm (x), where the choice of the proportionality constant (−1)m is a matter of convention. 5.7.3 Obtain the power-series solutions of each of the following homogeneous differential equations: (a) y + 2y = 0; (b) y + 4y − 5y = 0; (c) [x(1 − x)y − xy − y] = 0; (d) xy + (sin x)y = 0; (e) (cos x)y + xy + ay = 0.

5.8

Differential eigenvalue equations and orthogonal functions

The Legendre equation



 d (1 − x ) 2 − 2x + c y(x) = 0 dx dx 2

d2

(5.83)

has the unusual feature that at x = ±1 the first term disappears, leaving a firstorder DE. These points x = ±1 are said to be singular points of the DE. Something interesting happens at these points, as we shall show in this section. Let us first discuss its Frobenius-series solutions in a systematic way. The differential operator in Eq. (5.83) is parity invariant, as happens so often in DEs of interest to physicists. Thus there are as a rule two linearly independent solutions yeven (x) and yodd (x), one of each parity. The coefficients of the Frobenius series for the entire DE is of the form of Eq. (5.74) with bλ = aλ+2 (λ + s + 2)(λ + s + 1) − aχ [(λ + s)(λ + s + 1) − c] = 0,

(5.84)

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Differential equations in physics

where aλ = 0 if λ < 0. For λ = −2 and −1, we find the same result as for the DE (5.27), namely a0 s(s − 1) = 0,

a1 (s + 1)s = 0.

Hence a0  0 is permitted if s = 0 or 1, while a1  0 occurs only for s = 0. The recursion formula from Eq. (5.84) for λ ≥ 0 is aλ+2 = aλ

(λ + s)(λ + s + 1) − c . (λ + s + 2)(λ + s + 1)

(5.85)

The subscript is stepped up by 2, so that all terms of the Frobenius series have the same parity. For s = 0, we have both an even and an odd chain of coefficients: 1. Even chain:

   c 6−c a0 → a2 = a0 − → a4 = a2 → a6 . . . , 2 12

2. Odd chain:

 a1 → a3 = a1

   2−c 12 − c → a 5 = a3 → a7 . . . . 6 20

(5.86)

These yield the solutions   c 2 c c−6 4 yeven (x) = 1 − x + x + ..., 2 2 12   2 − c 3 2 − c 12 − c 5 yodd (x) = x + x + ..., x + 6 6 20

(5.87a) (5.87b)

where we have set a0 = 1 and a1 = 1, respectively. For s = 1, only an even chain is allowed. The solution turns out to be just yodd (x), as expected. Next, we must determine if the Frobenius solutions (5.87) converge. The ratio of successive terms is aλ+2 2 (λ + s)(λ + s + 1) − c 2 x = x −→ x2 . aλ (λ + s + 2)(λ + s + 1) λ→∞ Hence the series converges if |x| < 1 and diverges if |x| > 1. (It turns out to be divergent also for |x| = 1.) An important special case occurs when c = cl ≡ l(l + 1),

l = 0, 1, 2, . . . = any non-negative integer.

(5.88)

Differential eigenvalue equations

349

For each of these special values, one of the two chains of coefficients for s = 0 shown in Eq. (5.86) terminates at al , because al+2 = al

l(l + 1) − cl = 0 = al+4 = . . . . (l + 2)(l + 1)

(5.89)

In particular, the even chain terminates when l is even, while the odd chain terminates when l is odd. The corresponding Frobenius solution in Eq. (5.87) then simplifies to a polynomial. It is called the Legendre polynomial of degree l, or Pl (x). (The degree of a polynomial is the value of its highest power.) Legendre polynomials occur so frequently in mathematics that the choice of the arbitrary multiplicative constant al giving Pl (x = 1) = 1 has become standard: P0 (x) = 1, P2 (x) = 12 (3x2 − 1), P4 (x) = 18 (35x4 − 30x2 + 3),

P1 (x) = x, P3 (x) = 12 (5x2 − 3x), etc.

(5.90)

Legendre polynomials, being finite power series, are finite for all finite values of x. In particular, they are also defined at the singular points x = ±1 of the DE. This nice behavior at x = ±1 is the main reason for considering the special case c = cl , because when c  cl the function is infinite at x = ±1 and therefore cannot be used to describe bounded physical properties at these points. The other chain of coefficients in Eq. (5.86) does not terminate, because its parity is opposite to that of the chain that does, and no coefficient in its chain ever vanishes. The corresponding Frobenius solution will remain an infinite series. As discussed earlier, this series converges to a well-defined function for |x| < 1, and is called a Legendre function of the second kind, Ql (x). This function is also defined for |x| > 1 if by Ql (x) we mean the linearly independent second solution of Eq. (5.83). However, it is not given by the Frobenius solution (5.61), which does not converge for |x| > 1. It may be calculated either by the method of Section 5.5 or by a Frobenius series in inverse powers of x. At the singular points x = ±1, the function Ql (x) turns out to be infinite. The special nature of the Legendre polynomials Pl (x) at the singular points x = ±1 of Eq. (5.83) has a very useful description in the language of eigenvalue problems. An eigenvalue equation in mathematics involves a general equation for an unknown quantity y(c) of the form L y(c) = cy(c),

(5.91)

where L is a linear operator that satisfies the linearity property of Eq. (5.10) and c is a scalar constant. Sometimes Eq. (5.91) does not have a solution unless c = ci , i = 1, 2, . . . , n, is the one of a set of eigenvalues. That is the situation for the matrix eigenvalue equation (2.49). If L is a differential operator in the variable x, the solutions are of course functions of x, so that Eq. (5.91) should be written more specifically as L (x)y(x; c) = cy(x; c).

(5.92)

350

Differential equations in physics

It is possible that y(x; c) for any c is finite everywhere except at a number of singular points of the equation where it is finite only when c = cl (l = 0, 1, 2, . . .) is one of a set of eigenvalues. This is the situation for the Legendre DE. It is also possible that y(x; c) is defined everywhere for all c, but only for a set of eigenvalues c = ci will the function y(x; c = ci ) have a certain desirable property. This is the situation for the trigonometric functions appearing in the Fourier series for the interval −π ≤ x ≤ π. In this case, we are only interested in eigenfunctions that are periodic in x with the same period of 2π. A DE such as Eq. (5.92), in which the solution has been further selected by imposing an additional eigenvalue condition leading to the choice c = ci , i = 1, 2, . . ., is called a differential eigenvalue equation. Each of the allowed values ci is called an eigenvalue and the corresponding solution y(x; ci ) is its eigenfunction. Since the eigenvalues of both the Legendre polynomials and the Fourier-series eigenfunctions are real, the corresponding differential operators must be Hermitian. The concept of Hermiticity for a differential operator turns out to be much more complicated than that for matrices. For example, it depends on the boundary conditions satisfied by the eigenfunctions, as well as on the operator itself. Nevertheless, it is also true that the eigenfunctions of a Hermitian differential operator are orthogonal or can be orthogonalized. (The orthogonality between two functions is defined in terms of an inner product that is an integral of a product of the functions. We have come across such inner products in Chapter 4.) For this reason we call Fourier-series functions orthogonal function and the Legendre polynomials orthogono1 polynomials. An earlier discussion of orthogonal systems has been given in Section 4.10. Orthogonal functions are useful in the expansion of arbitrary functions, especially in connection with the solution of partial differential equations. This application will be described in Section 5.10. Differential eigenvalue equations will be described in more detail in Section 7.8. Problem 5.8.1 Obtain a solution of the Legendre differential equation, Eq. (5.83), with c = 0, which is linearly independent of P0 (x). (Answer: Q0 (x) = 12 ln[(1 + x)/ (1 − x)].)

5.9

Partial differential equations of physics

In physics, there are.differential equations of motion that describe the response of systems to external disturbances. There are also differential equations of states, or field equations, whose solutions give the space-time dependence of physical properties. These are, of course, partial differential equations (PDE) in the four variables x, y, z, and t. Some common PDEs of physics are shown in Table 5.1. All these equations are linear. They are of second order in the space variables x, y, z (or x1 , x2 , x3 ) and of first or second order in time. The use of differentia1 operators guarantees invariance with respect to space translations and time displacements. The scalar differential operator ∇2 is actually the simplest operator that will also respect invariance under rotation and the parity operation xi → −xi .

Separation of variables and eigenfunction expansions

351

Table 5.1 Some common partial differential equations in physics

Equation

PDE

Physical applications

1 ∂2 u(r, c2 ∂t2 ω2 − c2 u(r)

Wave

∇2 u(r, t) =

Helmholtz

∇2 u(r) =

Laplace

∇2 φ(r)

Poisson

∇2 φ(r) = ρ(r)

Diffusion

∇2 u(r, t) =

Schr¨odinger wave

# " 2  ∇2 + V(r) ψ(r) = Eψ(r) − 2m

Maxwell

∇ · E = ρ/0 , ∇ × E = −∂t B, ∇ · B = 0, ∇ × B = ∂t E/c2 + μ0 J.

t)

=0

1 ∂ D ∂t u(r,

t)

Wave motion Wave motion of frequency ω Electrostatic potential in free space Electrostatic potential with charge density Diffusion and heat conduction Equation of state for a quantum-mechanical system Electromagnetic fields E(r, t), B(r, t) in free space

It is also amusing to note that certain equations, such as the wave equation, is second order in time, so that they are invariant under time reversal, that is, the transformation t → −t. A movie of a wave propagating to the left run backwards looks just like a wave propagating to the right. In diffusion or heat conduction, the field equation (for the concentration or temperature field) is only first order in time. The equation does not, and should not, satisfy time-reversal invariance, since heat is known to flow from a high-temperature region to a low-temperature region, never the other way around. A movie of a pool of water solidifying into a block of ice on a hot day has obviously been run backwards. The Maxwell equations have already been introduced in Sections 1.9 and 4.13. They are all first-order differential equations. The two divergence equations depend on time only parametrically. They state that there are electric charges but not magnetic monopoles. The two curl equations are coupled equations with couplings provided by the time derivative ∂t = ∂/∂t terms. On eliminating the couplings between E and B, one finds that the transverse part of each field satisfies a second-order wave equation, as we have already shown in section 4.13. These equations describe electromagnetic waves. The following sections are concerned with the solution of PDEs.

5.10

Separation of variables and eigenfunction expansions

Under certain circumstances the solution of a PDE may be written as a sum of terms, each of which is a product of functions of one of the variables. This is called a solution by a separation of variables. The method is conveniently described by an example.

352

Differential equations in physics

Let us consider first the 1D wave equation describing for example the transverse vibrations of a string: 1 ∂2 ∂2 u(x, t) = u(x, t). ∂x2 c2 ∂t2

(5.93)

It can be verified by direct substitution that the general form of the solution is u(x, t) = f (x − ct) + g(x + ct),

(5.94)

where the two linearly independent terms on the right-hand side (RHS) represent waves propagating along the +x and −x direction, respectively. The method of separation of variables does not require such insights into the nature of the solution right from the beginning. One simply looks for solutions with the separable form u(x, t) = X(x)T (t).

(5.95)

A direct substitution of Eq. (5.95) into the wave equation gives ∂2 d 2 X(x) 1 d2 T (t) u(x, t) = T (t) = X(t) . ∂x2 c2 dx2 dt2 Thus 1 d2 T (t) 1 d 2 X(x) = = λ, X(x) dt2 c2 T (t) dt2

(5.96)

where the left-hand side (LHS) is a function of x only, while the RHS is a function of t only. If the original PDE is to be satisfied, these two sides must be equal. This is possible only if each side is equal to a function neither of x nor of t. That is, each side must be equal to a constant, say λ. We have as a result two separated ordinary DEs: d2 X(x) dx2 d2 T (t) dt

2

= λX(x),

(5.97a)

= c2 λT (t).

(5.97b)

The separated ordinary DEs are not completely independent each other, however, because the same separation constant λ must appear in both. The solution X(x) of Eq. (5.97a) has the general form: √ √ √ X(x) = A cos(x −λ) + B sin(x −λ)/ −λ, (5.98) while the solution for T (t) is rather similar and contains the constant c2 λ instead of λ. As a rule, all possible values of the separation constant λ are allowed unless explicitly forbidden. Certain values of λ could be forbidden when the corresponding solution X(x), which depends on λ as shown in Eq. (5.98), does not have the right

Separation of variables and eigenfunction expansions

353

properties. The properties in question are the boundary conditions that select one or more solutions from the infinitely many possibilities contained in the general solution. It can happen that one or more of these boundary conditions can be satisfied only when the separation constant takes on one of a set of special values. This set then contains the only permissible values, or eigenvalues, for the problem. The corresponding solutions are called their eigenfunctions. The following example illustrates this point. Suppose, as in the case of the Fourier series of chapter 4, we are interested in solutions with a period of L = 2π, that is, the solutions 1, cos nx, and sin nx, where n is a positive integer. Then the only permissible separation constants are λn = −n2 ,

n = 0, 1, 2, . . . .

(5.99)

For λ = λn , we have the wave function Xn (x)T n (t) = an cos nx cos cnt + bn sin nx cos cnt + cn cos nx sin cnt + dn sin nx sin cnt.

(5.100)

Since the 1D wave equation (5.93) is linear, the general solution periodic in x with period 2π is the linear superposition ∞  1 u(x, t) = a0 + Xn (x)T n (t) 2 n=1

(5.101)

of all the possible solutions. The double series defined by Eqs. (5.100) and (5.101) is called a double Fourier series. In a similar way, the general solution of the 3D wave equation that is periodic in r with a period of 2π is the quadruple Fourier series ∞  1 Xl (x)Ym (y)Zn (z)T lmn (t). u(r, t) = a0 + 2 l,m,n=1

(5.102)

The derivation of this expression is left as an exercise. Now that we have obtained a general solution (5.101) of the 1D wave equation (5.93), it would be nice to extract from it the physical insight contained in Eq. (5.94), namely that there is a wave traveling along e x and a wave traveling along −e x . This is readily done by rewriting Eq. (5.100) in the form Xn (x)T n (t) = An cos n(x − ct) + Bn sin n(x − ct) + Cn cos n(x + ct) + Dn sin n(x + ct),

(5.103)

where the coefficients An , Bn , Cn , and Dn can be related to those in Eq. (5.100). Thus we have gained the physical insight contained in Eq. (5.94) by the method

354

Differential equations in physics

of separation of variables when the situation might not have been obvious at the beginning. To recapitulate, we note that in the method of separation of variables we look for a solution that is made up of a sum of products of functions of single variables. One or more of the boundary or other conditions can be absorbed into the general solution, which now involves eigenfunctions of one or more separated, ordinary DEs. For this reason, the method is also referred to as an eigenfunction expansion. Problems 5.10.1 Derive the general solution shown in Eq. (5.102) of the 3D wave equation whose solutions are periodic in r with a period of 2π. 5.10.2 Obtain the separated equations of the Schr¨odinger wave equation (Table 5.1) in rectangular coordinates when the potential function V(r) is 1 1 V(r) = mω2 r2 = mω2 (x2 + y2 + z2 ). 2 2

5.11

Boundary and initial conditions

The complete determination of a solution of a PDE requires the specification of a suitable set of boundary and initial conditions. The boundaries may not be just points, but, depending on the dimension, they can be lines or surfaces. Just what constitutes a suitable set of boundary conditions is a rather complicated question for PDEs. The answer depends on the nature of the PDE, the nature of the boundaries, and the nature of the boundary conditions. It is not our intention to describe this complicated situation. Rather, we would like to illustrate how boundary conditions can be imposed on functions of more than one variable. We shall restrict ourselves to the wave equation solved in rectangular coordinates. 5.11.1

Vibrations of a string

Let us consider first the 1D vibrations of a string rigidly attached to a support at the points x = 0 and L. They are described by the 1D wave equation (5.93) or the separated ordinary DEs (5.97). The boundary conditions at x = 0 and L are satisfied if  nπ  Xn (x) = sin x , (5.104) L that is, if λ = λn = −

 nπ 2 L

,

n = 1, 2, . . . ,

(5.105)

are used for the separation constant. The eigenfunction (5.104) of Eq. (5.97a) belonging to the eigenvalue λn is said to describe the nth normal mode (or eigenmode) of the vibration of the string. Fig. 5.1 shows the first three normal modes. We note that there are points

Boundary and initial conditions

355

Fig. 5.1 The first three normal modes of a vibrating string with fixed ends.

xm =

mL , n

m = 1, . . . , n − 1,

at which the displacement is always zero; that is, Xn (xm ) = sin(mπ) = 0. These are called nodal points. The time factor T n (t) associated with Xn (x) is the solution of Eq. (5.97b) for the same separation constant λn of Eq. (5.105). Since λn < 0, we can see with the help of Eq. (5.98) that T n (t) = Cn cos ωn t + Dn sin ωn t,

(5.106)

where ωn = nπc/L is the frequency of vibration of the nth normal mode. Hence the wave function of a vibrating string fixed at x = 0 and L has the general eigenfunction expansion u(x, t) =

∞ 

Xn (x)T n (t)

n=1

=

∞  n=1

sin

 nπ  x (Cn cos ωn t + Dn sin ωn t). L

(5.107)

If we pluck a string at time t = 0, which normal modes will be excited? How strongly? The answer depends of course on how we pluck the string, that is, on the initial conditions at t = 0. Since the differential equation (5.97b) is second order in time, we need two initial conditions—the displacement u0 (x) and the velocity v0 (x) of the string at t = 0. Thus u(x, t = 0) = u0 (x) =

∞  n=1

sin

 nπ  x Cn . L

356

Differential equations in physics b

0

a (1, 2)

(2, 1)

(2, 3)

Fig. 5.2 A few normal modes of vibration of a rectangular drum.

This is a Fourier sine series for which the Fourier coefficients Cn can be shown to be  2 L  nπ  sin (5.108) x u0 (x)dx. Cn = L 0 L Similarly  ∞  nπ   ∂ Dn ωn sin u(x, t) = v0 (x) = x , ∂t L t=0 n=1 so that 2 Dn = ωn L



L 0

 nπ  sin x v0 (x)dx. L

(5.109)

These linear coefficients describe the excitation strength of various normal modes. 5.11.2

Vibrations of a rectangular drum

The vibrations of a 2D membrane fixed at the boundaries x = 0, x = a, y = 0, and y = b can be described in a similar manner. The resulting eigenfunction expansion is u(x, y, t) =

∞  m,n=1

sin

 mπ   nπ  x sin y (C mn cos ωmn t + Dmn sin ωmn t), a b

(5.110)

where  ωmn =

m2 n2 + a2 b2

1/2 πc

(5.111)

is the frequency of the (m, n) normal mode. Fig. 5.2 shows a few examples of the normal modes of vibration of a rectangular drum. The broken lines mark nodal lines where the displacement is zero at all times. The strength with which various normal modes are excited depends on the initial conditions  ∂ (5.112) u(x, y, t) = v0 (x, y). u(x, y, t = 0) = u0 (x, y), t=0 ∂t

Boundary and initial conditions

357

From these functions the coefficients of the eigenfunction expansion (5.110) can readily be calculated: ,  ,  a  b  mπ   nπ   4 Cmn u0 (x, y) . (5.113) dx dy sin = x sin y Dmn v0 (x, y) ab 0 a b 0 It should be pointed out, however, that, while the method of eigenfunction expansion is completely general, it tends to be rather cumbersome. In simple situations it might be possible to solve a problem without using it at all. An example of this is provided by the following problem. Example 5.11.1 Obtain the solution of the 1D wave equation satisfying the following boundary conditions: u(x, 0) = u0 (x) ∂ u(x, 0) = 0. ∂t According to D’Alembert, the general form of the 1D wave function is Eq. (5.94) u(x, t) = f (x − ct) + g(x + ct). At t = 0 u(x, 0) = u0 (x) = f (x) + g(x) ∂ ∂t u(x, 0)

= 0 = −c f  (x) + cg (x).

(5.114a)

That is, g (x) =

d g(x) = f  (x). dx

This means that g(x) = f (x) + C,

(5.114b)

where C is an integration constant. Eqs. (5.114) have the solution 1 1 f (x) = u0 (x) − C 2 2 1 1 g(x) = u0 (x) + C 2 2 1 1 1 1 ∴ u(x, t) = u0 (x − ct) − C + u0 (x + ct) + C 2 2 2 2 1 1 = u0 (x − ct) + u0 (x + ct).  2 2

(5.115)

358

Differential equations in physics

h L

0 x=a

Fig. 5.3 Initial displacement of a stretched string for Problem 5.11.3.

Problems 5.11.1 An infinite string has an initial displacement u0 (x) = exp(−x2 ) and zero initial velocity. Obtain the wave function u(x, t) for all times. 5.11.2 An infinite string has an initial displacement u0 (x) and an initial velocity v0 (x). Use D’Alembert’s method to show that the wave function at all times is 1 u(x, t) = [u0 (x − ct) + u0 (x + ct)] 2 1 + [−w0 (x − ct) + w0 (x + ct)], 2c  s v0 (x )dx . w0 (s) = 0

5.11.3 A stretched string of length L is fixed at each end. Suppose it has the initial displacement shown in Fig. 5.3 ⎧ "x# ⎪ ⎪ x < a, ⎨ h"a , # u0 (x) = ⎪ ⎪ L−x ⎩h x ≥ a, L−a , and zero initial velocity. Show that its motion is described by the function  ∞  nπx nπa nπc 2hL2  1 sin sin cos t u(x, t) = 2 2 L L L π a(L − a) n=1 n at subsequent times. 5.11.4 (a) A 1D rod of length a originally at temperature T 0 has its ends placed in contact with a heat reservoir at temperature T = 0. Show that its subsequent temperature is      2 lπ 4 sin T (x, t) = T0 x e−(lπ/a) κt , lπ a l=odd integer where κ is the thermal conductivity.

Separation of variables for the Laplacian

(i)

(ii)

(iii)

359

(iv)

Fig. 5.4 Nodal lines of four vibrations of the square drum described by the spatial eigenfunctions u(x, y) of Problem 5.11.5(c).

(b) A cube of side a originally at temperature T 0 is suddenly placed in a heat reservoir at temperature T = 0. Show that its subsequent temperature is  64 lπ mπ nπ T (x, t) = T 0 sin x sin y sin z 3 a a a lmnπ odd l,m,n × e−(l

2 +m2 +n2 )(π/a)2 κt

.

(c) What would be the answers if the temperature of the heat reservoir had been T 1 ? 5.11.5 Consider the vibrations of a square drum of side a. (a) Show that the (m, n) and (n, m) modes are degenerate in frequency; that is, ωmn = ωnm . (b) Obtain the eigenfunctions umn (x, y) of the (m, n) mode. (c) Show that the nodal line of the hybrid mode described by each of the following spatial functions is as sketched in Fig. 5.4. √ (i) u(x, y) = [u12 (x, y) + u21 (x, y)]/√ 2, (ii) u(x, y) = [u12 (x, y) − u21 (x, y)] 2, √ (iii) u(x, y) = [u12 (x, y) + 2u21 (x, y)]/ √5, (iv) u(x, y) = [u12 (x, y) − 2u21 (x, y)]/ 5.

5.12

Separation of variables for the Laplacian

After separating out the time dependence, many of the PDE of Table 5.1 can be reduced to the form (∇2 + k2 )u(r) = 0,

(5.116)

where the constant k2 is zero for the Laplace equation, positive for the Helmholtz equation, and negative for the diffusion equation. One exception is the Schr¨odinger equation (Table 5.1), where an additional term −v(r)φ(r) appears with v(r) = − proportional to a potential function V(r).

2m V(r), 2

(5.117)

360

Differential equations in physics

The Laplacian ∇2 , which appears in Eq. (5.116), is the simplest scalar differential operator of r that is invariant under rotation and parity transformations. It has simple forms when expressed in terms of the following common coordinate systems: ∂2 ∂2 ∂2 + + , rectangular coordinates, ∂x2 ∂y2 ∂z2   ∂2 ∂ 1 ∂2 1 ∂ cylindrical coordinates, ρ + 2 2 + 2, = ρ ∂ρ ∂ρ ρ ∂φ ∂z     ! ∂ ∂ 1 ∂2 ∂ 2∂ 1 r + sin θ + sin θ , = 2 ∂r ∂r ∂θ ∂θ sin θ ∂φ2 r sin θ

∇2 =

spherical coordinates.

(5.118)

The Laplacian ∇2 in rectangular coordinates is just a sum of the separate partial differential operators. Hence the development of Section 5.10 can be used to show that the solution of Eq. (5.116) is separable in rectangular coordinates. Even the Schr¨odinger equation is separable in rectangular coordinates if v(r) = v1 (x) + v2 (y) + v3 (z)

(5.119)

can be written as a sum of functions of x, y, or z individually. This is the case for a harmonic-oscillator potential for which v(r) is proportional to r2 = x2 + y2 + z2 . To separate Eq. (5.116) in cylindrical coordinates, we assume the separable form u(r) = u(ρ, φ, z) = R(ρ)Φ(φ)Z(z)

(5.120)

for the solution u. The z part of the PDE then separates out cleanly as d2 Z(z) = α2 Z(z) dz2 with the help of the separation constant α2 , leaving the PDE !   1 ∂ d 1 ∂2 2 2 ρ + 2 2 + k + α R(ρ)Φ(φ) = 0. ρ ∂ρ dρ ρ ∂φ This may be written as   ! 1 d d 1 d2 2 2 2 ρ ρ + ρ (k + α ) R(ρ) = − Φ(φ) = m2 . R(ρ) dρ dρ Φ(φ) dφ2

(5.121a)

(5.122)

(5.123)

The separation of R(ρ) from Φ(φ) requires that both sides of Eq. (5.123) be equal to the same constant. In this way, another separation constant m2 is introduced, leading to the separated ordinary DEs:

Separation of variables for the Laplacian

d2 Φ(φ) = −m2 Φ(φ), dφ2   d2 d ρ2 2 + ρ + (β2 ρ2 − m2 ) R(ρ) = 0, β 2 = k 2 + α2 . dρ dρ

361

(5.121b) (5.121c)

The separation in spherical coordinates proceeds in a very similar way when the separable form u(r) = u(r, θ, φ) = R(r)Θ(θ)Φ(φ)

(5.124)

is substituted into Eq. (5.116). This calculation is left as an exercise. The results are the three separated ordinary DEs: d2 Φ(φ) = −m2 Φ(φ), dφ2 ! + c Θ(θ) = 0,

  1 d d m2 sin θ − sin θ dθ dθ sin2 θ !   c 1 d 2d 2 r + k − 2 R(r) = 0, dr r2 dr r

(5.125a) (5.125b) (5.125c)

where m2 and c are the separation constants. The separated ordinary DE for Φ(φ) of the azimuthal angle φ turns out to be the same in both cylindrical and spherical coordinates:  2  d 2 + m Φ(φ) = 0. dφ2 This is the DE for harmonic oscillations. We note that, although the angle φ can increase monotonically, the point r in reality always returns to the same position after one or more complete turns of 2π. Since φ(r) is supposed to be a single-valued function of r, the angular function Φ(φ) must have a period of 2π. That is, Φ(φ + 2π) = Φ(φ).

(5.126)

This periodicity condition is satisfied if Φ(φ) = Φm (φ) = am cos mφ + bm sin mφ,

(5.127)

where m is an integer. In other words, Eq. (5.127) gives the two eigenfunctions of the eigenvalue equation (5.125a) belonging to the same eigenvalue m2 . Eq. (5.125b) is also an interesting differential eigenvalue equation. If we eliminate θ in favor of x = cos θ in this equation, we find   m2 d 2 d + c X(x) = 0. (1 − x ) − dx dx 1 − x2

362

Differential equations in physics

That is,

!  d2 d m2 X(x) = 0, (1 − x ) 2 − 2x + c − dx dx 1 − x2 2

|x| ≤ 1.

(5.128)

This equation is called an associated Legendre equation, which contains the Legendre equation (5.83) as the special case of m = O. Eq. (5.128) can be solved by the Frobenius method. The infinite Frobenius series turns out to be convergent only for |x| < 1. The solution at |x| = 1 turns out to be infinite as a rule, except when c = cl = l(l + 1), l = 0, 1, 2, . . .. At these eigenvalues the Frobenius series terminates (x). as a polynomial, the associated Legendre polynomial Pm l The points |x| = | cos θ| = 1 are, of course, the singular points of the DE (5.128). The reason its solutions tend to be ill behaved at these points is that they are the north and south poles of the sphere. Since every meridian passes through both poles, the longitude (or azimuthal angle φ) at the poles can have any value (−∞ ≤ φ ≤ ∞). This is, of course, a purely artificial result since the poles can be chosen to be elsewhere. Physical fields on the sphere must be bounded everywhere. This is possible only if we use the eigenfunction Pm (x) of Eq. (5.128) belonging to the eigenvalue c = cl , l l being a non-negative integer. Eq. (5.128), (5.121c), and (5.125c) are among the most common nonelementary differential equations we need in mathematical physics. They are so important that it is worthwhile to rewrite them in a standard form: !  2 d m2 2 d (1 − x ) 2 − 2x + l(l + 1) − Pm (x) = 0, x = cos θ; (5.129) dx dx 1 − x2 l   2 d 2 d 2 2 z = βρ; (5.130) z 2 + z + (z − m ) Jm (z) = 0, dz dz   2 d 2 d 2 z = kr. (5.131) z 2 + 2z + [z − l(l + 1)] jl (z) = 0, dz dz Eq. (5.130) is called a (cylindrical) Bessel equation, while Eq. (5.131) is a spherical Bessel equation. The corresponding solutions Jm (z) and jl (z) are the (cylindrical) Bessel function of order m and the spherical Bessel function of order l. The function Jm (z) is that solution of Eq. (5.130) that behaves like (z/2)m for small values of z, while  π 1/2 Jl+1/2 (z). jl (z) = 2z (See Problem 5.12.1.) These Bessel functions are among the most useful of the socalled higher transcendental functions. (A transcendental function is one that is not algebraic. An algebraic function of x is one that can be generated from x by a finite number of algebraic operations +, −, ×, ÷. For example, a polynomial is algebraic, but log x is not. The elementary transcendental functions are the exponential, logarithmic, circular and hyperbolic functions.)

Separation of variables for the Laplacian

363

The linearly independent second solution of Eq. (5.129) is denoted by the symbol (x), and is called an associated Legendre function of the second kind. Similarly, Qm l there is the Bessel function of the second kind Nm (z) and the spherical Bessel function of the second kind nl (z). The most general form of the solution of a linear PDE is, according to Eq. (5.13), a superposition of all permissible solutions. In the case of cylindrical coordinates this superposition is  u(r) = ui (ρ, φ, z), (5.132) i

where i denotes collectively the two separation constants α and m in Eqs. (5.121). There are three ordinary DEs in Eqs. (5.121), each with two linearly independent solutions. Hence each ui (ρ, φ, z) is a sum of eight terms: ui (ρ, φ, z) = uαm (ρ, φ, z) = [a1 Jm (βρ) + a2 Nm (βρ)](b1 cos mφ + b2 sin mφ) × [c1 Z1 (z, α) + c2 Z2 (z, α)], where we have made use of the fact that Eq. (5.121c) is a Bessel equation. Similarly we have in spherical coordinates  u(r) = ui (r, θ, φ),

(5.133)

(5.134)

i

where ui (r, θ, φ) = ulm (r, θ, φ) m = [a1 jl (kr) + a2 nl (kr)][b1 Pm l (cos θ) + b2 Ql (cos θ)]

× (c1 cos mφ + c2 sin mφ).

(5.135)

Here we have used the result that Eq. (5.125c) is a spherical Bessel equation. Since we are not yet familiar with some of the higher transcendental functions appearing in Eqs. (5.133) and (5.135), we shall not use these results until Chapter 7 except for one very special case. This involves the Laplace equation ∇2 u(r) = 0, for which the separated radial equation (5.125c) simplifies because k = 0. With c = l(l + 1) we can readily verify that the two linearly independent solutions are rl and r−l−1 . The, function Qm l is known to be singular at cos θ = ±1, that is, at the north and south poles of the sphere. Hence physically interesting solutions defined everywhere on the sphere do not contain Qm l . Thus ui (r, θ, φ) of Eq. (5.135) simplifies to ui (r, θ, φ) = (a1 rl + a2 r−l−1 )(c1 cos mφ + c2 sin mφ)Pm l (cos θ).

364

Differential equations in physics

The simplest of such solutions are those with axial symmetry, that is, solutions independent of the aximuthal angle φ. This occurs if m = 0 so that cos mφ = are just the Legendre polynomials of Eq. (5.90). 1, sin mφ = 0. The functions Pm=0 l Thus we have shown that axially symmetric solutions of the Laplace equation can be written as  u(r) = (a1 rl + a2 r−l−1 )Pl (cos θ). (5.136) l=0

As an example, we note that spherically symmetric solutions are independent of θ and therefore must involve the l = 0 terms only. Furthermore, if the solution is also finite as r → ∞, then a1 = 0 must be satisfied. Thus by successive eliminations we are left with only one function r−1 . This is just the Coulomb potential due to a point charge at the origin. We must postpone applications of these eigenfunction expansions to Chapter 7 because such applications require the calculation of the linear coefficients ai . This cannot be done without some knowledge of the orthogonality properties of these special functions. We also note that PDE (5.116) is known to be separable in the eight other coordinate systems shown in Table 1.2 in addition to the three common ones shown in Eq. (5.118). Discussions of these less common coordinate systems can be found in Morse and Feshbach (Chapter 5) and in Arfken (Second Edition, Chapter 2). Problems 5.12.1 Starting from Eqs. (5.130) and (5.131) show that 1 jl (z) = const × √ Jl+1/2 (z). z 5.12.2 Verify that the ordinary differential equations shown in Eq. (5.125) are obtained on separating the Helmholtz equation (5.116) in spherical coordinates.

5.13

Green functions for partial differential equations

Green functions for partial DEs are more complicated than those for ordinary DEs. The main reason is that the boundary conditions necessary and sufficient to define a unique solution are much more complicated. For example, they depend also on the nature of the DE itself. However, in many cases of physical interest, simple expressions can be obtained readily without a detailed knowledge of the theory of PDEs, as the following examples show. We shall first consider the Green functions for the Poisson equation in two and three dimensions: ∇21G(r1 , r2 ) = δ(r1 − r2 ).

(5.137)

Green functions for partial differential equations

365

The 3D case is perhaps more familiar. The 3D δ function can be expressed simply in rectangular coordinates as δ(r1 − r2 ) = δ(x1 − x2 )δ(y1 − y2 )δ(z1 − z2 ), or more generally in terms of the integral  δ(r1 − r2 )d3 r1 = 1

(or 0)

(5.138)

if the integration region includes (excludes) r2 . Thus r2 is a very special point of the equation. If the DE is defined over the entire 3D space, nothing is changed by putting r2 at the origin. Then the solution must be spherically symmetric in r = r 1 − r2 ; that is, G(r1 , r2 ) must be a function of r only Suppose we integrate the DE over the volume of a sphere of radius r. We obtain  ∇ · ∇G(r)d3 r = 1. This may be written, with the help of the Gauss theorem, as    d G(r) 4πr2 , 1 = ∇G(r) · dσ = dr

(5.139)

where 4πr2 is the area of the sphere. This gives (d/dr)G(r), which can next be integrated directly to give   r  r 1 d G(r) = dr G(r) dr = dr 4πr2 =−

1 + C, 4πr

(5.140)

where C is an integration constant. This shows that it is possible to make G vanish as r approaches infinity by choosing C = 0. Equation (5.140) appears in physics as the familiar Coulomb field surrounding a unit point charge at r2 . We should note that, while the Coulomb field is spherically symmetric about the source at r2 , it is not spherically symmetric in a coordinate system in which the source is not located at the origin: G(|r1 − r2 |) = −

1 . 4π|r1 − r2 |

(5.141)

366

Differential equations in physics

The problem in two dimensions differs from the above only because the 2D δ-function relation  (5.142) δ(r1 − r2 )d2 r1 = 1 involves a 2D integration. Equation (5.139) now reads   d G(r) 2πr, 1= dr where 2πr is the circumference of the circle around r2 (which is also located at the origin). As a result  r 1 1 G(r) = dr = ln r + C. (5.143) 2πr 2π This shows that a 2D G(r) cannot be finite as r approaches infinity. We now know what a function should look like near a point source where its Laplacian is a δ function. This information can be used to obtain the Green functions of many second-order PDEs. Consider, for example, the Helmholtz equation (∇2 + k2 )Gk (r) = δ(r).

(5.144)

If r  0, Gk (r) satisfies the homogeneous equation (∇2 + k2 )Gk (r) = 0. Hence Gk (r) must be that solution that behaves at the origin like (−4πr)−1 in three dimensions, or (2π)−1 ln r in two dimensions, as k → 0. Let us work out the 3D case. Since G k (r) depends only on r, we may “separate” the Laplacian in spherical coordinates to yield (for r  0) the spherical Bessel equation (5.131) with l = 0:   d2 d r2 2 + 2r + (kr)2 j0 (kr) = 0. (5.145) dr dr Although this equation does not look too familiar, the equation for u(r) = rG(R) can easily be shown by direct substitution into Eq. (5.145) to be  2  d 2 + k u(r) = 0. dr2 Thus the two linearly independent solutions for G(r) are (−4πr)−1 cos kr or (−4πr)−1 sin kr. The second solution, called a regular solution, is well behaved at the origin and does not give rise to a δ function when operated on by ∇2 . It is the irregular solution,

Green functions for partial differential equations

G(R) = (−4πr)−1 cos kr

367

(5.146a)

that is the Green function for the Helmholtz equation. This Green function can be written in two other alternative forms G±k (r) = −

1 ±ikr e , 4πr

(5.146b)

where the factors exp(±ikr) describe spherical waves since the wave fronts of constant phase (kr = const) are spheres. The direction of the wave motion cannot be determined in the absence of a time dependence. It is customary to add the time factor of exp(−iωt), so that exp(ikr) describes an outgoing spherical wave, while exp(−ikr) describes an ingoing spherical wave. An equivalent convention leading to the same result is obtained by examining the eigenvalue of the propagation operator K = −i∇, which is also the infinitesimal generator of space translation, as we have discussed in Section 2.9. Thus Ke±ikr = ±ker e±ikr .

(5.147)

The wave exp(ikr) thus propagates outward. It is therefore called an outgoing spherical wave. The solution (−4πr)−1 cos kr thus contains an equal linear combination of an outgoing and an ingoing spherical wave. These two waves in opposite directions will interfere to form a standing-wave pattern in the same way two waves running in opposite directions in a violin string do. Hence (−4πr)−1 cos kr is called a standingwave Green function. To change the boundary conditions, we can add the homogeneous solution ±i sin kr/(−4πr) to get back the spherical-wave Green functions. These Green functions are said to satisfy the outgoing or ingoing spherical-wave boundary conditions. Problems 5.13.1 Show that the 3D δ function can be written explicitly in spherical coordinates as δ(r1 − r2 ) =

1 δ(r1 − r2 )δ(cos θ1 − cos θ2 )δ(φ1 − φ2 ). r12

5.13.2 Express the 3D δ function explicity in an orthogonal curvilinear coordinate system using the notation of Section 1.10. 5.13.3 Construct a 1D function f (x) whose Laplacian (d2 /dx2 ) f (x) is the δ function δ(x). 5.13.4 Verify that the outgoing-wave Green function for the 2D Helmholtz equation is −(i/4)H0(1) (kr), where H0(1) is the Hankel function of the first kind

368

Differential equations in physics

and of order 0. It satisfies the Bessel equation (5.130) for m = 0, behaves like (2i/π) ln kr near r = 0, and like (2/πkr)1/2 ei(kr−π/4) as r → ∞.

Appendix 5 Tables of mathematical formulas 1 Linearity property If L is a linear operator in x and L y1 (x) = 0,

L y2 (x) = 0,

L yp (x) = R(x),

then L [y p (x) + c1 y1 (x) + c2 y2 (x)] = R(x)

2 Linear differential equations The first-order linear differential equations   d + p(x) yh (x) = 0, dx   d + p(x) yp (x) = R(x) dx have solutions

  yh (x) = yh (a) exp −

x

 p(x )dx , 



a

and yp (x) =

 x  a

 , R(x ) dx yh (x). yh (x )

The second-order linear differential equation   2 d d L y1 (x) = + P(x) + Q(x) yi (x) = 0, dx dx2

i = 1, 2

has two linearly independent solutions y1 (x) and y2 (x) with a nonzero Wronskian   y1 (x) y2 (x)    W(x) =  y (x) y (x)  1

2

   x   P(x )dx . = W(a) exp − a

Tables of mathematical formulas

369

If y1 (x) is known, the differential equations L y2 (x) = 0, L yp (x) = R(x)

have the solutions



x

y2 (x) = b

,   t $   2 P(t )dt y1 (t)dt y1 (x), exp − b

and yp (x) = v1 (x)y1 (x) + v2 (x)y2 (x), where

 v1 (x) = − a

x



y2 (t)R(t) dt, W(t)

x

v2 (x) = a

y1 (t)R(t) dt. W(t)

The homogeneous differential equation L y(x) = 0

often has one or more solutions of the form y(x) =

∞ 

aλ xλ+s ,

λ=0

The coefficients aλ can be obtained by requiring that the coefficients bμ of the power series  L y(x) = bμ xμ+s μ

are identially zero, that is, bμ = 0. 3 Green functions If G(x, x ) is a solution of the differential equation L G(x, x ) = δ(x − x ),

then the solution of the differential equation L y(x) = R(x)

is



b

y(x) = a

G(x, x )R(x )dx .

370

Differential equations in physics

This solution y(x) satisfies the same boundary conditions as G(x, x ). If L is an nth-order linear differential operator, the kth derivative G(k) (x, x ) =

dk dxk

G(x, x )

for k ≤ n is smooth everywhere except at x = x , where G(n−2) has a kink, G(n−1) has a jump discontinuity, and G(n) has a δ-function discontinuity. 4 Separation of variables and eigenfunction expansion In certain coordinate systems, a linear partial differential equation might be separable into a number of ordinary differential equations of the general form L f (x) = λ f (x).

The separation constant λ appears in more than one equation, thus permitting the relationship between different coordinates to be maintained. Each ordinary differential equation must satisfy certain boundary or regularity conditions. As a result, only a selection or spectrum of separation constants is allowed. These are called eigenvalues, while the corresponding solutions f (x) are their eigenfunctions. Under these circumstances, the linear partial differential equation has solutions in the form of a product of eigenfunctions, one for each separated coordinate. The most general solution is therefore a linear combination of such products, the linear coefficients being determined by fitting to suitable boundary and initial conditions in the manner of a Fourier series. 4(a) 1D wave equation ∂2 1 ∂2 u(x, t) = u(x, t): ∂x2 c2 ∂t2  2  d − λn Xn (x) = 0 dx2   2 d 2 − c λn T n (t) = 0 dt2 & & & Xn (x) = An cos(x −λn ) + Bn sin(x −λn )/ −λn & & & T n (t) = Cn cos(ct −λn ) + Dn sin(ct −λn )/ −λn  u(x, t) = Xn (x)T n (t). n=1

Tables of mathematical formulas

371

This eigenfunction expansion can be written symbolically as u(x, t) =



 An

n

, , √ √ cos(x cos(ct −λ ) −λ ) n n √ √ √ √ . sin(x −λn )/ −λn sin(ct −λn )/ −λn

 Initial conditions: If at t = 0, u(x, 0) = u0 (x), (d/dt)u(x, t)

t=0

= v0 (x):

1 u(x, t) = [u0 (x − ct) + u0 (x + ct)] 2 1 + [−w0 (x − ct) + w0 (x + ct)], 2c s w0 (s) = (x )dx 0

4(b) Helmholtz equation (∇2 + k2 )u(r) = 0:  , , √ √  cos(x cos(y −λ ) −λ ) m n √ √ √ √ Amn u(x, y, z) = sin(x −λm )/ −λm sin(y −λn )/ −λn m,n & ⎧ ⎫ ⎪ ⎪ ⎨ cos(z λm + λn&+ k2 ) ⎬ & ×⎪ ⎩ sin(z λ + λ + k2 )/ λ + λ + k2 ⎪ ⎭ m n m n & ⎫ ⎧ , , √  ⎪ ⎬ cos mφ cos(z ⎨ Jm ( &k2 + λn ρ) ⎪ −λ√ n) √ u(ρ, θ, z) = Amn ⎪ ⎭ sin mφ ⎩ N ( k2 + λ ρ) ⎪ sin(z −λn )/ −λn m n m,n , m , ,   Pl (cos θ) cos mφ jl (kr) u(r, θ, φ) = . Alm Qm sin mφ nl (kr) l (cos θ) l,m

The special case k = 0 gives the solutions to the Laplace equation. We note the simplification 

,  l , jl (kr) r → −l−1 . nl (kr) r

372

Differential equations in physics

5 Green functions for partial differential equations 5(a) Laplace equation 1 ln |r1 − r2 | 2π 1 =− 4π|r1 − r2 |

G(r1 − r2 ) =

in two dimensions in three dimensions.

5(b) Helmholtz equation The outgoing-wave Green functions are i G(r1 − r2 ) = − H0(1) (k|r1 − r2 |) 4 =−

eik|r1 −r2 | 4π|r1 − r2 |

in two dimensions,

in three dimensions.

6 Nonlinear systems∗ 6.1

Introduction

Nonlinear behaviors in physics show two strangely contradictory faces. On the one hand, nonlinearity causes irregularities, instabilities, turbulences and chaos that are often induced by small changes in certain parameters. The proverbial last straw that breaks the camel’s back is a good example of the sudden catastrophes that often appear in nonlinear systems. Apparent unpredictability too is a hallmark of many unstable nonlinear systems. The complex twisting dances of dust devils, those mini hurricanes that appear and disappear unpredictably in the desert, might even remind us of the events at the height of the French Revolution. Yet nonlinearity can also bring unexpected stability, regularity and coherence. A school of fish, a flock of birds and a stampede of wildebeests are examples of nonlinear behavior characterized by persistent patterns. Such unusual coherence is quite common in nonlinear physical and mathematical systems too. The great Sumatra earthquake of 2004 generated tsunamis that caused serious damage and loss of lives in many countries on the rim of the Indian ocean thousands of miles away. Tsunamis are persistent shock waves on the ocean surface that can carry the energy of destruction far from their source. In this chapter we shall show that both chaotic and persistent behaviors arise from nonlinearity. In unstable systems, nonlinearity ensures that they do not run like a clock—regular, predictable, commonplace, and linear. It gives a dynamical system a multiplicity of possible outcomes instead of unique predictability. For the discrete time evolution or mapping of a dynamical property from one time stage or generation to another, it induces bifurcations when circumstances are suitably changed. The process may look simple at first, but repeated bifurcations of even simple dynamical systems can lead to a very complicated quasi-random behavior that has been aptly described as deterministic chaos. Using a hand-held calculator in 1975, Feigenbaum discovered that the ratio of the intervals between successive bifurcations tends to a universal constant 4.6692. . . , now called the Feigenbaum constant. He then showed that the same constant appears in many different dynamical systems. The unique value of a physical attribute is a point of dimension 0 in the 1D space of possible values. The infinitely many possible solutions of the deterministic chaotic state form a space of fractional dimension between 0 and 1. Such objects of fractional dimensions are called fractals. Fractals are particularly interesting when embedded in multi-dimensional spaces. In the 2D H´enon map, it gives rise to a 2D trapping region to which the system is attracted, but inside which only certain paths are realized.

374

Nonlinear systems

This is the strange attractor inside which the mapping is stable in one direction but unstable in another direction. The possible paths inside the strange attractor form a fractal of dimension between 1 and 2. Dynamical systems such as the driven dissipative nonlinear pendula satisfy differential equations instead of finite difference mapping equations. In these differential equations, the unknown spatial variable changes continuously in time. Their chaotic behavior is then harder to discern. It can still be isolated, however, by examining the system only periodically, say once per driving period. In between two successive strobe views, the system evolves in a complicated manner according to the nonlinear differential equation, but the result is still a nonlinear map. One then finds qualitatively the same bifurcation behavior and quantitatively the same Feigenbaum constant as simpler nonlinear maps. In contrast, a very different nonlinear effect appears in certain partial differential equations (PDEs) in both space and time coordinates. The unexpected feature here is a persistent pattern that travels. Such a nonlinear wave can appear in a time-evolution equation where the time derivative is of first order, or in a hyperbolic-wave equation where the time derivative is of second order. In both cases, there is a distinctive wave form f (z = kx), say in one spatial dimension x at time t = 0, in the shape of a localized pulse or step-like wave front that is the solution of a nonlinear ordinary differential equation (ODE). It becomes a nonlinear traveling wave after a velocity boost to z = kx − ωt as the ODE is transformed into a PDE. Such localized wave forms differ from the simple linear wave forms f (z) = e±iz that are eigenfunctions of the linear differential operator d/dz. The latter are spread out all over space with the same absolute value 1 everywhere. The localization of a nonlinear wave form is achieved with the help of an additional nonlinear term that serves the same purpose as a confining potential. This type of mathematical system has found interesting applications in many diverse fields including solitons (solitary waves) in water canals, tidal bores and tsunamis, solitons in field theories of elementary particles, and solitons in optical fibers. In this chapter we shall explain how these two very different manifestations of nonlinearity can be given simple mathematical descriptions and explanations in spite of their unexpectedness, unfamiliarity and complexity.

6.2

Nonlinear instabilities

Systems far away from states of equilibrium are intrinsically nonlinear. Recent studies of nonlinear properties of physical systems have emphasized the concepts of catastrophe, bifurcation, chaos, and strange attractors. This section gives a brief introduction to these interesting ideas, and is based in part on a review by Martens (Martens 1985). 6.2.1

Catastrophe and hysteresis

Catastrophe refers to a sudden change in the response of a system to a small change in the value of a parameter. This behavior is exhibited by the first-order nonlinear differential equation (NLDE)

Nonlinear instabilities

375

p xc B

x˙ > 0 C x

D

x˙ < 0

Stable

A

Unstable

Fig. 6.1 The equilibrium or static solution of Eq. (6.1).

x˙ =

d x(t) = bx − x3 + p dt

(6.1)

called the Landau equation (and used by him in 1944 to study the stability of fluid flow). For a given b > 0, the system is in equilibrium (with zero velocity, x˙ = 0) along the cubic curve shown in Fig. 6.1: p = xe3 − bxe .

(6.2)

The equilibrium curve by its nature separates the px diagram into two regions, each with a definite sign for the velocity x˙. When p  bx > 0, x˙ = p − x3 ; the negative velocity region therefore appears to the right of xe , as shown by the direction of arrows in Fig. 6.1. Starting from a value xe  0 on the right of the curve, we see that the equilibrium or static √ solution is stable down to the local minimum A at (xA , pA ) = √ ( b/3, −2b/3) b/3. It is unstable from A to the local maximum B at (−x A , −pA ), as indicated by the broken curve in the figure. Below B, the equilibrium is stable again. The broken part of the stability curve is thus an instability curve. Suppose b is fixed and p is decreased slowly from a value p  0. The system will follow the equilibrium curve until the point A is reached. When p is further decreased, the system will jump to the point D on the left half of the stability curve. It then continues on to more negative values of p as p decreases further. If p is now increased from this p  0 value, the system does not trace its original path exactly. If goes all the way up to the local maximum B instead. After reaching B, it jumps over to the point C on the other stability curve before continuing upwards. The jumps A → D and B → C, where the system changes abruptly in response to quite small changes in p, are called catastrophes, while the history-dependent behavior described above is referred to as hysteresis.

376

Nonlinear systems x √ b

b

√ – b

Stable

Unstable

Fig. 6.2 Pitchfork bifurcation of a stable equilibrium curve into two stable branches as b increases through the value b = 0 at the fork of the bifurcation.

Given values of b and p and an arbitrary initial position x(0), the system will move to the stable equilibrium point nearest to it. Such a stable equilibrium point is called a point attractor, because it “attracts” the system for a variety of initial conditions. Another rather familiar example of a point attractor is an oscillating spring that comes to rest as a result of dissipation. 6.2.2

Bifurcations

√ At p = 0, Fig. 6.1 shows two stable solutions, at xe = ± b. These solutions approach each other as b → 0, collapsing into a common solution at b = 0. For negative values of b, the differential equation x˙ = bx − x3 = b(x − xe ) − (x3 − xe3 )

(6.3)

has only one static solution. Since x˙ < 0 if x > xe , this unique solution is stable. This situation is described graphically by the xb diagram of Fig. 6.2. The stable and unstable equilibrium curves separate regions in which the velocities have the same sign. The branching of the static solutions caused by a change of parameter is called a pitchfork bifurcation. Successive bifurcations play a crucial role in turbulence and chaos. The first-order NLDE (6.1) is particularly interesting in two dimensions (x, y). The structure of the DE remains simple if the complex representation z = x + iy = Reiθ is used: d z = (μ + iω0 )z − |z|2 z + Feiωt . dt

(6.4)

The complex constant b = μ + iω0 contains two real parameters μ and ω0 that control different properties of the system. Replacing the term p by a time-dependent forcing

Nonlinear instabilities

377

term of driving frequency ω and strength F greatly increases the complexity and usefulness of this NLDE. In the absence of the forcing and cubic damping terms, the DE is linear. It is a DE with constant coefficients. It therefore has the exponential solution z(t) = Re(μ+iω0 )t if initially z(0) = R is in the x direction. Actually we can always use the initial position to define the x-axis. When μ = 0, the solution behaves like the tip of a simple clock hand of constant length R rotating anticlockwise with constant angular speed ω0 . When μ < 0, the length R of the clock hand shrinks exponentially with time as it rotates. It shrinks to zero at large times. Hence R = 0 is a point attractor when μ < 0. That is, the system ends there for all initial positions. When μ > 0, the amplitude R grows exponentially with time. Seen in the full xy-plane, the motion is circular if μ = 0. It is a contracting spiral if μ < 0, and an expanding spiral if μ > 0. The situation is more interesting when the cubic damping term is also present. The general solution has the form z(t) = R(t)ei Ω t ,

(6.5)

where R(t) also depends on the time t. It happens frequently that a solution reaches a steady state where R has a nonzero terminal value that does not change further in time. These steady-state solutions can be found directly by looking for a solution z(t) = ReiΩt , where R is independent of time. The substitution of the assumed steadystate solution into Eq. (6.4) allows the common time dependence in every term to be factored out. One is then left with an algebraic equation that can be rearranged to read R2 = μ + i(ω0 − Ω).

(6.6)

We are now in a position to find the steady-state solution for R. √ Since R is real and non-negative by definition, Eq. (6.6) is solved if Ω = ω0 , R = μ, provided that μ > 0. The result z(t) =

√

μeiω0 t

(6.7)

describes a steady state where the exponential growth of R induced √by μ > 0 is exactly canceled out by the cubic damping. If μ < 0, however, R = μ becomes purely imaginary. Since R has to be real, the contradiction means that the solution (6.7) does not really exist. Given any μ > 0, the unforced NLDE can easily be solved numerically √ in a computer. Two solutions are shown in Fig. 6.3, with initial amplitudes R < μ and √ the same asymptotic steady R > μ, respectively. In both cases, the system reaches √ state (6.7), with the amplitude R stabilized at R∞ = μ. The resulting stable circular orbit of radius R∞ is called a limit cycle. It is an attractor because the system ends there for all initial positions. The growth of √the point attractor at μ = 0 into a stable circular attractor with increasing radius μ after the damping cubic term is turned on is called a Hopf bifurcation. The Hopf bifurcation provides a mechanism for stable time keeping in

378

Nonlinear systems y

1

0.5

–1

–0.5

0.5

1.5

2

x

–0.5

–1

√ Fig. 6.3 Hopf bifurcation with the system reaching the same steady-state limit cycle of radius R∞ = μ at large times. Here μ = 1 is used with R(t = 0) = 2.0 for the contracting spiral outside the limit cycle, and R = 0.4 for the expanding spiral inside the limit cycle.

neural and genetic clocks. It can also be used to turn off the clock as well by allowing the control parameter μ to become zero or negative. 6.2.3

Hopf resonator

The Hopf bifurcation is particularly versatile in the presence of a forcing term Feiωt , where F is real and nonnegative. The system becomes a resonator because its response R (the radius of its limit cycle) under forcing is maximal at a frequency that turns out to be just ω0 . The Hopf resonator is special among resonators because at the bifurcation point μ = 0, the maximum gain or amplification G max = ω0

Rmax F

(6.8)

is very large for small forcing strengths F. We shall show that the actual relationship is Gmax = ω0 F −2/3 .

(6.9)

As a result, the Hopf resonator is an extremely sensitive detector when operating at or near the bifurcation point μ = 0. The forced Hopf resonator owes its detection sensitivity to the cubic damping term −|z|2 z = −R2 z. In the steady state, R has become independent of time. So is R2 . The NLDE then has the simpler time dependence of a linear DE with constant

Nonlinear instabilities

379

coefficients. We know from experience that such a linear DE has a particular solution of the form z(t) = (Reiφ )eiωt , with a time dependence matching that of the forcing term. The NLDE (6.4) then simplifies to the algebraic equation F = Reiφ [(R2 − μ) + i(ω − ω0 )]. With F, R, φ, μ, ω0 , and ω all real, we must have % F = R (R2 − μ)2 + (ω − ω0 )2 , tan φ = −

ω − ω0 . R2 − μ

(6.10)

(6.11) (6.12)

The appearance of an extra phase φ when ω  ω0 is common to both nonlinear and linear resonators in the steady state. At the bifurcation point μ = 0, Eq. (6.11) can be written in the dimensionless form % ˜ ˜ F = R R˜ 4 + (ω ˜ − 1)2 , (6.13) 1/2 ˜ where the dimensionless variables are ω ˜ = ω/ω0 , F˜ = F/ω3/2 0 , R = R/ω0 . Hence the dimensionless gain is

1 R˜ . G˜ = = & ˜ F R˜ 4 + (ω ˜ − 1)2

(6.14)

At resonance (ω ˜ = 1), R˜ has the maximal value (as a function of ω) ˜ of R˜ max = F˜ 1/3 . −2 −2/3 , as asserted previously. So the gain there has the maximal value G˜ max = R˜ max = F˜ 2 ˜ Far from the resonance (meaning |ω ˜ − 1|  R ), the gain G˜ ≈ 1/|ω ˜ − 1| is inde˜ ˜ pendent of F. So the sensitivity to F develops only near ω ˜ = 1. The behavior of the Hopf resonance near resonance (ω ˜ ≈ 1) at constant R˜ is shown ˜ in Fig. 6.4(a) for three differing values of R. An increase in gain begins to develop as R˜ falls below 0.3. The resonance structure rapidly becomes narrower and taller as R˜ decreases further. ˜ as shown in The Hopf resonance is equally dramatic at constant forcing F, ˜ Fig. 6.4(b). The resonance becomes narrower and taller as F decreases below 10−3 . Mammalian hearing is so sensitive because the cochlear of the inner ear, or more specifically the hair cells in it, are Hopf resonators (Eguilux et al. 2004). Acuity in hearing confers a significant advantage to both hunters and preys in the survival of the fittest in Darwinian evolution. Proximity to the resonance is more quantitatively characterized by its FWHM (full width at half maximum) Γ˜ = 2|ω ˜ 1/2 − 1| in the resonance curve at constant forcing. Here ω ˜ 1/2 is the frequency where the response R˜ has fallen to R˜ max /2, half of its peak value. Γ˜ can readily be shown (Problem 6.2.6) to be √ √ 3 7 7 ˜ 2/3 3 2 Γ˜ = (6.15) R˜ max = F . 2 2

380 (a)

Nonlinear systems ~ G 104

(b)

~ R 1 10–1

103

10–2 102 10–3 10

10–4 10–5

1

0.9

0.95

1

1.05

1.1

~ ω

0.9

0.95

1

1.05

1.1

~ ω

˜ F˜ as a function of the Fig. 6.4 Two views of a Hopf resonance: (1) The dimensionless gain G˜ = R/ dimensionless frequency ω ˜ for R˜ = 0.02 (heavy solid curve), R˜ = 0.15 (light solid curve) and R˜ = 0.5 (dashed curve). (b) The dimensionless limit cycle radius R˜ as a function of the dimensionless frequency ω ˜ for F˜ = 10−6 (heavy solid curve), F˜ = 10−4 (light solid curve) and F˜ = 10−2 (dashed curve).

So when the forcing F˜ is weak, the resonance width is small, but the gain is large. When the forcing is strong, the resonance width is large, but the gain is small. These features are illustrated in Fig. 6.4(b).

Problems 6.2.1 (Saddle-node bifurcation) Consider the NLDE x˙ = a + x2 . (a) Show that there is no equilibrium solution if a > 0, one solution if a = 0, and two solutions if a < 0. (b) Draw its bifurcation diagram in the ax-plane, using a solid curve for stable equilibrium and a dashed curve for unstable equilibrium. Use arrows to display the sign of x˙ in the regions bounded by equilibrium curves. 6.2.2 (Subcritical pitchfork bifurcation) Consider the NLDE x˙(t) = bx + x3 . (a) Show that the null solution x(t) = 0 describes unstable equilibrium for b > 0, and stable equilibrium for b < 0. (b) Show that the remaining equilibrium curve is given by the bifurcated √ equilibrium solutions xe = ± −b = real on the parabola x2 = −b > 0 enclosing the negative b-axis. (c) Sketch the bifurcation diagram to show the basin of attraction to the stable equilibrium line that is the negative b-axis. Use arrows to indicate the direction of x˙ in each region between two equilibrium curves. Note: The bifurcation obtained here is called a subcritical bifurcation because bifurcated solutions appear when the parameter b falls below the

Nonlinear instabilities

381

critical value bc = 0. In contrast, the bifurcations shown in Fig. 6.2 are supercritical because they appear when b > bc = 0. 6.2.3 (Transformation to LDE) (a) Show that the nonlinear transformation x(t) → z(t) = 1/x2 (t) transforms the NLDE x˙(t) = bx − cx3 to the LDE z˙(t) + 2bz = 2c. (b) Verify that the exact solution for the initial condition x(t = 0) = x0 is ⎧ 1 ⎪ ⎪ + 2ct, if b = 0, ⎪ ⎪ ⎨ x02  −2  z(t) = x = ⎪ ⎪ c 1 c −2bt ⎪ ⎪ ⎩ + 2 − e , if b  0. b

x0

b

(c) For c = 1, verify that the solutions are attracted to the appropriate pitchfork parts for stable equilibrium shown in Fig. 6.2. 6.2.4 Show that the Hopf resonator equation (6.4) can be rewritten in the following 2D forms: (a) In circular coordinates z(t) = R(t)eiθ(t) and with F = 0, it is d R˙ ≡ R = μR − R3 , dt θ˙ = ω0 , where we have used Newton’s overdot notation for the more familiar time derivative d/dt first used by Leibniz. (b) In rectangular coordinates z(t) = x(t) + iy(t) and with F  0, it is x˙ = μx − ω0 y − (x2 + y2 )x + F cos ω0 t, y˙ = μy + ω0 x − (x2 + y2 )y + F sin ω0 t, 6.2.5 Let the cubic term −|z|2 z in Eq. (6.4) be replaced by −a|z|2 z, where a is a real parameter. Show that the resulting equation can be rescaled into one that looks exactly like the original Eq. (6.4). 6.2.6 (Resonance widths) (a) At constant forcing, Eq. (6.13) is satisfied at all ω ˜ values for the same ˜ ˜ including ω forcing F, ˜ = 1, 1 ± Γ/2. Show that as a result, Γ˜ satisfies the algebraic equation  6  2  2 Γ˜ 1 1 + = 1. 2 ˜ 2 2 2Rmax Verify that the width appearing in this expression is given by Eq. (6.15).

382

Nonlinear systems

˜ the gain resonance shown in Fig. 6.4(a) appears (b) At constant response R, because the forcing needed to produce R˜ is minimal at resonance (ω ˜ = 1). Show that the gain resonance has a FWHM of √ √ 2/3 Γ˜ g = 2 3R˜ 2 = 2 3F˜ min .

6.3

Logistic map and chaos

The general availability of electronic calculators and computers by the early 1970s led to many quantitative advances in our understanding of nonlinear (NL) systems. Among the surprising discoveries was the realization that even simple deterministic systems could become chaotic. The disorder of these chaotic states develops suddenly, in much the same way as chaos in more complicated dynamical systems. 6.3.1

Logistic map

Particularly interesting results were obtained by May, Feigenbaum and others by studying insect population dynamics. Many adult insects do not overlap from generation to generation, each generation lasting several months or years. Fertile female insects in the last generation laid a large number of eggs before the entire population died at the end of its natural cycle. At the beginning of the new generation, eggs hatch into infants. The number rNn of hatchlings is proportional to the population Nn of the last (nth) season. If these hatchlings could grow unchecked, the insect population will soon explode exponentially in time after many generations, as predicted by Malthus in his famous treatise, An Essay on the Principle of Population, which ran through six editions between 1798 and 1826. In real life, not every hatchling will grow into an adult. Food is always limited. The environment, too, is not always insect-friendly. The total mortality is likely to be a complicated function of the preceding population Nn , and almost certainly not a linear function. It could be a quadratic function proportional to Nn2 , describing in the simplest term the Darwinian fact of life that proportionally more insects will perish on the way to adulthood if the community is overpopulated. Then the adult population this season will be rNn − mNn2 = rNn [1 − (m/r)Nn ]. Here r (for birth rate) and m (for mortality) are positive parameters. By using xn = (m/r)Nn , one finds the standard expression xn+1 = rxn (1 − xn )

(6.16)

for the scaled population. This is the logistic or population map—a quadratic, iterative map of the population from one generation to the next. It is an example of 1D nonlinear maps or difference equations. The logistic map gives a simple model of population dynamics that contains a driving mechanism (the birth rate r) in competition with a dissipative mechanism (the mortality m). We shall see that the insect population evolves from season to season in a totally deterministic way that nevertheless shows an unexpectedly rich structure as

Logistic map and chaos

383

1

0.8

y

0.6

0.4

0.2 xe3 0

0.2

0.6

0.4

xe2 xe4 0.8

1

x

Fig. 6.5 The logistic map y(x) = rx(1 − x) for r = 3.1 (solid curve) and the bisector y = x (dashed line). The graphical construction of x0 = 0.1 → x1 = 0.279 → x2 = 0.623592 is shown by the rising steps. The fixed point xe2 = 1 − 1/r = 0.677419 is given by the intersection of the map y(x) with the bisector y = x. The period 2 equilibrium solutions xe3 = 0.558014 and xe4 = 0.764567 define the boundary of the closed square that describes the map xe3 → xe4 → xe3 .

the delicate balance between these two competing mechanisms is changed by varying the birth rate r for a population variable x scaled by the mortality m. For 0 ≤ x ≤ 1, the mapping function f (x) = rx(1 − x)

(6.17)

is positive for 0 < x < 1. It has a maximum of r/4 at x = 1/2, as shown in Fig. 6.5. By taking 0 ≤ r ≤ 4, the population of the next generation will also fall within the same bounds 0 ≤ xn ≤ 1. The quantity x is then a proper population variable expressed as a fraction of the maximum possible population of Nmax = r/m of the model. Nmax = 1 is realized only for r = 4 at x = 1/2. Fig. 6.5 can be used to find the population evolution in successive seasons. If x0 is the population last season, its value the next season, x1 = f (x0 ), is given by the intersection of the vertical line at x = x0 with the map f (x). The graphical solution can be iterated either by bringing the f value to the x-axis or equivalently by drawing the horizontal line y = x1 and follow it until it intersects the bisector y = x at x = x1 . At the intersection x = x1 , a vertical line can be drawn to find the intersection x2 = f (x1 ) with the map, etc. Graphical mapping is a useful conceptual tool. The adult population xe is in equilibrium when its value this generation is the same as that last generation. Such a solution, where f (x) = x itself, is called

384

Nonlinear systems

a fixed point of the function. It satisfies the quadratic equation obtainable from Eq. (6.16) rxe2 + (1 − r)xe = 0.

(6.18)

This algebraic equation has two solutions or fixed points: the trivial solution xe1 = 0, and the non-trivial solution 1 xe2 = 1 − . r

(6.19)

Since the population x cannot be negative, we require that r ≥ 1 as well. Hence our population model is one with 1 ≤ r ≤ 4. 6.3.2

Instability: period-doubling bifurcation cascade

The stability of an equilibrium population xe can be determined by examining a Taylor expansion of the mapping function near it: xn+1 = rxn (1 − xn ) ≈ rxe (1 − xe ) + (xn − xe ) f  (xe ),

(6.20)

where f  (x) =

df = r(1 − 2x). dx

(6.21)

(The prime symbol  is Lagrange’s notation for Leibniz’s d/dx.) At xe1 = 0, f  (xe1 ) = r. Any xn near this trivial xe1 will map to rxn which diverges from it more and more on iterations for all r > 1. So this equilibrium point is unstable. It is called a repellor. The second equilibrium solution xe2 is more interesting. Here f  (xe2 ) = 2 − r,

(6.22)

with the help of Eq. (6.19). Any xn near xe2 approaches it on iterations if |2 − r| < 1, i.e., 1 < r < r1 = 3. The equilibrium is then stable, and xe2 is an attractor. If r > r1 , the equilibrium is unstable. xe2 becomes a repellor instead. The instability that develops for r just above r1 = 3 is a period-doubling bifurcation (PDB) like the pitchfork bifurcation shown in Fig. 6.2. The system then oscillates between two distinct populations in successive generations, with each population reappearing after a period of two generations. Such period 2 solutions x p can be found directly from the period 2 iterative relation: xn+2 = rxn+1 (1 − xn+1 ) = r[rxn (1 − xn )][1 − rxn (1 − xn )].

(6.23)

That is, they satisfy the quartic equation x p − r[rx p (1 − x p )][1 − rx p (1 − x p )] = 0.

(6.24)

Logistic map and chaos

385

Table 6.1 Regions of different periodicities N for the logistic map. (PDB = period doubling bifurcation, F. = Feigenbaum.)

Region

N

Comment

r1 ≤ r ≤ r2 ≈ 3.45 r2 ≤ r ≤ r3 ≈ 3.54 r∞ = 3.569944 . . .

2 4 ∞

r = 3.6786 . . . r = 3.8284 . . . r=4

3

First PDB Second PDB F. accumulation point of the PDB cascade: chaotic region begins First odd cycle First 3-cycle Almost total chaos

In Eq. (6.24), the known equilibrium solutions xe1 = 0 and xe2 = 1 − 1/r can be factored out. We are left with a quadratic equation. By using the new variable y = r(1 − x p ), the quadratic equation can be simplified to y2 + (1 − r)y + 1 = 0. Its two solutions are ye3,e4

1 = (r − 1) ± 2

%

(6.25) !

(r −

1)2

−4 .

(6.26)

If r < 3, the discriminant (r − 1)2 − 4 is negative, giving a pair of complex roots. This means that there are no real, or actual, solutions of period 2 distinct from the known equilibrium solutions of period 1. For r > r1 = 3, however, the discriminant is positive. Eq. (6.26) then gives two real solutions ye3,e4 that are not period-1 solutions. They can be shown to be bi-stable near r = 3, i.e., stable with respect to a period 2 disturbances. (See Problem 6.3.5.) The old solutions xe1,e2 = 0, 1 − 1/r (or ye1,e2 = r, 1) are still present, but they have become unstable, as illustrated in Fig. 6.2. In fact, the PDB arises from the destabilization of the previous stable solution xe2 = 1 − 1/r by the increasing birth rate r. As a result, the population oscillates between two values ⎤ ⎡ 4 1 ⎢⎢⎢⎢ 4 ⎥⎥⎥⎥ 2 (6.27) xe3,e4 = 1 − ⎢⎣ xe2 ± xe2 − 2 ⎥⎦ , 2 r with xe3 < xe2 and xe4 > xe2 . These “straddling” inequalities can easily be verified numerically. See Problem 6.3.1 for an analytic proof. The calculation of period 2 solutions can be repeated for any integer period N. The result is partially summarized in Table 6.1. Fig. 6.6 shows schematically the branching structure of a period-doubling cascade in the bifurcation diagram for the logistic map, a plot of all the stable populations as a function of the birth rate r.

386

Nonlinear systems 1.0

x 0.5 2n Chaos Periods

1 stable fixed point 0.0 1

2

3 b

3.45 3.57 4 3.54

Fig. 6.6 Schematic representation of the branching structure of a period-doubling cascade that ends in chaos in the bifurcation diagram for the logistic map.

The complete bifurcation diagram for the logistic map is given in Fig. 6.7. We see that the first PDB begins when r exceeds r1 = 3. As r increases further, additional PDBs appear thicker and faster in such a way that the bifurcation interval Δrn = rn − rn−1

(6.28)

decreases rapidly with increasing n. Feigenbaum has found that the ratio δn =

Δrn Δrn+1

(6.29)

very rapidly approaches the limiting value called the Feigenbaum number or constant δ = lim δn = 4.669 201 609 102 990 9... n→∞

(6.30)

The first few bifurcation points are given in Table 6.2. The PD cascade is also called a Feigenbaum sequence or scenario. The Feigenbaum constant δ describes an interesting geometrical property as the number of iterations n → ∞. The PD tree seen in Fig. 6.7 near an (n + 1)-th bifurcation point when examined under a magnification δn looks exactly the same as that seen near an nth bifurcation point. The reappearance of the same structure at a smaller scale is called self similarity. Moreover, the successive reduction with increasing n of the bifurcation interval by the same factor 1/δ causes the cascade of PDBs to approach an accumulation point at r∞ . This is the Feigenbaum point, whose value is given in Table 6.1. 6.3.3

Fractals and universality

At the Feigenbaum point r∞ , the vertical population line 0 ≤ x ≤ 1 in Fig. 6.7 contains infinitely many isolated re points congregating only in certain neighborhoods. They

Logistic map and chaos

387

1.0

0.8

0.6

x 0.4

0.2

0.0 2.4

2.6

2.8

3.0

3.2

3.4

3.6

3.8

4.0

r Fig. 6.7 Bifurcation diagram for the logistic map, from Wikipedia (entry: Logistic map, 2012).

contain not only the stable equilibrium populations x s of period ∞, but also the unstable equilibrium populations of all preceding periods. As a result, the distribution of xe points has the same recurrent, iterative structure or self similarity at every magnification down to arbitrarily small distances. Such a collection of xe points on the population line is called a fractal, defined as a collection of fractured geometrical objects, here isolated points, having a self-similar structure at a constant iterative scale s. The fractured geometrical objects in a fractal do not cover all available space. Its spatial dimension can nevertheless be estimated as follows: Consider first a Table 6.2 Bifurcation points rn for Eq. (6.16) from Peitgen et al., 1992, p. 612. The numbers in the second column are given in the scientific notation a(n) = a × 10n .

n

rn

Δr n

δn

1 2 3 4 5 6 7

3 3.449489. . . 3.544090. . . 3.564407. . . 3.568759. . . 3.569692. . . 3.569891. . .

– 4.49489(–1) 9.4601 (–2) 2.0317 (–2) 4.3521 (–3) 9.3219 (–4) 1.9964 (–4)

– 4.7514

388

Nonlinear systems

solid line of unit length, a filled unit square, and a filled unit cube. When each side of the object is increased by an arbitrary scale factor s, the scaled object contains c = s, s2 , s3 copies, respectively, of the original unit object. That is, c satisfies the universal scaling law c = sD ,

D = log c/ log s.

or

(6.31)

Here D = 1, 2, 3 are the known dimensions of the line, square, cube, respectively. This scaling law can be used to define a scaling dimension of a fractal of broken geometrical objects where the linear scale change is not arbitrary but restricted to a single iterative scale s such as the Feigenbaum constant δ. For the PDB cascade of the logistic map, for example, c = 2 copies appear when the linear dimension is increased by δ. Hence its scaling dimension Dδ = log 2/ log δ ≈ 0.45. More sophisticated estimates give values between 0.49 and 0.54. The Feigenbaum constant δ appears not only in the quadratic map, but also in other NL 1-dimensional (1D) maps with a single maximum (where f  (xmax )  0). It also appears unexpectedly in many NL differential equations as well. The frequent appearance in many distinct NL systems of a PDB cascade terminating at an accumulation point attests to the universality of the phenomenon. 6.3.4

Chaos: instability and disorder

We are finally ready to discuss the concept of chaos. Its most important characteristic is a sensitivity to initial conditions. Any 1D map f (x) is exponentially expansive at x0 if two slightly different initial states near x0 will grow apart exponentially after many iterations. That is, Δn ≡ | fn (x0 + ) − fn (x0 )| = enλ ,

(6.32)

with λ > 0. Here  > 0 is the initial small difference in the states. More precisely, enλ = lim

→0

Δn Δ1 ... Δn−1 

= | f  (xn−1 )... f  (x0 )|.

(6.33)

One can then define the Lyapunov exponent as the average exponential growth constant per iteration 1 λi . n→∞ n i=0 n−1

λ = lim

(6.34)

Here λi = ln | f  (xi )| is the local Lyapunov exponent at xi . If λ > 0, the map at x0 is exponentially expansive, destabilizing or divergent. If λ < 0, the map is exponentially contractive, stabilizing or convergent. If λ = 0, the map is scale (or measure) preserving.

Logistic map and chaos

389

The simplest example of this scaling behavior can be seen in the linear scaling map f (x) = ax. Since f  (x) = a everywhere, its Lyapunov exponent λ = ln a is positive whenever a > 1. This simply means that the position x is scaled by the constant factor a > 1 everywhere, and therefore at every iteration. Such a behavior is predictable rather than chaotic. Hence the requirement λ > 0 of instability is necessary but not sufficient to cause chaotic motion. A general 1D map is more complicated only because the linear expansion factor f  (x) varies with x. The scale change at the i-th iteration is still given by the local scale  . The average of the local exponential growth constants λ in the regions sampled fi−1 i by the map as the number of iterations n → ∞ is just the Lyapunov exponent. Examined in this light, the logistic map shows a relatively simple sampling behavior when the birth rate r ≤ r∞ is in the range of PD bifurcations. In this parameter range, the map started from any initial state reaches the same 2m -cycle containing 2m distinct populations. Here m = 0, 1, 2, . . . The Lyapunov exponent then simplifies to just the average of local exponents in the 2m distinct populations of one cycle. For example, the 1-cycle appearing at 1 ≤ r ≤ 3 contains only the equilibrium population xe2 . Its local scale factor f  (xe2 ) = 2 − r as a function of r is a straight line joining the point f  = 1 at r = 1 to f  = −1 at r = 3. Hence its Lyapunov exponent λ(r) = ln | f  | is a smooth curve joining the point λ = 0 at r = 1 to λ = −∞ at r = 2. Beyond r = 2, the curve returns from ∞ as a mirror image in a “vertical” mirror placed at r = 2. The behaviors at other ranges of the PD cascade can be described similarly, but the description becomes complicated very rapidly. Finally at the Feigenbaum point, each cycle contains an infinitely long, non-periodic set of isolated populations. Its Lyapunov exponent has been estimated to be 0. Note that the 2m -cycle as a composite structure is an attractor because all mapping iterations end there. We therefore expect that in this range of birth rates, the Lyapunov exponent is non-positive and the system is non-chaotic. This expectation is confirmed by numerical calculations of λ. For r > r∞ , however, computer calculations find λ > 0 for most but not all values of r. Are these chaotic solutions? Chaos develops rapidly. For this we need sensitivity to initial conditions, or a positive λ. However, chaos specifically means disorder or the absence of regularity. A simple way to characterize disorder is to require that any point as it is mapped repeatedly will get arbitrarily close to any other point in the permissible interval. Such an idea of disorder can be quantified by determining a probability density P(x) of finding the system at x in any iteration when it is started anywhere. That is, P(x)dx is the probability of finding the system in an interval dx containing x. The result for r = 4 turns out to be the function P(x) =

1 . π x(1 − x) √

(6.35)

1 It is a bowl-shaped function properly normalized to 0 P(x)dx = 1. It is not zero for 0 ≤ x ≤ 1, and it is independent of the starting condition. It thus signifies

390

Nonlinear systems

quantitatively that the mapping is indeed disorderly. So the system is chaotic at r = 4. There are in addition periodic solutions at r = 4 as well, but these are not chaotic and therefore less interesting in the present context. The situation for r∞ < r < 4 is more complicated. First the mapping covers only a part of the x interval, namely 0 < xmin ≤ x ≤ xmax < 1. Then one or more reducedrange copies of the probability density (6.35) can appear that may be nested inside another P(x) structure. They may overlap instead, and they may also be cleanly separated from one another in ranges of x values. These finite P(x) subregions represent chaotic states of a more complicated structure. Fig. 6.7 shows that there are even intervals (“gaps”) of r where the mapping is not chaotic at all. The most prominent of these gaps is the one beginning at A ≈ 3.828 with the appearance of a 3-cycle. It then develops into a PDB cascade that terminates at the end of the gap. See these details clearly in the electronic figure in Wikipedia. To summarize, we have described a chaotic region that can be approached by following a period-doubling cascade. It is reached by crossing a threshold r∞ in the birth rate r that powers the insect population. This scenario is called the PD route to chaos. Returning to our insect model, the complexity of insect population dynamics might appear to be driven by the birth rate r alone, with the mortality m providing only a scale for the maximum permissible population r/m. In reality, the “dissipative” mortality term is indispensable by providing the counterbalancing mechanism that produces the many delicate interactions between birth and mortality underlying the very complicated NL phenomena only partially described here.

Problems 6.3.1 (Straddling) (a) Verify Eq. (6.25). (b) Show that for 3 < r < 4, √ xe3,e4 − xe2 = ∓

# √ r − 3 "√ r+1∓ r−3 . 2r

Hence show that xe3 < xe2 and xe4 > xe2 in Eq. (6.27). 6.3.2 (Feigenbaum ratio) (a) Use a calculator (like Feigenbaum) to complete the table of Feigenbaum ratios δn from the information given in Table 6.2. (b) Show that if the Feigenbaum ratio δn were exactly the Feigenbaum constant δ for all n, the accumulation point r∞ would have the value ≈ 3.572, a little bigger than the Feigenbaum constant 3.5699. . . 6.3.3 (Final-state time order for 4-cycles) The final state for any birth rate r is given by the bifurcation diagram of Fig. 6.7, where the first Feigenbaum sequence of period-doubling bifurcations between r1 = 3 and r∞ = 3.5699 . . . can be seen clearly. We are interested here in the time ordering of the population

Logistic map and chaos

391

evolution among the branches of a given r in the final state, starting (say) from the topmost branch. For the 2-cycle (r1 < r < r2 ), the two branches can be addressed as H (or xe4 ) and L (or xe3 ). Then the time order is obviously H → L → H . . . . For the 4-cycle, the time order can ber determined by using two general ordering rules for periodic final states: (i) The straddle rule requires that a H population maps onto a L population, and vice versa. (See Problem 6.3.1). (ii) The step-down rule requires that the highest available high maps onto the lowest available low. Derive these rules. Let the four stable populations xi , i = 8, 7, 6, 5 be addressed HH, HL, LH, LL in descending order along the f -axis. Show that the mapping will proceed in the time order HH → LL → HL → LH → HH. Note: The time ordering of the larger cycles is more complicated because the L to H mapping order depends on whether x > 1/2 or x < 1/2 due to the presence of the maximum of f (x) at x = 1/2. 6.3.4 (Numerical analysis of the 4-cycle in the logistic map) The 4-cycle at r = 3.5 contains the equilibrium populations xe =0.874997, 0.826941, 0.500884, 0.38282, in descending order. Use a calculator or PC to verify the following results: (a) The mapping order is that given in Problem 6.3.3: HH → LL → HL → LH → HH. (b) The Lyapunov exponent is λ = −0.873. In particular, show that the stability property of the two high populations HH, HL is different from that of the two low populations LH, LL. 6.3.5 (Bi-stability of period 2 solutions) (a) Show that near a period 2 solutions x p defined by Eq. (6.24) xn+2 ≈ x p + (4 + 2r − r2 )(xn − x p ).

(6.36)

Hint: Show that the needed Taylor coefficient is r2

d [r f (1 − f )] = (2y − r)(2y2 − 2ry + r), dx

(6.37)

where f (x) = rx(1 − x). Then use Eq. (6.25) to eliminate y2 twice to simplify to the final result. (b) Show that the period √ 2 solutions ye3,e4 are bistable for r1 < r < r2 , where r1 = 3 and r2 = 1 + 6. (Bi-stability means stable recurrence at period 2. That is, each 2-cycle is stable, but the population jumps up and down with the pattern H → L → H within each 2-cycle.) 6.3.6 (The symmetric quadratic map) Show that the logistic map (6.16), xn+1 = rxn (1 − xn ), x in [0, 1], can be changed into the symmetric quadratic map

392

Nonlinear systems

zn+1 = 1 − μz2n ,

z in [−w/2, w/2],

(6.38)

by the choice w=

4 , r−2

μ=

r . w

Note: The symmetric quadratic map is often called a logistic map too. Both maps can be used with parameters and variables outside the ranges used in the normalized insect population model discussed in the text.

6.4

Strange attractor

An attractor is so called because the system is attracted to it for a variety of initial conditions. Sensitivity to initial conditions is just the opposite. Two points on opposite sides of a point of unstable equilibrium will end up in very different places. Such a point of unstable equilibrium is called a repellor. A strange attractor is an attractor that is also sensitive to initial conditions. That is, it is also a repellor. How such contrary attributes could coexist can be illustrated by a simple example taken from Martens. A good labyrinth or trap is an attractor in that people who go in do not come out. It is also a repellor, since the people inside may not even see one another, although they all got in at the same gate. In other words, there is a trapping region with complicated instabilities inside. Strange attractors were discovered quite recently by E. Lorenz in 1963 in modeling turbulence in the atmosphere by coupled first-order NLDEs in 3D. The model was simplified into the 2D NL map T , T:

(xn+1 , yn+1 ) = (yn + 1 − axn2 , bxn ),

(6.39)

by H´enon in 1976. The model parameters a, b are usually taken to be positive. The trapping property is similar to that seen near a stable population in the 1D logistic map (6.16). Indeed the x part of the H´enon map obtained by eliminating the y variable is described by the second-order NL difference equation xn+1 = 1 − axn2 + bxn−1 .

(6.40)

The situation in 2D is more interesting. For the H´enon map T in particular, a vertical line segment (with x = const) in the xy plane is mapped onto a horizontal line segment with the original higher end on the right. A horizontal line segment −x0 ≤ x ≤ x0 , on the other hand, is mapped onto a left-opening parabola symmetric about the x-axis, as shown in Fig. 6.8(a) for the “standard” choice of parameters (H´enon, 1976) a = 1.4,

b = 0.3.

(6.41)

The parabola is folded down at the opening (cd separation) by a factor b, if b < 1, and is squashed into a horizontal line if b = 0. The value of b = 0.3 is chosen so that the

Strange attractor 0.8

0.6

B

A 0.6 D

C

A b c

0.2

0.2

y

y

0.4

c b

0.4

0

B

0 –0.2

–0.2

C d a

–0.4 (a)

(b) d a

–0.4 –1

393

–0.5

0

x

D

–0.6 0.5

1

1.5

2

–1.5

–1

–0.5

0

x

0.5

1

1.5

Fig. 6.8 H´enon maps for the standard choice of parameters a = 1.4, b = 0.3 (H´enon, 1976) after one iteration. (a) A higher horizontal line segment AB (−1 ≤ x ≤ 1 at y = 0.7) maps to the parabola ab (solid curve) on the right, while a lower horizontal line segment DC (−1 ≤ x ≤ 1 at y = 0.5) maps to the parabola dc (solid curve). (b) The trapping boundary of the quadrilateral ABCD maps to the boomerang abcd after one iteration.

folding is significant, but not too large to be followed easily in successive iterations. Such a middling choice of b bypasses a serious complication of the Lorenz equations where the effective b can be as small as 7 × 10−5 . The parabola has an x extension of ax02 . If a > 1, the original line is stretched longitudinally. When the original horizontal line is moved higher by Δy (here 0.2), the parabola is moved to the right by the same amount Δy with unchanged shape. As a result, the box ABCD of Fig. 6.8(a) is mapped into the boomerang figure on the next iteration. This “stretch and fold” process is similar to that used to make Chinese pulled noodle. More generally, one can show that any straight line of finite slope is mapped onto a parabola with its axis parallel to the y axis. The straight line separates the xy plane into two half planes that are mapped into the regions between and outside the “jaws” of the parabola, respectively. For the right choice of a, b parameters, the boomerang is entirely within a quadrilateral ABCD, shown in Fig. 6.8(b) for the standard parameters, with (clockwise from the upper left) A = (−1.33, 0.42), C = (1.245, −0.14),

B = (1.32, 0.133), D = (−1.06, −0.5).

(6.42)

See Problem 6.4.3 for more details. The quadrilateral ABCD from which the system cannot escape is called the H´enon trap. It is part of a larger region in the xy plane called the basin of attraction, made

394

Nonlinear systems

up of all points that will end up in the trap. What is interesting about the trap is that it contains a strange attractor. To appreciate the strangeness of the strange attractor, we need to study the stability of the H´enon map. For any choice of parameters, the map T has two equilibrium (fixed, invariant) points where (xn+1 , yn+1 ) = (xn , yn ) ≡ (xe , ye ), and ! % 1 xe1,e2 = −(1 − b) ± (1 − b)2 + 4a , 2a ye1,e2 = bxe1,e2 ,

(6.43)

where xe2 < xe1 . These points are real if the discriminant is positive, or 1 a > a0 = − (1 − b)2 . 4

(6.44)

The stability of these equilibrium points can be studied in the regions close to them by examining a linearized theory based on the mapping of the differential line segments     dxn dxn+1 =M , dyn+1 dyn ⎞ ⎛ ∂x ⎜⎜⎜ n+1 ∂xn+1 ⎟⎟⎟  −2axn 1  ∂yn ⎟ n ⎟⎠ = (6.45) M = ⎜⎜⎜⎝ ∂y∂xn+1 ∂yn+1 ⎟ b 0 ∂xn

∂yn

is the Jacobian matrix. M is only a function of xn . Its two eigenvalues at xe are % λ± = −axe ± (axe )2 + b. (6.46) An equilibrium point is stable only if both inequalities |λ± | < 1 are satisfied. (The linearized theory gives no information when |λ| = 1. A higher-order stability theory must then be used.) Fig. 6.9 shows both xe1,e2 and their eigenvalues for b = 0.3. We see that the equilibrium point e1 is stable (heavy solid curve) only between a0 = −0.1225 and a1 = 3|a0 |,

(6.47)

It is a point attractor there. The equilibrium point e2 is always unstable because its eigenvalue λ2+ > 1. The unstable part of the e1 curve (dotted curve in Fig. 6.9) is of great interest. Numerical studies (H´enon, 1976) show that for a < a0 or a > a3 , where a3 ≈ 1.55 when b = 0.3, the iterated solution always escapes to infinity. This means that there is no attractor. For a0 < a < a3 , the iterated solution either escapes to infinity or end at an attractor, depending on the starting point on the xy plane. The attractor is a point

Strange attractor

395

4 λ2+ 3 2

xe or λ

1

e1 λ1+ λ2-

0 –1

e2

–2

λ1-

–3

–0.25

0

0.25

0.5

0.75

1

1.25

1.5

a Fig. 6.9 Equilibrium points xe1,e2 of the H´enon map for b = 0.3 and the eigenvalues λ j± , j = 1, 2 of their Jacobian matrices. The stable part of xe1 is shown as a heavy solid curve. The dashed vertical line superposed on the y axis represents the values xe2 = ±∞. Its other values are shown as two light dashed curves. The possible values of xe1,e2 actually form a single closed curve when the parts outside the figure are also included. The eigenvalues with |λ| < 1 are shown as light solid curves. These eigenvalues are also joined together into a single curve.

attractor below a1 . Above a1 , the bifurcation diagram, giving x values of the finalstate solutions, is shown in Fig. 6.10. It begins as a period-doubling cascade (PDC) that ends at the accumulation/critical value a2 (≈ 1.058 for b = 0.3). This PDC is a Feigenbaum sequence described by the universal Feigenbaum constant δ. Up to a2 , an N-cycle is made up of N points, with N → ∞ as a → a2 . The dimension of these fractal sets remains less than 1, even though the points are distributed all over the xy plane. These stability properties are summarized in Table 6.3. Above a2 , some of the solutions shown in the bifurcation diagram are (stably or intermittently) periodic, like those in the 1D logistic map, but there are features unique to 2D maps. One is the appearance of small isolated islands (called “shrimps”) each made up of a PD cascade in miniature. (The shrimps and other details are not shown clearly in Fig. 6.10, but can be seen in Figs. 12.15 and 12.16 of Peitgens et al., 1992, pp. 675–6, and in the original figure in Wikipedia.) The most unusual feature appears when the motion becomes chaotic. Chaotic motion appears also in the 1D logistic map, but in 2D the chaos has another dimension to romp. The Jacobian eigenvector belonging to the “unstable” eigenvalue λ1− < −1 at each x point of the bifurcation diagram of a chaotic solution gives the direction along which the mapping is unstable and chaotic. This chaotic direction changes with

396

Nonlinear systems

1.5

0.5 x –0.5

–1.5 1.

1.1

1.2

1.3

1.4

1.5

a Fig. 6.10 Bifurcation diagram of the H´enon maps for 1.0 < a < 1.43 and b = 0.3, from Wikipedia (H´enon Map, 2012). The y values are not shown. They can be obtained from yn+1 = bxn if xn of the last iteration is known.

x as the mapping proceeds. Eventually, the mapped points form a series of orbits that are simple (i.e., dense or continuous) 1D curves in the 2D xy plane. Mathematicians call such simple curves 1D manifolds. These curves do not intersect, because the H´enon map is a one-to-one map. These curves are well separated from one another—the result of the fact that the system is stable in the direction of the eigenvector of the eigenvalue 0 < λ1+ < 1, but unstable in the direction of the eigenvector of the second eigenvalue |λ1− | > 1. This bundle of curves is our strange attractor. The strange attractor is strange because the bundle structure in the transverse direction is self-similar, and is repeated again and again at smaller scales. Its self Table 6.3 Some stability properties of the H´enon map for b = 0.3 as a function of the quadratic coefficient a (H´enon, 1976). Here a0 = −0.1225, a1 = 0.3675, a2 = 1.058, a3 ≈ 1.55, F. = Feigenbaum, PDC = period-doubling cascade. The H´enon parameter a = 1.4 is in the chaotic region a2 ≤ a ≤ a3 .

a a0 a1 a2 a3

Stability < a < a1 < a < a2 ≤ a ≤ a3 ≤a

xe1 is stable 1st PDC, a2 = 1st F. point Chaos appears frequently Solutions always escape to ∞

Strange attractor

397

0.4 0.20 0.2

0

y

y

0.18

0.16 –0.2 0.14 –0.4 –1.0

–0.5

0

0.5

1.0

1.5

0.50

x

0.55

0.60

0.65

0.70

0.75

x

0.1896 0.190 0.188

y

y

0.1894

0.1892

0.186 0.1890 0.184 0.1888 0.62 0.625 0.63 0.635 0.64 0.645

x

0.63 0.6305 0.631 0.6315 0.632 0.6325

x

Fig. 6.11 The self similarity of the H´enon strange attractor is illustrated by three successive enlargements of areal magnification 100. The first (upper left) plot of the full extent of the strange attractor shows that it is inside the boomerang-shaped area of Fig. 6.8(b). The small rectangle contains the equilibrium point xe1 = 0.631354 . . . , ye1 = 0.189406 . . . . Its area is 1% that of the plot containing it. The same points obtained in 106 iterations starting from (x, y) = (0, 0) are used, after omitting the first 300 points. Each succeeding plot contains fewer points. The last plot shows clearly the pointwise nature of the H´enon map.

similarity is illustrated in Fig. 6.11. A discrete set of infinitely many points or elements (in the transverse direction) that has a self-similar structure like a strange attractor is called a Cantor set. (The 1D logistic map at a Feigenbaum point also generates a Cantor set.) As H´enon succinctly summarizes it, a strange attractor “appears to be the product of a one-dimensional manifold by a Cantor set.” A strange attractor has also been described as a well-oiled pastry dough, made after infinitely

398

Nonlinear systems

many many stretchings and foldings. It is a bundle of Chinese pulled noodles after infinitely many pulls. A Cantor set is a fractal with a fractional dimension. The set at the Feigenbaum point has a scaling dimension of D = log 2/ log δ ≈ 0.45. The Feigenbaum constant δ there gives the magnification needed to scale the structure back to the previous size. The fractal dimension of the H´enon map is a function of the parameters defining it, such as the parameters a, b in T . The strange attractor of the map at a = 1.4, b = 0.3 has a dimension of about 1.26, depending on the definition of fractional dimensionality. There are many other parameter values where strange attractors also appear. Another interesting feature of the H´enon map is that it is in general not area preserving. That is, the differential area dxn dyn is changed to dxn+1 dyn+1 = |J|dxn dyn ,

(6.48)

where J = det M = −b is the Jacobian determinant, obtained from Eq. (1.123). The area changes by a factor |b| that is the same everywhere on the xy plane, increasing if |b| > 1, unchanged if |b| = 1, and decreasing if |b| < 1. In the last case, the area vanishes in the limit where the number of iterations n → ∞. The result is the bundle of curves that makes up a strange attractor. Let us summarize other interesting properties of these strange attractors without elaboration. If b = 0 is used in the H´enon map (6.39), the y variable simply disappears, leaving just the 1D symmetric quadratic map (6.38). It differs from the logistic map used to describe insect populations in that the x range is no longer restricted to a unit interval. The usual PDB cascade appears with the Feigenbaum ratios converging to the Feigenbaum constant δ. As b increases from 0, the area shrinkage per iteration is reduced. An area on the xy plane is mapped onto strange attractors with a fractional dimension that varies with the model parameters. The 1D map (6.40) obtained by eliminating the y variable shows a PDB cascade described by the universal Feigenbaum constant δ. When b = 1 is reached, however, the H´enon map becomes area-conserving or “conservative”. For this special case, the Feigenbaum ratios in the 1D PDB cascade converge to a larger δc = 8.7210... than the Feigenbaum constant and consequently a smaller scaling dimension. Studies of the Lyapunov exponent show no chaotic behavior in the PDB region of the x map. Indeed, the appearance of a PDB cascade in 1D means that the mapping there is non-chaotic.

Problems 6.4.1 (Baker’s map) The unit square in the xy-plane, in the 2D interval ([0, 1], [0, 1]), is mapped onto itself in the baker’s map ⎧ ⎪ if xn < 1/2 ⎪ ⎨(2xn , ayn ), (xn+1 , yn+1 ) = ⎪ ⎪ ⎩(2x − 1, ay + 1/2), if x > 1/2, n n n

Strange attractor

399

where 0 ≤ a ≤ 1/2. This map stretches the width of the square by a factor 2 and reduces its height by a factor a. It then cuts the result into two horizontal halves and stack them vertically into a square again, with the strip resting on the bottom of each vertical half squares and a gap of height a at the top of each half square. (a) If the original square is colored black, describe each of three successive mappings of a baker’s map with a = 3/8. (b) Show that for an arbitrary a value, the height of each black strip has the same value an for all strips in the nth map. Show that the gap heights are not all the same after the first mapping, and that the smallest gap height in the nth map is an−1 [(1/2) − a]. (c) Show that the scaling dimension of the baker’s map is D(a) = 1 + log 2/ log (1/a). Calculate the numerical values for a = (1/4, 3/8, 1/2). (The scaling dimension of a fractal is defined right after Eq. (6.31). Note: The black area shrinks with mappings when a < 1/2. Area-shrinking maps are said to be dissipative. 6.4.2 (One-to-one mapping) Show that the H´enon map (xn+1 , yn+1 ) = (yn + 1 − axn2 , bxn ) is one-to-one, and is therefore invertible. In contrast, the 1D quadratic map xn+1 = 1 − axn2 is in general two-to-one. Hint: The inverse H´enon map is (xn , yn ) = (b−1 yn+1 , xn+1 − 1 + a[b−1 yn+1 ]2 ). 6.4.3 (H´enon trap) (a) Show that in a H´enon map, the straight line y = c + sx in the xy plane is mapped onto an arc of the parabola K x − X = − (y − Y)2 , where 2   s2 sb (X, Y) = c + 1 + , 4a 2a is the vertex position and K = 2a/b2 is the curvature of the parabola at the vertex. (The curvature of a plane curve y(x) is K ≡ y /(1 + y2 )3/2 , where y = dy/dx.) Show that the parabola open to the left (right) if K > 0 (< 0), and that K is the same everywhere in the xy plane. (b) Show that the image of the boundary of quadrilateral ABCD of the trapping region of the strange attractor shown in Fig. 6.8 lies inside ABCD. Show that the interior of ABCD is mapped to the interior of its boomerang image. 6.4.4 (Stability of the H´enon map) (a) Verify Eqs. (6.43) and (6.44). (b) Verify Eq. (6.46). (c) The stable ↔ unstable transition point appears at λ = ±1. Show that this transition appears twice, at a = a0 and a1 . Verify Eq. (6.47).

400

Nonlinear systems

(d) Find the eigenvectors v± belonging to the eigenvalues λ± . Show that v+ · v− = const(1 − b). That is, they are orthogonal only when b = 1, the case of area-conserving maps.

6.5 6.5.1

Driven dissipative linear pendula Nonlinear complexity

We are familiar with the idea that a classical system disturbed slightly from a point of stable equilibrium experiences a linear restoring force that will move the system back to the equilibrium position. If the system is dissipative, its kinetic energy will be lost gradually to friction. It will eventually come to rest at the equilibrium position, which is often called an attractor because the system is attracted to it for a variety of initial conditions. Linear systems behave simply even when driven by an external force. The general solution of the resulting Newton’s equation is made up of a sum of a particular (i.e., any) solution and a complementary function that is a linear superposition of all the linearly independent solutions of the homogeneous differential equation (DE), as we have discussed in Section 5.5 for second-order LDEs. Solutions satisfying different initial conditions differ only in their complementary functions. As a result, the dynamics of linear systems is highly predictable in the sense that their behavior shows simple patterns that are intuitive and easy to understand. For large disturbances, however, the potential energy will no longer be quadratic in the displacement from equilibrium. The force is then no longer linear in the displacement. The resulting Newton’s equation of motion is said to be nonlinear (NL). The most notable feature of such an NLDE is that its solutions do not satisfy a superposition principle, i.e., linear combinations of solutions in general are not solutions of the equation any more. This nonlinearity is the basic reason why solutions of NLDEs are complicated and hard to describe quantitatively. Nonlinear systems that are dissipative but not driven may still be difficult to describe precisely, but their motions can be understood intuitively. Consider, for example, a dissipative pendulum made up of a mass attached to one end of a massless rigid arm that can rotate in a vertical plane about a pivot at the other end of the arm. When set in motion with a large initial velocity, the pendulum can be expected to rotate about its pivot, losing energy all the time. Soon it will not even make it to the top, but oscillates about the equilibrium point. It will always come to rest there eventually because that equilibrium point is an attractor. If an external force also acts on this pendulum, however, the situation changes drastically. The driving force can put energy into the system to counterbalance the energy lost to friction. The delicate interplay between these two opposing mechanisms then generates the intricate drama that is NL dynamics in physical systems. We shall describe here how chaos can develop in such NL systems through the perioddoubling bifurcation route analyzed in some detail in the preceding sections on NL maps. Complexities such as but not limited to chaos appear when all three essential elements—nonlinearity, driving force, and dissipation—are present.

Driven dissipative linear pendula

401

We shall present nonlinear complexity in two steps. We shall first review, in this section, why the motion of two kinds of driven dissipative linear pendula remains predictable and nonchaotic. Chaos in nonlinear pendula is then described for the parametrically driven nonlinear pendulum in the next section using numerical solutions obtained with a computer. 6.5.2

Externally driven dissipative linear pendulum

The rigid-arm pendulum when driven by an external periodic force satisfies a surprisingly simple inhomogeneous NLDE of motion: θ¨ + γθ˙ + ω20 sin θ = 2A cos ωt,

(6.49)

where θ = θ(t) is the angular displacement measured from an equilibrium position (that is well-defined & when the driving term is absent). In this DE, γ is the friction parameter, ω0 = /g is the natural angular frequency of the pendulum of length  at a location where the acceleration due to gravity is g, and ω is the angular frequency of the driving force. The strength parameter A is real, and for notational simplicity in the following discussion, taken to be nonnegative. We know by looking at unused playground swings that in the absence of a driving term (A = 0), the final state is just the motionless state in the equilibrium position θ = 0 (modulo 2π). For small angular displacements, sin θ ≈ θ. The DE can then be linearized to θ¨ + 2βθ˙ + ω20 θ = 0,

(6.50)

where β = γ/2. Its mathematical solution is well-known but worth repeating. Since the LDE has constant coefficients, its solutions are the exponential functions θ(t) = θ0 e st . Substitution into the LDE yields the algebraic equation and its solutions s2 + 2βs + ω20 = 0, s = −β ± iω1 ,

ω21 = ω20 − β2 .

(6.51)

Hence the undriven or homogeneous solutions are θh (t) = e−βt θ1 (t), where ⎧ ±iω t 1 , ⎪ if ω1 > 0 ⎪ ⎪ ⎪e ⎨ θ1 (t) = θ0 ⎪ 1, if ω1 = 0 ⎪ ⎪ ⎪ ⎩e∓|ω1 |t , if ω1 = i|ω1 |.

(6.52)

The critically damped solution θ0 e−βt also provides the envelope inside which the underdamped oscillations take place when ω1 > 0. The last case is an overdamped motion where the position returns to the equilibrium point most slowly because the frictional force is here the strongest. In all cases, the exponentially decaying envelope

402

Nonlinear systems

e−βt ensures that the homogeneous solution describes a transient state of motion that dies out at large t when dissipation is present and β > 0. We are now ready to add the driving term proportional to 2A cos ωt = A(eiωt + −iωt e ) to the external driven linear pendulum [Eq. (6.49)]. It is convenient to begin by using just the first term Aeiωt . We can then try a particular solution of the form θ p (t) = Deiωt dependent only on the driving frequency ω. Its substitution into the driven LDE yields an inhomogeneous algebraic equation linear in D [(−ω2 + ω20 ) + i2βω]D = A.

(6.53)

This equation can be be solved for the linear response D=

A (ω20

|D| = % tan δ =

−

ω2 )

+ i2βω

= |D|e−iδ

A (ω20 − ω2 )2 + (2βω)2

,

2βω , − ω2

ω20

θ p (t) = |D|ei(ωt−δ) .

(6.54)

Since the result holds for both positive and negative ω values, the solution for the complete driving term A cos (ωt) is, by linear superposition, θ p (t) = |D| cos (ωt − δ). That is, if Eq. (6.49) is written symbolically as L θ p1 = R1 and L θ p2 = R2 for two different driving terms R1 and R2 , then L (θ p1 + θ p2 ) = R1 + R2 because of the linearity of the differential operator L . The amplitude |D(ω)| (or its square |D|2 ) of the response as a function of the driving frequency ω is a bell-shape curve & called a resonance curve. The system is said to resonate at the frequency ωR = ω2 − 2β2 where |D(ω)| has a maximum, just like a struck bell that rings the loudest at its own resonance frequency. Although the response D of the particular solution is frequency dependent, it conforms exactly to the time independence e±iωt of the external driving term. This means that it describes a steady-state response of the system where the energy lost to dissipation is exactly counterbalanced by the energy gained from the external driving force at every instant of time. However, the homogeneous solution must also be included to match any given ˙ initial conditions θ(t = 0), θ(0). If the initial state of motion has more energy than the final state, the frictional loss after motion has begun must exceed the energy gain from the external force in order to reach the final balanced state. If the initial energy is too small, the frictional loss must first fall below the energy gain from the driving force until the system energy has reached the same final value. In short, the particular solution obtained is entirely independent of the initial conditions.

Driven dissipative linear pendula

6.5.3

403

Parametrically driven dissipative linear pendulum

If the driving mechanism is not a simple external force, but comes instead from a periodic variation in the length of the pendulum arm, the equation of motion takes the form of a homogeneous NLDE θ¨ + 2βθ˙ + (ω20 + 2A cos ωt) sin θ = 0,

(6.55)

first studied by E. L. Mathieu in 1868 in the linear limit and without dissipation (sin θ → θ, γ = 0, and now called the Mathieu equation) in connection with the vibrations of an elliptic membrane. A dissipative Mathieu pendulum is also obtained if it is the point of support that moves periodically (Landau and Lifshitz, p. 81). The Mathieu equation itself describes a nondissipative parameter-driven or parametric oscillator. The driving strength parameter A is real, and in the following discussion taken to be nonnegative. In the absence of the driving term, the dissipative Mathieu pendulum is just the dissipative undriven rigid-arm playground swing of Eq. (6.49) with A = 0. So the same solution obtained previously applies. With the driving term in place, the solution of the linearized equation can be written as θ(t) = e−βt θ1 (t), where θ1 (t) satisfies the Mathieu DE θ¨1 + (ω21 + 2A cos 2ωd t)θ1 = 0,

(6.56)

where ωd = ω/2. So θ1 is just a Mathieu function. The effect of dissipation resides only in the critical damping envelope and in the change of the natural frequency ω0 to the damped frequency ω1 . As the Mathieu equation is a second-order LDE, a general solution is a linear combination of two linearly independent solutions. The linearly independent solutions can be taken to be the Mathieu cosine and sine functions that become the ordinary cosine and sine in the limit A → 0. An approximate solution of these Mathieu functions can be found by trying a solution of the form θ1 (t) = a(t) cos ωd t + b(t) sin ωd t.

(6.57)

Such a form works because the periodic structure of the driving term can be simplified by the trigonometric identities 1 cos 2φ cos φ = (cos φ + cos 3φ), 2 1 cos 2φ sin φ = (− sin φ + sin 3φ). 2

(6.58)

With φ = ωd t, one finds two dominant or resonant terms of frequency ωd . The remaining terms involving the third harmonic 3ωd are way off the driving frequency. They can be dropped if A is sufficiently small.

404

Nonlinear systems

The assumed solution (6.57) can now be substituted into Eq. (6.56) to get an expression of the form C s sin (ωd t) + Cc cos (ωd t) = 0. Since the two trigonometric functions are linearly independent, solution requires that C s = Cc = 0. Most of the time dependence appears in the two trigonometric functions. What is left is contained in a(t), b(t) that can be expected to be only weakly t dependent. The second approximation is to drop the small a¨ , b¨ terms. The conditions C s = Cc = 0 then become two coupled first-order LDEs: Cc ≈ (A − Δ)a + 2ωd b˙ = 0, C s ≈ −(A + Δ)b − 2ωd a˙ = 0,

(6.59)

where Δ = ω2d − ω21 . We now look for a solution for a(t) and b(t) of the form a(t)/a0 = b(t)/b0 = eσt , because cos ωd t and sin ωd t are two parts of the same exponential function eiωd t . The result is a set of two homogeneous algebraic equations in the unknown functions a, b: −(A − Δ)a = 2ωd σb, −(A + Δ)b = 2ωd σa.

(6.60)

A solution is possible only if the coefficients appearing there satisfy the following consistency or determinantal equation (2ωd σ)2 − (A − Δ)(A + Δ) = 0.

(6.61)

The condition is satisfied if √ σ=±

A2 − Δ2 . 2ωd

(6.62)

σ is real if A2 > Δ2 . For the positive root, one finds the general solution θ1 (t) ≈ eσt (a0 cos ω1 t + b0 sin ω1 t), θ(t) ≈ e(σ−β)t (a0 cos ω1 t + b0 sin ω1 t),

(6.63)

where a0 , b0 are arbitrary constants that can be fitted to any given initial conditions. Thus when σ > σc = β = γ/2, the system is destabilized, and the attractor at θ = 0 disappears. The destabilizing effect of the driver A is greatest when Δ = 0 or ωd = ω1 . Then σ = σmax = A/2ω1 . Destabilization of the Mathieu pendulum thus appears when A > Ac = γω1 ≈ γω0 .

(6.64)

Driven dissipative linear pendula

405

When happens, the system is said to show a parametric resonance. Equation (6.63) also shows that the effective frictional parameter is γeff ≈ γ −

A , ω1

or

βeff ≈ β −

A . 2ω1

(6.65)

As Mathieu functions appear frequently in physics, we have included a number of problems dealing with them in the exercises.

Problems 6.5.1 (Resonance in externally driven linear (EDLP) pendula) Show that the quantity |D|2 of the complex response % D of the EDL pendulum given in

Eq. (6.54) has a maximum at ωR = ω20 − 2β2 and a FWHM (full width at half maximum) Γ = ω+ − ω− , where % 2 2 ω± = ωR ± 2β ω2R + β2 .

Hence, if β  ω0 , the resonance appears at ≈ ω0 with a width Γ ≈ γ or a half width Γ/2 ≈ β. 6.5.2 (Mathieu pendulum) Verify Eq. (6.59). 6.5.3∗ (Mathieu functions) (a) Show that the Mathieu Eq. (6.56) can be written in the standard form y (x) + (a + 2q cos 2x)y(x) = 0, where y = dy(x)/dx. The variable x and the parameters a, q are real in this problem. (b) (Operator symmetries) Any homogeneous linear DE can be written in the form L x y(x) = 0, where L x is the linear differential operator (DO) appearing in the DE. For the Mathieu equation, Lx =

d2 + (a + 2q cos 2x). dx2

A symmetry of L x is a transformation T under which T {L x } = L x remains unchanged or is invariant. The transformation can be one involving a change of variable x → xc under which L x → T {L x } = L xc . (→ is called a “maps-to” symbol.) Show that the Mathieu L x is invariant under (i) complex conjugation L x∗ , (ii) reflection xc = −x, and (iii) finite translation xc = x + π.

406

Nonlinear systems

By applying T to the Mathieu equation, show that if y(x) is a solution, so is T {y(x)} = y∗ (x) for case (1), and T {y(x)} = y(xc ) for cases (ii) and (iii). (c) (Symmetry invariant solutions) Under a symmetry transformation T , T {y(x)} of the solution y(x) of a second-order LDE is not always proportional to y(x). However, there is always a linear combination y(x, b) = y(x) + bT {y(x)},

for which

T {y(x, b)} = ky(x, b), k a complex constant, where the solution y(x, b) is invariant under the symmetry transformation. Show that the logarithmic derivative y /y of such a solution remains unchanged under the symmetry transformation. (d) (Complex conjugation) The two linear combinations of an arbitrary solution y invariant under complex conjugation are y + y∗ = 2yR and y − y∗ = 2iyI , or b± = ±1 with k = b. Show that as a result, the two linearly independent Mathieu functions (i) can be taken to be real, and (ii) are generated by the boundary conditions (y(t = 0), y (0)) = (1, 0) and (0, 1), respectively. Hint: Use the Wronskian determinant to test for linear independence. (e) (Reflection) Show that under the reflection x → xc = −x, k2 = 1 or k = ±1. Show that k = b. The resulting linear combinations are even or odd functions of x. They are called the Mathieu cosine C(x) and sine S (x), respectively. (f) (Finite translation) Let k = eiνπ . Show that ν is defined only up modulo 2m, m any integer. Show that the function P(x) = e−iνx y(x, b) satisfies the finite translation property P(x) = P(x + π). You have thus proved the Floquet theorem that the Mathieu equation has Floquet solutions of the form Fν (x) = y(x, b) = eiνx P(x), where P(x) is periodic with period π. (The envelope is more easily visualized by writing it as eiνx = cos νx + i sin νx.) Equivalently, the Floquet solutions satisfy the scaled periodic property Fν (x + mπ) = km F ν (x), where m is any integer and the scale k is a complex constant. The parameter ν is called the characteristic exponent, and k is called the Floquet multiplier. (g) (Stability of Mathieu functions) A general or complete solution containing both linearly independent solutions is said to be stable if it remains bounded (< ∞) as x → +∞. Show that the Mathieu function is stable if ν is real, and unstable if ν = β − iα is complex. (In this problem we shall assume that ν  an integer so that the solution linearly independent of y(x) can be taken to be taken to be y(−x).)

Chaos in dissipative nonlinear pendula

407

Table 6.4 Three types of characteristic exponents ν = β − iα. Here m is any integer.

ImC(x) = 0 if

C(x) is real and

α = 0, β = 2m, β = 1 + 2m,

|C(π)| ≤ 1 C(π) > 1 C(π) < −1

(h∗ ) (Mathieu cosines) The Mathieu cosine is defined as F ν (x) + Fν (−x) 2Fν (0) = φ+ (x) cos νx + iφ− (x) sin νx,

C(x) =

where φ± are the even and odd parts of φ(x) = φ+ + φ− = P(x)/P(0). Show that any Mathieu function does not contain a term that is the product of two odd functions in x. Hence the φ− term is absent. Hint: A product of two odd functions is even and ∝ x p near the origin, where p ≥ 2 is an even integer. Such a function cannot be a solution of the Mathieu equation. Show that C(0) = 1, C  (0) = 0, C(x) is real, and C(π) = cos νπ. The last expression can be used to compute ν from a numerical solution of C(π). Verify the behavior of the characteristic exponent ν = β − iα summarized in Table 6.4. Finally, show that the function Cm (x) = φ+ (x) cos [(±ν + 2m)x],

integer m,

has the same value Cm (π) = cos νπ at x = ±π.

6.6

Chaos in parametrically driven dissipative nonlinear pendula

Detailed numerical studies of NLDEs became possible when the high-precision of modern digital computers allowed the fine tuning of initial conditions. The reader interested in a quantitative understanding of the subject will eventually have to use a computer to generate computer simulations of NL dynamical behaviors. Many computer codes are available from commercial and non-commercial sources including books on computer simulation. Even videos of detailed solutions are available on the Web. We shall restrict ourselves here to a brief description of data “mining” of numerical solutions of NLDEs for their outstanding dynamical features. The process will be illustrated by an actual analysis of the motion of the parametric dissipative nonlinear

408

Nonlinear systems

(PDNL) pendulum for one driving frequency and in a small range of initial conditions where period-doubling bifurcations (PDBs) and strange attractors appear. The first step in data mining is to choose the data to mine. The most interesting solution has the pendulum swing over the top repeatedly in the final state. So θ(t) is roughly linear in t, with the pendulum slowing down and then speeding up in every ˙ would concentrate on these speed fluctuations, but even this does swing. A plot of θ(t) not show enough of the subtlety of its motion in full detail. We need a more precise characterization of its state of motion to select the most relevant data for quantitative analysis. The PDNL pendulum satisfies the second-order DE (6.55). Its state of motion at any time, like that for any second-order DE, is fully specified by the two state ˙ Each state of motion can be plotted as a single point in the 2D variables θ(t), θ(t). ˙ state space (θ(t), θ(t)), with the state point moving in time. Such motion, as a function of another parameter t on which both state variables depend, is said to be a parametric plot or representation of its time-dependent motion. A parametric plot can be made for the solution of any second-order DE, not just for the parametric pendulum. So the two parametric features are not related. In classical mechanics where masses and moments of inertia can change when the system breaks up into pieces, it is the momentum or angular momentum that is more fundamental than the velocity or angular velocity because the momentum is conserved in an isolated system. The 2D position-momentum or angle-angular momentum state space is called a phase space. For our pendulum, however, its moment of inertia does not change in time, so the angular velocity is just the angular momentum measured in units of the constant moment of inertia. For this special case, the state space is also a phase space. By mining data in phase space, we are sure that each datum so mined potentially carries a complete characterization of the state of the system. How should data be mined? Recall that when the pendulum is initially set into motion for a chosen set of initial conditions, the energy fed in by the driver and the frictional loss to dissipation are not in balance. Their balancing act will play out in time differently for different parameters: ω0 , ωd , A. Even more interesting are the steady-state or final-state behaviors. So we need to decide on a promising choice of parameters for an illustrative example. Recall that in the linear regime, the balancing act between the driven gain and the frictional loss takes a surprisingly simple form given by Eq. (6.65) when the driving frequency ω is set at the principal parametric resonance frequency 2ω1 ≈ 2ω0 . The choice ω = 2ω0 is likely to simplify the nonlinear dynamics of the system too. It has the even greater advantage of simplifying data mining to just one look of the motional orbit in phase space per driving period. If one driving period is equivalent to one insect generation in the logistic map, this is all it takes to uncover the basic nonlinear responses of the system. One look per driving period also means exactly two looks per natural period. So anything happening at exactly the natural period will be seen as well. We shall use the friction parameter γ = 0.2 (dimensionless for a dimensionless time variable) of McLaughlin (1981) to allow some mutual comparisons.

Chaos in dissipative nonlinear pendula

409

Many different strobe (or stroboscopic) views are possible. For example, the strobe can be turned on not according to time but when the pendulum returns to the same same angle θP (modulo 2π). The resulting view is called a return map, or a Poincar´e map. In general, fixing one variable at a constant value in a multidimensional phase space is said to generate a Poincar´e section of the motion in the full phase space. This return map is harder to generate than the simple periodic look in time, and shall not be used here. In picking the driving strengths A to study, we note that as it is increased gradually from 0, the final state is likely to differ only a little from the undriven final state where the equilibrium position θe = 0 (modulo 2π) is an attractor. This means that a stability analysis can be made in the linear regime sin θ ≈ θ. The result obtained in the last section is that the driving term can be combined with the dissipative term to give an effective dissipation parameter of γeff = γ − A/ω1 . Hence it is only necessary to examine the numerical solutions of the PDNL pendulum for A > Ac = γω1 . In Fig. 6.12, solutions of the PDNL pendulum are shown for the parameter choice ω0 = 1 (dimensionless), ω = 2 (dimensionless) and γ = 0.2. The initial conditions ˙ = 3 (dimensionless), meaning that the pendulum is given are θ(t = 0) = 0 and θ(0) an initial push as it hangs in equilibrium below the point of support. Four solutions are obtained, for A = 0.25, 0.5, 0.85 and 1.0, from top to bottom. The left figures (a) give strobe views of the velocity θ˙ at the end of each of n = 80 driving periods. These views show the overall trend of the dynamical evolution to a “steady” final state. The right figures (b) give phase plots for the next nP driving periods to show the complete final state reached. All these figures are functions of the angular displacement θ. The pendulum is at the top of its circular motion whenever θ reaches an odd integral multiple of π. Each of these eight figures will be referred to by its row location (1,2,3 or 4) and column location (a or b). In Fig. 6.12(1a), for A = 0.25 a little above Ac , the pendulum turns over once and then oscillates between two accumulation points about the average final-state position θ∞ = θ(t → ∞) = 2π as the initial “transient” dies down. Thereafter, driven energy gain appears to be in balance with the energy loss to friction, when averaged over a complete driving cycle. So our parametrically driven homogeneous NLDE behaves here a bit like an externally driven inhomogeneous NLDE. The motional period τ0 = 2π = 2τ can be read from Fig. 6.12(1a), as each repetition takes two strobe flashes. This is confirmed by the phase plot (1b) for the next nP = 1 driving period. The motion begins at the upper left corner and ends at the lower right corner during a driving period. It covers exactly half of the motional period. At the start of the driving period, the pendulum is at an angle about 0.84 ˙ radian (48◦ ) below θe . It has a maximum velocity of ≈ 0.94, about 1/3 of θ(0), a little before θ∞ = 2π. At θ∞ , the moving pendulum in the final state has an energy only 10% of its initial energy at t = 0. The fact that the motional period is τ0 shows that the the driving strength is not strong enough to dominate the natural period for the given initial conditions. The motion of the pendulum changes with the driving strength A. Fig. 6.12(2a) for A = 0.5 shows what happens when the pendulum is driven hard enough initially to turn over at the top of its swing in a steady manner once per driving period π

410

Nonlinear systems (b) Phase plot for A = 0.25, nP = 1

(a) Strobe view for A = 0.25, dθ/dt = 3 3

1

dθ/dt

dθ/dt

2

0 –1 0

2

4 6 θ (radian)

0.8 0.6 0.4 0.2 0 –0.2 –0.4 5.5 5.75

8

dθ/dt

dθ/dt

7.25

2.4

2.6 2.4 2.2

2.2 2 1.8

2

1.6

1.8 0

100

200 300 θ (radian)

400

500

502 503 504 505 506 507 θ (radian) (b) Phase plot for A = 0.85, nP = 2

(a) Strobe view for A = 0.85, dθ/dt = 3

2.9 dθ/dt

2.8 dθ/dt

7

2.6

2.8

2.7 2.6 2.5 2.4 0

100

200 300 θ (radian)

400

2.75 2.5 2.25 2 1.75 1.5 1.25 504 506 508 510 512 514 θ (radian)

500

(b) Phase plot for A = 1.01, nP = 4

(a) Strobe view for A = 1.01, dθ/dt = 3 3.25 3 2.75 2.5 2.25 2 1.75

3 2.5 dθ/dt

dθ/dt

6.25 6.5 6.75 θ (radian)

(b) Phase plot for A = 0.5, nP = 1

(a) Strobe view for A = 0.5, dθ/dt = 3 3

3

6

2 1.5 1

0

100

200 300 θ (radian)

400

500

505

510

515 520 θ (radian)

525

Fig. 6.12 (a) Strobe view (left column) of the velocity θ˙ at the end of each of n = 80 driving periods, and (b) phase plot (right column) of the next nP (denoted as nP in the figure) periods of the PDNL pendulum ˙ = 3 (dimensionless). The equation parameters used are for the initial conditions θ(t = 0) = 0 and θ(0) ω0 = 1, ω = 2, and friction parameter γ = 0.2. The strength parameter used is shown in each figure. The two figures in the nth row of the figure array are referred to in the text as figures (na) and (nb).

Chaos in dissipative nonlinear pendula

411

Table 6.5 Final-state periodicity T of the PDNL pendulum of Eq. (6.55) as a function of the driving strength A. The friction parameter γ = 0.2, natural period τ0 = 2π, and driving period τ = π are used. ˙ = 3 (dimensionless) are given under The results for the initial conditions θ(t = 0) = 0 radian and θ(0) the column heading “Here”. Those obtained by McLaughlin (1981) are shown under the column heading “McL”. The initial conditions used are not stated in McLaughlin’s paper, but probably vary with A. F. = Feigenbaum.

A

Here

McL

T

Comment

A < Ac = γω1 = Ac  A ≤ A1 = A1 ≤ A ≤ A2 = A2 ≤ A ≤ A4 = A4 ≤ A ≤ A8 = AF = A > Ad =

0.191 0.481 0.789 0.984 1.026 1.033 1.578

0.2 0.713 0.793 0.984 1.0198 1.033 1.58

τ0 τ 2τ 4τ ∞ τ

At rest Hopf 1-cycle 2-cycle 4-cycle F. point 1-cycle

in the final motional state. The driving term now dominates. The situation is conceptually equivalent to the appearance of a single stable 1-cycle in the logistic map of Eq. (6.16). In confirmation, the phase-space plot (2b) for nP = 1 driving period shows one complete motional period. Another interesting stage appears at A = 0.85. Fig. 6.12(3a) shows a final motional state that has bifurcated into a 2-cycle spanning two driving periods. The phase diagram (3b), plotted for n p = 2 driving periods, describes a periodic motion of period 4π, as one can check by the change in θ. In a similar way, Figs. (4a) and (4b) for the final state motion at A = 1.01 illustrate the nature of the pendulum motion in a 4-cycle final state. The strobe view displays the internal structure of the 4-cycle much more clearly than the phase diagram even if the latter were to be repeated many times. The stability of the pendulum motion in the final state as a function of the driving strength A can be shown very effectively by constructing a bifurcation diagram, a plot of an essential state property against the controlling parameter, here A. Since a strobe view of the final state gives both position and velocity, both quantities can be plotted in different bifurcation diagrams, or one diagram can be studied before the other. In our study, the velocity is more promising because the initial velocity is nonzero. However, in the resulting bifurcation diagram, two phase points can occasionally appear as one when they are actually separate in the 2D phase space. When in doubt, we examine the full strobe view or the complete phase space orbit. The result is the bifurcation diagrams of Fig. 6.13 calculated for the initial conditions ˙ = 1.5, (b) θ(0) ˙ = 3, and (c) θ(0) ˙ = 5. The branching points for θ(0) = 0 and (a) θ(0) case (b) are examined at higher magnification to obtain the numerical values given in Table 6.5. These three figures will be referred to as Fig. 6.13(a), (b) and (c), respectively. These figures show that for all initial conditions, the attractor in the resting state [θ˙ = 0 at θ∞ = 0 (modulo 2π)] is destabilized as A increases through Ac = γω1 =

412

Nonlinear systems Bifurcation diagram for θ(0)=0, dθ(0)/dt =1.5 4

dθ/dt

2

0

–2

–4 0.25

0.50

0.75

1.00

1.25

1.50

1.75

1.50

1.75

1.50

1.75

A Bifurcation diagram for θ(0)=0, dθ(0)/dt =3 4

dθ/dt

2

0

–2

–4 0.25

0.50

0.75

1.00

1.25

A Bifurcation diagram for θ(0)=0, dθ(0)/dt =5 4

dθ/dt

2

0

–2

–4 0.25

0.50

0.75

1.00

1.25

A Fig. 6.13 Bifurcation diagram for the PDNL pendulum showing the strobe view once per driving cycle of the angular velocity θ˙ as a function of the driving strength A for the initial position θ(0) = 0 and three ˙ = 1.5, (b) θ(0) ˙ = 3, and (c) θ(0) ˙ = 5. Solutions contain different initial velocities (top to bottom) (a) θ(0) ˙ of the last 200 periods are plotted in the figure for each 400 driving periods. The strobe views of θ(t) value of A.

Chaos in dissipative nonlinear pendula

413

0.191. When this happens, the pendulum continues to oscillate about the average final position θ∞ = 0 (modulo 2π) like a Hopf oscillator with an amplitude that is roughly independent of the initial conditions until A ≈ 0.42. We shall call the bifurcation at Ac a Hopf bifurcation even though its time dependence is not simple harmonic. The two velocities seen in two successive driving periods of the Hopf bifurcation in these bifurcation diagrams represent a simple (1-loop) closed Hopf orbit in phase space having exactly the natural period τ0 . For A  0.42, dependence on initial conditions begins to appear in the final-state ˙ range examined, most bifurcation diagrams are similar to strobe view. In the θ(0) those shown in Figs. 6.13(a) and (c), where the Hopf bifurcation [characterized by θ∞ = 0 (modulo 2π)] still appears intermittently for A as large as 0.75. The exception shown in Fig. 6.13(b) with a single curve for the Hopf final states and an almost completely continuous Feigenbaum sequence is seen only in the narrow ˙  3.056. range 2.707  θ(0) ˙ in the strobe The Feigenbaum sequence begins for A as weak as ≈ 0.42. Its θ(t) view often acquires a negative instead of a positive sign. The missing pieces of the Feigenbaum sequence consisting of solutions with the opposite signs can be found in the solutions generated by the initial velocity of the opposite sign. These two sets of solutions together form two identical and complete Feigenbaum sequences that differ from each other only in the sign of the angular velocity. It is interesting that in general the Hopf bifurcation curve and the Feigenbaum 1-cycle can appear in the same range of A values in an either/or basis. The Hopf curve contains other interesting structures which we shall touch on briefly. A magnification of the bifurcation diagram Fig. 6.13(c) of the region 0.60 < A < 0.624 is shown in Fig. 6.14. We are interested in the Hopf bifurcations closed to θ˙ = 0. All these Hopf solutions have θ∞ = 4π, meaning that the final state is realized after two complete rotations of the pendulum. The Hopf segments with two strobe points are the Hopf bifurcations forming 1-loop closed orbits in phase space of period τ0 . The segments with 6 strobe points are closed 3-loop Hopf bifurcations. For 0.6195 ≤ A ≤ 0.6209, the 3-loop Hopf orbit is unable to close. Examination of the phase space orbits suggests that this is probably a 3-cycle period-doubling bifurcation cascade (PDC) where closed orbits are formed with N = 6, 12, 24, . . . loops until N → ∞ is reached at an accumulation point near A = 0.6209. The next A point is a Feigenbaum 1-cycle. The solution at A = 0.6209 closest to the expected accumulation point is shown in the bifurcation diagram in four “pieces”. In phase space, the two inner pieces are actually well separated because they cover different positions. Each piece is made up of phase points of zero extensions, but they may appear closer together in displayed figures because each point must be given a spurious finite size. However, a 3-cycle PDC is known to end up in a chaotic state. So we presume without further ado that we have located a strange attractor of the Hopf bifurcation. In Fig. 6.13(a), a hint of a similar Hopf (strange) attractor can be seen. Magnification of the bifurcation diagram shows that it is indeed there, the point closest to the accumulation point being a little further, at A = 0.6211. This small change does not appear to be the result of a sensitivity to initial conditions for the following

414

Nonlinear systems Bifurcation diagram for θ(0) = 0, dθ(0)/dt = 5

1 0.75 0.5

dθ/dt

0.25 0 –0.25 –0.5 –0.75

0.605

0.610

0.615

0.620

0.625

A Fig. 6.14 Magnified view of the bifurcation diagram of Fig 6.13(c) for 0.60 < A < 0.63. Solutions ˙ of the last 400 periods are plotted in the figure contain 600 driving periods. The strobe views of θ(t) for each value of A.

reason. In this region of A, the Hopf bifurcation is playing hide and seek with the Feigenbaum 1-cycle, as one can see in Figs. 6.13(a) and (c). Which one appears for a given A depends on the initial conditions. In Fig. (b), for example, the Hopf 3-cycle is completely “hidden” and inaccessible. So the Hopf 3-cycle is there although we may not see any or all of it in a solution. Let us next return to the Feigenbaum period-doubling cascade in Fig. 6.13(b). Its 2-cycle first appears at A2 and its 4-cycle at A4 . The 8-cycle branching point A8 is a little hard to find because there is a range of A values where the balance of power between the two configurations goes back and forth. This is probably a consequence of the finite number of driving periods used (400 periods here) and the roundoff errors in the numerical calculation. The Feigenbaum point AF is more easily found, however. A mostly chaotic region then follows until Ad ≈ 1.58. In this region, the structure of the strange attractors becomes more complicated as A increases. For A > Ad , the driving frequency dominates unambiguously with a simple 1-cycle motion that does not bifurcate. These results are summarized in Table 6.5 to show the general agreement with the results obtained by McLaughlin (1981). See McLaughlin for solutions for other initial conditions where more complicated types of motion appear including those with intermittent backward rotations. In summary, the PDNL pendulum related to the Mathieu equation has Feigenbaum period-doubling and chaotic solutions dominated by the driving frequency that are independent of the initial conditions. In addition, there are solutions with dynamical features that appear in ways that dependent on the initial conditions. A particularly interesting feature is the Hopf bifurcations controlled by the natural frequency that

Solitons

415

form closed orbits around the point attractors of a resting pendulum. 3-cycle perioddoubling cascades can bifurcate from them to end up probably in strange attractors.

6.7

Solitons

A wave is a disturbance that travels, transferring momentum and energy from one location to another as a function of time. A wave form usually consists of many oscillations, but occasionally a single localized wave pulse can also travel alone. In 1834, Scott Russell saw a single or solitary wave pulse rolled away from the prow of a horse-drawn boat in a narrow canal when the boat stopped suddenly. The pulse was some 30 feet long and more than a foot high. It preserved its original shape with gradually diminishing height for the mile or two that he was able to follow on horseback. Thus began the saga of solitary waves or single wave pulses that travel. Intrigued by the unexpected persistence of solitary waves, Scott Russell studied its properties in wave tanks at home for many years afterwards. His work motivated the subsequent mathematical studies of Joseph Boussinesq and of Lord Rayleigh, in 1871 and 1876, respectively. The nonlinear partial differential equation (PDE) studied by Boussinesq in 1872 is now called the Boussinesq equation. A related PDE that appeared in Boussinesq’s 1877 Memoires is now named after Korteweg and de Vries for their definitive study of 1895. The name “soliton” was originally invented by Zabusky and Kruskal (Zabusky 1965) to refer to a “solitary-wave pulse” in a wave containing several pulses. However, shape-preserving wave pulses (localized wave forms) and their scattering properties in classical nonlinear field theory in nuclear and particle physics had already been studied earlier by Skyrme (1961) and Perring and Skyrme (1962), two of the nine papers Skyrme published on the subject in the period 1954–71. These special solitons are now called Skyrmions. Solitons are nonlinear pulses whose amplitude, shape and velocity remain unchanged even after passing through one another. Their stability in collisions comes from the conservation in time of important physical or mathematical properties. Skyrmions, for example, are topological solitons that are characterized by a topological property called a winding number that can be interpreted as a baryon number. Baryon numbers are conserved quantities assigned to baryons, which are strongly interaction particles such as protons and neutrons. Solitary waves with slowly changing shapes and amplitudes are sometimes called soft solitons. They appear when the system is not strictly integrable, i.e., when the system does not have properties that are strictly conserved in time. Solitons and soliton-like waves occur frequently in the physical, engineering and biological sciences. The primary but not only tool we shall use in our analysis of traveling solitons and their relatives the traveling kinks or wave fronts is the persistent wave form f (z), z = kx. When boosted to the velocity c = ω/k, the resulting wave f (z), z = kx − ωt, becomes dispersive when ω = ω(k) and ω (k) = d2 ω/dk2  0. We shall explain and further emphasize through examples and problems why these waves become persistent and last a long time when they are both dispersive and nonlinear.

416

Nonlinear systems

These persistent waves are solutions of partial differential equations (PDEs) in both time t and at least one spatial coordinate x. They appear in both evolution equations where only ∂t = ∂/∂t appears, and in hyperbolic PDEs where ∂tt = (∂t )2 appears in the combination ∂ xx − α2 ∂tt = (∂ x − α∂t )(∂ x + α∂t ). The factorization shows why the same wave form f (z) can appear in both classes of PDEs. However, the hyperbolic wave equations are richer in being able to describe standing waves made up of both wave components running in opposite directions. This involves a superposition of nonlinear waves which we shall also describe. Nonlinear PDEs can be very difficult to solve directly. Evolution PDEs involving only ∂t satisfies simpler initial conditions than PDEs containing ∂tt . With two or more variables in the problem, the boundary curve or surface is more complicated, however. For example, it can be closed (in one continuous piece) or open in the two or more dimensional space of variables. The specification of the boundary conditions (value, slope or both) needed for a unique solution will in general differ among the different types of PDEs: Parabolic PDEs containing: α∂t − ∂ xx , Hyperbolic PDEs containing: α∂tt − ∂ xx , Elliptic PDEs containing: α∂t + ∂ xx . These complications are not present for the special case of the wave form f (z). Its study is further simplified by first finding the nonlinear ODE (ordinary DE in one variable) it satisfies before turning the ODE into an evolution or hyperbolic wave equation. Other tools used in our study of nonlinear waves includes the separation of variables, and the superposition of nonlinear waves using the bilinear method of Hirota. 6.7.1

Korteweg–de Vries soliton

Let us consider the KdV equation: φt + αφφ x + δ2 φ xxx = 0,

(6.66)

where φt = ∂φ/∂t, etc. It is a nonlinear partial differential equation (PDE) in 1D space. We shall show that this nonlinear PDE have soliton solutions. The wave properties of the KdV Eq. (6.66) can be clarified by comparing it with the linear wave equation f xx −

1 ftt = 0. c2

(6.67)

The latter equation has solutions of the form f (x − ct) = exp i(kx − ωt), where on substitution into Eq. (6.67), the wave velocity can be shown to be c = ±ω/k.

Solitons

417

The KdV PDE can be made linear by leaving out the nonlinear term, i.e., with α = 0. Trying a solution of the form f (kx − ωt) gives the algebraic equation −i(ω + δ2 k3 ) f (kx − ωt) = 0.

(6.68)

Hence the wave velocity c ≡ ω/k = −(kδ)2 differs for different k values. A linear superposition of such harmonic waves is allowed for the PDE, as for any other linear PDE. However, with the wave velocity changing in different Fourier components, the wave shape changes in time. So the solution is not a soliton whose shape is unchanged in time. Waves with k-dependent velocities c(k) are said to be dispersive because a similar k dependence causes a beam of sunlight to disperse into beams of different colors after passing through a glass prism, as Isaac Newton found around 1670. For this reason, the φ xxx term is said to be dispersive, and the functional relationship such as ω(k) = −δ2 k3 is called a dispersion relation. What we have shown is that a linear dispersive wave cannot be a soliton. A linear nondispersive wave is also not a soliton even when its wave form remains unchanged in time. This is because solitons are defined to be nonlinear waves. They do not exist with arbitrary amplitudes, and they are not linear superpositions of distinct waves. Solitons also cannot appear in the KdV PDE if the dispersion term is absent. Then φt + αφφ x = 0

(6.69)

used with the trial solution φ(x, t) = f (x − ct) gives (−c + α f ) fx = 0.

(6.70)

One solution f = const is not interesting, while the other describes a wave with velocity c = α f (s = x − ct) that varies with the position s on the wave “envelope”. So the wave shape too will vary in time. This property can be called spatial dispersion to distinguish it from the optical dispersion in the reciprocal space of the inverse wavelength k. Such a wave is not a soliton because its shape does not remain constant in time. Thus dispersion and nonlinearity acting alone would distort the wave shape. Acting together they can conspire instead to preserve it. It is therefore interesting to see how they work together in the full KdV equation. We begin by looking for a solution of the form φ(x, t) = f (z), with z = k(x − ct), for a single pulse traveling to the right with speed c > 0. After taking care of the φt term as −kc f , f  = d f /dz, we can set t = 0 and ∂ x → d x to obtain from the KdV equation the ordinary DE (ODE) −kc f  + αk f f  + δ2 k3 f  = 0

(6.71)

for the initial pulse shape f (kx). Every term in this DE involves a derivative f (m) = (d/dz)m f (z), the lowest derivative being f  . So the DE can be simplified further by one integration with respect to z to give: α δ2 k2 f  = c f − f 2 , (6.72) 2 where the integration constant vanishes because both f, f  → 0 as |x| → ∞.

418

Nonlinear systems f(z) , gi(z) 1

0.8

0.6

0.4

0.2

–4

–2 Sech2

2 Sech

4

z

1/ (1+z2)

Fig. 6.15 Comparison of three simple pulse shapes: g2 (z) = sech2 (z), g1 (z) = sech(z) and f (z) = 1/(1 + z2 ).

There are only a few promising candidates for the pulse shape involving only elementary functions. An exponential function is not localized, because if it decreases exponentially for z > 0, it will increase exponentially for z < 0. A Gaussian function is localized, but differentiating it will generate additional factors that are polynomials of z. These factors are not present in the KdV equation. On the other hand, the derivative of an exponential function is still an exponential function (multiplied by an appropriate constant). So promising candidates are localized functions of exponential functions. The simplest candidates that are finite everywhere are the hyperbolic functions gn (z) = sechn (z) = [cosh(z)]−n .

(6.73)

Here z = kx is dimensionless, and the additional parameter k is an inverse distance. Now one can verify by direct differentiation that 2 2/n g ]. n ≡ dzz gn (z) = gn [n − n(n + 1)(gn )

(6.74)

So a second-order ODE of the type (6.72) where no fractional power of f appears can involve gn (x) with n = 1, 2 only. These “hyperbolic” pulse shapes are compared with another simple pulse shape in Fig. 6.15. It is interesting to note that without the nonlinear term, the solution of Eq. (6.74) would have been the exponential functions e±nz that are not localized pulses. So the nonlinear term serves the same purpose as a confining potential by shaping the wave form into a suitable pulse.

Solitons

419

For n = 2, the resulting nonlinear ODE 2 g 2 = 4g2 − 6g2 .

(6.75)

matches the structure of Eq. (6.72). Note that if g2 is a solution, the nonlinear DE is not solved by the function ag(z) with a  0, 1. g2 is thus a standing soliton. We therefore look for standing KdV solitons of the form f (z) = ag2 (z). Its substitution into Eq. (6.72) yields an equation for g2 (z) similar to Eq. (6.75). The coefficients c/(kδ)2 , −αa/2(kδ)2 appearing on the right are then matched to the equivalent coefficients 4, −6 of Eq. (6.75) to give √ k=

c , 2δ

a=

3c . α

(6.76)

a simple boost, x → x − ct, leads finally the moving KdV solitons ! √ 3c c 2 φ(x, t) = sech (x − ct) . α 2δ

(6.77)

This solution becomes infinite (or singular) when α → 0. It vanishes for all values of x − ct  0 when δ → 0 vanishes. Hence the soliton does not exist without both nonlinear and φ xxx terms, as we have already concluded in our previous qualitative analysis. The 1-soliton φ(x, t) has some unusual properties. Neither k nor c (or ω = kc) appear in the original KdV Eq. (6.66). They can therefore be chosen arbitrarily. There are thus infinity many such 1-solitons, with different velocities. Second, linear superpositions of 1-solitons with different k, c are not solutions of the KdV equation. One has to perform nonlinear superposition, a complicated matter that will be discussed in a Sections 6.9–10. Nonlinearity also prevents the overall height or amplitude of the soliton φ(x, t) to be varied arbitrarily once the parameters are chosen. In fact, the pulse height is proportional to the speed c, so that a taller pulse travels proportionally faster. Conversely, a slower pulse is shorter. The pulse stops moving when its height vanishes. Indeed, it has completely disappeared. All these speed/height correlations appear quite natural and reasonable. The mean-square width of the pulse is inversely proportional to c. So the taller and faster pulse is also narrower. Finally, the solution αφ of Eq. (6.77) is always positive. So φ has the same sign as α. Another 1-soliton solution can be built from h2 (kx) = csch2 (kx) = 1/sinh2 (kx). Unlike gn , the functions hn (z), n = 1, 2, are infinite at z = 0. Solutions of the the KdV equation of the form bh2 (kx) are singular. (See Problem 6.7.1.) They are not normally realized in most physical problems, but are needed for mathematical completeness and in the construction of waves containing two or more solitons.

420

Nonlinear systems

Another type of nonlinear wave equations can also be generated directly from the nonlinear ODE (6.74) for gn (z) with n = 1, 2. Let z = kx − ωt. Then   (gn ) xx − α2 (gn )tt = k2 − (ωk α)2 g n (z)   = k2 − (ωk α)2 Rn (gn ), (6.78) where Rn is the expression on the right of Eq. (6.74). The result is a nonlinear wave equation for all parameter values k2  (ωk α)2 . If k2 = (ωk α)2 , the wave equation reduces to the usual linear wave equation (6.67). 6.7.2

Conservation laws

Consider a mass distribution in a 1D space as described by the mass density ρ(x, t). The mass dm = ρdx located in the interval dx changes in time as d (dm) = (∂t + v∂ x )ρ(x, t)dx, dt

(6.79)

where v = dx/dt is the velocity of the mass dm. In the coordinate frame co-moving with dm, we have in Newtonian mechanics t = t, x = x − vt. Hence v = dx /dt = 0 and d d dm =  dm = ∂t ρ(x , t )dx = 0, dt dt

(6.80)

where the final result is obtained by noting that in the co-moving frame, the mass density distribution ρ(x , t ) is static and does not change in time if mass is conserved. Hence, the mass density ρ in an arbitrary frame satisfies the continuity equation for the conserved density ρ ρt + vρ x = ρt + J x = 0,

(6.81)

where J = vρ is the mass current density or flux. The local differential statement of mass conservation by the continuity equation can be restated as a global conservation law by integrating over all space  dM d ∞ ρ(x, t)dx = dt dt −∞ = −J(x = ∞) + J(−∞) = 0,

(6.82)

if there is no net inflow of mass at ∞, or J(∞) = J(−∞). Hence the total mass M of the system is conserved. The same procedure works for any PDE. To each continuity equation ρt + J x = 0 that can be constructed from the same PDE,there exists a corresponding global ∞ constant of motion or conserved quantity M = −∞ ρ(x, t)dx that does not change in time if there is no net inflow from ∞. For example, the dimensionless KdV equation ut − 6uu x + u xxx = 0

(6.83)

Solitons

421

can be rearranged readily to give two related continuity equations ut + (−3u2 + u xx ) x = 0, (u )t + (−4u + 2uu xx − 2

3

u2x ) x

= 0.

(6.84) (6.85)

The last two terms of Eq. (6.85) come from the term 2uu xxx . For water solitons, the wave function u describes the height  of the soliton above the water surface. Hence the first constant of motion M1 = udx is proportional to the soliton mass. The second constant of motion M2 = u2 dx is proportional to the soliton momentum if the second u is used to describe the soliton velocity, using the pulse height/speed connection present in the soliton. It turns out unexpectedly that the KdV soliton has infinitely many constants of motion, as one can show in Problem 6.7.9. The next constant does not come from (u3 )t alone because it does not satisfies a continuity equation:   9 4 3 2 (u )t = u − 3u u xx + 6uu x u xx , 2 x     9 1 1 (6.86) u3 + u2x = u4 − 3u2 u xx + 6uu2x + u2xx − u x u xxx . 2 t 2 2 x One has to add another term (u2x /2)t in order to produce a continuity equation, as shown explicitly in the second equation in Eq. (6.86). The resulting constant of motion M3 can be interpreted as the soliton energy. The physical meaning of the remaining infinitely many constants of motion is unknown.

Problems 6.7.1 (KdV scaling properties) Show that the following properties hold for the solution (6.77) and for all other solutions of the KdV Eq. (6.66): (a) φ(α) = (6/α)φ(6), (b) If f [k(x − ct)] is a solution of the KdV Eq. (6.66) with δ = 1, then the solution of the full KdV equation for any δ is f [(k/δ)(x − ct)]. Hint: Find the ODE satisfied by f (z), z = q(x − ct). 6.7.2 (Singular KdV 1-soliton) (a) If hn (z) = cschn (z) = 1/sinhn (z), show that   2 2/n . h n = dzz hn (z) = hn n + n(n + 1)(hn ) (b) Use the nonlinear ODE shown in part (a) with n = 2 to show that the the KdV equation (6.66) has another 1-soliton solution of the form ! √ 3c c 2 φ(x, t) = − csch (x − ct) . α 2δ

422

Nonlinear systems

Hint: You can match either to Eq. (6.72), or directly to the KdV equation after differentiating h n once more. The second alternative is probably easier. 6.7.3 (Modified KdV equation) (a) Verify Eq. (6.74). For n = 1, the equation reads 3 g 1 = g1 − 2g1 .

(b) Use the nonlinear ODE shown in part (a) to show that the 1-soliton solution of the modified KdV equation φt + α2 φ2 φ x + δ2 φ xxx = 0 that is finite for nonzero α is

√

6c φ(x, t) = sech α

! √ c (x − ct) . δ

This modified KdV equation appears in theories of electric circuits, plasma double layers and ion acoustic waves. 6.7.4 (Nonlinear Klein–Gordon equation) Use the nonlinear ODE (6.75) to show that ψ(x, t) = ag2 (kx − ωk t) is a solution of the nonlinear PDE α2 φtt − φ xx = β2 φ − γφ2 , for any real parameter k if a = 3β/2γ and ω2k = (β2 + 4k2 )/4α2 . 6.7.5 (Another simple 1-soliton) (a) Show that the function f (z) = 1/(1 + z2 ) satisfies the nonlinear ODE f  = 6 f 2 − 8 f 3 . (b) Use this ODE to show that ψ(x, t) = a f (kx − ωk t) is a solution of the nonlinear PDE α2 φtt − φ xx = β2 φ2 − γφ3 , for any real parameter k if a = 4β2 /3γ and ω2k = [2β4 /(9γ2 ) + k2 ]/α2 . 6.7.6 (Modified KdV equation) Show that the modified KdV equation vt − 6v2 v x + v xxx = 0 satisfies the following continuity equations for conserved densities: vt = (2v3 − v xx ) x , (v2 )t = (3v4 − 2vv xx + v2x ) x , (v4 + v2x )t = (4v6 − 4v3 v xx + 12v2 v2x + v2xx − 2v x v xxx ) x .

Solitons

423

The modified KdV equation, like the KdV equation, has infinitely many other constants of motion. 6.7.7 (Peregrine equation) Show that the Peregrine equation ut − uu x − u xxt = 0 satisfies the following continuity equations for conserved densities: ut = (u2 /2 + u xt ) x , (u2 + u2x )t /2 = (u3 /3 + uu xt ) x , (u3 )t /3 = (u4 /4 + u2 u xt − u2t + u2xt ) x . These are the only known conservation laws of the Peregrine equation. 6.7.8 (KdV equation) Show that the KdV equation ut − 6uu x + u xxx = 0 satisfies the following continuity equation where the conserved density depends explicitly on x, t: (xu + 3tu2 )t = (3xu2 − xu xx + u x + 12tu3 − 6tuu xx + 3tu2x ) x . 6.7.9∗ (Gardner series for the KdV equation) The KdV equation can be shown to contain infinitely many conservation laws as follows: (a) (Gardner equation) Let u = w + εw x + ε2 w2 , where ε is an arbitrary real number. Show that K u ≡ ut − 6uu x + u xxx

= (1 + ε∂ x + 2ε2 w)G w, G w ≡ wt − 6(w + ε2 w2 )w x + w xxx .

Hence u satisfies the KdV equation K u = 0 if w satisfies the Gardner equation G w = 0, but not necessarily vice versa. Hint: Show that −6uu x = −6(1 + ε∂ x + 2ε2 w)[(w + ε2 w2 )w x ] − 6ε2 wx wxx . (b) The Gardner equation G w = 0 satisfies a continuity equation of the form wt + J x = 0, J = w xx − 3w2 − 2ε2 w3 .

424

Nonlinear systems

(c) (Gardner series) Expand w = u − εw x − ε2 w2 ∞  = εn wn (u) n=0

into an infinite series in ε. Show that w0 = u, w1 = −w0x = −u x , w2 = −w1x − w20 = u xx − u2 , wn = uδn0 − wn−1 x −

n−2 

wi wn− j−2 ,

i=0

where wi = 0, if i < 0. (d) Write the conserved current J=

∞ 

εn Jn (u)

n=0

as an infinite series in ε. Show that J0 = w0xx − 3w20 , J1 = w1xx − 6w0 w1 , J2 = w2xx − 3w21 − 6w0 w2 − 2w30 , Jn = wnxx − 3

n 

wi wn−i − 2

i=0

n−2 

wi w j wn−i− j−2 .

i, j=0

The infinitely many conservation laws that appear are thus wnt + Jnx = 0,

n = 0, 1, 2, ...

(See Drazin, 1984, pp. 34–5.)

6.8

Traveling kinks

Flood waves, called hydraulic jumps in engineering, are formed when water is suddenly released from dams or in irrigation canals. The free surface of the released water has a recognizable front that separates the region to be flooded from that already flooded. The free surface of the flood is usually undulatory, but can be turbulent near the front. Tsunamis are hydraulic jumps in the ocean caused by earthquakes. They

Traveling kinks

425

Ki (z) 1

0.8

0.6

0.4

0.2

z –4

2

–2 Tanh

ArcTan

4

ArcTanExp

Fig. 6.16 Comparison of three simple kink shapes: K1 (z) = [tanh(z) + 1]/2, K2 (z) = [(2/π) f (πz/2) + 1]/2, and K3 (z) = (2/π) f [exp (πz/2)], where f (z) = arctan(z).

can travel as fast as many airplanes, but not as fast as sound. A tidal bore is a flood of sea water at high tide that rolls upstream into a river when the sea level is higher than the river level. The world’s largest bore appears at the mouth of the Quintang River in China. It can be as high as 30 feet, and travel upstream as fast as 25 mph. A traveling kink soliton is an idealized traveling jump in the shape of a smoothed step function that is unchanged in shape as the wave front travels in time. 6.8.1

Nonlinear wave equations

Three simple kink shapes with the same slope at z = 0 and the same asymptotic values of 1 (0) as z → ∞ (−∞) are compared in Fig. 6.16. The functions h(z) = tanh(z), θ(z) = arctan(z), ψ(z) = arctan(ez )

(6.87)

appearing in these kinks satisfy the nonlinear ODE: h (z) = −2h + 2h3 , 

(6.88)

θ (z) = − sin(2θ) cos θ, 1 ψ (z) = sin(4ψ), 4 2

respectively. They can all be turned into nonlinear wave equations.

(6.89) (6.90)

426

Nonlinear systems

Indeed, a wave pulse is the spatial derivative of a wave kink: h (z) = [cosh(z)]−2 = g2 (z), 1 , 1 + z2 ψ (z) = g1 (z)/2. θ (z) =

(6.91)

A related pair of pulse and kink satisfy related NLPDEs. We shall illustrate how the transformation to a PDE can be achieved for ψ. Suppose we want a wave equation similar to the left side of Eq. (6.67). Trying a solution of the form φ(x, t) = aψ(kx − ωk t), we find by direct substitution  a φ xx − α2 φtt = k2 − (ωk α)2 sin(4ψ) 4 = β2 sin(γφ),

(6.92)

where the middle expression has been obtained by using Eq. (6.90). The final expression is chosen (with β  0) to have the form of a nonlinear PDE called the sine– Gordon (sG) equation. Its name refers to the fact that for small γφ, the final expression on the right can be linearized to β2 γφ to give the Klein–Gordon equation of relativistic quantum mechanics. The same nonlinear PDE had already been studied in the 1880s in connection with transformations of surfaces in differential geometry. The sG equation (6.92) can be solved by comparing the coefficients of the middle and final expressions. The result is " # φ(x, t) = a arctan ekx−ωk t , a=

4 , γ

ω2k =

k2 − β2 γ . α2

(6.93)

Since the parameter k does not appear in the sG equation, the assumed functional form gives a family of 1-kink solutions for any choice of k that will give ω2k > 0. If ωk = 0, the kink is time-independent. It is then stationary, and does not travel. Let us use the notation that both k and ωk can be positive or negative, but c = |ωk /k| ≥ 0. The solutions obtained then have the form φ± (x, t; k) =

4 ψ[k(x ∓ ct)], γ

(6.94)

where γ > 0 is assumed. The solutions φ+ (x, t; k) [φ− (x, t; k)] describe waves traveling to the right [left]. Kinks are those with the step rising to the right, or k > 0, as shown in Fig. 6.16, while antikinks are those with the step rising to the left, with k < 0. For flood water, the wave motion must proceed in real time in the direction of high to low water levels. So a kink of flood water must move to the left in real time, but an antikink moves to the right because the force of gravity operating here gives preference to the

Traveling kinks

427

downward direction. In other properties where there is no directional preference, all four solutions may be realized naturally. Under time reversal (t → −t) alone, φ+ (x, t; k) is transformed into φ+ (x, −t; k) = φ− (x, t; k), so that a kink of flood water appears to move to the right in an unnatural way. What is seen then must be a movie of a real flood run backward. Both the natural real-time description and the unnatural time-reversed movie are contained in our mathematical solutions (6.94). All these solutions appear automatically because the PD operator ∂ xx − α2 ∂tt appearing in the PDE is even in t and unchanged under time reversal. However, the solution φ+ (x, t; k) does not have the time-reversal symmetry of the PDE, and changes instead to another solution φ− (x, t; k). The sG equation is also even in the space variable x, but the solutions are not. The solutions φ± (x, t; k) change under space inversion (x → −x) instead to φ± (−x, t; k) = φ∓ (x, t; −k).

(6.95)

So the mathematical solutions (6.94) can describe both superficially different situations seen by two observers facing each other watching the same flood kink pass by between them. A flood kink moving naturally to the left of one observer becomes an antikink moving naturally to the right of the other observer. 6.8.2

Evolution equations

Evolution equations are PDEs first order in the time derivative. Traveling kinks appear in many nonlinear evolution equations used to describe shock waves in fluids, chemical reactions and population dynamics. One such PDE is the dimensionless Burgers equation ut − u xx + uu x = 0,

(6.96)

It is a diffusion equation, an evolution PDE where the highest linear spatial derivative is ∂ xx . For simple traveling kink waves of the form u(z = kx − ωt), the PDE simplifies to the (still nonlinear) ODE −ωu − k2 u + ku u = Dz u(z = kx − ωt) = 0,

(6.97)

where u = (d/dz)u(z). We are interested in finding the kink solutions u(z). Note first that the tanh kink has the useful property that its mth derivative h(m) = m (d /dzm )h(z) is a polynomial of h of degree m + 1: h(z) = tanh(z) : h = 1 − h2 , h = −2h(1 − h2 ), . . .

(6.98)

This polynomial property makes it possible to look for solutions of the ODE (6.97) by matching polynomials in h. The tanh kink h goes from –1 for z  0 to 1 for z  0. If one is looking for a kink between 0 and 2, it has to be moved up or down by 1. A popular choice is the antikink function

428

Nonlinear systems

fn (z) = (1 − h)n

(6.99)

for a positive integer power n. fn is a polynomial of h of degree n. On applying d d d = h = (1 − h2 ) dz dh dh

(6.100)

to f (z), we see that d/dh acting on f (h) decreases its polynomial degree M by 1, but h = 1 − h2 increases it by 2, leading to a net gain of 1. Hence fn(m) (z) is a polynomial in h of degree n + m. Specifically fn = −n(1 − h)n (1 + h), fn = n(1 − h)n (1 + h)[n − 1 + (n + 1)h].

(6.101)

We now look for a solution of Eq. (6.97) of the form u = a fn by direct substitution of the necessary polynomials fn(m) . The first issue is to decide the power n to use. This can be done by examining the highest power (or polynomial degree) J of h in each term of the nonlinear ODE. The highest power for the first or linear term in u is J = n + 1 , for the second or linear term in u is J = n + 2, and for the third or nonlinear term u u is JNL = 2n + 1. The highest powers of the two linear terms are always different, thus showing that the linear diffusion PDE without the nonlinear term does not have a kink solution. When the nonlinear term is present, a solution is possible if its highest power matches the larger highest power Jd = n + 2 of the linear derivative terms, here u . We therefore need JNL = 2n + 1 = Jd = n + 2,

(6.102)

i.e., n = 1. No other choice of n will give a solution. Thus the stability of kink or shock waves in nonlinear diffusion equations comes from the cancellation between the nonlinear spatial dispersion and the dispersion caused by the restoring force present in the u term. To see if a solution is possible even for n = 1, we need the polynomial expansion of the ODE (6.97) Dz u = a

J NL =3

b j h j = 0.

(6.103)

j=0

Direct subsitution gives the matching (b3 , b2 , b1 , b0 ) = (−ka + 2k2 , −ω + ka, ka − 2k2 , ω − ka) = 0.

(6.104)

These four algebraic equations can be solved to give a = 2k,

ω = 2k2 ,

(6.105)

Traveling kinks

429

with each of the two distinct conditions given twice. Hence the solutions are   (6.106) u(x, t) = 2k 1 − tanh(kx − 2k2 t) for any choice of the inverse wavelength k. The velocity of the kink front in this solution is v = 2k. The wave form is an antikink for any k, positive or negative. The k > 0 waves traveling to the right are the physical solutions if u(x, t) describes the height of gravity waves on the surface of water. The k < 0 solutions are then the unphysical solutions. 6.8.3

Finite power-series methods

If one does not know what function f (h) to try for a given nonlinear ODE, one must look instead for the more general finite power-series solution f (h) =

M 

am hm .

(6.107)

m=0

Then the method is call a tanh method. After a change of variable z → h(z) = tanh(z), f ( j) (h) = (d/dz) j f (h) becomes a polynomial of h of degree M + j. The maximum power M needed in the solution series can be determined by matching the highest power JNL of the nonlinear terms (if more than one is present) to the highest power Jd = M + d of the highest order d of the the linear derivatives. For example, for the integrated ODE (6.72) derived from the KdV equation, d = 2 and JNL = 2M. Hence there are solutions for Jd = M + d = JNL = 2M.

(6.108)

So the choice is M = d = 2. The remaining steps of the power-series expansion then proceed in the standard way for power-series expansions, namely by writing the ODE as the algebraic polynomial equation Dz f (h) =

JNL 

b j h j = 0.

(6.109)

j=0

The resulting simultaneous nonlinear algebraic equations can be solved for the solution coefficients a j . The calculation of b j and the extraction of aj are straightforward but always a bit tedious. The process can be simplified slightly by writing the solution as u(z) = a f (z), as we have done previously. Then a0 = 1 can be chosen in f , and Eq. (6.109) simplifies to Dz f (h) = a

JNL  j=0

bj h j = 0,

(6.110)

430

Nonlinear systems

where the overall factor a appears again inside the sum in the nonlinear terms. The coefficients a j that appear now are actually the ratios a j /a0 of the original coefficients. Equation (6.110) can often be simplified by factoring out a power of 1 − h before the summation over the bj coefficients. If the ODE is very complicated, the complete power-series solution can be done by using a computer algebra program such as Mathematica or Maple. The tanh method can be generalized to other functions y(z): f (y) =

M 

am ym .

(6.111)

m=0

With d dy d = y , y = , dz dy dz    2 2 d   d  d =y y 2 +y , ..., dz dy dy

(6.112)

the power-series method will work if y itself is a polynomial of y: y =

J 

cjy j.

(6.113)

dy y

(6.114)

j=0

Then  z(y) =

can be evaluated and inverted to get y(z). For example, J J J J

= 1, c0 = 0, c1 = 1 : y = ez ; = 2, c0 = c1 = 0, c2 = −1 : y = 1z , = 2, c0 = c2 = 1, c1 = 0 : y = tan z, = 2, c0 = −c2 = 1, c1 = 0 : y = tanh z.

(6.115)

In fact, the indefinite integral z(y) can be expressed as a sum of J logarithmic functions. To simplify the result, it is now convenient to choose c J = 1 instead of c0 = 1 so that 

y = (y − y1 )(y − y2 )... =

J ' i=1

(y − yi ),

(6.116)

Traveling kinks

431

where y j is the j-th root of the algebraic equation y (y) = 0. Then 1  γi = , y i=1 y − yi J

z(y) =

J 

γi ln (y − yi ),

(6.117)

i=1

where the complex coefficients γi = γi ({c j }) can be calculated from the c j coefficients of Eq. (6.113). The analytic inversion of the solution z(y) to y(z) is in general nontrivial, but it is not needed in finding solutions of the finite power-series method. Even the solutions a f (z) of the nonlinear PDE can be plotted numerically in a computer without ever expressing y(z) in terms of known functions. This saves a lot of unnecessary work.

Problems 6.8.1 (ODEs) Verify that the functions h, θ, ψ satisfy the ODEs shown in Eqs. (6.88–6.90). 6.8.2 (Translation property) (a) Let ψ(z) be the function defined in Eq. (6.87). Find the ODE satisfied by ψ(z − z0 ). (b) Find the ODE satisfied by arctan (bez ). 6.8.3 Verify Eq. (6.115). 6.8.4 (Fisher’s NL diffusion equation) (a) The Fisher equation describing chemical reactions and population dynamics is the nonlinear diffusion equation ut − u xx − (1 − u)u = D u = 0. Show that there might be nontrivial tanh kink solutions of the form f2 = (1 − h)2 , where h = tanh(z) and z = kx − ωt, defined in Eq. (6.101). (b) Show that these tanh kink solutions satisfy a nonlinear ODE that can be written as the algebraic equation in powers of h: Dz u = a(1 − h)

2

2 

b j h j = 0,

j=0

where the coefficients are (b2 , b1 , b0 ) = (−6k2 + a, 2ω − 8k2 − 2a, 2ω − 2k2 + a − 1). Verify that the solutions are given by the unique choice of parameters a = 1/4, k2 = 1/24, ω = 5/12.

432

Nonlinear systems

Hence Fisher’s equation has the unique traveling kink solution  !2 5 1 1 u(x, t) = 1 − tanh √ x − t . 4 12 2 6 6.8.5 (Burgers’ NL wave/diffusion equation) (a) The Burgers equation is the nonlinear wave/diffusion equation ut + uu x − u xx = D u = 0. Show that there might be nontrivial tanh kink solutions of the form f1 = (1 − h), where h = tanh(z) and z = kx − ωt, defined in Eq. (6.101). (b) Show that these tanh kink solutions satisfy a nonlinear ODE that can be written as the algebraic equation in powers of h: Dz u = a(1 − h)

2 

b j h j = 0,

j=0

where the coefficients are (b2 , b1 , b0 ) = (−2k2 + ka, ω − 2k2 , ω − ka). Verify that the solutions are given by the family of parameters a = 2k, ω = 2k2 = ka. Hence Burgers’ equation has the family of traveling kink solutions of different k   u(x, t) = 2k 1 − tanh(kx − 2k2 t) . Explain why there is a solution for every real k  0. Note: The Burgers’ equation contains a diffusion term −u xx added to the simplest first-order “quasi-linear” wave equation ut + uu x = 0.

6.9

Nonlinear superposition of solitons

The sum of any two solutions of a nonlinear PDE is not another solution of the same PDE because of the presence of the nonlinear term. The sum is in general a solution of another nonlinear PDE that sometimes differs only slightly from the original PDE. The situation is of some interest physically in connection with the superposition of solitons. Mathematically, however, one is more interested in the situation where the same nonlinear wave equation has solutions containing multiple solitons in addition to solutions containing single solitons. We shall describe both situations in some detail in this section. 6.9.1

Superposed solitons satisfying different nonlinear wave equations

The wave function

Nonlinear superposition of solitons

433

ψ1 (x, t) = arctan[y1 (x, t)], y1 (x, t) = ekx sech(ωt),

(6.118)

for k, ω > 0 at t = 0 shows an arctan (or a standing sine-Gordon, sG) kink with a spatial shape that centers at x = 0 where ψ1 = π/4. The kink falls to 0 as x → −∞ and rises to π/2 as x → ∞. The traveling property of this arctan kink can be made explicit by writing y1 (x, t) =

2ekx−ωt . 1 + e−2ωt

(6.119)

The function differs from the corresponding function of the traveling sG kink given in Eq. (6.93) by the extra factor 2/(1 + e−2ωt ). As t increases from t = 0, the wave front of kx − ωt = 0 moves to the right with velocity c = ω/k. However, the outside factor 2/(1 + e−2ωt ) increases by ultimately a factor of 2 as t → ∞. As a consequence, the kink center (kc), if defined by ψ = π/4 or ykc = 1, lags more and more behind this wave front as t increases. We may call this kink a shifting kink. The shift is always within the kink, and therefore never by a large amount. We shall ignore this small shift in all qualitative discussions below. ψ1 is even in time. The behavior from t = −∞ to t = 0 can be obtained by running the movie of the kink from t = 0 to t = ∞ backward. This runback movie shows a kink moving in from x = ∞ with velocity −c. The movie ends with the kink centered at x = 0 at t = 0. Thereafter, with the movie run forward, the kink changes direction and moves out with velocity c to x = ∞. The wave function ψ2 (x, t) = arctan[−y1 (−x, t)] is a negative arctan antikink that is obtained from ψ1 by changing its sign and by a mirror reflection or space inversion x → −x. It describes a negative antikink moving in from x = −∞ at t = −∞ with velocity c. Its center reaches x = 0 at t = 0. At that moment, it bounces back suddenly and then travels with velocity −c back to x = −∞ at positive times. Each of these kinks does not satisfies the sG nonlinear PDE (6.92), because of the additional time dependence in y1 , but they are soft kink solitons satisfying another nonlinear wave equation. Ignoring their softness, each of these kinks is a patchwork of an sG kink traveling in one direction at t < 0 that suddenly reverses direction at t = 0 to travel in the opposite direction at t > 0. Adding these two kinks together in the form ψ12 (x, t) = arctan[y1 (x, t) − y1 (−x, t)],

(6.120)

gives us a wave function containing two kinks (more precisely, a kink and a negative antikink) that come together, meet at x, t = 0 and then move apart. They appear to bounce from each other, but this property is already built in because each kink will bounce in the absence of the other, as we have already seen. The two-kink soliton ψ12 also satisfies a nonlinear wave equation, but it is not the same as the wave equation each satisfies. So overlapping solitons will generate an interaction. The wave equation is also not in general an sG equation.

434

Nonlinear systems

ψ

–7.5

–5

1.5

1

1

0.5

0.5

–2.5

b = 0.5

ψ

1.5

2.5

5

7.5

x

–7.5

–5

–2.5

2.5

–0.5

–0.5

–1

–1

–1.5

–1.5

b = 1.0

b = 5.0

t=0

t=5

5

7.5

x

t = 10

Fig. 6.17 Shapes of (a) two-kink nonlinear waves of Eq. (6.121) at t = 0 for b = 0.5, 1.0, 1.5, and (b) the Perring–Skyrme 2-kink for t = 0, 5, 10.

Consider next the more general 2-kink wave function ψ2 (x, t) = arctan {(bc/2)[y1 (x, t) − y1 (−x, t)]} = arctan[bc sinh(kx)sech(ωt)],

(6.121)

where c = |ω/k|, and b is a real parameter. Fig. 6.17(a) shows that the wave functions at t = 0 have the same values at x = 0 and at large |x| for any b. Elsewhere in x, they change only a little for relatively large changes of b. However, wave functions with different values of b will satisfy different nonlinear wave equations. For the two choices b = ±1, the wave equation turns out to be the sG equation, as one can verify with a computer. This result was first discovered by Perring and Skyrme (PS) in 1962 from numerical calculations. Fig. 6.17(b) shows how the PS kinks (for b = 1) separate as t increases from t = 0. We would like to find out next why the wave equation satisfied by the PS 2-kink is an sG equation. First recall that a change of the strength bc/2 can be visualized readily in term of a constant shift in x to x s = x + (1/k) ln (bc/2) for y1 (x, t) alone. At t = 0 when the motion reversal occurs, the kink center xc = −(1/k) ln (bc/2) is negative if bc > 2 and positive if bc < 2. For y1 (−x, t), x s = x − (1/k) ln (bc/2) and the kink center at t = 0, the instant of motion reversal, is at xc = (1/k) ln (bc/2) instead. So the amount of overlap of the two colliding kinks and therefore their nonlinear interaction increases with increasing b. We now need to see why the nonlinear interaction for the choice b = ±1 is just right for the two kinks to satisfy the sG equation in all space and for all times! This is a very detailed requirement that can only be answered by a full mathematical analysis. We shall use the opportunity to treat the problem in a more general context.

Nonlinear superposition of solitons

6.9.2

435

Separation of variables for the sine-Gordon equation

Consider a solution ψ(x, t) = arctan [y(x, t)] of the dimensionless sG equation ψ xx − ψtt −

1 sin(4ψ) = 0. 4

(6.122)

Here ψ x = ∂ x ψ, the x, t variables are dimensionless, y = tan ψ, and 1 y(1 − y2 ) . sin(4ψ) = 4 (1 + y2 )2

(6.123)

In terms of y, dψ/dy = 1/(1 + y2 ). The sG equation can then be changed to        2 y x y2t 1 y xx ytt 1 +1 (6.124) − 2 2 − 2 − 2 − 1 = 0. − y y y2 y y y The variables x, t in y can be separated if y(x, t) = f (x)/g(t) :

then

yx f x y xx f xx = , = ; y f y f g2 gt ytt gtt yt =− , = − + 2 t2 . y g y g g The sG equation then simplifies to   f xx gtt 2 2 − 2 f x2 − 2g2t + f 2 − g2 = 0, + (f + g ) f g

(6.125)

(6.126)

where only the first term is a function of both x and t. Operating on each of the remaining terms with ∂ xt will give nothing. The first term too will become separable in x and t (or expressible as a sum of functions of only one variable x or t) if ∂ xt acting on it gives nothing too:     gtt f xx 2 2 ( f )x + (g )t = 0. (6.127) g t f x So separability requires that (Lamb, 1980)     1 1 gtt f xx =− 2 = 2μ, ( f 2 )x f x (g )t y t where the separation constant μ is independent of both x and t.

(6.128)

436

Nonlinear systems

Each of the separated DEs can be integrated once with respect to x or t to give the DEs f xx = 2μ f 3 + c1 f, gtt = −2μg3 + d1 g.

(6.129)

These DEs can be used to eliminate fxx , gtt from Eq. (6.126). The result can again be separated in their x, t dependences using another separation constant ν: f x2 = μ f 4 + k2 f 2 + ν, g2t = −μg4 + ω2 g2 − ν;

(6.130)

where 1 k2 = (c1 + d1 + 1), 2 ω2 = k2 − 1.

(6.131)

The parameters k, ω are integration constants. The solutions x( f ) and t(g) can be expressed in terms of the following elliptic integrals  x=

f (x) f (0)

 t=

&

df μ f 4 + k2 f 2 + ν

g(t) g(0)

&

,

dg −μg4

+ ω2 f 2 − ν

.

(6.132)

Some special cases are worthy of note. When μ = ν = 0, we recover the known solutions f = e±kx , g = e±ωt . The PS 2-kink solution is obtained by the choice μ = 0, ν = α2 : α α f (x) = ± sinh (kx), g(t) = ± cosh (ωt) : k ω ψPS (x, t) = arctan[c sinh(kx)sech(ωt)],

(6.133)

with ψ actually independent of α. These solutions can be verified by direct substitution into Eq. (6.130). We are now in a position to answer the question posed before this subsection. Why is b = ±1 needed for arctan [bc f (x)/g(t)] to becomes an sG wave function? The answer is that when b  ±1, the variables x, t do not separate in the sG PDE. Returning to the separated solutions of the sG equation, we note that for μ = 0, ν = −α2 , the solution can be obtained from Eq. (6.133) by the interchange sinh ↔ cosh. f (x) = ±(α/k) cosh (kx) is now even in x, while g(t) = ±(α/ω) sinh (ωt) is odd in t.

Nonlinear superposition of solitons

437

ψ 1.75 1.5 1.25 1 0.75 0.5 0.25

–10

–5

ωt = 0.2

5

ωt = 1.2

10

x

ωt = 5.0

Fig. 6.18 Comparison of the shapes of the gusher ψgush (x, t) of Eq. (6.135) for c = ω/k = 0.5, √ k = 1/ 1 − c2 , and ωt = 0.2, 1.0, 5.0.

The awkwardness of 1/g(t) near t = 0 can be avoided by using the cotangent relation to write ! π f (x) = − ψc (x, t), ψ(x, t) = arctan g(t) 2 ! g(t) ψc (x, t) = arctan (6.134) f (x) in terms of its complementary angle ψc (x, t). Any complementary function is interesting by itself because it is also a solution of the sG equation. (Verify this statement.) The wave shapes of ψc , ψgush (x, t) = arctan [(k/ω)sech(kx) sinh (ωt)],

(6.135)

for three different snapshots of ωt > 0 are shown in Fig. 6.18. There is nothing at t = 0. As t increases, a bell-shape pulse materializes at x = 0. It rises and broadens in time until the top reaches the limiting value of π/2. Thereafter the flooding continues as the kinky sides travel towards x = ±∞. The flooding at a depth of π/2 is complete as t → ∞. The description of the motion of this “gusher” for t < 0 is left as an exercise. The gusher is actually a 2-kink soliton because of the two kinks traveling to (or from) x = ∞ once it reaches its limiting top (or bottom) value. It has a relatively long interacting time near t ≈ 0, with the kinks annihilating each other completely at t = 0.

438

Ω2

Nonlinear systems

The special case where ω2 < 0 (or equivalently k2 < 1) is worthy of note. Then = −ω2 > 0 and g(t) = (α/Ω) sin (Ωt), without a factor of i, to give ψbreath (x, t) = arctan [(k/Ω)sech(kx) sin (Ωt)],

(6.136)

The spatial dependence is a single bell-shape pulse even in x centered at x = 0 for all t. Its height changes periodically with t, since g(t) providing an oscillatory time dependence. The resulting soliton ψc is called a stationary breather because of its height varies periodically between ± limits, but its center is fixed at x = 0. This stationary breather becomes a breather traveling with velocity v when seen by an observer traveling with velocity −v under an appropriate space-time transformation x, t → x , t that would preserve the PD operator ∂ xx − ∂tt = ∂ x x − ∂t t

(6.137)

appearing in the dimensionless sG equation. Here ∂xx = ∂2 /∂x2 . The needed transformation is the pseudo-Lorentz transformation (pLT) appropriate to the native wave speed cn = 1 of the dimensionless linear wave equation (∂ xx − ∂tt )φ(x, t) = 0: x = γ(x − vt), t = γ(t − vx),

(6.138)

with γ2 (1 − v2 ) = 1. This space-time transformation is mathematically similar to the Lorentz transformation, but the wave is not in general a light wave. The principle of relativity, stating that all space-time frames give equivalent descriptions, implies that if ψ(x , t ) is an sG solution in one spacetime frame, there must exist a solution ψ(x , t ) in another spacetime frame. Hence a traveling observer sees a stationary breather as the traveling breather (tb)  , k sin [Ωγ(t − vx)] ψtb (x, t) = arctan . (6.139) Ω cosh [kγ(x − vt)] 6.9.3

Rational linearization of nonlinear DEs

The linearization of certain nonlinear DEs can be achieved by using rational transformations u(z) = g(z)/ f (z). The procedure is described by examples. Example 6.9.1 Nonlinear Riccati ODE Consider the quadratic DE u = a(z) + 2b(z)u + u2 ,

(6.140)

where u = (d/dz)u(z). With u(z) = g(z)/ f (z), and therefore u = (g f − f  g)/ f g, the ODE can be rearranged first to the bilinear equation g f − f  g = a f 2 + 2bg f + g2 .

(6.141)

Nonlinear superposition of solitons

439

It can next be separated non-uniquely into two sides (or terms) (g − a f ) f = ( f  + 2b f + g)g.

(6.142)

This separation of terms is valid for any z. Hence both sides must be equal to a common function of z that can be written conveniently as λ f g, where λ = λ(z). The common separation function allows the result to be expressed as two coupled linear equations g − a f − λg = 0, f  + 2b f + g − λ f = 0.

(6.143)

are solutions of the simultaneous LDEs (6.143), so will One can show that if (gi , fi )  (g, f ), where g = ni gi , f = ni fi .  Example 6.9.2 Nonlinear Burgers equation Consider the PDE ut (x, t) = u xx + 2uu x

(6.144)

that has been used to describe gas dynamics and traffic flow. A functional change u(x, t) = w x (x, t) gives the DE w xt = w xxx + 2w x wxx . The result can be integration in x to wt = w xx + w2x + a(t),

(6.145)

where the integration constant a(t) is a function only of t. A final simplification is obtained by the functional change w = ln f . With wt = ft / f , w x = f x / f (= u), and w xx = ( f xx f − f x2 )/ f 2 , we find by simple substitution the linear PDE ft = f xx + a(t) f.

(6.146)

This n means that if fi (x, t) are distinct solutions, the linear superposition f (x, t) = i fi (x, t) is also a solution. Hence the function u(x, t) = ∂ x [ln f (x, t)], called a Cole– Hopf transformation, is a solution of the original nonlinear Burgers equation that contains a kind of nonlinear superposition of solutions.  The Cole–Hopf transformation can be used in reverse to change a linear PDE into a nonlinear one, as we shall describe in the next section.

Problems 6.9.1 (Time dependences of two 2-kink sG solutions) (a) Describe in words the time dependence for t < 0 of the complementary angle ψc (x, t) shown in Fig. 6.18 for t > 0.

440

Nonlinear systems

(b) Sketch the breather ψbreath (x, t) of Eq. (6.136) for c = Ω/k = 0.5 and three choice of Ωt = (0.1, 0.2, 0.5)π. Use these sketches to describe in words its time dependence from t  0 to t  0. 6.9.2 (Pseudo-Lorentz transformation) If ψ(x, t) is a solution of the sine-Gordon equation, verify the following results: (a) The complementary angle ψc (x, t) = π/2 − ψ(x, t) is also a solution. (b) Verify directly that ∂ xx − ∂tt = ∂ x x − ∂t t if x = γ(x − vt), t = γ(t − vx), and γ2 (1 − v2 ) = 1. (Note that if the soliton ψ(x , t ) travels with the velocity c  0, its total traveling velocity in the old pseudo-Lorentz frame (x, t) is not c + v, as part (c) shows.) (c) If ψ(x , t ) = φ(x − c t ) is a solution, then φ[a(x − ct)] is also a solution if x , t are those defined in part (b), a = γ(1 + c v), and c = (c + v)/(1 + cv) is the pseudo-Lorentz velocity addition formula. (d) Show that the inverse of the space-time transformation of part (b) is x = γ(x + vt ), t = γ(t + vx ). If ψ(x, t) = φ(x − ct) is a solution of the sG equation, then φ[a (x − c t )] is also a solution if a = γ(1 − cv), and c = (c − v)/(1 − cv). (Parts (c) and (d) together establish the relativity principle that all pseudo-Lorentz frames are equivalent for the description of sG wave motions.) Note: Parts (b, c, d) are related.

6.10

More general methods for multi-solitons∗

There exist a number of more general and systematic methods for finding multisoliton solutions of nonlinear PDEs. In this section, we shall describe two methods that can be understood in relatively simple mathematical terms, although the execution of these methods may not always be elementary. 6.10.1

B¨acklund transformation

For the purpose of our discussion, a B¨acklund transformation (BT) u(x, t) → v(x, t) of the solution u(x, t) of a PDE D1 u = 0 gives the solution v(x, t) of another PDE D2 v = 0. Here D1 is a PD operator that depends on u itself if the PDE is nonlinear. It is obvious that for any known BT u → v, the PDE D2 v = 0 can be found by direct substitution. This is how we can find out easily how some PDEs are related to one another. The problem becomes nontrivial if the unknown BT is to be found that will satisfy a specified PDE D2 v = 0. If D2 = D1 , the transformation is called a self or auto BT. The self BT can be used to construct a second solution from a known solution u of a given PDE. This can be done readily by looking for another solution of the form v(x, t) = w(x, t) + u(x, t). If the given PDE is nonlinear, its nonlinearity ensures that v and w are not both solutions in addition to u. This is because a linear superposition of solutions is not a solution of a nonlinear DE. Although w does not satisfy the original nonlinear PDE if v does, it satisfies a generally nonlinear PDE D3 w = 0 that can be

More general methods for multi-solitons

441

found by direct substitution. Solution of this PDE will complete our construction. We shall illustrate this application of the BT. Example 6.10.1 Nonlinear transformation of a linear PDE The simplest BTs are those where linear PDEs are transformed into nonlinear ones by nonlinear transformations u = f (v). Consider, for example, the linear diffusion equation ut = σu xx ,

(6.147)

where σ is the diffusion constant. If u(x, t) = ev(x,t) , the diffusion equation is turned into the nonlinear PDE vt = σ(v xx + v2x )

(6.148)

that is also a Burgers equation. The result can readily be verified by direct differentiation. The linear diffusion equation has the known solutions u(x, t) = √

A

2 /(4σt)

4πσt

e−x

,

(6.149)

where A is an arbitrary constant. Hence the nonlinear Eq. (6.148) has the solutions v(x, t) = B −

x2 1 ln t − , 2 4σt

(6.150)

where B is an arbitrary constant, as one can verify directly.  Example 6.10.2 Second solution from a known solution We turn next to finding a second solution v = w + u, where u is a known solution of yet another Burgers equation ut = σu xx − uu x

(6.151)

by using the self BT. Direct substitution shows that w satisfies the PDE wt = σw xx − w(w x + u x ) − uwx .

(6.152)

This looks complicated, but as we might have suspected, it can be simplified by a Cole–Hopf transformation w(x, t) = b∂ x (ln f ). The result can be reduced to another linear diffusion equation ft = σ f xx − u(x, t) f x + a(t) f,

(6.153)

when b = −2σ is used. (The reduction can readily be done using the hints given in Problem 6.9.3. It involves an x integration where a(t) appears as an integration “constant”.)

442

Nonlinear systems

For the special choice of the trivial solution u = 0 in Eq. (6.153) and an integration function a(t) = 0, f (x, t) is just the simple Gaussian diffusion function (6.149). Hence one solution of the Burgers equation (6.151) is x v(x, t) = w(x, t) = −2σ∂ x (ln f ) = .  t

(6.154)

A general B¨acklund transformation for a second order PDE can be specified in terms of the first order derivatives of the new solution v: vξ = vξ (u, v, uξ , uη , ξ, η), vη = vη (u, v, uξ , uη , ξ, η),

(6.155)

where ξ, η are linear functions of x, t chosen to simplify the original PDE. Such a BT will allow v to be found by integration, as we shall see for the sG equation. We begin by expressing the sG equation in the so-called characteristic form by the change of variables to ξ = (x − t)/2, ∂ x = (∂ξ + ∂η )/2,

η = (x + t)/2 : ∂t = (−∂ξ + ∂η )/2,

∂ xx − ∂tt = ∂ξη .

(6.156)

1 sin (4u). 4

(6.157)

Hence the sG equation reads uξη = The BT λ sin [2(v + u)], 2 1 sin [2(v − u)], vη = −uη + 2λ vξ = uξ +

(6.158)

where λ is an arbitrary real parameter, has the interesting property that vξη =

1 sin (4v), 4

(6.159)

as one can easily verify by direct substitution. So Eq. (6.158) is a self BT. Since we already know that sG solutions are of the form v = arctan w, further simplification can be realized by a functional change to w = tan v (Seeger et al., 1953). The left sides of the BT (6.158) can be written as vξ = (cos2 v)wξ , etc. The expressions on the right can be simplified after separating the two angles v, u as a sum of terms proportional to sin2 v, sin v cos v, or cos2 v. Dividing through by cos2 v and rearranging, we find

More general methods for multi-solitons

443

wξ = αw2 + βw + γ,

where

wη = α w2 + β w + γ , , α λ = uξ ∓ sin 2u, β = λ cos 2u, γ 2 ,  α 1 sin 2u, β = λ−1 cos 2u. = −uη ±  γ 2λ

(6.160)

(6.161)

This self BT can be used to generate a sG solution from the trivial solution u = 0. Then α = γ = α = γ = 0. The resulting first order DEs is readily integrated directly to give λξ = ln w + c(η), λ−1 η = ln w + d(ξ).

(6.162)

The integration constants c(η), d(ξ) are functions of the passive variables η and ξ, respectively. Comparison of the two solutions shows that c(η) = −ηλ − a, d(ξ) = −λξ − a, where a is a constant. So the sG solution found is " # v(x, t) = arctan bekx−ωt , (6.163) where

λ 1 + , 2 2λ & 1 λ = k2 − 1. ω= − 2 2λ b = ea ,

k=

This is just the 1-kink solution obtained in Section 6.8.1. Given a solution u = un−1 , a new solution v = un can be found by a repeated application of the BT (6.160). The integration of the Eqs. (6.160) to known functions is not easy, but if it can be done, we have a solution of the form Fn (ξ, η, wn ) = C,

(6.164)

where C is an integration constant. However, the PDEs (6.160) can always be solved numerically. It is obvious that the discovery of a BT for a nonlinear PDE gives a key to unlock many of its solutions. The BT of Eq. (6.158) was found by A. V. B¨acklund in the 1880s in connection with transformations in the differential geometry of surfaces of Gaussian curvature –1. 6.10.2

Hirota’s bilinear superposition of solitons

Many nonlinear PDEs D1 u = 0 can be transformed into a bilinear form D2 f · f = 0 under the transformation u(x, t) → f (x, t). The bilinear form has the unusual property that it allows the direct superposition of nonlinear solutions. For this reason, it is usually called a “direct method”.

444

Nonlinear systems

We shall begin by defining bilinearity obtained by a transformation of the Kordeweg–de Vries (KdV) equation ut + 6uu x + u xxx = 0.

(6.165)

Note that we have followed Hirota by using a form of the KdV equation where the nonlinear term carries a positive sign instead of the negative sign used in a previous section. This change of notation will make it easier to make contact with Hirota’s results (for example, in Hirota 2004). The transformation needed for the KdV equation and for many other PDEs as well is the logarithmic transformation   fx u = 2(ln f ) xx = 2 : f x   f xt f − f x ft , ut = 2 f2 x   DKdV f · f ut + 6uu x + u xxx = . (6.166) f2 x In the third equation, the result 2 DKdV f · f = 2( f xt f − f x ft + 3 f xx + f xxxx f − 4 f xxx f x )

(6.167)

is obtained by the direct substitution of u, u x , u xxx into the KdV PD equation. The calculation will be left as an exercise. Eq. (6.167) can be integrated with respect to x to the bilinear PDE DKdV f · f = c f 2 .

(6.168)

The term bilinearity refers to the fact that f or its derivatives appear twice in every term of the PD operator DKdV f · f . The solutions can be separated into two distinct groups according to whether the constant c of integration is zero. The solutions for c = 0 are much easier to find, and include the soliton solutions obtained previously and their multi-soliton generalizations. We shall restrict our study to only the c = 0 solutions. The bilinear differential operator in Eq. (6.168) can be written compactly as DKdV f · f = D x (Dt + D3x ) f · f,

(6.169)

by using the Hirota derivatives n Dm t Dx f · g ≡

lim (∂t − ∂t )m (∂ x − ∂ x )n f (x, t)g(x , t )

t →t,x →x

n = (−1)m+n Dm t D x g · f.

(6.170)

More general methods for multi-solitons

445

These are unusual non-local operators that are evaluated locally only after all differentiations. The following examples show how they actually work: Dx f · g = fx g − f gx , D2x f · g = D x ( f x g − f g x ) = f xx g − 2 f x g x + f g xx , D3x f · g = f xxx g − 3 f xx g x + 3 f x g xx − f g xxx , D x Dt f · f = D x ( ft f − f ft ) = 2 ( f xt f − f x ft ) , # " 2 . D4x f · f = 2 f xxxx f − 4 f xxx f x + 3 f xx

(6.171)

The last two equations involving f · f can be used to verify the Hirota form (6.169) of the KdV equation. Let the function f (x, t) be expanded into the infinite Hirota series f = 1 + ε f1 + ε2 f2 + ...,

(6.172)

where ε is an arbitrary real parameter. Then f · f = 1 + ε( f1 · 1 + 1 · f1 ) + ε2 ( f2 · 1 + f1 · f1 + 1 · f2 ) +ε3 ( f3 · 1 + f2 · f1 + f1 · f2 + 1 · f3 ) + . . . The KdV equation for c = 0 DKdV f · f =

∞  m=0

⎡ ⎤ m  ⎢⎢⎢ ⎥⎥⎥ εm ⎢⎢⎢⎣DKdV fi · fm−i ⎥⎥⎥⎦ = 0

(6.173)

(6.174)

i=0

becomes an infinite series in ε that must be satisfied for each power of ε. If the infinite series terminates at a finite power M, it will give an exact solution. The first term (m = 0) of the infinite series of PDEs is trivially satisfied, because f0 = 1. The next term is the linear PDE 2D x (Dt + D3x ) f1 · 1 = 0.

(6.175)

Its solutions are f1 (x, t) = eηi , ηi = ki x − ωi t + δi , ωi = ki3 .

(6.176)

This will give 1-soliton solutions, as we shall confirm shortly. Since f1 satisfies a linear PDE, the linear superposition of N solutions f1 =

N  i=1

eηi ,

(6.177)

446

Nonlinear systems

where the superposition coefficients have been absorbed into the strength parameters δi in ηi , is also a solution for f1 . Thus the key to Hirota’s direct construction of N-soliton solutions is the presence in the bilinear series (6.173) of a linear “window” f1 · 1 where the usual linear superposition of solutions is possible. Of course the KdV equation is nonlinear. Its nonlinearity shows up in the nonlinearity of the higher-order PDEs of the Hirota series (6.174). To see how these PDEs are handled, let us first complete the 1-soliton solution (6.176). It involves the ε2 PDE 2DKdV f2 · 1 = −DKdV f1 · f1 = 0.

(6.178)

The zero on the right is obtained from the Hirota identity n η1 η2 m n η1 +η2 Dm t D x e · e = (ω2 − ω1 ) (k1 − k2 ) e

(6.179)

when applied to the f1 · f1 term with η2 = η1 when only one soliton is present. The PDE for f2 can be satisfied by the trivial solution f2 = 0. Eq. (6.173) then shows that fi = 0, i ≥ 3, can be chosen as well. This gives the exact solution f = 1 + eηi , where ε has been absorbed into δi . The resulting 1-solitons are from Eq. (6.166) 2 ( f xx − f x2 ) f2   k2 2k2 eηi 2 ηi = . sech = 2 2 (1 + eηi )2

u1 = 2(ln f ) xx =

(6.180)

Note that the k of this section is half the k value defined by Eq. (6.76). Two-solitons come from the linear superposition f1 = eη1 + eη2 . They give the exact solutions f = 1 + eη1 + eη2 + eA12 +η1 +η2 , eA12 =

(k1 − k2 )2 . (k1 + k2 )2

(6.181)

The derivation of this result will be left to Problem 6.10.6. Fig. 6.19 shows a Hirota 2-soliton at three different times. Both solitons are traveling to the right, but the taller, narrower pulse (with k2 = 1.4) is faster, as we already know. It overtakes the shorter pulse (with k1 = 1.0) at t  0. Numerical calculations like those used to plot Fig. 6.19 greatly simplify all but the simplest nonlinear studies. Computer algebra too will make the manipulation of long expressions involving many PD operations less arduous. The serious student of nonlinear systems will sooner or later delegate all tedious algebra to the computer. However, analytic insights are also important in the conceptual understanding of the less familiar nonlinear techniques and results. To illustrate the point, let us explain why the unusual D operators appear when nonlinear PDEs are bilinearized (Hirota 2004).

More general methods for multi-solitons

447

u(x)

0.5 0.4 0.3 0.2 0.1

–20

–10

10

20

x

–0.1 t = –6

t=0

t=6

Fig. 6.19 One soliton passing another in Hirota’s two-soliton wave of Eq. (6.181), using k1 = 1.0, k2 = 1.4 for t = −6, 0, 6.

For one variable x, the D operator is Dnx f (x) · g(x) ≡ lim (∂ x − ∂ x )n f (x)g(x )  x →x

= (−1)n Dnx g · f.

(6.182)

The difference between Dx and ∂ x can be seen clearly in the translation formulas for Taylor expansions: eδ∂ x f (x) = [1 + δ∂ x + ...] f (x) = f (x + δ), eδ∂x f (x)g(x) = f (x + δ)g(x + δ) = (eδ∂ x f )(eδ∂x g).

(6.183)

In contrast, the D operator “knows” that the g factor comes from the denominator of a rational expression: eδDx f (x) · g(x) = lim [eδ∂x f (x)][e−δ∂ x g(x )]  x →x

= f (x + δ)g(x − δ).

(6.184)

448

Nonlinear systems

Rational expressions are generated by the logarithmic transformation u(x) = 2[ln f (x)] xx , as shown explicitly in Eq. (6.166). A generating function 2 cosh (δ∂ x ) ln f = ln [cosh (δD x ) f · f ]

(6.185)

relates the two types of differential operators for this log transformation. The formula holds because its LHS is (eδ∂x + e−δ∂ x ) ln[ f (x)] = ln f (x + δ) + ln f (x − δ) = ln [ f (x + δ) f (x − δ)].

(6.186)

The ∂nx u terms can be expressed in terms of D x operations on f · f by Taylor expanding both sides of Eq. (6.185). The Taylor expansion on the right can be shown (Problem 6.10.9) to be ln [cosh (δD x )F] = ln F +

δ2 2 δ4 [FD4x F − 3(D2x F)2 ] + . . . , (6.187) Dx F + 2!F 4!F 2

where F = f · f . Consequently one can generate from the RHS u= u xx

D2x F , F

 2 2 D F D4x F −3 x = ,... F F

(6.188)

The presence of other variables t, y, . . . can be handled in a similar way. Using these and other u derivatives, a nonlinear PDE can be bilinearized and the superposition of nonlinear solitons realized. Bilinearization can be achieved using other transformations, but the resulting bilinearized PDEs turn out to be more complicated that those from the logarithmic transformations (Hirota 2004).

Problems 6.10.1 (Reduction to Eq. (6.153)) (a) Eliminate from Eq. (6.152) the w function and all its partial derivatives in favor of f and its partial derivatives by using the relations w = b f x / f , wt = b( f xt f − f x ft )/ f 2 , etc. Remove a common factor b/ f from every term. (b) Four of the terms that appear can be simplified to L = f xt − σ f xxx + f x u x + f xx u = ψ x , ψ = ft − σ f xx + u f x .

More general methods for multi-solitons

449

The remaining terms can be moved to the right side of the equation and simplified by choosing b = −2σ to R=

fx ψ. f

(c) The resulting DE L = R can be integrated to ln ψ = ln f + c(t), or ψ = a(t) f, where a(t) = ec(t) . This is the PDE to be derived. 6.10.2 Verify Eq. (6.159), vξη = 14 sin (4v). 6.10.3 (B¨acklund transformation for harmonic functions in the xy plane) Consider two spatial variables x, y instead of x, t. A certain function u(x, y) is given. Its BT to new functions v(x, y) is known to be v x = −uy ,

vy = u x .

(6.189)

(a) By eliminating u, show that v satisfies the Laplace equation ∇2 v = 0 in the xy plane. Solutions of Laplace equations are called harmonic functions. Show that u itself is also a harmonic function. (b) Let f (z) = u(x, y) + iv(x, y) define a function of a complex variable (FCV) z = x + iy. Show that the BT arises from the requirement that the complex derivative df = f x = −i fy dz is independent of the direction in the complex plane (x, iy) along which the d f /dz is defined. The BT (6.189) itself is called a Cauchy– Riemann condition in the theory of FCV. 6.10.4 (Miura transformation) Let u = v2 + v x . Show that K u ≡ ut − 6uu x + u xxx

= (2v + ∂ x )K M v, K M v ≡ vt − 6v2 v x + v xxx .

Hence u satisfies the KdV equation K u = 0 if v satisfies the modified KdV equation KM v = 0, but not necessarily vice versa. Hint: Identify and simplify the terms contributing to ∂ x K M v, where ∂ x = ∂/∂x.

450

Nonlinear systems

6.10.5∗ (B¨acklund transformation for the Liouville equation) Any function u(x, y) = f (x) + g(y) satisfies the PDE u xy = 0. Consider the BT √ v x = − f x + 2e(v− f −g)/2 , √ vy = gy + 2e(v+ f +g)/2 . (a) Show that the new functions vx , vy satisfy the Liouville equation vxy = ev . (This PDE appears in the description of the metric of 2D surfaces.) A general solution of the Liouville equation will be found in the remaining parts of the problem. (b) Show that the BT can be re-written in the form √ (v+ f −g)/2 2e = e f (v + f − g) x = (v + f − g)X , √ (v+ f −g)/2 2e = e−g (v + f − g)y = (v + f − g)Y , where dx/dX = e f , dy/dY = e−g . (c) Show that the ODEs in part (b) can be integrated to √ − 2e(v+ f −g)/2 = X(x) + c(y), = Y(y) + d(x). These two solutions can be combined to give √ − 2e(v+ f −g)/2 = X(x) + Y(y) + a,  where X(x) = e− f (x) dx,  Y(y) = eg(y) dy, and a is an integration constant independent of x, y. (d) Show that the Liouville equation has solutions of the form √ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ 2 ⎢ ⎥⎦. v(x, y) = g(y) − f (x) + 2 ln ⎣− X(x) + Y(y) + a (See Drazin, 1984, pp. 109-10.) 6.10.6 (KdV equation) Verify Eq. (6.167) by direct substitution and simplification. Hint: Derive and use the expression 2 2 f + 12 f xx f x2 f − 6 f x4 )]. uxxx = 2∂ x [(1/ f 4 )( f xxxx f 3 − 4 f xxx f x f 2 − 3 f xx

Tables of mathematical formulas

451

6.10.7 (KdV 2-solitons) (a) Verify Eq. (6.181) for 2-solitons by showing that f2 satisfies the PDE −2DKdV f2 · 1 = 2(k1 − k2 )[ω2 − ω1 + (k1 − k2 )3 ] eη1 +η2 . Then find the solution aeη1 +η2 , where a = (k1 − k2 )2 /(k1 + k2 )2 . (b) Show that f3 , f4 satisfy the PDEs −2DKdV f3 · 1 = 2DKdV f2 · f1 = 0, −2DKdV f4 · 1 = DKdV f2 · f2 = 0, where the zeros on the right have to be verified by using Eq. (6.179). Hence show that the Hirota series contains only terms m ≤ M = 2. 6.10.8 (Kadomtsev–Petviashvili equation) 2D KdV solitons in shallow water are described by the KP equation (ut + 6uu x + u xxx ) x + 3uyy = 0. (a) Show that under the logarithmic transformation u(x, y, t) = 2[ln f (x, y, t)] xx , the KP equation can be written in the Hirota form [D x (Dt + D3x ) + 3D2y ] f · f = 0. (b) Show that the N-soliton solution first-order in the Hirota series is f1 =

N 

eηi ,

i=1

ηi = ki x + i y − ωi t, ωi = ki3 + 3i2 /ki . 6.10.9 (Generating function) Verify the generating function expansion (6.187) by an explicit Taylor expansion or any other method. 6.10.10 (Bilinearization by rational transformations) Consider the rational transformation u(x) = f (x)/g(x). The generating formula for the u derivatives turns out to be   exp (δD x ) f · g δ∂ x f e = . g exp (δD x )g · g Derive this identity by making explicit the result of the translation operations on both sides of the equation.

Appendix 6 Tables of mathematical formulas 1 Nonlinear instabilities Landau equation:

x˙(t) = bx − x3 + p.

452

Nonlinear systems

Equilibrium/fixed points at x˙e = 0 are attractors (repellors) if equilibrium is stable (unstable). Pitchfork bifurcation occurs if a point attractor at xe splits into two attractors on a change of parameter. Hopf resonator: is described by the complex solution z(t) = R(t)eiθ(t) of the nonlinear DE: z˙(t) = (μ + iω0 )z − |z|2 z + Feiωt . Hopf bifurcation occurs √ when the point attractor Re = 0 expands into a circular attractor of radius Re = μ > 0, when F = 0. If F  0, the particular solution is z(t) = (Reiφ )eiωt , with % F = R (R2 − μ)2 + (ω − ω0 )2 , tan φ = −

ω − ω0 . R2 − μ

Gain G = ω0 R/F is maximal (Gmax = ω0 F −2/3 ) at resonance frequency ω = ω0 . 2 Logistic map xn+1 = rxn (1 − xn ) = f (xn ) maps interval 0 ≤ x ≤ 1 into itself for 1 ≤ r ≤ 4. Equilibrium/fixed points where f (xe ) = xe are xe = 0, 1 − 1/r, . . . Scaling fractal dimension at a Feigenbaum point: Dδ = log 2/ log δ. Mapping f (x) is stable if its Lyapunov exponent 1 ln | f  (xi )| < 0. n→∞ n i=0 n−1

λ = lim

3 Strange attractors H´enon map:

a 2D map containing parameters a, b: (xn+1 , yn+1 ) = (yn + 1 − axn2 , bxn )

Straight line y = c + sx maps to a parabola parallel to the x-axis: K x − X = − (y − Y)2 , 2   s2 sb (X, Y) = c + 1 + , , 4a 2a

K =

2a . b2

Tables of mathematical formulas

Equilibrium/fixed points are xe1,e2

1 = −(1 − b) ± 2a

%

(1 −

b)2

! + 4a ,

453

ye1,e2 = bxe1,e2 .

Differential line segments change on mapping as       dxn+1 dxn −2axn 1 =M , M= . dyn+1 dyn b 0 The Jacobian matrix M has eigenvalues λ± = −axe ±

%

(axe )2 + b.

Strange attractors are associated with an unstable equilibrium point where the two Jacobian eigenvalues are of the type |λ> | > 1 (unstable mapping) and |λ< | < 1 (stable mapping). They occur for parameters where the unstable mapping in the direction of its eigenvector is chaotic. 4 Driven pendula Externally-driven dissipative nonlinear (EDNL) pendulum: satisfies an inhomogeneous DE: θ¨ + γθ˙ + ω20 sin θ = 2A cos ωt. Parametrically-driven dissipative nonlinear (PDNL) pendulum: satisfies a homogeneous DE: θ¨ + γθ˙ + (ω20 + 2A cos ωt) sin θ = 0. Nonlinear instabilities can be studied numerically. Pendula are linear if oscillation amplitudes are sufficiently small when sin θ → θ. Analytic solutions can be found: ED linear pendulum: Homogeneous solution (A = 0): θh (t) = e−βt θ1 (t),

β=

θ1 (t) = θ0 e±iω1 t ,

ω1 =

γ ; 2 %

ω20 − β2 .

Particular solution (A  0): θ p (t) = Deiωt , D=

A (ω20

−

ω2 )

+ i2βω

= |D|e−iδ .

454

Nonlinear systems

|D(ω)|2 is maximal at the resonance frequency ωR = width at half maximum) is Γ ≈ γ. PD linear pendulum: θh (t) = e−βt θ1 (t),

β=

% ω20 − 2β2 . Its FWHM (full

γ ; 2

θ1 (t) ≈ eσt (a0 cos ω1 t + b0 sin ω1 t),

σ=

A . 2ω1

θ1 (t) satisfies the Mathieu equation (with ωd = ω/2) θ¨1 + (ω21 + 2A cos 2ωd t)θ1 = 0. Linear parametric resonance appears when θ(t) becomes unstable (unbounded as t → +∞). This occurs when σ > σc = β, or equivalently A > Ac = γω1 . Mathieu functions: satisfy the standardized Mathieu equation y (x) + (a + 2q cos 2x)y(x) = 0, a homogeneous LDE with a period-π coefficient. It has Floquet solutions with the repetition property F ν (x + π) = kFν (x),

or

Fν (x) = e P(x), iνx

where k = eiνπ , and P(x) is periodic with period π. Mathieu cosine and sine functions are real, and even and odd, respectively, in x. 5 Solitons 1-soliton solutions of the Korteweg–de Vries equation ut − 6uu x + u xxx = 0 : c u(x, t) = − sech2 2

! c (x − ct) , 2

√

for any c.

1-soliton solutions of the modified Korteweg–de Vries equation vt − 6v2 v x +v xxx = 0 : √ √ v(x, t) = − c sech[ c(x − ct)], Continuity equation for a conserved density ρ: ρt + J x = 0.

for any c.

Tables of mathematical formulas

455

Both KdV equations have infinitely many conserved densities expressible as a Gardner series ρ = w ≡ u − εu x − ε2 w2 ∞  = εn wn (u); n=0

J = w xx − 3w2 − 2ε2 w2 ∞  = εn Jn (u); n=0

wnt + Jnx = 0. The function w(x, t) satisfies the Gardner equation wt − 6(w + ε2 w2 )w x + w xxx = 0. 6 Kinks 1-kink solutions of the Burgers equation ut − u xx + uu x = 0 : u(x, t) = 2k[1 − tanh(kx − 2k2 t)],

for any k.

1-kink solutions of the sine–Gordon equation ψxx − ψtt = 14 sin (4ψ) : " # ψ(x, t) = arctan ekx−ωk t , with ω2k = k2 − 1 for any k. The complementary angle ψc = π/2 − ψ(x, t) is also a soliton. Tanh method: If u = u(z), z = kx − ωt, PDE D u = 0 → ODE Dz u = 0. Solutions are often expressible as u(z) = f [h(z)] =

M 

am hm (z),

m=0

with h(z) = tanh z. The finite maximum power M is obtained by matching the highest power from a nonlinear term to the highest power from a linear derivative term. Power-series method: More generally, u(z) = f [y(z)] =

M 

am ym (z),

m=0

where

z(y) =

J  i=1

γi ln (y − yi ).

456

Nonlinear systems

7 Superposition of solitons Superposed solitons do not generally satisfy the same PDE. However, the original PDE may have solutions containing two or more solitons. Sine–Gordon equation: has separable solutions ψ(x, t) = arctan [ f (x)/g(t)] including the Perring–Skyrme, gusher, and breather solutions (all with ω2 = k2 − 1): ψPS = arctan [(ω/k) sinh (kx)sech(ωt)], ψgush = arctan [(k/ω)sech(kx) sinh (ωt)], ψbreath = arctan [(k/Ω)sech(kx) sin (Ωt)]. Pseudo-Lorentz transformation: ∂ x x − ∂t t = ∂ xx − ∂tt ⇒

x = γ(x − vt), t = γ(t − vx), % γ = 1/ (1 − v2 ); c = (c − v)/(1 − cv).

So the sG equation has solutions of the form (tb = traveling breather) 

ψkink = arctan [ek γ(x−ct) ],

c = (c + v)/(1 + c v),

ψtb = arctan [(k /Ω )sech[k γ(x − vt)] sin [Ω γ(t − vx)]]. Rational linearization: The Riccati ODE u (z) = a(z) + 2b(z)u(z) + u2 (z) for solutions of the rational form u = g/ f separates into two coupled linear ODEs g − a f − λg = 0 f  + 2b f + g − λ f = 0. The solution pairs ( fi , gi ) can be linearly superposed. Cole–Hopf transformation: u(x, t) = [ln f (x, t)] x for the Burgers equation ut = u xx + 2uu x gives the linear PDE ft = f xx + a(t) f , where a(t) is an integration constant for an x-integration. 8 More general methods B¨acklund transformation (BT): When u(x, t) → v(x, t), the PDE D1 u = 0 → D2 v = 0. If D2 = D1 , the BT is a self or auto BT.

Tables of mathematical formulas

Sine–Gordon equation:

In the characteristic form

1 sin (4u), 4

uξη =

457

1 ξ = (x − t), 2

1 η = (x + t), 2

the sG equation has a self BT λ sin [2(v + u)], 2 1 vη = −uη + sin [2(v − u)]. 2λ vξ = uξ +

Hirota’s bilinear superposition of solitons: If u(x, t) = 2(ln f )xx , many nonlinear PDEs D1 u = 0 → bilinear PDE D2 f · f = 0. KdV equation ut + 6uu x + u xxx = 0 becomes D x (Dt + D3x ) f · f = 0, n Dm t Dx f

·g≡

where lim (∂t − ∂t )m (∂ x − ∂ x )n f (x, t) · g(x , t )

t →t,x →x

n = (−1)m+n Dm t D x g · f.

If f = 1 + ε f1 + ε2 f2 + . . . , then Df · f =

∞  m=0

⎡ m ⎤ ⎢  ⎥⎥⎥ ε ⎢⎣D fi · fm−i ⎥⎥⎥⎦ = 0 m ⎢⎢⎢⎢

i=0

must of ε. First-order PDE D f1 · 1 = 0 is linear. So f1 =  ηi be satisfied for every power e , ηi = ki x − ωi t, ωi = ki3 , can be a linear superposition of 1-soliton solutions. Higher-order PDEs take care of the nonlinearity. Hirota 2-solitons for the KdV equation: f = 1 + eη1 + eη2 + eA12 +η1 +η2 , eA12 = (k1 − k2 )2 /(k1 + k2 )2 . Translation properties: eδ∂x f (x)g(x) = f (x + δ)g(x + δ) = (eδ∂ x f )(eδ∂x g). [eδ∂x f (x)][e−δ∂ x g(x )]. eδDx f (x) · g(x) = lim  x →x

= f (x + δ)g(x − δ) Generating function: for ∂nx u(x) for the logarithmic transformation u(x) = 2[ln f (x)] xx is 2 cosh (δ∂x ) ln f = ln [cosh (δD x ) f · f ]

7 Special functions 7.1

Introduction

A function is a rule defining the relation between one variable and another. Among the oldest functions is the sine function, which relates the length of a chord to the half-angle of the arc on a circle. A table of chords could already be found in a book compiled by Ptolemy of Alexandria in the second century. Most functions are of relatively recent origin, however. The logarithmic function was invented in the early seventeenth century, and the exponential function in the late seventeenth century. The idea of a power is older, but its algebraic representation began with Descartes in the early seventeenth century. Indeed our modern understanding of functions can be said to begin with Descartes. It gained much cohesion with the publication of the first treatise on functions by Euler in 1748. Lagrange, Fourier, Cauchy and Riemann, among others, gave the subject its modern form in the early and mid-nineteenth century. Special functions are functions in frequent use in the mathematical and physical sciences. Functions of particular interest in physics are those that describe simple systems. They are almost always the solutions of differential equations—the equations of motion or of state of simple systems. Perhaps not surprisingly, the most useful functions are those associated with the description of wave motion.

7.2

Generating function for Legendre polynomials

We have seen in Eq. (5.136) that an axially symmetric solution of the Laplace equation ∇2 u = 0 regular at the poles of the sphere and finite at infinity is just the “Coulomb” potential r−1 . In Section 5.13 we have seen that this is actually the solution of the inhomogeneous Poisson equation ∇2 u(r) = ρ(r) for a δ-function, or point, source ρ(r) = −4πδ(r).

(7.1)

Generating function for Legendre polynomials

459

Suppose the source is placed not at the origin, but at the point r s = (0, 0, 1) on the z axis. The resulting Coulomb potential  1 1 Al rl Pl (cos θ) u(r) = = = |r − rs | |r − ez | l=0 ∞

(7.2)

will still be axially symmetric about the z axis, but not spherically symmetric about the origin. It is finite at the origin where u(0) = 1; hence near the origin it can be written in the form of Eq. (7.2), according to the method of separation of variables. To determine the coefficients Al , we examine the field along the z axis. Since cos θ = 1 and all Pl (1) = 1 along the z axis, we have u(r = rez ) =

∞ 

Al rl .

l=0

Now the field along the z axis between the origin and the source is just 1/(1 − r), according to one of the middle expressions in Eq. (7.2). Therefore u(r) =

1 1−r

= 1 + r + r2 + . . . =

∞ 

rl .

(7.3)

l=0

This shows that Al = 1 for all l. The series shown in Eq. (7.3) is convergent for r < 1. At r = 1, that is, at the source position r = rs , it becomes infinite or singular. We shall see in Chapter 8 that such a singular point, or singularity, of a function is a source of the function all over space, in the same sense that a point charge is a source of the electrostatic potential. The series shown in Eq. (7.3) is divergent for r > 1. This means that the power series in r shown on the right-hand side of Eq. (7.2) is valid only for r < 1, although the Coulomb potential on the left-hand side is valid for all r. Now that Al is determined, we have for r < 1 and any angle θ (1 − 2r cos θ + r2 )−1/2 =

∞ 

rl Pl (cos θ).

l=0

A function like this, written in the form 2 −1/2

G(x, t) = (1 − 2xt + t )

=

∞ 

tl Pl (x),

t < 1,

(7.4)

l=0

in which a set of functions, here Pl (x), appears as the coefficients of the power-series expansion in t is called their generating function (GF). The name arises from the fact that the function Pl (x) can be generated from G(x, t) by differentiation

460

Special functions

1 ∂l Pl (x) = G(x, t) l! ∂tl

! (7.5) t=0

A GF is very usefull. It can be used to find the analytic form of the functions themselves: P0 (x) = G(x, t = 0) = 1, ! ∂ = x, P1 (x) = G(x, t) ∂t t=0 1 P2 (x) = (3x2 − 1), 2 1 P3 (x) = (35x4 − 32x2 + 3), . . . . 8 It can also be used to derive many properties of the functions it generates. For example, the GF in Eq. (7.4) has the property G(−x, t) = (1 + 2xt + t2 )−1/2 = G(x, −t). As a result, ∞  l=0

tl Pl (−x) =

∞ 

(−t)l Pl (x) =

l=0

∞ 

tl (−1)l Pl (x).

l=0

Hence Pl (−x) = (−1)l Pl (x).

(7.6)

Thus Legendre polynomials of even (odd) degrees are said to have even (odd) parity. Legendre polynomials of different degrees are related to each other. For example, since P0 (x) = 1, we have P1 (x) = x = xP0 (x).

(7.7)

Relations between these functions of different degrees, or between functions and their derivatives, are called recursion formulas, or recurrence relations. These can be derived readily from the GF in Eq. (7.4). As illustrations, we shall derive the following two recursion formulas: (2l + 1)xPl (x) = lPl−1 (x) + (l + 1)Pl+1 (x),

(7.8)

xPl (x) −

(7.9)

lPl (x) =

Pl−1 (x),

where the primes denote differentiations with respect to x. Eq. (7.8) involves no derivatives of Pl but there are changes in the degree of the polynomials, that is, in the power of t in Eq. (7.4). This suggests that we have to differentiate G(x, t) with respect to t:

Generating function for Legendre polynomials

461

  ∂G x−t = l tl −1 Pl (x). = ∂t (1 − 2xt + t2 )3/2 l =0 ∞

That is, (x − t)G(x, t) = (1 − 2xt + t ) 2

≡

∞ 



l tl −1 Pl (x)

l =0

= (x − t )

∞ 



tl Pl (x).

l =0

We now rearrange the power series so that they will eventually carry the same power t . As a visual aid, terms that will be added together are underlined the same way: 

(2l + 1)xPl t = l

∞ 



(l +

 1)Pl tl +1

l =0

l=0

=

∞ 

+

∞  l =0

lPl−1 t + l

l=1

= P1 +

∞ 



l Pl t l −1 ≡

(l + 1)Pl+1 tl

l=0 ∞ 

[lPl−1 + (l + 1)Pl+1 ]tl .

l=1

Equating coefficients of tl , we obtain Eq. (7.7) for l = 0, and Eq. (7.8) for l > 0. Equation (7.8) is very useful for calculating the numerical values of a Legendre polynomial. This is because, if we know Pl−1 (x) and Pl (x) for a given value of x, the next Legendre polynomial must be Pl+1 (x) = [(2l + 1)xPl (x) − lPl−1 (x)]/(l + 1).

(7.10)

Since we already know P0 (x) = 1 and P1 (x) = x, all other Legendre polynomials can thus be generated by successive applications of Eq. (7.10). This is particularly easy to do on a digital computer. When a recursion formula is used like this in the direction of increasing degree, it is called a forward recursion formula. (Occasionally a recursion formula is used backwards in the direction of decreasing degrees in order to improve numerical precision by minimizing roundoff errors in a computer.) For Eq. (7.9), it is obvious that a differentiation with respect to x is needed: t ∂ t ∂ = G(x, t) = G(x, t), 2 3/2 ∂x x − t ∂t (1 − 2xt + t ) or t

∂ ∂ G(x, t) = (x − t) G(x, t). ∂t ∂x

462

Special functions

In terms of the power series in tl , this equation reads ∞ 

Pl ltl = (x − t)

l=0

∞ 

Pl tl

l=0

=

∞ 

xPl tl

−

l=0

∞ 

Pl−1 tl .

l=1

Hence Eq. (7.9) follows for l > 0. For l = 0, the relation obtained is xP0 = 0, which can be considered to be a special case of Eq. (7.9) if P−1 (x) = 0 is used. The Legendre polynomials form a family of orthogonal polynomia satisfying the orthogonality relation  1 2 Pl (x)Pl (x)dx = (7.11) δll . 2l + 1 −1 This relation can also be derived from the GF by squaring it and integrating with respect to x:  1  1 2 G (x, t)dx = (1 + t2 − 2tx)−1 dx −1

−1

1 = [ln(1 + t) − ln(1 − t)] t ∞  2 2l = t , 2l + 1 l=0

(7.12)

where use has been made of the Taylor expansion ln(1 + t) =

∞ 

tn (−1)n+1 . n n=1

The same integral can be expressed in terms of integrals of the Legendre polynomials, if one uses the right-hand side of Eq. (7.4):  1  1 ∞ ∞    G2 (x, t) dx = tl+l Pl (x)Pl (x) dx. (7.13) −1

l=0 l =0

−1

The double sum in Eq. (7.13) can be made equal to the single sum in Eq. (7.12) only if the integral of two Legendre polynomials is proportional to δll . Equation (7.11) then follows on equating coefficients.

Generating function for Legendre polynomials

463

Pn(x) 1

P2

0.5

P3

–1

P4

0

P5

1

x

–0.5

Fig. 7.1 Legendre polynomials Pn (x), n = 2 − 5 (from Abramowitz and Stegun).

Explicit formulas for some Pl (x) can also be found in Section 4.9. The polynomials P2 to P5 are plotted in Fig. 7.1.

Problems 7.2.1 Use the generating function of Eq. (7.4) to derive the Legendre differential equation ! 2 d 2 d (1 − x ) 2 − 2x + l(l + 1) Pl (x) = 0. dx dx Hint: Repeated use of the identity t∂t G = (x − t)∂ xG, where ∂t = ∂/∂t. 7.2.2 Obtain the recursion formula Pl+1 (x) − Pl−1 (x) = (2l + l)Pl (x) by using the generating function. Hint: Start by showing that (1 − 2xt + t2 )∂ xG = tG, where ∂ x = ∂/∂x. 7.2.3 Obtain the recursion formula (1 − x2 )Pl (x) = l[Pl−1 (x) − xPl (x)].

464

Special functions

Hint:See Hint for 7.2.2. 7.2.4 Calculate P4 (x) at x = 0.1, 0.5, −0.1, by using a forward recursion formula.

7.3

Hermite polynomials and the quantum oscillator

A family of functions can be defined by the differential equations they satisfy (Section 5.12). They can be defined by means of a generating function (Section 7.2). Functions can also be defined in terms of a differential formula called a Rodrigues formula. For example, the Rodrigues formula for Hermite polynomials is Hn (x) = (−1)n e x

2

d n −x2 e , dxn

n = 0, 1, 2, . . .

(7.14)

Direct evaluation of this formula gives H0 (x) = 1,

H1 (x) = 2x,

H2 (x) = 4x2 − 2, . . . .

The generating function for Hn (x) can itself be deduced from Eq. (7.14). This can be done by first introducing a “dummy” variable t into Eq. (7.14) to generate the necessary powers of t: ! ! n ∂n 2x−t2 x2 ∂ −(x−t)2 e = (e ) Hn (x) = e ∂tn ∂tn t=0 t=0 There is thus a Taylor expansion about t = 0 of the function 2

e2xt−t =

∞ 

Hn (x)

n=0

tn = G(x, t). n!

(7.15)

This simple GF can now be used to derive recurrence and orthogonality relations such as the following: Hn+1 (x) = 2xHn (x) − 2nHn−1 (x), Hn (x) Hn (x)

= 2nHn−1 (x),

Hn+1 (x) − 2xHn (x) + = 0,  ∞ √ 2 e−x Hm (x)Hn (x) dx = 2n n! πδmn = hn δmn . −∞

(7.16) (7.17) (7.18) (7.19)

The actual derivation of these relations using the GF in Eq. (7.15) will be left as exercises. Equation (7.18) itself can also be obtained directly from the Rodrigues formula (7.14) as follows: n+1 2 d 2 dn −x2 + (−1)n e x n+1 e−x ne dx dx = 2xHn (x) − Hn+1 .

Hn (x) = (−1)n (2xe x ) 2

Hermite polynomials and the quantum oscillator

465

On differentiating Eq. (7.18) once more, we find  . Hn = 2Hn + 2xHn − Hn+1  A differential equation for Hn can then be obtained by eliminating Hn+1 in favor of Hn with the help of Eq. (7.17):

Hn (x) − 2xHn (x) + 2nHn (x) = 0.

(7.20)

This is just the Hermite differential equation, which can also be used to define these Hermite polynomials. Hermite polynomials appear in the quantum-mechanical description of conservative systems near equilibrium. The statement of energy conservation p2 + V(r) = E, 2m

(7.21)

for a particle of mass m and momentum p that is experiencing a potential energy V(r) at the position r holds in quantum mechanics as in classical mechanics. However, atomic systems do not behave in the same way as familiar systems in classical mechanics. It was Niels Bohr who first realized that an electron bound electrostatically to an atomic nucleus in a circular orbit could exist only in certain “stationary” states, while other rather similar orbits were not allowed. Bohr discovered that the states permitted by nature satisfied the “quantum condition” that the orbital angular momentum of the electron was an integral multiple of  = h/2π, the reduced Planck’s constant. When quantum mechanics was eventually formulated by Heisenberg, Born and Jordan, and by Schr¨odinger, it was realized that this selectivity was a consequence of a change of meaning of the momentum p in Eq. (7.21), namely that it is not a number, but the differential operator p = −i∇.

(7.22)

In quantum mechanics, even the potential energy V(r) should be treated as an operator. We have in fact discussed in Section 4.8 why this should be the case. It is useful to refer to the operator nature of the left-hand side of Eq. (7.21) in quantum mechanics by giving it the special name of Hamiltonian and denoting it by the symbol H: H=

p2 + V(r). 2m

(7.23)

The statement (7.21) for energy conservation now becomes H = E in quantum mechanics. But surely this cannot be literally true, because an operator H cannot be equal to a number such as E. It is conceivable, however, that the operator H operating on a certain function ψ(r) could give a result equal to the number E times the same function ψ(r). That is, energy conservation in quantum mechanics might take the form of a differential equation  2  p + V(r) − E ψ(r) = 0. (7.24) 2m

466

Special functions

This is the famous Schr¨odinger wave equation. The wave function ψ(r) that solves the equation provides a physical description of the system. The mathematical structure of Eq. (7.24) reminds us of the rotation of a rigid body. The angular momentum of a rigid body, given by the matrix product of its 3 × 3 inertial matrix and its angular velocity vector, can actually be parallel to its angular velocity, but this occurs only when the angular velocity is along one of its three principal axes of inertia. The situation is described by Eq. (2.48), which can be seen to be similar to Eq. (7.24) here. For the same reason, only for very special functions will the Schr¨odinger equation be satisfied. The energies E at which this occurs are called its energy eigenvalues, while the wave functions ψ(r) so obtained are called its energy eigenfunctions. Consider now a 1D quantum system near a potential minimum at x = 0. Near this equilibrium point, the force is linear restoring and the potential is quadratic 1 V(x)  mω2 x2 . 2 The Schr¨odinger equation can then be simplified to ⎡ ⎛ ⎞ ⎤ 2⎟ ⎢⎢⎢ 1 ⎜⎜⎜ d2 e x ⎟ ⎟⎟⎟ − ⎥⎥⎥⎥⎥ ψ(x) = 0, ⎢⎢⎣− ⎜⎜⎝ − 2 dx2 x04 ⎠ x02 ⎦ where



 x0 = mω

(7.25)

(7.26)

1/2

has the dimension of a distance. It gives a characteristic size to the system, while e=

E ω

(7.27)

is a “dimensionless” energy, that is, energy in units of the characteristic energy ω. Equation (7.26) can be further simplified by writing it as an equation for the dimensionless distance ξ = x/x0 :   ! 1 d2 2 − ξ − e ψ(ξ) = 0. (7.28) − 2 dξ 2 It is now appropriate to ask what exactly is the physical information contained in the wave function ψ(ξ). The answer first proposed by Born is that |ψ(ξ)|2 gives the probability density for finding the system at the position ξ. That is, the integral  ∞ |ψ(ξ)|2 dξ (7.29) P = x0 −∞

gives the total probability of finding the system somewhere. If we know for a certainty that the system is oscillating about the point of equilibrium, this total probability must

Hermite polynomials and the quantum oscillator

467

be 1. This means that the wave function ψ(ξ) must be square integrable. To be square integrable, ψ(ξ) must vanish sufficiently rapidly for large values of |ξ|. At large |ξ|, the Schr¨odinger equation (7.26) simplifies to   2 d 2 (7.30) − ξ ψ(ξ) = 0. dξ 2 Its solutions can be found easily: ψ(ξ) ∼ e±(1/2)ξ . 2

(7.31)

ξ→∞

# " The solution exp( 12 ξ 2 ) is not square integrable; so we are left with only exp − 12 ξ 2 . This suggests that there might be a solution of the complete Eq. (7.26) having the form ψ(ξ) = H(ξ)e−(1/2)ξ . 2

(7.32)

The DE satisfied by H(ξ) can be obtained by direct substitution. We find ψ (ξ) = −ξψ + e−(1/2)ξ H  , 2

ψ (ξ) = −ψ − ξ(−ξψ + e−(1/2)ξ H  ) − ξe−(1/2)ξ H  + e−(1/2)ξ H  2

2

2

= e−(1/2)ξ (H  − 2ξH  − H + ξ2 H). 2

The result is



 d2 d − 2ξ + (2e − 1) H(ξ) = 0. dξ dξ2

(7.33)

Comparison with Eq. (7.20) shows that the solutions are just the Hermite polynomials Hn (ξ) if 2e − 1 = 2n. That is, the dimensionless energy eigenvalues are 1 e=n+ . 2

(7.34)

What happens when e  n + 12 ? One can show by doing Problem 7.3.4 that Eq. (7.33) then has solutions that are infinite series rather than polynomials. These solutions all behave like exp(ξ2 ) at large ξ. The resulting wave functions are then exp(ξ2 /2) at large distances. They are not square integrable, and are therefore not realizable as physical states. The requirement of square integrability thus selects certain functions Hn (ξ) out of the original set H(ξ). Certain quantum characteristics of atomic systems have their origin in the discreteness of their eigenvalues caused by this selection. The wave functions ψn (ξ) = cn Hn (ξ)e−(1/2)ξ

2

(7.35)

468

Special functions

–5.0

–3.0

ψ0(ξ) 0.8

ψ2(ξ) 0.8

0.4

0.4

–1.0

1.0

3.0

5.0

ξ

–5.0

–3.0

–1.0

–0.4

–0.4

–0.8

–0.8

ψ1(ξ) 0.8

5.0

3.0

5.0

ξ

0.4

–3.0 –1.0

3.0

ψ3(ξ) 0.8

0.4

–5.0

1.0

1.0

3.0

5.0

ξ

–0.4

–5.0

–3.0

–1.0

1.0

ξ

–0.4

–0.8

–0.8

Fig. 7.2 Some normalized wave functions ψn (ξ) of the quantum oscillator.

of the quantum oscillator are plotted in Fig. 7.2. The normalization constants cn are chosen so that  ∞ ψ2n (ξ)dξ = 1. (7.36) x0 −∞

Problems 5.3.1 Show that Hn (−x) = (−1)n Hn (x). 5.3.2 Use the generating function shown in Eq. (7.15) for Hermite polynomials to derive the recursion formulas Eq. (7.16)–(7.18). 5.3.3 Use the generating function shown in Eq. (7.15) for Hermite polynomials to derive their orthogonality relation Eq. (7.19).

Orthogonal polynomials

469

5.3.4 Solve the Hermite differential equation (7.33) by looking for a Frobeniusseries solution of the form H(ξ) = ξ s (a0 + a1 ξ + a2 ξ2 + . . .), with a0  0 and s ≥ 0. Show explicitly that (a) The Frobenius series terminates as a polynomial when e = n + 12 ; and (b) The infinite series can be summed to exp(ξ2 ) for large ξ when e  n + 12 . √ 2 5.3.5 If ψn (ξ) = (2n n! π)−1/2 Hn (ξ)e−(1/2)ξ are the square-integrable solutions of the oscillator Eq. (7.28), show that # " d + ξ ψ0 (ξ) = 0; (a) dξ # " √ d (b) √1 dξ + ξ ψn (ξ) ≡ aψn (ξ) = nψn−1 (ξ); 2 " # √ d (c) √1 dξ − ξ ψn (ξ) ≡ −a† ψn = − n + 1ψn+1 (ξ); 2

(d) [a, a† ] ≡ aa† − a† a = l;   d2 2 = a† a + 1 ; − ξ (e) H ≡ − 12 dξ 2 2 (f) (H − en )ψn (ξ) = 0, with en = n + 12 , is the oscillator equation (7.28) for physically realizable states; (g) [H, a† ] = a† , [H, a] = −a. (h) If the solution ψn (ξ) is said to have n oscillator “quanta,” show that a† ψn has n + 1 quanta and aψn has n − 1 quanta. The first-order differential operator a(a† ) is called a destruction (creation) operator for oscillator quanta. It is also called a lowering (raising) operator, or a step-down (step-up) operator, or a ladder operator. H is the Hamiltonian of the quantum oscillator, and en its energy eigenvalue in the physical state with n oscillator quanta.

7.4

Orthogonal polynomials

We have seen in Section 4.10 that a family of polynomials { fn (x), n = 0, 1, . . . , ∞}, where fn (x) = kn xn + kn xn−1 + . . .

(7.37)

is a polynomial of degree n, is said to be orthogonal in the closed interval [a, b], with respect to the weight function w(x) if  a

b

w(x) fm (x) fn (x)dx = hn δmn .

(7.38)

470

Special functions

Table 7.1 Rodrigues’ formula for several orthogonal polynomials fn (x) =

dn 1 n an w(x) dxn {w(x)[s(x)] }.

fn (x)

Name

w(x)

s(x)

an

Pn (x) Ln (x) Hn (x)

Legendre Laguerre Hermite

1 e−x 2 e−x

1 − x2 x 1

(−1)n 2n n! n! (−1)n

T n (x)

First Chebyshev

(1 − x2 )−1/2

1 − x2

N.B.: Γ(n + 12 ) = (n − 12 )Γ(n − 12 ), Γ( 12 ) =

(−1)n 2n+1

Γ(n+ 12 ) √ π

√ π.

Although the fn in different families are all polynomials of degree n, they have different coefficients kn , kn , . . . . These differences arise from differences in the weight functions and in the intervals of integration. Tables of a, b, w(x) and hn for four common orthogonal polynomials (Legendre, Laguerre, Hermite polynomials, and Chebyshev polynomials of the first and second kinds have been given in Table 4.1. For example, the Legendre and Chebyshev polynomials are defined in the same interval x = −1 to 1, but differ in their weight functions. As a result, orthogonality means different things to different families of orthogonal polynomials, even for those defined in the same interval. This point was emphasized in Section 4.10 A table of the first four polynomials of each family has been given in Table 4.2. These orthogonal polynomials satisfy a generalized Rodrigues formula of the same general form fn (x) =

6 1 dn 5 w(x)[s(x)]n , an w(x) dxn

(7.39)

where w(x) is the weight function of Eq. (7.38), s(x) is a polynomial, of degree ≤ 2, and an is a normalization constant. These quantities are shown in Table 7.1. Table 7.2 gives the generating functions G(x, t) =

∞ 

bn fn (x)tn ,

for

|t| < 1.

(7.40)

n=0

Table 7.2 Generating functions G(x, t) =

∞

n=0 bn fn (x)t

n,

for |t| < 1.

fn (x)

G(x, t)

bn

Note

Pn (x)

R−1  xt  1 1−t exp t − 1

1

|x| < 1

exp(2xt − t2 )

1/n!

Ln (x) Hn (x)

(1 −

T n (x)

1/R2

U n (x) N.B: R =

xt)/R2

√ 1 − 2xt + t2 .

1

1

|x| < 1

1

|x| < 1

Orthogonal polynomials

471

Table 7.3 Recursion formulas for orthogonal polynomials: fn+1 (x) = (bn x + cn ) fn − dn fn−1 , with f0 = 1.

fn (x)

bn

cn

dn

fi

Pn (x) Ln (x) Hn (x) T n (x) Un (x)

(2n + 1)/(n + 1) −1/(n + 1) 2 2 2

0 (2n + 1)/(n + 1) 0 0 0

n/(n + 1) n/(n + 1) 2n 1 1

x 1−x 2x x 2x

From these one can derive recursion formulas of the form fn+1 (x) = (bn x + cn ) fn − dn fn−1,

(7.41)

where the coefficients bn , cn , and dn are given in Table 7.3. One can also obtain the differential relations p2 (x) fn (x) = p1 (x) fn (x) + p0 (n) fn−1 (x),

(7.42)

where fn (x) = d f (x)/dx. The functions pi (x) are shown in Table 7.4. Finally, the orthogonal polynomials satisfy second-order differential equations of the form g2 (x) fn (x) + g1 (x) fn (x) + g0 (n) fn (x) = 0,

(7.43)

where the functions gi (x) are given in Table 7.5. Plots of some Legendre polynomials have already been given in Fig. 7.2. Some of the other polynomials are shown in Fig. 7.3. Orthogonal polynomials can be used to represent arbitrary functions, including solutions of differential equations, in their interval. This is done by writing the function to be represented as R(x) =

∞ 

cn fn (x).

(7.44)

n=0

Table 7.4 Differential relation for orthogonal polynomials, p2 (x) fn (x) = p1 (x) fn (x) + p0 fn−1 (x).

fn (x)

p2 (x)

p1 (x)

p0 (n)

Pn (x) Ln (x) Hn (x) T n (x) Un (x)

1 − x2 x 1 1 − x2 1 − x2

−nx n 0 −nx −nx

n −n 2n n n+1

472

Special functions Table 7.5 The differential equation satisfied by orthogonal polynomial fn (x): g2 (x) fn + g1 (x) fn + g0 (n) fn = 0.

3

fn (x)

g2 (x)

g1 (x)

g0 (n)

Pn (x) Ln (x) Hn (x) T n (x) Un (x)

1 − x2 x 1 1 − x2 1 − x2

−2x 1−x −2x −x −3x

n(n + 1) n 2n n2 n(n + 2)

L2

Ln(x)

8

H4 64

Hn(x)/n3

H3 27

H5 125

2

2

H2 8

4

1

L3 1

0

x 1

3

4

5

6

0

x 3

–1 –2

L5

L4

(a)

(b)

Tn(x) T2

T3

T4

T1

0.5

x –1

0

1

–0.5

(c)

Fig. 7.3 (a) Laguerre polynomials Ln (x), n = 2 − 5; (b) Hermite polynomials Hn (x)/n3 , n = 2 − 5; (c) Chebyshev polynomials T n (x), n = 1 − 4 (from Abramowitz and Stegun).

Orthogonal polynomials

473

The expansion coefficients cn can be extracted by multiplying both sides by fm (x)w(x) and integrating from a to b. It is convenient to write the resulting integral in the compact notation of a functional inner product:  b fm (x)w(x)R(x)dx ( fm , R) ≡ a

=

∞ 

cn ( fm , fn ),

(7.45)

n=0

where all quantities have been taken to be real. Since ( fm , fn ) is just Eq. (7.38), we have ∞  ( fm , R) = cn hn δmn = cm hm . (7.46) n=0

Example 7.4.1 Expand R(x) = eax in the Hermite series R(x) =

∞ 

cn Hn (x).

(7.47)

n=0

According to Eq. (7.46) the coefficients are ax cn = h−1 n (Hn , e ).

√ Using hn = π2n n! and Hn from Tables 4.2 and 4.3, respectively, we find  ∞ 2 1 eex e−x dx c0 = √ π −∞  2 1 a2 /4 ∞ −[x−(1/2)a]2 = √ e e dx = ea /4 , π −∞  ∞ 2 1 2xeax e−x dx c1 = √ 2 π −∞  ∞ 1 d 2 = √ eax e−x dx π da −∞ =

d a2 /4 a a2 /4 = e , e da 2

(7.48)

(7.49)

(7.50)

and so on. The calculation increases in complexity when the degree n of the polynomial increases.  An equivalent method is to use the Rodrigues formula and integrate by parts. In many cases, the calculation of cn can be greatly simplified by using the generating function, as the following example illustrates.

474

Special functions

Example 7.4.2 Calculate the Hermite coefficients cn in Eq. (7.47) with the help of the generating function 2

G(x, t) = e2xt−t =

∞ n  t n=0

n!

Hn (t).

We have by direct substitution (G, R) = =

∞ n  t

n! n=0 ∞ n  t n=0

n!

(Hn , R) hn cn .

(7.51)

If (R,G) can be calculated and expanded in power of tn , the coefficients cn can be extracted. Now  ∞ 2 2 ax (G, e ) = e−x e2xt−t eax dx −∞

−t2



=e

∞ −∞

e−x

−t2 +(t+a/2)2

2 +2(t+a/2)x



=e

∞ −∞

√ 2 = π eat+a /4 .

dx

e−|x−(t+a/2)] dx 2

The expansion in powers of t turns out to be easy in this example: (G, eax ) =

√

2 /4

π ea

∞  (at)n n=0

n!

.

Comparison with Eq. (7.51) shows that √ 2 2 cn = π an ea /4 /hn = (a/2)n ea /4 /n!.

(7.52)

(7.53)

The generating function allows us to obtain all the coefficients “wholesale,” thus avoiding the laborious calculation of coefficients one after another. 

Problems 7.4.1 Laguerre polynomials Ln (x): (a) Show that their generating function 1 −xt/(1−t)  = Ln (x)tn e G(x, t) = 1−t n=0 ∞

Orthogonal polynomials

475

satisfies the following relations 1−t−x G, (1 − t)2 t ∂xG = − G. 1−t ∂t G =

(b) Use these relations to derive a recursion formula and two differential relations (n + 1)Ln+1 = (2n + 1 − x)Ln − nLn−1 ,  Ln − Ln−1 = −Ln−1 ,

xLn = nLn − nLn−1 . (c) Use the results of part (b) to derive the Laguerre differential equation xLn + (1 − x)Ln + nLn = 0. 7.4.2 Chebyshev polynomials (CPs), Cn (x), can be conveniently defined by their recursion formula Cn+1 (x) − 2xCn (x) + Cn−1 (x) = 0, with C0 (x) = 1. Two kinds of polynomial can be obtained for different choices of C1 (x): (a) First CPs, T n (x), are obtained by the choice T 1 (x) = x. Use the given information to show that these polynomials are generated by GI (x, t) =

∞ 

T n (x)tn =

n=0

1 − xt . 1 − 2xt + t2

(b) Second CPs, Un (x), are obtained by the choice U 1 (x) = 2x. Use the given information to show that these polynomials are generated by GII (x, t) =

∞  n=0

Un (x)tn =

1 . 1 − 2xt + t2

Hint: Find an analytic expression in terms of G, x, t for each term of the sum ∞  n=0

[Cn+1 (x) − 2xCn (x) + Cn−1 (x)] tn = 0.

476

Special functions

7.4.3 The generating function G(x, t) for the second Chebyshev polynomials Un (x) is  1 = Un (x)tn . 1 − 2xt + t2 n=0 ∞

G(x, t) ≡

Verify that the following recursion and differential formulas/equations are consistent with G by multiplying each equation by tn and summing from n = 0 to ∞: Un+1 − 2xUn + Un−1 = 0, (1 − x

2

)Un

+ nxUn (x) − (n + 1)U n−1 (x) = 0,

(1 − x2 )Un − 3xUn + n(n + 2)Un = 0. Note: These formulas/equations can thus be derived from G by working backwards from the penultimate expressions in terms of G. This reverseengineering method should also work for the first Chebyshev polynomials T n (x). If you want to do this, you might want to derive and use a more convenient form of their generating function:   ∞  1 1 − t2 n GI (x, t) ≡ T n (x)t = +1 . 2 1 − 2xt + t2 n=0 7.4.4 Give the Hermite representation Eq. (7.47) of (a) x3 ; (b) x4 ; (c) exp(−a2 x2 ). 7.4.5 Give the Laguerre representation of exp(−ax). 7.4.6 Show that ∞  2 (−1)n H2n (x) 2n et cos 2xt = t , (2n)! n=0 2

et sin 2xt =

∞  (−1)n H n=0

7.5

2n+1 (x) 2n+1

(2n + 1)!

t

.

Classical orthogonal polynomials∗

It turns out to be possible to define families of orthogonal polynomials by the generalized Rodrigues formula (7.39) fn (x) =

1 dn 5 n6 n w(x)[s(x)] an w(x) dx

(7.39)

Classical orthogonal polynomials

477

if three conditions are imposed (Dennery and Kryzwicki, 1995): 1. f1 (x) is a polynomial of first degree; 2. s(x) is a polynomial of degree ≤ 2 and has real roots; 3. The weight function w(x) is real, positive, and integrable in the interval [a, b]. It satisfies the boundary conditions (BC) w(a)s(a) = w(b)s(b) = 0

(7.54)

at the boundary points a and b. We shall first show that the function fn (x) is a polynomial p≤n of degree ≤ n if f1 is a first-degree polynomial satisfying the equation d (ws) = a1 wf 1 = wp≤1 . dx

(7.55)

This can be done by first considering d d ds (wsn ) = sn−1 (ws) + ws(n − 1)sn−2 dx dx dx = wsn−1 p≤1 + (n − 1)wsn−1 p≤1 = wsn−1 p≤1 ,

(7.56)

where we have used the information that s = p≤2 . The different p≤1 that appear are not the same function, but they all are polynomials of degree not exceeding one. Repeated differentiation or an inductive argument can be used to show that dm (wsn ) = wsn−m p≤m , dxm

m ≤ n.

(7.57)

Eq. (7.57) used with m = n states that fn is a polynomial of degree ≤ n. We next show that fn is orthogonal to any polynomial pl of degree l < n. It is therefore not of degree < n. The overlap between these two polynomials is from Eqs. (7.38) and (7.39) b (pl , fn ) ≡

w(x)pl (x) fn (x)dx a

=

1 an



b

pl (x) a

dn (wsn )dx. dxn

We now integrate by parts n times. Each time there appear boundary terms that vanish because of the BC (7.54). Hence  (−1)n b n d n (pl , fn ) = ws n (pl )dx. (7.58) an dx a

478

Special functions Table 7.6 The three classes of classical orthogonal polynomials.

fn (x)

s(x)

an

f1 (x)

w(x)(α, β > −1)

Interval

Hn Ln(α) (α,β) Pn

1 x 1 − x2

(−1)n n! (−1)n 2n n!

2x −x + α + 1 See (a)

e−x xα e−x (1 − x)α (1 + x)β

(−∞, ∞) [0, ∞) [−1, 1]

2

(a) f1 (x) = α + 1 + [ 12 (α + β) + 1](x − 1)

This also vanishes for l < n. Since fn is a polynomial of degree ≤ n [from Eq. (7.57)], but not of degree < n from Eq. (7.58), it must be a polynomial of degree exactly n. We have thus verified that given s(x) and w(x), the polynomials fn (x) defined by Eq. (7.42) form a family of orthogonal polynomials. The three condition given near Eq. (7.54) are actually very restrictive. Only three distinct possibilities exist. They give rise to three classes of polynomials called classical orthogonal polynomials. They are obtained by choosing s(x) = 1, x, 1 − x2 , respectively, as shown in Table 7.6. Once s(x) is chosen, w(x) can be determined by solving the first-order DE obtained from Eq. (7.55),   ds dw , (7.59) = w a1 f 1 − s dx dx subject to the BC (7.54) and with a choice of a1 f1 . Since a1 is a normalization for the first-degree polynomial f1 ∝ x − b, different choices of a1 actually generate the same set of polynomials with related normalization and intervals [a, b]. So it is sufficient to use the standardized choice shown in Table 7.6. For s = 1, a1 = −1, f1 = 2x, Eq. (7.59) simplifies to the DE dw = −2xw. dx

(7.60)

w(x) = Ae−x .

(7.61)

Its solution is 2

To satisfy the BC (7.54) for s = 1, the boundary points should be taken to be (−∞, ∞). (The points ±∞ are always considered to be outside the interval.) Used in the Rodrigues formula, the solution generates the Hermite polynomials Hn (x). Since this formula and therefore the generated polynomials are independent of A, we may for simplicity use A = 1. For this choice, w(x) ≥ 0. The solutions for the other two cases are left as exercises. The results are summarized in Table 7.6. The polynomials constructed for the choice s = x are the generalized Laguerre polynomials Ln(α) (x). Those for s = 1 − x2 are the Jacobi polynomials. They include the Legendre and first Chebyshev polynomials of Table 7.1:

Classical orthogonal polynomials

Pn (x) = P(0,0) n (x), √ n! π (−1/2,−1/2) (x). Pn T n (x) = Γ(n + 12 )

479

(7.62)

Since fn (x) is a polynomial of degree n, x fn (x) must be a polynomial of degree n + 1. We can therefore write x fn (x) =

n+1 

anl fl (x).

l=0

The fact [from (7.58)] that (pl , fn ) = 0 if l < n can now be used to show that (xf n , fl ) = ( fn , xf l )  0

(7.63)

only if n − 1 ≤ l ≤ n + 1. This means that there is a recursion formula of the form fn+1 (x) = (bn x + cn ) fn (x) − dn fn−1 (x).

(7.64)

The reader can verify by direct construction and with the help of Eqs. (7.37) and (7.38) that b = kn+1 /kn ,

" # 7 8  cn = (kn+1 /kn ) kn+1 /kn+1 − kn /kn dn = (hn /hn−1 )kn+1 kn−1 /kn2 .

(7.65)

(This is Problem 7.5.2.) Since fn (x) is a polynomial of degree n, df n (x)/dx must be a polynomial of degree n − 1. Now Eq. (7.56) can be written in the more general form (see Problem 7.5.3) d 7 n 8 ws pm = wsn−1 p≤m+1 . dx This shows that

(7.66)

  1 d d ws fn = p≤n (x) w dx dx =−

n 

λnl fl (x).

l=0

The expansion coefficients on the right-hand side are just the overlaps       b d d d 1 d λnl hl = − fl , ws fn = − ws fn dx. fl w dx dx dx dx a

(7.67)

480

Special functions

We now integrate twice by parts, using the boundary condition (7.54) to eliminate all boundary terms:     b d d fl ws fn dx λnl hl = dx a dx  !   b d d =− fn fl ws dx dx dx a = λln hn .

(7.68)

Eq. (7.67) can now be applied to the last integrand to obtain  b fn wp≤l dx. λln hn = − a

According to Eq. (7.58), this vanishes if l < n. We have thus shown that the righthand side of Eq. (7.67) contains only one term involving the coefficient λnn . In other words, the polynomial fn (x) satisfies the second-order differential equation   d d (7.69) ws fn (x) + wλnn fn (x) = 0. dx dx We may note with some satisfaction that both Eqs. (7.64) and (7.69) are derived by applying simple arguments cleverly. The undaunted reader may want to derive the actual expression for λnn , which is   df 1 1 d2 s (7.70) + (n − 1) 2 . λnn = −n a1 dx 2 dx where a1 is the normalization constant used in Eq. (7.39). Problem 7.5.4 gives a hint on how this can be done.

Problems 7.5.1 Verify the weight function w(x) and the choice of interval shown in Table 7.6, for (a) the generalized Laguerre, and (b) the Jacobi, polynomials. 7.5.2 (a) Show that in the notation of Eq. (7.37),   kn+1 L(x) ≡ fn+1 (x) − x fn (x) kn    kn+1 kn n − x + .... = kn+1 kn+1 kn (b) Show that L(x) = R(x) = cn fn (x) − dn fn−l (x) , where cn , dn are real coefficients. (c) Use these results and the orthogonality property (7.38) to verify Eq. (7.65). 7.5.3 Use the method of Eqs. (7.55) and (7.56) to verify Eq. (7.66).

Associated Legendre polynomials

481

7.5.4 Verify Eq. (7.70) by first showing that     df n df 1 1 d d2 s sw = a1 n + n(n − 1) 2 wkn xn + . . . , dx dx dx 2 dx where kn is the coefficient shown in Eq. (7.37).

7.6

Associated Legendre polynomials and spherical harmonics

Other orthogonal polynomials can be obtained from the classical ones by additional manipulations. For example, the associated Legendre polynomials Plm (x) of degree l and order m are related to the Legendre polynomials (order m = 0) by the differential relation 2 m/2 Pm l (x) = (1 − x )

dm Pl (x), dxm

0 ≤ m ≤ 1.

(7.71)

It is possible to show (for example by induction) that the function um (x) =

dm Pl (x) dxm

(7.72a)

satisfies the DE   2 d 2 d (1 − x ) 2 − 2(m + 1)x + [l(l + 1) − m(m + 1)] um (x) = 0. dx dx

(7.72b)

The use of um (x) = (1 − x2 )−m/2 Plm (x) will then give the associated Legendre DE: !  2 d m2 2 d (1 − x ) 2 − 2x + l(l + 1) − (7.73) Pm l (x) = 0. dx 1 − x2 dx (This is Problem 7.6.1.) These polynomials satisfy the orthogonality relation "

Plm , Plm

#

 ≡ =

1

−1

Plm (x)Plm (x)dx

(l + m)! 2 δll . (l − m)! 2l + 1

(l + m)! = (l − m)!

1 Pl (x)Pl (x)dx

(7.74a)

−1

(7.74b)

This can be proved by using Eq. (7.71) in Eq. (7.74a) and by integrating by parts: m " #  1 d m Pl  m m 2 m d Pl Pl , Pl = (1 − x ) dx m dxm −1 dx    1 m−1 m d Pl d 2 m d Pl =− (1 − x ) dx, m−1 dx dxm −1 dx

482

Special functions

where zero boundary terms have been discarded. The required differentiation in the integrand can be performed explicitly with the help of Eq. (7.72b)     m d d 2 m d Pl 2 m d = (1 − x ) (1 − x ) um−1 dx dxm dx dx = (1 − x2 )m

d 2 um−1 dx

= −(1 − x )

2 m−1

Hence "

m Pm l , P l

#

 = (l + m)(l − m + 1)

2

− 2mx(1 − x2 )m−1

dum−1 dx

(l + m)(l − m + 1)um−1 .

1

dm−1 Pl

−1

dxm−1

(1 − x2 )m−1

# " , Plm−1 . = (l + m)(l − m + 1) Plm−1 

d m−1 Pl dxm−1

dx (7.75)

Successive applications of this reduction formula for the order m yield Eq. (7.74a). Eq. (7.74b) then follows from the orthogonality relation for Legendre polynomials. Explicit formulas for Plm (x) for m > 0 can be obtained directly from Eq. (7.71). They are conveniently expressed in terms of cos θ = x and sin θ = (1 − x2 )1/2 : P11 (x) = (1 − x2 )1/2 = sin θ P12 (x) = 3x(1 − x2 )1/2 = 3 cos θ sin θ P22 (x) = 3(1 − x2 ) = 3 sin2 θ,

etc.

(7.76)

Note that Plm always vanishes at x = ±1 for m  0. The generating function for Plm can be obtained by applying Eq. (7.71) to both sides of the GF (7.4) for Pl :  (2m)! l−m = Pm . l (x)t m 2 m+1/2 2 m!(1 − 2tx + t ) l=m ∞

(1 − x )

2 m/2

(7.77)

The resulting expression is often too complicated to be of much use. In many cases, it might be simpler to use Eq. (7.71) directly. If we go back all the way to the Rodrigues formula for Pl (x) =

1 dl 2 (x − 1)l , 2l l! dxl

it is clear that associated Legendre polynomials of negative orders m = −|m| can also be defined down to m = −l by extending Eq. (7.71) to negative m: P−|m| (x) = (1 − x2 )−|m|/2 l

d l−|m| dx

l−|m|

(x2 − 1)l

1 , 2l l!

(7.78)

Associated Legendre polynomials

483

with l 2 l/2 l P−l l (x) = (−1) (1 − x ) /(2 l!).

itself: This turns out, perhaps unexpectedly, to be proportional to Pm l m P−m l (x) = (−1)

(l − m)! m P (x). (l + m)! l

(7.79)

This result can be proved by showing that Pl−m is not orthogonal to Plm The key is the derivation of a reduction formula for the overlap (Plm , Pl−m ) in the order m by integration by parts in a way similar to the derivation of Eq. (7.75): #  1 dm Pl d l−m " 1 m −m Pl , Pl = (x2 − 1)l l dx m l−m 2 l! dx −1 dx  1 m−1 d Pl dl−m+1 2 1 =− (x − 1)l l dx m−1 l−m+1 2 l! dx −1 dx # " −m+1 . (7.80) = − Pm−1 l  , Pl (The boundary terms in the integration by parts vanish because Pl−m vanishes at x = ±1.) Equation (7.80) can be used repeatedly to give # " −m = (−1)m (Pl , Pl ) Pm l , Pl = (−1)m

2 δll . 2l + 1

(7.81)

Comparison with Eq. (7.74b) gives Eq. (7.79). Associated Legendre polynomials appear frequently in physics as parts of functions called spherical harmonics. Harmonic functions are functions that satisfy the Laplace equation ∇2 ψ = 0. The spherical harmonics Ylm (θ, φ) are those that satisfy it on the surface of a sphere. That is, the Laplace equation in spherical coordinates has solutions of the form ψ(r, θ, φ) =

l ∞  

" # Ylm (θ, φ) Al rl + Bl r−l−1 .

(7.82)

l=0 m=−l

According to the results of Section 5.12, Ylm (θ, φ) = Φ(φ)Θ(θ), where Φ(φ) = Φ±m (φ) = exp(±imφ)/(2π)1/2

(7.83)

484

Special functions

satisfies Eq. (5.125a), while Θ(θ) satisfies the associated Legendre equation Eq. (5.129) or Eq. (7.73). Hence the normalized spherical harmonics are 

Ylm (θ, φ)

(l − m)! 2l + 1 = (l + m)! 4π

1/2 eimφ Pm l (cos θ).

(7.84)

Spherical harmonics behave simply under complex conjugation and parity operations:  1/2 (l − m)! 2l + 1 m∗ e−imφ Pm Yl (θ, φ) = l (cos θ) (l + m)! 4π = (−1)m Yl−m (θ, φ),

(7.85)

where use has been made of Eq. (7.79). Under the parity transformation r → −r or (r, θ, φ) → (r, π − θ, π + φ). Since cos θ then transforms into − cos θ, we find m 2 m/2 Pm l (x) → Pl (−x) = (1 − x )

dm Pl (−x) d(−x)m

= (1 − x2 )m/2 (−1)m

dm (−1)l Pl (x) dxm

= (−1)l+m Pm l (x) with the help of Eq. (7.6). As a result, 

Ylm (θ, φ)

→

Ylm (π

(l − m)! 2l + 1 − θ, π + φ) = (l + m)! 4π

1/2 eim(π+φ) (−1)l+m Pm l (x)

= (−1)l Ylm (θ, φ).

(7.86)

Thus it has the same parity property as Pl (x). Explicit formulas for Yim (θ, φ) are listed below for the reader’s convenience: √ Y00 = 1/ 4π; √ √ Y1±1 = ∓ 3/8π sin θe±iφ ; Y10 = 3/4π cos θ, √ √ Y20 = 5/16π (3 cos2 θ − 1), Y2±1 = ∓ 15/8π cos θ sin θe±iφ , √ Y2±2 = 15/32π sin2 θe±2iφ ; . . . .

(7.87)

Occasionally, the solid spherical harmonics, or (spherical) harmonic polynomials Ylm (r) = rl Ylm (θ, φ)

(7.88)

Associated Legendre polynomials

are used because they are functions of the x,y,z coordinates themselves: 9√ Y00 = 1 4π ; & & Y10 = 3/4π z, Y1±1 = ∓ 3/8π (x ± iy); & & Y20 = 5/16π (2z2 − x2 − y2 ), Y2±1 = ∓ 15/8π z(x ± iy), & Y2±2 = 15/32π (x ± iy)2 ; . . . . The functions (7.83) satisfy the orthonormal relation  2π Φ∗m (φ)Φm (φ)dφ = δm m .

485

(7.89)

(7.90)

0

It therefore follows that the orthonormal relation for the spherical harmonics is  ∗ Ylm (θ, φ)Ylm (θ, φ)d cos θdφ = δl l δm m . (7.91) We recall that the discussion of Section 5.12 makes clear that Ylm (θ, φ) satisfies the periodicity condition Ylm (θ, φ + 2π) = Ylm (θ, φ), everywhere and the regularity condition (Ylm finite) at the poles θ = 0 and π. As a result, any function on the sphere satisfying these conditions can be expanded in spherical harmonics F(Ω) =

l ∞  

clm Ylm (Ω),

(7.92)

l=0 m=−1

where the notation Ω = (θ, φ) is used. The expansion coefficients can be calculated readily from F(Ω) with the help of Eq. (7.91);  clm = dΩYlm∗ (Ω)F(Ω), (7.93) where dΩ = d cos θdφ, as in Eq. (1.89). Substitution of Eq. (7.93) into Eq. (7.92)   Ylm∗ (Ω )Ylm (Ω) F(Ω) = dΩ F(Ω ) lm

486

Special functions

yields the completeness relation:  Ylm∗ (Ω )Ylm (Ω) = δ(Ω − Ω ) = δ(cos θ − cos θ )δ(φ − φ ),

(7.94)

lm

where the Dirac δ function satisfies the usual properties discussed in Section 4.4. δ(Ω − Ω ) = 0 if Ω  Ω ;  δ(Ω)dΩ = 1.

(7.95a) (7.95b)

The completeness relation (7.94) can be further simplified into a very useful identity known as the addition theorem for spherical harmonics. The possibility of a simplification arises because the angular difference Ω − Ω between two directions on a sphere can be characterized by a single angle α between them. If we choose one of these directions as the new z axis, any function of Ω − Ω is a function only of the new colatitude angle α and can therefore be expanded in terms of Pl (cos α) In particular,  δ(Ω − Ω ) = dl Pl (cos α), (7.96) l

where

 2l + 1 d cos α δ(Ω − Ω )Pl (cos α) 2  2l + 1 = dΩ δ(Ω − Ω )Pl (cos α) 4π 2l + 1 2l + 1 = Pl (1) = . 4π 4π

dl =

Comparison between this and Eq. (7.94) yields the desired result l 4π  m∗  m Y (Ω )Yl (Ω). Pl (cos α) = 2l + 1 m=−l l

(7.97)

Problems 7.6.1 Verify (a) Eq. (7.72b), and (b) Eq. (7.73) with the help of the Legendre differential equation (7.83). Hint for part (a): Try a recursive/inductive method that contains the following crucial step: Let um (x) = a† um−1 (x), where um−1 (x) satisfies the DE Lm−1 um−1 (x) = 0. The differential operator Lm−1 is in general m-dependent. Then Lm um = 0 if the operators satisfy the recursion condition a† Lm−1 = Lm a† = Lm−1 a† + (Lm − Lm−1 )a† .

Bessel functions

487

7.6.2 Obtain explicit expressions for Pm 3 (x), m = −3, . . . , 3 7.6.3 Use the generating function given in Eq. (7.77) to show that l − m = odd, (a) Plm (0) = 0, (l+m−1)!! (l−m)/2 = (−1) l − m = even; (l−m)!! , " #1/2 δm0 . (b) Plm (1) = δm0 , Ylm (0, φ) = 2l+1 4π 7.6.4 Obtain the following recurrence relations with the help of Eqs. (7.8, 7.9, 7.71): (a) Pm (x) = (1 − x2 )1/2 (2l + 1)Pm−1 (x) + Pm (x). l+1 l−1 l m m (b) (l + 1 − m)Pm l+1 (x) = (2l + 1)xPl (x) − (l + m)Pl−1 (x). m

m

m−1

d d d Hint: dx m (xPl ) = x dxm Pl + m m−1 Pl . dx 7.6.5 Expand f (x, y, z) = x + y + z in spherical harmonics. 7.6.6 Express the Coulomb potential field u(r) in spherical coordinates [See Eq. (7.2)] at the field poult r due to a source −4πδ(r − r ) at an arbitrary source point r in terms of the spherical harmonics Ylm (Ω) and Ylm (Ω ). ! l ∞ rl  4π  m∗ (Ω )Y m (Ω) . < Y Answer: u(r) = l+1 2l+1 l l l=0 r>

7.7

m=−l

Bessel functions

Orthogonal polynomials are well suited to the expansion of functions, or the description of physical systems, that are localized near the origin of coordinates. Such an expansion is quite similar to the Taylor expansion. The further we go away from the origin, the more terms we need for a reasonable approximation. Orthogonal polynomials are not so useful if the system is not localized, but’is spread out all over space. For example, solutions of the wave equation describing waves over all space are more conveniently expanded in terms of trigonometric functions (as in Fourier series) and other functions containing infinite series of powers rather than polynomials of finite degrees. Indeed, solutions of the wave equation (including the Helmholtz equation of Section 5.l2) in various coordinate systems give a class of special functions particularly suited to the description of physical phenomena that can propagate to infinity. The Bessel function Jn (x) of an integral order n is the solution of the Bessel differential equation (5.130) x2 Jn + xJn + (x2 − n2 )Jn (x) = 0

(7.98)

obtained by separating the 2D Helmholtz equation in circular coordinates, or the 3D equation in cylindrical coordinates. These Bessel functions can also be defined in terms of the generating function !  ∞  1 1 G(x, t) = exp x t − = tn Jn (x). (7.99) 2 t n=−∞

488

Special functions

Both sides of this equation diverge at t = 0. The infinite series on the right-hand side is not a Taylor series, which has only positive powers. Infinite series containing negative powers like this are called Laurent series. They will be discussed in Section 8.8. For the present, it is sufficient to note that Bessel functions of negative integral orders are associated with negative powers of t in Eq. (7.99). To generate explicit expansions for Jn (x), we expand G(x,t) in powers of t: ⎡∞ ⎤⎡ ∞ ⎤  x r tr ⎥⎥ ⎢⎢  x  s t−s ⎥⎥  x  1 ! ⎢⎢ - x  . ⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ − t exp − = ⎢⎢⎢⎣ exp ⎥⎦ ⎢⎣ ⎥⎦ 2 2 t 2 r! 2 s! r=0 s=0  x r+s tr−s  . = (−1) s 2 r!s! r,s A change of the summation index to n = r − s (or r = n + s, r + s = n + 2s) yields G(x, t) =

∞  ∞  n=−∞ s=0

(−1) s

 x n+2s 2

tn . (n + s)!s!

Comparison with Eq. (7.99) shows that (−1) s  x n+2s (n + s)!s! 2 s=0   x n  1 x2 /4 (x2 /4)2 = − + − ... , 2 n! 1!(n + 1)! 2!(n + 2)!

Jn (x) =

∞ 

(7.100)

if n ≥ 0. The expression for Jn (x) with n < 0 is slightly more complicated. This is because the condition r = n + s = s − |n| ≥ 0 requires that s ≥ |n|. Hence J−|n| (x) =

∞  s=|n|

 x 2s−|n| (−1) s . (s − |n|)!s! 2

(7.101)

If we now change variables to s = s − |n| or s = s + |n|, we see that J−|n| (x) =

 ∞     (−1) s +|n| x 2s +|n| = (−1)n J|n| (x)  !(s + |n|)! 2 s s =0

(7.102)

is proportional to J|n| (x). The mathematically facile reader will recognize this relation by simply inspecting Eq. (7.99) (Problem 7.7.1).

Bessel functions

489

A number of recursion formulas can be obtained easily from G(x,t). Two are of special interest (see Problem 7.7.2): 1 n (Jn−1 + Jn+1 ) = Jn , 2 x 1 (Jn−1 − Jn+1 ) = Jn . 2

(7.103) (7.104)

The difference and the sum of these equations give the ladder operators for Jn (raising operator Rn and lowering operator Ln ):   n d Rn Jn = Jn = Jn+1 , − x dx   n d Ln Jn = (7.105) Jn = Jn−1 . + x dx The expression Ln+1 Rn Jn raising the order once and lowering it once must be equal to Jn itself. Hence (Ln+1 Rn − 1)Jn = 0.

(7.106)

This is just the Bessel equation (7.98). One can easily verify by direct differentiation (in Problem 7.7.3) that Eq. (7.100) with n replaced by any real number v defines a Jv (x) that also satisfies the recursion formulas (7.103) and (7.104). Hence it is also a solution of the Bessel equation (7.98) with n replaced by v. We should note that the factorial (v + s)! appearing in Eq. (7.100) has the usual property that z! = z(z − 1)!.

(7.107)

It follows from this that 1! =1 1 0! (−1)! = = ∞ = (−2)!, 0 0! =

etc.

(7.108)

Thus μ! is finite except at the negative nonzero integers. Because of Eq. (7.107), we need to know z! only over the range −1 to 0. A useful definition turns out to be the integral:  ∞ z! ≡ e−t tz dt, Re z > −1 0

= Γ(z + 1).

(7.109)

490

Special functions

It is actually valid for complex values of z provided that Re z > −1. The factorial function is also called a gamma function, but for argument z + 1, as shown in Eq. (7.l09). One gamma function of fractional argument is particularly worthy of note:      ∞ √ 2 1 1 e−u du = π. (7.110) = − !=2 Γ 2 2 0 Eq. (7.100) shows that Jv (x) and J−v (x) are linearly independent if v is nonintegral, because they differ in at least some terms. If v becomes integral, their difference vanishes, and they become linearly dependent. This difference in behavior is a minor annoyance that can be avoided by using the Neumann function Nv (x) =

cos vπJv (x) − J−v (x) sin vπ

(7.111)

instead. (Nv is often denoted Yv ) It is obvious that Nv is linearly independent of Jv when v is nonintegral. When v → n becomes integral, both numerator and denominator vanish, but the quotient remains well defined, as one can verify by applying l’Hospital’s rule. One finds (Problem 7.7.7) that Nn (x) then takes the form ! 1 ∂Jv (x) n ∂J−v (x) Nn (x) = . (7.112) − (−1) π ∂v ∂v v=n One can show (Problem 7.7.4) that this Nn (x) also satisfies the recursion formulas (7.103) and (7.104). Hence it is a solution of the Bessel DE (7.106). That it is linearly independent of Jn can be demonstrated by first calculating their Wronskian W = Jv Nv − Jv Nv

(7.113a)

This can be calculated easily by using the asymptotic forms as x → ∞ of Jv and Nν , as described in Problem 7.7.8. The result W=

2 πx

(7.113b)

is independent of v. It should also hold for v = n, thus showing that Nn is linearly independent of Jn . Some Bessel and Neumann functions are shown in Fig. 7.4. Separation of the 3D Helmholtz equation in spherical coordinates yields the spherical Bessel equation (5.131) 

 d 2 z 2 + 2z + [z − l(l + 1)] fl (z) = 0, dz dz 2

d2

(7.114)

Bessel functions 1.0

491

Jn(x) J0

0.5

J1 J0

J1 x

0 5

10

15

–0.5 (a) 1.0

Nn(x)

N0

0.5

N1 N0

0

x 5

10

15 N1

–0.5

–1.0 (b)

Fig. 7.4 (a) Bessel functions J0 (x) and J1 (x); (b) Neumann functions N0 (x) and N1 (x).

where z = kr is a dimensionless variable. Physical applications often involve integrals of the form   Fl (z)F l (az)z dz, fl (z) fl (az)z2 dz = where Fl (z) = z1/2 fl (z)

(7.115)

is the “equivalent” function in cylindrical or circular coordinates. Indeed, a direct substitution of fl = z−1/2 F l into Eq. (7.114) shows that Fl satisfies the Bessel equation   2 d - 2 " 1 #2 . 2 d (7.116) Fl (z) = 0. z 2 +z + z − l+ 2 dz dz

492

Special functions

Hence Fl is proportional to Jl+1/2 or Nl+1/2 . A standard choice of the proportional constant then gives the spherical Bessel and Neumann functions: jl (z) =

 π 1/2

Jl+1/2 (z), 2z  π 1/2 nl (z) = Nl+1/2 (z). 2z

(7.117)

Eqs. (7.100) and (7.111) can now be used to obtain the power series   1 z2 /2 (z2 /2)2 l jl (z) = z − + − ... (2l + 1)!! 1!(2l + 3)!! 2!(2l + 5)!! and   (2l − 1)!! z2 /2 (z2 /2)2 1− + − ... , nl (z) = − 1!(1 − 2l) 2!(1 − 2l)(3 − 2l) zl+1 where the double factorial of a positive integer n is  3.1, n!! = n(n − 2) × . . . × 4.2,

odd even.

(7.118)

(7.119)

Recursion formulas for fl follow from those for Jv . For example, Eqs. (7.103) and (7.104) give fl−1 + fl+1 =

2l + 1 fl z

(7.120)

and l fl−1 − (l + 1) fl+1 = (2l + 1) fl .

(7.121)

The resulting ladder operators are 

 1 d Rl fl = fl = fl+1 − z dz   l+1 d Ll fl = fl = fl−1 . + z dz

(7.122) (7.123)

The spherical Bessel equation (7.114) for l = 0 can be cast into a familiar equation for Z(z) = zf 0 (z).

Bessel functions

493

The result (Problem 7.7.11) is 

d2 dz2

 + 1 Z(z) = 0.

(7.124)

This means that Z(z) is proportional to sin z or cos z. Eq. (7.118) then shows that j0 (z) =

sin z , z

n0 (z) = −

cos z . z

(7.125)

Finally, functions of higher orders can be generated by using the raising operator of Eq. (7.122): sin z cos z − , z z2   3 3 1 j1  − j1 = 3 − sin z − 2 cos z; j2 = z z z z

j1 = − j0 =

cos z sin z − , z z2   3 1 3 n2 = − 3 − cos z − 2 sin z. z z z

(7.126)

n1 = −

(7.127)

Some spherical Bessel and Neumann functions are shown in Fig. 7.5.

Problems

# " 7.7.1 Show that Eq. (7.102) follows from the symmetry property G x, − 1t = G(x, t). 7.7.2 Derive the recurrence relations (7.103) and (7.104). 7.7.3 Show that the function Jv defined by Eq. (7.100) with n replaced any real number v also satisfies Eqs. (7.103) and (7.104). 7.7.4 Show that the Neumann functions Nv (x) defined by Eq. (7.111) satisfies Eqs. (7.103) and (7.104) for any real v. 7.7.5 Use the product of generating functions G(x + y, t) = G(x, t)G(y, t) to derive the addition theorem Jn (x + y) =

∞ 

Jk (x)Jn−k (y).

k=−∞

7.7.6 Use the product of generating functions G(x, t)G(−x, t) = 1

494

Special functions j1(x) 0.6 j0 0.4

j1

0.2

j2

0

8

4

x

12

–0.2 (a)

n1(x)

n0 n1

0.2

n2 13

9 0

1

x

5

–0.2

–0.4 (b)

Fig. 7.5 (a) Spherical Bessel functions jl (x), l = 0 − 2; (b) Spherical Neumann functions nl (x), l = 0 − 2 (from Abramowitz and Stegun).

identity to derive the identity 1=

J02 (x)

+2

∞ 

Jn2 (x).

n=1

7.7.7 Verify Eq. (7.112) for the Neumann function of integral order n by differentiating both numerator and demoninator of Eq. (7.111) before setting v = n. 7.7.8 Show that the Neumann function Nv (x) of Eq. (7.111) is linearly independent of the Bessel function Jv (x) as follows:

Sturm-Liouville equation and eigenfunction expansions

495

(a) Show that the Bessel DE (7.98) with n → v can be written in the Sturm– Liouville form ! ,  d d p(x) + q(x) u(x) = 0, dx dx where p(x) = x and u(x) = Jv (x) or Nv (x). (b) Let u(x) and v(x) be two solutions of the Bessel DE of part (a). Show that 0 d / p(x)(uv − vu ) = 0, or dx A W = uv − vu = , p(x) where A is a constant independent of x. Show that u and v are linearly independent if A  0. (c) The Bessel and Neumann functions have the asymptotic forms as x → ∞ of 4   2 1 1 Jv (x) ∼ cos x − νπ − π , πx 2 4 4   2 1 1 sin x − νπ − π . Nv (x) ∼ πx 2 4 Use these asymptotic forms without derivation to show that W = 2/πx. 7.7.9 Verify Eq. (7.116). 7.7.10 Verify Eq. (7.118). Show that jl (z) can also be expressed in the form jl (z) = (2z)l

∞  (−1)n (n + l)! 2n z . n!(2n + 2l + 1)! n=0

7.7.11 Verify Eq. (7.124). 7.7.12 Obtain explicit expressions for j3 (z) and n3 (z).

7.8

Sturm-Liouville equation and eigenfunction expansions

Ordinary differential equations obtained by separating the Helmholtz equation (∇2 + k2 )ψ = 0 in various coordinate systems can be written in the general form   ! d d p(z) + q(z) ψ Lψ ≡ dz dz = −λw(z)ψ,

a ≤ z ≤ b,

(7.128)

496

Special functions

which is called a Liouville equation. Here λ is the separation constant (Section 5.12). The function p(z) could have zeros where the term involving ψ disappears. These points are called singular points of the equation. The behavior of the equation changes across a singular point. Hence it is convenient to separate a region into intervals in each of which p(z) has the same sign. Thus, by design, singular points when they appear coincide with one or both of the boundary points a and b of Eq. (7.128). The functions p(z), q(z) and w(z) are determined by the chosen coordinate system, and are real and well behaved inside each interval. Furthermore, the weight (density) function w(z) by choice does not change sign in the interval, and can be taken to be positive definite. A detailed discussion of these properties can be found in Morse and Feshbach (Sections 5.1 and 6.3). In the Sturm-Liouville theory, we study how λ affects the values of ψ and ψ at certain points in the interval. Consider two DEs of the form (7.128): L ψj = −λj wψj , L ψ∗i = −λ∗i wψ∗i ,

(7.129)

where we have allowed for complex solutions of DEs containing only real functions p(z), q(z), and w(z). Complex solutions are always possible because, if ψ1,2 are two linearly independent real solutions of a DE, then ψ1 + iψ2 are two linearly independent complex solutions. These equations can be combined into the expression ψ∗i L ψj − ψj L ψ∗i = (λ∗i − λj )wψ∗i ψj . Let us integrate both sides:  b  b (λ∗i − λj ) wψ∗i ψj dz = (ψ∗i L ψj − ψj L ψ∗i )dz. a

(7.130)

a

The first integral on the right-hand side can be integrated twice by parts:    b d ∗ d ψi p ψj dz dx dz a   b d ∗ d ∗ b ψi p ψj dz = ψi pψj |a − dz a dz =

p(ψ∗i ψj

−

b ψ∗ i ψj )|a

b +

!  d d ∗ p ψ ψj dz. dz dz i

(7.131)

a

The boundary terms vanish for each of the following choices of “homogeneous” boundary conditions (BCs) at z = a and b: ψi = 0,

Dirichlet BC,

(7.132a)

ψi ψi

= 0,

Neumann BC,

(7.132b)

+ cψi = 0,

mixed BC.

(7.132c)

Sturm-Liouville equation and eigenfunction expansions

497

In each case, the two integrals on the right-hand side of Eq. (7.130) become equal:  b  b ψi∗ L ψj dz = ψj L ψi∗ dz a



a

b

=

(L ψi )∗ ψj dz

a

 =

a

b

(ψj∗ L ψi )∗ dz,

(7.133)

where we have written the right-hand side in three equivalent forms. An operator L satisfying Eq. (7.133) is said to be Hermitian or self-adjoint. Thus an operator is Hermitian partly because of the BC (7.132) satisfied by the functions ψi∗ , ψj appearing in Eq. (7.l33). For other functions for which the boundary terms in Eq. (7.131) do not vanish, the same operator L is not Hermitian. There is no reason to believe that one of the BCs (7.132) will always be satisfied for any value of λ. It is possible, however, that it is satisfied for selected values (called a spectrum) of λ = λi , i = 0, 1, 2, . . . , N. We call these special values the eigenvalues of the DEs, and the associated solutions ψi their eigenfunctions. The eigenvalue spectrum satisfies a number of simple properties (Morse and Feshbach, p. 274): 1. They are infinite in number, that is, N = ∞. 2. Their distribution is discrete if b − a is finite. That is, λn+1 − λn  0 between two neighboring eigenvalues. The distribution becomes continuous, like points on a line, when b − a becomes infinite. Now if L is Hermitian, Eq. (7.130) gives  b ∗ (λi − λj ) wψ∗i ψj dz = 0.

(7.134)

a

The integrand is non-negative for j = i, since both w(z) and |ψi , (z)|2 are non-negative. The resulting integral does not vanish, except for the trivial case of ψi (z) = 0, which we shall exclude. Equation (7.134) then requires that the eigenvalue λ∗i = λi is real. On the other hand, if λj  λi , Eq. (7.134) can be satisfied only if the integral itself vanishes. Hence  b wψi∗ ψj dz = δij hi . (7.135) a

This shows that the eigenfunctions ψi form a system of orthogonal functions in the interval (a, b). 7.8.1

Eigenfunction expansion

In exact analogy to the Fourier-series expansion, any piecewise-continuous function F(z) between a and b can be expanded in terms of these eigenfunctions

498

Special functions

F(z) =

∞ 

ci ψi (z).

(7.136)

i=0

The expansion coefficient ci can be extracted as usual by calculating the inner product or overlap  b wψ∗i F dz = ci hi . (7.137) (ψi , F) ≡ a

Indeed, the similarity with the Fourier-series expansion can be taken quite literally: If ψi are ordered with increasing eigenvalues (i.e., λi+1 > λi ), the difference between an eigenfunction series and the Fourier series for the same function over the same interval, taken to the same number n of terms, can be shown to be uniformly convergent as n → ∞. (This is demonstrated in Morse and Feshbach, p. 743.) Eigenfunction expansions have been used in Sections 4.10, 5.8, 5.10–5.12, 7.4, and 7.5.

Problems 7.8.1 Show that the wave function u(r, θ, t) describing the vibrations of a circular drum of radius R has an eigenfunction expansion of the form u(r, θ, t) =

∞ ∞  

aim Jm (kim r)(sin mθ + bm cos mθ)

m=0 i=1

×(sin ωim t + cim cos ωim t). How are the eigenvalues m, kim , and ωim determined? 7.8.2 Identify the eigenfunctions of the following Sturm-Liouville equations. Explain why the eigenvalue spectrum is as shown, with n = 0, 1, 2, . . ..

(a) (b) (c) (d)

p(x)

q(x)

w(x)

a

b

λ

1 − x2 (1 − x2 )1/2 xe−x 2 e−x

0 0 0 0

1 (1 − x2 )−1/2 e−x 2 e−x

−1 −1 0 −∞

1 1 ∞ ∞

n(n + 1) n2 n 2n

Appendix 7 Tables of mathematical formulas 1 Orthogonal polynomials Tables 7.1–7.5 in Section 7.4 give Rodrigues’ formulas, generating functions, recursion formulas, differential relations and differential equations for the Legendre, Laguerre, Hermite and Chebyshev polynomials. Tables 4.1 and 4.2 of Section 4.10 give orthogonality relations and explicit formulas for polynomials of low degrees.

Sturm-Liouville equation and eigenfunction expansions

Expansion of R(x) in the orthogonal polynomials fn (x): R(x) =

∞ 

cn fn (x),

n=0

where ( fn , R) , ( fn , fn )  b ( fn , R) = fn (x)w(x)R(x) dx. cn =

a

2 Quantum oscillator  !   1 1 d2 2 ψn (ξ) = 0, − ξ − n + 2 dξ 2 2 √ 2 ψn (ξ) = (2n n! π)−1/2 Hn (ξ)e−(1/2)ξ ,   √ 1 d † − ξ ψn (ξ) = n + 1 ψn+1 (ξ), a ψn (ξ) ≡ − √ 2 dξ   √ 1 d aψn (ξ) ≡ √ + ξ ψn (ξ) = n ψn−1 (ξ), 2 dξ −

[a, a† ] = aa† − a† a = 1, (a† a − n)ψn = 0. 3 Spherical harmonics 

Ylm (θ, φ)  

(l − m)! 2l + 1 = (l + m)! 4π

1/2 eimφ Pm l (cos θ).



Ylm ∗ (θ, φ)Ylm (θ, φ)d cos θdφ = δl l δm m , 1

−1

m Pm l (x)Pl (x) dx =

Pl (cos α) =

2 (l + m)! δl l . (l − m)! 2l + 1

l 4π  m∗  m Y (Ω )Yl (Ω). 2l + 1 m=−l l

δ(Ω − Ω ) = δ(cos θ − cos θ )δ(φ − φ ) =

 lm

Ylm∗ (Ω )Ylm (Ω).

499

500

Special functions

4 Bessel functions !   ∞ 1 1 = tn Jn (x). x t− 2 t n=−∞   n d Rn Jn ≡ Jn = Jn+1 , − x dx   n d Ln Jn ≡ Jn ≡ Jn−1 , + x dx

exp

(Ln+1 Rn − 1)Jn = 0. J−n (x) ≡ (−1)n Jn (x). cos vπJv (x) − J−v (x) Nν (x) = sin vπ 2 . W = Jv Nv − Jv Nv = πx 5 Spherical Bessel functions  π 1/2 Jl+1/2 (z), nl (z) = Nl+1/2 (z). 2z 2z   l d Rl fl = − fl = fl+1 , z dz   l+1 d fl = fl−1 , + Ll fl = z dz jl (z) =

 π 1/2

(Ll+1 Rl − 1) fl = 0,

fl = j l

or

nl .

Explicit formulas for some spherical Bessel and Neumann functions are given in Eqs. (7.125)–(7.127). 6 Sturm-Liouville equation   ! d d p(z) + q(z) ψi = −λi w(z)ψi . L ψi = dz dz The operator L is Hermitian for each of the following homogeneous boundary conditions:

Sturm-Liouville equation and eigenfunction expansions

ψi = 0,

Dirichlet,

ψi = 0,

Neumann,

ψi

mixed,

+ cψi = 0,

501

at the boundaries z = a and b. Then the eigenvalues λi are real, and the eigenfunctions are orthogonal with respect to the density function w(z):  b w(z)ψ∗i (z)ψj (z) dz = δij hi . a

8 Functions of a complex variable 8.1

Introduction

The real number system is incomplete. For example, it does not include all the roots of algebraic equations such as x2 + 1 = 0 It was in connection with the study of the roots of cubic equations when the concept of the square root of a negative number became widely used in the early sixteenth century. However, the mathematics of complex numbers was not devel√ oped for another two centuries. It was Euler who introduced the symbol i = −1 in 1777. Years later, Gauss first used the notation a + ib to denote a complex number. In physics, we deal mostly with measurable attributes that can be quantified in terms of real numbers only. Complex numbers are tolerated, and even welcome, for the simplicity and convenience they bring to mathematical manipulations. The real significance of complex numbers in physics is not a matter of convenience, however, but of completeness. The use of complex numbers makes the description of nature mathematically complete. It is this completeness that makes the resulting physical theory elegant and aesthetically satisfying. It also ensures that nothing of importance is accidentally left out, including unexpected relations between distinct physical observables. How this happens is the story of this chapter.

8.2

Functions of a complex variable

A real number can be represented conveniently as a point on a straight line. A complex number, containing two real numbers, can be represented by a point in a 2D plane, the complex plane. A complex variable z = x + iy can be plotted in the complex plane by using its real and imaginary parts x = Re z and y = Im z as its two rectangular coordinates, as shown in Fig. 8.1. This complex variable can also be represented by the circular polar coordinates (r, θ) with the help of the trigonometric relations x = r cos θ,

y = r sin θ.

(8.1)

Functions of a complex variable

503

y = Im z z r θ

x = Re z

–θ

z* Fig. 8.1 The complex plane.

The dependence of z on the angle θ can be analyzed with the help of the Taylor series cos θ = 1 −

θ2 θ4 + − ..., 2! 4!

sin θ = θ −

θ3 θ 5 + − .... 3! 5!

(8.2)

As a result cos θ + i sin θ = 1 + iθ +

(iθ)2 (iθ)3 + + ... 2! 3!

= eiθ .

(8.3)

This Euler formula gives rise to the polar (or Wessel–Argand) representation z = reiθ

(8.4)

involving the modulus (amplitude or magnitude) r and the phase (or argument) θ of z. For example, on the unit circle we find the points eiπ/2 = i,

eiπ = −1,

ei3π/2 = −i,

ei2π = 1.

Other examples are ei = cos 1 + i sin 1 = 0.54 + i0.84; π   πe   πe  ie = exp i e = cos + i sin . 2 2 2 The multiplication of complex numbers can be done readily by using the rectangular representation, and even more easily by using the polar representation. For example, if

504

Functions of a complex variable

z1 = x1 + iy1 = r1 eiθ1 z2 = x2 + iy2 = r2 eiθ2 , then z1 z2 = (x1 + iy1 )(x2 + iy2 ) = x1 x2 − y1 y2 + i(x1 y2 + x2 y1 ),

(8.5)

z1 z2 = r1 r2 ei(θ1 +θ2 ) .

(8.6)

or

By equating the real and imaginary parts of these two equations for the special case of r1 = r2 = 1, we obtain the trigonometric identities cos(θ1 + θ2 ) = cos θ1 cos θ2 − sin θ1 sin θ2 ,

(8.7)

sin(θ1 + θ2 ) = sin θ1 cos θ2 + cos θ1 sin θ2 .

(8.8)

Trigonometric relations involving multiple angles can be similarly obtained: (cos θ + i sin θ)n = (eiθ )n = cos nθ + i sin nθ,

(8.9)

a result known as de Moivre’s Theorem. The replacement i → −i changes a complex expression into its complex conjugate. For example, the complex conjugate of z = x + iy is z∗ = x − iy = re−iθ .

(8.10)

This is just the mirror reflection of z about the real axis in the complex plane. The product zz∗ = r2 = z∗ z

(8.11)

is real and non-negative. The magnitude r is commonly denoted |z|. Functions can be defined for a complex variable in much the same way as for a real variable. We recall that by a real function of a real variable we mean a relation or mapping whereby one or more real numbers f (x) can be assigned to each value of the real variable x. This functional relationship is shown schematically in Fig. 8.2. Such a functional relation can be generalized to complex planes by assigning to each point of the complex plane of a complex variable z = x + iy one or more points in a different complex plane, the complex plane of the functional values. This mapping is represented schematically in Fig. 8.3. Each functional value is itself a complex number: f (z) = u(x, y) + iv(x, y).

(8.12)

This single-valued (or multivalued) function f(z) is called function of a complex variable (FCV).

Functions of a complex variable

505

f (x)

Mapping

x

Fig. 8.2 Function of a real variable.

Example 8.2.1 (a) The function w = az describes a scale change if a is real and a scale change plus a rotation if a is complex. (b) The function w = z + b describes a translation of the origin of the complex plane.  Example 8.2.2 f (z) = z2 = r2 e2iθ .

(8.13)

This is a single-valued function. The mapping is unique, but not one to one. It is a two-to-one mapping, since there are two variables z1 and z2 = −z1 with the same square. This is shown in Fig. 8.4(a), where we have used for simplicity the same complex plane for both z and f (z).  Example 8.2.3 f (z) = z1/2 . (a)

(8.14)

(b) y

v (x, y)

Mapping z f (z) x

u (x, y)

Fig. 8.3 Complex function of a complex variable. (a) Plane of the complex variable z = x + iy. (b) Plane of the complex function f (z) = u(x, y) + iv(x, y).

506

Functions of a complex variable (a)

(b) 2 1 2 2

f(z) = z =z

y

y z z1

f1(z) = z ½

2θ θ

x

z2

x

f2(z) = z ½ = –f1(z)

= –z1

Fig. 8.4 The square and square-root functions, (a) f (z) = z2 ; (b) f (z) = z1/2 .

There are two square roots: f1 (reiθ ) = r1/2 eiθ/2 , f2 = − f1 = r1/2 ei(θ+2π)/2 .

(8.15)

The function is double valued, and the mapping is one to two, as shown in Fig. 8.4(b).  In a similar way, the pth root z1/p = r1/p ei(θ+2πn)/p ,

p a positive integer,

(8.16)

gives a one-to-p mapping. Example 8.2.4 f (z) = ez = e x+iy = e x (cos y + i sin y).

(8.17)

The complex exponential function is periodic in y with a period of 2π. The maximum amplitude |ez | = e x increases monotonically from zero at x = −∞, through the value of 1 at x = 0 to infinity at x = ∞. Since its real and imaginary parts do not vanish at the same point, ez does not vanish for any finite value of z.  Example 8.2.5 f (z) = sin z = sin x cos iy + cos x sin iy. With the help of the results 1 1 cos iy = (ei(iy) + e−i(iy) ) = (e−y + ey ) = cosh y, 2 2 1 −y sin iy = (e − ey ) = i sinh y, 2i

(8.18)

Functions of a complex variable

507

the complex sine function can be written as sin z = sin x cosh y + i cos x sinh y.

(8.19)

In a similar way cos z = cos x cos iy − sin x sin iy = cos x cosh y − i sin x sinh y.

(8.20)

Both functions are periodic in x with a period of 2π. Since sinh(y = 0) = 0, these functions are real everywhere on the real axis, as we would expect. Away from the real axis, the real and imaginary parts do not vanish at the same time. As a result, these functions have zeros only along the real axis.  Example 8.2.6 tan z =

sin z , cos z

cot z =

cos z . sin z

Both functions are periodic with a period of π. Both functions are finite in the entire complex z plane except on the real axis at the zeros of the function in the denominator. 

Problems 8.2.1 Verify that (a) (1 + i)4 = −4; (b) ii = e−π/2"; # " # " # i (c) ie = exp − π2 sin 1 cos π2 cos 1 + i sin π2 cos 1 ; # " # " # " e πe πe (d) ei = exp cos πe 2 cos sin 2 + i sin sin 2 . 8.2.2 A Verify that (a) (1 + 2i) + i(2 + i) = 4i; (b) (1 + 2i)(2 + i) = 5i; (c) 1+2i 2−i = i; 1 (d) [(1 + i)(1 + 2i)(1 + 3i)]−1 = − 10

√ 8.2.3 Show that the roots of the quadratic equation z2 + z + 1 = 0 are (1 ± i 3)/2. What are the roots of z∗2 + z∗ + 1 = 0? 8.2.4 Find all the zeros of (a) sin z; (b) cosh z. 8.2.5 If z = (x, y) is considered a vector in the xy plane, show that (a) z1 · z2 = Re(z∗1 z2 ) = Re(z1 z∗2 ); (b) z1 × (z2 ) = k Im(z∗1 z2 ) = −k Im(z1 z∗2 ).

508

Functions of a complex variable

8.2.6 If (i − 1)z = −(i + 1)z∗ , show that z must lie on one of the two lines making an angle of 45◦ with the x axis. Hint: Use the polar representation. 8.2.7 Prove by purely algebraic means that |z1 | − |z2 | ≤ |z1 + z2 | ≤ |z1 | + |z2 | for any two complex numbers zi and z2 . Hint: The inequalities hold for the square of each expression. 8.2.8 Use de Moivre’s theorem to show that (a) cos 3θ = 4 cos3 θ − 3 cos θ; (b) sin 3θ = 3 sin θ − 4 sin3 θ. 8.2.9 Use de Moivre’s theorem to show that (a)   n n cos nθ = cos θ − cosn−2 θ sin2 θ 2   n cosn−4 θ sin4 θ − . . . ; + 4 (b)

    n n n−1 cos θ sin θ − sin nθ = cosn−3 θ sin3 θ 1 3   n cosn−5 θ sin5 θ − . . . , + 5

" # where mn = n!/(n − m)!m! are binomial coefficients. N inx e of the geometric series is 8.2.10 Show that the partial sum S N+1 = n=0   S N+1 = ei(N+1)x − 1 /(eix − 1). Use the result and Euler’s formula to show that N 

cos nx =

sin 12 (N + 1)x sin 12 x

n=0

Obtain an analogous result for 8.2.11 Show that the equation

cos

Nx . 2

N

n=0 sin nx.

(z − a)(z∗ − a∗ ) = R2 is the equation of a circle of radius R centered at the point a in the complex plane.

Multivalued functions and Riemann surfaces y

B A

v A B

z

x

509

f

u f2 in LHP

B A

f1 in RHP

Fig. 8.5 The two roots of z1/2 .

8.2.12 Describe the functions (a) z−1 ; (b) (cz + d)−1 ; and (c) (az + b)/(cz + d), where a, b, c and d are complex constants. 8.2.13 Show that (a) sinh(x + iy) = sinh x cos y + i cosh x sin y, cosh(x + iy) = cosh x cos y + i sinh x sin y. (b) | sinh z|2 = sinh2 x + sin2 y, | cosh z|2 = sinh2 x + cos2 y. sinh 2x + i sin 2y . (c) tanh z = cosh 2x + cos 2y (Hint: First find expression for sin 2y, cos 2y, sinh 2x and cosh 2x.) 8.2.14 Show that −1 (a) (z/z∗ )i = e−2 tan (y/x) ; " # ew − e−w −1 1+z 1 (b) z = tanh w = w , where w = tanh z = ln 2 1−z ; e +√e−w (c) sinh−1 z = ln(z ± z2 + 1), √ cosh−1 z = ln(z ± z2 − 1); √ √ (d) sin−1 z = −i ln(iz + 1 − z2 ) = i ln(−iz + 1 − z2 ), √ √ + z2 − 1) = i ln(z − z2 − 1), cos−1 z = −i ln(z " # tan−1 z = 2i ln i+z i−z .

8.3

Multivalued functions and Riemann surfaces

If the angle θ is restricted (arbitrarily) to −π ≤ θ ≤ π in z = reiθ , its first square root f1 = r1/2 eiθ/2 covers the right half-plane (RHP) (−π/2 ≤ θ/2 ≤ π/2) of f , while f2 = −r1/2 eiθ/2 covers the left HP. This is shown in Fig. 8.5. Two points A and B with zA = rei(−π+δ) and zB = rei(π−δ) , where δ > 0 is small, are actually close together in the complex z plane. But their square roots are widely

510

Functions of a complex variable y

v

z

f

A2 B 1 B2 B1 A2 A1

C1

C2

x

C2

C1

u

B2 A1

Fig. 8.6 Mapping of points for the function z1/2 .

separated in the complex f plane. This is because A is close to the lower boundary of θ = −π/2, while B is close to the upper boundary θ = π/2. To cover the left HP of the second square root f2 , it is necessary to go around the z plane once more, as shown in Fig. 8.6. We see that, although all four points A1 B1 A2 , B2 are close together on the z plane, only A1 and B2 , and only B1 and A2 , are close together in the complex f plane, because of our choice of the boundaries for θ. This is not a problem of FCV alone; it appears also for a real function of a real variable. For example, x1/2 has a positive root |x1/2 | and a negative root −|x1/2 |, as shown in Fig. 8.7. Two points C and D that are close together in x can become widely separated in x1/2 , as shown by the pairs C1 , D2 , and C2 , D1 . The distinct values of a multivalued function are called its separate branches. For example, the first branch of the complex square root, as defined in Fig. 8.6, is on the RHP, while the second branch is on the left HP. In order to differentiate the function at the boundary points such as A1 or B1 , it is necessary to include another branch of the function, corresponding to the points B2 and A2 , respectively. Yet in the complex plane of the variable z, the points A1 and A2 are indistinguishable if they are at the same location. This causes problems in differential calculus.

x 1/ 2

C1 D1

C

D

C2 D 2

Fig. 8.7 The two branches of x1/2 .

x

Multivalued functions and Riemann surfaces

8.3.1

511

The Riemann Surface

To distinguish between identical points in z that give rise to different branches of a function, Riemann (Doctoral Thesis, 1851) suggested the following procedure: For a double-valued function such as z1/2 , take two sheets or copies of the complex z plane. Make a cut, the branch cut, on both sheets from r = 0 to ∞ along any suitably chosen boundary value of θ. In our example, this is just the negative real axis, as shown by the wavy line on Fig. 8.6. Next, cross-join together the edges of the cut so that B1 is joined to A2 , and B2 is joined to A1 . The resulting Riemann surface is a continuous two-storied structure with two sets of stairs at the cross-joints. We shall refer to a line of cross joints as a branch line. It coincides in location with that of the branch cut. To explore this structure, we may start at A1 on the first, say bottom, sheet. Going around the origin once, we reach B1 on the first sheet. This point is separated from A1 by an impassable crevice, which is our chosen branch cut. We can, however, take the cross-joining stairs up to A2 on the second, or top, sheet. A second turn around the origin brings us to B2 on the second sheet across the abyss from A2 . We see a second set of cross-joining stairs that now takes us down to our starting point A1 on the first sheet. A complete exploration of this Riemann surface thus requires two complete turns around the origin. In this way, the area of the complex z plane has been doubled. The mapping is thereby changed from the original one-to-two mapping to two one-to-one mappings. The double-valued function has been changed to a continuous single-valued function on one two-sheeted Riemann surface. Thus the antidote for the discontinuities of a multivalued function is a multisheeted Riemann surface for the complex z “plane”. A branch cut has to begin or end at a point that of necessity belongs to more than one sheet of the Riemann surface. This point is called a branch point of the function. The function is either zero or infinity at a branch point. It is then immaterial whether it is considered to be single, or multiple, valued there. The position of a branch point is a property of the function, and is independent of the choice of branch cuts. For z1/2 , the branch points are at z = 0 and ∞. The point at ∞ is sometimes not considered a branch point, although a branch cut can terminate there. Since one and only one branch of a function can be located on each Riemann sheet, only one branch cut can be drawn through a branch point. Again an exception has to be made for z = ∞, where any number of branch cuts can terminate. It can happen that two branch cuts both go to z = ∞, but that point is not a branch point. In that case, there is only one branch cut that connects two branch points by way of z = ∞. While a branch cut has to terminate at a branch point, its exact position and shape can be quite arbitrary. The latter depend on the arbitrary choice of the boundary for θ. Fig. 8.8 shows other equally acceptable choices of the branch cut for z1/2 . The feature common to all these choices is that neighboring points on the Riemann surface (such as the points B1 and A2 , and the points A1 and B2 in Fig. 8.6) are always mapped onto neighboring points on the complex f (z) plane. Concerning the arbitrariness of branch cuts, it has been said that God created branch points, but men made the cuts.

512

Functions of a complex variable

(a)

(c)

(b)

Fig. 8.8 Equally acceptable choices of the branch cut for f (z) = z1/2 . (a) 0 ≤ θ ≤ 2π; (b) −π/2 ≤ θ ≤ 3π/2; (c) θ of branch cut is r dependent.

The construction of the Riemann surface can be represented symbolically by a simple notation. We shall denote by S i , the ith Riemann sheet on which the function f (z) is in its ith branch fi (z). For example, z1/2 has two branches, and its Riemann surface has two sheets: S1 :

f = f1 = r1/2 eiθ/2 ,

S2 :

f = f 2 = − f1 .

For a branch cut we usually use just a straight line L connecting the two branch points P1 , and P2 . Then L may be specified by the symbol (P1 , P2 ). For example, a branch cut connecting the origin and infinity is (0,∞) if it is along the positive real axis, (−∞, 0) if it is along the negative real axis, (0, i∞) if it is along the positive imaginary axis, and (−i∞,0) if it is along the negative imaginary axis. We shall denote by S i± the upper and lower edges of a branch cut on the sheet − ) or S i , The joining of neighboring sheets will be denoted by the pairing (S i+ , S i+1 + − (S i+1 , S i ). Thus a Riemann surface for f (z) = z1/2 may be characterized as follows: Branch points: Branch cut: Riemann surface:

z = 0, ∞ (i.e., including ± ∞, ±i∞). L = (−∞, 0). is made by the following joinings of sheets: at L : (S 1+ , S 2− ),

(S 2+ , S 1− ).

These details are shown in Fig. 8.9, where U(L)HP stands for the upper (lower) halfplane. A number of examples are given below to help the reader become familiar with the concept of Riemann surfaces. Example 8.3.1 f (z) = (z + 1)1/2 + (z − 1)1/2 .

Multivalued functions and Riemann surfaces Left view

Top view y z 2 x

Left

1

Right

UHP LHP S +2 S –2 S +1 S –1

513

Right view LHP

Sheet

UHP

2

2

2

1

1

1

Fig. 8.9 Three views of a Riemann surface of z1/2 .

This function has four branches: f1 (z) = r11/2 eiθ1 /2 + r21/2 eiθ2 /2 , f2 (z) = r11/2 eiθ1 /2 − r21/2 eiθ2 /2 , f3 (z) = −r11/2 eiθ1 /2 + r21/2 eiθ2 /2 ,

(8.21)

f4 (z) = −r11/2 eiθ1 /2 − r21/2 eiθ2 /2 , where, r1 , θ1 , r2 and θ2 are defined in Fig. 8.10. The branch points are −1, 1, and ∞. The branch cuts may be chosen to be L1 = (−∞, −1)

and

L2 = (1, ∞),

these being the branch cuts of the first and second square roots, respectively. The resulting Riemann surface is then obtained by the following joining of sheets: at L1 :

(S 1+ , S 3− ),

(S 3+ , S 1− ),

(S 2+ , S 4− ),

(S 4+ , S 2− ),

at L2 :

(S 1+ , S 2− ),

(S 2+ , S 1− ),

(S 3+ , S 4− ),

(S 4+ , S 3− ),

Fig. 8.10 gives three views of this Riemann surface. To calculate the actual value of f (z) at each point on its four-sheeted Riemann surface, we must remember that our choices of branch cuts imply that −π ≤ θ1 ≤ π 0 ≤ θ2 ≤ 2π. For example, at the origin (z = 0) where r1 = (r1 , θ1 ) = (1, 0), Left view

y z r1 L1

θ1 –1

r2 = (1, π),

1

UHP

r2 θ2 x

L2

Right view

LHP

LHP

UHP

4 3 2 1

Fig. 8.10 Three views of a Riemann surface for (z + 1)1/2 + (z − 1)1/2 .

514

Functions of a complex variable Left view

y z

UHP

r θ

Right view

LHP

LHP

UHP

3 2 1

x

Fig. 8.11 Three views of a Riemann surface for z1/3 .

we find f1 (0) = 1 + ei(π/2) = 1 + i, f3 (0) = −1 + i,

f2 (0) = 1 − i,

f4 (0) = −1 − i. 

Example 8.3.2 g(z) = z1/3 , f (z) = z2/3 = g2 (z). The function g(z) has three branches ! θ + 2π(n − 1) . exp i 3 

gn (z) = r

1/3

(8.22)

Thus g1 (z) = r 1/3 ei(θ/3) , g2 (z) = kg1 (z), g3 (z) = k2 g1 (z),

with k = ei(2π/3) .

The branch points are at 0 and ∞. As we circle the origin in the z plane in a positive direction, we see these three branches of the function repeatedly in the order g1 → g2 → g3 → g1 , etc. Hence if the branch cut is taken to be L = (−∞, 0), the Riemann surface should be formed by joining the cut edges as follows (S 1+ , S 2− ),

(S 2+ , S 3− ),

(S 3+ , S 1− ).

The angular range in g1 is −π ≤ θ ≤ π. The result is shown in Fig. 8.11. The mapping g = z1/3 involves three Riemann sheets in z mapping onto a single g plane. The function f = g2 maps the single g plane onto two copies of the f plane. Hence z2/3 maps three sheets of the z plane onto two sheets of the f plane, as shown in Fig. 8.12. For the chosen branch cuts, we find that the top half of the first sheet plus the second sheet of z maps onto the first sheet of f , while the bottom half of the first sheet plus the third sheet of z maps onto the second sheet of f . 

Multivalued functions and Riemann surfaces Left view z

UHP

515

g

LHP

Right view f

LHP

g

UHP

Fig. 8.12 Riemann surfaces for z2/3 .

Example 8.3.3 

z+1 f (z) = z−1

1/3 .

This function has two branch points (z1 = −1 and z2 = 1) and three distinct branches (for n = 1, 2, and 3):  1/3 !  θ − θ  2π r1 1 2 + (n − 1) , exp i (8.23) fn (z) = r2 3 3 where ri and θi , are the polar coordinates of z as measured from the branch point zi . Note that n = n1 − n2 , mod 3, where ni is the number of counterclockwise turns around the branch point zi . The point z = ∞ is not a branch point, since the function is still multivalued there. a branch cut connecting the two branch points can be made by way of infinity, leading to the Riemann surface shown in Fig. 8.13. The connection on the right is dictated by the requirement that, on circling z2 = 1 in a positive (counterclockwise) direction, we see the different branches backwards, that is, in the order g1 → g3 → g2 → g1 , etc. Left view

y

–1

z

1

UHP

x

Right view

LHP

LHP

UHP

3 2 1

Fig. 8.13 Three views of a Riemann surface for [(z + 1)/(z − 1)]1/2 .

516

Functions of a complex variable v(x,y)

y z f1(z) B A

C

f0(z)

x

f–1(z)

B C

u(x,y)

A

Fig. 8.14 Different branches of the function ln z.

This is because z − 1 is raised to a negative power, so that increasing n2 decreases n = n1 − n2 . The actual functional values are easily calculated with the help of Eq. (8.23), with −π ≤ θ1 ≤ π and 0 ≤ θ2 ≤ 2π. For example: f1 (0) = e−i(π/3) , f2 (0) = k f1 = ei(π/3) , f3 (0) = k2 f1 = e+iπ = −1. where k = ei(2π/3) is a cube root of 1.  Example 8.3.4 f (z) = ln z. The complex logarithmic function has infinitely many branches: fn (z) = ln r + i(θ + 2πn),

(8.24)

where n is any (positive or negative) integer. The function f0 (z) is called its principal branch. Different branches of ln z occupy different strips of the complex f plane; each strip has a height 2π along the y direction, and an infinite width stretching from x = −∞ → ∞. The branch points are at z = 0 and ∞. Fig. 8.14 shows three branches of the function for the choice L = (−∞, 0) for the branch cut. The Riemann surface should be joined together as follows + , S 0− ), (S 0+ , S 1− ), (S 1+ , S 2− ), . . . , . . . , (S −1

since in Eq. (8.24) the branch number n increases when θ increases through 2π. The angular range for f1 is −π ≤ θ ≤ π. The result is shown in Fig. 8.15.  Example 8.3.5 f (z) = ln[z/(z + 1)], √ g(z) = z + 1 ln z.

Multivalued functions and Riemann surfaces Left view y

UHP

517

Right view

LHP

LHP

UHP

2 1 0 –1

x

–2

Fig. 8.15 Three views of a Riemann surface for ln z.

Both functions have branch points at z = −1, 0, and ∞. The branch cuts can be chosen to be the same for both functions, as shown in Fig. 8.16(a) and (b). both functions have an infinity of branches because of the ln. For f (z), the Riemann surface for Fig. 8.16(a) can be so chosen that the effects of the two ln branch cuts along the positive real axis cancel each other. The function then remains on the same branch on crossing the positive real axis. This is the situation described by Fig. 8.16(c). The situation is different for g(z), which has a double set of ln branches due to the square root. The branch cut from z = −1 to infinity separates two square-root branches of the same ln function, while the branch cut from 0 to ∞ separates two ln branch of the same square-root function. These two branch cuts have different effects. They do not cancel each other out even if they lie together along the positive x axis.  y

x

0

–1

(a) y

–1

(b)

0

y

x

–1

0

x

(c)

Fig. 8.16 Branch lines for ln[z/(z + 1)]; (a) and (b), but not (c), are also valid choices for

√

z + 1 ln z.

518

Functions of a complex variable y

Left view UHP

L1 Inside Ö

Right view

LHP

LHP

UHP

4 3 2 1

L x 2 1 Outside Ö for branches 3 and 4

Fig. 8.17 Three views of a Riemann surface for (1 +

√

z)1/2 .

Example 8.3.6 f (z) = (1 +

√ 1/2 z) .

There are four branches: f1 (z) = (1 + r1/2 eiθ/2 )1/2 , f2 (z) = −(1 + r1/2 eiθ/2 )1/2 , f3 (z) = (1 − r1/2 eiθ/2 )1/2 , f4 (z) = −(1 − r1/2 eiθ/2 )1/2 ,

(8.25) −π ≤ θ ≤ π.

The inside square root gives branch points at z = 0 and ∞ on all four sheets. The outside square root has a branch point 1 and ∞, but only on the third and fourth sheets. The associated branch cuts can be taken to be the real axes at x ≥ 1 on these two sheets. The Riemann surface can then be obtained by cross-joining the cut edges as follows: at L1 :

(S 1+ , S 3− ), (S 3+ , S 1− ), (S 2+ , S 4− ), (S 4+ , S 2− ),

at L2 :

(S 3+ , S 4− ), (S 4+ , S 3− ),

This Riemann surface is illustrated in Fig. 8.17.  Example 8.3.7 f (z) = arcsin z. This function is not as simple as it might look. First, it is necessary to define it in terms of more familiar functions: & f (z) = i ln( 1 − z2 − iz) = i ln[h(z) − iz] = i ln g(z).

Multivalued functions and Riemann surfaces

519

This shows that the function has infinitely many branches, which can be arranged in the form f2n (z) = i ln |g1 (z)| − φ1 (z) − 2πn, f2n−1 (z) = i ln |g2 (z)| − φ2 (z) − 2πn,

(8.26)

√ where gi (z) = hi (z) − iz is defined in terms of the ith branch of h(z) = 1 − z2 , and φi (z) is the polar angle of gi (z). The determination of branch points and the choice of branch √ cuts are also more involved. We first note that the Riemann surface for g(z) = 1 − z2 − iz has two sheets joined at the branch lines, which may be taken to be L1 = (−∞, −1)

and

L2 = (1, ∞).

These two sheets are cross-joined at these branch lines in the usual way for the squareroot function. Like the simple square root, the two cross-joints at the same branch line are not equivalent because they map onto different lines in g(z). Consider for example the two copies of the branch line L2 . On that copy lying at the cross joint (S 1+ , S 2− ), the function g(z) has the value of its first branch & g1 (z) = g1 (x) = i x2 − 1 − ix. Hence this copy of L2 maps onto that part of the negative imaginary axis of the complex g plane from −i to 0. At the cross-joint (S 2+ , S 1− ), g(z) has the value of its second branch & g2 (z) = g2 (x) = −i x2 − 1 − ix. This second copy of L2 therefore maps onto the remaining part of the negative imaginary axis from −i to −i∞. In a similar way, the branch line L1 maps onto the positive imaginary axis. We now consider f (z) as a logarithmic function of g. The Riemann surface for i ln g has infinitely many sheets. That is, the complex g plane is a multisheeted structure with branch points at g = 0 and ∞. We may choose the branch cut for i ln g to be the negative imaginary g axis. If so, this coincides with two maps of the branch line L2 in the complex z plane. The complex z plane is now made up of infinitely many copies of the two-sheeted Riemann surface of the last paragraph. They can be joined together into one continuous surface by cutting open both branch lines at L2 and cross-joining together the edges belonging to different logarithmic branches. To summarize, a Riemann surface can be constructed according to the scheme: at L1 :

+ − + − (S 2n , S 2n−1 ), (S 2n−1 , S 2n );

at L2 :

+ − + − (S 2n , S 2n+1 ), (S 2n−1 , S 2n+2 ).

520

Functions of a complex variable Left view

y

L1

–1 ÷

x 1 ÷ and ln

Right view 4 3 2 1 0 –1 –2 –3

L2 +÷ –÷ + – + –

UHP

LHP

+ – + – + –

LHP

UHP

Fig. 8.18 Three views of a Riemann surface for arcsin z.

This Riemann surface is shown in Fig. 8.18, where the two square-root branches belonging to the same ln branch are indicated. Across L1 , the ln part of the function remains on the same branch, while across L2 , both the ln and the square root change branch. 

Problems 8.3.1 The branch cuts of the function f (z) = (z2 − 1)1/p , where p is a positive integer, have been chosen to be L1 = (−∞, −1),

L2 = (1, ∞),

as shown in the Fig. 8.19. Obtain the value of f (z) for the four points shown, namely z = 2 ± iε, −2 ± iε, with  → 0+ (i.e.,  approaches zero along the positive real axis) on the Riemann sheet specified below: (a) For p = 2 or f (z) = (z2 − 1)1/2 , on that Riemann sheet on which f (0) = ei(3π/2) ; (b) For p = 3 or f (z) = (z2 − 1)1/3 , on that Riemann sheet on which f (0) = ei(5π/3) ; y

–2 + i ε

2+iε

–2 – i ε

2–iε

x

Fig. 8.19 Branch points and cuts for f (z) = (z2 − 1)1/p to be used in Problem 8.3.1.

Complex differentiation: Analytic functions

521

(c) For p = 4 or f (z) = (z2 − 1)1/4 , on that Riemann sheet on which f (0) = ei(5π/4) . 8.3.2 Construct the Riemann surfaces for (a) z1/2 + z1/3 (6 branches); (b) z5/6 (6 branches); (c) (z − 1)1/3 + (z + 1)1/3 (9 branches); (d) (z2 − 1)1/3 (3 branches); (e) arctan √ z (∞ branches); (f) ln( z + 1) (∞ branches); (g) (z1/3 + 1)1/2 (6 branches). Verify the given number of branches by finding an explicit expression for fn (z), the function on the n-th branch. 8.3.3 On which sheet of your chosen Riemann surface does the denominator of f (z) =

1 (z + 1)1/2 + 2i

have a zero?

8.4

Complex differentiation: Analytic functions and singularities

Given a continuous, single-valued FCV f (z) = u(x, y) + iv(x, y), we define as usual the derivative f  (z) =

f (z + Δz) − f (z) d f (z) = lim . Δz→0 dz Δz

This means that f  (z) =

lim

Δx,Δy→0

u(x + Δx, y + Δy) − u(x, y) + i(same for v) . Δx + iΔy

(8.27)

There are of course infinitely many ways to approach a point z on a 2D surface, but there are only two independent ways—along x or along y: Along x: f  (z) = Along y:

∂ ∂u ∂v f (z) = +i ∂x ∂x ∂x

  1 ∂ 1 ∂u ∂v . f (z) = +i f (z) = i ∂y i ∂y ∂y 

522

Functions of a complex variable

A unique derivative can appear only if these two results agree. Hence the CauchyRiemann conditions ∂u ∂v = , ∂x ∂y

∂v ∂u =− ∂x ∂y

(8.28)

must be satisfied. These conditions turn out to be both necessary and sufficient. If f (z) has a unique and finite derivative, it is said to be differentiable. If f (z) is differentiable at z0 , and in a small neighborhood around z0 in the complex plane, f (z) is said to be an analytic function (also called regular or holomorphic) at z = z0 . If f (z) is analytic everywhere in the complex plane within a finite distance of the origin, it is an entire function. If f (z) is not differentiable at z = z0 , it is said to be singular there. The point z0 is then called a singular point or singularity of f (z). Example 8.4.1 For f (z) = z2 = x2 − y2 + 2ixy: ∂u ∂v = 2x = , ∂x ∂y ∂v ∂u = −2y = − . ∂y ∂x Hence the Cauchy-Riemann conditions are satisfied. In addition, the partial derivatives are finite except at z = ∞, while the function itself is single valued. Hence the function is analytic for all finite z. That is, it is an entire function.  A much faster check can be made by first assuming the validity of complex differentiations, as illustrated in the following example. Example 8.4.2 For f (z) = 2zn u = Re f (z) = zn + z∗n , v = Im f (z) = (zn − z∗n )/i. ∴

(8.29)

∂z∗

∂u ∂z = nzn−1 + nz∗n−1 = nzn−1 + nz∗n−1 , ∂x ∂x ∂x ∂u ∂v 1 n−1 = [nz i − nz∗n−1 (−i)] = , ∂y i ∂x ∂u ∂v = nzn−1 i + nz∗n−1 (−i) = − . ∂y ∂x

The function is single-valued for integral values of n. It is then analytic at every point where the partial derivatives are finite. For positive integers n, this occurs for all finite z. Hence the positive power zn is an entire function, but there is a singularity at z = ∞. For negative integers n, the function zn is analytic everywhere except at the origin. Finally, for n = 0 the function is analytic everywhere including z = ∞. It is actually just a constant [ f (z) = 2]. 

Complex differentiation: Analytic functions

8.4.1

523

Singularities

Functions of a complex variable have relatively simple structures in regions where they are analytic. Therefore many problems reduce to a study of these functions at singular points where they are not analytic. If a function is singular at the point z = a, but is analytic in all neighborhoods of a, the point z = a is called an isolated singular point of the function. If near an isolated singular point z = a, the function behaves like (z − a)−n , where n is a positive integer, then z = a is a pole, or more specifically an nth-order pole, of f (z). It is a simple pole if n = 1. If the order n of the pole goes to infinity, the singularity is said to be an essential isolated singularity. A classic example is exp(1/z) at z = 0: ∞  1 −n e =1+ z . n! n=1 ⎧ ⎪ positive real axis : exp(1/z) → exp(∞) → ∞ ⎪ ⎪ ⎪ ⎪ ⎨ negative real axis : → exp(−∞) → 0 If z → 0 along the ⎪ ⎪ imaginary axis : → exp(±i∞) → ⎪ ⎪ ⎪ ⎩ oscillatory. 1/z

Indeed an essential singularity, whether isolated or not, is one at which the function behaves in a wild manner. If an essential singularity is isolated, it satisfies Picard’s theorem: A function in the neighborhood of an essential isolated singularity assumes every complex value infinitely many times, except perhaps one particular value. While the function sin(1/z) has an essentially isolated singularity at z = 0, its reciprocal [sin(1/z)]−1 behaves even more wildly at and near z = 0. Indeed it shows an infinity of poles as we approach their limit point z = 0. We call this point an essential singularity rather than an essential isolated singularity. A branch point is another singularity that is not isolated. A branch point belongs to two or more branches; it is a point in the neighborhood of which the function is multivalued. A branch of a function is not continuous across a branch cut on a Riemann sheet, although the function itself is continuous if we go continuously on the Riemann surface across a branch line to another Riemann sheet. Among the simplest functions is the power zn . It is an entire function, analytic everywhere within a finite distance of the origin. However, it has an nth-order pole at infinity in the same way that the inverse power z−n has an nth-order pole at the origin. An entire function that behaves at infinity more weakly than zn is a polynomial of degree less than n. A function that is everywhere finite and analytic must be a constant. The last result is known as the Liouville Theorem. Thus singularities are the sources of functional behavior. Since physics describes phenomena that are changing in spacetime, we must conclude that singularities are also sources of physical phenomena. For this reason, we are more interested in singularities than in analyticity, although conceptually the two must go together.

524

Functions of a complex variable

Problems 8.4.1 Which of the following are analytic functions of z? (a) Im z = y; (b) e|z| ; 2 (c) ez ; (d) z∗n , integer n > 0; (e) f (z∗ ), an analytic function of z∗ . 8.4.2 Show that z1/2 is analytic in the complex z plane within a finite distance of the origin except at the branch point z = 0 by using (a) complex differentiation, (b) real differentiation. For the latter, use the Cauchy–Riemann conditions in circular polar coordinates given in Problem 8.4.4. 8.4.3 Locate and identify the singularities of the following functions (a) (z + i)2/3 (b) tan(1/z); 1 ; (c) √ z + i + 2i z2 (z − 1) (d) . sin2 πz 8.4.4 Show that in circular polar coordinates the equality of the two independent derivatives of an analytic function f (z = reiθ ) requires that ∂f 1 ∂f = . ∂r ir ∂θ Hence if f = ReiΘ , the Cauchy-Riemann conditions read ∂R R ∂Θ = , ∂r r ∂θ ∂Θ 1 ∂R R = . ∂r r ∂θ 8.4.5 Show that, if an analytic function is everywhere real, it can only be a constant. 8.4.6 Show that f (z) = sinh(x2 − y2 ) cos(2xy) + i cosh(x2 − y2 ) sin(2xy) is an entire function.

8.5

Complex integration: Cauchy integral theorem and integral formula

A function f (z) of a complex variable z can be integrated along a path C on the complex z plane   f (z)dz = [u(x, y) + iv(x, y)](dx + idy). C

C

Complex integration: Cauchy integral theorem

525

The results are particularly simple when f (z) is an analytic function, because of the Cauchy integral theorem: If f (z) is analytic in a simply connected domain D, and C is a piecewise smooth, simple closed curve in D, then  f (z)dz = 0, (8.30) C

where the circle on the integration sign denotes the closure of the integration path C. To prove this theorem, we note that    f (z)dz = (udx − vdy) + i (vdx + udy) C

C

 =

 A(r) · dr + i

C

C

B(r) · dr. C

if expressed in terms of the 2D vector fields A(r) = ui − vj,

B(r) = vi + uj.

Stokes’s theorem can now be applied to yield    f (z)dz = dσ · ( ∇ × A + i∇ × B) C

S

  =

  ! ∂u ∂v ∂v ∂u +i . dx dy − + − ∂x ∂y ∂x ∂y 

Since f (x) satisfies the Cauchy-Riemann conditions, both the real and the imaginary parts of the integrand vanish, thus proving the Cauchy theorem. The simply connected domain referred to in the Cauchy theorem is one with no hole in it. A simple curve is one that does not intersect itself. A curve is piecewise smooth if it is made up of a finite number of pieces, each of which is smooth. The closed path C is always taken in a positive, that is, counterclockwise, direction, unless otherwise stated. Because of the Cauchy theorem, an integration contour can be moved across any region of the complex plane over which the integrand is analytic without changing the value of the integral. It cannot be moved across holes or singularities, but it can be made to collapse around them, as shown in Fig. 8.20. As a result, an integration contour C0 enclosing n holes or singularities can be replaced by n separately closed contours Ci each enclosing one hole or singularity:  n   f (z)dz = f (z)dz. C

i=1

Ci

526

Functions of a complex variable

C0

C1

C2

Fig. 8.20 Collapsing a closed contour around a hole (the shaded area) and a singularity (the dot) without changing the value of an integral.

If the integrand is analytic over the whole region enclosed by a closed contour, the contour can be shrunk to a point. Consequently the integral vanishes. This is just the result stated by the Cauchy theorem. Nontrivial contour integrals involve functions that are not everywhere analytic in the enclosed region. The simplest such functions are of the type (z − a)−n , where n is a positive integer (i.e., functions with an nth-order pole at z = a). A closed-path integration around such a pole can be evaluated easily by shrinking the contour C to a small circle c of radius  around a:    2π 1 0, n1 iθ 1−n dz = (εe ) idθ = (8.31) n 2πi, n = 1, (z − a) 0 C since the periodic function eiθ(1−n) reduces to 1 only for n = 1. More generally, if f (z) is analytic inside and on a closed curve C, and if the point z = a is inside C, then the integral  f (z)dz I= C z−a can again be evaluated by collapsing C into a small circle c around a:   f (z) − f (a) 1 I= dz + f (a) dz z−a C C z−a  = d f (z) + 2πi f (a). C

Since df (z) is a total differential, the first term vanishes, leaving  f (z)dz = 2πi f (a), C z−a a result known as the Cauchy integral formula. It is useful to write Eq. (8.32) in the form

(8.32)

Complex integration: Cauchy integral theorem

1 f (z) = 2πi

 C

f (z )dz z − z

527

(8.33)

to emphasize the fact that z can be any point inside the closed path C. Furthermore, it states that an analytic function at z has a value that is a certain average of functional values on any closed path C surrounding z. Knowledge of an analytic function f (z) on a closed path is thus sufficient to determine its value anywhere inside the path where the function remains analytic. Equation (8.33) can be differentiated n times to yield the formula  dn f (z) f (z )dz n! = . (8.34) dzn 2πi (z − z)n+1 This result is useful in understanding Taylor expansions of analytic functions in Section 8.7.

Problems 8.5.1 Verify the following contour integrals along the triangular closed contour (0, 0) → (1, 0) → (1, 1) → (0, 0) by direct integration for each segment.  i (a) Re z dz = ; 2  (b) z dz = 0. Interpret these results in the context of the Cauchy integral theorem. Hint: Impose any constraint on dz = dx + idy appropriate to each segment. 8.5.2 For integration once around the unit circle in the positive direction, show that  z e 2πi (a) dz = ; 5 4!  z 1/z e (b) dz = 0. z5 8.5.3 If the Legendre polynomial of degree n is defined by the Rodrigues formula  n 1 d (x2 − 1)n , Pn (x) = n 2 n! dx show that Pn (z) =

1 1 2n 2πi



(t2 − 1)n dt, (t − z)n+1

called the Schlafti integral representation. If the contour is next chosen to be a circle of radius |(z2 − 1)1/2 | centered at t = z by making a change of variables from t to φ with & t = z + z2 − 1 eiφ ,

528

Functions of a complex variable

show that the integral representation can be written in the form  & 1 π Pn (z) = (z + z2 + 1 cos φ)n dφ, π 0 first given by Laplace. (Hint: First show that (t2 − 1)/(t − z) = 2(z + √ z2 − 1 cos φ).

8.6

Harmonic functions in the plane

In the region of the complex plane z where f (z) = u(x, y) + iv(x, y)

(8.12)

is analytic, the Cauchy-Riemann equations can be used to write     ∂2 u ∂ ∂v ∂ ∂v ∂2 u = = = − ∂y ∂x ∂x2 ∂x ∂y ∂y2     ∂2 v ∂u ∂ ∂u ∂2 v ∂ − = − = − = . ∂y ∂x ∂x2 ∂x ∂y ∂y2 Thus both u(x, y) and v(x, y) separately, and hence f (z) as a whole, satisfy the 2D Laplace equation  2  ∂ ∂2 + [u(x, y) + iv(x, y)] = 0, ∂x2 ∂y2 or ∇2 f (z) = 0.

(8.35)

Solutions of Laplace equations are called harmonic functions. Hence the real part or the imaginary part of an analytic function is a harmonic function in the plane. More specifically, these two parts of f are said to form a pair of conjugate harmonic functions that are joined together by Eq. (8.12). Given one harmonic function in the plane, say u(x, y), its conjugate harmonic function v(x, y) can be constructed with the help of one of the Cauchy-Riemann equations if the answer is not obvious on inspection:  r v(x, y) = dv(x, y) + v(x0 , y0 ) r0

 r

 ∂v ∂v = dx + dy + v(x0 , y0 ) ∂y r0 ∂x   r ∂u ∂u = − dx + dy + v(x0 y0 ). ∂y ∂x r0

(8.36)

Harmonic functions in the plane

529

According to the Cauchy integral theorem, the result is path-independent provided that no singularity is crossed as the path is deformed. Example 8.6.1 (a) Show that u(x, y) = x2 − y2 − y is harmonic. (b) Obtain its conjugate function v(x, y) and the analytic function f = u + iv. ∂2 u ∂2 u = 2, = −2, ∴ ∇2 u(x, y) = 0 2 ∂x2 ∂y r (b) v(x, y) = r [(2y + 1)dx + 2xdy] + v(x0 , y0 ) (a)

0

Each of these integrals is path-dependent, but their sum is not. For a path parallel first to the x axis and then to the y axis, x0 , y0 → (x, y0 ) → (x, y), we find v(x, y) = (2y0 + 1)(x − x0 ) + 2x(y − y0 ) + v(x0 , y0 ) = 2xy + x. Therefore

f (z) = x2 − y2 − y + i(2xy + x) = z2 + iz. 

Many 2D systems satisfy the Laplace equation in the plane and are described by analytic functions. For example, if u(x, y) is a potential function, the curve along which u is constant is called an equipotential. Along an equipotential 0 = du =

∂u ∂u dx + dy = ∇u · dr. ∂x ∂y

This shows that the vector field ∇u = i

∂u ∂u +j ∂x ∂y

gives a vector that is everywhere perpendicular to the equipotential passing through that point. A vector field ∇v can be constructed in a similar way for the harmonic function v conjugate to u. Now the Cauchy-Riemann conditions can be used to show that the two gradient vectors ∇u and ∇v are everywhere perpendicular to each other: ∂u ∂v ∂u ∂v + ∂x ∂x ∂y ∂y     ∂v ∂v ∂v ∂v + − = 0. = ∂y ∂x ∂x ∂y

(∇u) · (∇v) =

Since ∇v is also perpendicular to curves of constant v everywhere, we see that 1. ∇u is everywhere tangent to a curve of constant v. 2. Curves of constant v are everywhere perpendicular to curves of constant u. If u(x, y) is a potential function, a curve of constant v whose tangent vector gives the direction of ∇u is called a line of force or a field (flow, flux, or stream) line. The

530

Functions of a complex variable

function v(x,y) itself is called a stream function. The function f (z) is sometimes called a complex potential. Some of the results described above are applied in the following example. Example 8.6.2 Calculate the complex electrostatic potential of an infinite line charge placed at the origin perpendicular to the xy plane. By symmetry, the potential looks the same on any 2D plane perpendicular to the line charge. If r, θ are the circular coordinates on this plane, the electric field “radiates” from the perpendicular line charge in the form E = E r er . An application of the Gauss law of electrostatics over the surface of a cylinder of unit length surrounding the line charge yields the result  q E · dσ = 2πrE r = , ε0 where q is the charge per unit length of the source. Therefore, Er =

q d = − u(r). 2πε0 r dr

Integration of this expression gives the electrostatic potential u(r) = −

q ln r. 2πε0

(8.37)

The complex potential that can be constructed from Eq. (8.37) is q ln z 2πε0 q (ln r + iθ). =− 2πε0

f (z) = −

(8.38)

It contains an imaginary part v(x, y) = Im f (z) = −

q θ, 2πε0

which describes the field lines of constant θ radiating from the line charge. 8.6.1

(8.39) 

Conformal mapping

It is interesting that the real and imaginary parts of an analytic function f (z) = u(x, y) + iv(x, y) are everywhere perpendicular to each other. Thus u and v form a 2D orthogonal curvilinear coordinate system. Indeed the map z → f (z) is conformal, or anglepreserving, in the following sense. If two curves intersect at z = z0 with an angle α,

Harmonic functions in the plane

531

their images in the complex f plane will intersect at f (z0 ) with the same angle α. This result arises because the derivative of an analytic function f  (z0 ) = | f  (z0 )|eiφ(z0 )

(8.40)

if nonzero is independent of the direction in the complex plane along which it is evaluated. The image of every short segment Δ f in the complex f plane: Δ f  | f  (z0 )|eiφ(z0 ) Δz is rotated from Δz by the same angle φ(z0 ) independently of the orientation of Δz itself. This means that the angle between two intersecting segments or curves does not change under the mapping. Composite transformations are also useful. For example, the transformations z → (maps to) f (x), and z → g(z) can be combined to produce the transformation f → g: g(z( f )) = G( f ),

or

g(x( f ), y( f )) = G(u, v),

(8.41)

where G is in general a different function of its arguments if it is to have the same numerical image as g. This transformation property of functions underlies many applications of conformal mapping in physics problems, as we shall illustrate in the problems. If you are interested in acquiring a working knowledge of conformal mapping, you will need to do many of the problems in the subsection.

Problems 8.6.1 Determine if the function f (z) = u(x, y) + iv(x, y) can be analytic. If yes, find f (z) (a) given only its real part u(x, y) = x2 − y4 ; (b) given only its imaginary part v(x, y) = e−y sin x. 8.6.2 Two infinite line charges of equal strength q and opposite signs ±q are placed at x = ±a, respectively, perpendicular to the xy plane. (a) Show that the equipotentials and field lines are given by the equations r1 /r2 = const and θ1 − θ2 = const. Show that the zero-potential plane (where u = 0) intersects the xy plane at the y axis. (b) Show in particular that the equipotentials of value u are two circles of radius R centered at ±x0 : R = a|csch α|,

x0 = a| coth α|,

where α = 2πε0 u/q. These circles do not overlap. The circle centered at x0 (−x0 ) lies entirely in the half-plane with x > 0 (x < 0).

532

Functions of a complex variable

(c) A parallel pair of infinite cylinders of radius R centered at ±x0 are used as a capacitor. Show that its capacitance per unit length is C=

πε0 . arc cosh(x0 /R)

Hint: The complex potential is f (z) = u + iv = −(q/2π0 )[ln(z − a) − ln(z + a)]. C = q/V, where V is the potential difference between the cylinders. 8.6.3 (Conformal maps) Show that each of the following maps f (z) is conformal. Find the domain in z and the range in f of the map (from where to where). Describe the nature of each map. Here α, β, . . . are complex constants, while a, b, . . . are real constants. (a) αz + β: Show that the map involves a scaling, a rotation and a translation. (b) In z: Show that the entire z plane is mapped to an infinite strip of the f plane of height 2π in the v direction (say from v = −π to v = π). (c) l/z: The equation a(x2 + y2 ) + bx + cy + d = 0 describes both straight lines and circles for appropriate choices of the constants a, b, . . .. Show that under the transformation z → f = 1/z = u + iv, it can be rewritten as d(u2 + v2 ) + bu − cv + a = 0. Show that (i) if a, d = 0: a straight line passing through Oz , (origin of the z plane) → a straight line passing through Of ; (ii) if a, d  0: a circle → a circle, provided that 

b 2a

2 +

 c 2 d − = R2z > 0; 2a a

(iii) if d = 0, a  0: a circle → a straight line not passing through Of ; and (iv) if a = 0, d  0: a straight line not passing through Oz → a circle. (d) (M¨obius or linear fractional transformation) β αz + β = α +  , γz + δ z with γ  0: Find z , α , β . Show that the map involves a scaling + rotation, an inversion and two translations. (e) (z − ia)/(z + ia), where a > 0 is real: Show that the function maps the upper-half z plane into the interior of the unit circle in the f plane. Find the points z that map into f = 1, i, −1, −i, respectively. Hence show that the positive (negative) x axis maps into the lower (upper) semicircle in the f plane. Find the function that maps the lower-half z plane into the interior of the unit circle in the f plane.

Taylor series and analytic continuation

533

8.6.4 (Invariance of the Laplace equation) Let φ(x, y) be a harmonic function of x, y in a certain domain D of the complex z = x + iy plane. Let f (z) = u(x, y) + iv(x, y) defines a conformal mapping of D into (a certain region of) the complex f plane. Let Φ(u, v) = φ(x, y) reproduce all the numerical values of φ and provide a map of φ in f . (a) Show that φ x = Φu u x + Φv v x , (Φu u x ) x = Φuu u2x + Φuv u x v x + Φu uxx , where φ x = ∂φ/∂x, u x = ∂u/∂x, etc. (b) Show that ∇2z φ ≡ φxx + φyy = | f  (z)|2 ∇2f Φ, where f  (z) = df /dz, ∇2f Φ = Φuu + Φvv . That is, Φ(u, v) is a harmonic function of u, v if φ(x, y) is a harmonic function of x, y. 8.6.5 (Electrostatics by conformal mapping) Since a harmonic function remains harmonic under a conformal mapping (according to Problem 8.6.4), a known electrostatic solution for a simple geometry can be used to generate the solution in a more complicated geometry. (a) Show that the function φ(x, y) = θ/π = (1/π) arctan(y/x) is a harmonic function of x, y with the value φ = 0 (1) on the positive (negative) real axis. (b) Show that the electrostatic potential in the interior of a unit sphere in the complex f plane is Φ(u, v) =

  2 1 u + v2 − 1 arctan π 2v

with Φ = 0 (1) on the lower (upper) half semicircle. Hint: Use the conformal map f (z) = (z − ia)/(z + ia) with a > 0 of Problem 8.6.3(e) to map φ(x, y) in the upper-half z plane to Φ(u, v) in the interior of the unit sphere in the f plane.

8.7

Taylor series and analytic continuation

We are familiar with the Taylor expansion of a real function f (x) of a real variable x about the point x = a: f (x) = f (a) + (x − a) f  + . . . +

(x − a)n (n) f (a) + . . . , n!

(8.42)

534

Functions of a complex variable

where f (n) (a) =

dn f (x)| x=a . dxn

For example, the Taylor expansion of (1 − x)−1 about the origin x = 0 can be obtained with the help of the derivatives   d 1  1     f (0) = = = 1, dx 1 − x  x=0 (1 − x)2  x=0  2    f (0) = = 2, etc. (1 − x)3  x=0 The result is  1 = 1 + x + x2 + . . . + xn + . . . = xn . 1−x n=0 ∞

Since the ratio of succeeding terms is  n+1   x   n  = |x|, x

(8.43)

(8.44)

we see that this Taylor series converges for |x| < 1, but fails to converge for |x| > 1. For example, at x = 1/2 the function is 2, While the partial sums of its Taylor series (1, 1.5, 1.75, 1.875, 1.90625, . . . .) rapidly approach the right answer. On the other hand, at x = 2 where the function is −1, the partial sums (1, 3, 7, 15, 31, . . . ) are divergent and has the wrong sign. The convergence of the complex series obtained from Eq. (8.43) by replacing x by z is also determined by the ratio test based on Eq. (8.44), namely that |z| < 1. Because of this condition, the resulting complex series is said to have a radius of convergence of 1. It is easy to see that the Taylor series for (1 − z)−1 diverges at z = 1, since there is a pole there. The difficulty of the Taylor series elsewhere on the circumference of the circle of radius 1 is not as apparent, but is nevertheless real. For example, at z = −1 the partial sums are 1, 0, 1, 0,1, . . . , so that the series is also divergent. If we now keep clear of this boundary inside the circle of convergence, in the sense that |z| < R0 < 1, the complex Taylor series for (1 − z)−1 is convergent everywhere in the region. The series is then said to be uniformly convergent in this region. A uniformly convergent infinite series can be differentiated or integrated any number of times. Hence the Taylor series for (1 − z)−1 is analytic inside the circle of convergence. Indeed, it coincides there with the analytic function (1 − z)−1 . Thus convergence implies differentiability. Conversely, differentiability implies convergence. That is, given a function that is differentiable any number of times at z = a, the Taylor series

Taylor series and analytic continuation

z z'

a

535

C

Fig. 8.21 The contour C for the complex integration needed to evaluate the nth derivative.

 1 d n f (z)   S = (z − a) n! dzn z=a n=0 ∞ 

n

(8.45)

should converge in some neighborhood of a. To see this, we first write the Taylor series in the form  ∞  (z − a)n n! f (z )dz , S = n! 2πi c (z − a)n+1 n=0 where we have used the Cauchy integral formula, Eq. (8.34), for the nth derivative. If the integration contour C encloses the circle of radius |z − a| around a, as shown in Fig. 8.21, then |z − a| > |z − a| is guaranteed for any point z on C. Hence the geometrical series ∞  (z − a)n 1 1 1 =  =    n+1 z − a 1 − (z − a)/(z − a) z − z (z − a) n=0

converges, and can be summed to (z − z)−1 . As a result, the Taylor series converges to  1 f (z )dz = f (z), S = 2πi c z − z that is, the function f (z) itself. We see that every function f (z) analytic at z = a can be expanded in a convergent Taylor series S of Eq. (8.45) in some neighborhood of the point a. This series is unique, because the nth derivative  dn f (z)   = n!cn dzn z=a is unique for an analytic function. Further, the radius of convergence of this Taylor series is |z0 − a|, where z0 is the singularity of f (z) nearest a. Since complex differentiation works in exactly the same way as real differentiation, a complex Taylor series can be obtained in a familiar way. The only difference

536

Functions of a complex variable

is that the point a may not lie on the real axis. The following examples illustrate how the knowledge of a Taylor series about the origin (i.e., a Maclaurin series in x) can be used. Example 8.7.1 Expand (1 − z)−1 about a: 1 1 1 1 = = 1 − z (1 − a) − (z − a) 1 − a 1 − (z − a)/(1 − a) ∞ 1   z − a n = .  1 − a n=0 1 − a Example 8.7.2 Expand ln(1 + z) about a: Suppose we know the Maclaurin series ln(1 + x) = x − Then

x2 x3 + − ... : 2 3

:  z − a .; ln(1 + z) = ln(1 + a + z − a) = ln (1 + a) 1 + 1+a  z − a . = ln(1 + a) + ln 1 + 1+a  z − a  1  z − a 2 − + ....  = ln(1 + a) + 1+a 2 1+a

Example 8.7.3 Expand sin z about a: Since sin x = x −

x3 + ..., 3!

cos x = 1 −

x2 + ..., 2!

we have sin z = sin[a + (z − a)] = sin a cos(z − a) + cos a sin(z − a)   (z − a)2 = sin a 1 − + ... 2!   (z − a)3 + cos a (z − a) − + ... .  3!

Taylor series and analytic continuation

8.7.1

537

Analytic continuation

Frequently we know an infinite series S (z), but not the function f (z) itself for which it is a Taylor series. The function f (z) could be analytic outside the circle of convergence of S (z). If so, it is of greater interest than S (z). To discover its behavior outside, it is necessary to “extend” S (z) to outside its circle of convergence. The process is called an analytic continuation of S (z). More generally, if we have two functions g(z) and G(z) satisfying the following properties: 1. g(z) is defined on a set (i.e., a part) E of the z plane; 2. G(z) is analytic in the domain D containing E; 3. G(z) coincides with g(z) on E, we call G(z) the analytic continuation of g(z) to the domain D. Example 8.7.4 The real exponential function’ g(x) = ex =

∞ n  x n=0

n!

converges for all finite real values of x, while the complex exponential function G(z) = ez =

∞ n  z n=0

n!

converges for all finite complex values of z. Therefore ez is the analytic continuation of e x .  Example 8.7.5 The series S =

∞ 

zn

n=0

has a radius of convergence of 1. Hence it cannot be used at a point, say 1 + i, outside its circle of convergence. We remember, however, that S is the Taylor series of f (z) =

1 1−z

about the origin, and that f (z) is analytic everywhere except at z = 1, where it has a pole. Hence f (z) is the analytic continuation of S valid everywhere except at z =1. This shows that the analytic continuation of S to 1 + i is f (1 + i) =

1 = i. 1 − (1 + i)



The analytic continuation of a Taylor series can be made by changing the point of the Taylor expansion. If the original Taylor series is

538

Functions of a complex variable

a z0 b

Fig. 8.22 The circles of convergence for two Taylor series S a (z) and S b (z) of the same function.

S a (z) =

∞ 

cm (z − a)m ,

m=0

we look for a Taylor series about another point b, S b (z) =

∞ 

dn (z − b)n .

n=0

The new Taylor coefficients can be calculated in terms of the old ones by choosing b to be within the circle of convergence of S a (z):  ∞  m! 1 dn  = 1 dn = S (z) cm (8.46) (b − a)m−n . a  n n! dz n! m=n (m − n)! z=b The new Taylor series S b (z) has a different circle of convergence, part of which is outside the circle of convergence of S a (z). In this way, S a (z) has been analytically continued to this new region. In other words, we can find Taylor expansions all around an isolated singularity z0 , as shown in Fig. 8.22.  n Example 8.7.6 Analytically continue the series S a = ∞ n=0 z to z = −3/2 with the help of another Taylor series. The point b about which the second Taylor series is expanded has to be inside the circle of convergence of S a , which turns out to be the unit circle centered at a = 0. Its distance from −3/2 must be less than its distance to the singular point z = 1 of S a . Both conditions are satisfied by the choice b = −1/2 Hence  m−n ∞ 1  m! 1 dn = − n! m=n (m − n)! 2    ∞ 3 Sb − = dn (−1)n . 2 n=0

Taylor series and analytic continuation

539

The calculation needed is a little tedioous and has to be done with a computer. For example, the sum for d4 = 0.131687 requires about fifty terms to achieve an accuracy of six significant figures. Since we already know the analytic function f (z) for this problem, we can bypass this step by using dn = 1/(1 − b)n+l , and concentrate on the analytic continuation S b (z) by Taylor series. The exact result at z = −3/2, known from f (z) to be 2/5, can be reproduced to an accuracy of 6 significant figures by the Taylor series S b with around thirty-five terms. Analytic continuation by Taylor series always works, but it is not usually an elegant or practical method for analytic continuations. A wall of singularities can completely obstruct analytic continuation across it. For example, the series S (z) = 1 + z + z2 + z4 + z8 + . . . ∞  z2n =z+ n=0

is convergent for |z| < 1, but it has a singularity at z = 1. Actually it has singularities everywhere on the unit circle |z| = 1. (See Problem 8.7.4.) As a result, there is no analytic continuation across it. Some common Maclaurin series useful in analytic continuations by Taylor series are: ∞  1 zn , = 1 + z + z2 + . . . = 1−z n=0  1 2 = 1 + 2z + 3z + . . . = (n + 1)zn , (1 + z)2 n=0 ∞

 zn z2 e = 1 + z + + ... = , 2! n! n=0 ∞

z

 z2 z3 zn ln(1 + z) = z − + − . . . = (−1)n+1 , 2 3 n n=1 ∞

 zn+1 z3 z5 (−1)n + − ... = . 3 5 2n + 1 n=0 ∞

arctan z = z −

(8.47)

The derivation of the first three terms of each series is left as an exercise

Problems 8.7.1 Expand the following functions in Taylor series about z0 and determine the radius of convergence in each case. 1 (a) , z0 = 4; z+2

540

Functions of a complex variable

(b) eaz , z0 = 2i; (c) sin(z + i), z0 = −2; (d) ln z, z0 = 2. " # n ∞  2n/2 sin 14 nπ zn! 8.7.2 Show that ez sin z = n=0

8.7.3 Show that the following pairs of series are analytic continuations of each other: ∞ ∞   (z+a)n zn (a) and , where a  −3 is any complex number; n+1 3 (3+a)n+1 n=0 n=0 " # ∞ ∞   n (−1)n+1 z−a n (−1)n+1 zn and ln(1 + a) + (b) n 1+a . n=1

n=1

Hint: They sum to the same analytic function 8.7.4 Show that, if the series f (z) = 1 + z + z2 + z4 + z8 + . . . = z + f (z2 ) = z + z2 + f (z4 ) has a singularity at z = a on the unit circle, then it has singularities at the points a1/2 , a1/4 , a1/8 , etc. Hence there is a wall of singularities on the unit circle. 8.7.5 If the Taylor expansion ∞  cn zn f (z) = n=0

converges for |z|< R, show that for any point z0 inside a circle C of radius r < R, centered at the origin, the partial sum n  sn (z0 ) = ck zk0 k=0

is given by the integral 1 sn (z0 ) = 2πi

8.8

 C

n+1 n+1 f (z) z − z0 dz. zn+1 z − z0

Laurent series

It is even possible to make a power-series expansion of a function about a point z = a at which the function has a kth-order pole, and even an essential isolated singularity. This is achieved by allowing negative, as well as positive, powers of (z − a), as described by the following theorem: Every function f (z) analytic in an annulus R1 < |z − a| < R2 can be expanded in a series of positive and negative powers of (z − a)

Laurent series

–Γ2

a

541

Γ1

z

Fig. 8.23 The doubly closed curve specifying Γ1 and −Γ2 originates as a single closed curve excluding the hole of the annulus.

f (z) =

∞ 

cn (z − a)n

n=−∞

=

∞  n=0

cn (z − a)n +

−1 

cn (z − a)n .

(8.48)

n=−∞

The series is called a Laurent series. It can be separated into two parts: the regular part made up of the positive powers of (z − a), and the remaining principal part made up of negative powers. If the principal part vanishes, the Laurent series reduces to a Taylor series. Then f (z) is analytic also at z = a. The coefficient cn of the Laurent series is given by the same formula  f (z)dz 1 cn = , (8.49) 2πi Γ (z − a)n+1 as that for the Taylor series, with the difference that Γ is a closed curve around z = a lying entirely in the annulus. It is instructive to derive this integral formula for cn because of the insight it gives on the origin of the regular and principal parts of the Laurent expansion. We start by observing that according to the Cauchy integral formula, f (z) at a point z in the annulus can be expressed as the integral   f (z )dz f (z )dz 1 1 f (z) = + , 2πi Γ1 z − z 2πi −Γ2 z − z where the second term is needed because the region has a hole in it, as shown in Fig. 8.23. Note the negative sign of the path symbol −Γ2 of the second term used to denote the negative (clockwise) direction of integration. It is convenient to choose for Γ1 and Γ2 circles centered at a. Then for a point z on Γ1  (z − a)n 1 1 , = = z − z (z − a) − (z − a) n=0 (z − a)n+1 ∞

542

Functions of a complex variable

while for a point z on Γ2 z

1 1 1 =− =−  −z z−z (z − a) − (z − a) =−

∞ −1   (z − a)n (z − a)n = − . n+1  − a)n+1 (z − a) (z n=−∞ n=0

Both geometrical series converge uniformly; they can be integrated term by term. Hence    ∞  f (z )dz n 1 (z − a) f (z) = 2πi Γ1 (z − a)n+1 n=0 −1 



1 (z − a) + 2πi n=−∞



n

Γ2

 f (z )dz , (z − a)n+1

where we have used the negative sign in front of the second geometrical series to reverse the direction of integration over Γ2 . Since f (z ) is analytic in the annulus, the integration contour can be deformed into an arbitrary contour Γ around z = a lying entirely in the annulus. Thus the coefficients of the Laurent series are just those of Eq. (8.49). The coefficients for negative powers n = −m, m  1  1 f (z )(z − a)m−1 dz c−m = 2πi Γ have integrands with non-negative powers of z − a. They vanish whenever f (z ) (z − a)m−1 is analytic in the hole of the annulus. For example, if f (z ) has a kthorder pole at a, all negative-power coefficients vanish for m  k + 1. A function f (z ) analytic also in the hole region will involve no negative power in its Laurent series, which is then just a Taylor series. On the other hand, if c−m does not vanish when m → ∞, we say that f (z ) has an essential isolated singularity at a. Finally, we should note that if f (z) has more than one singularity, it will have more than one annular region about any point a. The Laurent series is not the same in different annular regions. Example 8.8.1 In making power-series expansions of f (z) =

1 (z − 2)(z + 1)

(8.50)

about the origin, we see first a circular region up to the first pole at z = −1, then an annular region up to z = 2, and finally an annular region outside z = 2, as shown in Fig. 8.24.

Laurent series

543

y

III II I –1

x

2

Fig. 8.24 Different regions for power-series expansions of [(z − 2)(z + 1)]−1 about the origin.

By expanding f (z) into the partial fractions   1 1 1 , − f (z) = 3 z−2 z+1 we see that convergent power-series expansions can be made separately for (z − 2)−1 and (z + 1)−1 . These are ∞ 1 1 1   z n , |z| < 2, =− =− z−2 2(1 − z/2) 2 n=0 2 ∞  1 2n 1 , = = z − 2 z(1 − 2/z) n=0 zn+1

 1 (−z)n , = z + 1 n=0

|z| > 2;

∞

|z| < 1,

and  (−1)n 1 1 , = = z + 1 z(1 + 1/z) n=0 zn+1 ∞

|z| > 1.

They show that the convergent power-series expansion for region I of Fig. 8.24 is the Taylor series   ∞ 1 n 1 1 n z − n+1 − (−1) , |z| < 1. = (z − 2)(z + 1) 3 n=0 2 For region II, it is a Taylor series for (z − 2)−1 and Laurent series for (z + 1)−1 .

544

Functions of a complex variable

 ∞  1 (−1)n zn 1 − n+1 , − = (z − 2)(z + 1) 3 n=0 2n+1 z

1< |x| < 2.

Finally, in region III, we have the simple Laurent series 1 1 1 [2n − (−1)n ], = (z − 2)(z + 1) 3 n=0 zn+1 ∞

|z| > 2.

The last series can also be considered as a Taylor series in positive powers of z−1 . This is possible because f (z) has no singularity for |z| > 2. In a similar way, we can verify that the same function has two different Laurent expansions about the point z = −1: ⎡ ⎤ n ∞  1 1 ⎥⎥⎥⎥ 1 ⎢⎢⎢⎢ 1  z + 1 ⎥⎥, |z + 1| < 3, − (8.51a) = ⎢⎢− (z − 2)(z + 1) 3 ⎣ 3 n=0 3 z + 1⎦ and n ∞   1 1 3 , = (z − 2)(z + 1) (z + 1)2 n=0 z + 1

|z + 1| > 3. 

(8.51b)

Let us return for a moment to the Cauchy integral formula (8.49) for Laurent coefficients. Besides telling us all about these coefficients, it also has many practical applications. One good example concerns the definition of functions. We are familiar with functions defined as algebraic functions of more elementary functions, with functions defined as power series, both finite (i.e., polynomials) and infinite (Taylor or Laurent series), and with functions defined as derivatives of more elementary functions (e.g., the Rodrigues formula for Pn (x) shown in Problem 8.5.3]. It is also useful to define functions as integrals of more elementary functions, since many of their mathematical properties become more transparent when such integral representations are used. Integral representations can sometimes be constructed by using the Cauchy integral formula (8.33) for analytic functions (e.g., in Problem 8.5.3). Eq. (8.49) now extends the procedure to functions that are singular inside the path of integration, as the following example shows. Example 8.8.2 An integral representation of Bessel functions Jn (t) can be calculated from their generating function !  ∞  1 1 exp t z − Jn (t)zn = f (z), z  0. (8.52) = 2 z n=−∞ This shows that Jn (t) are just the Laurent coefficients of f (z) in a Laurent-series expansion about the origin. Hence from Eq. (8.49)

Laurent series

545



f (z)dz n+1 C z  exp  1 t(z − 1/z) 1 2 = dz. 2πi C zn+1

1 Jn (t) = 2πi

If we now take the closed path C to be the unit circle, then z = eiθ , dz = eiθ idθ. Hence

 2π exp  1 t(eiθ − e−iθ ) eiθ idθ 1 2 Jn (t) = 2πi 0 eiθ(n+1)  2π 1 = ei(t sin θ−nθ) dθ. 2π 0

(8.53)

This expression can be simplified further to other interesting expressions, as illustrated in Problem 8.8.2. 

Problems 8.8.1 Obtain all possible power-series expansions of the following functions about the point a: 1 about a = −2; (a) (z − 2)(z + 1) 1 (b) about a = 2; (z − 2)2 (z + 1)2 Hint: Not necessary to decompose into partial fractions. sin z (c) 2 about a = 1; z 8.8.2 The Bessel function Jn (t) can be defined in terms of the generating function (valid for z  0) !  ∞  1 1 exp t z − = Jn (t)zn . 2 z n=−∞ Show that Jn (t) has the following equivalent integral representations:  π cos(nθ − t sin θ)dθ, (a) Jn (t) = π1 0  π −n (b) Jn (t) = i π eit cos φ cos nφdφ. 0

Hints: (a) Show that because of even/odd symmetry about θ = π,

546

Functions of a complex variable



2π



0



2π

cos(nθ − t sin θ)dθ = 2

cos(nθ − t sin θ)dθ

0 2π

sin(nθ − t sin θ)dθ = 0.

0

(b) Use φ = π/2 − θ. Change the φ integration range from [−3π/2, π/2] first to [−π, π] and finally to [0, π]. 8.8.3 Show that the following partial fraction decompositions are unique: c2 c1 1 + , = (a) (z − z1 )(z − z2 ) z − z1 z1 − z2 where zi are distinct and ci are complex constants. n n +  1 ci = , (b) z − z z − zi i=1 i i=1 c2,1 c2,2 1 c1 (c) = + + , 2 z − z1 z − z2 (z − z2 )2 (z − z1 )(z − z2 ) where ci,k are complex constants. m c2,k  c1 1 = + . (d) m (z − z1 )(z − z2 ) z − z1 k=1 (z − z2 )k

8.9

Residues

Many applications of the theory of functions of a complex variable involve complex integrations over closed contours. In the simplest situation, the closed contour encloses one isolated singular point z0 of the integrand f (z). Let the integrand be expanded in a Laurent series about z0 . f (z) =

∞ 

cn (z − z0 )n .

n=−∞

Integrating term by term, we find that the nth power term gives z  (z − z0 )n+1  1 n cn (z − z0 ) dz = cn  = 0, n + 1 z 1

provided that n  −1. If n = −1, the result is   dz reiθ idθ c−1 = c−1 = c−1 2πi. z − z0 reiθ Hence 1 2πi

 f (z)dz = c−1 = Res[ f (z0 )].

(8.54)

Residues

547

z1

z3 z2

Fig. 8.25 Closing a contour around each isolated singularity.

This integral is called the residue Res[ f (z0 )] of f (z) at z0 . Its value is just the coefficient c−1 of (z − z0 )−1 in the Laurent expansion of f (z) about z0 . If the contour encloses n isolated singularities {zi }, it can be deformed into separate contours, each surrounding one isolated singularity, as shown in Fig. 8.25. Hence  n  f (z)dz = 2πi Res[ f (zi )], (8.55) C

i=1

a result known as the residue theorem. 8.9.1

Calculation of residues

Given a Laurent series about z0 , the coefficient c−1 , which is Res[ f (z0 )], can be found by inspection. The resulting residue is for an integral over a closed contour surrounding z0 in a region where the Laurent series converges. If the contour encloses more than one isolated singularity, c−1 gives the sum of the enclosed residues. To avoid possible ambiguities, we shall use the symbol Res[ f (z0 )] to denote the contribution from z0 alone. In other words, the Laurent series of interest is the one valid in the annular region between z0 and the next-nearest singularity. However, it is not necessary to perform a complete Laurent expansion in order to extract c−1 . This one coefficient in a series can be picked out with the help of the simple formula   m−1 1 d m Res[ f (z0 )] = [(z − z0 ) f (z)] (8.56) lim (m − 1)! z→z0 dzm−1 if f (z) has an mth-order pole at z0 . This formula can easily be derived by noting that Res[ f (z0 )] is also the coefficient dm−1 of the term dm−1 (z − z0 )m−1 in the Taylor expansion about z0 of the analytic function (z − z0 )m f (z). In the case of a simple pole, Eq. (8.56) reads Res[ f (z0 )] = lim (z − z0 ) f (z). z→z0

(8.57a)

A special case of this formula is worth knowing. If f (z) = P(z)/Q(z), where P(z) is analytic at z0 and Q(z) has a simple zero there, then

548

Functions of a complex variable



P(z) Res Q(z)

 =" z0

P(z0 ) # . d dz Q(z) z0

(8.57b)

If z0 is an essential isolated singularity, Eq. (8.56) cannot be used. Then a direct Laurent expansion is needed, although one does not have to go beyond the c−1 term. Example 8.9.1 f (z) = ez / sin z has a simple pole at z = 0. Therefore  z  e = 1. Res[ f (0)] = lim z z→0 sin z Alternatively,

 ez   = 1.  Res[ f (0)] = cos z z=0

Example 8.9.2 f (z) =

1 (z+1)(z−2)

Res[ f (−1)] =

at

z=  −1.  = −1.

1 (z+1)+(z−2) z=−1

3

This is just the coefficient c−1 of Eq. (8.51a). The coefficient c−1 = 0 of Eq. (8.51b) is actually the sum of residues at z = −1 and 2 enclosed by a closed contour around both poles of f (z).  Example 8.9.3 f (z) =

etz (z + 2)(z − 1)4

at

z = 1.

There is a fourth-order pole at z = 1. Hence  ! d 3 etz 1 Res[ f (1)] = lim 3! z→1 dz3 z + 2   2 !  1 d d 1 d3 tz 1 tz (e ) + 3 (e ) = lim 3 z→1 z + 2 dz3 dz z + 2 dz2  ! 3  1 , d2 1 d tz tz d +3 2 (e ) + e dz z + 2 dz dz3 z + 2  3  2t 3! −t2 t 1 +3 2 +3 3 − 4 , = et 3! 3 3 3 3 where we have used the Leibniz formula  n  m    m−n  dm d m d [ f (z)g(z)] = f (z) g(z) , n dzm−n dzm dzn n=0

(8.58)

Residues

where

549

  m! m = n n!(m − n)!

is a binomial coefficient.  Example 8.9.4 Locate the singularities of f (z) =

z2 ez , 1 + e2z

and evaluate their residues. The singularities are located at the zeros of the 1 + e2z , where e2z = −1 = ei(2n+1)π . Hence

  1 zn = i n + π 2

on the imaginary axis are the singularities. Near zn 1 + e2z  1 − e2(z−zn )  −2(z − zn ). So zn is a simple pole, and " #2 i2 n + 12 π2 z2n ezn " #  Res[ f (zn )] = 2z = 2e n 2 exp i n + 12 π 2  1 π2 n .  = i(−1) n + 2 2 Example 8.9.5 f (z) = e1/z at z = 0. The function e1/z = 1 +

1 1 + ... + z 2!z2

has an essential isolated singularity at z = 0. The Taylor expansion in powers of 1/z shows that c−1 = 1. Hence Res(e1/z ) at z = 0 is 1.

Problems 8.9.1 Locate the singularities of the following functions and evaluate their residues:

550

Functions of a complex variable

1 1 (a) sin ; (b) cos ; z z sin z sin z (d) ; (e) 2 ; (z + 2)2 z z+2 cot z (g) 2 ; (h) 3 . z −z−1 z

ez ; (z − 1)3 z e ; (f) sin2 z (c)

8.9.2 (Partial fraction) Let f (z) = P(z)/Q(z) be a rational function defined in terms of the polynomials P(z) and Q(z) of z. Let Q(z) = (z − z1 )(z − z2 ) . . . , and m  c2,k c1 + + ... f (z) = z − z1 k=1 (z − z2 )k be the factorized form for Q(z) and the partial fraction expansion for f (z), respectively, with other factors or terms not shown explicitly. Show that (a) c1 = lim (z − z1 ) f (z); z→(z1 )

=

P(z1 ) , Q (z1 )

where Q (z1 ) = dQ/dz|z1;

(b) c2,m = lim (z − z2 )m f (z); z→z2 ! 1 d m−k m (c) c2,k = (z − z2 ) f (z) , lim (m − k)! z→z2 dzm−k for

k < m.

(d) Find the partial fraction expansion of f (z) =

1 (z + 1)(z − 2)2

by longhand, using the expansion

  1 1 1 1 = − (z + 1)(z − 2) 3 z − 2 z + 1

from Example 8.8.1. Repeat the calculation by using the formulas given in parts (a)–(c).

8.10

Complex integration: Calculus of residues

The use of the residue theorem of Eq. (8.55) to evaluate complex integrals is called the calculus of residues. One serious restriction of this theorem is that the contour must be closed. Many integrals of practical interest involve integrations over open curves,

Complex integration: Calculus of residues y

551

y C+ –R

x –R

R

x

R C–

(a)

(b)

Fig. 8.26 Closure in the (a) upper or (b) lower half-plane if |z f (z)| → 0 as z → ∞.

for example, over parts of the real axis. Their paths of integration must be closed before the residue theorem can be applied. Indeed, our ability to evaluate such an integral depends crucially on how the contour is closed, since it requires knowledge of the additional contributions from the added parts of the closed contour. A number of techniques are known for closing open contours. A common situation involves the integral along the real axis  ∞ I= f (x)dx. −∞

Suppose f (z), the analytic continuation of f (x), is a single-valued function so that there is no branch point or branch line. The following method can often be used. Method 1 If |z f (z)| → 0 as |z| → ∞, the contour can be closed by a large semicircle either in the upper, or in the lower, half-plane (HP), as shown in Fig. 8.26. In either case  Res( f ) − IR , I = ±2πi enclosed

where

 IR = I(semicircle) = lim

R→∞

θ=±π θ=0

[z f (z)]

dz z

≤ lim ±πi Max[z f (z)] = 0. R→∞

Hence I = ±2πi



Res( f ),

enclosed

where the +(−) sign is used for closure in the upper (or lower) HP. The final result is the same in either case. This is because    Res( f ) = Res( f ) + Res( f ) = 0, all

UHP

LHP

because there is no contribution from a large circle as R → ∞.

552

Functions of a complex variable

Example 8.10.1

 I=

We first verify that

∞ −∞

dx . +i

x3

  z    = 0. lim  z→∞  z3 + i 

We also find that f (z) has three simple poles at z = i, e−iπ/6 and ei7π/6 . Hence  dz I= = 2πi Res[ f (z = i)] 3 C+ z + i   2π 1 = − i, = 2πi 2 3 3z z=i if the contour is closed in the upper HP. For closure in the lower HP, I = −2πi{Res[ f (e−iπ/6 )] + Res[ f (ei7π/6 )]} =−

2πi 2πi iπ/3 + e−iπ/3 ) = − (e , 3 3

as expected.  Method 2 Complex Fourier integrals have the form  ∞ g(x)eiλx dx, I= −∞

(8.59)

where λ is a real constant. The integration contour of these integrals can often be closed with the help of Jordan’s Lemma: If lim|z|→∞ g(z) = 0, the integration contour in Eq. (8.59) can be closed by a large semicircle in the upper HP if λ > 0, and in the lower HP if λ < 0. The contribution on this semicircle will vanish, giving  I = ±2πi Res[g(z)eiλz ]. (8.60) enclosed

Since Fourier integrals are rather important in physics, it is instructive to prove this useful result. It is sufficient to consider the case of λ > 0. The idea is to show that the contribution IR on the large semicircle of radius R in the upper HP does not exceed an upper bound which itself vanishes as g(z) → 0. On this large semicircle z = Reiθ = R cos θ + iR sin θ,

dz = Reiθ idθ.

Complex integration: Calculus of residues

553

2θ π 1.0 sin θ

π 2

0

θ

Fig. 8.27 Graphical demonstration of sin θ > 2θ/π for 0 < θ < π/2.

Therefore



π

IR =

g(z)eλR(i cos θ−sin θ) Reiθ idθ.

0

If  = Max g(z =

Reiθ )

is the upper bound of g on this semicircle, then  π e−λR sin θ dθ IR ≤ εR 0



π/2

= 2εR

e−λR sin θ dθ,

0

since sin θ is symmetric about π/2. Now between 0 and π/2, sin θ > 2θ/π, as shown in Fig. 8.27. Because of this inequality, the upper bound of the integral can be simplified to  π/2 |IR | ≤ 2εR e−λR(2θ/π) dθ 0

π (1 − e−λR ) = 2εR 2λR πε ≤ . λ Hence lim |IR | ≤

R→∞

(8.61)

π lim |g(z)| = 0. λ |z|→∞

With λ > 0, the contour cannot be closed in the lower HP, because the upper bound in Eq. (8.61) would then become an exponentially increasing function of R. Example 8.10.2 ∞ (a) I = −∞ [eiλx /(x + ia)]dx, where λ and a are both real and positive. The contour can be closed in the upper HP, giving

554

Functions of a complex variable

 I=

 eiλz Res( f ) = 0, dz = 2πi UHP z + ia UHP

(8.62a)

since the pole is in the LHP. ∞ (b) I = −∞ [e−iλx /(x + ia)]dx, where λ and a are both real and positive. The contour can be closed in the LHP, giving  e−iλz (8.62b) dz = −2πie−aλ . I= z + ia LHP (c) Since the sine function can be written in terms of exponential functions, we have  ∞  sin λx 1 ∞ eiλx − e−iλx I= dx = dx. 2i −∞ x + ia −∞ x + ia The two Eqs. (8.62) can now be used to give I=−

1 (−2πie−aλ ) 2i

= πe−aλ . The integral



∞

0

sin kx 1 dx = lim a→0 2 x



(8.63) ∞

−∞

sin kx π dx = x + ia 2



is a special case of Eq. (8.63).

Method 3 If the integrand f (z) along another path is proportional to its value along the original path, these two paths can often be joined to form a useful closed contour. This technique is best described by examples. Example 8.10.3



∞

I= 0

1 I= 2



x2 ∞ −∞

dx . + a2 dx 1 = 2 2 2 x +a

 z2

π dz = , 2 2a +a

where closure with a large semicircle can be made in either the upper or lower HP. This is a rather trivial case, but it does illustrate the idea, which in this case consists of adding the contribution from the negative real axis.  Example 8.10.4

 I= 0

∞

dx . 1 + x3

Complex integration: Calculus of residues

555

y

I

x

I

C

Fig. 8.28 Closing a contour along re−i2π/3 .

The identity z3 = r3 holds not only along the positive real axis, but also along the lines rei2π/3 and re−i2π/3 . Either of these can be added to the original path. The contour can then be closed with a large arc on which the contribution vanishes, as shown in Fig. 8.28. We then find that  dz = I + I 1 + z3 where the contribution along the radius re−i2π/3 is  0 −i2π/3 e dr  I = = −e−i2π/3 I. 3 1 + r ∞ Hence



dz 2πi = − −iπ/3 2 , 3 1 + z 3(e ) C π 2π 1 2πi = = √ . ∴ I= 3 ei2π3 − e−i2π/3 3 sin(2π/3) 3 3

(1 − e−i2π/3 )I =

(8.64)

This example gives a rather nice illustration of the technique.  Example 8.10.5  I=

∞

eax dx, x −∞ 1 + e

0 < a < 1.

The complex exponential function ez is periodic in y with a period of 2π. Hence the analytic continuation of the integrand satisfies the periodicity property f (x + i2πn) = ei2πna f (x).

(8.65)

556

Functions of a complex variable y Cn

I' i2nπ

I–

I+ x

I

Fig. 8.29 A possible closed contour that makes use of the periodicity property Eq. (8.65).

This means that if Cn is the closed path shown in Fig. 8.29, then  f (z)dz = I + I  + I+ + I− , Cn

where I =



−∞+i2πn

∞+i2πn

f (x + i2πn)dx = −ei2πna I.

On the two short sides of the rectangular contour  (a−1)(x+iy) ea(x+iy) e , f (z) = = a(x+iy) x+iy , e 1+e

x→∞ x → −∞.

Hence f (z) → 0 as |x| → ∞, and both I+ and I− vanish. This leaves  1 f (z)dz. I= 1 − ei2πna Cn The residue theorem can now be applied. The integrand has poles at the solutions of the equation ezn = −1 = ei(2n+1)π , where n is any integer; that is, zn = i(2n + 1)π. Hence I=

n−1 2πi  (−1)ei(2m+1)πa . 1 − ei2πna m=0

The result is actually independent of n. The simplest expression is obtained with n = 1, for which

Complex integration: Calculus of residues

I=

557

2πi π (−1)eiπa = . sin πa 1 − ei2πa

The same result is obtained for other values of n, even in the limit n → ∞. In this limit, the term exp(i2πna) in the denominator oscillates wildly, and may be replaced by its average value of 0. Hence ∞ 

I = −2πi

ei(2m+1)πa

m=0

= −2πi

eiπa π = . i2πa sin πa 1−e

This result can also be obtained by simply closing the contour by a large semicircle at infinity in the upper (or lower) HP.  Example 8.10.6

 I= 0

∞

√

x dx . 1 + x2

On the negative real axis f (−x) = i f (x) is in one of its two branches. Hence  ∞ √ 1 x dx 1 I= = 2πi Res[ f (z = i)] 2 1 + i −∞ 1 + x 1+i if we close in the upper HP. Therefore

√  z 1 π 2πi I= = √ . 1+i 2z z=i 2



The presence of branch cuts may not be a hindrance to complex integration. They are often of help when the function just below the cut differs from, but is simply proportional to, the function just above the cut, as the following example shows. Example 8.10.7 The integral in the previous example can be evaluated with the help √ of the closed contour shown in Fig. 8.30. The branch cut for x has been chosen to be (0, ∞). Below the cut, but on the same branch, the square root changes sign & √ z(below cut) = − x. Hence the contribution from the path just below the cut is  0 √ (− x) dx = I = I+ . I− = 1 + x2 ∞ There is no contribution from either the small circle (I0 ) or the large circle (I∞ ) because ⎧  3/2  z ⎪ ⎪ ⎨  z2  → 0, |z| → ∞ |z f (z)| → ⎪ ⎪ ⎩ |z3/2 | → 0, |z| → 0.

558

Functions of a complex variable y I∝ i

I+

R x

I0 –i

I–

Fig. 8.30 A closed contour around a branch cut.

Hence 1 I= 2

 √ C

z dz = πi {Res[ f (i)] + Res[ f (−i)]} 1 + z2   iπ/4 π ei3π/4 e = √ + = πi  2i −2i 2.

Method 4 A branch cut is so useful that it is sometimes manufactured for the purpose. For example, the integral  ∞ f (x) dx I= 0

can sometimes be evaluated by first considering another integral  J= f (z) ln z dz = J+ + J− + J0 + J∞ along the same closed contour shown in Fig. 8.30. The contributions J0 and J∞ on the small and large circles vanish if |z f (z) ln z| → 0 as |z| → 0 and ∞, respectively. We are then left with the contributions above and below the branch cut:  ∞  0 J+ = f (x) ln x dx and J− = f (x)(ln x + 2πi) dx. 0

∞

The first term in J− exactly cancels J+ , leaving  0 f (x)dx = −2πiI. J = 2πi ∞

Complex integration: Calculus of residues

Hence I=−



Res[ f (z) ln(z)].

559

(8.66)

enclosed

Example 8.10.8  I= 0

∞

  3  dx ln z =− Res , 1 + x3 1 + z3 z=zn n=1

where zn are the three roots of 1 + z3 = 0: z1 = eiπ/3 ,

z2 = eiπ = −1,

z3 = ei5π/3 .

The residues turn out to be (iπ/9) exp(−i2π/3), iπ/3, and (i5π/9) √ exp(−i10π/3) for n = 1, 2, and 3, respectively. Equation (8.66) then gives 2π/ 27, in agreement with Eq. (8.64). Exactly the same result is obtained if the cube roots exp(i7π/3), exp(i9π/3), and exp(i11π/3) are used. This choice actually means that ln z is evaluated on the next-higher sheet of the Riemann surface. The result is actually the same on every sheet of the Riemann surface. (See Problem 8.10.2.)  Method 5 Another common situation involves angular integrations of the type  2π I= G(sin θ, cos θ)dθ. 0

This can be converted into a contour integration over a unit circle with the help of the change of variable z = eiθ , dz = eiθ idθ,      1 1 1 1 z− , z+ = f (z) G 2i z 2 z into

 I=

f (z)

dz . iz

Example 8.10.9  I= 0



2π

dθ , a + b sin θ

a > |b| > 0.

dz 1 iz a + b(1/2i)(z − 1/z)  2 dz = . 2 b (z − 1) + 2i(a/b)z

I=

(8.67)

560

Functions of a complex variable

The poles of the integrand are at ⎡ 1/2 ⎤  2 ⎥⎥ ⎢⎢⎢ a a z± = ⎢⎢⎣− ± 2 − 1 ⎥⎥⎥⎦ i. b b Since z+ z− = −1 and a > |b|, one of these poles is inside the unit circle, while the other is outside. For example, if b > 0, a/b is greater than 1. Hence 1/2  2 a a a < − < −1; −iz− = − − 2 − 1 b b b that is, z− is outside the unit circle, and z+ is inside. In a similar way, if b < 0, z− is inside. In either case, the result is 2 I = 2πi Res[ f (zinside )] b 2π .  = 2 (a − b2 )1/2

Problems 8.10.1 Use contour integration to verify the following results:  ∞ dx π (a) = √ ; 4+1 x 2 2 0 ∞ dx π = ; (b) 2 2 4 0 ∞ (x + 1) dx π (c) = ; 2 2 2 18 0 (x + 1) (x + 4)  ∞ dx 5π = ; (d) 2 2 2 288 0 (x + 1)(x + 4) √  ∞ x6 3π 2 ∗ (e ) dx = ; 4 2 16 0 (x + 1) (Hint: Close contour in the first quadrant. Verify that 

Res [ f (eiπ/4 )] = ∞

(f) 0

 (g) 0

√ 3 2 i 8 (1+i)3 .) ∞

sin2 x 1 dx = lim 2 a→0 2 x

0 ∞

π cos πx dx = e−π ; 2 2 x +1

1 − cos 2x π dx = ; 2 2 2 x +a

Poles on the contour and Green functions

561

y

I– x I+

Fig. 8.31 Two possible paths around a pole on the contour of integration.



∞

xa−1 π dx = , for 0 < a < 1; 1 + x sin aπ 0 (Hint: Relate it to Example 8.10.5.)  ∞ a−1 x ln x π2 cos aπ (i) dx = − , for 0 < a < 1; 1+x sin2 aπ 0 a a (Hint: First show that dx da = x ln x.)  1 πa(1 − a) ∗ xa (1 − x)1−a dx = , for −1 < a < 2; (j ) 2 sin πa 0 (Hint: First change variable x → 1/t.) √  2π 2 sin θdθ a2 − b 2 ) 2π(a − (k∗ ) , for 0 < b < a; = b2 0 a + b cos θ  ∞ 2m x π/2n (l) dx = , for integers n > m > 0. 2n sin{[(2m + 1)/2n]π} 0 1+x (Hint: Sum a geometric series.) 8.10.2 Show that the integral of Example 8.10.8 has the same value on every sheet of the Riemann surface of ln z. (h)

8.11 8.11.1

Poles on the contour and Green functions Simple pole on the contour

If there is a simple pole located right on a closed contour of integration, the integral is not uniquely defined. Two different results are possible dependent on whether the small semicircle around the pole z0 on the contour is completed in the positive (counterclockwise) or negative direction:   I± = f (z)dz = − f (z)dz + I± (z0 ). (8.68) 

±

Here the symbol denotes the common contribution exclusive of the small semicircle around z0 , while I± (z0 ) denotes the contribution from the semicircle:

562

Functions of a complex variable

 I± (z0 ) = lim

z→z0

±

[(z − z0 ) f (z)] 

= Res[ f (z0 )]

±

dz z − z0

i dθ = ±πiRes[ f (z0 )].

The final result is obtained by noting that the polar angle θ of z − z0 goes through a range of ±π on the semicircle. The situation is illustrated in Fig. 8.31. The unique quantity is not I± , but the Cauchy principal value   f (z) dz = P f (z)dz = I± − I± (z0 ) ⎞ ⎛ ⎟⎟⎟ ⎜⎜⎜   1 ⎟ ⎜ Res( f ) + Res( f )⎟⎟⎟⎟. = 2πi ⎜⎜⎜⎜ ⎠ ⎝ 2 on enclosed

(8.69)

contour

We should note that the on-contour contribution is either (2πi − πi) Res or (0 + πi) Res for either the positive or negative semicircle at z0 . Thus in a sense the on-contour pole is counted as half inside and half outside the contour for the principal-value integral. Example 8.11.1 Show that  ∞ ikx e − dx = sgn(k)πieikx0 , x − x 0 −∞ where sgn(k) is the sign of k. The contour can be closed at ±∞ with the help of an infinite semicircle into  eikz I=− dx. z − x0 If k > 0, Jordan’s lemma requires closure in the UHP. Hence  ikz  e = πieikx0 . I = 0 − I− = πi Res z − x0 If k < 0, we must close in the LHP with the result  ikz  e = −πieikx0 . I = 0 − I+ = −πi Res z − x0 The special case (k = 1, x0 = 0)  ∞ ix e − dx = πi −∞ x

Poles on the contour and Green functions

yields the results



and

563

∞

cos x dx = 0, −∞ x



∞

sin x dx = π. −∞ x

The former integral vanishes because the integrand is odd. The latter agrees with Eq. (8.63).  8.11.2

Moving a pole off the contour

The integral I+ (or I− ) also gives the result when the simple pole on the contour is moved to inside (or outside) the enclosed region. If the simple pole is originally on the real axis, the result can be written explicitly as   g(x) g(x) dx ± iπg(x0 ) (8.70) dx = − I± = lim ε→0 x − (x0 ± iε) x − x0 if we substitute f (x) = g(x)/[x − (x0 ± i)] in Eq. (8.68) Eq. (8.70) appears frequently in physics, where all observables are real quantities. For this reason, it is instructive to derive it by another procedure that illuminates the origin of the on-contour contribution. We start with the identity 1 ε x − x0 ±i . = 2 2 x − (x0 ± iε) (x − x0 ) + ε (x − x0 )2 + ε2 In the limit  → 0 1 P ± iπD(x − x0 ), = ε→0 x − (x0 ± iε) x − x0 lim

where the quantity D(x − x0 ) = lim

ε→0

ε/π (x − x0 )2 + ε2

has the properties that D(x − x0 ) ∝ and



∞

:

ε → 0, 1/ε → ∞,

ε D(x − x0 )dx = π −∞

 z2

x  x0 x = x0 , dz = 1, + ε2

564

Functions of a complex variable

a result obtained by closing the contour in either the upper or the lower HP. This is just the properties of a Dirac δ function first studied in Chapter 4. Hence D(x − x0 ) = δ(x − x0 ), and 1 P ± iπ δ(x − x0 ). = ε→0 x − (x0 ± iε) x − x0 lim

(8.71)

The fact that the value of a contour integral is changed on moving a pole off the contour has some interesting consequences in the theory of certain inhomogeneous linear differential equations. 8.11.3

Inhomogeneous linear differential equations

Consider, for example, the 1D wave equation  2  ∂ 1 ∂2 − u(x, t) = 0. ∂x2 c2 ∂t2

(8.72)

By separating variables, we can write the solution in the form u(x, t) = f (x)e−iω0 t ,

(8.73)

provided that the spatial factor f (x) (called a time-independent wave function) satisfies the 1D Helmholtz equation   2 d 2 + k0 f (x) = 0, where k0 = ω0 /c. (8.74) dx2 Eq. (8.72) or (8.74) describes a wave propagating in free space. If we would like to include the source of the wave disturbance in our description, we must solve an inhomogeneous differential equation such as  2  d 2 + k0 f (x) = ρ(x), (8.75) dx2 with a nonzero source function ρ(x). A rather similar situation occurs in the classical mechanics of a point mass m moving under the influence of a linear restoring force and an additional driving force   2 F d (t) d 2 + ω0 x(t) = . (8.76) 2 m dt In this case, we may refer to the right-hand-side term as the driving term.

Poles on the contour and Green functions

8.11.4

565

Green functions

In solving an inhomogeneous linear DE, it is useful to write it compactly in the form L (x) f (x) = ρ(x),

(8.77)

where L (x) is a linear differential operator such as d2 /dx2 + k02 in Eq. (8.75). We now show that the solution f (x) has the simple integral representation  ∞ G(x − x0 )ρ(x0 )dx0 , (8.78) f (x) = −∞

where G(x − x0 ) satisfies the inhomogeneous linear DE L (x)G(x − x0 ) = δ(x − x0 ).

(8.79)

This is easily done by direct substitution into Eq. (8.77):  ∞  ∞ L (x) G(x − x0 )ρ(x0 )dx0 = L (x)G(x − x0 )ρ(x0 )dx0 −∞

 =

−∞ ∞ −∞

δ(x − x0 )ρ(x0 )dx0

= ρ(x). The function G(x − x0 ) is called a Green function. It is a solution arising from a point source at x0 , or an impulsive driving force if we are dealing with an equation of motion in classical mechanics such as Eq. (8.76). According to Eq. (8.78), it is that part of the solution f (x) that is completely independent of the source function ρ(x). As a result, the same Green function can be used with different sources to generate different solutions by superposing point-source solutions in different ways, as shown in Eq. (8.78). This nice feature is a consequence of the linearity of the differential equation. A unique solution of a differential equation, whether homogeneous or inhomogeneous, requires the specification of a number of boundary (or initial) conditions. For example, a second-order DE requires two boundary conditions such as the value and derivative of the solution at one point. Each choice of these two boundary conditions gives rise to a unique solution, and hence also to a unique Green function. However, all these solutions can be expressed in terms of a particular solution and a linear combination of the two linearly independent solutions of the homogeneous equation, as shown explicitly in Section 5.2. As a result, there are only two linearly independent Green functions for the inhomogeneous equation. To illustrate these remarks, let us calculate the Green functions for the inhomogeneous Helmholtz equation   2 d 2 + k0 G(x − x0 ) = δ(x − x0 ). dx2

566

Functions of a complex variable

We can first solve this equation with the point source at the origin (i.e., x0 = 0):   2 d 2 (8.80) + k0 G(x) = δ(x). dx2 The general solution is then obtained afterwards by simply substituting x − x0 for x in G(x). This method works because the differential operator involved is invariant under a translation in x; that is, L (x − x0 ) =

d2 d(x − x0 )2

+ k02 = L (x).

Eq. (8.80) can be handled conveniently by the method of Fourier transformation, which transforms the differential equation into an algebraic equation. In our case, it works in the following way: We start by recalling the “derivative” property of a Fourier transform ,  2 d F G(x − x0 ) = (ik)2 F {G(x − x0 )} dx2 tabulated in Table 4.1. Hence the left-hand side of Eq. (8.80) transforms as  2  , d 2 ˜ F + k0 G(x) = (−k2 + k02 )G(k), dx2 while the right-hand side becomes 1 F {δ(x)} = √ . 2π Therefore 1 ˜ = − √1 G(k) . 2 2π k − k02 The Green function G(x) is the inverse Fourier transform of this:  ∞ 1 ikx ˜ dk G(k)e G(x) = √ 2π −∞  ∞ ikx 1 e =− dk. 2 2π −∞ k − k02

(8.81)

The integrand in Eq. (8.81) has poles at k = ±k0 on the real axis. The integral is therefore of a type discussed in this section. The contour can be closed in the upper (or lower) HP if x > 0 (or x < 0) with the help of Jordan’s lemma, since for complex k    1   = 0. lim  |k|→∞  k2 − k2  0

Poles on the contour and Green functions

567

GP(x – x0) 2k0

–π

π

0

k0(x – x0)

Fig. 8.32 The principal-value Green function for the 1D Helmholtz equation.

If we evaluate the integral in Eq. (8.81) as a Cauchy principal value, we get the principal-value Green function:  1 eikx GP (x) = − − 2 dk 2π k − k02 i ˜ 0 )]}. ˜ 0 )] + Res[G(−k = − sgn(x){Res[G(k 2 Here sgn(x) appears because we must close the contour in the upper HP if x > 0, and in the lower HP if x < 0. The result is  ik x  i e−ik0x e 0 1 − sin k0 |x|, (8.82) = GP (x) = − sgn(x) 2 2k0 2k0 2k0 where |x| = x if x > 0, and −x if x < 0. The function GP (x − x0 ) =

1 sin k0 |x − x0 | 2k0

(8.83)

is sketched in Fig. 8.32. We see that the function is continuous but not smooth at the position x = x0 of its point source. Rather its slope is discontinuous at x = x0 , increasing by exactly 1 unit as x increases through x0 . Without this discontinuity the point-source term in the DE cannot be reproduced. This important feature of the slope of the Green function of a second-order linear DE can be seen directly from Eq. (8.80) even before G(x) itself is calculated:   ε  ε 2 d 2 1= δ(x)dx = + k0 G(x)dx 2 −ε −ε dx ε d G(x) = (8.84) dx −ε Before discussing the boundary conditions under which GP (x) appears, let us first calculate two other Green functions obtained by giving k0 a small positive or negative imaginary part:

568

Functions of a complex variable

k± = k0 ± iε. The resulting Green functions 1 G± (x) = − 2π



∞

eikx. dk 2 2 −∞ k − k±

can be evaluated by noting that G+ involves a pole k+ in the UHP and a pole −k+ in the LHP, while G− involves a pole k− in the LHP and a pole −k− in the UHP. As a result, G± (x) = ±

1 ±ik± |x| e . 2ik±

In the limit  → 0, G± (x) = ±

1 ±ik0 |x| e 2ik0

(8.85)

will thus satisfy Eq. (8.79), including the requirement that its slope has a unit discontinuity at x = 0. Both these Green functions differ from GP (x). The physical situation described by these Green functions can be understood readily by first constructing the actual solution for a given source function ρ(x):  ∞ f+ (x) = G+ (x − x )ρ(x )dx −∞

=

1 2ik0



x −∞



eik0 (x−x ) ρ(x )dx +



∞

  eik0 (x −x) ρ(x )dx .

(8.86)

x

For example, if ρ(x ) is localized near the origin in the sense that ρ(x ) = 0 if then

|x | > X,

⎧  X ⎪  ⎪ ⎪ ik x 0 ⎪ e e−ik0 x ρ(x )dx ⎪ ⎪ 1 ⎨ −XX f+ (x) = ×⎪ ⎪  2ik0 ⎪ ⎪ −ik x ⎪ ⎪ eik0 x ρ(x )dx , ⎩e 0 −X

x>X (8.87) x < −X.

To get a physical picture, we go all the way back to the time-dependent wave function u(x, t) of Eq. (8.72) by reinserting the time dependence exp(−ik0 ct) to obtain ⎧ ik (x−ct) ⎪ ⎪ , x>X ⎨e 0 (8.88) u+ (x, t) = const × ⎪ ⎪ ⎩ e−ik0 (x+ct) , x < −X. This result shows that a point of constant phase of the wave satisfies the equation x − ct = const for x > X, and x + ct = const for x < −X. The wave is therefore traveling with the velocity

Poles on the contour and Green functions

 dx c, v= = −c, dt

569

x>X x < −X.

That is, it is moving away from the source (i.e., the inhomogeneity term) near the origin. We call this solution the outgoing-wave solution, and the Green function G+ (x − x ) the outgoing-wave Green function. In a similar way, the solution f− (x) obtainable from G − (x − x ) is proportional to exp(−ik0 x) for x > X, and to exp(ik0 x) for x < −X. When used with the time factor exp(ik0 ct), it describes a wave traveling toward the source. It therefore gives an ingoing-wave solution, and G− (x − x ) is the ingoing-wave Green function. Why do we use a time factor of exp(−iωt) and not exp(iωt)? If is just a convention, but one that is useful in special relativity, where the combination kx − ωt appears naturally. If we had used exp(iωt) instead, the solution f− (x) would have been the outgoing-wave solution. What about the principal-value Green function? According to Eqs. (8.82) and (8.85), it is half ingoing wave and half outgoing wave. It is a standing-wave Green function, and it gives rise to a standing-wave solution. The two traveling-wave Green functions G± are linearly independent of each other in the sense that one cannot be expressed in terms of the other. On the other hand, the standing-wave Green function GP can be expressed in terms of G± , and is therefore linearly dependent on them. This is not unexpected, since a second-order linear DE has only two linearly independent solutions. As a result, any solution or Green function can be expressed as a linear combination of two linearly independent solutions or Green functions. For wave motion, it is convenient to express an arbitrary Green function in terms of the traveling-wave Green functions G(x − x ) = aG+ (x − x ) + (1 − a)G− (x − x ),

(8.89)

because of the simplicity of the physical situations they describe. Note that this G(x − x ) also has one unit step discontinuity in its slope at x = x . Other choices of the two linearly independent Green functions might be more useful for other differential equations. These correspond to other ways of moving the poles off the contour. For example, the DE  2  d 2 + ω0 G(t − t ) = δ(t − t ) (8.90) dt2 satisfied by a mechanical system under an impulsive driving force is mathematically identical to Eq. (8.80). As a result, it can be written in a form analogous to Eq. (8.81):  ∞ −iω(t−t ) e 1  G(t − t ) = − dω. (8.91) 2π −∞ ω2 − ω20 Eq. (8.91) is actually identical to Eq. (8.81) except for the use of −i instead of i in the Fourier transform. This is the result of the convention mentioned earlier that is designed to lead naturally to the combination kx − ωt of significance in special relativity. If Eq. (8.91) is evaluated after moving both poles to just below the real axis, we get

570

Functions of a complex variable

⎧ 1   ⎪ ⎪ ⎨ ω0 sin ω0 (t − t ), t > t Gr (t − t ) = ⎪ ⎪ ⎩0 t < t . 

(8.92)

This is called a retarded Green function, because it describes a response of the system occurring after the disturbance at time t . In physics, we insist that the results of a disturbance cannot be seen before the disturbance itself takes place. This common-sense requirement that effects always follow causes is referred to as the principle of causality. Causality is guaranteed if the transformed Green function, in our case 1 1 G˜ r (ω) = − √ , 2π (ω − ω0 + iε)(ω + ω0 + iε)

(8.93)

is analytic in the upper half ω plane. Of course, G(ω) should not be analytic in the lower half ω plane also, otherwise it would simply be a constant, which gives rise to no response at all for t > t . In other words, measurable responses of a physical system to disturbances come from singularities in the lower half ω plane. There is more than meets the eye, however. The physical or retarded Green function is only one of the two linearly independent Green functions needed to provide a complete solution of the second-order linear DE (8.90). The second Green function may be taken to be the advanced Green function obtained from Eq. (8.91) by moving both poles to above the real axis. The result is: ⎧ ⎪ t > t ⎪ ⎨ 0,  (8.94) Ga (t − t ) = ⎪ ⎪ ⎩ − ω1 sin ω0 (t − t ), t < t . 0 It describes a science-fictional situation in which the response takes place entirely before the disturbance. We may call this an unphysical situation that gives rise to unphysical solutions. Like a movie run backwards, the scenario described by Ga is unreal, in the sense that we can never see it in real life. It is nevertheless important in helping to provide a mathematically complete description of the system. For this reason, the properties of a system in unphysical regions are also of interest in the mathematical description of a physical theory.

Problems 8.11.1 Use contour integration to verify the following principal-value integrals where a is real: ∞ x (a) −−∞ cos x−a dx = −π sin a;  ∞ cos x π sin a. (b) −0 x2 −a2 dx = − 2a 8.11.2 Show that the Green function for the first-order DE d G(x) + aG(x) = δ(x), dx

Laplace transform

for real a > 0 is

 G(x) =

0, e−ax ,

571

x 0.

8.11.3 Show that the retarded Green function for the driven oscillator equation with dissipation d2 d G(t) + 2β G(t) + ω20G(t) = δ(t) dt dt2 is

 Gr (t) =

0, t 0, ω−1 e 1 1

where ω1 = (ω20 − β2 )1/2 .

8.12

Laplace transform

A Fourier transform 1 F { f (x)} = √ 2π



∞

−∞

f (x)e−ikx dx

may not converge because f (x) does not vanish sufficiently rapidly at infinity. This means that often a problem in which f (x) appears cannot be treated by the method of Fourier transforms. If f (x) is such that it does not diverge more rapidly than exp(σ0 x), with σ0  0, as x → ∞, then it can be rendered well behaved by adding a convergence factor exp(−σx), σ > σ0 . Of course, this factor will diverge for negative values of x, so we must restrict ourselves to x > 0. This restriction can be achieved by the multiplication with a Heaviside step function ⎧ ⎪ 0, x < 0 ⎪ ⎪ ⎨ 1, x > 0 (8.95) Θ(x) = ⎪ ⎪ ⎪ ⎩ 1 , x = 0. 2

The Fourier transform of the modified function  ∞ f˜(k) = [ f (x)e−σx Θ(x)]e−ikx dx −∞ ∞

 =

f (x)e−(σ+ik)x dx

0

= g(s = σ + ik)

(8.96)

572

Functions of a complex variable

is then well defined. The transform (8.86) can be inverted with the help of the Fourier inversion formula  ∞ 1 −σx f˜(k)eikx dk. f (x)e Θ(x) = 2π −∞ Moving the factor e−σx to the right, we find  σ+i∞ 1 g(s)e sx ds, f (x)Θ(x) = 2πi σ−i∞

s = σ + ik.

(8.97)

We call g(s) the Laplace transform (LT) of f (x), and Eq. (8.97) the Laplace inversion formula. We have assumed that the function f (x) does not have other divergence problems elsewhere. Otherwise its Laplace transform may not exist. For example, the integral  ∞ t−n dt 0

diverges at t = 0 if n  1. Hence We shall use the notation

t−n , n

 1, does not have an LT. 

L { f (t)} =

∞

e−st f (t)dt

(8.98)

0

for an LT. Example 8.12.1 The calculation of Laplace transforms is illustrated below with the help of a number of simple examples.  ∞ 1 (a) L {1} = e−st dt = . s 0  ∞  ∞ d L {t} = te−st dt = − e−st dt (b) ds 0 0 1 = 2. s  ∞ 1 (8.99) (c) L {eiλt } = eiλt−st dt = .  s − iλ 0 8.12.1

Properties of Laplace transforms

The basic properties of Laplace transforms are summarized in Table 8.1. Both translation and attenuation properties can be derived by a change of variables. In translating the argument of f from t to t − a, we must require that a > 0. This is because if a = −|a| < 0, the beginning of the time integration at t = 0 would correspond to the point t − a = −a = |a|, as shown in Fig 8.33. As a result, that part of f (t − a) between

Laplace transform

573

Table 8.1 Basic properties of Laplace transforms.

Property

F(t)

G(s)

Conditions

Definition Translation Attenuation

f (t)Θ(t) f (t − a)Θ(t − a) e−at f (t)

g(s) e−at g(s) g(s + a)

Re s > σ0 , if |e−σ0 t f (t)| < M a > 0, Re s > σo Re s > σ0 − a

Derivatives

d dt f (t) d2 f (t) dt2

sg(s)) − f (0) * sL dtd f (t) − f  (0) = s2 g(s) − s f (0) − f  (0)   n  n−k sn g(s) − sk−1 dtd n−k f (t)

dn dtn

f (1)

k=1

t=0

0 and |a| would be excluded from the Laplace transform. The remaining part then differs from the original function f (t). The derivative property arises from an integration by parts:   ∞ ∞  ∞  d −st  −st d −st e f (t)dt = e f (t) − e f (t)dt dt dt 0 0 0 = − f (0) + sg(s).

(8.100)

Repeated applications of Eq. (8.100) lead to the formulas for the higher derivatives. A number of related Laplace transforms can be deduced from Eq. (8.99): ,    1 1 iλt 1 s 1 −iλt L {cos λt} = L , = 2 (e + e ) = + 2 2 s − iλ s + iλ s + λ2   ,  1 λ 1 1 1 iλt −iλt = 2 L {sin λt} = L , (e − e ) = − 2i 2i s − iλ s + iλ s + λ2 s L {cosh κt} = L {cos iκt} = 2 , s − κ2 f(t – a) t=0

0

|a|

t–a

Fig. 8.33 Exclusion of the shaded part of f (t − a) from the Laplace transform if a < 0.

574

Functions of a complex variable

1 κ L {sinh κt} = L { sin iκt} = 2 , i s − κ2  , d s 2 − λ2 d  λ  = L {t cos λt} = L . sin λt = dλ dλ s2 + λ2 (s2 + λ2 )2 8.12.2

(8.101)

Solution of differential equations

An important application of LTs is in the solution of DEs. We illustrate this by working out a simple example. Example 8.12.2 Solve the differential equation   2 d d − 3 + 2 y(t) = at2 dt dt2

(8.102)

using an LT. Let us denote L {y(t)} = y˜ (s).

The LT changes the differential equation into the algebraic equation [s2 y˜ (s) − sy(0) − y (0)] − 3[s˜y(0) − y(0)] + 2˜y(s) =

2 a, s3

or (s2 − 3s + 2)˜y(s) + (3 − s)y(0) − y (0) =

2a . s3

This can be solved to get y˜ (s) =

2a/s3 + y (0) − (3 − s)y(0) . s2 − 3s + 2

(8.103)

This result contains a number of interesting features: 1. The first term on the right is a particular solution of the original inhomogeneous DE, Eq. (8.102). In particular, it satisfies the boundary conditions y (0) = 0 and y(0) = 0. 2. The second term gives the homogeneous solution for the boundary conditions y (0)  0 and y(0) = 0. 3. The third term gives the homogeneous solution for the boundary conditions y (0) = 0 and y(0)  0. 4. y¯ (s) is the LT of the general solution satisfying the boundary conditions y (0)  0 and y(0)  0. 

Laplace transform

575

This example shows that the Laplace transform has an advantage over the Fourier transform in treating differential equations by making explicit the dependence on the boundary conditions. The solution y˜ (s) must be inverted to obtain the actual solution y(t) = L −1 {˜y(s)}. This will be done in the next section.

Problems 8.12.1 Verify the following Laplace transforms where real a, b > 0: 1 1 ; (a) L { 3 (at − sin at)} = 2 2 a s (s + a2 ) 1 1 ; (b) L { 3 (sin at − at cos at)} = 2 2a (s + a2 )2   1 s−a (c) L { (ebt − eat )} = ln ; t s −b  π 1/2 ; (d) L {t−1/2 } = s √ √ √ 1 (e) L { 3/2 (ebt − eat )} = 4π( s − a − s − b); t   2 1 s + a2 1 ; (f) L { (1 − cos at)} = ln t 2 s2   √ √ 1 k (g) L { √ cos(2 kt)} = π/s exp − ; s t Hint: You may use the result of part (d).   2 , √ √ 1 k = π/s exp(−k s), k  0. (h) L √ exp − 4t t You may use without verification the integral  ∞ √ √ 2 2 e−x −b/x dx = e−2 b π, for real b > 0. −∞

8.12.2 If f (x) is periodic with a period of T, f (x + T ) = f (x), show that  T $ " # −sx L { f (x)} = e f (x)dx 1 − e−sT . 0

8.12.3 With the help of the integral 

∞

e 0

−x2

√ dx =

π , 2

576

Functions of a complex variable

show that √ (a) L {t−1/2 } = π/s, (2n − 1)!! √ π/s, integer n > 0. (b) L {tn−1/2 } = (2s)n √ 8.12.4 Show that L {J0 (t)} = 1/ s2 + 1, where J0 , the Bessel function of order 0, can be defined by the integral  2π 1 cos(t cos θ)dθ. J0 (t) = 2π 0 8.12.5 The parametric integration for t > 0  ∞ 1 e−pt d p = t 0 can be used to add a factor 1/t to f (t) in a Laplace transform. Use this method to derive the following relations: ) * ∞ (a) L f (t) t * =  s g(v)dv, ) f (t) ∞ ∞ (b) L t2 = s dv1 v g(v2 )dv2 , 1

where g(s) = L { f (t)}. 8.12.6∗ (Convolution theorem) (a) Show that in the limit a → ∞, the following expression gives no contribution from the upper right triangular half (where x + y > a) of the square of area a2 in the first quadrant of the xy plane:  a  a −sx g1 (s)g2 (s) = lim e f1 (x)dx e−sy f2 (y)dy a→∞



0 a

= lim

a→∞

0

e−sx f1 (x)dx



0 a−x

e−sy f2 (y)dy.

0

(b) By changing variables to t = x + y, z = y, or equivalently x = t − z, z = y, show that  ∂x ∂y   ∂t ∂t  dtdz, dxdy =  ∂x  ∂y  ∂z ∂z ,  t f1 (t − z) f2 (z)dz = L { f1 ∗ f2 (t)}. g1 (s)g2 (s) = L 0

That is, the lower left triangular half of the square in the xy plane is mapped into the lower right triangle in the first quadrant of the tz plane bounded by the t axis, the bisector (z = t) and the vertical line t = s. The integral in the argument of the Laplace transform is called the convolution integral, and is often denoted f1 ∗ f2 (t).

Inverse Laplace transform

577

Table 8.2 A table of basic Laplace transforms

f (t)

g(s) = L { f (t)Θ(t)}

Restrictions

tn

n! sn+1 1 s−a √ (2n−1)!! π 1 2n " # sn+1/2 ln s−a s−b " # √ π/s exp − ks

s > 0, n > −1

eat tn−1/2 1 bt at t (e − e ) √ 1 √ cos 2 kt t " 2# 1 √ exp − k4t t

√

s>a n = 1, 2, 3, . . . Re s > a, b

√ π/s exp(−k s)

k≥0

a, b, k are real constants.

8.13

Inverse Laplace transform

The inverse Laplace transform f (t) = L

−1

1 {g(s)} = 2πi



σ+i∞

σ−i∞

g(s)e st ds

(8.104)

can be obtained by consulting a table of LTs or by evaluating the complex integral (called a Bromwich integral) with the help of the calculus of residues. 8.13.1

Using a table of Laplace transforms

This is not as trivial as it appears at first sight. The reason is that a table such as Table 8.2 gives the transforms for only a limited number of basic functions. A certain amount of manipulation might be needed before the table entries can be used. Example 8.13.1 (a) L

−1



 , , 1 at −1 1 =e L s−a s  at e , t>0 at = e Θ(t) = 0, t < 0,

with the help of the attenuation property shown in Table 8.1. (b) L

−1



,  , 1 at −1 1 =e L (s − a)n sn = eat

tn−1 Θ(t), (n − 1)!

where use has been made of the first entry of Table 8.2.

578

Functions of a complex variable

(c) L −1 {arctan(k/s)}: We first note that the polar form of the complex number s + ik gives 1 ln e2i arctan(k/s) 2i   1 s + ik = ln . 2i s − ik

arctan(k/s) =

The fourth entry of Table 8.2 can now be used with a = −ik, b = ik, to give 1 ikt (e − e−ikt )Θ(t) 2it 1 = sin kt Θ(t).  t

L −1 {arctan(k/s)} =

Many LT g(s) involve the inverse of a polynomial in s. These can readily be handled by decomposition into partial fractions:  ci 1 1 = n = , pn (s) + s − ai i=1 (s − ai ) n

g(s) =

(8.105)

i=1

if pn (s) has no multiple root. [Note that the coefficients ci are just the residues of g(s) at s = ai .] If one of the roots, say a1 , occurs m times, its contribution in the final partial fraction decomposition should be replaced by the expression m 

c1,k , (s − a1 )k k=1

which is just the principal part of the Laurent expansion of g(s) near a1 . Similar changes should be made for other multiple roots. We now have to find the inverse LT of a linear combination of basic functions. This can be done easily because both the LT and its inverse are linear operators satisfying the linearity property ⎧ ⎫ ⎪ ⎪  ⎪ ⎪ ⎨ ⎬  = L −1 ⎪ c g (s) ci L −1 {gi (s)} ⎪ i i ⎪ ⎪ ⎩ ⎭ i i  ci fi (t). (8.106) = i

This procedure is illustrated in the following example. Example 8.13.2 Find the inverse Laplace transforms of the following functions of s:

Inverse Laplace transform

579

1 1 = − 3s + 2 (s − 2)(s − 1) 1 1 = − , s−2 s−1

(a)

g1 (s) =

s2

where the partial-fraction decomposition has been made by comparing the coefficients appearing in the numerator functions on both sides of the equation. Therefore,    , 1 1 −1 −1 −1 f1 (t) = L {g1 (s)} = L −L s−2 s−1

(b)

(c)

= (e2t − et )Θ(t). 1 s−3 2 =− + g2 (s) = 2 s−2 s−1 s − 3s + 2 f2 (t) = L −1 {g2 (s)} = (−e2t + 2et )Θ(t).   1 1 3 1 71 1 1 2a 1 = 2a + + + − gp (s) = 3 2 2 s3 4 s2 8 s 8 s − 2 s − 1 s (s − 3s + 2) !   1 t2 3 7 1 2t −1 t + t + + e − e Θ(t). fp (t) = L {gp (s)} = 2a 2 2 4 8 8

The three functions considered above are just those appearing in the three terms of Eq. (8.103). Hence the general solution to the DE (8.102) is just y(t) = L −1 {˜y(s)} = L −1 {gp (s) + y (0)g1 (s) + y(0)g2 (s)} = fp (t) + y (0) f1 (t) + y(0) f2 (t). This completes the solution of Eq. (8.102) by LT. It can further be verified that the solution obtained is also valid for negative t. Hence the step function can be dropped.  Many computer algebra software programs have function calls for Laplace transforms and inverse LTs. These programs can also integrate given functions formally or numerically. The function calls in both cases are equivalent to table lookups that are actually easier to use than printed tables. The working scientist now does not have to be expert in integrations, but some familiarity with integrals will make it easier to spot nonsensical results caused by typing and other errors in a big computer program. 8.13.2

Bromwich integrals

When an available table of LTs turns out to be inadequate, the complex integral  σ+i∞ 1 −1 f (t) = L {g(s)} = g(s)e st ds 2πi σ−i∞ might have to be evaluated exactly or approximately.

580

Functions of a complex variable Im s

σ s0

Re s

Fig. 8.34 The integration path used in the inverse Laplace transform should be to the right of the rightmost singularity at s = s0 of g(s) in the complex s plane.

The path of integration is a line in the complex s plane parallel to the imaginary axis. It should be chosen to the right of the rightmost singularity s0 of g(s); that is, σ > Re s0 = σ0 ,

(8.107)

as shown in Fig. 8.34. This will ensure that f(t) vanishes for t < 0. To see this result, let us close the contour of integration and evaluate the integral by residue calculus. If |g(s)| → 0 as |s| → ∞, Jordan’s lemma can be used for this purpose. Writing est = ei(−is)t , we see that if t < 0, we should close in the lower HP of (−is). This is the HP to the right of the vertical line of integration. Since there is no singularity in this right HP, the integral vanishes for all t < 0. For t > 0, the contour may be closed in a large semicircle to the left, thus enclosing all the singularities responsible for the functional behavior of f (t). This feature is illustrated in the examples below. Example 8.13.3 ) * (a) L −1 1s =

 σ+i∞ 1 1 st 2πi σ−i∞ e s ds.

Since s−1 has a simple pole at s = 0 with residue 1, we choose σ > 0 to get  ,  1, t > 0 −1 1 L = 0, t < 0. s : ; :  ;  k2 k2 e st −1 = Θ(t) Res s(s2 +k2 ) at s = 0, ±ik , (b) f (t) = L s(s2 +k2 ) where σ > 0 has been used. Hence   eikt k2 e−ikt k2 f (t) = Θ(t) 1 + + ik(2ik) −ik(−2ik) = Θ(t)(1 − cos kt). 

Inverse Laplace transform

581

Im s

1+

–1 σ

1–

Re s

Fig. 8.35 Closing the contour for the Bromwich integral of Eq. (8.108).

Example 8.13.4 f (t) = L

−1



,

1 = √ 2πi s+1 1



σ+i∞ σ−i∞

√

e st s+1

dx.

(8.108)

We shall first look up the result for the third entry of Table 8.2: L −1 {s−1/2 } = (πt)−1/2 Θ(t).

With the help of the “attenuation” property shown in Table 8.1, this yields (with a = 1) L −1 {(s + 1)−1/2 } = (πt)−1/2 e−t Θ(t).

Let us now evaluate the Bromwich integral directly. The branch point s = −1 is placed to the left of the integration path by choosing σ > −1. If the contour is closed in the way shown in Fig. 8.35, there is no contribution from the large semicircle or the small circle around s = −1. The closed contour also does not enclose any singularity. As a result, f (t) = −(I+ + I− )   −1  −∞ 1 e st e st =− ds + ds . √ √ 2πi −∞ s + 1 −1 s+1 These integrals can be evaluated by the following changes of variables: For I+ above the branch cut on the negative real axis: s + 1 = reiπ , ds = −dr,  0 −(r+1)t e 1 (−dr), I+ = 2πi ∞ r 1/2 eiπ/2  ∞ −(r+1)t 1 e =− dr. 2π 0 r1/2

582

Functions of a complex variable

For I− below the branch cut: s + 1 = re−iπ , ds = −dr,  ∞ −(r+1)t e 1 I− = (−dr), 1/2 2πi 0 r e−iπ/2  ∞ −(r+1)t 1 e =− dr = I+ . 2π 0 r1/2 Hence f (t) = −2I+  e−t ∞ e−rt = dr π 0 r1/2  ∞ 2 e−t e−u t du, 2 = π 0

(8.109)

where u = r1/2 is a real variable. We now need to evaluate a real integral of the type 

∞

−x2

e 0

1 dx = 2



∞ −∞

−x2

e

  dx

∞

−∞

e

−y2

!1/2 dy

  ∞ 1/2 1 −r2 2 = π e dr 2 0 √ π = . 2

(8.110)

Hence f (t) =

e−t & e−t π/t = √ , π πt

for t > 0.

This result agrees with that extracted from the third entry of Table 8.2.

Problems 8.13.1 Verify the following relations by parametric or partial integration involving inverse Laplace transforms, with L −1 {g(s)} = f (t):  t −1 f (u)du; (a) L {g(s)/s} = 0 ) ∞ *  t f (u) 1 −1 (b) L du; s 0 g(r + s)dr = u 0

Inverse Laplace transform

583

∞ (c) L −1 {G(s)} = f (t) t , where G(s) = s g(v)dv. Then G(∞) = 0, G (s) = −g(s). 8.13.2 Verify the following inverse Laplace transforms by evaluating Bromwich integrals.  , # 1 1 " −at −1 e − e−bt ; = (a) L (s + a)(s + b) b−a ) " #* (b) L −1 In 1 + s12 = 2t (1 − cos t);  ,  " # 1 1 du 1 1 ∗ −1 ; = J0 (t) ≡ (c) L e 2 u− u √ 2πi u s2 + 1 (d)∗



, ln(s) L = −γ − ln t, where s ,   ∞ −1 ln(st) γ = −L ≡− e−r ln rdr s 0 −1

is the Euler’s constant that can be defined as the given real integral in r. Hints: (b) First show by partial integration that L −1 {g (s)} = −t f (t) when g(s = ∞) = 0. (c) Change variable s → u, where s = (1/2)(u − 1/u). Then show that a circle in the complex u plane maps into an ellipse enclosing the two square root singularities of g(s) in the complex s plane. (d) The complication is in the final step of expressing γ as a real integral in r. This can be done by using a branch cut for ln z, z = st, along the negative x = Re z axis and a Bromwich contour that excludes the branch cut extending from the branch point at z = 0. ∗ 8.13.3 The 1D diffusion equation for thermal conduction is 

 ∂2 1∂ T (x, t) = 0, − ∂x2 κ ∂t

where κ is the thermal conductivity. (a) Show that a general solution of this equation for x > 0 can be written in the form √ √ T (x, t) = L −1 {g1 (s)e− s/κx + g2 (s)e s/κx }, if T (x, t = 0) = 0 for x > 0.

584

Functions of a complex variable

(b) Show by evaluating a Bromwich integral that   ,  √ x −1 1 − s/κx L = erfc √ , e s 2 κt where 2 erfc(y) = √ π



∞

e−u du 2

y

is the complementary error function. Hint: There is a branch point and a pole, both at s = 0. The contour is similar to Fig. 8.35. There are contributions from the small circle around the pole at the origin, and on the negative real axis both above and below the branch cut. (c) One end (at x = 0) of an infinitely long 1D conductor is suddenly brought into contact with a heat reservoir at a constant temperature T 0 at t = 0. Show that g2 (s) = 0, g1 (s) = 1/s, and that the subsequent (t > 0) temperature distribution of the conductor (x > 0) is  , √ −1 1 − s/κ x e T (x, t) = T 0 L s   x = T 0 erfc √ . 2 κt (d) Use a table of error functions such as that found in AMS 55 (1964) to sketch erfc(y) for y = 0 − 1.5. Describe the temperature distribution obtained in (c) when x2  κt and when x2  κt.

8.14

Construction of functions and dispersion relations

Research in physics has some similarity to detective work. In both professions, one tries to reconstruct a situation from clues. In physics, the situations of interest are functions describing physical attributes. Since singularities are the sources of functional behavior, we must examine properties of functions near their singularities. In physics, this is done by performing meaningful experiments! The mathematical construction of functions from their singularities can be achieved with the help of the Cauchy integral formula  f (z )  1 dz f (z) = 2πi Γ z − z over a contour surrounding z and enclosing a region in which f (z ) is analytic. Although all singularities are thus excluded, they are nevertheless important, as the following example shows. Example 8.14.1 Construct a function f (z) satisfying the following properties:

Construction of functions and dispersion relations

585

y G1

G2

x

Fig. 8.36 A closed contour around a pole and a branch cut.

(a) f (z) is analytic except for a simple pole of residue R at z = a and a branch cut (0, ∞) at which the function has a discontinuity f (x + iε) − f (x − iε) = 2πig(x),

x ≥ 0.

(b) | f (z)| → 0 as |z| → ∞, and |z f (z)| → 0 as |z| → 0. The closed contour appearing in the Cauchy formula comes in two pieces, Γ1 and Γ2 , as shown in Fig. 8.36. There is no contribution from the small circle in Γ2 . The contribution from the large circle also vanishes by virtue of property (b). Hence   ∞ f (z )  f+ (x ) − f− (x ) 1 1 + f (z) = dz 2πi Γ1 z − z 2πi 0 x − x  ∞ R g(x )  = (8.111) + dx .  z−a 0 x −z The clockwise direction of Γ1 comes from the fact that it would form a continuous path with Γ2 if connected to the latter by a corridor between two paths going in opposite directions. It contributes −R/(a − z) to f (z).  Thus the sources of functional behavior for f (z) are the residue at the pole and the discontinuity across the branch cut. The corresponding terms in Eq. (8.111) are sometimes called the pole contribution and the continuum contribution to f (z). If R and g(x ) are known, we can perform the integration and complete the task of finding f (z). Example 8.14.2 Construct a function f (z) satisfying the following properties: (a) f (z) is analytic except for a simple pole at z = −1 with residue 12 , and a branch cut (0, ∞) where the function has a discontinuity: f (x + iε) − f (x − iε) =

2πi . 1 + x2

586

Functions of a complex variable

(b) | f (z| → 0 as |z| → ∞, and |z f (z)| → 0 as |z| → 0. According to Eq. (8.111), the answer is 1 + f (z) = 2(z + 1)

 0

∞

x

1 1 dx . − z 1 + x 2

The integral can be evaluated with the help of Eq. (8.66):    ∞  1 ln z 1 1  dx = − Res   2 z − z 1 + z2 0 x −z1+x enclosed  1  π =− ln0 z + (z − 2i) , 2 1 + z2 where ln0 z is the principal branch of the ln function on which ln(−i) = i3π/2. The result turns out to be the same on every sheet of the Riemann surface of ln z or ln z (Problem 8.14.1). The reader should also verify explicitly that  π 1  1 ln z + − (z − 2i) (8.112) f (z) = 0 2(z + 1) 1 + z2 2 is analytic at z = ±i. 8.14.1



Dispersion relations

In physics, information on the R and g(x ) of a function f (z) describing a physical attribute of interest is to be deduced from meaningful measurements. This is not easy to do, since our experimental knowledge of a physical situation is usually quite incomplete. In many cases where detailed physical theories have not yet been formulated, we may not even know the possible singularity structures of f (z). Under the circumstances, a much more indirect procedure might have to be used. On the other hand, the functions of physical interest are usually relatively simple. In many cases they are complex functions of a real variable f (x), −∞ ≤ x ≤ ∞. Suppose its analytic continuation f (z) vanishes at infinity, say in the UHP. The Cauchy integral formula can then be applied over the contour shown in Fig. 8.37 to give  ∞ f (x )  1 1 f (x) = − dx + f (x), (8.113) 2πi −∞ x − x 2 where 12 f (x) on the right is the contribution of the semicircular contour below the point x on the real axis. Eq. (8.113) gives two relations between two real functions Re f (x) and Im f (x) of the real variable x:

Construction of functions and dispersion relations

587

z'

x

Fig. 8.37 The closed contour for Eq. (8.113) in the complex z plane.

 1 ∞ Im f (x )  Re f (x) = − dx , π −∞ x − x  1 ∞ Re f (x )  Im f (x) = − − dx . π −∞ x − x

(8.114)

Thus, knowledge of Im f (x) for all real x will give us Re f (x), or vice versa. Two functions Re f (x) and Im f (x) that are related to each other in this way are said to be Hilbert transforms of each other. In a sense, they are two versions of the same mathematical structure. Although Re f (x) and Im f (x) involve the same mathematical structure, they may have distinct physical manifestations, each of which can be measured independently of the other. In this way, knowledge of one property permits information to be deduced on the second. This possibility was first recognized by Kronig and by Kramers in connection with the study of the dispersion of light by a medium. For this reason, they are known in physics as dispersion relations. It is useful to see how Re f (x) and Im f (x) could affect two separate physical properties. In the case of optical dispersion, we are interested in the refractive index n(ω) for a plane wave of frequency ω. This function appears in the optical wave function in the form . -   ω . -  z z (8.115) exp iω n(ω) − t = exp iω Re n(ω) − t − Im n(ω)z , c c c if the wave propagates in the +z direction. Here c is the speed of light in vacuum. The actual speed in the medium is not c, but the phase velocity, that is, the velocity of the wave front of constant phase Re n(ω) z − t = const. c

(8.116)

Differentiation of this expression with respect to t gives v(ω) =

dz c = . dt Re n(ω)

(8.117)

588

Functions of a complex variable

It is a physical fact that ν(ω) decreases below c in a medium. This decrease causes a light ray to bend as it enters the medium from the vacuum at an angle. The phase velocity differs for different frequencies or colors. This causes colors to disperse when lights of different frequencies j are refracted to different angles by a prism. Eq. (8.115) also shows that Im n(ω) is concerned not with wave propagation or refraction, but with its attenuation in the medium. This attenuation of the wave intensity (i.e., the square of the wave function) is described by an absorption coefficient α(ω) =

2ω Im n(ω). c

(8.118)

Refraction and absorption can be measured independently. Thus we have independent probes into the same mathematical structure. To construct dispersion relations for n(ω), we also need to known how n(ω) behaves as ω → ∞. Now light is known to be just an electromagnetic (EM) radiation in a certain range of frequencies. In the limit ω → ∞, the EM radiation becomes penetrating, and its speed then approaches the vacuum value. That is, n(ω) → 1 as ω → ∞. This means that dispersion relations of the form of Eq. (8.114) cannot be constructed for Re n(ω), but only for Re n(ω) − 1, which does vanish as ω → ∞. [The atomic theory of the index of refraction shows that Re n(ω) − 1 is proportional to ω2 for large ω.] Hence the dispersion relations should be written as  1 ∞ Im n(ω )  dω , Re n(ω) = 1 + − π −∞ ω − ω  1 ∞ Re n(ω ) − 1  Im n(ω) = − − (8.119) dω . π −∞ ω − ω One final complication is that Eqs. (8.119) involve negative frequencies. All physically measurable quantities such as Re n(ω) and α(ω) are independent of the sign of ω. This means that Re n(ω) must be even in ω, while Im n(ω) is odd in ω. As a result  ∞  0 Im n(ω )  Im n(|ω |) = dω d|ω |,  |ω | + ω −∞ ω − ω 0 and    1 ∞ 1 1  Re n(ω) = 1 + − Im n(ω )  dω + π 0 ω − ω ω + ω  2 ∞ ω Im n(ω )  =1+ − dω π 0 ω2 − ω2  c ∞ α(ω )dω =1+ − . π 0 ω2 − ω2

(8.120)

Construction of functions and dispersion relations

589

In a similar way, we obtain α(ω) = −

 (2ω)2 ∞ Re n(ω ) − 1  dω . − πc 0 ω2 − ω2

(8.121)

The first of these relations is the original Kramers-Kronig dispersion relation connecting refraction to absorption. Dispersion relations are not substitutes for physical theories. They only make manifest the connection between related observables. Of course, a great deal of physical insight is needed just to recognize the relation (if any) between two distinct observables. Once recognized, these relations can be checked experimentally by measuring both related observables. Alternatively, if a theoretical relation is assumed, experimental measurements of theoretically related quantities can be checked against each other for consistency in the theoretical description. Finally, quantities that have not yet been measured can sometimes be constructed from known quantities with the help of dispersion relations. The important observation made in this section is that these dispersion relations between distinct physical properties arise solely from the analyticity of f (z) on and inside the chosen contour of integration. The function f (z) describing these physical properties must be nontrivial, and therefore has singularities. But these singularities must lie entirely outside the chosen integration contour.

Problems 8.14.1 Show that f (z) of Eq. (8.112) is the same function on every sheet of the Riemann surface of ln z or ln z. Show that f (z) is finite at z = ±i. 8.14.2 Construct a function f (z) having the following properties: (a) f (z) is analytic except for (i) a simple pole of residue 5 at z = −3, (ii) a branch cut from 0 to ∞ along the real axis where f (x + iε) − f (x − iε) = 2πi[(x + 1)(x + 2)]−1 . (b) | f (z)| → 0 as |z| → ∞, and |z f (z)| → 0 as |z| → 0. # " ln z 5 iπ 2 0 Answer: f (z) = z+3 − (z+1)(z+2) − 1+z + iπ+ln 2+z . 8.14.3 From an appropriate dispersion relation, obtain the sum rule  1 ∞ Im f (x )dx Re f (0) = − π −∞ x if | f (x)| → 0 as |x| → ∞. Evaluate the sum rules  1 ∞ Im f (x )  dx , n = 2, 3, . . . . − π −∞ x n  d n−1 1 Answer: (n−1)! dxn−1 Re f (x) . x=0

590

Functions of a complex variable

8.14.4 Construct dispersion relations for (1/x) f (x). What are the least restrictions that can be placed on f (z) to ensure that there is no contribution on the large and small circles in the complex z plane? Use your results to derive the “subtracted” dispersion relations  Im f (x ) Re[ f (x) − f (x0 )] 1 ∞ = − dx x − x0 π −∞ (x − x0 )(x − x)  Im[ f (x) − f (x0 )] Re f (x ) 1 ∞ =− − dx . x − x0 π −∞ (x − x0 )(x − x) Answer: f (z) → const as |z| → ∞, f (z) → 0 as z → 0. 8.14.5 Construct a function f (z) having the following properties: (a) f (z) is analytic except for (i) a single pole of residue R at z = a; (ii) branch cuts from −∞ to −1 and from 1 to ∞ on the real axis. (b) f (z) → 0 as |z| → ∞ and (z − b) f (z) → 0 as |z| → b = ±1. (c) f (x + iε) − f (x − iε) = 2πi(x2 + 4)−1 across both branch cuts. R Answer: z−a − z21+4 [ln(z − 1) − ln(−z − 1) − z(π − tan−1 2)]

8.15

Asymptotic expansions∗

In approximation theory, one looks for simpler approximations under controllable circumstances to an original function of interest. In this section, we shall describe an important tool in approximation theory called asymptotic expansions. It is used to approximate functions appearing in mathematics, science and engineering. In physics, it is frequently called on to treat systems having many, many degrees of freedom such as statistical mechanics and to find approximate solutions in quantum mechanics and quantum field theory. To begin, let us consider the Laplace transform  , 1 g(s) = L 1 + t2 = −si(s) cos s + Ci(s) sin s,  ∞ sin t si(s) = − dt, t s  ∞ cos t dt. Ci(s) = − t s

where

(8.122)

The transform can be expressed in terms of the sine and cosine integrals si and Ci used in the mathematical theory of Fresnel diffraction. To readers unfamiliar with these functions, however, the result is worse than the original Laplace integral

Asymptotic expansions

591

The original integral looks simple enough. Let us write it as a function g(z) of the complex variable z:  ∞ −zt e dt g(z) = 2 0 1+t ∞  1 2! 4! =? − 3 + 5 − ... = gn (z). (8.123) z z z n=1 If Re z > 0, the Laplace integral is everywhere smaller than the first term 1/z of the infinite series on the second line. The rest of the series is obtained by a geometric series expansion of the integrand factor 1/(1 + t2 ) = 1 − t2 + t4 − .... The n-th term of the resulting integrated series can be obtained by parametric differentiation 2n−2    1 d n−1 gn (z) = (−1) − dz z = (−1)n−1

(2n − 2)! . z2n−1

(8.124)

The original geometric series diverges when t2 ≥ 1, because of the poles at z = ±i. The strong convergence factor e−zt (when Re z > 0 is large) in the integral might help. A ratio convergence test for the infinite series for g(z) gives    2 gn+1 (z)  2n   = lim   → ∞. (8.125) lim  n→∞ n→∞ z gn This shows that no matter how large z is, |gn (z)| increases monotonically with n when 2n exceeds |z|. So the question posed in Eq. (8.123) can be answered: The infinite series shown is everywhere divergent and appears useless. The strong convergence factor e−zt with Re z > 0 that ensures the convergence of the Laplace integral itself simply delays the onset of divergence of the infinite series, but cannot eliminate it altogether. If the number N of terms is kept fixed, however, one gets a well-defined finite sum S N (z) if z  0 and a remainder RN (z) that vanishes as Re z = |z| cos θ → ∞, |θ| < π/2: g(z) = S N (z) + RN (z),

where

1 2! − + ... + gN (z), z z3  ∞ −zt e (−t2 )N dt. RN (z) = 2 0 1+t

S N (z) =

The remainder has an upper bound  |RN (z)| ≤ 

0

∞

e

 t dt

−zt 2N

(8.126)

592

Functions of a complex variable

 ≤  =

∞

 t dt

−(Re z)t 2N

e 0

(2N)! = |gN+1 (Re z)|. (Re z)2N+1

(8.127)

Thus the strong convergence factor e−zt with Re z → ∞ ensures that the partial sum S N (z) gives a controllable approximation to g(z) in the sense that RN (z) = O[gN+1 (z)],

meaning

|RN (z)| ≤ AN |gN+1 (z)|,

(8.128)

where real AN > 0 is finite. S N (z) is then called the asymptotic expansion of g(z) to N terms:  ∞ −zt e dt ∼ S N (z) 2 0 1+t 1 2! = − 3 + . . . + gN (z) (|z| → ∞). (8.129) z z The asymptotic symbol ∼ is read “is asymptotic to”, while the O-symbol is read “of big O order” of its argument. An asymptotic expansion without qualification means N = ∞, while the lowest approximant (with N = 1) is called an asymptotic representation or form of the integral. Since divergence of the infinite series (8.123) can be said to begin when 2N first exceeds Re z, the optimal asymptotic approximant is roughly S N (z) where 2N is the largest even integer contained in Re z. The Laplace integral in Eq. (8.123) can be re-written, after a change of variable to u = zt, in a form where the convergence factor does not appear in the Laplace form e−zt :  ∞ −u e g(z) = z du. (8.130) 2 2 0 u +z Thus many integrals containing an integrand factor e−u have asymptotic expansions. Asymptotic expansions to the same N terms may be added, multiplied, divided and integrated. Adding a constant term a0 , we have for z → ∞: If

f (z) ∼

N 

an z−n ,

g(z) ∼

n=0

α f (z) + βg(z) ∼

N 

N 

bn z−n :

n=0

(αan + βbn )z−n ;

n=0

f (z)g(z) ∼

N  n=0

−n

cn z ,

cn =

N  k=0

ak bn−k ;

Asymptotic expansions

593

1 1  −n dn z , ∼ g(z) b0 n=0 N

d0 = 1, d1 = −bˆ 1 , d2 = bˆ 21 − bˆ 2 , . . . ,

where

b0  0, bˆ n = bn /b0 ;  I(z) =

∞-

z

(8.131)

N−1  an+1 a1 . f (t) − a0 − dt ∼ z−n . t n n=1

(8.132)

Differentiation is more complicated, however. If the function f (z) is known to be differentiable and its derivative f  (z) has an asymptotic expansion, then f  (z) ∼ −

N 

nan z−n−1 .

(8.133)

n=1

Asymptotic properties of functions are ultimately justified by the behavior of the remainder terms associated with their asymptotic expansions to N terms. The situation can be described concisely by using the notation { f (z)} to denote an operation on f (z): f (z) = S N { f (z)} + RN { f (z)}, S N{ f } =

N 

an z−n ,

RN { f } = O(z−(N+1) ),

bn z−n ,

RN {g} = O(z−(N+1) ).

n=0

S N {g} =

N 

(8.134)

n=0

The partial sums found in Eq. (8.131) then show that RN {α f + βg}, RN { f g}, RN {1/g} are O(z−(N+1) ), while RN−1 {I(z)}, RN−1 { f  (z)} are O(z−N ). Hence all these expansions are indeed asymptotic expansions. 8.15.1

Asymptotic expansions for integrals of a real variable

The evaluation of integrals as asymptotic series was described by Laplace in his 1812 book Th´eorie analytique des probabilit´es. If a function q(t) is infinitely differentiable for 0 ≤ t < ∞, its Laplace transform defines a function that has simple asymptotic expansions. This can be seen by repeated integrations by parts:  ∞ e−xt q(t)dt I(x) = L {q(t)} = 1 = q(0) + x

0 ∞



−xt 

e 0

q (t)dt

!

594

Functions of a complex variable

=

q(0) q (0) q(N−1) (0) + RN (x), + 2 + ... + x xN x

where the remainder term is RN (x) =

1 xN



∞

e−xt q(N) (t)dt.

(8.135)

0

Suppose q(N) (t) = O(eσt ) for 0 ≤ t < ∞, where σ is real and finite, meaning that |q(N) (t)| ≤ AN eσt , then  AN ∞ −xt σt |RN (x)| ≤ n e e dt x 0 =

AN , − σ)

xN (x

(8.136)

where AN is real, positive and finite. Hence RN (x) = O(x−(N+1) ) → 0 as x → ∞, thereby showing that I(x) ∼

∞ (n)  q (0) n=0

xn+1

(x → ∞)

(8.137)

is an asymptotic expansion. This is the simplest way to obtain an asymptotic expansion when q(t) has the prescribed properties. This is also how the asymptotic expansion (8.129) can also be obtained. Laplace also studied the integral 

b

I(x) =

e−xp(t) q(t)dt,

(8.138)

a

where t, a, b, p, x are real, and x > 0. If p(t) has a minimum at t = t0 within the integration interval (a, b), then e−xp(t) peaks at t0 . As x → ∞, the peak becomes extremely sharp, with important contributions only from the neighborhood of t0 . q(t) and p(t) in the integrand can then be approximated by their leading terms q(t0 ) and p(t0 ) + [p (t0 )/2](t − t0 )2 , respectively. The integration limits can be extended to (−∞, ∞). The result is called the Laplace method for the asymptotic form of I(x). Example 8.15.1 In the interval (a, b), p(t) has a single simple minimum at t0 where q(t0 )  0. Then as x → ∞

Asymptotic expansions



b

I(x) ≈

 (t )/2](t−t )2 } 0 0

e−x{p(t0 )+[p

a

≈ q(t0 )e

−xp(t0 )



b

 (t

e−[xp

595

q(t0 )dt,

0 )/2](t−t0 )

2

dt

a

(

2π xp (t0 )

∼ q(t0 )e−xp(t0 )

where p (t0 ) > 0, is the integral’s asymptotic form.

(x → ∞),

(8.139)



Example 8.15.2 (Stirling’s formula) Find an asymptotic form for Euler’s integral for the gamma function (valid when n + 1 > 0)  ∞ Γ(n + 1) = n! ≡ e−t tn dt. (8.140) 0

A change of variable to v = t/n gives



∞

Γ(n + 1) = n

n+1

 =n

e−nv vn dv

0 ∞

n+1

e−n(v−ln v) dv.

(8.141)

0

The result suggests an asymptotic expansion for large n. The function p(v) = v − ln v has a minimum at d p(v)/dv = 1 − 1/v = 0, or v = 1, where 1 p(v) − p(1) ≈ (v − 1)2 > 0. 2

(8.142)

Thus the integrand of Eq. (8.141) has a maximum at v = 1. This means that an asymptotic expansion for large n is indeed possible. Changing variable to y2 = (n/2)(v − 1)2 , we find √  ∞ −y2 Γ(n + 1) ∼ nn e−n 2n e dy n −n

=n e

√

−∞

2πn

(n → ∞).

(8.143)

The same result can be obtained directly from Eq. (8.139).  Example 8.15.3 (Watson’s lemma) An asymptotic series q(t) ∼ tλ−1

∞  n=0

qn t n ,

Re λ > 0,

(8.144)

596

Functions of a complex variable

in Eq. (8.138) can be integrated term by term to give an asymptotic expansion of the integral I(x). Here λ is a complex parameter: 

∞

e−xt q(t)dt ∼

0

∞ 

qn x−(n+λ) Γ(n + λ) (x → ∞),

(8.145)

n=0

where the integration to a gamma function has been obtained by a change of variable to u = xt:   ∞  ∞ −xt n+λ−1 −(n+λ) −u n+λ−1 e t dt = x e u du . 0

0

The integral representation for the gamma functions is valid if Re (n + λ) > 0, or Re λ > 0 if n ≥ 0; otherwise the integral is divergent.  Example 8.15.4 Find the asymptotic expansion for  ∞ e−x cosh t dt. I(x) = 0

using Watson’s lemma. Under two changes of variable, u = cosh t and then v = u − 1, we find  ∞ dt I(x) = e−xu du du 1  ∞ 1 e−xv √ √ dv, = e−x 0 2v 1 + (v/2) √ √ where the derivative (d/du) arc cosh u = 1/ u2 − 1 = 1/ v2 + 2v has been used. The Maclaurin series for q(v) is asymptotic: ∞  v −1/2 1  1+ q(v) = √ = v−1/2 qn vn , 2 2v n=0

where

(2n − 1)!! 1 1 qn q1 =− = (−1)n . , q0 = √ , q0 2·2 q0 n!22n 2 √ With λ = 1/2, Γ(1/2) = π, Γ(3/2) = (1/2)Γ(1/2), we get from Eq. (8.145) 4   ∞ ∞ 1 π qn Γ(n + 2 ) −n −x cosh t −x e dt ∼ e x 2x n=0 q0 Γ(1/2) 0 4  ∞ π [(2n − 1)!!]2 −x =e (−1)n (x → ∞).  2x n=0 n!(8x)n

Asymptotic expansions

Our final topic is concerned with integrals of the form  b eixp(t) q(t)dt, I(x) =

597

(8.146)

a

where t, a, b, x, p(t) are all real. For large x, the exponent of eixp changes rapidly. This causes the integrand to oscillate rapidly and to contribute very little to the integral except near the boundaries and the stationary points of p(t) where it is slowly varying because p (t) = 0. When x → ∞, the contributions of the stationary points will dominate over those from the boundaries. Let us consider for simplicity an interval (a, b) containing one and only one stationary point t0 in its interior. Near t0 , p(t) ≈ p(t0 ) + c2 (t − t0 )2 , c2 = p (t0 )/2.

(8.147)

Assuming that p(t) is twice differentiable and q(t) is continuous, we find that as x→∞  t0 + 2 2 I(x) ≈ q(t0 )eixp(t0 ) eixc (t−t0 ) dt t0 −

∼ q(t0 )e

ixp(t0 )

2π = xp (t0 )

2 √

c x

!1/2



∞

ir2



e dr 0

q(t0 )eixp(t0 ) eiπ/4 ,

(8.148)

√ where the final integration variable r = xc(t − t0 ) is real if c is real. The method used to find the asymptotic form (8.148) of the integral (8.146) is called the method of stationary phase. It is related to Laplace’s method, and was suggested by Cauchy in 1827 in his study of wave propagation. Many extensions of Laplace’s method are now known that can handle more complicated integrals. However, the availability of computer algebra software in recent years has freed the practicing scientist or engineer from the need to master complicated techniques of integration. An acquaintance with the basic asymptotic methods illustrated here does increase a user’s appreciation of such computer results. 8.15.2

Complex integration: method of steepest descent

It is often of interest to find an asymptotic form or expansion for large |z| of an integral of a complex variable t along a contour Γ of the form  I(z) = ezp(t) q(t)dt, (8.149) Γ

where p(t) and q(t) are analytic functions of t in a part D of the complex t plane containing the contour Γ. Here z = |z|eiφ is a complex variable. For large |z|, the

598

Functions of a complex variable

integrand contains a factor eiIm[zp(t)] that oscillates more rapidly as |z| increases. These rapid oscillations make it difficult to evaluate the integral along a given contour Γ. Much of these violent oscillations can be avoided, however, if the domain D of analyticity of the integrand permits the contour Γ to be deformed so that along the new path Γ , Im[zp(t)] = Im[zp(t0 )] = const at least in the neighborhood of t0 where the integrand is largest. That is,  iIm[zp(t0 )] I(z) ≈ e eRe[zp(t)] q(t)dt, (8.150) Γ

where Γ passes through a stationary point t0 of p(t) where p (t0 ) = 0. Is such a choice of Γ possible? Let us examine the function p(t) close to t0 . In the integrand, just q(t) ≈ q(t0 ) is adequate if q(t0 )  0. In the exponential function, however, one has to do better because of its exponentially stronger functional dependence. Let ! 1  [p(t) − p(t0 )] ≈ p (t0 ) (t − t0 )2 2 2  = −ρ0 r2 e2iθv = − veiθv , where

−p (t0 )/2 = ρ0 eiψ0 , ψ0 v2 = ρ0 r 2 , θv = θ + . 2 f (t) = z[p(t) − p(t0 )]

t − t0 = reiθ ,

Then

(8.151)

= −|z|ρ0 r2 eiχ , χ = φ + 2θv = φ + ψ0 + 2θ = 2(θ − θ0 ) :  Im 2 sin χ, f (t) ≈ −|z|ρ0 r Re cos χ. ,

(8.152)

Consider a circle of radius |t − t0 | = r centered at t0 in the complex t plane. At θ = θ0 ≡ −(φ + ψ0 )/2, one finds χ = 0, sin χ = 0, cos χ = 1. Along the radius of the circle at this orientation, Im f (t) = 0, and therefore eiIm[zp(t)] is constant, while Re f (t) has a maximum at t0 at the center of the circle where r = 0. After a half turn around this circle to θ = θ0 + π, one finds χ = 2π and the same behavior for f (t). So the behavior holds along the the entire diameter at the orientation defined by θ0 and θ0 + π. In the perpendicular orientation where the directions are θ = θ0 + π/2 and θ0 + 3π/2, one has χ = π, 3π. So Im f (t) is also constant along this diameter of the circle. However, Re f (t) has a minimum instead at the center t0 of the circle. The topology of the integrand of I(z) in the complex t plane is thus that of a saddle, mountain pass or col. The mountain pass connecting two valleys occupy the two quadrants in the angular ranges

Asymptotic expansions

|θ − θ0 | <

π , 4

|θ − θ0 − π| <

π . 4

599

(8.153)

They form the stirrup or leg sides of the saddle. The remaining two quadrants contain the topological hills or the pommel/cantle parts of the saddle. A contour map of such a saddle point has been given in Fig. 1.5. So it is possible to deform Γ to a path Γ along the mountain pass just where it is steepest and the maximum of the integrand at t0 sharpest, namely the steepest pass along θ − θ0 = 0, π. Two directions of travel are possible along this path, from valley 1 to valley 2, or from valley 2 to valley 1, with the directions measured from the saddle point. These directions give two different overall signs, (sign) = ±1, to the integral. The final result is obtained by using the complex probability integral  ∞ 2 2iθ Iθ = e−r e eiθ dr 0

√ π/2, if |θ| < π/4 √ = , − π/2, if |θ − π| < π/4

(8.154)

evaluated at θ = −π/4. Integral (8.154) can be derived along the following lines: (a) Along the positive and negative real  ∞axis2 (θ = 0√and π, respectively), it reproduces the real probability integral −∞ e−r dr = π. (b) When |θ| < π/4, one integrates along a closed pie-shape contour O → A → B → O, where O is the origin of the complex reiθ plane, and A, B are points on the circumference of a large circle of radius R → ∞ centered at the origin at angles θ and 0, respectively. The angular condition ensures that there is no contribution from the arc of the large circle from A to B. Since the integrand is an entire function, the complete contour integral vanishes, thus showing that the contribution along the radius OA is the same as that along OB on the positive real axis. (c) When |θ − π| < π/4, the contour is closed along the negative real axis (θ = π). The direction of integration along the negative real axis is responsible for the negative sign shown in Eq. (8.149). As φ = arg z increases, it causes the whole mountain pass to rotate in the complex t plane. For the integral (8.150) in particular, the mountain pass rotates once around the saddle point t0 for every two turns around z = 0 in the complex z plane, where φ increases by 4π. Once every half turn around t0 , Γ finds itself right on the ridge line perpendicular to the line of steepest paths. It can then be deformed into one or the other valley so that the integral can have either signature (sign). That is, the integral becomes double-valued under the circumstances, a situation called the Stokes phenomenon, first described by Stokes in 1850. This means that the integral has jump discontinuities at these Stokes values of φ where the path of steepest descent reverses direction as a mountain ridge is rotated through the original direction θΓ of the path Γ.

600

Functions of a complex variable

As z → ∞, the pass in the complex t plane becomes so steep that it is necessary to include only a very short segment (−, ) of Γ , the path of steepest descent, of the integral just where the Taylor approximant (8.153) is best. Furthermore the integration range can be extended to infinity,  → ∞, to give a real probability integral. The integration variable can now be changed to the real variable v dv dt = eiθsp dr = eiθsp √ ρ0

(8.155)

to give I(z) ≈ (sign)

q(t0 )ezp(t0 ) iθsp 2e √ ρ0



∞

e−|z|v dv. 2

(8.156)

0

The asymptotic form is thus  I(z) ∼ (sign)q(t0 )e

zp(t0 )

2π z[−p (t0 )]

,1/2 (z → ∞).

(8.157)

This asymptotic form has a square root singularity at z = 0 that is spurious, not being present in the original integral. It is an artifact of the quadratic Taylor approximation used in the exponent of the exponential function in the integrand. More specifically, it arises because this quadratic Taylor approximant is linear in z for the integral (8.150). The asymptotic form (8.157) thus resides on a Riemann surface made up of two sheets of the complex z plane. This is consistent with the previously noted fact that the mapping from t to z in the approximation is one-to-two. The technique of integrating along the steepest pass at a saddle point is called the method of steepest descent, or a saddle-point method. It is an extension to contour integrals of Laplace’s quadratic approximation (1774) at the maximum of the integrand. It was used by Riemann as long ago as 1863, and was first published by Debye in 1909. The method works also for integrals with z dependences different from Eq. (8.150). The details involved in executing the method may differ from those given here, but the idea remains the same. 8.15.3

Complex integration: series inversion

We finally consider asymptotic expansions of integrals that require a functional inversion (or reversion) of w(t) to t(w). This can now be done by computer algebra by matching coefficients of appropriate series. The method of complex integration gives important insights on the structure involved in series inversion from the viewpoint of asymptotic expansions. The result is particularly transparent for integrals of the form  ∞ I(z) = ezp(t) dt, −∞

where

p(t) = p(t0 ) + (t − t0 )μ v(t), ∞  v(t) = vn (t − t0 )n , n=0

(8.158)

Asymptotic expansions

601

where μ is real, while all other variables are in general complex. p(t) has a maximum at t0 if μ > 1. It is useful to write

where

I(z) = ezp(t0 ) J(z),  ∞ μ dt e−zw (t) dw, J(z) = dw −∞

wμ (t) = p(t0 ) − p(t) = (t − t0 )μ v(t).

(8.159)

The specific functional relation needed is dt  an wn . = dw n=0 ∞

(8.160)

We shall show that the an coefficients can be found readily by residue calculus. Once found, they can be used to calculate the integral as the series I(z) = e

zp(t0 )

 where

Jn (z) =

∞ 

an Jn (z),

n=0 ∞

−∞

μ

e−zw wn dw

(8.161)

is not a Bessel function. The result is simplest when μ = 2m, m = 1, 2, . . . , is a positive even integer. μ The integrand factor e−zw is then even in w. With the integration interval symmetric about w = 0, only even-power terms in dt/dw will contribute. The J integrals needed are  " # 2 2n+1 ∞ −η 2n+1 J2n (z) = z− μ e η μ −1 dη, (8.162) μ 0 where a change of variable to η = zwμ has been made. The integral in the final expression is Γ((2n + 1)/μ), valid for all real μ > 0. Hence  2n+1 ∞   a2n Γ( μ ) −2n/μ zp(t0 ) 2a0 Γ(1/μ) (|z| → ∞). (8.163) I(z) ∼ e z a0 Γ(1/μ) μz1/μ n=0 The an coefficients in Eq. (8.160) can be found by using the Cauchy integral formula applied to dt/dw:  dt(w) 1 [dt(w )/dw ]  = dw dw 2πi w − w   ∞ wn 1 = dt(w ), (8.164) 2πi n=0 (w )n+1

602

Functions of a complex variable

where w is a variable, not the derivative of w. The nth term on the right can be matched to the same term in Eq. (8.160) to give !  dt(w ) 1 1 an = = Res n+1 2πi [w (t)]n+1 w (t) t=t0 ⎤ ⎡   n+1 ⎥ 1 ⎢⎢ d n t − t0 ⎥⎥⎥ = ⎢⎢⎢⎣ n ⎥⎦ . n! dt w(t) t=t

(8.165)

0

So series inversion has been reduced to the calculation of residues, one for each an . The function to be differentiated to find an is the power series in t from Eqs. (8.158, 8.159) 

t − t0 w(t)

n+1

−α = v−α (t) = v−α 0 [1 + u(t)]

! 1 2 = 1 − αu(t) + α(α + 1)u (t) + . . . , 2 ∞    n+1 vn n α= t . , u(t) = μ v n=1 0 v−α 0

where

(8.166)

Example 8.15.5 Find the asymptotic expansion of Euler’s integral for the gamma function of Example 8.15.2 now for a complex variable z with Re (z + 1) > 0  Γ(z + 1) = z

z+1

∞

e−z(t−ln t) dt.

0

With p(t) = ln t − t in Eq. (8.158), we have p (t) = t−1 − 1, t0 = 1 and p(1) − p(t) = (t − 1)2 v(t),

μ = 2,

1 t − 1 (t − 1)2 − + + ... 2 3 4 1 v1 2 v2 1 v0 = , =− , = . 2 v0 3 v0 2

v(t) =

(8.167)

The remaining steps of the calculation will be left to the reader as an exercise. The results are

Asymptotic expansions

a0 = Γ(z + 1) ∼

√ 2, √

a2 1 = , a0 6 z+ 12 −z

2π z

e

1 1+ + ... 12z

603

! (|z| → ∞).



(8.168)

We are now ready to consider more complicated integrals of the form  ∞ ezp(t) q(t)dt, I(z) = −∞

where

q(t) = q0 (t − t0 )λ−1 .

(8.169)

Once this integral is known, the result for the more general case of q(t) ∼ (t − t0 )λ−1

∞ 

qn (t − t0 )n

(8.170)

n=0

can be written as an asymptotic expansion where each term has the same form as Eq. (8.169) with the parameter λ increased by a positive integer. The key to the basic integral is the infinite series for t(w) obtained by integrating Eq. (8.160) term by term: t(w) − t0 =

∞  an−1

n

n=1

wn :

q[t(w)] ≡ Q(w) λ−1

= q0 (a0 w)

∞ 

Qn wn .

(8.171)

n=0

The coefficients Qn are those in the series expansion of λ−1

[1 + u(w)]

=

∞ 

Qn wn ,

n=0

where

u(w) =

∞  (an /a0 ) n=1

n+1

wn ,

1 (1 + u)α = 1 + αu + α(α − 1)u2 + . . . , 2 α = λ − 1, Q0 = 1.

(8.172)

Finally, we need to multiply the two infinite series dt(w)/dw and Q(w) and integrate the resulting infinite series in w to get

604

Functions of a complex variable ∞ 

I(z) = q0 aλ0 ezp(t0 )

bn Jn (z),

n=0

where

bn = Jn (z) = =

n  ak

Qn−k , b0 a k=0 0  ∞ −zwμ n+λ−1 −∞

e

w

fn (z) Γ(ρ), μ

ρ=

= 1, dw n+λ , μ

fn (z) = zρ − (−1)n+λ [(−1)μ z]ρ , if Re {μ, λ, z, (−1)μ z} > 0.

(8.173)

Note that the Jn integrals in this paragraph are not Bessel functions. Our purpose here is to illustrate series inversion, and the fact that like most scientific and engineering systems, a mathematical integral too can have a complex structure made up of many simple parts.

Problems 8.15.1 Let f (z) ∼ g(z) for complex z → ∞, where g(z) = S N { f (z)} =

N 

an z−n ,

a0  0,

n=0

= a0 (1 + u), u=

N 

aˆ n z−n ,

n=1

aˆ n =

an ; a0

RN { f } = O(z−(N+1) ). Determine the validity of the following asymptotic relations as z → ∞: (a) [ f (z)] p ∼ [g(z)] p , real p > 0? (b) e f (z) ∼ eg(z) ? (c) ln [ f (z)] ∼ ln [g(z)]? 8.15.2 Justify the asymptotic expansion  ∞ −t 1 1 2! e dt ∼ − 2 + 3 − . . . , z z z 0 z+t as Re z → ∞.

Asymptotic expansions

605

8.15.3 (Incomplete gamma function) Verify the following results:  ∞ (a) e−t q(t)dt, real x > 0, I(x) = x

= S N (x) + RN (x), where   S N (x) = e−x q(x) + q (x) + . . . + q(N−1) (x) ,  ∞ e−t q(N) (t)dt, RN (x) = q(n) (t) =

dxn

q(t).

RN (x) = O(e−x x−(b+N) ),

(b) if (c)

x dn

q(n) (t) = O(t−(b+n) ), 

∞

Γ(α, x) ≡

real b > −1.

e−t tα−1 dt,

Re α > 0,

x −x

∼e

p p(p + 1) 1 − p+1 + + ... p x x x p+2 (real x → ∞),

!

p = 1 − α.

The final expression gives the asymptotic expansion for the incomplete gamma function Γ(α, x). Hint for part (b): Change the integration variable to t − x in RN (x). 8.15.4 The complementary error function 2 erfc(x) = 1 − erf(x) = √ I(x), π  ∞ 2 I(x) = e−t dt, x

is defined in terms of an incomplete integral I(x). Verify the following results:  ∞ 2 −x2 (a) I(x) = e e−u e−2xu du 0

= e−x

2

! 1 4! 2! + − . . . , − 2x 1!(2x)3 2!(2x)5

606

Functions of a complex variable

where the infinite series has been obtained by a term-by-term integration 2 of a convergent Maclaurin series for e−u . Show that the resulting series for I(x) is divergent. (b) Assume that a Maclaurin expansion for f (y) to N terms has the reminder formula f (y) = S N { f (y)} + RN { f (y)}, where

S N { f (y)} = f (n) (y) = RN { f (y)} =

N−1  yn n=0 dn

dyn

n!

f (n) (0),

f (y),

yN (N) f (αy), N!

0 ≤ α ≤ 1.

Show that I(x) = S N {I(x)} + RN {I(x)}, where

S N {I(x)} =

N−1 

(−1)n

n=0

Hence

(2n)! , n!(2x)2n+1

RN {I(x)} = O(x−(2N+1) ). ∞  −x2 erfc(x) ∼ e (−1)n n=0

(2n)! n!(2x)2n+1

(x → ∞).

(c) Derive the remainder formula for the Maclaurin expansion as follows:  y f  (t)dt f (y) = f (0) + 0 



= f (0) + y f (0) +



y

0

= f (0) + . . . +

t

dt

yN−1 (N − 1)!

f  (t )dt

0

f (N−1) (0) + RN { f (y)},

RN { f (y)} = VN (y) f (N) (αy), 0 ≤ α ≤ 1,  y  y where V1 (y) = dt = y, Vn (y) = Vn−1 (t)dt. 0

0

8.15.5 (Asymptotic representations of Bessel functions) The generating function for Bessel functions Jn (z) of integer order n is

Asymptotic expansions

G(z, t) = ez(t−t

−1 )/2

=

∞ 

607

tn Jn (z),

n=−∞

where z is complex. Use it to obtain the following results: Jn (z) has the integral representation  −1 1 Jn (z) = t−(n+1) ez(t−t )/2 dt, 2πi Γ where Γ is any closed counter clockwise contour enclosing the origin. The function p(t) = (t − t−1 )/2 has saddle points at t± = ±i. In the limit z → ∞, their contributions to Jn (z) can be written separately as 4 1 2 ±i[z−(2n+1)π/4] Jn± (z) ∼ ∓(sign)± , e 2 πz where the upper signs are for the saddle point t+ = i. The sign function (sign) is that defined in Eq. (8.155). The sign dependence makes the asymptotic form of Jn (z) multivalued, with a 4π periodicity in φ = arg z: ⎧ ⎪ if |φ| < π/2 cos [ fn (z)], ⎪ 4 ⎪ ⎪ ⎪ ⎪ 2 ⎪ ⎨ −i sin [ fn (z)], if |φ − π| < π/2 Jn (z) ∼ , ⎪ ⎪ − cos [ fn (z)], if |φ ∓ 2π| < π/2 πz ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ i sin [ f (z)], if |φ + π| < π/2 n

π fn (z) = z − (2n + 1) , 4

(|z| → ∞).

These asymptotic forms are discontinuous whenever φ = (2m + 1)π/2, integer m, is a half odd integral multiple of π. Note: The discontinuities or Stokes phenomena of many asymptotic representations of functions of a complex variable z comes from the direction reversals in the paths of steepest descent as φ goes through certain values. These discontinuities are not present in the original Bessel functions that are analytic functions of √z, as they can be defined in terms of convergent Taylor series. The z factor too is an artifact of the steepest-descent approximation coming from a change of variable leading to the real probability√integral in Eq. (8.156). The 4π periodicity in φ = arg z arises because z is single-valued if spread out on two Riemann sheets of z. Our angular ranges agree with those given by Morse and Feshbach (1953, pp. 622, 609-611). They disagree with that given by AMS55 (1964, #9.2.1, p. 364). 8.15.6* (WKB approximation) Consider the 1D Schr¨oedinger wave equation (where all quantities are dimensionless and x is real)

608

Functions of a complex variable

d2 ψ(x) + k2 (x)ψ(x) = 0, where dx2 ⎧ ⎪ > 0, if x < 0 ⎪ ⎪ ⎪ ⎨ 2 2 k (x) = k0 − V(x) ⎪ = 0, if x = 0 . ⎪ ⎪ ⎪ ⎩ < 0, if x > 0

(8.174)

The point x = 0 where k2 (x) = 0 is called the classical turning point. This point is where a classical particle of kinetic energy k02 hits the potential hill V(x) and bounces back elastically. A quantum particle satisfying a wave equation can penetrated some distance into the potential hill. For potentials with |k (x)|  |k2 (x)|,

|x|  0,

k = dk/dx,

the mathematical problem involves two very different scales, as we shall describe. Verify the following description. (a) Define a (generalized) phase function φ(x) by ψ(x) = eiφ(x) . Then φ(x) satisfies the nonlinear differential equation (DE) iφ + (iφ )2 + k2 = 0, where φ = dφ(x)/dx, etc. The approximate solution obtained by neglecting the φ term when |φ |  |φ |2 is, for x  0  x φ(x) ≈ φ0 (x) = ± k(x )dx . 0

Here as elsewhere in this problem, the variable x is placed at the upper limit of integration. If the φ term is included approximately and itera tively as φ ≈ φ 0 = ±k , the improved solution is φ(x) ≈ φ0 (x) + φ1 (x),

φ1 (x) =

i ln [k(x)]. 2

Hence the resulting two linearly independent solutions (called Liouville– Green or LG functions) can be taken to be 1 e±iφ0 (x) , ψa± (x) ≈ √ k(x)

x  0,

where the subscript a denotes the “above” solution where the energy k02 is above the potential V(r). The solution below the potential, for x  0, can√be found by analytic √ continuation, i. e., by writing k2 = −κ2 , where κ = −k2 and all .. s are positive real quantities:

Asymptotic expansions

ψb± (x) ≈ √ 

1 e±Φ0 (x) , κ(x) x

Φ0 (x) =

609

x  0,

κ(x )dx .

0

In these solutions, all constant phase factors have been left out, including one (−1)−1/4 = e−iπ/4 in ψb (x) that will materialize later. We have now two disjoint general solutions ψa (x) ≈ a− ψa− (x) + a+ ψa+ (x),

x  0,

ψb (x) ≈ b− ψb− (x) + b+ ψb+ (x),

x  0,

in two non-overlapping spatial domains that have to be joined together. These solutions are not valid near the classical turning point where they are in fact singular and therefore incorrect. The problem is to find a way to connect them by overcoming this obstacle. (b) The WKB approximation gives connection formulas to stitch together the two disjoint sets of LG functions. This is done by noting that near the classical turning point, k2 ≈ −a3 x ≈ 0, a > 0, is very different in magnitude from the values |k2 |  0 on the asymptotic regimes. Keeping only this linear term, the wave equation (8.174) simplifies to the Airy DE w (y) − yw(y) = 0,

y = ax.

(8.175)

Its general solution is the linear combination w(y) = αAi(y) + βBi(y) of Airy functions Ai(y) and Bi(y) valid in all regions of x. These Airy functions have normal asymptotic forms as well as asymptotic representations by LG functions. Since asymptotic forms are unique in the limit x → ∞, the two disjoint sets of LG functions for the Airy equation can be related via the normal asymptotic forms of Airy functions. The WKB method is to just use the Airy connection formulas, exact only for linear potentials, as an approximation to connect the LG functions of nonlinear potentials. The resulting approximate is often surprisingly good for many potentials of physical interest. Assume without verification that the Airy functions of real arguments have the normal asymptotic forms " # ⎧ 2 3/2 π ⎪ 1 ⎪ ⎨ cos 3 |y| − 4 (y → −∞) Ai(y) ∼ &4 , ⎪ ⎪ 1 − 2 y3/2 (y → ∞) π2 |y| ⎩ 2 e 3 # " ⎧ 2 3/2 π ⎪ 1 ⎪ ⎨ − sin 3 |y| − 4 (y → −∞) Bi(y) ∼ &4 . (8.176) ⎪ ⎪ 2 y3/2 (y → ∞) π2 |y| ⎩ e 3

610

Functions of a complex variable

The LG functions for the Airy equation are 1 ± 2 (ax)3/2 (x → ∞), ψb± (x) ∼ √4 e 3 a3 x 2 3/2 1 ψa± (x) ∼ &4 e±i 3 (a|x|) (x → −∞). a3 |x| Hence show that the WKB/Airy connection formulas (denoted by ) are ψb∓ (x) ∼ &4

1 |k2 (x)|

e∓Φ0 (x)

(x → ∞),

⎧   π ⎪ ⎪ 2 cos φ (x) − ⎪ 0 1 ⎨ 4   √ ⎪ ⎪ π ⎪ k(x) ⎩ − sin φ0 (x) − 4  |x| & Φ0 (x) = |k2 (x )| d|x |,

(x → −∞),

0

 φ0 (x) =

x

k(x )dx .

(8.177)

0

8.15.7∗∗

The definition of Φ0 (x) has been generalized so that it applies also when the potential hill rises on the left. In that case, the asymptotic limits in the connection formulas must be changed appropriately: ±∞ → ∓∞. Historical note: The WKB method was first established independently by Joseph Liouville and by George Green in 1837. The connection formula was found for the Schr¨odinger wave equation independently by Wentzel, Kramers and Brillouin (after whom the method is named in the physics literature), all in 1926, and two years earlier in a more general mathematical context by Harold Jeffreys. (Airy functions) The asymptotic forms (8.176) of the Airy functions needed in the WKB approximation of the last problem can be found from integral representations of these functions. If the Airy differential equation (DE) for a complex variable z w (z) − zw(z) = 0,

where w (z) =

d2 w(z) , dz2

is solved by Laplace transform, the Airy functions are inverse Laplace transforms of the type  1 w(z) = ezt v(t)dt, 2πi Γ where Γ is a suitable contour in the complex t plane. Verify the following description of the Airy functions.

Asymptotic expansions

611

(a) The term zw(z) in the Airy DE has the integral representation  ,  1  zt B zt d zw(z) = e v(t) − e v(t)dt , A 2πi dt Γ where A, B are the end points of the contour Γ. The end-point term vanishes if the contributions at the end points are equal, and more specifically if each contribution vanishes. Many choices of Γ are thus possible. Before proceeding further, verify that under the Laplace transform, the Airy equation reduces to v (t) + t2 v(t) = 0

⇒

v(t) = e−t

3 /3

.

This shows that the complex t plane can be separated into three equal sectors on each of which an identical copy of v(t) can be found. The sector boundaries can be denoted by their end points A, B,C at |t| → ∞. The most common choice of sector boundaries are ) * {A, B,C} = lim R k2 , k, 1 , k = ei2π/3 . R→∞

With this choice, v(t) = e−|t| /3 decreases most rapidly along the sector boundaries as |t| → ∞. Three different Airy functions can then be defined with the two sector end points as the end points of the contour Γ:  B 1 Ai(z) = e f (t) dt, 2πi A 3

f (t) = zt − t3 /3,  C 1 e f (t) dt, kAi(kz) = 2πi B  A 1 e f (t) dt. k−1 Ai(k−1 z) = 2πi C

(8.178)

The contour is counter clockwise in each integral and can be deformed subject to the condition that the contour direction θΓ at a saddle point is independent of φ = arg z. The sector for the first integral, called the first sector, contains the negative real t axis. The positive real t axis is also the sector boundary OC, where O is the origin, between the second and third sectors. It is shared by the second and third integrals that have been defined by the same integral representation but different contours. These Airy functions satisfy the sum rule Ai(z) + kAi(Jz) + k−1 Ai(k−1 z) = 0

612

Functions of a complex variable

obtained by using the closed contour integral A → B → C → A. Hence at most two of the three Airy functions are linearly independent. (b) The asymptotic form of the first Airy integral in Eq. (8.178) can be found by the method √ of steepest descent. The function f (t) has two saddle points at t± = ± z, near which √ 2 f (z) ≈ ± z3/2 ∓ z(t − t± )2 . 3 For √ z = x (real) > 0, the saddle √ point located in the first section is t− = − x. For complex z, t− = − z is not on the real axis. The method of steepest descent gives the asymptotic form 2 3/2 1 Ai(z) ∼ (sign) √4 e− 3 z (|z| → ∞), 2 2 π z  1, if |φ| < 2π . (sign) = −1, if |φ − 4π| < 2π

(8.179)

Note: (i) The mathematical solution accepts both positive and negative overall signs for any φ. Our choice of signs comes from the convention that for real x ≥ 0, Ai(x) ≥ 0. This is the same convention that defines the exponential function to be e±x ≥ 0. (ii) The steepest pass direction θsp (φ) in the complex t plane makes one complete turn around the saddle point as φ makes 4 turns around z = 0 to cover the √ four-sheeted Riemann surface once. (iii) 4 x is the positive real root for real x ≥ 0. (c) For Ai(−z), the entire steepest descent procedure must√be repeated with f (t) = −zt − t3 /3. The saddle points are now at t± = ±i z. They are located on the imaginary t axis when z = x is real and nonnegative and out of the first sector, but off the axis more generally. The contour Γ must be deformed to pass both of them. The contour direction θΓ = π/2 at the saddle points remains unchanged from part (b). The resulting asymptotic form is ⎧ ⎪ ⎪ sin η, if |φ| < π ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎨ i cos η, if |φ − 2π| < π , Ai(−z) ∼ √4 ⎪ ⎪ ⎪ 2 − sin η, if |φ ∓ 4π| < π ⎪ π z⎪ ⎪ ⎪ ⎩ −i cos η, if |φ + 2π| < π 2 π η(z) = z3/2 + , 3 4

(|z| → ∞).

(8.180)

The phase shift √ π/4 in η comes from the factor ±i in the saddle point position t± = ±i z appearing in e f (t± ) .

Asymptotic expansions

613

(d) The corresponding asymptotic forms for the second Airy function Bi can be obtained from those for Ai by simply changing φ = arg z until z3/2 → −z3/2 , z → k±1 z,

or

φ→φ±

2π 3

k = ei2π/3 .

Direct substitution shows that Ai(k±1 z) ≡ c e∓iπ/6 Bi(z), ∓iπ/6

∼e

2 3/2 1 (sign) √4 e3z 2 π2 z

Ai(−k±1 z) ≡ e∓iπ/6 Bi(−z), ⎧ ⎪ ⎪ sin η , if ⎪ ⎪ ⎪ ⎪  1 ⎪ if ⎨ i cos η , Bi(−z) ∼ √4 ⎪ , ⎪ ⎪ 2 − sin η if ⎪ π z⎪ ⎪ ⎪ ⎩ −i cos η , if 2 π η (z) = − z3/2 + , 3 4

!

|φ| < π |φ − 2π| < π , |φ ∓ 4π| < π |φ + 2π| < π

(|z| → ∞).

One can see by inspection that the Bi functions are linearly independent of the Ai functions. We have chosen to define the analytic continuation of the oscillatory Ai(−z) to the variable −kz to be Bi(−z)e−iπ/6 . That leaves the normalization c of the non-oscillatory Bi(z) function still undetermined. The normalization c can be fixed by the requirement that the Wronskian between any Ai and its Bi partner is the same for both oscillatory and non-oscillatory solutions. Verify that W(−z) = Ai(−z) =

1 , π

W(z) = Ai(z) = Hence c = 1/2.

d d Bi(−z) − Bi(−z) Ai(−z) dz dz

2c . π

d d Bi(z) − Bi(z) Ai(z) dz dz

614

Functions of a complex variable

8.15.8 Verify the asymptotic expansion of the gamma function given in Eq. (8.168) by using the residue formula (8.165) for the coefficients an of the inverted series (8.160).

Appendix 8 Tables of mathematical formulas 1 Function of a complex variable f (z) = f (x + iy) = u(x, y) + iv(x, y). The FCV f (z) = (z + 1)1/2 + (z − 1)1/2 has four branches f1 = r11/2 eiθ1 /2 + r21/2 eiθ2 /2 f2 = r11/2 eiθ1 /2 − r21/2 eiθ2 /2 f3 = −r11/2 eiθ1 /2 + r21/2 eiθ2 /2 f4 = −r11/2 eiθ1 /2 − r21/2 eiθ2 /2 residing on a single, continuous Riemann surface z made up of four Riemann sheets. The branch points are at −1, 1, and ∞, where all four branches have the same value. The four Riemann sheets can be cut from or to the branch points at ±1. They are then cross-joined together at these cuts to form a continuous Riemann surface. The actual value of f (z) in each of its four branches depends on the angular ranges of θ1 and θ2 , that is, on the way the branch cuts are made. 2 Complex differentiation and analytic functions ∂u ∂u ∂v ∂v = , =− . ∂x ∂y ∂x ∂y Isolated singularities: mth-order pole: R/(z − a)m , essential isolated singularity: sin(1/z). Essential singularity: [sin(1/z)]−1 . Entire function: zn . Liouville theorem: a function everywhere finite and analytic is a constant. Singularities are sources of functional behavior.

Cauchy-Riemann conditions:

3 Complex integration

 f (z)dz = 0.

Cauchy integral theorem: Cauchy integral formulas:

c

f (z) =

1 2πi

dn f (z) n! = n dz 2πi

 c



f (z )dz , z − z

f (z )dz .  n+1 c (z − z)

Tables of mathematical formulas

615

4 Harmonic functions in the plane   2 ∂2 ∂ + f (x + iy) = 0, ∂x2 ∂y2 f (z) = u(x, y) + iv(x, y). u = const (equipotentials) ⊥ ∇u (lines of force), ν = const ⊥ ∇ν, ∇ν ⊥ ∇u. 5 Taylor series and analytic continuation S a (z) = f (a) + (z − a) f  (a) + . . . + =

∞ 

(x − a)n (n) f (a) + . . . n!

cn (z − a)n

n=0

is convergent in a circular region from a to the nearest singularity of f (z) if a is not a singularity of f (z). Similarly, S b (z) =

∞ 

dn (z − b)n

n=0

has a circle of convergence around b. If these circles overlap and S b (z) = S a (z) in the region of intersection of the two circles, the two Taylor series are analytic continuations of each other. 6 Laurent series and residues A Laurent series is a power series f (z) =

∞ 

cn (z − a)n

n=−∞

with negative powers. It describes an analytic function in an annular region surrounding an isolated singularity a of f (z). The coefficient c−1 of a Laurent series is   1 Res[ f (zi )]. f (z)dz = c−1 = 2πi enclosed The residue at zi

 m−1  d 1 m Res[ f (zi )] = [(z − zi ) f (z)] lim (m − 1)! z→zi dzm−1

616

Functions of a complex variable

is the coefficient c−1 of a Laurent series in the annular region immediately surrounding the mth-order pole at zi . For simple poles Res[ f (zi )] = lim [(z − zi ) f (z)] z→zi

$

  ! d Q(z) . dz zi

= lim P(zi ) z→zi

7 Complex integration: Calculus of residues Integrals can be evaluated by calculating residues if the integration path can be closed in the complex plane. Method 1  I=

∞

−∞

 f (x)dx =

f (z)dz

in the UHP or LHP if |z f (z)| → 0 as z → ∞. Method 2 (Jordan’s lemma).  I=

∞ −∞

 g(x)eiλx dx =

g(z)eiλz dx c

if |g(z)| → 0 as z → ∞. Close C in the UHP (LHP) if λ > 0 (< 0). Method 3 Clever return paths. Method 4 Adding a branch cut 



∞

I=

f (x)dx, 0

I=−



J=

ln z f (z)dz,

/ 0 Res f (z) ln z .

enclosed

Method 5 Use a unit circle  I=  =

0

2π

G(sin θ, cos θ) dθ

   ! 1 1 1 1 dz G z− , z+ . 2i z 2 z iz

Tables of mathematical formulas

8 Poles on contour and Green functions  I± = − f (z)dz + I± (z0 ) P 1 = ± iπδ(x − x0 ). ε→0 x − (x0 ± iε) x − x0 lim

If L (x) f (x) = ρ(x), then

 f (x) =

G(x − x )ρ(x )dx ,

where L (x)G(x − x ) = δ(x − x ). If L (x) = d 2 /dx2 + k02 , 1 G(x) = − 2π



∞

eikx dk : 2 2 −∞ k − k0

G± (x) = ± 2ik1 0 e±ik0 |x| , ingoing- (outgoing-) wave GF; G p (x) =

1 2k0

sin k0 |x|,

principal-value (standing-wave) GF.

If L (t) = d 2 /dt2 + ω20 ,



∞

e−iωt dω : 2 2 −∞ ω − ω0 : 1 Θ(t), retarded GF Gr,a (t) = sin ω0 t × −Θ(−t), advanced GF, ω0 1 G(t) = − 2π

where Θ(t) is the unit step function. 9 Laplace transform



∞

L { f (x)} =

f (x)e−sx dx = g(s)

0

L

−1

{g(s)} = f (x)Θ(x)  σ+i∞ 1 g(s)e sx ds (Bromwich integral). = 2πi σ−i∞

The integration path is to the right of all singularities. 10 Construction of functions and dispersion relations  1 f (z )  f (z) = dz . 2πi z − z

617

618

Functions of a complex variable

Example

 ∞ g(x )  R + dx ,  z−a 0 x −z where R is the residue at the pole at a and 2πig(x) is the discontinuity across the branch cuton the positive real axis. ∞ (b) f (x) = (1/πi) − [ f (x )/(x − x)]dx gives the Hilbert transforms (a) f (z) =

−∞

 1 Im f (x )  dx , Re f = −  π x −x  1 Re f (x )  Im f = − −  dx . π x −x

(c) Kramers-Kronig dispersion relations for the complex index of refraction n(ω) in exp[iωn(ω)z/c] with n(ω) → 1 as ω → ∞:  c ∞ α(ω )dω , Re n(ω) = 1 + − π 0 ω2 − ω2 where α(ω) ≡ (2ω/c) Im n(ω); α(ω) = −

 (2ω)2 ∞ Re n(ω ) − 1  dω . − πc 0 ω2 − ω2

11 Asymptotic expansions An asymptotic expansion to N terms f (z) = S N + O(z−N−1 ) ∼ S N SN =

N 

(z → ∞),

an z−n ,

n=1

is the partial sum S N of a series when N is kept fixed as z → z0 = ∞ and the remainder is known to be of order z−N−1 . The asymptotic limit z0 is sometimes finite instead of ∞. It is an asymptotic form or representation if N = 1. Laplace integrals for a real variable x:  ∞ ∞ (n)  q (0) I(x) = e−xt q(t)dt ∼ (x → ∞), xn+1 0 n=0  dn q(n) (0) = n q(t) . dt t=0

Tables of mathematical formulas

619

Complex integrals for a complex variable z = |z|eiφ along contour Γ in the complex t plane: Asymptotic form by the method of steepest descent  I(z) = ezp(t) q(t)dt Γ

,1/2 2π (z → ∞), z[−p (t0 )] ! 1  f (t) = z[p(t) − p(t0 )] ≈ z p (t0 ) (t − t0 )2 2 

∼ (sign)q(t0 )ezp(t0 )

= −|z|ρ0 r2 eiχ , t − t0 = re , iθ

where χ = φ + ψ0 + 2θ,



−p (t0 )/2 = ρ0 eiφ0 ,

at a saddle point t0 where d p(t0 )/dt = 0 and contour Γ has direction θΓ . Here  φ + ψ0 1, if |θsd | < π/2 (sign) = , θsd = − − θΓ . −1, if |θsd ± π| < π/2 2 Additional terms of the asymptotic expansion of  ∞ ezp(t) dt I(z) = ∞

=e μ



zp(t0 )

∞ ∞

e−zw

μ (t)

dt dw, dw

w (t) = z[p(t0 ) − p(t)] = (t − t0 )μ v(t), can be calculated after the series inversion dt  an wn : = dw n=0 ⎡  n+1 ⎤ ⎥⎥⎥ 1 ⎢⎢⎢⎢ dn t − t0 ⎥⎥⎦ , an = ⎢⎣ n n! dt w(t) t=t N

0

I(z) ∼ e

zp(t0 )

∞  

 2n+1 2a0 a2n Γ( μ ) −2n/μ Γ(1/μ) z a0 Γ(1/μ) μz1/μ n=0

(|z| → ∞).

Appendix A Tutorials A.1

Complex algebra

A complex√number c = a + ib = (a, b) is an ordered paired of real numbers a and b, where i = −1 stands for the positive square root of –1. The number a is called the real part of c, denoted by the mathematical expression a = Re c. Similarly, b = Im c is the imaginary part of c. The imaginary term ib of c is as different from the real term a as apple is from orange. Yet there is a simple connection between them on multiplication because i2 = −1 is real. An appreciation of the role played by i is the key to the understanding of complex numbers and complex algebra. The algebraic operations of addition, subtraction and multiplication of real numbers are carried over naturally to complex numbers: (a1 + ib1 ) + (a2 + ib2 ) = (a1 + a2 ) + i(b1 + b2 ), (a1 + ib1 ) − (a2 + ib2 ) = (a1 − a2 ) + i(b1 − b2 ), (a1 + ib1 ) × (a2 + ib2 ) = a1 a2 + ia1 b2 + ia2 b1 + i2 b1 b2 = (a1 a2 − b1 b2 ) + i(a1 b2 + a2 b1 ).

(A.1)

To handle division, we shall need a new algebraic operation on an expression called a complex conjugation. It is represented by an asterisk ∗ , and it only changes every i in the expression into −i: i∗ = −i,

a∗ = a, b∗ = b :

c∗ = (a + ib)∗ = a∗ + i∗ b∗ = a − ib, c∗ c = (a − ib)(a + ib) = a2 + b2 = |c|2 .

(A.2)

Here |c|, the positive square root of |c|2 , is real. |c| is called the absolute value or magnitude of c. Complex division can then be done by first defining the inverse of a complex number: c−1 =

1 1 = c a + ib

Complex algebra

621

  1 a − ib = a + ib a − ib =

1 (a − ib). |c|2

(A.3)

Two complex numbers c1 = a1 + ib1 , c2 = a2 + ib2 can then be divided as follows: 1 c1 = c1 c∗2 2 c2 |c2 | 1 = [(a1 a2 + b1 b2 ) + i(a2 b1 − a1 b2 )] , |c2 |2

c1 /c2 =

(A.4)

where |c2 |2 = a22 + b22 . We now have all the rules needed to function in the complex number world. A.1.1

The complex plane

A few more ideas and shortcuts will make complex calculations and manipulations easier. A real number a can be considered a point in an infinite one-dimensional (1D) line. In a similar way, a complex number or variable can be visualized as a point in a 2D xy plane z = x + iy = r(cos θ + i sin θ) = reiθ .

(A.5)

Here x = Re z and y = Im z are the Cartesian coordinates of a chosen coordinate system. The radius (or a clock hand of length) r and the polar angle θ are the polar (or circular) coordinates in the plane. The polar angle is measured in the counterclockwise direction from the polar axis, usually taken to be the x axis, as shown in Fig. A.1. To establish the relation between these two coordinate systems as shown in Eq. (A.5), we need the ancient trigonometric relations x cos θ = , r

y sin θ = , r

(A.6)

and the latter-day identity eiθ = cos θ + i sin θ.

(A.7)

This identity, called Euler’s formula (1748), will be derived in Tutorial A.7 on exponential functions. Two special cases of the Euler formula are worthy of note. Under complex conjugation

622

Tutorials y = Im z

z = x + iy y

= reiθ

r

θ x

0

x = Re z

Fig. A.1 The complex z = x + iy plane or Wessel-Argand diagram (1797, 1806). The imaginary unit i is automatically attached to a y coordinate when read. It is usually left out when labeling the y or imaginary axis. The x axis is called the real axis. The polar form given by the Euler formula z = reiθ is also shown.

z∗ = x − iy = cos θ − i sin θ = (reiθ )∗ = re−iθ = cos (−θ) + i sin (−θ).

(A.8)

Hence the trigonometric functions satisfy the reflection (θ → −θ) properties cos (−θ) = cos θ, sin (−θ) = − sin θ.

(A.9)

That is, cos θ is an even function of θ, while sin θ is an odd function. The second special case is made up of the special values eiπ/2 = cos (π/2) + i sin (π/2) = i, eiπ = −1,

ei3π/2 = −i,

ei2π = 1.

(A.10)

These values are generated by rotating a unit vector (or clock hand) r = 1 counterclockwise from the x axis by the angle θ = π/2, π, 3π/2, and 2π, respectively. The clock hand will then point respectively along the positive y axis, the negative x axis, the negative y axis, and the original positive x axis. In particular, ei2nπ = 1n = 1,

integer n,

causes the clock hand to return to the x axis after n complete rotations.

(A.11)

Complex algebra

623

The multiplication property of exponential functions (also from Tutorial A.7) e s et = e s+t can then be used to obtain |z|2 = zz∗ = r2 ei(θ−θ) = r2 ,

(A.12)

using e0 = 1. Hence the radius r is just the absolute value |z|. Multiplications, including inversion and division, also simplify in the polar representation: z1 z2 = (r1 eiθ1 )(r2 eiθ2 ) = r1 r2 ei(θ1 +θ2 ) , 1 e−iθ 1 = iθ = , z re r z1 r1 i(θ1 −θ2 ) = e . z2 r2

z−1 =

(A.13)

Much of the advantage of using complex numbers when it is not obligatory to do it comes from the simple elegance of these complex multiplications in the polar form. This is the key to the magical word of complex numbers. However, additions are easier to do in the Cartesian form. Also to get back to reality with real numbers, it is necessary to convert complex polar results to the Cartesian form. The word “imaginary” in connection with complex numbers was first used by Descartes (in La G´eometrie, 1637) in a derogatory sense. It now connotes the richness of imagination, as in “imaginative”. Complex algebra can be confusing when the expression involved has multiple values. Such ambiguity already appears in real algebra where 41/2 = ±2. We often have to depend on the context to decide if we are considering one or both roots. Multiple values appear much more frequently in complex algebra because many of the values are complex and therefore excluded from real algebra. Examples of many common issues of this type in complex algebra will be given in the Exercises. Historical note

Square roots of negative numbers might have been considered by the Greek mathematician Heron of Alexandria in the first century. They definitely appeared in solutions of cubic equations obtained by Italian mathematicians in the sixteenth century, √ notably by Cardano in 1545. √ The notation i = −1 was first used by Gauss in 1831. In electrical engineering, −1 is denoted j because i is reserved for the electric current. Example A.1.1 (2 + i)∗ (2 − i)(1 + i3) 1 = = (1 + i), 1 − 3i 10 2 1 1 |c|2 = c∗ c = (1 − i)(1 + i) = , 4 2 c=

624

Tutorials

1 c = √ eiπ/4 (polar form); 2 √ 1 i = eiπ/4 = √ (1 + i), 2 i1/2 = ±eiπ/4 ; i1/3 = eiπ/6 , ei(π/6+2π/3) , and ei(π/6+4π/3) . 

(A.14)

Problems A.1.1 c1 = 3 − i4, c2 = 1 + i are two complex numbers. (a) Represent the sum c1 + c2 in the complex z plane by first drawing the c1 directed arrow from the origin and then drawing the c2 arrow from the c1 arrow head. Represent the sum c2 + c1 by first drawing the c2 arrow from the origin. Show graphically that the results give the same composite arrow c1 + c2 = c2 + c1 = 4 − i3. (b) Show that c3 = c1 c2 = 7 − i, c1 1 c4 = ∗ = (7 − i). c2 2 Use the polar forms of these complex numbers to explain why c3 and c4 point in the same direction. That is, why is θ3 = θ4 ? A.1.2 (Square roots) The complex number c = |c|eiθ has two distinct complex square roots: & c1/2 = ± |c|eiθ/2 . √ To handle these two roots unambiguously, we shall use the notation √|c| for the positive square root of the nonnegative real number |c| and i = −1  √ 1/(−1) = 1/i. Show that for two real numbers a, b: √ √ (a) (ab)1/2 = a1/2 b1/2 = ± |a| |b|; a1/2 (b) (−a)1/2 = (−1) 1/2 . (c) Find a convention that will avoid the paradoxes 4 √ √ √ a a −a = i a  = , −1 i & √ √ √ √ √ √ −a −b = i2 a b  (−a)(−b) = a b. √ Show that your convention gives an unique result for ab.

Complex algebra

625

A.1.3 (Cube roots)√The complex number c = |c|eiθ has three distinct complex cube roots ( 3 |c|eiθ/3 )ei2πn/3 , n = 0, 1, 2. √ To handle these three roots unambiguously, we shall use the notation 3 |c| for the positive cube root of the nonnegative real number |c| and √3 k = −1 = eiπ/3 ,

√3

1 = 1;

√ while c1/3 refers to all three cube roots {1, k2 , k4 } 3 c. Show that for two real numbers a, b: (a) 11/3 = {1, k2 , k4 }; (−1)1/3 = {k, k3 , k5 }. √ √3 (b) (ab)1/3 = a1/3 b1/3 = {1, k2 , k4 } 3 a b. 1/3

a (c) (−a)1/3 = (−1) 1/3 , √ 3a √3 √3 −a = k a  √3 = −1

√ 3

a k .

(d) Find a convention to avoid paradoxes like those stated in Problem A.1.2(c). Note: These results √ can be extended to higher roots of complex numbers by using c1/m = ( m |c|eiθ/m )ei2πn/m , n = 0, 1, . . . m − 1, and to irrational powers p, namely c p = (|c| p eiθp )ei2πnp , n = 0, 1, . . . , ∞. A.1.4 Complex powers: (a) Show that for any real p (cos θ + i sin θ) p = cos (pθ) + i sin (pθ). The special case p = integer n is known as the De Moivre’s formula (1730). (b) For a purely imaginary power p = iq, show that 1 p = 1iq = e−(2πq)n , n = 0, 1, . . . , ∞, has infinitely many values if n  0 values are allowed. (c)∗ Resolve the Clausen paradox (1827) e = e1+i2πn = (e1+i2πn )1+i2πn 2 n2

= e1+i4πn−π

= ee−4π

2 n2

?

Hint: One of the steps is illegitimate. (d) Show that ii = e−(π/2+2πn) , integer n, is really multivalued. A.1.5 Show that cos(A + B) = cos A cos B − sin A sin B, sin(A + B) = sin A cos B + cos A sin B;

626

Tutorials

cos(A − B) = cos A cos B + sin A sin B, sin(A − B) = sin A cos B − cos A sin B. Hint: Use z1 = eiA , z2 = e±iB . A.1.6 (Square roots) Find the square roots ±(α + iβ) of c = a + ib, where a, b are real. Calculate their numerical values when a = 3, b = 4. Plot the numerical values of c and its two roots in the complex plane. Hint: The answer is much easier to √find in polar form: α = √ √ (a + r)/2, β = sign(b) (r − a)/2, where r = a2 + b2 . A.1.7 If a = α + iγ, b = β + iδ, where α, β, γ, and δ are real, find Re (ab ) and Im ab in terms of α, β, γ, and δ. " #b " #b Hint: ab = reiθ , and more generally ab = reiθ ei2πn . A.1.8 (Principal value of ln z) (a) Use the polar form of the complex variable z to show that Im (ln z) = θ + 2πn, integer n. Here ln is the natural logarithm. So ln z is multivalued. To avoid referring to this multiplicity of values in a complex expression, we can use its principal value, denoted Ln z as that value where −π < Im (Ln z) ≤ π. This restriction can often be confusing when comparing two expressions differing by i2πm, integer m. If both expressions refer to a natural logarithmic function, they could be considered equal, modulo i2π, but sometimes there are subtle differences between them.  √  (b) Verify that Ln e3 (1 − i)/ 2) = 3 − i π4 . (c) By calculating explicit results, show that the two expressions in each of the following sets are related, and that all four expressions have the same principal value: (i) ln [(1 + i)(1 + i)], 2 ln (1 + i). (ii) ln [(−1 − i)(−1 − i)], 2 ln (−1 − i). Can you find any significance to the differences between these four expressions? Hint: There are positive and negative square roots. (d) Show that if c, d are complex, (i) c ln d = c(Ln d + i2πm), integer m, is a subset of (ii) ln (dc ) = c(Ln d + i2πm) + i2πn, integer m, n. A.1.9 (Cardano equation) The following exercises describe why Cardano proposed to use the square root of a negative number in 1545. (a) If z = i1/3 + (−i)1/3 , calculate z3 − 3z. Hint: Expand z3 = (c + d)3 into a polynomial before handling complex numbers. (b)∗ If z = (q + w)1/3 + (q − w)1/3 , show that it satisfies the Cardano equation z3 − 3pz = 2q, where p, q are both real, if w = (q2 − p3 )1/2 . Cardano accepted solutions even when p3 > q2 . Then w in the Cardeno solution z = (q + w)1/3 + (q − w)1/3 is the square root of a negative number. (For the full story, see Penrose, 2004, pp. 75–6.)

Vectors

A.2

627

Vectors

A vector (“to carry” in Latin) r = rer is a directed line segment. It carries a point a distance r ≥ 0 in the direction defined by the unit vector er , of length or magnitude 1. A 2D vector can be specified by its two Cartesian coordinates as the ordered pair r = (x, y): x = r cos θ, y = r sin θ, % r ≡ |r| = x2 + y2 , er = (cos θ, sin θ).

(A.15)

Each coordinate is called a component of the vector. A single number, such as the component of a vector if considered alone outside the context of the ordered array, does not have a direction. It is then called a scalar. Because of its unit length, a unit vector, say er , can be specified by a direction angle, or equivalently by its direction cosine cos θ relative the x axis, as shown in Eq. (A.15). If the unit vector is parallel to the ith coordinate axis, it is called the basis vector ei . Basis vectors can be used to “represent” r as an algebraic expression rather than an ordered array of numbers: e x = e1 = (1, 0),

ey = e2 = (0, 1) :

r = xe x + yey = r1 e1 + r2 e2 =

2 

ri ei .

(A.16)

i=1

It is often more convenient to label the basis vectors by the ordinal numbers 1, 2 rather than the coordinates x, y. The ith component of a vector r is then simply ri . This ordinal notation leads to more compact expressions that facilitate many manipulations of vectors. The fact that we need two coordinates to specify a location in 2D space is of course familiar to most people, from pirates burying their treasures to Polynesians sailing to the next island over the horizon. The big surprise is the time it took scientists to use 3D vectors in their work. It was not until Gibbs in 1881 and Heaviside also in the 1880s when 3D vector analysis began to be used as a professional tool. They did so not by a straightforward generalization of 2D vectors to 3D, but by the reduction of the complicated 4-component generalization of complex numbers called quaternions to three components. The reason for this devious route is that there are unique operations involving 3D vectors that do not exist in 2D, but they appear in 4D. A good example is the vector product of two vectors A × B that is defined and studied in Chapter 1. If A and B lie on a 2D plane, the vector product is defined to lie in the third dimension perpendicular to the AB plane. Such a vector product has no meaning in 2D. There is thus a conceptual disconnection between the limited 2D world of our everyday experience and the third dimension of height. In 3D vector analysis, one has to overcome this limitation and visualize the uncommon consequences of the presence of an extra dimension.

628

Tutorials

These and other aspects of 3D vectors are covered in Chapter 1 of the text. So the purpose of this tutorial will be limited to providing additional exercises covering the more elementary aspects of vectors that appear in both 2D and 3D, including the scalar product. We shall concentrate on 3D vectors, treating 2D vectors as special cases where one of the three components vanishes. Thus we have e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1); A = A1 e1 + A2 e2 + A3 e3 =

3 

Ai e i

i=1

= AeA .

(A.17)

Addition and subtraction of two vectors A and B involve their respective components treated as simple numbers or scalars: A=B±C

⇒

Ai = Bi ± Ci ,

A=B

⇒

Ai = Bi ,

A=0

⇒

Ai = 0,

(A.18)

where ⇒ means “implies”. Multiplication or division by a scalar (say A) involves the operation for each and every component, as typified by the expressions we have already used without explanation: A = AeA and its inverse eA = A/A. Products of two vectors are a different matter altogether. In a scalar or “dot” product, the result is defined to be a scalar: A · B ≡ A1 B1 + A2 B2 + A3 B3 =

3 

Ai Bi .

(A.19)

i=1

This includes A · A = |A|2 = A21 + A22 + A23 , the length squared of A. Let us list some useful properties involving scalar products: (a) Since every component of a zero vector is zero, its length is also zero. (b) A vector A can be represented geometrically by a directed line segment (or an arrow) drawn from the origin O = (0, 0, 0) of coordinates to the point A = (A1 , A2 , A3 ). (c) For any two vectors A and B: A · B = ABcos θAB ,

(A.20)

where θAB is the angle between the line A and the line B. (See Problem A.2.3.) The Cartesian components of the unit direction vector er are the direction cosines er · ei = cos θri . Here θri is the angle between the vector r and the basis vector ei .

Vectors

629

(d) The vector difference A − B is a directed line pointing from B to A. Its length is |A − B|. The point C of the vector sum C = A + B can be located geometrically by starting from the point A and moving a distance B (length of B) along a line parallel to the line B. The length of the line C is |A + B|. (e) If A · B = 0 when both vectors are nonzero (A, B  0), they are said to be orthogonal or perpendicular to each other. For example, the unit vectors ei , i = 1, 2, 3, along the Cartesian coordinate axes are orthogonal to one another. Example A.2.1 If

A = (4, 3), B = (1, 2) : & & √ A = 42 + 32 = 5, B = 12 + 22 = 5; C = A − 2B = (4, 3) − 2(1, 2) = (2, −1), & √ |C| = 22 + 12 = 5;

A · B = (4, 3) · (1, 2) = 4 + 6 = 10,     2 A·B = cos−1 √ = 26.6◦ . θAB = arccos AB 5 If

er = (cos θ, sin θ) : sin θ = cos θr2 = cos (π/2 − θ).



(A.21)

Problems A.2.1 (2D vectors) A = (4, 3), B = (3, 4) are 2D vectors. Calculate |A|, |B|, C = A − B in Cartesian coordinates and in polar coordinates (call them rC , θC ), A · B, and the angle θAB (in degrees) between A and B. Sketch the vectors A, B and C as directed line segments or arrows in the xy plane. A.2.2 (2D straight line) (a) Express r the position vector of a point in the xy plane that lies on the infinite straight line joining the 2D points A and B as an algebraic function of A and B. Sketch the vectors A, B and r as arrows in the xy plane. Is your expression valid in nD, n ≥ 3? (b) Verify that the result agrees with the usual intercept-slope form of a 2D straight line: y = c + sx, if r = (x, y). (Partial answer: r − A = a(B − A), s = (By − Ay )/(Bx − A x ), c = Ay − sA x .) A.2.3 (Scalar product) (a) A and B are two 3D vectors. Show that A · B = ABcos θ AB , where A, B are the vector lengths, and θAB is the angle between the two vectors. Hint: Choose B = Bez to be the z axis of a spherical coordinate system where r = z + r⊥ , z = r cos θ, r⊥ = r sin θ.

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(b) Show that the expression continues to hold in nD space (with n > 3) where r⊥ is an (n − 1)D vector perpendicular to z. A.2.4 (Spherical and hyperspherical coordinates) (a) The spherical coordinates of the position r are the radius r, the colatitude angle θ and the longitude or azimuthal angle φ. Show that the unit direction vector er contains the direction cosines er · ei as its Cartesian components if the Cartesian components are ordered as (x, y, z). However, a more natural ordering is (z, x, y) where the component containing the cosine of the angle measured from the highest ranking remaining polar axis comes before or takes precedence over the components with the sines of the same angle. In the spherical coordinates, the primary polar axis is the z axis, while the secondary polar axis is the x axis. This explains why the x component follows the z component but precedes the y components. Re-write er with the components written in the natural order. Verify that er has unit length. (Partial answer: er = (sin θ cos φ, sin θ sin φ, cos θ) for the ordering (x,y,z), but this component list is not in natural order.) (b) The nD hyperspherical coordinates of dimension n consist of one radius r and n − 1 angles θ, φ, ψ, η, . . . in the order shown. Complete the Cartesian specification of the unit direction vector of er for hyperspherical coordinates in 4D: er = (cos θ, sin θ cos φ, . . .). Fill in the remaining two components in the natural order of precedence. Verify that er has unit length. (c) Give the Cartesian specification of the unit direction vector er of the 5D position vector r in hyperspherical coordinates in its natural order of precedence. Verify that er has unit length.

A.3

Simple and partial differentiations

Differentiation is the act of finding the derivative of a continuous function f (x). The derivative of f (x) at the point x = a is defined as the ratio of the difference Δ f = f (a + Δx) − f (a) to a change Δx in the variable x from a to a + Δx in the limit Δx → 0: d Δf df f (x) = lim = . Δx→0 Δx dx dx

(A.22)

It is useful to think of this difference quotient in the limit Δx → 0 as the quotient of two “infinitesimally small” but nonzero quantities d f and dx called differentials. The notation d f /dx was used by Leibniz who discovered calculus after Newton but published his results before Newton. Newton used the overdot notation f˙, and his variable is the time variable t. If the variable is not the time t, the Lagrange notation f  (x) = d f /dx is a useful short hand. Another notation D x f (x), from Euler, is occasionally used. We shall also use d x instead of D x .

Simple and partial differentiations

631

f (x)

f(x + Δx) Tangent Δf

Δx

x

f(x)

x + Δx

x

Fig. A.2 Geometrical picture of the limiting process that defines the derivative (d/dx) f (x) as the slope of the tangent to the curve f (x) at x.

It is useful to visualize the limiting process defining the derivative geometrically as the slope of the line segment connecting f (a) and f (a + Δx) as the latter point on the curve f (x) approaches the former point. The situation is illustrated in Fig. A.2. The limit coincides with the line that is tangent to the curve at x = a. The concept of a line “touching” (tangens in Latin) a curve at only one point was known to the Alexandrian Greeks. For example, Euclid (ca. 300 BC) stated in his Elements that that no other line can be drawn between the tangent to a circle and the point on the circumference it touches. However, the quantitative understanding of the limiting process described by Eq. (A.22) began only in the seventeenth century. When the limit in Eq. (A.22) is well defined, the derivative at x = a exists, and the function f (x) is said to be differentiable there. It can happen that the limit exists only when the limit is approach either from the right (Δx > 0) or from the left (Δx < 0), or both, but not at x = a itself. The right-hand derivative f  (a+ ) is then defined for a+ = a + 0 just to the right of a, or the left-hand derivative f  (a− ) is then defined for a− = a − 0 just to the left of a, or both, but not exactly at x = a. This happens if f (x) is discontinuous at x = a, or if it has a kink (sharp turn) there. The derivatives f  (a+ ) and f  (a− ) can be different. For example, f (x) = |x| has a derivative of f  = 1 for x > 0, but f  = −1 for x < 0. Thus f  (x) is discontinuous at x = 0, and it cannot be differentiated there. That is, its second derivative f  (x) does not exist at x = 0. When the same derivative (A.22) is obtained from both sides of a, the continuous function (C-function in short) is said to be C1 -smooth there. If it can be differentiated n times, it is Cn -smooth. For example, the function |x| is C0 -smooth at x = 0, while xn is Cn -smooth everywhere. One of the most obvious, elementary and useful properties of a differentiable function is that 1/ f  (x) = dx/d f is also the derivative (d/d f )x( f ) of the inverse

632

Tutorials Table A.1 Calculating df/dx by the limiting process defined in Eq. (A.22).

(a) (b) (c) (d)

f (x)

Δf

d f /dx

x x2 x3 x−1

(x + h) − x = h (x + h)2 − x2 ≈ 2xh (x + h)3 − x3 ≈ 3x2 h (x + h)−1 − x−1 ≈ −h/x2

1 2x 3x2 −1/x2

functional relation x( f ). This result is self evident in the Leibniz notation. This functional inversion property is particularly useful when one of the derivatives is much easier to compute than its functional inverse, as we shall see later in the Examples and Problems. Some functions are simple enough for their derivatives to be calculated directly from Eq. (A.22). The key to its use is that Δ f has to be calculated only to order h = Δx. Examples are shown in Table A.1. The approximation 1/(x + h) = (x − h)/(x2 − h2 ) ≈ −h/x2 is used in the last entry. The calculation of derivatives is simplified by the use of the following rules: ( f g) = f  g + f g ,   1 f = 2 ( f  g − f g ), Quotient rule : g g

Product rule :

Chain rule :

d f (g(x)) d f dg = . dx dg dx

(A.23)

In the chain rule, the Leibniz notation makes the relationships more obvious. These rules are easily demonstrated by noting that if the derivatives exist, then near x f (x + h) ≈ f (x) + f  (x)h, g(x + h) ≈ g(x) + g (x)h, q(x + h) ≈ q(x) + q (x)h,

(A.24)

where q is the function on the left side of each rule. The details are left to an exercise. The derivatives of many elementary functions are given in Table A.2. The use of these entries to generate other derivatives will be illustrated in the Examples. A.3.1

Partial differentiation

Partial differentiation means the differentiation of an expression with respect to only one of the two or more variables present in the expression. The variables not being differentiated on are treated as constants. The symbol for partial differentiation is ∂/∂x where the variable x being differentiated on appears explicitly. For example,

Simple and partial differentiations

633

Table A.2 Table of derivatives if used from left to right, and of indefinite integrals if used from right to left. p, a are complex parameters.

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k)

f (x)

d f /dx

xp ln x e px a x = e x ln a sin x cos x tan x sinh x cosh x arctan x arctanh x

px p−1 1/x pe px (ln a)a x cos x − sin x 1/ cos2 x cosh x sinh x 1/(1 + x2 ) 1/(1 − x2 )

f (x, y) = x + xy + y2 : ∂ f (x, y) = 1 + y, ∂x ∂ f (x, y) = x + 2y. ∂y

(A.25)

Partial differentiations are always unambiguous. Different partial derivatives involve different dependences in f (x, y). The total differential of f (x, y) is the total but still infinitesimally small change in f (x, y) when its variables are changed by the infinitesimals dx and dy. The total differential is given by d f (x, y) =

∂f ∂f dx + dy. ∂x ∂y

(A.26)

The expression is valid whether or not the two variables are independent of each other. If y is actually a function y(x) of x, Eq. (A.26) gives the total derivative, ∂ f ∂ f dy df = + , dx ∂x ∂y dx

(A.27)

where the term involving ∂ f /∂y comes from the combined product and chain rules of differentiation for those terms of f dependent on y(x) that are not included in the ∂ f /∂x term. For the function f (x, y) used in Eq. (A.25), if y = 2x is additionally imposed after the total derivative d f /dx of Eq. (A.27) has been calculated, one gets a total contribution made up of a contribution of 1 + 2x from ∂ f /∂x and a contribution of

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10x from the ∂ f /∂y term. The sum 1 + 12x of these two terms always agrees with the result obtained by first eliminating y = 2x in favor of x in f (x, y(x)) = x + 6x2 = g(x) before its derivative g (x) = 1 + 12x is calculated in a single step. Mathematical expressions can be differentiated formally in a computer with computer algebra software. The software also serves as mathematical tables. Hence it is no longer necessary to reproduce long mathematical tables in book form. It would be awkward, however, if a trained scientist cannot determine if the result from the software is not wrong due to an unsuspected error either in using the software or in the software itself. A.3.2

Examples

(1) The special case (eix ) = ieix can be used with the Euler formula for eix to find the derivatives of the two trigonometric functions (cos x + i sin x) = ieix = i(cos x + i sin x).

(A.28)

Alternatively, we can use the fact that sine comes from the odd part of eix : 1 ix (e − e−ix ) 2i 1 = [ieix − (−i)e−ix ] = cos x.  2i

(sin x) =

(A.29)

(2) An example of the quotient rule:   sin x  (tan x) = cos x =1+

sin2 x 1 = .  2 cos x cos2 x

(A.30)

(3) For y(x) = arctan x, we have x = tan y and therefore −1  dy d tan y (arctan x) = = dx dy 

=

1 1 .  = 2 1 + tan y 1 + x2

(4) Differentiation of a long expression: [9 + x1/3 − csc x + x2 e−x − 4 cos (3 ln x)] 1 1 = 0 + x(1/3−1) + 2 (sin x) + [(x2 ) e−x + x2 (e−x ) ] 3 sin x

(A.31)

Simple and partial differentiations

+ 4 sin (3 ln x)(3 ln x) 1 cos x 12 = x−2/3 + 2 + [2xe−x − x2 e−x ] + sin (3 ln x). 3 x sin x



635

(A.32)

Problems A.3.1 Verify the following: (a)

(b) (c)

d ix e = ieix = ei(x+π/2) , dx dn ix e = ei(x+nπ/2) , dxn dn cos x = cos (x + nπ/2), dxn dn sin x = sin (x + nπ/2)). dxn d cot x = − 12 . sin x dx If If

df = f gh. dx f = ln g, g = ln h, h = ln x : f  = (ghx)−1 . f = eg , g = eh , h = e x : f  ≡

A.3.2 Use the table entries and differentiation rules given in the text to differentiate √ 2 f (x) = 5 + 4 x − x sin x + 3e−x − sech(3x) + tan (ln x). Hints: The chain rule is needed in the last three terns. Also sech x = 1/ cosh x. A.3.3 (Quotient rule) Show (recursively, inductively or by any other means) that f f f ( f1 f2 . . . fn ) = 1 + 2 + ... + n . f1 f2 . . . fn f1 f2 fn

(A.33)

A.3.4 Derive the table entry for (e x ) from the table entry (ln x) = 1/x. A.3.5 (Functional inversion) (a) Show that (tanh x) = 1 − tanh2 x. Derive (arctanh(x)) = 1/(1 − x2 ) from it. √ (b) Show that (sec x) = sin x/ cos2 x. Derive (arcsec x) = 1/(x x2 − 1) from it. A.3.6 Derive the three differentiation rules stated in Eq. (A.23).

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A.3.7 (Leibniz rule) Derive the Leibniz rule for the nth derivative of the product f (x)g(x):  n n    n (n−k) (k) d f ( f g) ≡ ( f g)(n) = g k dx k=0 = f (n) g + n f (n−1) g(1) 1 + n(n − 1) f (n−2) g(2) + . . . + f g(n) . 2

A.4

Simple and multiple integrals

Integration or anti-differentiation is the reverse operation to differentiation. (This statement is called the fundamental theorem of calculus.) We represent the process symbolically as   d f (x) dx = d f (x) = f (x) + C, (A.34) dx where the arbitrary constant C that contributes nothing on differentiation is called an integration constant. It is needed because the integral in Eq. (A.34) is an indefinite integral. That is, it is equal to f (x) only up to a constant, here denoted C. In contrast, the definite integral  b d f (x) dx = f (b) − f (a) = f (x) |ba (A.35) dx a is fully specified by giving the limits a, b of the integration interval. (Eq. (A.35) is called the second fundamental theorem of calculus.) Table A.2 when used backwards (from right to left) is a table of indefinite integrals. If the interval h = b − a of integration in Eq. (A.35) is sufficiently small, we get  a+h f  (x)dx = f (a + h) − f (a) ≈ f  (a)h. (A.36) a

The result is the area of a very narrow rectangular strip of height f  (a) and width h. An integral over a finite interval b − a then represents the area under the curve f  (x) from x = a to x = b, as illustrated in Fig. A.3. It is given by the sum  b f  (x)dx = f (b) − f (a) a

= lim

M→∞

where

M−1 

g(xm )h

m=0

g(x) = f  (x),

xm = a + mh,

(A.37)

Simple and multiple integrals

637

g(x)

a

b

x

b Fig. A.3 Geometrical picture of the area contained in the integral a g(x)dx = f (b) − f (a) =  M−1 m=0 g(xm )h, where g(x) = d f (x)/dx, xm = a + mh, h = (a − b)/M. The case of M = 10 rectangular strips of width h is shown.

in the limit where the number M of strips of width h = (b − a)/M becomes infinite.  The integration symbol , an elongated S , was introduced by Leibniz to refer specifically to the sum shown on the right of Eq. (A.37). The sum is now called a Riemann sum. If the sum exists and is independent of where f  is evaluated within each subdivision of width h, the function is said to be integrable. The resulting integral is called a Riemann integral. (The reader should consult a book on integral calculus in those cases where the Riemann integral is not defined, but other types of integrals are possible.) While integration is the operational inverse of differentiation, they differ qualitatively in many ways: (a) Differentiation involves a small neighborhood of a function, but integration is an intrinsically global concept. As a result, integration is less sensitive to the detailed local behavior. For example, it is defined for discontinuous functions, whereas the derivative is not defined at the discontinuities. More specifically, a C n function integrates to a Cn+1 function, but it differentiates to a C n−1 function. Integrations are more intuitive to understand and easier to estimate numerically than differentiations. (b) For known functions, however, differentiations are as a rule easier to find than integrations. So one’s first objective as a practitioner of integral calculus is to accumulate a sufficiently large personal working table of derivatives to use for integrations in the opposite direction. The procedure works because the local properties of most known functions also determine their global properties. As you make up your own table of derivative/integrals, you will soon find that many integrals are related to each other by a change of variable, including trigonometric substitutions. (c) As an alternative, one can learn to use a computer algebra software to “look up” integrals expressible in terms of known functions. Past experience of analytic

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Tutorials

work on integrations is very helpful in developing the ability to use such computer help intelligently. (d) If the function you are integrating is sufficiently unique, even a computer algebra program may not give you a closed form in terms of known functions, especially functions you are familiar with. It is then time to use the computer to calculate the numerical value of a Riemann sum instead. Even here, it will help to know of certain complications where the integral becomes “improper”. This happens if the integrand or a limit of integration is unbounded. When this happens, the unboundedness has to be examined more carefully. We shall clarify some of these issues by examples. Further explanations follow these examples. Example A.4.1  (1)

1

lim

a→0

# 1 " 1 − a p+1 a→0 p + 1

x p dx = lim

a

= 

b a

1 , p+1

if

p > −1;

dx = ln b − ln a. x

(A.38)

The first integral is divergent at the lower limit x = a → 0 if p < −1. If p = −1, the result is the second integral instead. The second integral is divergent when a, b = 0, ±∞. These are the well-known logarithmic divergences.    x+1 (2) dx → (y4 + 6)4y2 dy (x − 5)1/4 4 (A.39) → (x − 5)3/4 (x + 9) + C. 7 The third integral requires a substitution or change of variable x − 5 = y4 to handle the denominator. After the y integration, y is changed back to x. Some intermediate steps have been left out, but they are all straightforward.   ∞  ∞ dx dy (3) √ →2 2 1 (x + 1) x 1 y +1 = 2 arctan y |∞ 1 π π π = 2 − = . 2 4 2

(A.40)

The fourth integral requires a change of variable x = y2 . The integral is improper at the upper limit x = ∞, but the result there is finite since arctan ∞ = π/2. That there is no trouble at the upper limit of integration can be seen in the original x integral as

Simple and multiple integrals

639

follows. The integrand is bounded by the function x−3/2 . This upper bound integrates to −2x−1/2 , which is –2 at x = 1 and 0 at x = ∞. The integrated upper bound 2 is indeed greater than π/2, and is finite.      1 1 −2 4 dx (4) (tan x) sec (x)dx = + cos2 x sin2 x = − cot x + tan x + C. (A.41) The fifth integral requires a geometric simplification using an extra factor 1 = sin2 x + cos2 x in the numerator. The middle expression is then integrated by using the derivatives (tan x) and (cot x) . Of course, one needs to remember these derivatives or know where to find them.  (5) In integration by parts, one uses the product rule for differentiation in reverse to replace one integration by its partner if the latter is easier to integrate: 

b a

f g = ( f g) − f  g ⇒   b f (x)g (x)dx = f (x)g(x) |a −

b

f  (x)g(x)dx.

(A.42)

a

For example, 

   n+1 xn+1 1 x (ln x)x dx = (ln x) − dx n+1 x n+1   1 xn+1 ln x − .  = n+1 n+1 n

(A.43)

(6) In parametric differentiation, one generates one integral from another by differentiating the latter with respect to a parameter it carries. For example,  ∞ 1 e−sx dx = ⇒ f (s) = s 0    ∞ d 1  xe−sx dx, f (s) = 2 = − f (s) = − ds s 0  n d n! (−1)n f (n) (s) ≡ − f (s) = n+1 ds s  ∞ = xn e−sx dx. (A.44) 0

The last integral, with s = 1 and changes of variable x → t, n → z − 1, can be used to define an important function called the gamma function:

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Tutorials

 Γ(z) = (z − 1)! =

∞

tz−1 e−t dt

(A.45)

0

of the complex variable z = x + iy. Defining a function by an integral is often useful in both formal and practical applications. Another good example of an integral representation of a function is  x dt .  (A.46) ln x = 1 t (7) The additional powers of x generated by differentiating de−sx /ds under the integration sign can also be used to get rid of awkward powers of 1/x in the original integrand. For example,  ∞ sin x −sx g(s) = e dx ⇒ x 0  ∞  ∞  −sx −g (s) = sin xe dx = Im e(−s+i)x dx 0

0

1 1 = Im . = s − i 1 + s2 Finally, g (s) can be integrated to get

(A.47)



1 ds 1 + s2 = − arctan s + C.

g(s) = −

(A.48)

The integration constant C is found by noting that g → 0 when s → ∞ because the intgrand is killed by the huge exponential decrease. Hence C = arctan ∞ = π/2. We are now in a position to calculate the integral  ∞ sin x dx = 2g(0) −∞ x  π = 2 arctan 0 + = π.  (A.49) 2 (8) As a final example, we consider a 1D dynamical system whose mass density ρ(x, t) depends on both the position x and the time t. The total time derivative is dρ(x, t) ∂ρ ∂ρ dx = + dt ∂t ∂x dt ∂ρ ∂J + , = ∂t ∂x

(A.50)

where J = ρ(dx/dt) = ρv is the mass current density. dρ/dt contains not only a direct contribution due to the explicit time dependence of ρ, but also a separate motional contribution caused indirectly by the velocity dx/dt of the position x itself.

Simple and multiple integrals

641

Consider now the mass 

x+Δx

Δm(x, t) =

ρ(x, t)dx

(A.51)

x

residing in the interval x to x + Δx. Integrating Eq. (A.50) term by term over this interval, we find dΔm ∂Δm = + ΔJ, where dt ∂t  x+Δx ∂J ΔJ = dx ∂x x

(A.52)

is called the net outflow of mass from the interval Δx. In Newtonian mechanics, mass is conserved, meaning dΔm/dt = 0. Hence the mass Δm in Δx can increase only by a net inflow of mass, or ΔJ < 0 of the right amount. Under mass conservation, dρ/dt = 0. The resulting Eq. (A.50) then reads ∂ρ/∂t + ∂J/∂x = 0.

(A.53)

This equation is called a continuity equation for a conserved density ρ. It illustrates the distinction between a total time derivative and a partial time derivative. More specifically, it describes how the idea of conservation extends from the integrated mass Δm to its density ρ = dm/dx. 

A.4.1

Multiple integrals

An integrations of n variables spanning an nD space can also be defined. The simplest examples are concerned with the calculation of areas and volumes of simple geometrical figures. These results have been of great practical importance since ancient times. Empirical formulas for them based on practical experience were already known when ancient Egyptians built their first big pyramids (before 2600 BC). An existing papyrus called the Moscow Mathematical Papyrus (dated ca. 1800 BC and now the property of a museum in Moscow) contains recipes giving the correct surface area of a (presumably hemispherical) basket and the correct volume of a truncated pyramid. With integral calculus, both can be calculated easily. The area of the curved part of the hemispherical basket of radius R is of course 2πR2 . The pyramid volume will be left to an exercise. Example A.4.2 The area between the parabola y = x2 and the line y = 1 and the volume of the solid of revolution generated by rotating this area about the y axis are, respectively,

642

Tutorials Table A.3 Functions f  (x) to be integrated in Problem A.4.1.

f  (x)

f (x)

(a) (x + a) p (b) 1/[(x − a)(x − b)] (c) tan√x (d) 1/ √ 1 − x2 (e) 1/ √ 1 + x2 (f) 1/ x2 − 1

(x + a) p+1 /(p + 1) γ ln[(x − a)/(x − b)], γ = 1/(a − b) − ln (cos x) arcsin x, or − arccos x arcsinh x arccosh x

 A=

1

0

 V=

0

1

4 √ 2 ydy = , 3 πydy =

π , 2

(A.54)

where only the final y integrations are shown. The average squared distance of a point in the area A to the origin is given by the 2D integral  √y    1 3 1 2 2 2 r = (x + y )dxdy = dy (x2 + y2 )dx A 2 0 0 A  3/2   y 22 3 1 5/2 dy = . +y (A.55) = 2 0 3 35 The key to such multiple integrals is to find the correct limits of integration for each integrated variable. 

Problems A.4.1 (Change of variable) Find the antiderivatives f (x) of Table A.3 using only the entries of Table A.2. The answers, given in the second column, are not known before the problem is solved. Hints: For (b), first show that   1 1 1 . =γ − (x − a)(x − b) x−a x−b For (d), change to the variable x = cos y or sin y. A.4.2 (a) Check the result  xeix dx = (1 − ix)eix by differentiating the expression on the right.

Matrices and determinants

643

Integrate the expression on the left directly by the method of (i) parametric differentiation, and (ii) integration by parts.   Find x sin x dx, and x cos x dx. (b) Find x2 eix dx by parametric differentiation. (Answer: [2x + i(2 − x2 )]eix .) A.4.3 (Volume of a spherical segment) A sphere of radius R is cut into two segments by a plane. Show by evaluating an integral that the volume of the smaller segment of height h < R is V = πh2 (R − h/3). A.4.4 (nD pyramid) (a) If the height h = αa of an isosceles triangle (2D right pyramid) is proportional to its base length a, show that the triangle has area A = αa2 /2. (b) Show that in n dimensions (nD), a right pyramid of height h = αa based on an (n − 1)D hypersquare or hypercube of side a has a hypervolume Vn = αan /n. (c) Consider a 3D right pyramid of height h = αa with a square base of side a and a volume Va = αa3 /3. If its top is truncated where the side is b < a, show that the truncated pyramid has volume V = h(a2 + ab + b2 )/3, where h = ha − hb is the height of the truncated pyramid. This is the 4000-year old formula written in hieroglyphics in the Moscow papyrus. Give the volume formula for the truncated right pyramid in nD. (Do not expand it into a sum of powers.) (d) A hemisphere of radius R is placed base down inside a cylindrical shell of the same radius and height h = R. Show that the space between the hemisphere and cylindrical wall has the same volume as a right cone of height h = R on the same circular base of radius R. State all assumptions and formulas used in the demonstration. Derive all the volume formulas you need. Note: A picture of a cylinder circumscribing a sphere is supposed to have been carved on the tombstone of Archimedes, together with the inscription of 3/2 (for the ratio of their volumes), as he wished. ∞ √ 2 A.4.5 (2D integral) Verify that −∞ e−x dx = π by evaluating its square, a 2D integral, in circular coordinates. This is another simple but memorable result.

A.5

Matrices and determinants

The term matrix (Latin for womb) for a rectangular array of numbers was coined by Sylvester in 1850 to refer to these arrays from which “determinants may be engendered as from the womb of a common parent”. However, its use in the solution of simultaneous algebraic equations already appeared in Chapter 8, Rectangular Arrays, of a Chinese mathematical text The Nine Chapters on the Mathematical Art, a second century compilation of solutions of mathematical problems of practical importance found by mathematicians of the tenth to second century BC. Determinants also appear

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there to determine if the system of equations has a unique solution. The equations are then solved by the method of Gauss elimination or sequential elimination of the unknown variables. (The method was named after Gauss for the notation he used but was already found by Newton in 1670 and published in 1707). Many of the matrix properties we use were established by Cayley in 1858. A 2 × 3 matrix   11 12 13 A = (aij ) = (Aij ) = (A.56) 21 22 23 contains two rows and three columns. Its matrix element aij or Aij (here ij itself) is the number appearing at the ith row and jth column. The notation aij makes a long expression easier to read, while the notation Aij refers directly to the original matrix. Matrices of the same dimensions can be added or subtracted locally, element by element. A negative matrix contains the negative element everywhere, as in −A = (−Aij ). Two matrices are equal A = B only if Aij = Bij for all i, j. A zero matrix is one with zero elements everywhere: A = 0 if every Aij = 0. A square matrix is one where the row and column dimensions are the same: m = n. A square matrix with nonzero elements only along its diagonal positions i = j is a diagonal matrix. A diagonal matrix with elements 1 along the diagonal is a unit or identity matrix:   1 0 I= = Diag(1, 1), (A.57) 0 1 where it is only necessary to specify the diagonal elements. The elements of the unit matrix I is called the Kronecker delta symbol  1, when i = j (A.58) Iij ≡ δij = 0, when i  j. A matrix made up of a single row or column is called a row vector, or column vector, respectively. In multiplication and division, both by a scalar α, and in complex conjugation, the operation is executed globally for each and every element: αA = (αaij ), ∗

A/α = (aij /α),

A = (a∗ij ).

(A.59)

Another basic matrix operation is transposition, denoted by a superscript T , which interchanges rows and columns: " # (A.60) AT = Aji = aji , ij

where the parentheses in the expression on the left has no special meaning other than defining AT as a single matrix. If A is m × n, then AT is n × m. All these matrix operations can be executed in any order.

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645

Matrix multiplication between matrices is a different matter. Each matrix element of the product C = AB of two matrices A and B is defined to be the sum of products of their matrix elements of the form cik = ai1 b1j + ai2 b2j + . . . + ain bnk =

n 

aij bjk .

(A.61)

j=1

The sum goes across the ith row of the first matrix A and down the jth column of the second matrix B. Consequently, the column dimension of A must match the row dimension of B. If A is m × n and B is n × p, the matrix dimension of AB is m × p. (The expression m × p is usually pronounced “m by p”. It looks neater the first way, but it sounds better the second way.) So BA may not even have the same dimensions as AB, unless they are both square matrices, with m = n = p. Even then BA  AB in general, unless both matrices are diagonal. A square matrix is diagonal if its nonzero matrix elements lie only on the diagonal ( j = i) locations in the array. Consequently, Eq. (A.61) gives cii = aii bii = bii aii and ciki = 0. Then C = AB = BA is also diagonal. More generally, a matrix product depends on the order of appearance of the matrices. Hence matrices do not generally commute under matrix multiplication. Such an algebra is said to be non-commutative under multiplication. The only general exception to this is the multiplication by the unit matrix I. Such a multiplication, whether done on the left or right has no effect on A. So I commutes with all matrices: AI = IA. The explicit demonstration of this result will be left as an exercise. Under transposition, (AB)T = BT AT for two square matrices is also order-specific. This result can be demonstrated as follows: n n   T [(AB) ]ik = (AB)ki = Akj Bji = (AT )jk (BT )ij j=1

=

n 

j=1

(BT )ij (AT )jk = (AT BT )ik .

(A.62)

j=1

In the second to last step, the product of two ordinary numbers are transposed as they are permitted to do in order to rearrange their indices in the correct matrix multiplication order. A square matrix S is symmetric if it is unchanged under transposition: S T = S . An antisymmetric matrix A changes sign under transposition: AT = −A. Example A.5.1  11 21

  0 12 13 + 3 22 23

1 4

  11 + 0 2 = 21 + 3 0  11 13 = 24 26

12 + 1 13 + 2 22 + 4 23 + 0  15 . 23



(A.63)

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i 0 0 5+i

3 6 

T ∗

1 4

0 5

  

A.5.1

1 4 0 2

⎛ ⎛ ⎞ 0 ⎟⎟∗ ⎜⎜ −i ⎜⎜⎜ i ⎜ ⎟ = ⎜⎜⎜⎝ 0 5 + i ⎟⎟⎟⎠ = ⎜⎜⎜⎝ 0 3 6 3 ⎛ ⎞  ⎜0 1⎟  ⎜⎜ ⎟⎟ 3 12 × ⎜⎜⎜⎝ 2 3 ⎟⎟⎟⎠ = 0 10 4 0  12 = 10     0 0 1 0 × = 5 2 3 10     1 1 0 4 × = 3 4 5 14

⎞ 0 ⎟⎟ ⎟ 5 − i ⎟⎟⎟⎠ 6 1 4 + 15



 1 , 19  1 19  5 .  15

(A.64)

Inverse matrix

The division B/A of two n × n square matrices can be defined as BA−1 , where the inverse matrices A−1 = L and R satisfy the multiplication properties LA = I = AR giving the n × n unit matrix I. That is, L is the left inverse and R is the right inverse. Since I commutes with any square matrix, we have IA = (AR)A = AI = A(LA), or L = R = A−1 The matrix inverse A−1 for an arbitrary n × n matrix A can be found directly from the matrix equation AA−1 = I. The n × n unit matrix I contains n unit column vectors ei with elements (ei )j = δji . This unit vector is the same as the unit vector ei , but now written as a column matrix. It has a component of 1 along the ith axis/direction and zero components elsewhere. If the n column vectors of A−1 are denoted vi , then matrix inversion requires the solution of n simultaneous algebraic equations of the type Avi = ei ,

i = 1, 2, . . . , n.

(A.65)

These equations can be solved directly for the column vectors vi of A−1 when n is small. For n = 2, for example,       1 0 ab e1 = , e2 = , A= . (A.66) 0 1 cd The resulting matrix equations Avi = ei , i = 1, 2, can be solved readily to give   1 d −b −1 A = (v1 , v2 ) = , det A −c a   a b    = ad − bc. (A.67) det A =  c d

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647

The details will be left to Problem A.5.3. Determinant In Eq. (A.67), the number det A (also denoted |A|) is called the determinant of the matrix A. The determinant det A is said to have the same order n as that of the n × n matrix A. It can be defined rather simply and recursively in terms of the determinants of submatrices of A of order n − 1: det A =

n 

AkqCkq ,

(A.68)

k=1

where the column index q is a spectator. The formula holds for any choice of q. The formula is called the cofactor expansion (or Laplace development) of the qth column of A. Ckq , an element of an order n matrix called the cofactor matrix C, is defined as Ckq = (−1)k+q Mkq ,

(A.69)

where Mkq , called a minor of det A, is the order n − 1 determinant of the order n − 1 submatrix of A obtained by deleting the kth row and qth column of A. All these definitions seem very abstract, but Eq. (A.68) is really a very elegant expression once we get to know it. We shall do so in the natural direction, the direction of increasing order. The simplest determinants are the second order determinants already defined in Eq. (A.67). A third order determinant by a cofactor expansion of the first column is det A = A11C11 + A21C21 + A31C31 , C11 = M11 , C21 = −M21 , C31 = M31 .

(A.70)

Only the elements of the first columns of A and C (or M) are used. The explicit expression in terms of the elements of a general third order matrix ⎛ ⎞ ⎜⎜⎜ a b c ⎟⎟⎟ A = ⎜⎜⎜⎝ d e f ⎟⎟⎟⎠ is gh i      b bc e f   − d   + g  det A = a  e hi h i

(A.71)  c  . f

(A.72)

The remaining elements of the third order matrices C and M do not appear here. They appear in cofactor expansions of the second and third columns of A. They also appear in different combinations in the cofactor expansion of the kth row det A =

n  q=1

AkqCkq ,

(A.73)

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where the column index q is now summed and the row index k is a spectator. There are a total of 2n cofactor expansions all giving the same det A, with each expansion involving the same column (or row) of the matrices A and C. The structure of an order n determinant can now be seen. The second order determinant of Eq. (A.67) contains two terms, and the third-order determinant in Eq. (A.72) contains 3 × 2 = 6 terms. So by extension, an order n determinant contains n! terms. Each term for the order n determinant is made up of a product of exactly n elements Aij selected in a special way. These elements come from different rows and columns in such a way that row or column indices are never repeated in the same term, as one can verify for determinants of orders 2 and 3. Since there are only n indices 1 to n, every index appears once and only once. We can order the product in each term so that the column indices appear in the standard or reference 12. . . order, then the row indices appear as a permutation ij . . . of the standard ordering. Each term carries a sign, + for an even permutation (12 in an order 2 determinant), and – for an odd permutation (here 21). (A permutation is even if it takes an even number of transpositions of next neighbors to reach the permutation from the standard ordering.) The signature can be denoted by a permutation (or Levi–Civita) symbol εij··· of the n indices ij . . . . The given prescription for row-column deletion in defining the matrix element of the minor matrix M when followed to the lowest orders enforces the no-duplication rule on row and column indices. The additional signature in the cofactor matrix element maintains the signature rule for each of the n! terms of the determinant. As a result,  det A = εP Ai1 Aj2 . . . Aln , (A.74) P

where εP is the permutation symbol for the permutation P = ij . . . l. We now have all the tools for constructing the inverse matrix A−1 . Eq. (A.73) can be written as 1=

n 

Akq

q=1

Ckq = Ikk . det A

(A.75)

It suggests that Ckq / det A can be taken to be (A−1 )qk after the indices kq are transposed. The identification would be perfect if the inversion relation Iik = 0 =

n  q=1

Aiq

Ckq det A

(A.76)

also holds when i  k. This turns out to be true, but its demonstration is nontrivial and is the subject of a problem in Section 2.3. The key step is to realize that this expression with i  k comes from the replacement of the original kth row of A by its ith row. The modified A matrix thus contains the ith row twice, once in its original row

Matrices and determinants

649

and again in the kth row. It is known that such a matrix always has zero determinant. So the problem reduces to a demonstration of this well-known result, which is the objective of the problem. To summarize, we have used a definition of the determinant to construct the matrix inverse of A as A−1 =

CT . det A

(A.77)

The result shows clearly that such an inverse does not exist if det A vanishes. When this happens, the matrix A is said to be singular. The inverse matrix A−1 can be used to solve the linear equation Ax = c

⇒

x = A−1 c,

(A.78)

where A is n × n square, and x, c are both n × m. There is then a unique solution only if A−1 exists, or det A  0. On the other hand, the homogeneous equation Ax = 0 has only the trivial solution A = 0 if det A  0. To get a nonzero solution, it is necessary but not sufficient to impose the condition det A = 0. This topic is covered in Chapter 2, and will not be repeated here.

Problems A.5.1 Calculate 

1 0 3 i

5 2+i

T

⎛ ⎞ 4 ⎟⎟∗ ⎜⎜⎜ 1 ⎟ + ⎜⎜⎜⎝ 2 + i5 2i ⎟⎟⎠⎟ 0 6

A.5.2 (Unit matrix) Show that IA = A = AI, where I is the n × n unit matrix and A is any n × n matrix. A.5.3 Show that any matrix B = S + A can be written as the sum of a symmetric matrix S and an antisymmetric matrix A. A.5.4 (Matrix inversion) Show by solving two sets of two simultaneous algebraic equations that A−1 for a general 2 × 2 matrix A is given by Eq. (A.67). A.5.5 (Determinants) (a) Verify that the signs of the terms of det A of a general second-order square matrix A = (aij ) satisfy the even/odd permutation symmetry rule. (b) Write down det A as the cofactor expansion of the first row of the general third order matrix A = (aij ). Verify that the signs of the six terms present satisfy the even/odd permutation symmetry rule. A.5.6 (Cofactor matrix) Write down explicitly the complete cofactor matrix C or all its elements Cij of order 3 in terms of the elements of a general third order matrix A given in Eq. (A.71).

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A.6

Tutorials

Infinite series

A series is the sum S N = a1 + a2 + . . . + a N =

N 

an

(A.79)

n=1

of terms a1 , a2 , . . . , an . A finite series has a finite number n of terms. If the number n is infinite, the series is an infinite series: S = S ∞ = lim S N = N→∞

∞ 

an .

(A.80)

n=1

It is convenient to write S = S N + RN+1 ,

RN+1 =

∞ 

an ,

(A.81)

n=N+1

by separating a partial sum S N of N terms from a remainder RN+1 . If RN+1 → 0 in the limit N → ∞, the series is said to converge to a limit S . If not, it is a divergent series. The idea of counting to infinity needed to construct an infinite series was understood by ancient Greeks and Indians. The philosopher Zeno of Elea (in Italy) (ca. 490–430 BC) is best remembered for his motion paradoxes involving infinities of subdivided intervals. One of his paradoxes states that it is impossible to take a first step in a journey because there is a half step before it, a quarter step before the half step, etc. Such an infinity of tasks cannot be completed, so argued Zeno. A clearer way to present the paradox is to replace a single step first by a half step, then a step half of the remaining half, etc. After an infinity of halved steps, one still cannot reach the original single step. The paradox illustrates the difficulty people had in Zeno’s time and place in understanding the difference between counting infinity and the infinity of magnitude. The added time aspect of motion makes the issue more confusing. However, Aristotle (384–320 BC) was able to distinguish between the infinity in divisibility from the infinity in magnitude or extension. We know now that the time it takes to complete all these partial steps and the distance they cover are both described by the geometrical series to be given in Eq. (A.85): 1/2 1 1 1 + + + ... = = 1. 2 4 8 1 − 1/2

(A.82)

Archimedes (287–212 BC), the greatest mathematician of the ancient world, had phenomenal geometric intuition. He found a geometrical result (the equal triparitition of areas) that is the generalization to two dimensions of the infinite sequence of linear bisections. The result amounts to the sum of another infinite geometric series 1 1 1 1 + + + ... = . 4 16 64 3

(A.83)

Infinite series

651

This geometric sum is actually easy to demonstrate for areas of simple shapes such as a square or an equilateral triangle, as describe further in Problem A.6.1. Indian mathematical texts of ca. 400 BC distinguishes between the nearly infinite, the truly infinite and the infinitely infinite, and between countless infinity and endless infinity. The Indian mathematician–astronomer Madhava of Kerala (ca. 1350–1425) appeared to have been the first to obtain infinite series of functions, including estimates of the remainders of certain infinite series. He also developed differentiation and integration more than two centuries before Newton and Leibniz. Example A.6.1 (1) Some infinite series can be summed easily after simplification. For example, a partial fraction expansion gives ∞  1 1 1 1 = − ⇒ = 1. n(n + 1) n n + 1 n(n + 1) n=1

(A.84)

The final result is obtained by noting that the two separated infinite series cancel completely except for the first term of the first infinite series.  (2) Geometric series: The geometric series G s (x) and the function G(x), G s (x) = 1 + x + x2 + x3 + . . . , 1 G(x) = , 1−x

(A.85)

are related. Their relationship can be found easily either by long division of 1/(1 − x) or by noting its recurrent structure in G s (x): G s (x) = 1 + xG s (x) ⇒ G s (x) =

1 . 1−x

(A.86)

This result seems to suggest that G s (x) = G(x), but in Eq. (A.85) they are obviously different. We shall explain in the following that appearance has not deceived us, and that the complete solution of Eq. (A.86) is really the function G(x) and not just the infinite series G s (x). The infinite series G s (x) turns out to be well behaved over a more restricted range of x than G(x). Equivalently, G is actually a generalization of G s , although both satisfy the same algebraic equation (A.86) derived from G s . The partial sum G N (x) of the first N terms of G s (x) is always finite since each term an is finite. For x < 1, the geometric series can be shown to converge to the sum 1/(1 − x) by showing that the partial sum G N (x) converges to 1/(1 − x) in the limit N → ∞. Problem A.6.2 gives more details on the demonstration. At x = 1, both the infinite series G s (1) = 1 + 1 + 1 + . . . and G(x) = 1/(1 − x) = 1/0 are infinite. For x > 1, G(x) is always finite. The infinite series G s (x), on the other hand, is clearly infinite, since it is greater than G s (1) term-by-term after the common first term. For all x  1, the function 1/(1 − x) is not only finite, but differentiable any finite number of times.

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Differentiable functions are called analytic functions when the variable x is generalized to a complex variable z = x + iy in the complex plane. We need not worry about such complicated functions in this tutorial. The procedure of replacing a series where it is divergent and therefore defective in some sense, by a betterbehaved function, here G(x) = 1/(1 − x), that agrees with the series where the series is convergent, is called the analytical continuation of the series. It is sometimes referred to as the “summation of a divergent series”. In reality, we end up with something better than the original infinite series. This is what happens in Eq. (A.86) after G s is replaced by G. The formula for the remainder term RN (x) in 1 = G N (x) + RN (x), 1−x ∞  xN RN (x) = xn = , 1−x n=N

(A.87)

is useful in determining the convergence of the geometric series and many other series related to it. See Problem A.6.2.  (3) Riemann zeta function ζ(p) = 1 +

1 1 1 + + + ... : 2p 3p 4p

(A.88)

The series for p = 1, ζ(1) = 1 + 1/2 + 1/3 + . . . , is called the harmonic series. It is a divergent series because the ratio an+1 n lim = lim = 1. (A.89) n→∞ an n→∞ n + 1 So for large n, the harmonic series is proportional to the geometric series with x = 1, and is therefore divergent. Another way to see this divergence is to re-group the terms so that each group has a known lower bound. Taking groups of 2m terms, with integer m = 1, 2, . . ., we have 1 1 1 + > , 3 4 2

8  1 n=5

1 > , n 2

...

(A.90)

The rearranged series is thus divergent for being greater than 1/2 + 1/2 + . . . = ∞. So for an infinite series of positive terms, the fact that an → 0 as n → ∞ confers no protection against divergence. If an tends to a finite value for large n, the series is obviously divergent. For p < 1, ζ(p) > ζ(1), and is also divergent by comparison. For p > 1, the series is convergent. Its convergence is easily shown by the integral test, described in the next section.   (4) Types of convergence:  A series n an (x) is said to be absolutely convergent if the series of absolute values n |an (x)| is convergent. If it is convergent, but not

Infinite series

653

absolutely convergent, it is semi-convergent, or conditionally convergent. It is pointwise convergent if it converges for each point x for which an (x) is defined, x being an extra parameter or variable in an (x).  The following sections are more advanced. They are included here for the interested reader and for future reference. A.6.1

Convergence tests

An infinite series may be divergent because it does not converge to a definite value, or because it is actually endless and unbounded. The simplest tests for divergence include the following: (a) Integral test: Suppose the series term an is a positive and monotonically decreasing function of n. Let S N be the partial sum of the first N terms and RN+1 = S − S N =

∞ 

an

(A.91)

n=N+1

the remainder of the series. Let B± be the bounds of RN+1 : B− < RN+1 < B+ .

(A.92)

 The series an is convergent if these bounds are finite. To find these bounds, we first change n into the continuous variable ν. Then an becomes the continuous function a(ν) that passes through the series terms as ν increases through the counting numbers n = 0, 1, 2, . . .. As a result, the following inequalities hold for the integrals over two neighboring unit intervals:  an+1 < n

n+1

 a(ν)dν < an <

n

a(ν)dν.

Using the last two inequalities and summing the terms in RN+1 , we find  ∞  ∞ B− = a(ν)dν < RN+1 < B+ = a(ν)dν. N+1

(A.93)

n−1

(A.94)

N

When the infinite series is unbounded, both bounds B± are infinite. The infinity comes from the upper limit of integration as ν → ∞. It is present for any choice of the lower limit N. A convergent series must give finite bounds. The actual values of the bounds are of interest only if the remainder RN+1 is to be estimated. A rough estimate is  ∞ 1 a(ν)dν. (A.95) RN+1 ≈ (B− + B+ ) ≈ 2 N+1/2

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Table A.4 The Riemann zeta function ζ(2) = π2 /6 = 1.644934067 . . . evaluated as a partial sum S N plus an approximate remainder RN+1 ≈ 1/(N + 1/2) for N = 0 to 5.

N 0 1 2 3 4 5

SN

RN+1

0 1 1.25 1.3611 1.42361 1.463611

2 0.666667 0.4 0.2857 0.22222 0.181818

S N + RN+1 2 1.667 1.650 1.6468 1.64583 1.645429

Error 0.36 0.022 0.0052 0.0019 0.00090 0.00049

For the Riemann zeta function ζ(p > 1), the remainder is  ∞ dν RN+1 ≈ p N+1/2 ν =

(N + 1/2)−(p−1) , for p > 1. p−1

(A.96)

The result is finite, showing that the infinite series is convergent. For the harmonic series ζ(p = 1), the same integral test gives a logarithmic divergence at ν = ∞. So the harmonic series in divergent, as we have already determined by comparison with the geometric series. The details of this integral test are left to an exercise. It is also of interest to see how well the approximate remainder RN+1 of Eq. (A.96) works out quantitatively. Table A.4 shows the results for ζ(2) evaluated at N = 0 to 5. The remainders are indeed overestimated by Eq. (A.96). Even so, the rough estimate still takes care of 99.7% of the actual remainder for N = 5. Another kind of integral test can be made when RN itself can be written as an integral. Then the series is convergent if |RN | is less than a simpler finite integral. The series is divergent if |RN | is greater than a simpler infinite integral. See Problem A.6.5.  (b) Comparison test: If the  series cn is absolutely convergent,  and |an /cn | tends converges absolutely. If |dn | diverges, and to a constant C as n → ∞, then an also if |an /dn | tends to C > 0 as n → ∞, then |an | also diverges. We have used this test previously in connection with the Riemann zeta function. For another example, consider the series of terms an = 1/(2n − n2 ) from n = 5 to ∞. We begin by noting that n2 /2n tends to 0 as n → ∞. (This limit is more obvious if the numerator and denominator are first replaced by their logs. A more definitive demonstration makes use of l’Hospital’s rule which states that when lim x→∞ x2 /2 x is not obvious, one should differentiate both numerator and denominator the same number of times until their new ratio has an obvious value.) So the comparison series can be taken to be the geometric series with cn = 1/2n . It can now be seen that an /cn tends to 1 as n → ∞. Hence the series converges like the geometric series.

Infinite series

655

The comparison test  can be used to establish the uniform convergence of a function series F(x) = fn (x). Uniform convergence means the convergence of the function series for all values of x in a specified interval I. Consider the series  sin (nx)/n2 . Compare it with the convergent series of terms cn = 1/n2 that is the Riemann zeta function ζ(2). Clearly | fn (x)/cn | ≤ 1 for all x. This establishes the uniform convergence of F(x) for all values of x. The comparison series used in this context is independent of x. It is called a convergent majorant for F(x).  (c) Ratio test: If the ratio |an+1  /an | tends to a constant C < 1 as n → ∞, then an converges absolutely. If C > 1, |an | diverges. If C = 1, the test is indeterminate. We have used this test previously in connection with the geometric series. For another example, consider the exponential function  xn x2 + .. = . 2! n! n=0 ∞

ex = 1 + x +

(A.97)

The ratio |an /an−1 | = |x|/n vanishes as n → ∞ for all finite values of x. Hence  the exponential function is absolutely convergent for all finite x. The base e = 1/n! of the exponent function is often used as a convergent majorant for function series because of its convenient factor 1/n! in the nth term of its series expansion.  1 for all sufficiently large n, then an (d) Root test: If the ratio of |an |1/n ≤ C <  converges absolutely. If |an |1/n ≥ 1, the series |an | diverges. This test may be related to the ratio test.  (e) Alternating series test: The series (−1)n an is alternating if an ≥ 0. If an is monotonically decreasing and decreases to 0 as n → ∞, the alternating series converges. These conditions are sufficient but not necessary for convergence. We have left out a number of more sensitive, but also more complicated, tests. A.6.2

Asymptotic series

 Asymptotic series are infinite function series of the form S (x) = n fn (x) that is usually divergent, but still useful for formal and numerical applications when used properly. For a brief introduction to this fascinating subject, let us describe some basic properties of a divergent series first studied by Euler in 1754: y(x) ≡ 1 − 1!x + 2!x2 − 3!x3 + . . . x ∞  = (−1)n n!xn .

S (x) =

(A.98)

n=0

The ratio test gives |an /an−1 | = n|x|. This ratio cannot be reduced to below 1 as n → ∞ because no matter how small |x| is, there are terms beyond which n|x| > 1. So the series must eventually diverge. The source of this difficulty is the n! factor that is trouble when it appears in the numerator. The situation is just the opposite of that in

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the convergent series expansion of the exponential function where an n! appearing in the denominator confers protection against divergence for any finite x no matter how large. Euler finds a differential equation (DE) that the function y(x) considered as an unknown function will satisfy: y (x) = 1! − 2!x + 3!x2 − . . . =

x − y(x) , x2

x2 y (x) + y(x) = x.

or (A.99)

A DE for y(x) is an equation containing derivatives of y(x), here y (x) = dy/dx. Eq. (A.99) is a first-order DE because it involves no derivatives higher than the first derivative. It is a linear DE because it contains no powers of y higher than the first power. It is an inhomogeneous DE because it contains a term independent of y. The solution y(x) of such a DE can be expressed as an integral of certain function of x. The theory of such DEs is given in Chapter 5. We shall concentrate here on extracting a solution using a method described there in more detail. An inhomogeneous linear first-order DE can be solved in two steps. The homogeneous DE, without the inhomogeneous term, is first solved. This is done by rearranging the equation so that one side is a function only of y and the other side is a function only of x. Hence the two sides can be integrated independently to give an intermediate result denoted yh (x) where ‘h’ stands for homogeneous: yh (x) = −

yh (x) , x2

  1 dx dyh = d(ln yh ) = − 2 = d yh x x ⇒

yh (x) = e1/x .

(A.100)

In the second step, a solution of the full DE of the form y(x) = c(x)yh (x) is looked for because this is known to work. Direct substitution shows that dc(x) 1/x e−1/x = = , to give dx yh (x) x  x −1/t e dt y(x) = e1/x t 0  ∞ −u e du. = e1/x 1/x u

(A.101)

The lower limit of the t integration has been so chosen that y(x) → 0 as x → 0, a result that is perhaps not obvious without using l’Hospital’s rule for further clarification. The u = 1/t integral form of the solution y(x) given in Eq. (A.101) shows that it is well-behaved for all values of x, unlike the original divergent series. This u integral is

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now called the exponential integral and denoted E1 (1/x). That is, y(x) = e1/x E1 (1/x). So y(x) is the sum, generalization or analytic continuation of the divergent series into a well-defined function. The situation is qualitatively similar to the analytic continuation of the geometric series into the more general function G(x) = 1/(1 − x) by solving an algebraic equation. Here a DE is solved instead. There is actually a real quantitative connection between the two analytic continuations. To display their relationship, we make another change of variable to v = u − 1/x to give  ∞ −v e y(x) = x dv 0 1 + xv  ∞ e−v [1 − xv + (xv)2 − (xv)3 + . . .]dv. (A.102) =x 0

In the last step, we have expanded the “geometric” function 1/(1 + xv) into a geometric series. We then find an infinite series on evaluating the v integrals by using the method of parametric differentiation given by Eq. (A.44). The result is just the divergent series shown in Eq. (A.98). So the divergence of the series comes from th divergence of the geometric expansion in the integrand, not from any pathological behavior of the original integral itself.  Going back to a general divergent infinite function series S (x) = fn (x), the focus of our present discussion, it is clearly counter productive to concentrate on its defective divergence as n → ∞. Indeed this divergence should be avoided at all costs. This is done by considering another limit called the asymptotic limit. The series is first separated into a partial sum S N (x) and the remainder RN (x) associated with it: S N (x) =

N−1 

fn (x),

n=0

RN (x) = S (x) − S N (x) =

∞ 

fn (x).

(A.103)

n=N

Keeping N fixed, we look for a limit point x = x0 such that   RN (x)    → 0. lim  x→x0  fN−1 (x) 

(A.104)

That is, the remainder is nothing compared to the last term kept in S N (x). Under the circumstances, we write S (x) ∼ S N (x), or ∞  S (x) ∼ fn (x) n=0

(x → x0 ).

(A.105)

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These equations are read as “S (x) is asymptotic to S N (x)” and “S (x) is an asymptotic series or expansion about x0 ”, respectively. For the special case where fn (x) = an x−n , the limiting property of an asymptotic series can be stated equivalently in a form more familiar to physicists S (x) ∼ S N (x) + O(x−N ),

(x → x0 ),

(A.106)

or “S (x) is asymptotic to S N (x) to order xN near x0 ”. The order (or Big O) expression describes the x dependence of the dominant term in RN (x). For Euler’s series for example, the ratio of successive terms is | fn (x)/ fn−1 (x)| = n|x|. This requires that  1/N for the use of S N . The safest choice is to take just one term S 1 (x) = 1, but this choice is not very useful because its does not have any x dependence. The optimal number of terms N to use in an application is the subject of a branch of study called superasymptotics. Euler’s series is perhaps too simple in structure to be of general interest in physics. In physics, one often has to deal with large numbers N such as the Avogadro number 6.02 × 1023 , the number of molecules per gram mole of materials. In statistical mechanics for example, one has to count the number of states populated by a collection of N molecules in terms of the number of permutations N! of N distinct objects. N ln n, this is not a very workable expression in a complicated Although ln (N!) = n=1 formula. Intuitively, one might try N ln N, equivalent to N! ≈ N N , but this is clearly a vast overestimate, even though it describes the behavior of N! to the leading order. Fortunately, Stirling had already found in 1730 a divergent asymptotic series for ln (x!) valid as x → ∞:   ∞  √ 1 a2k ln (x!) ∼ x + ln x − x + ln 2π + . (A.107) 2 x2k−1 k=1 It is useful to have the first few terms of the Stirling series for x! itself for large x    x x √ 1 139 1 2πx 1 + − − ... , (A.108) + x! ∼ e 12x 288x2 51840x3 where the first term (or S 1 ) is now called Stirling’s approximation. Note that the term indices of the asymptotic expansion have been changed from n, N to k, K to avoid conflict with the integers n, N used in factorials. How good are these first few partial sums for x as large as x = 1 where the expression can be calculated in a hand calculator. The exact result is just 1! = 1. The first four partial sums are are S 1 , S 2 , S 3 , S 4 = 0.92214, 0.99898, 1.00218, 0.99971.

(A.109)

We see that S 3 overshoots the actual value. So the next term is negative. From then on, it would be necessary to take the terms two at a time to smooth out the oscillations present in an alternating series. Soon, as more terms are added, the divergence will appear like the evil Mr Hyde emerging from the respectable Dr Jekyll.

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Other well known asymptotic series are those for the error function erfc (x) and the Riemann zeta function ζ(p): 2 erfc (x) = √ π



∞

e−t dt 2

x

⎡ ⎤ ∞  ⎢⎢⎢ ⎥⎥⎥ (2n)! e ⎥⎥⎥ (−1)n ∼ √ ⎢⎢⎢⎣1 + n!(2x)2n ⎦ πx −x2

(x → ∞),

n=1

ζ(p) ∼

N−1  n=1

n−p +

∞  N 1−p b2m (p) + N −p p−1 N 2m−1 m=1

(x → ∞).

(A.110)

The series coefficients b2m for ζ(p) and a2k for ln (x!) in Eq. (A.107) are left unspecified. The interested reader can find them in many handbooks and in the Wikipedia article “Asymptotic expansion”. The Riemann zeta function of Eq. (A.110) can be used for any argument p, even a complex one. Of course, the series is divergent if |p| ≤ 1. It has no simple integral representation in terms of elementary functions. Problems A.6.1 (Equal tripartition of areas) Verify the Archimedes geometric sum given in Eq. (A.83) as follows: (a) For a square: Subdivide a square into four identical sub-squares or quadrants. Take the upper-right quadrants and repeat the sub-square division. Repeat the subdivision process with the latest available upper-right quadrants. Show that the subdivided squares form into three identical sets. Each set contains exactly one quadrant at each scale of subdivision. What is the total area of each set of subdivided squares? (b) For an equilateral triangle: Use the same idea as the tripartition of the square described in part (a) to divide the area of an equilateral triangle into three identical sets, each set containing a subdivided equilateral triangle at each scale of subdivision. Illustrate your result by a drawing showing two or more subdivisions. Note: Archimedes described how he summed the infinite series (A.83) in his treatise Quadrature of the Parabola. The problem is to calculate the area bounded by a parabola (say y = x2 ) and a straight line that intersects the parabola at two points (such a bounding line segment being called a chord). His problem is harder to solve than the equal tripartition of squares and equilateral triangles. (c) Sum the Archimedes geometric series analytically. A.6.2 (Telescopic cancellation)

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(a) Use partial fractions to show that ∞ 

1 1 = . (2n − 1)(2n + 1) 2 n=1

(b) Show that the partial sum of the geometric series is G N (x) = (1 − xN )/(1 − x). Show that in the limit N → ∞, G N (x) exists only for x < 1. What is this limit? A.6.3 Use an integral test to show that the harmonic series ζ(1) is logarithmically divergent. A.6.4 (Factorization of sin x) (a) Show that x is a factor of sin x by writing down a convergent infinite series for sin x/x. (b) Show that sin x satisfies the periodicity properties sin (x − 2πn) = sin x, sin [x − (2n − 1)π] = − sin x, for any integer n. Show that as a result, x − nπ, any integer n, is also a factor of sin x. (c) Show that the infinite product  ∞  ' x2 P(x) = x 1− 2 2 n π n=1 containing all the zeros x = xn = nπ (integer n) of sin x is also an +N infinite series. Here a product of N factors an is denoted n=1 an = a1 · a2 · . . . · aN . Assuming that two functions with the same roots and the same derivatives at x = 0 are equal (certainly true for polynomials), find P(x) = sin x. P(x) is the Euler product (1735, 1743) for sin x. (d) Find a special case of the Euler product that gives the Wallis product (1655)      ∞  ' 1 π 2 2 4 4 6 6 1− 2 . = · · · ... = 2 1 3 3 5 5 7 n n=1 (e) Show that the coefficient of the term in P(x) of part (c) linear in x2 is   N 1  1 . p2 = lim − 2 N→∞ π n=1 n2 Compare p2 with the corresponding coefficient in sin x/x to get the sum π2  1 1 1 = 1 + 2 + 2 + ... . = 2 6 n 2 3 n=1 ∞

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661

This result was obtained by Euler in 1735 in the same paper as the Euler product. A.6.5 (Almost Stirling approximation) (a) Show that ln n and ln (N!) satisfy the inequalities:  n+1  n ln νdν < ln n < ln νdν; ln (n − 1) < n−1

 B− = (b) Verify the results:  m

N 1

n



ln νdν < ln (N!) < B+ =

N+1

ln νdν.

2

ln νdν = m ln m − n ln n − (m − n),

n

1 ln (N!) ≈ (B− + B+ ) 2     3 1 − ln 2 . = N + ln N − N + 2 2 Hint: Integrate the first integral by parts. Note: The rough estimate ln (N!) ≈ (B− + B+ )/2 gives the same Ndependent terms appearing in the Stirling approximation. The very small N-independent term √3/2 − ln 2 = 0.807 is less than the corresponding Stirling term ln 2π = 0.919. To derive the Stirling term √ ln 2π, it will be necessary to actually evaluate the integral form of N! = Γ(N + 1) to a sufficient degree of accuracy. This is done in Example 8.15.2 in Chapter 8. A.6.6∗ (Integral representation) Use the given integral representation, the geometric series, and term-by-term integration to find the given infinite series. Use the remainder formula (A.87) to obtain an integral representation of the remainder RN (x) to establish the given convergence bounds. (a) Arctangent: Gregory’s series (1671) for  x dt x 3 x5 = x − + − . . . , |x| ≤ 1. arctan x = 2 3 5 0 1+t Use this result to get Leibniz’s formula (1674) for π: π 1 1 1 = 1 − + − + ... . 4 3 5 7 Hint: arctan (−|x|) = − arctan |x|. So the convergence property has to be established only for x ≥ 0.

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Note: Leibniz’s formula is quite useless for a fast numerical computation of π. It would take some 100,000 terms to reach the accuracy of 22/7 obtained by Archimedes. (b) Newton’s series (ca. 1665) for 

x

ln (1 + x) = 0

dt x2 x3 = x− + − ..., 1+t 2 3

−1 < x ≤ 1.

Hints: (i) An integral test may work for −1 < x ≤ 1. (ii) Use a comparison test for x ≤ −1.

A.7

Exponential functions

The exponential function is the functional inverse of the logarithmic function. The logarithmic function is introduced to reduce a product to a sum: ln (ab) = ln a + ln b. So the exponential function of a sum is a product of their exponential functions: exp (a + b) = (exp a)(exp b).

(A.111)

The exponential function, when used with the “common” base 10, is called an antilogarithm, and denoted antilog x or log−1 x. The exponential function can be introduced independently by way of the compound interest. A principal of $1 earning interest at the annual rate of 100% will yield a total of $2 after one year at simple interest. Compounded monthly, the total is (1 + 1/12)12 = 2.613. The total is 2.715 if compounded daily, and n  1 e = lim 1 + = 2.7182818284 . . . , n→∞ n

(A.112)

if compounded continuously, a result first obtained by Jacob Bernoulli in 1683. The symbol e was first used by Euler in 1728 (when he was only 21) in a private manuscript that was only published posthumously. Two definitions of the exponential function still in use were first given by Euler. The first definition uses a generalization of the limiting process for e to give  nx 1 e = lim 1 + n→∞ n  p x = lim 1 + . p→∞ p x

(A.113)

Used with Newton’s generalized binomial theorem for an arbitrary real power p, the second expression gives

Exponential functions

   p  2 x x p(p − 1) x =1+ p + ... 1+ + p p 2! p  k p(p − 1) . . . (p − k + 1) x + ..., + k! p  x  n(n − 1/x)  x 2 + ... + ≈1+n n 2! n n(n − 1/x) . . . [n − (k − 1)/x]  x k + + ... . k! n

663

(A.114)

The limit p → ∞ for fixed x is the same as the limit n → ∞. One then finds Euler’s two definitions for e x as the limit of a single power and as an infinite series, as follows:  x n e x = lim 1 + (A.115) n→∞ n ∞ k  x2 x3 x4 x =1+x+ + + + ... = . (A.116) 2! 3! 4! k! k=0 The infinite series for e x used with the purely imaginary variable x = iθ gives the Euler formula   θ2 θ4 θ3 iθ + + ... + i θ − + ... e =1− 2! 4! 3! = cos θ + i sin θ. (A.117) √ The notation i = −1 was introduced by Euler, while the infinite series for the sine and cosine were known since ca. 1670. A special case of this formula eiπ + 1 = 0

(A.118)

was considered by Feynman “the most remarkable formula in mathematics”. This short equation contains five fundamental numbers (1, 0, π, e, i) and four basic algebraic operations (+, =, ×, exponentiation). The first systematic treatment of the trigonometric functions was given by Euler in a prize paper in 1748. The exponential function e x itself can also be separated into even and odd parts in x: e x = cosh x + sinh x, where 1 1 cosh x = (e x + e−x ), sinh x = (e x − e−x ). 2 2

(A.119)

The first comprehensive account of the hyperbolic functions was given by Lambert in 1768.

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A term-by-term differentiation of the infinite series (A.116) for e x gives d x e = ex . dx Hence



x

et dt = e x − 1.

(A.120)

(A.121)

0

A.7.1

Differential equations

The function f (x) = eax satisfies the first-order linear differential equation (DE) d f (x) − a f (x) = 0. dx

(A.122)

If a < 0, eax = e−|a|x decreases monotonically and “exponentially” with increasing x, with a decay constant |a|. This means that after a distance of X = 1/|a|, f (x) has decreased to f (X) = e−1 ≈ 0.368 of its value f (0) = 1 at x = 0. If a > 0, f (x) = eax increases monotonically and “exponentially” with f (X = 1/a) = e ≈ 2.72, just like the growth of a principal amount of money in one year at an annual interest rate of 100% when compounded continuously if x is a time variable. The DE (A.122) states that exponential growth/decay is continuous growth/decay, with the rate of change f  proportional to the instantaneous or present value f (x). Consider the DE dn d n−1 f (x) + b f (x) + . . . + b1 f (x) = 0. n−1 dxn dxn−1

(A.123)

It is called an nth order DE because the highest derivative that appears is the nth derivative. It is linear because every term depends at most linearly on f (x). It is homogeneous because every term depends on f (x). If the coefficients b1 , . . . , bn−1 are constants independent of x, its solutions are all of the form f (x) = eax because differentiations with respect to x do not change the x-dependence of eax . This is perhaps the most important property of exponential functions, a property that makes them the most important class of functions in science. The differential operators dn /dxn used in a Taylor expansion allow the functional value f (x ) at a distant point x to be generated when its derivatives f (n) (x) = dn f (x)/dxn are known at another point x. Since all the derivatives of eax are proportional to the function itself, exponential functions are very useful for describing physical phenomena occurring at different points in space. This point of view is taken up in more details in Chapter 2 of the text. In Chapter 4 on Fourier series and integrals, the use of exponential functions to describe arbitrary functions in science and mathematics is described. Let us return to the solution of the DE (A.122). We need to find the permissible values of the parameter a in the exponential function eax that satisfy the DE. This can

Exponential functions

665

be done by simply substituting the assumed form of the solution into the DE to give the algebraic equation (an + bn−1 an−1 + . . . + b1 ) f (x) = 0.

(A.124)

Since f (x)  0, the x-independent coefficient factor has to vanish. It is convenient to write the resulting algebraic equation for the coefficient factor in the factored form (a − a1 )(a − a2 ) . . . (a − an ) = 0.

(A.125)

The expression shows that there are exactly n solutions for a, a = ak , k = 1, 2, . . . , n. They are not necessarily all distinct. Let us suppose for simplicity that the ak s are all distinct. There are then n distinct solutions fk (x) = ck eak x each of which satisfies the original DE if ck is a constant. A sum of any two such  solutions with different ak will also satisfy the DE. So is a sum of all n solutions nk=1 ck eak x . Such a sum is called a linear superposition of solutions. Chapter 5 on DEs gives more details on how the unknown coefficients ck can be determined. The fact that linear systems described by linear DEs such as Eq. (A.122) have solutions that are linear superposition of a small number of possible solutions means that the physical phenomena they describe show a surprising degree of regularity and predictability. Predictability is the hallmark of science. No other mathematical function has a structure as simple, transparent and useful as the exponential functions. A.7.2

Transcendental functions

A transcendental function is a function that “transcends” algebra. That is, it cannot be expressed in terms of a finite number of algebraic operations (addition, multiplication, and roots). The elementary transcendental functions are the logarithmic, exponential, circular, inverse circular, hyperbolic and inverse hyberbolic functions. They are all related to the exponential function. The exponential function itself appears in the integral definition of many higher (or advanced) transcendental functions: the gamma (or factorial) function Γ(z) of Eq. (A.45), the exponential integral E1 (z) in Eq. (A.101), the error function erfc (z) of Eq. (A.110), the Bessel functions Jn (x) of Section 7.7, the Hermite polynomials Hn (x), and the Laguerre polynomials Ln (x). Here n if present is the order of the function used to denote different functions of each family. If present, n is either any integer or any nonnegative integer. The exponential function also appears in many generating functions. A generating function of a higher transcendental function Fn (x) of order n is an infinite series in tn of the form ∞  bn F n (x)tn , (A.126) G(x, t) = n=m

where m = 0 or −∞, and bn is an n-dependent constant such as n!. Examples of generating functions are given in Table A.5. The last entry of the table gives the

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Name

Fn (x)

Bessel function Hermite polyn. Laguerre polyn.

Jn (x) Hn (x) Ln (x)

Bernoulli #

Bn

G(x, t)

bn

m

exp [(t − 1/t)x/2] exp (2xt − t2 ) exp [xt/(x − 1)]/(1 − t)

1 1/n! 1

−∞ 0 0

t/(et − 1)

1/n!

0

generating function of a set of numbers called Bernoulli numbers that appear in many functions. So their generating function depends only on t. A generating function G(x, t) is useful because it can generate the function Fn (x) it contains. If G contains no negative powers of t, the function Fn (x) can be extracted by differentiations:  1 dn Fn (x) = G(x, t) . (A.127) bn n! dtn t=0 In other cases, a power series expansion often works. A.7.3

Logarithmic function

The word “logarithm” is a combination of two Greek words logos and arithmos meaning ratio-number. It refers to the number ln x to which a number called its base, say the constant e of Eq. (A.112), must be raised in order to produce a given number x: x = eln x .

(A.128)

The resulting function ln x is called the natural logarithm. It was discovered and named in 1614 by Napier. At around 1615, Briggs proposed to Napier to change the base from e to 10 to give the common logarithm log x. The idea behind the logarithmic function originates in the fact that the powers or exponents of terms in the geometric progression 1, r, r2 , . . . form the arithmetic progression 0, 1, 2, . . .. This is why Napier called the power a ratio-number. The quotient of two terms of the former is the difference of the corresponding two terms of the latter. This property was already noted by Chuquet in 1484, and again in 1544 by Stifel who extended the idea to negative and fractional powers. In modern notation, it is expressed by the basic logarithmic relation loga (bc) = loga b + loga c, where a is the base of loga . The logarithmic function was originally invented to simplify astronomical calculations. With its help, multiplications of long numbers had become as easy as additions plus table look ups. The slide rule that performs multiplication or division by adding or subtracting two segments on two adjacent sliding rulers were first

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developed in the early 1620s. It remained in use by engineers and scientists until 1974 when electronic calculators became available at affordable prices. The linearization of powers caused by the logarithmic function also lies at the heart of its importance in science. The use of the logarithmic scale to cover an extended range of values is so commonplace now that one seldom pauses to reflect on its elegance. The Richter scale of earthquake energy, the apparent magnitude of star brightness, the pH value of ionic solutions, and the decibel unit of the loudness of sound are all based on the logarithmic scale. Problems A.7.1 At an annual interest rate of 1/4%, the rate charged by many central banks during the Great Recession of 2008, how many years will it take a principal of $1 to grow to $e if the interest is compounded continuously? How many generations will this take, if each generation spans 20 years? A.7.2 A radioactive atomic nucleus decays randomly at a rate λ so that a number N(t) of these nuclei decreases in time as dN(t)/dt = −λN(t). Obtain the following results: (a) The solution of this first-order differential equation is N(t) = N0 e−λt , where N0 is the initial number at time t = 0. (b) The mean lifetime of these decaying nuclei is ∞ tN(t)dt 1 = . τ ≡ t = 0 ∞ λ N(t)dt 0

(c) The halflife t1/2 is the time it takes a population N(t) to decrease to N0 /2, half of its original value N0 . Show that t1/2 = (ln 2)τ = 0.693τ. Explain why t1/2 < τ. (d) If the nucleus has two independent decay modes of different decay rates or constants λ1 and λ2 , show that the nuclear population decays with the following characteristics: λ = λ1 + λ2 and τ = τ1 τ2 /(τ1 + τ2 ). A.7.3 (Second-order linear DE) In classical dynamics, one is interested in the position of a system at time t. Let its position in one spatial dimension be x(t). An object of constant mass m satisfies Newton’s force law in the form F = ma(t) = md 2 x(t)/dt2 , where a(t) is its acceleration. For small displacements from equilibrium, Hooke’s law states that the restoring force F = −kx(t) is proportional to the displacement x(t) from equilibrium. A pendulum displaced from the vertical equilibrium position will thus execute an oscillatory motion when released. A grandfather clock started this way will not oscillate forever, however, because of friction at the point of support. A fully winded grandfather clock will come to a stop in around 8 days. It is designed to be rewinded once a week. The frictional term is proportional to the velocity −bdx(t)/dt. Hence one ends up with a secondorder linear DE in x(t) of the form

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m

d2 d x(t) + b x(t) + kx(t) = 0. 2 dt dt

(A.129)

(a) Eq. (A.129) is a second-order linear DE with constant coefficients. There are solutions of the form eiωt , where the unknown constant ω can be determined by direct substitution into the DE. Show that the substitution gives a quadratic algebraic equation mω2 − ibω − k = 0.

(A.130)

Show that the quadratic equation can be written in the form m(ω − ω1 )(ω − ω2 ) = 0, where  & 1  2 ω1,2 = ib ± 4mk − b . 2m are its two roots. (b) Each of the two functions eiω1,2 t by construction satisfies the original DE (A.129). Show that the linear superposition x(t) = c1 eiω1 t + c2 eiω2 t for any coefficients c1 , c2 also satisfies it. If the initial conditions of motion are x(t = 0) = x0 , dx(t = 0)/dt = v0 , show that the two coefficients ci satisfy two equations that can be written as the matrix equation      c1 x 1 1 = 0 . iω1 iω2 c2 v0 Find the condition under which this matrix equation has a unique solution. Show that the resulting unique solution is     1 c1 iω2 x0 − v0 = . c2 i(ω2 − ω1 ) −iω1 x0 + v0 √

A.7.4 The two solutions e(−β±i 4mk−b /2m)t of Eq. (A.129) have a common time envelope e−βt , where β = b/2m. This suggests a slightly different approach to the solution that is useful for more complicated DEs. (a) Under time reversal (t → −t), the position x(−t) satisfies a time-reversed DE that differs from Eq. (A.129) in the sign of the friction term. Derive this time reversed DE. (b) Let x(t) = e−βt y(t). Show that y(t) satisfies the simple second-order linear DE y¨ (t) + ω21 y(t) = 0, where y¨ = (d 2 /dt2 )y(t) is Newton’s overdot notation for time derivatives. This is a more convenient notation when differentiating a product of time functions such as e−βt y(t). Show that the DE for y(t) is invariant under time reversal, and that if y(t) is a solution, so is y(−t). Show that ω21 = (k/m) − β2 , and that the DE has two distinct solutions y(t) = e±iω1 t if k/m  β2 . 2

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A.7.5 Show by long division by an infinite series. (a) (b)

sin x x3 2x5 = x+ + 15 + . . . . cos x 3 2 t t t =1− + − ... t e −1 2! 2 · 3!  tn B . = ∞ n n=0 n!

tan x =

Thence find the first three Bernoulli numbers B0 = 1, B1 = −1/2, B2 = 1/6. Note: The Wikipedia article “Long division” describes the method. A.7.6 (ln x) (a) Show that the functional inverse of the Euler limit expression (A.115) for e x is " # ln x = lim n x1/n − 1 . n→∞

(b) Use this definition of ln x with the binomial theorem to find the infinite series ln (1 + x) = x −

x2 x3 + − ... . 2 3

(c) The actual expression Napier used in 1614 is 1/e ≈ (1 − 10−7 )10 . Use a ten-digit scientific calculator to find the numerical value 1/e and Napier’s approximate value. Find the fractional error in his value. Find a mathematical expression for his error. Use it to determine if your calculator has the claimed 10-digit accuracy. A.7.7 loga x is defined by the expression aloga (x) = x. Derive the following results: (a) loga (b x ) = x loga (b). (b) [loga (b)][logb (a)] = 1. (c) [loga (b)][logb (x)] = loga (x). 7

Appendix B Mathematica and other computer algebra systems Computer algebra (CA) refers to the use of a computer for symbolic manipulations of mathematical expressions. The basic algebraic operations include simplification, symbol substitution, expansion in products and powers, factorization, partial fraction and equation solving. In CA systems, other capabilities are added including numeric mathematics and many of the following features: • symbolic and numerical differentiation, integration and solutions of differential

and difference equations in real and complex variables • operations on complex numbers, series, matrices, strings, data arrays and • • • • • • • • • •

sometimes tensors arbitrary-precision and integer arithmetics symbolic and numeric manipulations of special functions symbolic and numeric optimization plot and draw figures, diagrams and charts edit equations and manipulate strings data analysis and mining animation and sound, image processing and recognition provide a programming language to give integrated solutions to complex projects import and export resources and link to external programs automatic theorem proving

There are about sixty readily available CA systems of which about forty are free to download and use. The Wikipedia entry http://en.wikipedia.org/wiki/Comparison of computer algebra systems gives tables comparing them as to cost, latest version, history, capabilities and computer operating systems supported. Even hand-held CA-capable calculators have been available since 1998. This Wiki entry is an indispensable starting point for a beginning user of CA in search of a CA system to use. For more information, search the Web for “computer algebra systems”. Two popular commercial CA systems are Mathematica and Maple. Both have been developed and expanded extensively since the 1980s. In this Appendix, we shall use Mathematica 8 to illustrate how CA works at an introductory level. We give specific actionable instructions so that readers with the software can reproduce the

Mathematica and other computer algebra systems

671

stated results. There are some additional comments on error messages produced by older versions for readers using them. The readers reading the book without PCs can pass over these parts and assume that the stated results have actually been obtained from the software. Wolfram Research, the maker of Mathematica, has extensive online help for beginners. The most useful web site is likely to be the Learning Center for higher education located at http://www.wolfram.com/support/learn/higher-education.html. One finds here video tours, online tutorials of interest to beginners. Mathematica 8 is the latest version that comes with unusually helpful user interfaces and excellent online documentation. IT-literate users can probably begin to try out the program after a quick-start description. This rest of this Appendix gives a brief overview of CA systems in general and Mma (our nickname for Mathematica) in particular. Mathematica 8 has an very helpful interface called “free-form linguistic input” that understands plain English. Its use is illustrated in the following example. Example B.1 (Plotting a function) In this example, user action is described by a capitalized and parenthesized word or phrase: (Click) = left clicking a ‘choice’ in a toolbar or a dropdown menu. (Activate cell) = click while the cursor is in the cell body or at its right border line. (Evaluate cell) = either (a) (Shift+Enter), or (b) (Click) ‘Evaluation’ → ‘Evaluate cells’. A dialog between user and Mma can then be written out as in a play with the Mma response following the identifier Mma for the response from Mma: 1. (Click) ‘Insert’ → ‘Inline Free-form Input’. 2. (Type) plot pi sinx/x. 3. Mma: Plot[Pi∗Sin[x]/x, { x, -9.4,9.4} ] 4. (Activate cell) 5. (Evaluate cell) 6. Mma: (returns a graph of the oscillatory function π sin x/x from x = −9.4 to x = 9.4).  We see that in Line 3, the free-form interface translates an ordinary phrase into a grammatically correct Mma command. Here is a short summary of Mma grammar: • All Mma built-in objects (commands or specifications) are capitalized English

words, strings of words or letters representing standard abbreviations. They include actions: Plot, Integrate, Derivative (or D, Dt), Expand, Simplify, Show, Eigensystem, . . . functions: Sin, Cos, Tanh, Re, Im, . . . constants: Pi, E, I, Infinity, Degree, . . . for π, e, i, ∞,◦ , . . . , respectively. specifications: Integer, Real, Complex, Symbol, Condition,. . .

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Mathematica and other computer algebra systems

Any font can be used for the user input typed in the Mma window, but in the text part of this Appendix outside the simulated user ↔ Mma dialog, Mma phrases are typeset in Italics and Mma commands are in typewriter font for greater readability. • Variables can appear in a mathematical expression, such as x in Line 3. A variable named can be capitalized, or not. For example, X can be used instead of x in Example B.1. However, to avoid conflict with Capitalized objects, the convention is to use only uncapitalized names. There is no limit how long the name can be, but a shorter name is easier to use. The name can end with a number treated as a subscript but it cannot start with a number, because a starting number is treated as a multiplicand unless it is 0 (zero). Then the whole name is considered to be the integer 0. Thus 0x = 0, 056x = 56∗x, 5.6x = 5.6∗x. The multiplication sign used in the Mma window is ∗ in the text mode and the usual × symbol in the typeset mode. It can thus be omitted if there is no ambiguity, but using it or leaving a blank space gives the expression greater clarity. • The phrase Pi∗Sin[x]/x in the argument (the part . . . enclosed by the square brackets [. . .]) of Plot in Line 3 is all symbolic. Assignment of numerical values to x in order to plot the resulting numerical function from x = −9.4 to 9.4 is made by the last part of the argument of the object by using a range assignment, {x, −9.4, 9.4}. The plot is made using default values of the many options needed to define the plotting. The user can override these default values by adding optional commands to the argument of Plot. For example, the default printing often chops off the top or bottom of the generated curve in the range of variable under consideration in many older versions of Mma. If one wants the complete curve, one will then have to add the option PlotRange → {ymin, ymax}, where ymin, ymax are numerical values chosen to ensure that everything of interest within the y-range is shown. The grammar and punctuation must be rigidly followed before the computer would perform the requested action. • If you already know the name of the command you want to use, you can type it in the Mma window and ask for its function template by clicking on ‘Edit’ → ‘Make Template’. Example B.2 (Assigning value to a variable) In Example B.1, Mma is used in the single-command or calculator mode. The grammar is relatively simple and the result is essentially foolproof. However, if one is writing a long program of many commands using the same functions in different places to solve a typical problem in science or engineering, the situation becomes complicated. A frequent source of errors occurs on assigning values to variables because they can be symbolic or numeric depending on the circumstances. As a result, they do not always take the values we want them to have, as illustrated in the following. Suppose Line 3 of Example B.1 is just one of the many places where the function involved is used. It then makes sense to define the function separately and call it wherever it is needed, as follows:

Mathematica and other computer algebra systems

673

1. (Type) Clear[x, y, t]; 2. 3a. y[t ] = Pi∗Sin[t]/t; 4. 5. Plot[ y[x], {x, -9.4, 9.4}] The defined function y[t ] is a mathematical function in the traditional sense, but generalized to a symbolic function of the variable t that is itself a symbolic expression. Expressions are called patterns in Mma, meaning structures and qualities. Patterns of different types (in structure or quality) belong to different “classes”. Class specification is given by the Head of the pattern. Examples of heads include Integer, Real, Complex, Symbol, Condition—names that are self explanatory. The head of the pattern z is shown by the command Head[z]. The pattern variable t in the pattern function y[t ] is pronounced “t blank”, and means “any expression named t”. This means that the pattern function y[t ] can be called by a pattern variable of any class, including a purely symbolic variable. • Running this program in Mma will give the same graphical output as in Example

B.1. The purpose of the command Clear in Line 1 is to strip all numerical values from the variable x, y, t, and make them symbolic. That is, their heads are Symbol. Line 3a immediately assigns (= or Set) to y[t ] the RHS expression symbolically. The resulting function can then be used later in Line 5 to plot the curve y[x] when x takes on specified real values in the range {x, −9.4, 9.4} of Plot. The called functions y[x] are then real numbers that can be used to plot a curve. An alternative method uses instead of Line 3a the delayed function assignment denoted := or SetDelayed 3b. y[t ] := Pi∗Sin[t]/t; It differs from Line 3a by evaluating the function only when the function call is made eventually in Line 5, and not immediately at Line 3b. In the example given, both definitions work equally well, because both definitions are evaluated numerically at the function calls of y[x] in Plot. • Difficulties can arise if something happens at Line 2 to interfere with the Plot

command in Line 3. For example, suppose Line 2 read t = 2.0;. Then Line 3a immediately evaluated y[t] to a real value of around 1.4. Line 5 then plots this same numerical value at different x values of the range of Plot to give a horizontal straight line at y ≈ 1.4, instead of the intended oscillatory function. So the program executes correctly, but does not produce the same curve. Even with t = 2.0 in Line 2, the error can be avoided by using the SetDelayed function 3b. This works because y[t] is then not evaluated until it is finally called as y[x] inside the Plot command using its own real variable x in the range of Plot. So Line 2 is totally ignored in this usage.

• If Line 2 contains the expression “y = 3.0;”, Mma will return the error message

For Line 3a: Set::write : Tag Real in 3.[t ] is Protected. 

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Mathematica and other computer algebra systems

For Line 3b: SetDelayed::write : Tag Real in 3.[t ] is Protected.  Mma will draw the coordinate axes, but no curve for y[x].  is a link that opens a window labeled “General::write” giving the general system message (hence “General”) on “write”. The reason for omitting the curve is then given, namely that after Line 2, Mma has replaced y[t ] by a real expression 3.[x ] that is the product 3.0*[x ]. The real number generated by the function call cannot be used to replace 3.[x ] because they belong to different classes. 3.[x ], not being a simple real number, cannot be plotted. So the wanted curve does not appear. In the older version 4, additional error messages appear y[x] is not a machine-size real number at x = −9.4 y[x] is not a machine-size real number at x = −8.63734 y[x] is not a machine-size real number at x = −7.80559 Further output . . . will be suppressed during this calculation. These error messages describe the same problem, again in highly technical terms. These additional error messages have been eliminated in more recent versions. Note that Mathematica 8 is not 100% backward compatible with all older versions. • What happens if Line 2 is empty, but y = 3.0; is added at Line 4? The conflict

between the two y definitions remains, and y[x] remains undefined and cannot be plotted. So the Tag Real error disappears. The coordinate axes are again plotted without the curve y[x]. There is no error message from Mathematica 8, but in older versions, the y[x] errors appear before the curveless plot is returned. Both difficulties may seem obvious and trivial in this simple example, but if Line 2 is made up of hundreds of lines of codes, it is not always easy to find the error.  Mma error messages have continued to improve. Explanations of these errors remain rather technical and require a certain amount of knowledge of and experience in using Mma in the reader. It is unavoidable that a technical subject has its own technical language to be learned like a foreign language or interpreted by a translator. As a general rule, if nothing is returned by Mma when you expect something, then your program is faulty and has to be corrected. If you use Mma only in the calculator mode, you are unlikely to run into any serious problem. On the other hand, if you must write extensive programs, the best way to proceed is to build up your program from the bottom up, and not from the top down. This means that you start with the most basic calculations and check out your code step by step before you combine them into bundles called functions, procedures, subroutines or modules, then bundles of bundles, and finally the whole code for a complex project. The best attitude for a programmer to take is that any result obtained in a complex code is incorrect unless you can explicitly demonstrate that it is correct for as many cases as possible where you already know the correct answers.

Mathematica and other computer algebra systems

675

The simplification of a complex code into reusable functions and modules is called procedural programming. It produces a very flexible programming environment where data, constants and variables can either be restricted locally to one function or module or be available globally to other modules as well. However, when the code becomes very large and is worked on by many people at different times, the global quantities that can be changed by two or more modules tend to become unmanageable. Unintentional changes or corruptions of these quantities often happen accidentally, leading to unpredictably wrong results. This problem can be avoided by eliminating all global quantities. Such an approach is called functional programming because the result depends only on the function arguments and nothing else. However, it is not always convenient to eliminate all global quantities. For example, one might want to manipulate the same data in different functions. Moving huge amounts of data to and from function arguments is costly in computer time. To avoid any unintentional change in these data without moving them, one can sequester them into independent sets called objects. Each object contains its own data that are operated on by its own programs. The result is object-oriented programming. The programs and functions used in different objects can be the same or they can be different. Mma has commands that facilitate different programming styles and techniques, including procedural, functional and object-oriented programming. It has commands tailored to these and other programming styles. There are commands based on patterns, rules and recursion, and commands operating on lists. A simple list is an array of elements that are numeric or symbolic quantities, or simple objects. For example, v = {x1, x2, x3} defines a simple list of three elements. If x1, x2, x3 are the symbolic or numeric Cartesian coordinates of a 3D space, then v represents a 3D vector in this space. More generally, the list elements can be any expressions. In particular, they can be simple lists themselves. Then v becomes a nested list, or a 2D array that can be used to represent a matrix. This nesting can be repeated to define multidimensional arrays. The head of a list variable is List. Rules are substitutions of one expression by another, symbolically lhs− > rhs. The most elementary and useful substitution is ReplaceAll, also denoted “/.”, where the expression lhs is replaced by rhs no matter where it appears in the original expression expr. Another substitution Replace works only if lhs is the whole expression expr itself. The following example illustrates their usage. Example B.3 (Substitution rules) User ↔ Mma dialog: 1. (Type) ClearAll[x, y, a]; 2. ReplaceAll[x2 + y3 , x –> 2 + a] 3. x2 + y6 /. x –> 2 + a /. a –> 3 4. Replace[x2 , x2 –> y3 ] 5. Replace[x2 , x –> 2] 5a. (Evaluate cell) 6. Mma: (2 + a)2 + y3 7. 25 + y6 8. y3 9. x2

676

Mathematica and other computer algebra systems

We see in Line 9 that Line 5 has failed to execute because Mma has returned the original expression unchanged.  The reader should check the publisher’s higher-education book website http://www.oup.com/us/catalog/he/ to see if any electronic Mma files (called notebooks) giving some of the programs used in writing this book have become available online. Mathematica is a commercial software whose cost to users is least for students while they are in school or college. Once you have some CA experience, it is relatively easy to switch from one system to another. If you want to try out a free CA system first to see how you like the CA experience, you may want to try one of the many available systems such as Maxima and Reduce for Windows, Mac and Linus operating systems, and Sage for Mac and Linus. You can also ask for a free download of a trial copy of commercial software from its publisher: Wolfram Research (for Mathematica) and Maplesoft (for Maple), for example.

Appendix C Computer algebra (CA) with Mathematica C.1

Introduction to CA

CA uses a computer software such as Mathematica to manipulate mathematical expressions. It includes symbolic differentiations, integrations, series expansions, and evaluations of Fourier and Laplace transforms. This appendix shows most of the contents of the Mathematica notebook ComputerAlgebra.nb, an executable elecronic file that can be found in the book’s web page at Oxford University Press online (oup.com). In the following examples, starred parentheses enclose explanations useful to new users of Mathematica. This appendix can be used as a tutorial on Mathematica after first reading Appendix B. It can be read without using the software because most of the text output form the software is also given. It can be read casually for an overall qualitative view of CA, or attentively to learn the language of Mathematica. The graphics output is given in the electronic notebook. The Mathematica programs used to draw Figs. 6.9, 6.13 and 6.19 are also listed in this appendix. C.1.1 Derivatives Clear[u, x];

(∗ Clear = erase any values or definitions from the variables. Ending semicolon supressess output. ∗) D[ArcSinh[x], x] (∗ D[..., x] = d/dx ∗) Dt[ArcSinh[u], x] (∗ Dt = total derivative, with undefined symbols such as u here treated as functions of x. ∗)

Output 1 √ 1 + x2 Dt[ u,x] √ 1 + u2

678

Computer algebra (CA) with Mathematica

C.1.2

Integrals

Clear[x]; Integrate[Cot[x]∗Sin[4∗x], {x, 0, Pi}] Output π

C.1.3

Series

Series[Log[(Cos[x]+ Sin[x]) / (Cos[x]- Sin[x])], {x, 0, 6}] y = 1 + b1 ∗ x + b2 ∗ xˆ2 + b3 ∗ xˆ3; (∗ Define y ∗) Series[1/y, {x, 0, 5}] z = b1 ∗ x + b2 ∗ xˆ2 + b3 ∗ xˆ3 + 0[x]ˆ4; (∗ Define z ∗) ax = InverseSeries[z] z (∗ Print z to show it has not changed ∗) Output 4 x 3 4 x5 + + O[x]7 3 3 1 - b1 x +(b12 - b2)x2 +(-b13 + 2 b1 b2 - b3)x3 + (b14 - 3 b12 b2 + b22 + 2 b1 b3)x4 +(-b15 + 4 b13 b2 - 3 b1 b22 3 b12 b3 + 2 b2 b3)x5 + O[x]6 x b2 x2 (2 b22 - b1 b3)x3 + + O[x]4 b1 b15 b13 b1 x + b2 x2 + b3 x3 + O[x]4 2x+

The input is z, a power series in x. The output ax of InverseSeries in shown as as power series in x. It is actually meant to be x in powers of z. This is the inverse series, also called the reverted series. This interpretation can be verified by a substitution using ReplaceAll: FullSimplify[ReplaceAll[ax, x → z]] (∗ Get a series in powers of z after the substitution x → z. Its sum is the original x ∗) Output x + O[x]4

C.1.4

Fourier transform

Clear[x, b, k]; FullSimplify[FourierTransform[Sin[b∗xˆ2], x, k]]

Equation solvers

Output i e−

ik2 4b

2

√

√

b2 + i b e

ik2 2b

679



2(ib)3/2

The last expression can be simplified to Cos[kˆ2/(4∗b) + Pi/4]/Sqrt[2∗b], the result given in many tables of Fourier transforms. C.1.5

Laplace transform

Clear[t, s]; LaplaceTransform[tˆ4 Sin[t], t, s] Output 24(1 − 10 s2 + 5 s4 ) (1 + s2 )5

C.2

Equation solvers

CA can be used to solve algebraic and differential equations, and to verify equation solutions, symbolically. C.2.1

Solving algebraic equations

Solve[a ∗ xˆ2 + b ∗ x + c == 0, x] Output ::

x→

−b −

C.2.2

√ b2 − 4 a c ; : −b + b2 − 4 a c ;; , x→ 2a 2a

√

Solving ordinary differential equations(ODEs)

Clear[x, y]; DSolve[y'[x] + y[x] == a Sin[x], y[x], x] Clear[x, y]; FullSimplify[ DSolve[y'[x] + y[x] Sqrt[1 - y[xˆ2] == 0, y[x], x]] Output

:: y[x] → {{ y[x] →

;; 1 a(−Cos[x] + Sin[x]) 2 Sech[x − C[1]]}}

e− x C[1] +

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Computer algebra (CA) with Mathematica

The arrow in the output means that y[x] should be replaced by the expression that follows. The entire construct (sumbol + arrow + expression) is called a rule. C[1] in both solutions is an arbitrary constant. C.2.3

Constructing an Euler spiral

An Euler (or Cornu) spiral is a curcve whose curvature (inverse of the radius of curvature) is proportional to the length of the curve from the origin. The Euler spiral appears in the theory of Fresnel diffraction. sol = DSolve [{x'[s] == Cos[t[s]], y'[s] == Sin[t[s]], t'[s] == s, x[0] == 0, y[0] == 0, t[0] == 0}, {x, y, t}, s] (∗ The unknowns x[s], y[s], t[s] are functions of the same variable s. The six equations specified in the argument of DSolve define three differential equations and three initial conditions for three variables. The solutions are the Fresnel cosine and sine integrals. ∗) ParametricPlot[Evaluate[{x[u], y[u]}/.sol], {u, -10, 10} (∗ Note the special way the solution sol is used to generate numerical values for plotting. Evaluate basically removes the double sets of curly braces in the output of DSolve. ∗) Output ::

s2 . , t → Function {s}, 2 - s .. √ x → Function {s}, π FresnelC √ , π - s ..;; √ y → Function {s}, π FresnelS √ π

Function in Mathematica defines a pure (nameless) function. The Fresnel integrals are defined as follows:  z  z cos(π t2 /2)dt, S (z) = sin(π t2 /2)dt C(z) = 0

0

See the electronic notebook for all graphics output. C.2.4

Solving a partial differential equation (PDE): Laplace equations

ClearAll[u, x, y]; DSolve[D[u[x, y], x, x] + D[u[x, y], y, y] == 0, u[x, y], {x, y}] Output {{u[x, y] → C[1][i x + y] + C[2][-i x + y]}}

Equation solvers

681

Here C[1][i x + y] + C[2][−i x + y] are any functions of their arguments. Since the PDE is linear, any linear combination of these two solutions is also a solution. C.2.5

Solving a PDE: Wave equations

ClearAll[u, x, t, c]; DSolve[D[u[t, x], t, t] - cˆ2∗D[u[t, x], x, x] == 0, u[t, x], {t, x}] Output

√ √ {{u[t, x]→ C[1][- c2 t + x] + C[2][ c2 t + x]}}

If c > 0, the first solution C[1][x − c t] for any function C[1] describes a wave traveling towards x → ∞. The second solution C[2][x + c t] for any function C[2] describes a wave traveling towards x → −∞. Since the PDE is linear, any linear combination of these two solutions is also a solution. C.2.6

Solving a PDE: Diffusion equations

ClearAll[u, x, t, k]; DSolve[{D[u[x, t], t] - k∗D[u[x, t], x, x] == 0, u[0, t] = t0}, u[x, t], {x, t}] Output DSolve[{u(0,1) [x, t] - k u(2,0) [x, t] == 0, u[0, t] = t0}, u[x, t], {x, t}]

If the output is the same as the input, it means the command cannot be executed. The reason here is because DSolve cannot solve this type of PDEs. DSolve gives the general solution for only a restricted type of homogeneous linear second-order PDEs of the form a

∂2 u ∂2 u ∂2 u + b = 0, + c ∂x∂y ∂x2 ∂y2

where a, b, and c are constants. The Wikipedia entry “Partial Differential Equation” gives a nice introduction to the subject. The solution of a PDEs with specified boundary conditions is technically quite complicated. However, the verification that a given function might solve a certain PDE is very easy if done by CA, as the following examples illustrate. C.2.7

Verifying a solution: Linear diffusion equation

Clear[a, x, t]; u = Exp[-xˆ2/(4∗a∗t)]/Sqrt[4∗Pi∗a∗t]; FullSimplify[D[u, t] - a∗D[u, x, x]] Integrate[u, {x, - Infinity, Infinity}]

682

Computer algebra (CA) with Mathematica

Output 0 ConditionalExpression

. -4 1 √ 1 ]>0 a t, Re[ at at

This solution is particularly interesting if u is the 1-dimensional density of a conserved quantity that does not chang in time. Then its integral over x will be constant in time. The last step verifies that the integral is in fact 1. The condition that Re(at) > 0 is needed to ensure that the function vanishes at ±∞, for otherwise the integral is divergent. The function u is interesting also for another reason. In the limit at → 0+, it becomes a sharp spike at x=0 that integrates to 1. In other words, it is a Dirac δfunction in this limit. The next three examples involve nonlinear PDEs studied in Chapter 6. C.2.8

Verifying a solution: Nonlinear diffusion equation

phi = Sech[x]; ode = phi ∗ D[phi, x, x] - D[phi, x]ˆ2 + phiˆ4 FullSimplify[ode] Output Sech[x]4 - Sech[x]2 Tanh[x]2 + Sech[x] (-Sech[x]3 + Sech[x] Tanh[x]2 ) 0

Mathematica does not always give the simplest expression for the output without an explicit command. C.2.9

Verifying a solution: Korteweg–deVries equation

Clear[c, d, k, alfa]; k = Sqrt[c]/(2 ∗ d); w = k ∗ c; u = (3 ∗ c/alfa) ∗ Sech[k ∗ x-w ∗ t]ˆ2; FullSimplify[D[u, {t, 1} + alfa ∗ u ∗ D[u, {x, 1}] + dˆ2 D[u, {x, 3}]] Output 0

C.2.10

Verifying a solution: Sine-Gordon equation

Clear[b, c, x, t, k]; w = Sqrt[kˆ2-1]; c = w/k

Drawing figures and graphs

683

psi = ArcTan[Sech[k∗x]∗Sinh[w∗t]/c]; FullSimplify[D[psi, {x, 2}]-D [psi, {t, 2}]-Sin[4∗psi]/4] Output 0

C.3 C.3.1

Drawing figures and graphs Drawing Fig. 6.9: Equilibrium points and Jacobian eigenvalues for the Henon map

Clear[a]; style1 = {AbsoluteThickness[1]}; style2 = {AbsoluteThickness[1], AbsoluteDashing[{2, 2}]}; style3 = {AbsoluteThickness[1], AbsoluteDashing[{3, 3}]}; style4 = {AbsoluteThickness[1], AbsoluteDashing[{4, 4}]}; style5 = {AbsoluteThickness[1], AbsoluteDashing[{5, 5}]}; style6 = {AbsoluteThickness[2]; (∗ Defining different curve styles ∗) b = 0.3; amin = -0.3; amax = 1.5; xep[a ]:= (-(1-b) + Sqrt[(1-b)ˆ2+4∗a])/(2∗a); (∗ Defining functions ∗) xem[a ]:= (-(1-b) - Sqrt[(1-b)ˆ2+4∗a])/(2∗a); lambdap[xe , a_]:= -a∗xe + Sqrt[(a∗xe)ˆ2 + b]; lambdam[xe , a_]:= -a∗xe - Sqrt[(a∗xe)ˆ2 + b]; (∗ Module = collection of codes that can be used repeatedly with a new value of the function argument ∗) f[a ]:= Module[{tp, tm}, tp = xep[a]; tm = xem[a]; {tp, tm, lambdap[tp, a], lambdam[tp, a], lambdap[tm, a], lambdam[tm, a]} (∗ Last line of Module specifies output ∗) ]; (∗ End of Module ∗) f[1.4] (∗ Testing Module ∗) (∗ Plotting results calculated by f[a] ∗) Plot[Evaluate[f[a]], {a, amin, amax}, PlotStyle→{style6, style1, style2, style3, style4, style5}, PlotRange→{{amin, amax}, {-5, 5}} ] Output {0.631354, -1.13135, 0.155946, -1.92374, 3.25982, -0.0920296}

684

C.3.2

Computer algebra (CA) with Mathematica

Drawing Fig. 6.19: One soliton passing another

Needs["PlotLegends"] (∗ Load the PlotLegends package ∗); style1 = {AbsoluteThickness[1]}; style2 = {AbsoluteThickness[1], AbsoluteDashing[{2, 2}]}; style3 = {AbsoluteThickness[1], AbsoluteDashing[{3, 3}]}; style4 = {AbsoluteThickness[1], AbsoluteDashing[{4, 4}]}; style5 = {AbsoluteThickness[1], AbsoluteDashing[{5, 5}]}; style6 = {AbsoluteThickness[2]}; styleAR = {style3, style6, style1}; k1 = 1.0; w1 = k1ˆ3; k2 = 1.4; w2 = k2ˆ3; a12 = (k1-k2)ˆ2/(k1+k2)ˆ2 u2s[t , x ] = D[Log[1+Exp[k1∗x-w1∗t]+Exp[k2∗x-w2∗t]+ a12∗Exp[k1∗x-w1∗t]∗Exp[k2∗x-w2∗t]], {x, 2}]; t1 = -6.0; t2 = 0.0; t3 = 6.0; ps = Plot[{u2s[t1, x], u2s[t2, x], u2s[t3, x]}, {x, -20, 20}, BaseStyle → {FontFamily-> "Times", Bold, 11}, AxesLabel → {"x", "u(x)"}, PlotStyle → styleAR, PlotRange → {-0.1, 0.6}, PlotLegend → {t = -6", "t=0", "t=6"}, LegendPosition-> {-1.3, .25}, LegendShadow → {0, 0}, LegendSize → {0.5, 0.3}] C.3.3

Meissner exclusion of magnetic field from the interior of a superconductor

A figure is made up of graphics elements such as arrows, lines, circles and text. In older versions of Mathematica, each graphics element has to be placed at the desired location, as shown in the example given in the electronic notebook. This is a time-consuming process. In recent versions, a Drawing Tools palette makes it straightforward to modify existing plots or illustrations, or to create free-form ones form scratch, interactively.

C.4

Number-intensive calculations

Mathematica has extensive numerical capabilities. It distinguishes between two types of noninteger real numbers: machine-precision and arbitrary-precision numbers. Machine-precision numbers have sixteen significant digits, also called doubleprecision floating-point numbers. (Single-precision numbers with only eight digits were used in early digital computers.) Numerical constants containing less than sixteen are filled to sixteen digits with trailing zeros. Calculations using machineprecision numbers are speedy because they match the built-in format of the CPU

Number-intensive calculations

685

(central processing unit) in the computer. Such calculated results have sixteen digits, even when they are not accurate to all sixteen digits because of round-off errors. Arbitrary-precision numbers (also called bignums) contain more than sixteen significant digits each. Bignum calculations keep track of the significant digits of the results in software, but this refinement also slows down the calculations substantially. In Mathematica, constants such as Pi, Sqrt[7], 7/3 actually have infinite precision. The actual number of digits that can be printed out is limited by the time spent on the calculation and the storage limit of the computer. Many numerical calculations in physical sciences and engineering are capable of only limited precision. For this reason, Mathematica normally shows only six digits in numerical outputs if one does not specify the number of digits. All commands with numerical outputs start with N. The command N[x, n] truncates the number x to n digits. The operating default value of n, if not given, is sixteen. C.4.1

Arbitrary- and machine-precision numbers

MachineNumberQ[Pi] (∗ Q = ?, a Boolean command ∗) Precision[Pi] mach = N[Pi] (∗ Define mach ∗) OutputForm[Pi] OutputForm[mach] N[Pi, 20] RealDigits[mach, 10, 20] InputForm[mach] Output False ∞ Pi 3.14159 3.1415926535897932385 {{3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, Indeterminate, Indeterminate, Indeterminate, Indeterminate}, 1} 3.141592653589793

RealDigits[x, base, digits] gives the actual digits appearing in the number x. Its output is in the (shifted) scientific notation that is read as 0.3141592653589793 × 101 , the indeterminate digits being insignificant. C.4.2

Drawing Fig. 6.13: Bifurcation diagrams for the PDNL pendulum

Clear[x, v, xvList, avList]; nForce = 1; w0 = 1; w = 2;

gam = 0.2;

686

Computer algebra (CA) with Mathematica

(∗ nForce = 1 for parametrically drive pendulum, not= 1 for externally driven pendulum ∗) (∗ Fig. 6.13: use A0 = 0.151: dA = 0.005; nA = 320; ∗) A0 = 0.151; dA = 0.15; nA = 10; Amax = A0 + nA∗dA; yMax = 5.0; nstep = 401; dnstep = 200; mstep = nstep + dnstep; / 2π 0 nAccur = 16; delt = N , nAccur ; w nmode = 0; gBDiag[initP , initV ]:= Module[{A, initPos, initVel], (∗ A Module is a subprogram that can be used repeatedly with arguments that can be changed. ∗) A = A0; avList = {}; Do[force = If[nForce == 1, -(w02 +2A Cos[wt]) Sin[x[t]], -w02 Sin[x[t]]+2A Cos[wt]]; (∗ Begin outside Do loop ∗) eqs = {x’[t] == v[t], v’[t] == -gam v[t] + force}; tmin = 0; initPos = initP; initVel = initV; Do[tmax = tmin + delt; (∗ Begin inside Do loop ∗) allEqs = Join[eqs, {x[tmin] == initPos, v[tmin] == initVel}]; sol = NDSolve[allEqs, {x, v}, {t, tmin, tmax}, AccuracyGoal→nAccur]; (∗ NDSolve is the numerical differential eq. solver ∗) initxv = First[Evaluate[{x[t], v[t]}/.sol]/. t→ tmax]; initPos = If[nmode == 0, initxv1, N[Mod[initxv1 + π, 2π] -π, nAccur]]; initVel = initxv2; avList = If[m < nstep, avList, Append[avList, {A, initVel}]]; tmin = tmax; ClearAll[x, v], {m, mstep}]; (∗ End inside Do loop ∗) A = A+dA, {i, nA}]; (∗ End outside Do loop ∗) ListPlot[avList, PlotRange→{{A0, Amax}, {-yMax, yMax}}, PlotStyle → AbsolutePointSize[1], BaseStyle → {FontFamily-> "Times", Bold, 12}, PlotLabel → StyleForm["Bifurcation diagram for θ(0) = "ToString[initP]

Number-intensive calculations

", dθ(0)/dt = "ToString[initV]], AspectRatio → 0.5, Frame → True, FrameLabel → {"A", "dθ/dt"}] ]; (∗ End of Module ∗) Timing[ gBDiag[0.0, 5.0] ] (∗ Show time used for nA=10 A values ∗) Show[ GraphicsGrid[ {{gBDiag[0, 1.5]}, {gBDiag[0, 3]}, {gBDiag[0, 5]}}]] (∗ Test: Show an array of 3 figures with different initial velocities ∗) (∗ For Fig. 6.13, run the following program. The output will not be shown. ∗) A0 = 0.151; dA=0.005; nA=320; Amax = A0 + nA∗dA; Show[ GraphicsGrid[ {{gBDiag[0, 1.5]}, {gBDiag[0, 3]}, {gBDiag[0, 5]}}]]

687

Resources for students

Books The following list of books and sources have been selected with the needs of students in mind. Each book reference is described in full in the Bibliography.

General Arfken, Boas, Bradbury, Chow, Dettman (2011), Harper, Jones, Kraut, Margenau/Murphy, Mathews/Walker, Morse/Feshbach, Nearing, Pearson, Potter/Goldberg, Vaughn, Wikipedia Chapter 1 Vectors and fields in space Marion, Pearson (Ch. 3), Spiegel (Vector Analysis) Chapter 2 Transformations, matrices and operators Ayres, Pearson (Chs. 15, 16), Pettofrezzo Chapter 3 Relativity and square-root spaces Pearson (Ch. 4), Sakurai (1967) Chapter 4 Fourier series and Fourier transforms Carslaw, Pearson (Ch. 11) , Spiegel (1974), Tolstov Chapter 5 Differential equations in physics Hochstadt (l975), Pearson (Chs. 6, 8, 9) Chapter 6 Nonlinear systems Drazin (1984), Strogatz Chapter 7 Special functions Hochstadt (l986), Pearson (Ch. 7) Chapter 8 Functions of a complex variable Brown/Churchill, Dettman (1984), Pearson (Chs. 5,12), Spiegel (Complex Variables)

Tables Abramowitz/Stegun

Dictionaries and Handbooks Abramowitz/Stegun, Gellert+, Handbook of Chemistry and Physics, James/James, Pearson

Biographies Debus, Encyclopaedia Britannica (1985), Gillispie, Google, Kline, Physics Today, Rouse Ball, Struik, Wikipedia

Resources for students

689

Internet resources A web search by Google or other search engines on a mathematics or physics topic will give you Wikipedia entries, free eBooks and lecture notes, sites selling published books and eBooks, journal articles, graphics and other resources. Wikipedia entries are always worth a look. They almost always contain information of interest to you. Learn how to rapidly scan an article for information of immediate interest, and to ignore advanced and exotic treatments that can be left to the future. Look at the figures, references, external links especially to visualization/animation sites before you leave the article. The many visualization links available from Wikipedia entries will not be repeated here. Particularly helpful entries are marked by asterisks. Wikibooks is an open-content textbooks collection that anyone can edit. In February 2012 it contains 2,432 books with 40,336 pages. There are several interesting wikiBooks mathematical subjects, some in mid-development. There are many free eBooks and lecture notes around, usually in the pdf format, from author’s web sites. Read a page or two to see if the material is written in a style accessible to you before saving the file to your PC. If necessary, add words to the filename to help you identify the subject matter in the future. The filename format I like to use is SubjAuthYearTopic.pdf. For journal articles I use the filename spRelEinstein05AP17p891 for Einstein’s 1905 paper on special relativity published in vol. 17, p. 891, of Analen der Physik. If you download materials from the Internet you are responsible for following all applicable copyright rules in the country where you live. If you are downloading the material for your personal educational use and not for profit, it seems safe to download certain files on educational materials, including the following: • Posted files from an author’s own web page. • Posted files from government agencies, companies, publishers and web pages

that have been explicitly or implicitly been placed in the public domain and not c marked by the word copyright or the tag . • Online resources including chapters of published eBooks obtained through the online catalogs of your university libraries when you are a registered user of the library. You are on dangerous ground if you try to download files from file-sharing sites whose files come from user uploads. Many of these files are still under copyright protection that disallows file-sharing. The US copyright rules are complicated. A summary chart of US copyright laws can be found at http://copyright. cornell.edu/resources/publicdomain.cfm To locate a publication in the public domain as a downloadable file: • Search the Digital Book Index (for both free and for sale eBooks) at

http://www.digitalbookindex.org/search001a.htm

• Search the author’s name index at the Online Book Page at

http://onlinebooks.library.upenn.edu/.

• Search for author/title in Google eBook at

http://books.google.com/help/ebooks/content.html.

690

Resources for students

The Internet is a dynamic structure where links are frequently moved or removed without notice. None of the web links quoted here is guaranteed to work. Web searches are so good now that it would be easy to locate the same or similar resources anew. The links given illustrate the wealth and quality of information posted on the Web. For visualization/animation resources, we suggest looking first in the visualization section near the end of appropriate Wikipedia entries, and then searching the web with a search engine such as Google.

WikiBooks Many of the following titles are in development in early 2012: Introduction to Mathematical Physics, Mathematical Methods of Physics, Mathematics for Chemistry, Computational Physics, Introduction to Theoretical Physics, Waves, Special Relativity. Linear Algebra, Linear Algebra with Differential Equations, Real Analysis, Applied Mathematics, Complex Analysis, Differential Geometry, Famous Theorems of Mathematics, Fractals, Functional Analysis, Numerical Methods, Ordinary Differential Equations, Partial Differential Equations, Applied Mathematics, CLEP College Algebra (CLEP = College Level Examination Program). Home page: http://en.wikibooks.org/wiki/Main Page University mathematics page: http://en.wikibooks.org/wiki/Subject:University level mathematics books.

Notable web sites Miller, Jeff, Earliest Known Uses of Some of the Words of Mathematics, http://jeff560.tripod.com/mathword.html.

General eBooks on mathematical physics Abramowitz, Milton, and Irene A. Stegun (eds.), Handbook of Mathematical Functions, With Formulas, Graphs, and Mathematical Tables, 10th printing, with corrections, 1972. Page images at http://people.math.sfu.ca/∼cbm/aands/. Mauch, Sean, (Introduction to Methods of Applied Mathematics, unpublished, 2004) is a work in progress containing a wealth of examples and solutions. pdf file at http://www.cacr.caltech.edu/∼sean/applied math.pdf. Nearing, James, Mathematical Tools for Physics, (Dover, New York, 2010), has a free pdf version at http://www.physics.miami.edu/∼nearing/mathmethods/. NIST Digital Library of Mathematical Functions (DLMF), 2010, a revision of Abramowitz’s and Stegun’s Handbook of Mathematical Functions. Copyrighted by NIST (National Institute of Standards and Technology) but available free for reading online at http://dlmf.nist.gov/. Stephenson, G., Mathematical methods for science students, (New York, Wiley, (c. 1961)). Page images at HathiTrust: http://catalog.hathitrust.org/Record/ 0093246l7.

Resources for students

691

Chapter-specific Internet resources Chapter 1 Vectors and fields in space Google search: space, vector calculus, curvilinear coordinates, Stokes’ theorem, divergence theorem, Green’s theorem. Wikipedia entries: vector calculus∗ , vector field∗ , curvilinear coordinates∗ , Stokes’ theorem, divergence theorem, Green’s theorem. Wikibooks: Calculus. eBooks Brannon, Curvilinear Analysis in an Euclidean Space, 2004: http://www.mech.utah.edu/∼brannon/public/curvilinear.pdf. Corral, M., Vector Calculus, 2008: www.mecmath.net/calc3book.pdf. Nearing, Chapters 9 and 13 cover vector calculus: http://www.physics.miami.edu/∼nearing/mathmethods/vector calculus.pdf; http://www.physics.miami.edu/∼nearing/mathmethods/vector calculus-2.pdf. Polnarev, A. G., Calculus III class notes: http://www.maths.qmul.ac.uk/∼agp/calc3; pdf file of Chapter 8 on integral theorems: http://www.maths.qmul.ac.uk/∼agp/calc3/notes8.pdf. Chapter 2 Transformations, matrices and operators Google search: matrix, determinant, eigenvalue problem, wave equation, infinitesimal generators, rotation matrix, group theory. Wikipedia entries: matrix∗ , determinant∗ , eigenvalues and eigenvectors∗ , wave equation∗ , rotation matrix∗ . Wikibooks: Linear Algebra eBooks Cohen, A., R. Ushirobira, and J. Draisma, Group Theory for Maths, Physics, and Chemistry Students, 2007, at http://www.win.tue.nl/∼amc/ow/gpth/reader.pdf. Dawkins, Paul, Paul’s Online Math Notes, Lamar University, is a very good tutorial site for basic college mathematics. Home page: http://tutorial.math.lamar.edu/. Felippa, Carlos A., Introduction to Finite Element Methods (ASEN 5007), Fall 2011, Department of Aerospace Engineering Sciences, University of Colorado, Boulder, Appendix C on matrix algebra of an online class note: http://www.colorado.edu/ engineering/cas/courses.d/IFEM.d/IFEM.AppC.d/IFEM.AppC.pdf. Petersen, K. B., and M. S. Petersen, The Matrix Cookbook, 2008, a collection of useful formulas for matrices: http://orion.uwaterloo.ca/∼hwolkowi/matrixcookbook.pdf. Vvedensky, D., Groups, a course note. The second link points to the chapter on continuous groups and infinitesimal generators: http://www.cmth.ph.ic.ac.uk/people/d.vvedensky/groups/, http://www.cmth.ph.ic.ac.uk/people/d.vvedensky/groups/Chapter7. pdf.

692

Resources for students

Chapter 3 Relativitic square-root spaces Google search: special relativity, relativistic kinematics, quaternion, Dirac equation, spinors, Dirac spinor, Weyl spinor, Majorana spinor, spacetimes symmetries, symmetry violations, Cartan spinor, Lorentz group, Cartesian tensor, tensor analysis. Wikipedia entries: special relativity∗ , quaternion, Dirac equation∗ , spinor, Dirac spinor, spacetime symmetries, Lorentz group, tensor, covariant derivative. WikiBook: Special Relativity. eBooks de Vries, Hans, Understanding Relativistic Quantum Field Theory, book in progress, Chapter 16 on Dirac equation: http://physics-quest.org/Book Chapter Dirac.pdf. Smirnov, A. T., Introduction to Tensor Calculus, draft, 2004: http://faculty.gg.uwyo.edu/dueker/tensor curvilinear relativity/tensor analysis intro.pdf Chapter 4 Fourier series and Fourier transforms Google search: Fourier series, Fourier transform, quantum mechanics, orthogonal functions, Dirac delta function, Bessel inequality, Parseval equation, completeness, Hilbert space, Fourier spaces, Maxwell equations, Helmholtz decomposition theorem. Wikipedia entries: Fourier series∗ , Fourier transform∗ , quantum mechanics∗ , Dirac delta function∗ , Bessel inequality, Parseval theorem, Hilbert space∗ , Maxwell equations∗ , Helmholtz decomposition. eBooks Carslaw, H. S., Introduction to Theory of Fourier’s Series and Integrals, Dover, New York, c1930. Page images at HathiTrust (3rd ed., 1936) and U. Michigan (2nd ed., 1921), respectively: http://catalog.hathitrust.org/Record/000312211, http://quod.lib.umich.edu/cgi/t/text/text-idx?c=umhistmath; idno=ACR2399.0001.001. Chapter 5 Differential equations in physics Google search: ordinary differential equation, Green’s function, partial differential equation. Wikipedia entries: differential equation, Dirac delta function∗ , Green’s function∗ , Wronskian, partial differential equation∗ , separation of variables∗ , eigenfunction expansion. Chapter 6 Nonlinear systems Google search: nonlinear system, logistic map, chaos, strange attractor, self similarity, fractal, soliton, traveling kink, multi-solitons. Wikipedia entries: nonlinear system, parametric oscillator, logistic map∗ , Henon map, chaos theory, attractor∗ (including strange attractor), Mathieu function, soliton, sine-Gordon equation∗ , Korteweg-de Vries equation.

Resources for students

693

Chapter 7 Special functions Google search: special functions, Legendre polynomials, Hermite polynomials, classical orthogonal polynomials, associated Legendre polynomilas, spherical harmonics, Bessel functions, Sturm–Liouville equation, eigenfunction expansion. Wikipedia entries: special functions, Legendre polynomials∗ , Hermite polynomials∗ , orthogonal polynomials, classical orthogonal polynomials∗ , associated Legendre polynomials∗ , spherical harmonics∗ , Bessel functions∗ , Sturm-Liouville equation∗ , eigenfunction∗ . eBooks previously mentioned: Abramowitz/Stegun, NIST Digital Library of Mathematical Functions. Chapter 8 Functions of a complex variable Google search: complex variable, Riemann surfaces, complex integration, Laurent series, calculus of residues, Laplace transform, dispersion relations, asymptotic series, complex analysis, WKB approximation, Airy function. Wikipedia entries: complex analysis, Riemann surfaces∗ , methods of contour integration∗ , Laurent series, Laplace transform∗ , dispersion relations, asymptotic expansion, WKB approximation∗ , Airy function∗ . eBooks Ash, Robert B. and W. P. Novinger, Complex Variables, 1971/2004: http://www.math.uiuc.edu/∼r-ash/CV.html.

Bibliography

Abramowitz, Milton, and Irene A. Stegun (eds.), Handbook of Mathematical Functions, National Bureau of Standards, Applied Mathematics Series 55, U.S. Government Printing Office, Washington, D.C., 1964. Often referred to in the text as AMS-55. Arfken, George, Mathematical Methods for Physicists, Academic Press, New York, 2nd ed., 1970; 3rd ed., 1985; 7th ed., 2012 (with H. J. Weber and F. E. Harris). Ayres, Frank, Jr., Matrices, Schaum’s Outline Series, McGraw-Hill, New York, 1967. Baker, G. L., and J. P. Gollub, Chaotic Dynamics, Cambridge University Press, Cambridge, 1990. Bjorken, James D., and Sidney D. Drell, Relativistic Quantum Mechanics McGraw-Hill, New York, 1964. Boas, Mary L., Mathematical Methods in the Physical Sciences, Wiley, New York, 2nd ed., 1983, 3rd ed., 2005. Bohr, Aage, and Ben R. Mottelson, Nuclear Structure vol. I, Benjamin, New York, 1969. Bountis, T. C., Physica D3, 577 (1981). Bradbury, T. C., Mathematical Methods with Applications to Problems in the Physical Sciences, Wiley, New York, 1984. Bransden, B. H., and C. J. Joachain, Quantum Mechanics, Prentice Hall, Harlow, 2000. Brown, James W., Complex Variables and Applications, McGraw-Hill, New York, 8th rev. ed., 2008. Latest edition of a well-known text by Churchill and Brown. Butikov, E. I., J. Phys. A 35, 6209 (2002). Butikov’s Nonlinear Oscillations page: http://faculty.ifmo.ru/butikov/Nonlinear/index-html. Carslaw, H. S., An Introduction to the Theory of Fourier Series and Integrals, Dover, New York, 3rd ed., 1950. Cheng, Ta-Pei and Ling-Fong Li, Gauge Theory of Elementary Particle Physics, Clarendon, Oxford, 1984. Chow, Tai L., Mathematical Methods for Physicists: A Concise Introduction, Cambridge University Press, Cambridge, 2000. Commins, Eugene D., and Philip H. Bucksbaum, Weak Interactions of Leptons and Quarks, Cambridge University Press, Cambridge, 1983. Davis, H. T., Introduction to Nonlinear Differential and Integral Equations, Dover, New York, 1962. Debus, Allen G. (ed.), World Who’s Who in Science, Marquis Who’s Who, Chicago, 1968. Dennery, Philippe, and Andre Krzywicki, Mathematics for Physicists, Dover, New York, 1996. Dettman, John W., Applied Complex Variables, Dover, New York, 1984. Dettman, John W., Mathematical Methods in Physics and Engineering, Dover, New York, 2011. D’Inverno, R., Introducing Einstein’s Relativity, Clarendon, Oxford, 1992. Dirac, P. A. M., The Principles of Quantum Mechanics, Clarendon Press, Oxford, 4th ed., 1958.

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Drazin, Philip D., Nonlinear Systems (Cambridge Texts in Mathematics), Cambridge University Press, Cambridge, 1992. Drazin, Philip D., Solitons (London Mathematical Society lecture note series: 85), repr. with corrections, Cambridge University Press, Cambridge, 1984. Edmonds, A. R., Angular Momentum in Quantum Mechanics, Princeton University Press, Princeton, 1957. Einstein, A., H. A. Lorentz, H. Weyl, and H. Minkowski, The Principle of Relativity, Dover, New York, 1952. Encyclopaedia Britannica, Encyclopaedia Britannica, Chicago, 1985. Enns, Richard H., and George C. McGuire, Nonlinear Physics with Mathematica for Scientists and Engineers, Birkhauser, Boston, 2001. Equilux, V. M., et al., Phys. Rev. Lett. 22, 5232 (2000). Erd´elyi, A. et al. (eds.), Table of Integral Transforms (Bateman Manuscript Project), McGrawHill, New York, 1954. Erd´elyi, A., Asymptotic Expansions, Dover, New York, 1956. Evgrafov, M. A., Analytic Functions, Dover, New York, 1966. Fan, E. G., J. Phys. A: Math. Gen. 36, 7009 (2003). Feigenbaum, M. J., J. Stat. Phys. 19, 25 (1978). Ford, John R., Classical Mechanics, University Science Books, Sausalito, 2005. French, A. P., Special Relativity, Norton, New York, 1968. Gellert, W. et al. (eds.) The VNR Concise Encyclopedia of Mathematics, Van Nostrand Reinhold, New York, 1975. Gillispie, Charles Coulson (ed.), Dictionary of Scientific Biography, Charles Schribner’s Sons, New York, 1980. Goldstein, Herbert, Classical Mechanics, Addison-Wesley, Reading, 1950; 3rd ed., 2001 (with C. P. Poole and J. L. Safko). Handbook of Chemistry and Physics, CRC Press, Baton Raton, annual edition. Available online at http://www.hbcpnetbase.com/. Harper, Charlie, Introduction to Mathematical Physics, Prentice-Hall, Englewood Cliffs, 1976. Hassani, Sadri, Mathematical Physics, Springer, New York, 1999. H`enon, M., Comm. Math. Phys. 50, 69 (1976). Hirota, Ryogo, The Direct Method in Soliton Theory (Cambridge Tracts in Mathematics), Cambridge University Press, Cambridge, 2004. Hochstadt, Harry, Differential Equations, Dover, New York, 1975. Hochstadt, Harry, The Functions of Mathematical Physics, Dover, New York, 1986. Infeld, Eryk, and George Rowlands, Nonlinear Waves, Solitons and Chaos, Cambridge University Press, Cambridge, 1990. Jones, Lorella M., An Introduction to Mathematical Methods of Physics, Benjamin/ Cummings, Menlo Park, 1979. Klein, Morris, Mathematical Thought from Ancient to Modern Times, Oxford University Press, New York, 1972. Kobayashi, M. and T. Maskawa, 2008 Nobel Prize lectures: http://www.nobelprize.org/nobel prizes/physics/laureates/2008/kobayashi lecture.pdf, http://www.nobelprize.org/nobel prizes/physics/laureates/2008/maskawa lecture.pdf

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Kraut, Edgar A., Fundamentals of Mathematical Physics, McGraw-Hill, New York, 1967; Dover reprint, 2007. Lam, Lui (ed.), Introduction to Nonlinear Physics, Springer, New York, 1997. Lamb, G. L., Elements of Soliton Theory (Pure & Applied Mathematics), Wiley, New York, 1980. Landau, L. D., Comptes Rendus Acad. Sci., U.S.S.R. (Dolclady) 44, 311 (1944). Also Collected Papers, D. ter Haar (ed.), Pergamon, Oxford, pp. 445–60. Landau, L. D., and E. M. Lifshitz, The Classical Theory of Fields, Addison-Wesley, Reading, 1951; 4th rev. ed., Pergamon, Oxford, 1975. Landau, L. D., and E. M. Lifshitz, Mechanics, Pergamon Press, Oxford, 1960; 3rd ed., 1976. Lorenz, E. N., J. Atmos. Sci. 20, 130 (1963). Malfliet, M., Am. J. Phys. 60, 650 (1992). Margenau, Henry and George M. Murphy, The Mathematics of Physics and Chemistry, Van Nostrand, New York, 2nd ed., 1976; Young Press, 2009. Marion, Jerry B., Principles of Vector Analysis, Academic Press, New York, 1965. Martens, P. C. H., Phys. Rep. 115, 315 (1985). Mathews, Jon, and Robert L. Walker, Mathematical Methods of Physics, Benjamin, New York, 1970. May, Robert M., Nature 261, 459 (1976). McLaughlin, J. R., J. Stat. Phys. 24, 375 (1981). McQuarrie, Donald A., Mathematical Methods for Scientists and Engineers, University Science Books, Sausalito, 2003. Miura, R. M., SIAM Review 18, 412 (1976). Morse, Philip M., and Herman Feshbach, Methods of Theoretical Physics, McGraw-Hill, New York, 1953; reprinted by Feshbach Publ., 2005. Nearing, James, Mathematical Tools for Physics, Dover, New York, 2nd corrected printing (ISBN 048648212X), 2010. Pearson, C. E. (ed.), Handbook of Applied Mathematics Selected Results and Methods, Van Nostrand Reinhold, New York, 2nd ed., 1990. Peitgen, H.-O., H. J¨urgens, and D. Saupe, Chaos and Fractals, Springer, New York, 1992. Penrose, Roger, The Road to Reality, A. A. Knopf, New York, 2004. Penrose, Roger, and Wolfgang Rindler, Spinors and Space-time, Cambridge University Press, Cambridge, 1984. Perring, J. K., and T. H. R. Skyrme, Nucl. Phys. 31, 550 (1962). Pettofrezzo, Anthony J., Matrices and Transformations, Dover, New York, 1978. Polyanin, Andre D., and Valentin F. Zaitsev, Handbook of Exact Solutions for Ordinary Diflerential Equations, Chapman & Hall/CRC, Bota Raton, 2nd ed., 2003. Potter, Merle C., and Jack Goldberg, Mathematical Methods, Prentice-Hall, Englewood Cliffs, 2nd ed., 1987. Pound, R. V. and G. A. Rebka, Phys. Rev. Lett. 4, 337 (1960). Rose, M. E., Elementary Theory of Angular Momentum, Dover, New York, 1995. Rouse Ball, W. W., A Short Account of the History of Mathematics, 6th ed., Dover, New York, 1960. Sakurai, J. J., Advanced Quantum Mechanics, Addison-Wesley, Reading, 1967. Sakurai, J. J., Invariance principle and Elementary Particles, Princeton University Press, 1964.

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Simmons, G. F., Calculus Gems, McGraw-Hill, New York, 1992. Skyrme, T. H. R., Proc. Roy. Soc. Lon. A260, 127 (1961). Spiegel, M., and S. Lipschutz, Vector Analysis, Schaum’s Ouline Series, McGraw-Hill, New York, 2nd ed., 2009. Spiegel, M., S. Lipschutz, J. Schiller and D. Spellman, Complex Variables, Schaum’s Ouline Series, McGraw-Hill, New York, 2nd ed., 2009. Spiegel, Murray R., Fourier Analysis, Schaum’s Ouline Series, McGraw-Hill, New York, 1974. Strogatz, S. H., Nonlinear Dynamics and Chaos, Addison-Wesley, Reading, 1994. Struik, Dirk J., A Concise History of Mathematics, Dover, New York, 1957. Tolstov, Georgi P., Fourier Series, Dover, New York, 1976. Vaughn, Michael T., Introduction to Mathematical Physics, Wiley-VCH, Weinheim, 2007. Wald, R. M., General Relativity, University of Chicago Press, Chicago, 1984. Weinberg, Steve L., Gravitation and Cosmology, Wiley, New York, 1972. Whitham, G. B., Linear and Nonlinear Waves, Wiley, New York, 1973. Zabusky, N. J. and M. D. Kruskal, Phys. Rev. Lett. 15, 240 (1965). Zee, A., Quantum Field Theory in a Nutshell, Princeton University Press, Princeton, 2003.

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Name index

Abel (1802–1829) Niels Henrik 121, 296 Abramowitz (1915–1958) Milton 463, 472, 494, 690, 694 Airy (1801–1892) George Biddell 609–13 Amp`ere (1775–1836) Andr´e Marie 51, 55 Apollonius (ca. 260–200 BCE) of Perga 1 Archimedes (287–212 BCE) of Syracuse 1, 14, 643, 650, 659, 662 Arfken (1922– ) George B. 64, 364, 694 Argand (1768–1822) Jean Robert 503, 622 Aristotle (384–320 BCE) 650 B¨acklund (1845–1922) Albert Victor 440–3 Bernoulli (1700–1782) Daniel 248 Bernoulli (1654–1705) Jakob 662, 666 Bessel (1784–1846) Friedrich Wilhelm 293–4, 487–93 Bohr (1885–1962) Niels 244–5, 465 Born (1882–1970) Max 245, 247, 465, 466 Boussinesq (1842–1929) Joseph Valentin 415 Brahe (1546–1601) Tycho Ottesen 1 Briggs (1561–1630) Henry 666 Brillouin (1889–1969) L´eon Nicolas 610 Bromwich (1875–1929) Thomas John I’Anson 577, 579–4 Burgers (1895–1981) Johannes Martinus 427, 432, 439 Cabbibo (1935–2010) Nicola 185 Cardano (1501–1576) Gerolamo 623, 626 Carslaw (1870–1954) Horatio Scott 294, 692, 694 ` Cartan (1869–1951) Elie 165, 195–6 Cauchy (1789–1857) Augustin Louis 250, 458, 522, 524–6, 562, 597 Cayley (1821–1895) Arthur 200, 644 Chebyshev (1821–1894) Pafnuty Lvovich 290–1, 470–2 Christoffel (1829–1900) Elwin Bruno 224, 227–8 Chuquet (c. 1445–1488) Nicolas 666 Clairaut (1713–1765) Alexis Claude 248 Clifford (1845–1879) William Kingdon 170 Copernicus (1475–1543) Nicolas 2 Coriolis, de (1792–1843) Gaspard Gustave 129 Coulomb (1736–1806) Charles Augustin 55, 458 Cramer (1704–1752) Gabriel 92 Cronin (1931– ) James Watson 183

D’Alembert (1717–1783) Jean Le Rond 248, 358 Darwin (1809–1882) Charles Robert 379, 382 Davisson (1881–1958) Clinton Joseph 246 De Broglie (1875–1960) Louis C´esar Victoir Maurice 245–6 De Moivre (1667–1754) Abraham 504, 625 Descartes (1596–1650) Ren´e du Perron 1, 458, 623 De Vries (1866–1934) Gustav 323, 415–6 Dirac (1902–1984) Paul Adrien Maurice 138, 165–70, 172–8, 183, 186–7, 193, 258–61, 694 Dirichlet (1805–1859) Peter Gustav Lejeune 496 Doppler (1803–1853) Christian 151, 155 Drazin (1934–2002) Philip Gerald 424, 450, 695 Einstein (1879–1955) Albert 3, 124, 139–40, 146, 158, 168, 244–6, 695 Erd´elyi (1908–1977) Arthur 314, 695 Euclid (ca. 330–275 BCE) of Alexandria 1, 631 Euler (1707–1783) Leonhard 85, 162–3, 248, 458, 502–3, 630, 655–6, 660–3, 680 Faraday (1791–1867) Michael 2, 55, 301, 310 Feigenbaum (1944– ) Mitchell Jay 373, 382, 386–90, 395–8, 695 Feshbach (1917–2000) Herman 64, 202, 364, 496–8, 607, 696 Feynman (1918–1988) Richard Phillips 178, 663 Fischer (1875–1954) Ernst Sigismund 295 Fisher (1890–1962) Ronald Aylmer 431–2 Fitch (1923– ) Val Logsdon 183 Fourier (1768–1830) Jean Baptiste Joseph 245, 247–50, 265, 300–1, 458 Fresnel (1788–1827) Augustin Jean 246, 590 Frobenius (1849–1917) Georg Ferdinand 161, 342 Galilei (1564–1642) Galileo 1, 3, 121 Gardner (1924– ) Clifford Spear 423–4 Gauss (1777–1855) Carl Friedrich 44–6, 301, 502, 623, 643–4 Germer (1896–1971) Lester Halbert 246 Gibbs (1839–1903) Joseph Willard 202, 297, 627 Gordon (1893–1939) Walter 124, 166, 426 Gram (1850–1916) Jorgen Pedersen 288 Green (1793–1841) George 46–7, 53, 270–2, 309–12

700

Name index

Hamilton (1805–1865) William Rowan 14, 138, 159, 166 Heaviside (1850–1925) Oliver 571, 627 Heisenberg (1901–1976) Werner 245, 247, 276, 278, 465 Helmholtz, von (1821–1894) Hermann 13, 53–4, 122, 278, 304–10 H´enon (1931– ) Michel 392–8, 685 Hermite (1822–1901) Charles 80, 101, 108, 290, 350, 470–4 Heron (or Hero) (c. 10–70 AD) of Alexandria 623 Hertz (1857–1894) Heinrich Rudolph 2, 302 Hilbert (1862–1943) David 294–5, 587 Hirota, Ryogo 444–8, 695 Hooke (1635–1703) Robert 204, 667 Hopf (1902–1983) Eberhard 377–81, 411, 413–5, 439–41 Huggins (1824–1910) Sir William 155 Huygens (1629–1695) Christiaan 246 Jackson (1925– ) John David 216 Jacobi (1804–1851) Carl Gustav Jacob 14, 59–60, 75, 129, 221, 394, 478 Jeans (1877–1946) Sir James 244 Jordan (1902–1980) Ernst Pascual 245, 247, 465 Jordan (1838–1921) Marie Ennemond Camille 552–3 Kadomtsev (1928–1998) Boris Borisovich 451 Kepler (1571–1630) Johannes 1, 20 Klein (1849–1925) Felix 200 Klein (1908–1992) Morris 695 Klein (1894–1977) Oskar Benjamin 124, 166, 426 Kobayashi (1944– ) Makoto 183–5 Korteweg (1848–1941) Diederik Johannes 323, 415–6 Kramers (1894–1952) Hendrik Anthony 587, 589, 610 Kronecker (1823–1952) Leopold 5, 17, 206 Kronig (1904–1995) Ralph de Laer 587, 589 Kruskal (1925– ) Martin David 415, 697 Lagrange (1736–1813) Joseph Louis 248, 458, 630 Laguerre (1834–1886) Edmond Nicolas 291, 470–2, 478 Lambert (1728–1777) Johann Heinrich 663 Landau (1928–1968) Lev Davidovitch 225, 375, 403, 696 Laplace (1749–1827) Pierre Simon 24, 88, 248, 320, 351, 359, 483, 528–9, 572–3, 590–4, 610

Laurent (1813–1854) Pierre Alphonse 488, 541–2 Lebesgue (1875–1941) Henri L´eon 250 Lee (1926– ) Tsung-Dao 182–3 Legendre (1752–1833) Adrien Marie 248, 283–4, 347–9, 470–2, 481–3 Leibniz (1646–1716) Gottfried Wilhelm 1, 20, 247, 381, 548, 630, 632, 636, 637 Levi-Civita (1873–1941) Tullio 15, 87 L’Hospital, de (1661–1704) Guillaume Francois Antoine 20, 490, 654 Lie (1842–1899) Marius Sophus 132–3, 189, 238 Lifshitz (1915–1985) Evgeny Mikhailovich 225, 403, 696 Liouville (1809–1882) Joseph 450–1, 495–6, 523, 608, 610 Lorentz (1853–1928) Hendrik Antoon 101, 138–46, 188–93, 197–8, 300–1 Lorenz (1917–2008) Edward Norton 392–3, 696 Lyapunov (1857–1918) Aleksandr Mikhailovich 388–9, 398 Maclaurin (1698–1746) Colin 536, 539, 596 Madhava (c. 1350–1425) of Kerala 651 Majorana (1906–1938) Ettore 186–7 Malthus (1766–1834) Thomas Robert 382 Martens, Petrus C. H. 374, 392, 696 Maskawa (1940– ) Toshihide 183–5 ´ Mathieu (1835–1890) Emile L´eonard 403–5, 414 Maxwell (1831–1879) James Clerk 3, 56, 299–305, 351 May, Baron (1938– ) Robert McCredie 382, 696 Michelson (1852–1931) Albert Abraham 2 Minkowski (1864–1909) Hermann 140–5, 695 Miura (c. 1938– ), Robert M. 449, 696 Morley (1838–1923) Edward Williams 2 Morse (1903–1985) Philip McCord 64, 202, 364, 496–8, 607, 696 Napier (1550–1617) John 666, 669 Neumann (1832–1925) Carl Gottfried 490–5, 496 Newton (1642–1727) Sir Isaac 1–3, 20, 121, 246, 247, 417, 630, 644 Parseval, des Chˆenes (1755–1836) Marc-Anthoine 294 Pauli (1900–1958) Wolfgang 81–2, 132–3, 159, 161–2, 165–70, 177, 190–2 Pearson, Carl E. 688, 696

Name index Penrose (1931– ) Roger 200, 626, 696 Peregrine (1938–2007) Howell 423 Petviashvili (1936–1993) Vladomir Iosifovich 451 ´ Picard (1856–1941) Charles Emile 523 Planck (1858–1947) Max Karl Ernest Ludwig 123, 155, 244, 465 Poincar´e (1854–1912) Jules Henri 138 Poisson (1781–1840) Sim´eon Denis 54–5, 204, 308–9, 364, 458 Pound (1919–2010) Robert Vivian 158, 696 Ptolemy (c. 100–170) 2, 458 Rayleigh, Lord: Strutt (1875–1947) Robert John 244, 415 Rebka (1931– ) Glen Anderson, Jr. 158, 696 Riccati (1676–1754) Jacopo Francesco 438, 456 Ricci-Curbastro (1850–1925) Gregorio 221 Riemann (1826–1866) Georg Friedrich Bernhard 217, 228, 231–2, 250, 458, 511–23, 528–9, 600, 607, 612, 637, 652 Riesz (1880–1956) Frigyes 295 Rodrigues (1794–1851) Olinde 464, 470, 476, 482 Sakurai (1933–1982) 696 Schl¨afli (1814–1895) Ludwig 527 Schmidt (1876–1959) Erhard 288 Schr¨odinger (1887–1961) Erwin 124, 245–6, 465–6 Schwarz (1843–1921) Hermann Amandus 278 Scott Russell (1808–1882) John 415

701

Skyrme (1922–1987) Tony Hilton Royle 415, 434, 696, 697 Stegun (1919–2008) Irene A. 463, 472, 494, 690, 694 Stifel (1487–1567) Michael 666 Stirling (1692–1770) James 658 Stokes (1819–1903) George Gabriel 50–2, 69, 526, 599, 607 Sylvester (1814–1897) James Joseph 202, 643 Taylor (1685–1731) Brook 114, 118, 260, 327–9, 384, 447–8, 487–8, 533–44, 600, 664 Thomson (1892–1975) Sir George Paget 246 Tolstov, Georgi P. 313, 697 Voigt (1850–1919) Woldemar 155 Wessel (1745–1818) Casper 503, 622 Weyl (1885–1955) Hermann Klaus Hugo 179–83, 695 Whitham (1927– ) Gerald Beresford 697 Wien (1864–1928) Wilhelm Carl Werner Otto Fritz Franz 244 Wigner (1902–1995) Eugene Paul 175 Wronski (Hoene-Wro´nski) (1776–1853) J´ozef Maria 330 Yang (1922– ) Chen-Ning 182–3 Young (1773–1829) Thomas 246 Zabusky (1929– ) Norman J. 415, 697 Zeno (c. 490–430 BCE) of Elea 650

Subject index

Abelian group 121 addition theorem, Bessel functions 493 spherical harmonics 486 adjoint, see Hermitian conjugate Airy functions 609–12 algebra: Clifford 170 complex 503–5, 620–3 Lie 132–3, 189 matrix 79–80 noncommuting 82 quaternion 159–64 tensor 222–3 vector 4–19, 127, 201, 207, 281, 628–9 al-jabr (algebra) 14 Ampere’s, circuital law 51 law 51, 55 analytic continuation 533, 537–9, 586, 613, 657 analytic function 522; see also function of a complex variable angular momentum 97, 176, 188–9, 192–4, 212–4, 245, 408, 465 antikink 426–9, 433 antiunitary operator for time reversal 175 approximation: asymptotic expansions 590–600 Fourier series 297 linear 322 steepest descent 597–600 Stirling’s 658 WKB 607–10 Arabic mathematicians 1, 14 associated Legendre equation 362, 481 functions 363 polynomials 362, 482–3 asymptotic expansions 590–600 asymptotic series 590–3, 657–9 Airy functions 610–3 Bessel functions 606–7 stationary phase 596 steepest descent 597–600 attractor 376, 400 basin 380, 393 destabilized when driven 401, 411 fixed point 383–4, 386 limit cycle 377

logistic map 384, 389 point 376–7 strange 374, 392–5, 397–8, 413 axial (pseudo) vector 210 B¨acklund transformation 440–3 BAC rule 12–13, 17 operator form 31, 54 Bernoulli numbers 666, 669 Bessel equation 345, 362, 489–92 Bessel functions 362, 487–9, 666 addition theorem 493 generating function 487–8, 665–6 integral representation 544–5, 576, 582, 607 negative order 488 order 487 raising/lowering operators 489, 492 recursion formula 489, 492 spherical 362–3, 492 Wronskian 490 Bessel inequality 293 bifurcation 376–9 diagram for H`enon map 395–6 logistic map 385–7 parametric pendulum 411–14 Hopf 377–8, 413 period doubling 385–6, 395–8, 410–14 pitchfork 376 saddle-node 380 binomial coefficient 508, 549 Bohr’s atomic model 244–5 bore (tidal) 425 Born interpretation of wave function 466 boson 176, 191–2, 238 boundary conditions 247, 280, 329, 338–9, 350, 367 Dirichlet 496 first-order differential eq. (DE) 325–7 mixed 497 Neumann 496 partial DE 353–4, 364–5 second-order DE 329, 354–6, 367, 565, 575 branch, cut/line 511 of a function 510–20 point 511, 523 principal (ln function) 516

Subject index breather 438 Bromwich integral 579–82 Cabbibo angle 185 calculus: differential 20–30, 630–6 integral 31–47, 636–42 residue 547–60 vector 20–71 Cantor set 397–8 catastrophe 374 Cauchy, integral formula 526, 584, 601 integral theorem 525 principal value of integral 562 Cauchy–Riemann conditions 449, 522, 524–29 causality 272, 312, 570 Cayley-Klein parameters 200 centrifugal force 129 chain rule of differential calculus 31, 632 change of variables: complex integration 559–60 curvilinear coordinates 56–62 differential eq. 324–5 Dirac δ function 261 Fourier series 256–7 integration 638 Jacobian 59–60, 65 nonlinear transformation 440 chaos, in H´enon map 396 in logistic map 386, 388–90 in parametric pendulum 414 period-doubling route to 386–90 characteristic eq. 99 Chebyshev polynomials 290–1, 470–2, 478–9 chirality 179–82 chiral projection 179–80 Christoffel symbol 224, 227 circulation 42, 48, 50, 51, 69 Clifford algebra 170 cofactor matrix 90–1, 647–8 coherence 139, 373 commutation relation (commutator) 119–20, 126, 132–3, 161–4, 190–1, 278 conjugate variables 275–8 Jacobi identity 129 Lie algebra 132–3, 189 Lorentz group 190–1 Pauli matrices 89, 132 quaternions 159–60 rotation operators 126, 132, 163–4 space translation 120 time displacement 119 wave mechanics 278–9

703

completeness relation 7, 294, 486 complex algebra 620–6 complex differentiation 521–2 complexification of, complex numbers 160 real coordinates 195–6 complex integration 524–7, 550–60, 579–82 convergence factor 571 complex number 502–4, 620–3 complex conjugate 504 complexification 160 imaginary/real part 502, 620 modulus 503 phase 503 polar (Wessel-Argand) form 503, 621–2 complex plane 502–3, 621–2 complex potential 530 complex variable 502–8; see also function of a complex variable component of a vector 4–5, 627–8 computer algebra 670–6 conditions, sufficient and necessary 295–6, 522 conformal mapping 530–1 conjugate form of an operator 277–8 conjugate transformation 266 conjugate variables 275–8 constant of motion (conserved quantity) 302, 420–1 continuity equation (conserved density or current) 45–6, 55, 302, 420–1, 423, 641 contour closure, Jordan’s lemma 552–4 large semi-circle 551–2 ln z factor 558–9 return path 554–7 unit circle 559–60 contour, integration 525–7, 551–60, 579–82 lines 22 pole, on 561–2 pole moved off 563–4 convergence: absolute 652 mean-square 292 of Fourier series 295–8 pointwise 653 radius of 534 semi 653 uniform 249, 534, 655 convergence test for infinite series 653–5 alternating series 655 comparison 654–5 integral 653–4 ratio 534, 591, 652, 655 root 655 convolution 270, 277, 308

704

Subject index

convolution theorem 270–1 coordinate (or x-) representation 273–6 coordinate system, curvilinear 56–71 cylindrical 61 spherical 57–8, 60–2, 66–70 Coriolis force 129 correspondence principle 245 Coulomb potential 363–5, 458–9 CP-violating KM phases 183–5 Cramer’s rule 92 cross product, see vector product curl (rotation) 26–7, 49–51 curvilinear coordinates 67–8 cylindrical coordinates 73–4 Helmholtz’s theorem 54–5 Maxwell eqs. 299, 351 spherical coordinates 74–5 Stokes’s theorem 50–2 curve, simply connected 526 curvilinear coordinates 56–62, 73 table of 63 cylindrical coordinates 61 data mining 408–9 degenerate eigenvalues 99, 103, 108–9, 113, 122, 359 degree associated Legendre polynomials 481 Cartesian tensors 241 differential equation 321 Legendre polynomials 283, 349 multispinors 192 polynomials 349 spherical harmonics 213–4 square matrix 79 del operator, see gradient operator delta-function source 269–70, 337–41, 364–7, 458–9 de Moirve theorem 504, 625 determinant 9–10, 87–90, 647–49 cofactor 88–90, 647–8 Laplace development 88–92, 647 minor 89 order 87–8 secular (characteristic) 98 diagonalization of a square matrix 109–11 differentiability 26, 249, 522, 534, 568, 593, 631, 651 differential equation (DE) 319–67, 400–51 boundary conditions 327–9, 339, 350, 354–6, 496–7, 565, 574–5 complementary function 333, 400 coupled 56, 104, 299, 351

degree 321 eigenfunction expansion 351–7, 370–1 eigenvalue 118–20, 166, 167, 169, 181–2, 246–7, 349–50, 352, 354, 361–2, 367, 497–8 first-order 324–8 homogeneous linear 322, 325, 328, 332 inhomogeneous linear 322–3, 326–7, 333–5, 565–7, 574 initial condition 340, 354–6, 376, 388–9, 392, 400, 403, 407–14 in physics 122–4, 319–21, 350–1 irregular solution 366 order 321 ordinary 321 partial 56, 321; see also partial differential eq. separable 324 differential operator Hermitian (self-adjoint) 279–80, 350, 496–7 infinitesimal generator 118, 121–8, 131–3, 166, 169, 188–91, 194, 367 inverse, see Green function ladder (raising/lowering) operators 469, 489, 492–3, 500 quantum mechanics 123–4, 246 vector, see vector differential operators differentiation 630–6 diffusion equation 48, 175, 320, 427, 441 dimension of space 3–4 Dirac δ function 259–64, 316, 365–6, 486, 563–4 spherical coordinates 367, 385–6 Dirac, bracket notation 176, 193 equation 165–9, 172–8, 186–7 negative-energy state 177 matrices 169–71 sea 177 spinors 169, 171, 174, 179–80 directional derivative 23–4 direction cosine 6, 66, 77–8, 162, 164, 194, 627–8 direct product 131, 139, 170, 191–2, 200, 208, 211–12, 223 discontinuity, jump 251, 296–7, 338–9, 568–70, 585 kink 296, 338–9 dispersion, of light 587 of values 275 term in differential eqs. 417 dispersion relations 417, 588–90 Kramers–Kronig 588–90 distribution (Dirac δ function) 261 divergence 26–7, 44, 53–5, 68–70, 351 curvilinear coordinates 68–9

Subject index cylindrical coordinates 74 Gauss’s theorem 43–7 Helmholtz’s theorem 54–5 Maxwell equations 299, 351 spherical coordinates 69–70, 74 Doppler effect 151, 154–5 double factorial 492 dyadic 28, 200–6 dyad 200 field 28 stress 202–6 unit 28, 202 eigenfunction 119–20, 123, 246–7, 351–7, 362, 495–8 expansion 352–7, 370–1, 497–8 eigenvalue 98–9, 104–5, 122, 176, 181–3 boundary conditions 354 energy 181–2, 246–7, 467 finite series 347–50, 362 Hermitian matrix 101–3, 108–10 Hermitian operator 279–80, 350, 496–7 see also differential equation; eigenvalue Jacobian matrix 394–5 periodicity 361 spectrum 370, 496–8 continuous 115 eigenvalue problem: differential eq. 122–4, 349–50, 352 mass term in 104–6, 112 matrix 98–112 Sturm–Liouville theory 495–8 eigenvector 103, 105, 110, 112, 394–7 Einstein, energy–mom. relation 124, 150 postulates for special relativity 139 summation convention 168 twin puzzle 146 velocity addition 143 elasticity tensor (dyadic) 204 element: arc length 34, 59 group 121 line 34, 58–9, 204 matrix 79, 247, 644 surface 35, 59, 60, 202 tensor 218, 223 volume 59, 60 entire function 523–4 equation: characteristic 99 continuity 45–6, 55, 302, 420–1, 423, 641 differential, see differential equation diffusion 48, 175, 320, 427, 441

705

driven damped oscillator: linear 271–2, 339–41, 401–2 nonlinear 401 field 345, 351 indicial 343 integral 320–1 matrix eigenvalue 48–9, 104–7 nonlinear difference 382 of motion 76, 116, 122–4, 271, 319–20, 351, 403, 565 of state 122, 319–21, 345, 351 parametric damped oscillator linear 403–5 nonlinear 403, 407–15 partial differential, see partial differential eq. quantum oscillator 466–8 secular 98–9, 105 simultaneous algebraic 91–6 wave, see wave equation equation by name: associated Legendre 362, 481 Bessel 345, 362–3, 368, 489–90 Burgers 427, 439–42 Fisher 431–2 Helmholtz, see Helmholtz equation Hermite 466–9 Kadomtsev–Petviashvili 451 Klein–Gordon 124, 426 Korteweg–de Vries 323, 416–21, 423–4 Laplace, see Laplace equation Legendre 347–50 Liouville 450–1, 496 Mathieu 403, 405–7 Maxwell, see Maxwell equations Newton’s 76, 271, 319 Peregrine 423 Poisson 54–5, 308–9, 320, 364, 458 Schr¨odinger 246, 466 sine-Gordon 323, 426–7, 435–8 spherical Bessel 362–3, 366, 491–4 Sturm–Liouville 496–9 equilibrium, configurations 320 in elastic bodies 202–4 in nonlinear systems 374–6, 384–7 in pendula 322 equilibrium point: stable (attractor) 376, 394–8 unstable (repellor) 384–7 equipotential 22–3, 529 error function (erfc) 584, 605, 659 essential singularity 523 Euler, angles 85, 130, 198–200 formula 163, 164, 503, 621–2, 634, 663

706

Subject index

expectation value 275 exponential function 662–5 exponential population growth 382 factorial 489–90 double 492 function, see gamma function Stirling’s series 658 Faraday’s induction 55, 310 Feigenbaum: constant (non-universal) 398 constant (universal) 373–4, 386–8, 398 point in H`enon map 397–8 logistic map 386–9 parametric pendulum 411–14 fermion 153, 176, 191–2 field 2, 3 covector 222, 225 dyadic 28 in physics 2, 319–20 scalar 20 tensor 211, 223–32 vector 21, 25–6 field equation 345, 351 fixed point (attractor) 383–4, 386 Fischer–Riesz theorem 295 flow 42–3 flow curve 23 flux 42–3 Fourier series (representation) 244–98 arbitrary interval 256–7 coefficient 250–3 complex 258–9 convergence 295–8 double 353 even (odd) part 253–4 generalized 282–4 modified 253–5 periodic extension 255–6 summability 298 summation by arithmetic mean 298 Fourier transform 265–72, 300, 305–10 3D 305–6 4D 300 convolution theorem 270–1, 277, 308 Helmholtz decomposition theorem 309–10 inversion formula 265, 300, 305, 572 Maxwell eqs. 300–2 properties 267 fractal 373–4, 387–8, 395–8 Frobenius series 342–50 indicial equation 343

parity property 344–5 recurrence relation 343 function 20, 25, 458–98, 502–604 absolutely integrable 251 analytic 522–31 delta, see Dirac δ function differentiable 20, 26, 249, 521–2, 534, 631, 651 entire 522–3 exponential 662–6 harmonic 483, 449, 528 higher transcendental 362, 665 logarithmic 666–7 piecewise continuous 294, 296–8 piecewise smooth 296 regular 522, 541 special 458–98 step 272, 336, 338, 571 stream 529–30 transcendental 362, 665 function of a complex variable 502–614 analytic 522–31 analytic continuation 537–9 branch 510–20, 523 branch cut/line 511 branch point 511, 523 construction from singularities 584–6 continuum contribution 585 contour integration 525–6, 551–60, 579–82 differentiation 521–2, 534 dispersion relations 586–9 entire 522–3 Green function 565–70 harmonic function 528–30 Hilbert transform 587 Jordan’s lemma 552 Laplace transform 571–82 Laurent series 488, 540–5 multivalued 505–20 pole contribution 585 principal part 541 principal-value integral 562 regular part 541 residue calculus 547–60 Riemann surface 509–20; see also Riemann surface singularity 523; see also singularity Taylor series 533–6 gamma (Γ) function 490, 595–6, 602, 605, 639 gamma (Dirac) matrix 169–71 γ0 169, 181 γ5 170, 181

Subject index gauge transformation 102 Gauss’s law of electrostatics 44–5 Gauss’s theorem 44–47 operator form 46 generalized coordinates 57–63, 65–70 generalized work 41 generating function 448, 459–64, 665–6 associated Legendre polynomials 482 Bessel 487–8, 544–5, 665–6 Chebyshev polynomials 470, 475–6 Hermite polynomials 464–5, 470, 474, 666 Hirota’s differential operator 448 Laguerre polynomials 470, 474–5, 666 Legendre polynomials 459–61, 470 generator 118, 120, 122, 188–91, 230–1, 367 continuous group 131 Lorentz group 188–91 parallel transport 231 rotation 125–7 space translation 119–20, 367 SU(2) 131–5 time displacement 117–8 geometric series 508, 591, 650–1; see also infinite series Gibbs phenomenon 297 gradient operator 21, 41–2 curvilinear coordinates 66, 70, 73 spherical coordinates 66, 70, 74 Gram-Schmidt orthogonalization 288–91 Green function 269–72, 308–9, 310–2, 336–41, 364–7, 566–72 advanced 311–2, 570 driven oscillator eq. 569–70 Helmholtz eq. 366–7, 566–70 in/out-going spherical wave 366–7, 568–9 Laplace equation 364–5 partial differential eq. 364–7 Poisson equation 364–6 principal value 567 retarded 311–2, 570 standing wave 367, 569 wave equation 310–2 Green’s theorem 46 in the plane 47, 53 group 77, 120–34, 163–4, 167, 170, 184, 188–98, 211–4 Abelian 121 continuous 131–2 cover 198 definition 120–1 GL(2,C) 198 GL(n) 129 Lie 132–3 Lorentz 139, 188–91

707

matrix 129–34 non-Abelian 121 O(1, 3) 197 O(3) 125–26 O(n) 129–30 permutation symmetry 213 quaternion 163–4 representation (rep) 191–3 adjoint 167 fundamental 167 irreducible (irrep) 193, 208, 212–4 spinor 191, 197–8 SU(2) 191–3 tensor 211–4 SL(2,C) 195, 198 SL(n,C) 129 SO(1, 3) 197–8 SO(n) 130, 184 SU(2) ⊗ S U(2) 170 SU(n) 130–1 subgroup 130, 184, 197–8 U(n) 130 transformation 121 Hamiltonian 123–4, 166–9, 174, 181–2, 246–7, 465–7 Hankel function 367 harmonic, function 447, 484, 528–30, 533 polynomials 484–5 spherical 213–4, 306, 484–6 harmonic oscillator 271, 334–6, 401–3, 466–8 helicity 181, 186 Helmholtz decomposition theorem 53–6, 308–10 Helmholtz equation 122, 247–8, 359, 366–7, 371–2, 487, 490, 495, 564–7 H´enon map 392–7 stretch and fold 393, 397–8 Hermite equation 466–9 Hermite polynomials 290–1, 464–74 ladder operators 469 Hermitian conjugate 80 Hermitian matrix 80, 101 eigen values/vectors 108–12 Hermitian operator 280, 350, 496–8 eigen values/functions 279–80, 497 Hilbert space 294–5 Hilbert transform 587 Hirota’s superposition of solitons 444–9 2-solitons 447 bilinear form 444–6 derivatives 445 direct method 444–9 series 445

708

Subject index

Hopf bifurcation 377–8, 413–5 resonance 378–80 hysteresis 375 impulsive force 271, 340 indicial equation 343 inertia, matrix of 97 principal axes of 98–100 principal moment of 98–100 inertial frame 3, 128, 140–2, 146, 148–9 infinite series: Bessel functions as 487–8 exponential function as 665 Fourier 249–50 Frobenius 342–4, 349 Gardner 424 generalized Fourier 282–4 generating function 459; see also generating function geometric 650–1 Hirota 445 Laurent 488, 541 Legendre 283–6 Maclaurin 536, 539, 606 Riemann zeta (ζ) function as 652 Taylor 533–6 tutorial 650–9 uniform convergence 249, 655 infinitesimal generator 118–20, 122; see also generator initial conditions: for differential eqs., see differential equation sensitivity to 388–9, 413–4 steady state 377, 402, 408 transient 402, 409 inner product 273–7, 281–3, 294, 473–4, 498 inner product space 282, 294 integrability absolute 251, 297 square 251, 294–5, 467–8 integral contour 525–7, 551–60, 579–82 operator 46, 52, 265, 320, 571–2 path-dependent 31–4 principal-value 562 vector line 34–5 vector surface 35–41 integral representation of functions 270, 544–5 Airy function 610–1 Bessel function 544–5, 607 gamma (Γ) function 595 Legendre polynomials 527–8

integral theorem: Cauchy’s 526 Gauss’s 44–6 Green’s 46 Green’s (in the plane) 47, 53 Stokes’s 50–2 Internet resources 689–93 intrinsic spin 133, 192 invariance principle 122, 141, 169, 172–8, 320–1 inverse: complex number 620–1, 623 differential operator, see Green function Fourier transform 265, 300, 305 group element 121 Laplace transform 572, 578–83 matrix 90–1 quaternion 162 series 600–2 irreducible tensor 213 irrotational field 27, 42 isolated singular point 523, 547–9 Jacobi identity 14, 129 Jacobi polynomials 478–9 Jacobian, determinant 59–60, 64–5, 75, 221, 394 eigen value/vector 394–5 matrix 394–5 Jordan’s lemma 552 kink, discontinuity 296, 338–9 soliton 425–9, 433, 437 Klein-Gordon equation 124, 426 Kronecker δ symbol 4–5 ladder (raising/lowering) operators 469, 489, 492, 500 Laguerre polynomials 470–4, 478 Landau equation 374–5, 696 Laplace development (determinant) 88–92, 647 Laplace equation 320, 359, 363–4, 371, 483, 528–9 Laplace transform 571–82 Bromwich integral 579–82 convergence factor 571 inverse 577–82 inversion formula 572 properties 573–4 Laplacian 24, 70–5, 359–60 curvilinear coordinates 70, 73 cylindrical coordinate 73–4, 360 separation of variables 359–64 spherical coordinates 70, 75, 360 Laurent series 487–8, 541–5

Subject index law: Ampere’s 51, 55 Coulomb’s 45, 55, 307 Faraday’s 55, 301, 310 Gauss’s 44, 301 Hooke’s 204 Newton’s 76, 319 of physics 76–7, 121–2 of universal gravitation 1–2, 20 Rayleigh–Jeans 244 Legendre: associated L. equation 362 assoc. L. function, 2nd kind 363 assoc. L. polynomials 481–4 coefficients 284 equation 347 function of the second kind 349 polynomials 281–5, 291, 349–50, 459–63, 471–2 series 283, 383–4 Leibniz formula 548 Levi-Civita permutation symbol 15–17, 201, 206 n indices 87, 648 l’Hospital’s rule 490, 654, 656 Lie algebra 132–3, 189, 238 light 2–3, 138–9, 244–6 Doppler effect 151, 154–5 quantum 244 speed 2, 3, 139, 143 wave-particle nature 246 light cone 148, 195–8 spinor (Cartan) 195–8 vector 195 linear differential equation 270, 322–47, 664 boundary conditions 329, 333 complementary function 333 Frobenius series solution 342–4, 349 indicial equation 343 recurrence relation 343 linearly independent solutions 329–30, 333 linearity property 321–3 particular solution 322, 326, 333–7 second homogeneous solution 332–3 second-order 328–47 singular point 347, 496–7 source 269–71, 565 uniqueness of solutions 327–8 variation of constant 326, 334 linearizing nonlinear differential equation: parametric pendulum 403 pendulum 322–3, 401 rational transformation 439–40

709

sine-Gordon 426 stability properties 394 linear operator 321–3, 344, 349 linear superposition 321, 353, 439, 445–6, 665 line integral 33, 34–5 line of force 23, 529 Liouville theorem 523 logarithmic, function 666–7 scale 667 logistic map 382–90 2-cycle (period 2) 383–5 bifurcation diagram 385–7 chaos in 388–90 Feigenbaum point 385–6 fixed point (attractor) 383–5 Lyapunov exponent 388–9 period-doubling bifurcation 384–6 Lorentz: group 139, 188–91 invariance 145, 169 scalar 144, 150–3, 168–9, 300 transformation 101, 141–3, 145–7 vector 143–4, 169 lowering operator 469, 489, 492 Lyapunov exponent 388–9 Maclaurin series 536, 539, 606 map 382–98 area-conserving 398 baker’s 398–9 conformal (angle-preserving) 530–1 function of a complex variable 505–6 H´enon 392–8 logistic 382–90 Poincar´e (return) 409 quadratic 388, 398 Maple 430, 670, 676 Mathematica 430, 670–6, 677–87 Mathieu, equation 403 function 403–5 matrix 78–113, 643–9 adjoint (Hermitian conjugate) 80 algebra 79–80 angular momentum 97, 188–9 anti-Hermitian 86 determinant 10, 87–90, 647–9 diagonalization 109–11 Dirac 169–71 eigenvalue, equation 98 problem 98–112 function of a matrix 111, 113 Hermitian 80, 101, 108–11 inversion 90–2, 648

710

Subject index

matrix (cont.) nonsingular (invertible) 90 normal 113 operations 79–80 operator 76, 78, 82–5, 126, 169 orthogonal 80, 126, 141, 184 Pauli spin 81, 133, 159, 161, 170, 191 rank 96 rotation 78–9, 125–6, 133 singular 80 skew-Hermitian 86 skew-symmetric 86 square, order or degree 79 trace 80, 99, 102, 113 transformation of coordinates 78, 83–6 transformation of matrix 109–10, 113 unitary 80, 109–11 matrix mechanics 245, 247 matter wave 245–6 Maxwell equations 2, 56, 299–301, 351 Maxwell’s displacement current 55, 301 mean-square convergence 292–5 mean-square error (deviation) 292–3 metric, coefficient 59–60 tensor/matrix 168, 197, 218–22 Minkowski space 140–4, 239 minor (determinant) 87–9 modulus (complex number) 503 modulus (elastic), bulk 205 shear 206 Young’s 205 momentum in: classical mechanics 319, 408, 415, 421 quantum mechanics 123, 168–77, 181, 186, 245–6, 276, 465 relativity 150–6 soliton 415, 421 momentum (or k) representation 273–6 multipole expansion 120, 459 N-dim. scalar and vector products 210 Neumann function 490 spherical 492–4 Newtonian mechanics 2–4, 276, 420, 641 Newton’s equation of motion 76, 271, 319 Newton’s law of universal gravitation 1, 20, 121 Newton’s laws of mechanics 1, 121 nodal, lines 356, 359 points 355 nonlinear differential equations driven pendula 401 parametric pendulum 403, 407–15 bifurcation 411–14

bifurcation diagram 412, 414 chaotic regions 414 soliton (traveling pulse) 416–20 coherence 416–8 conservation law 420–1 Korteweg-de Vries eq. 416–24 traveling kink 425–9 Burgers eq. 427–9, 432, 439, 441–2 finite power series 429–31 sine-Gordon eq. 426–7, 434–8, 440, 442 nonlinear instabilities 374–8, 384–98, 410–15 nonlinear map, H´enon 392–8 logistic 382–90 nonlinear superposition of solitons: B¨acklund transformation 440–3 Hirota’s direct method 443–8 pseudo Lorentz transformation 438 rational linearization 439–40 normal (natural) coordinate 106 normal (eigen) mode 248, 354–6 nuclear radioactive decay 667 O (big O order of magnitude) 592, 658 operator: creation/destruction 469 differential 21, 24, 117, 119; see also differential operator frequency 119 gradient, see gradient operator Hermitian 279–80, 350, 496–7 integral 46–52, 265, 269–72, 320, 571–2 ladder 469, 489, 492, 500 linear 321–3, 344, 349 Lorentz transformation 188–91 matrix 76, 78, 82–5, 126, 169 momentum 123–4, 169 noncommuting 82, 278 propagation 123 quantum mechanical 123–4, 166, 469 rotation, matrix 78, 125–6, 133–4 quaternion 161–4 self-adjoint, see Hermitian operator space translation 120, 122 spin matrices 81, 198 Sturm-Liuville 495–8 time displacement 117 order: associated Legendre polynomials 481 Bessel function 362, 487–8 determinant 87–8 differential equation 321 permutation (Levi-Civita) symbol 89, 209 pole singularity 523

Subject index spherical harmonics 213, 481–6 square matrix 79 orthogonal complement space 94 orthogonal curvilinear coordinates 60–3, 530 vector differential operators in 65–71 orthogonal, functions 281–91, 497 matrix 80, 126, 141, 184 orthogonal polynomials 287–91, 470–80, 499 associated Legendre 363, 481–3 Chebyshev 290–1 classical 476–80 degree 213, 348, 428, 460, 469, 477–81 differential equation 472, 480 differential relation 471, 479 generating function 470 Hermite 290–91, 464–74 Laguerre 291, 470–4, 478 Legendre 281–5, 291, 349–50, 459–63, 471–2 recursion formula 471 Rodrigues formula 464, 470, 476–8, 481 solid spherical harmonics 484–5 spherical harmonics 213–4, 483–6 weight function 287–91, 469–70, 477, 496 orthogonal transformation 110, 113, 126, 207 orthogonal vectors 4, 60, 103, 108, 288 orthogonality relation: associated Legendre polynomials 481 functions 281, 287, 469 Hermite polynomials 290 Legendre polynomials 283, 462 matrices 80, 129–30, 184 spherical harmonics 485 vectors 4, 60, 102, 108–9, 112, 288 orthogonalization, Gram-Schmidt 288 oscillator classical, see pendulum Hopf 376–80, 413 quantum 244, 465–8 outflow 44 parity 24, 173, 344, 460 C (charge conjugation) 177, 186–7 conserving operator 24, 344–6, 360 even/odd permutation 15, 87 intrinsic 173–4 of operators 173, 217 of physical quantities 216, 345 of pseudotensors 210–11 of solutions of diff. eq. 344–50, 461, 485 of spinors 174 of tensors 211 operator 82–3, 173 orbital 174, 460 violation 182, 184

Parseval, equation (identity) 294 theorem 294 partial differential eq. 123, 350–67, 416 boundary conditions 354 eigenvalue equation 353 Green function: Helmholtz equation 366–7 Poisson equation 364–6 initial conditions 355–6 in physics 350–1 separation constant 352, 360–2 separation of variables 352–3; see also separation of variables superposition of solutions 353 partial differentiation 632–4 partial fraction 546, 548, 550, 578 particular solution of DE 322–3, 326, 333–6 path integral 32–35 Pauli spin matrices 81, 132–3, 159, 161, 170, 191 pendulum: linear 271–2, 339–41, 401–2, 403–5 nonlinear 401, 403, 407–15 period-doubling bifurcation cascade in: H`enon map 395–8 logistic map 384–90 parametric pendulum 410–14 periodicity condition 230 period-three (3-cycle) 386, 390, 413–5 permutation, cyclic 11 symbol 15–7, 87 phase: arbitrary overall 102, 173–8, 184 complex number 504, 623 gauge field 4 KM phases (CP violation) 183–5 lag in oscillator response 379 stationary phase method 597 U(1) group 130–1 velocity 116, 587–8 wave front 116, 367, 567–8 WKB phase function 607–8 phase space 408–11, 413 plot (orbit) 409–11 photon, energy/momentum 151, 155 gravitational redshift 157–8 photoelectric effect 244 relativistic Doppler effect 151–2, 154–5 rocket 148 Picard’s theorem 523 piecewise continuous function 294, 296, 497 smooth curve 525 smooth function 296

711

712

Subject index

quantization rules 123, 155, 245 quantum mechanics 123–4, 165–9, 174–8, 192–3, 213–4, 244–7, 465–8 quantum oscillator 244, 465–8 quantum state 244–5 quaternion 159–64

rank of a matrix 96 Rayleigh–Jeans law 244 recursion (recurrence) formula, backward 461 Frobenius series 343, 346 Hermite polynomials 466, 471 orthogonal polynomials 471 regular (analytic) function 522, 541 regular part of Laurent series 541 relativity, general theory 3, 225 special theory 3, 139–56 repellor 384, 392 representation: 4D Fourier 300 basis vectors 4 Fourier 250, 255–6, 259 k (momentum) & x (coordinate) 274–8 Legendre 283–4, 287 quaternions by matrices 162 residue 547 calculus 547–60 theorem 547 response function 270–2 rho (ρ) matrix 169–70 Riemann-Christoffel curvature tensor 228 Riemann √ surface 509–20 (1 + z)1/2 518 arcsin z 518–9 ln z 516 z1/2 512 z1/3 514 z2/3 514 [(z + 1)/(z − 1)]1/3 515–6 (z + 1)1/2 + (z − 1)1/2 512–4 (z + 1)1/2 ln z 516–7 Riemann zeta (ζ) function 652, 654–5, 659 right-hand rule 5, 15, 203 Rodrigues formula 464, 470, 476–8, 482 table 470 rotation 77–85, 125–34 Cayley-Klein parameters 200 complex multiplication 623 curl 26 generator 125–6, 131–4 invariance/scalar under 127, 350 Lorentz, group 188–91 transformation 141–2 matrix 77–9 operator 126 Pauli spin, SU(2) 131–5 quaternion 161–4 SO(n), SU(n) 184

radius of convergence 534–5 raising operator 469, 489, 492, 500

saddle point 23, 599–600 scalar 4–11, 127

Planck’s constant 123, 155, 244–6, 465 Poincar`e, map 409 section 409 Poisson equation 54–5, 308–9, 320, 364–6, 458 polar form (complex number) 504, 622 polar (true) vector 210 pole 523, 526, 540–2, 561–4, 585 polynomials 213–14, 282–3, 287–91, 469–81; see also orthogonal polynomials population dynamics, exponential growth 382 logistic map 382–90 potential flow 42, 51 power series, see series principal: axes 98, 110, 203 moment of inertia 98 stress (pressure) 203 principal-axis transformation 110, 203 principal part of Laurent series 541, 578 principal-value, Green function 567 integral 562 principle Archimedes 14 causality 272, 312, 570 correspondence 245 equivalence principle (masses) 157 exclusion 177 invariance 122–3 mass-energy equivalence 157 pseudo relativity 438 relativity 3, 139 rotational invariance 127 superposition 106, 115, 322–3, 336, 351 uncertainty 275–6, 278–9 probability density, of mapped states 389–90 of wave function 175, 466 projection, chiral 180, 188 orthogonal functions 288–91 vector 6–7, 13, 37, 309 pseudo, Lorentz transformation 438 scalar 171, 210 spin 192–3 tensor 171, 210–11, 217, 218, 241 (axial) vector 171, 210, 216

Subject index scalar field 20–4 scalar potential 54–5, 308–9 scalar product 6–11 Hermitian 101–2 of functions, see inner product triple 10 scalar source 44, 54, 309 scale factor 59–62, 69, 388–9 Schl¨afli integral representation of Pn 527–8 Schr¨odinger equation 246–7, 465–6 Schwarz inequality 278 secular determinant equation 98–9, 105 self-adjoint operator, see Hermitian operator self similarity 386–7, 397 separable coordinate systems 63, 364 separation of variables 351–4, 359–64, 370–1, 435–8, 459 curvilinear coordinates 364 cylindrical coordinates 360–3 rectangular coordinates 353, 360 sine-Gordon equation 435–8 spherical coordinates 360, 363–4 series, see also infinite series asymptotic 590–3, 657–9; see also asymptotic series Frobenius 342–4, 349 terminated 347–9 Hirota 445–7 terminated 446 simple curve 525 simply connected region 525 simultaneous algebraic equations 91–5 sine-Gordon equation 426–7 separation of variables for 435–8 singularity: branch point 511, 523 differential equation 347, 497 essential 523 essential isolated 523 isolated 523 pole 523, 526, 540–2, 561–4, 585 source of function 458–9, 584–6 singular point, see singularity solenoidal field 27, 44 solid angle 45 solitary (persistent) wave 415 soliton 415–48 breather 438 gusher 437 kink 424–31 Korteweg-de Vries 416–21 Perring-Skyrme 415, 434–6, 696 Skyrme 415, 697

713

source in differential eq. 54, 458–9, 565 scalar field 44–5, 54–5, 299, 309 vector field 54–5, 310 space: concepts of 1–4 Fourier 299–301 Hilbert 294–5 infinite-dimensional 282, 292–4 inner-product 282, 292–4 linear vector 288 Minkowski 140–1 n-dimensional 129, 288 orthogonal complement 94 separable 294 square-root 165–9, 186, 191, 195–8 special relativity 3, 139–40, 145, 569 Doppler effect 151, 154–5 gravitational redshift 157–9 Einstein’s postulates 139 energy-momentum relation 150 kinematics 150–6 length contraction 144 Lorentz scalar 144 Lorentz transformation 140–2 proper frame, time 144–5 time dilation 144–5 gravitational 157–9 twin puzzle 146–7 spectrum 370, 497–8 continuous 115 spherical Bessel, eq. 362–3, 366, 490–3 function 362–3, 492 raising/lowering operators 492 spherical coord. 57–62, 64, 69–70, 360–4, 366 spherical harmonics 213–4, 483–6 addition theorem 486 orthogonality relation 485 solid 484–5 spinor 169–98 Cartan 195–8 Dirac 169, 174–8 Majorana 186–7 Weyl 179–82 spin transformation 198 square integrability 251, 294–5, 467–8 stability diagram (H`enon map) 395 state, space 408 variables 408 vector 133 stationary state 244, 246–7 steady (final) state 377–9, 402, 408 step function 272, 336, 338, 571 Sterling’s series, factorial 658

714

Subject index

Stokes’s theorem 50–51 operator form 52–3 strange attractor 374, 392–5, 397–8, 413 stream function 530 stress dyadic 202–6 structure const., Lie algebra 132–3, 189, 238 Sturm-Liouville problem 495–8 summation (Einstein) convention 168 superposition: linear 321, 353, 439, 445–6, 665 nonlinear, of solitons 432–51 superposition prin. 106, 115, 322–3, 336, 353 surface integral (path-dep.) 35–41 symmetry charge conjugation 177–8 Dirac equation 172–8 parity 173–4 time reversal 174–6 twin puzzle 146–7 symmetry violations 182–7 CP 183–5 charge conjugation 186–7 parity 182–3 time reversal 174–5 table: 3D, 4D Fourier transforms 314–5 clock readings in twin problem 147 cylindrical/spherical coordinates 61 Fourier cosine/sine series 313 Fourier transforms 314 properties 267 gamma matrix products 171 Laplace transforms 577 properties 573 matrix operations 79–80 orthogonal curvilinear coordinates 63 orthogonal polynomials: classical 478 differential equation 472 differential relation 471 generating functions 470 orthogonality properties 291 recursion formula 471 Rodrigues formula 470 special cases 291 partial DEs of physics 351 spherical harmonics 214 transformations of physical quantities 216 table of mathematical formulas: 1D wave equation 370–1

3D Fourier transform 318 analytic functions 614 asymptotic expansions 618–9 steepest descent 619 calculus of residues 616 Cartesian tensors 240–1 complex differentiation 614 complex integration 614, 616 conjugate variables & operators 316 construction of functions 617–8 coordinate transformations 135 CP violation 236 cylindrical coordinates 73–4 determinant 135 differential eigenvalue eqs. 136 Dirac δ function 316 Dirac eq., spinor & matrices 234–5 symmetries 235 dispersion relations 618 driven pendula 453–4 dyadic 239–40 eigenfunction expansion 370–1 Fourier, series 315 transform 316 function of a complex variable 614 Green function 369 partial differential eqs. 372 harmonic function in the plane 615 Helmholtz, decomposition theorem 318 equation 371 infinitesimal generators 136 Jacobian 75 kinks 455 Laplace transform 617 Laurent series & residues 615–6 linear differential eqs. 368–9 linearity property 368 logistic map 452 Lorentz group 237–8 matrix, eigenvalue problem 135–6 equations 135 relations 137 Maxwell eqs. in Fourier space 317 nonlinear instabilities 451 orthogonal curvilinear coordinates 73 orthogonal polynomials 498–9 poles on contour & Green functions 617 quantum oscillator 499 quaternions 234 separation of variables 370–1 solitons 454 special relativity, kinematics 233 Lorentz transformation 232–3

Subject index spherical coordinates 74–5 spherical harmonics 499 spinor, Cartan 239 Dirac 234–5 Majorana 237 Weyl 235–6 strange attractor 452 Sturm-Liouville equation 500–1 superposition of solitons 456–7 B¨acklund transformation 456–7 Hirota’s bilinear superposition 457 Taylor series & analytic continuation 615 tensor analysis 241–3 vector 72 vector differential operators 72 tangent vector 58–60, 529 tanh method 429–30 Taylor, expansion 118, 384, 448, 464, 487, 533–7, 547, 664 series 114, 118–20, 328, 503, 534–45 tensor, Cartesian 206–14 contravariant vector 218–20 covariant vector 218–20 differential operations 223–5 in the abstract 221–3 spherical 213–14 tensor analysis 217–32 Christoffel symbol 224, 227 coordinate transformation 220–2 dual basis 219–20 parallel transport 229–31 Ricci, notation 221 tensor 231 Riemann curvature tensor 228–9, 231–2 spherical coordinates 226–7 vector differential operator 223–6 time reversal as antiunitary operator 175 transcendental function 362, 665 transform: 3D Fourier 305–10 4D Fourier 300–3, 310–12 Fourier 265–8 Hilbert 587 Laplace 571–83 transformation, see also map: affine 224 charge conjugation 177–8 conjugate 266, 278 continuous 131 coordinate 77–78, 82–3, 110, 209–11, 217, 220–1 discrete 83–4, 211 gauge 102 group 120–1

logarithmic 444, 448 matrix: congruent 113 conjunctive 113 equivalent 113 orthogonal 80, 82, 113, 207 similarity 113 unitary 109–11, 113 parity 82–3, 173–4, 218 rational 438, 451 rotation 77–8, 125–6, 163 spacetime 117–23 spin 195–200 successive 77, 83–5 time reversal 174–6, 218 transformation by name: Abelian 84, 121, 215 B¨acklund 441–3, 450 Cole-Hopf 439 Lorentz 139–45 Miura 449 pseudo Lorentz 438, 440 transient 402, 409 translation operator 119–22 transposition 15, 79, 89, 644–5 trapping region 373, 392–3 triple product of vectors: scalar product 10–11, 210 vector product (BAC rule) 10, 12–13, 18 uncertainty 275 uncertainty principle 275–9 uniqueness of solution: first-order differential eq. 327–8 second-order differential eq. 329–36 unit, dyadic 202 function 281–2 matrix 79 tangent vector 58, 68–9 vector 4–6, 58, 68, 77 unitarity 80, 130, 184 universality 388 vector 4–5, 629–30 axial (pseudo) 210 components 5, 7 length 6 orthogonality 4 polar (true) 210 unit 4–5 unit tangent 58, 68–9 vector algebra 5–11 in matrix form 127–8 vector analysis 20–71

715

716

Subject index

vector differential operators 20–31, 65–71 curvilinear coordinates 65–71 cylindrical coordinates 73–4 spherical coordinates 74–5 vector field 21, 25–31 irrotational 27, 53–5 solenoidal 27, 44, 53–5 vector potential 54–5, 216, 310 vector product 9, 16–18 triple (BAC rule) 10, 12–13, 18 vibration 106, 248; see also wave equation drum 356–7 string 354–56 vorticity (curliness) 51, 55, 67 wave: absorption coefficient 588 front 116, 367, 374, 425, 567–8, 587 infinite series description 487 in/out-going spherical 310–11, 367, 568–9 mechanics 245–7, 276–9

motion 76, 122–3, 247–9, 367, 416–17, 459, 568–9 phase 116, 367, 568, 587 phase velocity 116, 367, 568–9, 587–8 standing 367, 569 theory of light 246 wave equation 76–7, 319, 374 derivation 114–17, 122–3 electromagnetic 303–5 Fourier transform solution 267–8 Green function 310–12, 565–70 nonlinear 374, 417–20, 426, 435–9 standing wave 367, 569 traveling wave 115–17, 357, 568–9 wave-particle duality 246 wedge product 170 weight function 288–91, 469–70, 477 Wessel-Argand representation 503, 622 Wien’s radiation formula 244 winding number 415 Wronskian 330–5, 490, 613