1948Vector and Tensor Analysis
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VECTOR and TENSOR ANALYSIS By
LOUIS BRAND, Ch.E., E.E., Ph.D. PROFESSOR OF MATHEMATICS UNIVERSITY OF CINCINNATI
New York JOHN WILEY & SONS, Inc. London CHAPMAN & HALL, Limited
VECTOR AND TENSOR ANALYSIS. By Louis Brand. 439 pages. 5Y2 by 8%. Cloth.
VECTORIAL MECHANICS. By Louis Brand. 544 pages. 5% by 8%. Cloth.
Published by John Wiley & Sons, Inc.
VECTOR and TENSOR ANALYSIS By
LOUIS BRAND, Ch.E., E.E., Ph.D. PROFESSOR OF MATHEMATICS UNIVERSITY OF CINCINNATI
New York JOHN WILEY & SONS, Inc. London CHAPMAN & HALL, Limited
COPYRIGHT, 1947 BY
Louis BRAND
All Rights Reserved This book or any part thereof must not
be reproduced in any form without the writ+en permission of the putlisher.
THIRD PRINTING, NOVEMBER, 1948
PRINTED IN THE UNITED STATES OF AMERICA
To My Wife
PREFACE
The vector analysis of Gibbs and Heaviside and the more general tensor analysis of Ricci are now recognized as standard tools in mechanics, hydrodynamics, and electrodynamics. These disciplines have also proved their worth in pure mathematics, especially in differential geometry. Their use not only materially simplifies and condenses. the exposition, but also makes mathematical concepts more tangible and easy to grasp. Moreover tensor analysis provides a simple automatic method for constructing invariants. Since a tensor equation has precisely the same form in all coordinate systems, the desirability of stating physical laws or geometrical properties in tensor form is manifest.. The perfect adaptability of the tensor calculus to the theory of relativity was responsible for its original renown. It has since won a firm place in mathematical physics and engineering technology Thus the British analyst E. T. Whittaker rates the discovery of the tensor calculus as one of the three principal mathematical advances in the last quarter of the 19th century. The first volume of this work not only comprises the standard vector analysis of Gibbs, including dyadics or tensors of valence two, but also supplies an introduction to the algebra of motors, which is apparently destined to play an important role in mechanics
as well as in line geometry. The entire theory is illustrated by many significant applications; and surface geometry and hydro
dynamics * are treated at some length by vector methods in separate chapters. For the sake of concreteness, tensor analysis is first developed in 3space, then extended to space of n dimensions. As in the case of vectors and dyadics, I have distinguished the invariant tensor from its components. This leads to a straightforward treatment of the affine connection and of covariant differentiation; and also to a simple introduction of the curvature tensor. Applications of tensor analysis to relativity, electrodynamics and rotating elec* For a systematic development of mechanics in vector notation see the author's Vectorial Mechanics, John Wiley & Sons, New York, 1930. vii
PREFACE
viii
tric machines are reserved for the second volume. The present volume concludes with a brief introduction to quaternions, the source of vector analysis, and their use in dealing with finite rotations. Nearly all of the important results are formulated as theorems,
in which the essential conditions are explicitly stated. In this connection the student should observe the distinction between necessary and sufficient conditions. If the assumption of a certain property P leads deductively to a condition C, the condition is necessary. But if the assumption of the condition C leads deductively to the property P, the condition C is sufficient. Thus wehave symbolically
P * C (necessary),
C (sufficient) > P.
When P C, the condition C is necessary and sufficient. The problems at the end of each chapter have been chosen not only to develop the student's technical skill, but also to introduce new and important applications. Some of the problems are mathematical projects which the student may carry through step by step and thus arrive at really important results. As very full cross references are given in this book, an article as well as a page number is given at the top of each page. Equations are numbered serially (1), (2), . . . in each article. A reference to an equation in another article is made by giving article
and number to the left and right of a point; thus (24.9) means article 2.4, equation 9. Figures are given the number of the article
in which they appear followed by a serial letter; Fig. 6d, for example, is the fourth figure in article 6. Boldface type is used in the text to denote vectors or tensors of higher valence with their complement of base vectors. Scalar components of vectors and tensors are printed in italic type. The rich and diverse field amenable to vector and tensor methods is one of the most fascinating in applied mathematics. It is hoped
that the reasoning will not only appeal to the mind but also impinge on the reader's aesthetic sense. For mathematics, which Gauss esteemed as "the queen of the sciences" is also one of the great arts. For, in the eloquent words of Bertrand Russell: "The true spirit of delight, the exaltation, the sense of being more than man, which is the touchstone of the highest excellence,
is to be found in mathematics as surely as in poetry. What is
PREFACE
is
best in mathematics deserves not merely to be learned as a task, but also to be assimilated as a part of daily thought, and brought again and again before the mind with everrenewed encouragement. Real life is, to most men, a long secondbest, a perpetual compromise between the real and the possible; but the world of pure reason knows no compromise, no practical limitations, no barrier to the creative activity embodying in splendid edifices the passionate aspiration after the perfect from which all great work springs."
The material in this book may be adapted to several short Thus Chapters I, III, IV, V, and VI may serve as a
courses.
course in vector analysis; and Chapters I (in part), IV, V, and IX as one in tensor analysis. I But the prime purpose of the author was to cover the theory and simpler applications of vector and tensor analysis in ordinary space, and to weave into this fabric such concepts as dyadics, matrices, motors, and quaternions.
The author wishes, finally, to express his thanks to his colleagues, Professor J. W. Surbaugh and Mr. Louis Doty for their help with the figures. Mr. Doty also suggested the notation used in the problems dealing with air navigation and read the entire page proof. Louis BRAND
University of Cincinnati January 15, 1947
CONTENTS PREFACE .
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PAGE Vii
CHAPTER I VECTOR ALGEBRA ARTICLE
1. Scalars and Vectors . 2. Addition of Vectors .
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5 6 7 8
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3. Subtraction of Vectors
4. Multiplication of Vectors by Numbers . 5. Linear Dependence . . . . . . . . . . . . . . 6. Collinear Points . . . . 7. Coplanar Points .
8. Linear Relations Independent of the Origin. 9. Centroid . . . . . . . . . . . . . . . . 10. Barycentric Coordinates . . 11. Projection of a Vector . . . 12. Base Vectors . . . . . . . 13. Rectangular Components 14. Products of Two Vectors . . 15. Scalar Product . . . . . . 16. Vector Product . . . . . 17. Vector Areas . . . . . . . 18. Vector Triple Product . . . 19. Scalar Triple Product . . . 20. Products of Four Vectors . 21. Plane Trigonometry . . . 22. Spherical Trigonometry . . 23. Reciprocal Bases . . . . 24. Components of a Vector . . 25. Vector Equations . . . . 26. Homogeneous Coordinates . 27. Line Vectors and Moments 28. Summary: Vector Algebra . Problems . . . . . . . .
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24 24 26 29 29 34 37 40
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41
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43 44 44 46 48 50
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51
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55 57 59
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12 18 19 23
CHAPTER II MOTOR ALGEBRA
29. Dual Vectors . 30. Dual Numbers
31 Motors
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xi
63 64 65
CONTENTS
xii ARTICLE
PAGE
32. Motor Sum . . . . . . 33. Scalar Product . . . . . 34. Motor Product . . . . . 35. Dual Triple Product . . . 36. Motor Identities . . . 37. Reciprocal Sets of Motors 38. Statics . . . . . . . . 39. Null System . . . . . . 40. Summary: Motor Algebra Problems . . . . . . . .
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67 68 70 72 73 74 75 78 80 82
CHAPTER III VECTOR FUNCTIONS OF ONE VARIABLE
41. Derivative of a Vector . . . . . . 42. Derivatives of Sums and Products 43. Space Curves . . . . . . . . . 44. Unit Tangent Vector . . . . . . 45. Frenet's Formulas . . . . . . . 46. Curvature and Torsion . . . . . 47. Fundamental Theorem . . . . . 48. Osculating Plane . . . . . . . . 49. Center of Curvature . . . . . . 50. Plane Curves . . . . . . . . . . 51. Helices . . . 52. Kinematics of a Particle . . . . . 53. Relative Velocity . . . . . . . 54. Kinematics of a Rigid Body . . . 55. Composition of Velocities . . . . 56. Rate of Change of a Vector . . . 57. Theorem of Coriolis . . . . . . . 58. Derivative of a Motor . . . . . . 59. Summary: Vector Derivatives . . Problems . . . . . . . . . . .
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98 99 100 105 108 110 114 120
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121
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84 86 88 90 92 95 97
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123 126 128 130
CHAPTER IV LINEAR VECTOR FUNCTIONS
60. Vector Functions of a Vector . . . 61. Dyadics . . . . . . . . . . . . 62. Affine Point Transformation . . . 63. Complete and Singular Dyadics . . 64. Conjugate Dyadics . . . . . . . 65. Product of Dyadics . . . . . . . 66. Idemfactor and Reciprocal . . . . 67. The Dyadic 4) x v . . . . . . . . 68. First Scalar and Vector Invariant . 69. Further Invariants . . . . . . .
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135 136 138 139
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141
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142 144 146 147 148
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. .
CONTENTS
xiii
ARTICLE
PAGE
70. Second and Adjoint Dyadic . . . . 71. Invariant Directions . . . . . . . 72. Symmetric Dyadics . . . . . . . . 73. The HamiltonCayley Equation . . 74. Normal Form of the General Dyadic . 75. Rotations and Reflections . 76. Basic Dyads . . . . . . . . . 77. Nonion Form . . . . . . . . . . 78. Matric Algebra . . . . . . . 79. Differentiation of Dyadics . . . . . . . . . 80. Triadics . . . . . . 81. Summary: Dyadic Algebra . . . . . .
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Problems
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153 156 160 162 164 166 167 169
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171
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172 172
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174
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151
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CHAPTER V DIFFERENTIAL INVARIANTS
82. Gradient of a Scalar . . . . . . . 83. Gradient of a Vector . . . . 84. Divergence and Rotation . . . . 85. Differentiation Formulas . . . . 86. Gradient of a Tensor . . . . . . 87. Functional Dependence . . . . . 88. Curvilinear Coordinates . . . . . . . 89. Orthogonal Coordinates . . 90. Total Differential . . . . . . . 91. Irrotational Vectors . . . . . . . 92. Solenoidal Vectors . . . . . . . 93. Surfaces . . . . . . . . . . . . 94. First Fundamental Form . . . 95. Surface Gradients . . . . . . . . 96. Surface Divergence and Rotation . 97. Spatial and Surface Invariants . 98. Summary: Differential Invariants . Problems . . . . . . . . . . . .
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178
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181
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183 186 187 190 191 194 197 198 201
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203 204 206 207 209
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211
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213
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CHAPTER VI INTEGRAL TRANSFORMATIONS
99. Green's Theorem in the Plane . . . . . 100. Reduction of Surface to Line Integrals . 101. Alternative Form of Transformation . . 102. Line Integrals . . . . 103. Line Integrals on a Surface . . . . . . 104. Field Lines of a Vector . . . . . . . . 105. Pfaff's Problem . . . . . . . 106. Reduction of Volume to Surface Integrals 107. Solid Angle . . . . . . . . . . . . .
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216 218
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221
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222 224 226 230 233 236
CONTENTS
xiv
PAGE
ARTICLE
108. Green's Identities . . . . . . . . . 109. Harmonic Functions . . . . . . . 110. Electric Point Charges . . . . . . 111. Surface Charges . . . . . . . . . 112. Doublets and Double Layers . . . . 113. Space Charges . . . . . . . . . . 114. Heat Conduction . . . . . 115. Summary: Integral Transformations . Problems . . . . . . . . . . . .
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237 239 241 242 243 245 247 248 250
CHAPTER VII HYDRODYNAMICS
116. Stress Dyadic . . . . . . . . . . . . 117. Equilibrium of a Deformable Body . . . 118. Equilibrium of a Fluid . . . . . . . . . 119. Floating Body . . . . . . . . . . . 120. Equation of Continuity . . . . . . . . 121. Eulerian Equation for a Fluid in Motion. 122. Vorticity . . . . . . . . . . . . . . 123. Lagrangian Equation of Motion . . . 124. Flow and Circulation . . . . . . . . . 125. Irrotational Motion . . . . . . . . . . 126. Steady Motion . . . . . . . . . . . 127. Plane Motion . . . . . . . . . . . . 128. KuttaJoukowsky Formulas . . . . . . 129. Summary: Hydrodynamics . . . . . . . Problems . . . . . . . . . . . . . .
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253 255 257 257 258 260 262 263 266 267 268 270 273 278 280
CHAPTER VIII GEOMETRY ON A SURFACE
130. Curvature of Surface Curves . . . 131. The Dyadic On . . . . . . . . 132. Fundamental Forms . . . . . . 133. Field of Curves . . . . . . . . . 134. The Field Dyadic . . . . . . . . . 135. Geodesics . . . . . . . . . 136. Geodesic Field . . . . . . . . . 137. Equations of Codazzi and Gauss . 138. Lines of Curvature . . . . . . . 139. Total Curvature . . . . . . . . 140. Bonnet's Integral Formula . . . . 141. Normal Systems . . . . . . . . 142. Developable Surfaces . . . . . 143. Minimal Surfaces . . . . . . . . 144. Summary: Surface Geometry . . . . . . . Problems . . . . . . .
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283 285 289 293 293 297 299 300 302 306 308 313 314 316 319 321
CONTENTS
xv
CHAPTER IX TENSOR ANALYSIS ARTICLE
PAGE
145. The Summation Convention . . . . . . . . . . . . 146. Determinants . . . . . . . . . . . . . . . . . 147. Contragredient Transformations . . . . . . . . . 148. Covariance and Contravariance . . . . . . . . . . . 149. Orthogonal Transformations . . . . . . . . . . . 150. Quadratic Forms . . . . . . . . . . . . . . . . 151. The Metric . . . . . . . . . . . . . . . . . 152. Relations between Reciprocal Bases . . . . . . 153. The Affine Group . . . . . . . . . . . . . . 154. Dyadics . . . . . . . . . . . . . . . . . . 155. Absolute Tensors . . . . . . . . . . . . . . . . . 156. Relative Tensors . . . . . . . . . . . . . . . 157. General Transformations . . . . . . . . 158. Permutation Tensor . . . . . . . . . . . . . . 159. Operations with Tensors . . . . . . . . . . . . . 160. Symmetry and Antisymmetry . . . . . . . . . 161. Kronecker Deltas . . . . . . . . . . . 162. Vector Algebra in Index Notation . . . . . . . 163. The Affine Connection . . . . . . . . . . 164. Kinematics of a Particle . . . . . . . . . . . 165. Derivatives of e` and E . . . . . . . . . 166. Relation between Affine Connection and Metric Tensor 167. Covariant Derivative . . . . . . . . . . . 168. Rules of Covariant Differentiation . . . . . . 169. Riemannian Geometry . . . . . . . . . . 170. Dual of a Tensor . . . . . . . . . . . . . . . . . 171. Divergence . . . . . . . . . . . . . . . . . . . 172. Stokes Tensor . . . . . . . . . . . . . . . . . 173. Curl . . . . . . . . . . . . . . . . . . . . 174. Relation between Divergence and Curl . . . . . . 175. Parallel Displacement . . . . . . . . . . . . . 176. Curvature Tensor . . . . . . . . . . . . 177. Identities of Ricci and Bianchi . . . . . . . . . . . 178. Euclidean Geometry . . . . . . . . . . . . . 179. Surface Geometry in Tensor Notation . . . . . . . . 180. Summary: Tensor Analysis . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . .
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328 329 332 334 336 337 339 339 340 342 343 345 346 350 350 352 353 354 356 359 360
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361
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362 365 366 370
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371
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372 374 376 376 380 384 385 388 392 396
CHAPTER X QUATERNIONS
181. Quaternion Algebra . . . 182. Conjugate and Norm . 183. Division of Quaternions .
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403 406 409
CONTENTS
xvi
PAGE
ARTICLE
184. Product of Vectors . . . . . 185. Roots of a Quaternion . . . . . 186. Great Circle Arcs . . . . . . . 187. Rotations . . . . . . . . . . 188. Plane Vector Analysis . . . . . 189. Summary: Quaternion Algebra . Problems . . . . . . . . . .
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410 412 414 417
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421
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426
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427,
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431
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INDEX
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...
CHAPTER I VECTOR ALGEBRA 1. Scalars and Vectors. There are certain physical quantities, such as length, time, mass, temperature, electric charge, that may be specified by a single real number. A mass, for example, may be specified by the positive number equal to the ratio of the given mass to the unit mass. Similarly an electric charge may be specified by a number, positive or negative, according as the charge is "positive" or "negative." Quantities of this sort are called scalar quantities; and the numbers that represent them are often called scalars.
On the other hand, some physical quantities require a direction as well as magnitude for their specification. Thus a rectilinear displacement can only be completely specified by its length and direction. A displacement may be represented graphically by a segment of a straight line having a definite length and direction. We shall call such a directed segment a vector. Any physical quantity that involves both magnitude and direction, so that it may be represented by a line segment of definite length and direction, and which moreover conforms to the parallelogram law of addition (§ 2), is called a vector quantity. Velocity, acceleration, force, electric and magnetic field intensities are examples of vector quantities. It is customary, however, in applied mathematics, to speak of vector quantities as vectors. DEFINITIONS: A vector is a segment of a straight line regarded as having a definite length and direction. Thus we may represent a
vector by an arrow. The vector directed from the point A to the
point B is denoted by the symbol AB. With this notation AB

and BA denote different vectors; they have the same length but opposite directions. Besides the proper vectors just defined, we extend the term vector
to include the zero vector, an "arrow" of length zero but devoid of direction. 1
2
VECTOR ALGEBRA
§1
If the initial point of a vector may be chosen at pleasure, the vector is said to be free. If, however, its initial point is restricted to a certain set of points, the vector is said to be localized in this set. When the set consists of a single point (initial point fixed), the vector is said to be a bound vector. If the vector is restricted to the line of which it forms a part, it is called a line vector. For example, the forces acting upon rigid bodies must be regarded as line vectors; they may only be shifted along their line of action. Two vectors are said to be equal when they have the same length and direction. Vectors are said to be collinear when they are parallel to the same line. In this sense two parallel vectors are collinear. Vectors are said to be coplanar when they are parallel to the same plane. In this sense any two vectors are coplanar. A unit vector is a vector of unit length.
In addition to the foregoing notation, in which we denote a vector by giving its end points, we also shall employ single letters in heavy (boldface) type to denote vectors. Thus, in Fig. 2b the vectors forming the opposite sides of the parallelogram are equal (they have the same length and direction) and may therefore be represented by the same symbol,
AB=DC=u,
AD=BC=v.
A vector symbol between vertical bars, as I AB I or I u 1, denotes the length of the vector. We shall also, on occasion, denote the lengths of vectors, u, v, F by the corresponding letters u, v, F in italic type. Bars about real numbers denote their positive magnitudes: thus I 3 1 = 3. The preceding definition of a vector is adequate for the elementary applications to Euclidean space of three dimensions. For
purposes of generalization, however, it is far better to define a vector as a new type of numbera hypernumberwhich is given by a set of real numbers written in a definite order. Thus, in our ordinary space it will be seen that, when a suitable system of reference has been adopted, a vector can be represented by a set of three real numbers, [a, b, c], called the components of the vector. With this definition the zero vector is denoted by [0, 0, 0]. In order to complete this definition of a vector, a rule must be given to enable us to compute the components when the system of reference is changed.
ADDITION OF VECTORS
§2
3
2. Addition of Vectors. To obtain a rule for adding vectors, let us regard them, for the moment, as representing rectilinear displacements in space. If a particle is given two rectilinear displace
ments, one from A to B, and a second from B to C, the result is the same as if the particle were given asingle displacement from A to C. This equivalence may be represented by the notation,
AB + BC = AC.
(1)
We shall regard this equation as the definition of vector addition. The sum of two vectors, u, v, therefore is defined by the following triangle construction (Fig. 2a) : Draw v from the end of u; then the vector directed from the beginning of u to the end of v is the sum of u and v and is written u + v. C
Fia. 2b
FIG. 2a
Since any side of a triangle is less than the sum of the other two sides,
lu+vl _ lul +lvl, the equal sign holding only when u and v have the same direction. Vector addition obeys both the commutative and associative laws: (2)
u+v = v+u,
(3)
(u + v) + w = u + (v + w).
In the parallelogram formed with u and v as sides (Fig. 2b),
u+v=AB+BC=AC,
v+u=AD+DC=AC.
This proves (2). In view of this construction, the rule for vector addition is called the parallelogram law. To find u + v when u and v are line vectors whose lines intersect at A, shift the vectors along their lines so that both issue from A (Fig. 2b) and complete the parallelogram ABCD; then 3
3
9
u+v=AB+AD= AC.
VECTOR ALGEBRA
4
§2
Since the diagonal of the parallelogram on u, v gives the line of action of u + v, the term "parallelogram law" is especially appropriate for the addition of intersecting line vectors. We shall speak of the addition of line vectors as statical addition. The associative law (3) is evident from Fig. 2c:
(u + v) + w = (AB i BC) f CD = AC i CD = AD,
u + (v F w) = AB f (BC + CD) = AB + BD = AD. Since the grouping of the vectors is immaterial, the preceding swa
is simply written u + v + w.
B
Fia. 2c
FIG. 2d
From the commutative and associative laws we may deduce the following general result: The sum of any number of vectors is independent of the order in which they are added, and of their grouping to form partial sums. To construct the sum of any number of vectors, form a broken
line whose segments, in length and direction, are these vectors taken in any order whatever; then the vector directed from the beginning to the end of the broken line will be the required sum. The figure formed by the vectors and their sum is called a vector polygon. If A, B, C, , G, H are the successive vertices of a vector polygon (Fig. 2d), then (4)
When the vectors to be added are all parallel, the vector "polygon" becomes a portion of a straight line described twice. If, in the construction of a vector sum, the end point of the last vector coincides with the origin of the first, we say that the sum of the vectors is zero. Thus, if in (4) the point H coincides with A, we write (5)
4  3 3 AB+BC++GA=O.
SUBTRACTION OF VECTORS
§3
5
This equation may be regarded as a special case of (4) if we agree >
that AA = 0. The zero vector AA (or BB, etc.) is not a vector in the proper sense since it has no definite direction; it is an extension of our original vector concept. From  f
*
3
AB+BB=AB,
3 3 * AA+AB=AB
we have, on writing AB = u,
u+0=u,
0+u=u.
We shall refer to vectors which are not zero as proper vectors. 3. Subtraction of Vectors. The sum of two vectors is zero when, and only when, they have the same length and opposite directions:
AB + BA = 0. If AB = u, it is natural to write BA = u in order that the characteristic equation for negatives,
u + (u) = 0,
(1)
will hold for vectors as well as for numbers. Hence by definition: The negative of a vector is a vector of the same length but opposite direction.
Note also that  (u) = u. The difference u  v of two vectors is defined by the equation,
(u  v) + v = U.
(2)
Adding v to both sides of (2), we have
uv=u+(v);
(3)
that is, subtracting a vector is the same as adding its negative. The
construction of u  v is shown in Fig. 3a. A
0 Fia. 3a
Fia. 3b
If 0 is chosen as a point of reference, any point P in space may
be located by giving its position vector OP. Any vector AB may
VECTOR ALGEBRA
6
§4
be expressed in terms of the position vectors of its end points (Fig. 3b),
AB = AO + OB = OB + (OA), and, from (3), (4)
 4 4 AB = OB  OA.
4. Multiplication of Vectors by Numbers. The vector u + u is
naturally denoted by 2u; similarly, we write u + (u) = 2u. Thus, both 2u and 2u denote vectors twice as long as u; the former has the same direction as u, the latter the opposite direction. This definition is generalized as follows: The product au or ua of a vector u and a real number a is defined as a vector a times as long as u, and having the same direction as u, or the opposite, according as a is positive or negative. If a = 0,
au=0.
In accordance with this definition,
a(u) = (a)u = au,
(a)(u) = au.
These relations have the same form as the rules for multiplying numbers. Moreover, the multiplication of a vector by numbers is commutative (by definition), associative, and distributive: (1)
au = ua,
(2)
(a/3)u = a(/3u),
(3)
(a + a)u = au + (3u.
The product of the sum of two vectors by a given number is also distributive:
a(u + v) = au + av.
(4)
The proof of (4) follows immediately from the theorem that the corresponding sides of similar triangles are proportional. Figure 4 applies to the case when a > 0. The quotient u/a of a vector by a number av u+V a (not zero) is defined as the product of u v
au by 1/a.
U
Fia. 4
The developments thus far show that: As far as addition, subtraction, and multi
plication by numbers are concerned, vectors may be treated formally in accordance with the rules of ordinary algebra.
LINEAR, DEPENDENCE
§5
7
5. Linear Dependence. The n vectors u1, u2i , un are said to be linearly dependent if there exist n real numbers X1, X2, , An, not all zero, such that X1U1 + X2U2 + ... + XnUn = 0.
(1)
If the vectors are not linearly dependent, they are said to be linearly independent. Consequently, if a relation (1) exists between n linearly independent vectors, all the constants must be zero.
, um are linearly dependent, any greater If m vectors ul, U2, number n of vectors including these are also linearly dependent. For , um satisfy if u1i u2, X1U1 + A2U2 + ... + XrUm = 0,
we can give ?1, X2, , Xm the preceding values (at least one of these is not zero) and take X,n+1 = X m+2 = _ X n = 0. Then
(1) is satisfied, and the n vectors ui are linearly dependent. If Xu = 0 and X 0, u = 0; hence one vector is linearly dependent only when it is the zero vector. Hence the vectors of any set
that includes the zero vector are linearly dependent.
Conse
quently, we need only consider sets of proper vectors in the theorems following.
If X1u1 + X2u2 = 0 and X1 0 0, we can write ul = au2 i hence ul and u2 are collinear. Conversely, if ul and u2 are collinear, U1 = au2 (a 0 0). Therefore: A necessary and sufficient condition that two proper vectors be linearly dependent is that they be collinear.
If X1u1 + A2u2 + X3u3 = 0 and X1 0, we can write ul = aU2 + ,13u3; the parallelogram construction (Fig. 5a) now shows that ul is parallel to the plane of u2 and u3. Conversely, if U1, U2, u3 are coplanar, they are linearly dependent. For (a) if two of the vectors are collinear, they are linearly dependent, and
the same is true of all three; and (b) if no two vectors are collinear, we can construct a parallelogram on ul as diagonal whose sides are parallel to u2 and u3 (Fig. 5a), so that
ul =AC=AB+BC=au2+$u3. Therefore: A necessary and sufficient condition that three proper vectors be linearly dependent is that they be coplanar.
VECTOR ALGEBRA
8
§6
In space of three dimensions, four vectors ul, u2i u3, u4 are always linearly dependent. For (a) if three of the vectors are co
planar, they are linearly dependent, and the same is true of all four; and (b) if no three vectors are coplanar", we can construct a
FIG. 5a
FIG. 5b
parallelepiped on ut as a diagonal whose edges are parallel to U2i n3, U4 (Fig. 5b), so that
3 3 a  3 ut = AD = AB + BC } CD = au2 + 9u3 I yu4.
Therefore: Any four vectors are linearly dependent.
6. Collinear Points. If A, B, P are points of a straight line, P is said to divide the segment AB in the ratio X when
>
3
AP=APB.
(1)
As P passes from A to B (Fig. 6a), X increases through all positive values from 0 to infinity. If P describes the line to the left of A,
X varies from 0 to 1; and, as P describes the line to the right of B, X varies from oo to 1. Thus X = 0, X = =L00 , A = 1 correspond, respectively, to the points A, B, and the in0 FIG. 6a
finitely distant "point" of the line. The ratio A is positive or negative, according as P lies within or without the segment AB.
To find the position vector of P, relative to an origin 0, write (1) in the form, Then (2)
OP  OA = X(OB  OP).
OA + A OB OP = 1 + A
COLLINEAR POINTS
§6
9
or, if we write X = (3/a,

OPaOA +aOB =
(3)
a+0
In particular, if X = 1, P is the midpoint of AB.
In the following we shall denote the position vectors of the , p. Thus if C divides AB , P by a, b, c,
points A, B, C, in the ratio /3/a, (4)
c=ash(3b
a+0
Thus the midpoint of AB has the position vector (a + b). 2 When the points C, D divide a segment AB internally and externally in the same numerical ratios ±X, we have
a+Ab c =1+x,
d=
a  Xb
1x
If we solve these equations for a and b, we find that the points A, B also divide the segment CD in the same numerical ratios f(1  X)/(l + X). Pairs of points A, B and C, D having this property are said to be harmonic conjugates; either pair is the harmonic conjugate of the other. A useful test for collinearity is given by the following: THEOREM.
Three distinct points A, B, C lie on a straight line
when, and only when, there exist three numbers a, /3, y, different from zero, such that (5)
aa+/3b+yc=0,
a+$+y=0.
Proof. If A, B, C are collinear, C divides AB in some ratio /3/a; hence on putting y =  (a + ,B) in (4) we obtain (5). Conversely,
from (5) we can deduce (4) since a + y 0; hence C lies on the line AB. From (5) we conclude that C, A, B divide AB, BC, CA, respectively, in the ratios S/a, y/l3, a/y whose product is 1. If an equation of the form (5) subsists between three distinct non
collinear points, we must conclude that a = (3 = y = 0. For at least one coefficient y = 0; and from
as+3b=0,
a+/3=0,
we have a = b (A coincides with B) unless a = 0 = 0.
§6
VECTOR ALGEBRA
10
Another criterion for collinearity may be based on the statical addition of line vectors (§ 2). THEOREM. The points A, B, C are collinear when the line vectors
AB, BC, CA are statically equal to zero:
AB +BC +CA =O.
(6)
Proof. If we use = to denote statical equivalence, AB + BC
BD, a vector through B; and, since BD + CA = 0, B, C, A are collinear. Example 1. In the parallelogram ABCD, E and F are the middle points of the sides AB, BC. Show that the lines DE, DF divide the diagonal AC into thirds and that AC cuts off a third of each line (Fig. 6b).
E
"
B Fia. 6c
Fia. 6b
The hypotheses of our problem are expressed by the equations:
d  a=c  b,
2e=a+b,
2f =b+c.
Let DE cut AC at X. To find x, eliminate b from the first and second equations. Thus d + 2e 2a + c = =a;
da+2e=a+c and
3
3
for the first member represents a point on DE, the second member a point on AC, and, since the points are the same, the point is at the intersection X of these lines. Comparison with (4) now shows that X divides DE in the ratio 2/1, AC in the ratio 1/2. Let DF cut AC at Y. To find y, eliminate b from the first and third equations. Thus
da+2f=2c and d+2fa+2c_ 3 3
Hence Y divides DF in the ratio 2/1, AC in the ratio 2/1. Example 2. In a plane quadrilateral ABCD, the diagonals AC, BD intersect at P, the sides AB CD intersect at Q (Fig. 6c). If P divides AC and BD in the ratios 3/2 and 1/2, respectively, in what ratios does Q divide the segments A B, CD?
§6
COLLINEAR POINTS
11
By hypothesis 2a + 3c
2b + d
5
3
p hence
6a + 9c = 10b + 5d and

5d = lOb  6a _ 4
for the first fraction represents a point
on CD, the second a point on AB, and both points are the same, that is, the point Q. Therefore Q divides CD
in the ratio 5/9, AB in the ratio 10/6. Example 3. Prove that the midpoints of the diagonals of a complete
quadrilateral are collinear.
In the complete quadrilateral ABCDLM (Fig. 6d) let P, Q, R be the midpoints of the diagonals AC, BD, LM. The sum of the line vectors,
,
>
AB + AD = 2 AQ,
,

i
CB + CD = 2 CQ,
> > QA + QC = 2 QP;
hence we have the statical equivalence,
AB +AD +CB +CD =4PQ, for the quadrilateral ABCD with diagonals AC, BD. Similarly, for the rilateral BLDM with diagonals BD, LM,
ad
9  4 + BL + BM + DL + D31 = 4QR; >
and, for the quadrilateral LA.11C with diagonals LM, AC,
,
4 4 + + LA + LC + MA + MC = 4 RP. On adding these three equations, we find that the entire left member is statically equal to zero; for +  > * 
9 DM+MA = 0,
=0,
+ CB +BM +MC0,
+ CD +DL +LC O, 4
are statical equations, since the vectors in each are collinear.
PQ +QR+RP =0, and I', Q, R are collinear.
Hence
VECTOR ALGEBRA
12
§7
7. Coplanar Points. THEOREM. If no three of the points A, B, C, D are collinear, they will lie in a plane when, and only when, there exist four numbers a, /3, y, S, different from zero, such that (1)
as+$b+yc+Sd = 0,
a+0+y+S = 0.
Proof. If A, B, C, D are coplanar, either AB is parallel to CD,
or AB cuts CD in a point P (not A, B, C, or D). In the respective cases, we have
b  a=K(d  c);
a+Ab
c+A'd
1+A = 1+A, P,
where A, A' are neither 0 nor 1. In both cases, a, b, c, d are connected by a linear relation of the form (1). Conversely, let us assume that equations (1) hold good. If a + /3 = 0 (and hence
y + S = 0), we have
a(a  b) + y(c  d) = 0 and the lines AB and CD are parallel. If a + /3 5,16 0 (and hence
y+6
0),
as+/3b
(2)
a+/i
yc+Sd
y+6 P
where P is a point common to the lines AB and CD. In both cases, A, B, C, D are coplanar.
Note that (2) states that the point P in which AB, CD intersect divides AB and CD in the ratios #/a, 6/y. Similarly, if
a+y; 0, (3)
as + yc
/3b + 6d
a f.y
#+ 6
 qi
thus Q, the point in which AC and BD intersect, divides AC and BD in the ratios y/a, S//i. What conclusion can be drawn if a + 3 F6 0? If an equation of the form (1) subsists between four distinct noncoplanar points, we must conclude that a = a = y = S = 0. For at least one coefficient 6 = 0; then, from
as+/3b+yc = 0,
a+0+y = 0,
and the fact that A, B, C are not collinear, we deduce (§ 6) that
a=0=y=0.
COPLANAR POINTS
§7
Example 1. The Trapezoid. DC (Fig. 7a), then
13
If ABCD is a trapezoid with AB parallel to
AB=XDC or ba=A(cd). Hence we have
b+Xd=a+Xc or bXc=aXd. These equations may be written
b+ad_a+Xc_P 1+a +a
b  Ac_a  Ad
q;
for the former expressions represent the point P where the diagonals BD, AC meet, and the latter the point Q where the sides BC, AD meet. Evidently P divides both BD and AC in the same ratio A; and Q divides both BC and AD in the same ratio A. In what ratio does the line PQ divide AB? In particular, if A = 1, the trapezoid becomes a parallelogram. The diagonals then bisect each other at P, while Q recedes to infinity. Q
C
Fia. 7a Example 2. Theorem of Menelaus. If a line s cuts the sides BC, CA, AB of the triangle ABC in the points P, Q, R, respectively, the product of the ratios in which P, Q, R divide these sides equals  1. Conversely, if P, Q, R divide the sides of the triangle in ratios whose product is  1, the points are collinear. Proof. (Fig. 7b.) We lose no generality if we assume that P, Q divide
BC, CA in the ratios y//3, a/y: (i)
(/3y)P=$byc,
(ii)
(ya)q=ycaa.
In order to locate R, which lies on the lines PQ, AB, we seek a linear relation
between p, q, a, b. Add (i) and (ii) to eliminate c and divide by S  a; then
(R  y)P + (y  a)q
Sb  as
Thus R divides AB in the ratio S/a. The product of the division ratios
y//3, a/y, /3/a is 1.
VECTOR ALGEBRA
14
§7
Conversely, let us assume that P, Q, R divide BC, CA, AB in the ratios
y/S, a/y, a/a whose product is 1. Then we have equations (i), (ii) and also
(a  fl)r = as  13b.
(iii)
From these we deduce the linear relation,
(0y)p+(ya)q+(a$)r=0,
(iv)
in which the sum of the coefficients is zero. The points P, Q, R are therefore collinear. * Note. From (i), (ii),
(iii) it is easily proved that the three pairs of lines BQ, CR; CR, AP; AP, BQ meet in the points A', B', C' given by
(a+0+y)a' _ aa+ftb+yc, (aS+y)b' =aa$b+yc,
(a+9y)c'=as+9byc. If we add 2aa, 2gb, 2yc, respectively, to these equations, we find that the point S given by s =
(v)
as +Ab+yc
a+#+y
is common to the lines AA', BB', CC'. Thus to every line s given by (iv) we have a corresponding point S given by (v)
the pole of s relative to the triangle ABC.
C
Example 3. Theorem of Ceva. If S is a point in the plane of the triangle ABC, and the lines SA, SB, SC cut the sides opposite in the points A', B', C', then the product of the ratios
C,
Fte.7c
in which A', B', C' divide the sides BC, CA, AB equals 1. Conversely, if A', B', C' divide the sides BC, CA, AB in ratios whose product B is 1 , the lines AA', BB', CC meet in a p oint . Proof. (Fig. 7c.) Since A, B, C, S are coplanar,
aa+pb+yc+as =0,
a+0+y+S =0.
Hence
_
(i) a
(ii)
b' = c'
Sb + ye
as + SS
0+y
a+s
yc
as _ Rb + as
y+a
+b '
=as+Sb yc+Ss. a+R y+S
*This conclusion is obvious; for the line PQ must meet AB in the point R for which the product of the division ratios is 1.
§7
COPLANAR POINTS
15
These equations state that A', B', C' divide BC, CA, AB in the ratios y/fl, a/y, fl/a, whose product is 1. Incidentally, A', B', C' divide SA, SB, SC in the ratios a/a, fl/S, y/3 whose sum is 1. Conversely, let us assume that A', B', C' divide BC, CA, AB in the ratios y/$, a/j, fl/a whose product is 1. Then we have equations (i), (ii), (iii).
From these we find that the vectors as + (0 + y)a', fib + (y + a)b', yc + (a + fl)c' are all equal to as + lb + yc; the point, s
(iv)
as+pb+yc a + fl + y
is therefore common to the lines AA', BY, CC'. Note. From (i), (ii), (iii) it is easily proved that the three pairs of lines BC, B'C'; CA, C'A'; AB, A'B' meet in the points P, Q, R given by
(Qy)p=Rbyc, (y  a)q = yc  aa,
(v)
(afl)r=as/3b. From these equations we deduce the linear relation, (vi)
(fiy)p+(ya)q+(aa)r=0,
in which the sum of the coefficients is zero. The points P, Q, R therefore lie
on a line s. Thus to every point S given by (iv) we have a corresponding line s whose points P, Q, R are given by (v)the polar of S relative to the triangle ABC. Example 4. Let ABC and A'B'C' be two triangles, in the same or different
planes, so that the vertices A, B, C correspond to A', B', C', and the sides AB, BC, CA correspond to A'B', B'C', C'A'. We then have (Fig. 7d) P
FIG. 7d DESARGUES' THEOREM. If the lines joining the corresponding vertices of two
triangles are concurrent, the three pairs of corresponding sides intersect in collinear points, and conversely.
Let the lines AA', BY, CC' intersect at S; then as + a'a' = fib + fl'b' = yc + y'c' = s,
a+a =t3+fl' ='Y+y'=1.
VECTOR ALGEBRA
16
§7
From these equations we find in the usual manner the points P Q, R in which BC, B'C'; CA, C'A'; AB, A'B' intersect: (i)
Jr
=
(ii)
q
=
(iii)
'6',
fib
fl'b'
yC  as
y'C'  a'a'
 yc = fl_y
y_a
 y'c/
1
y,_«,
_«a3b «a'fl'b' at  of
Hence (iv) (v)
(l;  'y)p + (y  a)q + (a  6)r = 0,
'y')p + ('y'  a')q + (a'  6')r = 0;
either equation shows that P, Q, R are collinear. To prove the converse, we may start with the expressions (i), (ii), (iii) for p, q, r. These ensure that P, Q, R are collinear; but, in order that (iv) and (v) determine the same division ratios for P, Q, R, we must have
,.y' 'y'a' y
y  a
a 
= h,
or
(vii)
a'ha =fl'hfi=y'hy=k,
h and k representing the values of the equal members of (vi) and (vii). In the usual way we now find from (i), (ii), (iii) that
aa'haa=6'b'hfib=y'c'hyc=ks, where S is a point common to AA', BB', CC'.
FIG. 7e
Example 5. The Complete Quadrangle. A complete quadrangle consists of four coplanar points, its vertices, no three collinear, and the six lines, its sides,
which join them. The three pairs of sides which do not meet at a vertex
COPLANAR POINTS
37
17
arc said to be opposite; and the three points in which they meet are called rliagon 0, the reverse if X < 0. Vectors may be added, subtracted, and multiplied by real numbers in conformity with the laws of ordinary algebra. The n vectors ui are linearly dependent if there exist n real num
bers Ai, not all zero, such that EAiui = 0. When n = 2, 3, linear dependence implies that the vectors are collinear or coplanar, respectively; and conversely. In space of three dimensions any four vectors are linearly dependent.
A point P divides a segment AB in the ratio X = /3/a when
3
3
AP = APB; then
3
>
3
(a + $)OP = aOA + SOB. 3
A linear relation2;A1OP1 = 0 connecting n position vectors will
hold for any origin when and only when :Ai = 0. When n = 2, 3, 4, the points Pi are coincident, collinear, or coplanar, respectively.
A set of weighted points miPi has a unique centroid P if 2;mi $ 0; its defining equation is
3 * XmiOPi and OP* =
lmiP*Pi = 0;
Emi
If all mi = 1, P* is called the mean center. The scalar product u v is defined as
Jul Ivlcos(u,v). Laws: V. Up
uv
U. (v+w) =
uv uv=lu
is a e
a dextral set.
I vl
u
v
VECTOR ALGEBRA
58
§ 28
Laws :
UXV = `VxU,
Ux(V+w) = uxv+uxw.
If u, v 0, u x v = 0 implies that u l v and conversely. Expansion rule: l
ux(vxw) = Cross multiplication is not associative. The box product, u x v w or [uvw], is numerically equal to the volume of a "box" having u, v, w as concurrent edges; its sign is + or  according as u, v, w form a dextral or sinistral set. The value of u x v w is not affected by a change in cyclical order of the vectors, or by an interchange of dot and cross. If u, v, w 54 0, [uvw] = 0 implies that u, v, w are coplanar, and conversely. A set of three vectors ej forms a basis if E = [ele2e3] 0. To
every basis there corresponds a unique basis ei such that ej e' = S1; such bases are called reciprocal. If the indices i, j, k form a cyclical permutation of 12 3, ei x ek
e,xek
ei = [ele2e3] ;
ei =
[ele2e3][e'e2e3] = 1.
[e1e2e3]
Both bases are dextral if E > 0; sinistral, if E < 0. Given a basis e1, a vector u may be written as Euiei or ruiei; the numbers ui are contravariant components of u, ui covariant components.
u+v = E(ui'+vi')ei = E(ui+vt)ei; Xu = 2:xuiei = 2;xuie`;
uV=
uIv1
+ u2v2 + u3v3 = u1v1 + u2v2 + u3v3; el
U x v = E1
e2
e3
I
I
e3
v2
v3
E ul
U2 V3
[uvw] = E1 I
e2 el
vl U1
u2
U3
V1
V2
V3
W1
W2
W3
Ut
E
u2
1vi
u3 I
w2
w3 When the basis ej is selfreciprocal (ei = ei), its vectors form a mutually orthogonal set of unit vectors; E = 1, if the basis is dextral, E = 1, if sinistral. A dextral selfreciprocal basis is written
PROBLEMS
59
i, j, k. For such a basis ui = ui; these rectangular components are given by ui = I u I cos (ei, u),
and
= (ul)2 + (u2)2 + (u3)2. The two formulas previously given for u + v, Au, u  v, u x v, u x v w in each case become identical. I U I2
The equations of a point, line, and plane, in terms of their homogeneous coordinates (a, ao), (b, bo), (c, yo), are
ra=ao, respectively.
rxb=bo,
r  c=yo,
When the origin is shifted to P,
ap = ao+POa,
by = bo+POxb,
yP =
The moment of the line vector (f, fo) about the point P is fp = r x f, where r is any vector from P to its line of action. The moment of (f, fo) about an axis through P is the component of fp on this axis; if the axis is given by the unit line vector (e, eo), this axial moment
PROBLEMS
1. If ABC is any triangle and L, M, N are the midpoints of its sides, show that, for any choice of 0, a  3
3
3
4
3
3
OA + OB + OC = OL + OM + ON.

, >
9
2. If OA' = 3 OA, OB' = 2 OB, in what ratio does the point P in which AB and A'B' intersect divide these segments? 3. Show that the midpoints of the four sides of any quadrilateral (plane or skew) form the vertices of a parallelogram.
4. P and Q divide the sides CA, CB of the triangle ABC in the ratios 4 3 x/(1  x), y/(1  y). If PQ = X AB, show that x = y = X. 5. E, F are the midpoints of the sides A B, BC of the parallelogram A BCD.
Show that the lines DE, DF divide the diagonal AC into thirds and that AC cuts off a third of each line. 6. OAA', OBB', OCC', ODD' are four rays of a pencil of lines through 0 cut by two straight lines ABCD and A'B'C'D'. If C and D divide AB in
the ratios r and s, C' and D' divide A'B' in the ratios r' and s', prove that
r/s = r'/s'. 7. The points A, B, C and A', B', C' lie, respectively, on two intersecting Show that BC', CB'; CA', AC'; AB', BA' intersect in the collinear points P, Q, R (Pascal's Theorem). lines.
VECTOR ALGEBRA
60
8. Lines drawn through a point P and the vertices A, B, C, D of a tetrahedron cut the planes of the opposite faces at A', B', C', D'. Show that the sum of the ratios in which these points divide the segments PA, PB, PC, PD
is 1. [Equation (10.2), with e =  (a + 0 + y + S) may be written
a+i9+y+s+e =0.
as+8b+yc+ad+ep =0,
From this we conclude, as in § 7, ex. 3, that a' _ (aa + ep)/(a +,E), etc.] 9. The line DE is drawn parallel to the base AB of the triangle ABC and is included between its sides. If the lines AE, BD meet at P, show that the line CP bisects AB. 10. The points P, Q, R divide the sides BC, CA, AB of the triangle ABC in the ratios a/(1  a), l3/(1  0), y/(1  y). If P, Q, R are collinear,
x+y+z=0;
xp+yq+zr=0,
putting p = (1  a)b + ac, etc. in this equation, we obtain a linear relation in a, b, a whose coefficients have a zero sum. Show that this implies that the separate coefficients vanish; hence deduce the Theorem of Menelaus (§ 7, ex. 2):
afy/(1  a) (1  p)(1  y) = 1. 11. Using an argument patterned after that in Problem 10, prove Carnot's Theorem:
If a plane cuts the sides AB, BC, CD, DA of a skew quadrilateral ABCD
in the points P, Q, R, S, respectively, the product of the ratios in which P, Q, R, S divide these sides equals 1.
12. Id a plane quadrilateral ABCD, the point P in which the diagonals AC, BD intersect divides these segments in the ratios 4/3 and 2/3, respectively. In what ratio does the point Q, in which the sides AB, CD meet, divide these segments?
13. If el = O 1, e2 = OE2i e3 = 0
0
3E
form a basis, the reciprocal set
el, e2, e3 are vectors perpendicular to the planes OE2E3, OE3E1i OE1E2 and
having lengths equal to the reciprocals of the distances of El, E2, E3 from these planes, respectively. Prove this theorem.
14. Prove that a necessary and sufficient condition that four points A, B, C, D be coplanar is that [dbc] + [add] + [abd]  [abc) = 0. 15. Show that the shortest distance from the point A to the line BC is
Iaxb +bxc +cxal/Ib  cI.
16. If the vectors el, e2, e3; el, e2, e3 form reciprocal sets, show that
the vectors e2 x e3, e3 x el, el x e2; e2 x e3, e3 x e1, el x e2 do likewise.
17. Prove the formulas: (a)
[a x b, bxc, c x a] = [abc]2,
(b)
(bxc) (a x d) +(c x a) (b x d) +(a x b)
(c)
(bxc) x (a x d) +(c x a) x (b x d) +(a x b) x (c x d) = 2[abc]d,
(c x d) =0,
(d)
(da) (bc)+(db) (ca)+(dc) (ab)=0,
(e)
(ad) x (bc)+(bd) x (ca)+(cd) x (ab) =2(a x b+b x c+c x a).
PROBLEMS
61
18. Find the shortest distance between the straight lines AB and CD when (a)
A(2, 4, 3), B(2, 8, 0);
(b)
A(2, 3, 1),
B(0, 1, 2);
C(1, 3, 5),
D(4, 1, 7),
C(1, 2, 5),
D(3, 1, 0).
19. Show that the lines AB, CD are coplanar, and find the point P in which they meet:
A(2, 3, 4),
B(2, 3, 0);
C(2, 3, 2),
D(2, 0, 1).
4
[CD is parallel to CP = AP  AC = XAB  AC; find X and then P from
4
>
OP = OA + XAB.] 20. If the points P, Q, R divide the sides BC, CA, AB of the triangle ABC in the same ratio, show that A, B, C and P, Q, R have the same mean center. 21. If the vectors OA, OB, OC lie in a plane and are equal in length, prove
that OA + OB + OC = OH, where H is the orthocenter of the triangle ABC. [§ 15, ex. 4.] 22. The power of a point P with respect to a sphere of center C and radius r is defined as (CF)2  r2. Prove that the power of P with respect to a sphere
having AB as diameter is PA I'B. , P,,, 23. Prove that the sum of the n2 powers of n given points, P1, P21 with respect to the n spheres having for diameters the n segments joining the given points to a variable point P in space, is constant. [Am. Math. Monthly, vol. 51, p. 96.1
24. The lines (a, ao), (b, bo)'are coplanar; then a bo + b ao = 0 (26.14). Show that they determine the plane (a x b, ao b) if a x b 0 0; and the point
00.
aoxbo) if
Solve Problem 19 using these results.
25. Show that the three planes (a, ao), (b, $o), (c, yo) meet in the point, ([abc],
aobxc +Rocxa +yoaxb),
provided [abc] 0 0. 26. Show that the three points (a, ao), (0, bo), (y, co) determine the plane, (a bo x co + 0 co x ao + y ao x bo,
[aobocol),
provided aboxco +3coxao +yaoxbo 0 0. 27. The points P, Q, R divide the sides BC, CA, AB of the triangle ABC in the ratio of 1/2. The pairs of lines (AP, BQ), (BQ, CR), (CR, AP) intersect at X, Y, Z, respectively. Show that the area of the triangle XYZ is 1 /7 of the area of ABC. [Twice the vector area of XYZ is y x z + z x a +
xxyl
When P, Q, R divide the sides in the ratio t/1, the area of XYZ is
(1  t)2/(1 + t + t2) times the area of ABC.
VECTOR ALGEBRA
62
28. Using (26.16), prove that (a) If four points (a, ao), (0, bo), (y, co), (S, do) lie on a plane, the vectors ao/a, , do/S are connected by a linear relation in which the sum of the coefficients is zero;
(b) If four planes (a, ao), (b, 3o), (c, yo), (d, So) pass through a point, the vectors a/ao, , d/So are connected by a linear relation in which the sum of the coefficients is zero, provided ao, Qo, 'yo, So 54 0.
29. If a sphere S with a fixed center cuts two concentric spheres S1, S2i prove that the distance between the planes of intersection SS, and SS2 is independent of the radius of S. Extend this theorem to cover the case when S fails to cut one or both of the concentric spheres. [Consider the radical planes SS1, SS2.1
30. Prove that the radical planes of the three spheres (r  c1)2 = P1 (1 = 1, 2, 3) meet in the line J
`C1 X C2 + C2 X C3 + C3 X c1,
2c1(e2  e3
 P2 + A3) + cycl.;,
their radical axis. Consider the case when e1 X C2 + C2 X C3 + C3 X cl = 0. 31. A plane system of forces Fi acting through the points ri is equivalent to a single force F = 2;Fi. When all the forces Fi are revolved about these points through an angle B, show that their resultant F also revolves through B about the point f (the astatic center) given by
t=
F X M  (11ri Fi)F 2 F
where M = 'ri x Fi.
When the forces are parallel (Fi = Xie), show that the astatic center is the centroid of the weighted points Xiri. 32. P* is the centroid of a set of n weighted points miPi for which Tmi = in. Prove the Theorems of Lagrange: (a) (b)
1mi(OPi)2 = m(OP*)2 + 2:mi(P*Pi)2.
Vmim1(PPi)2 = in mi(P*Pj)2,
where ij ranges over 2n(n  1) combinations.
[OPi = OP* + P*Pi;
PiP7 = P*Pi  P*Pi.]
33. Deduce the following results from Prob. 32. (a) If ABCD are the vertices of a square in circuital order, (OA)2 + (OC)2 _ (OB)2 + (OD)2 for any 0. Generalize. (b) If r is the radius of a sphere circumscribed about a regular tetrahedron of side a, r2 = 3x2/8.
(c) The mean square of the mutual distances of all the points within a sphere of radius r is 6r2/5.
34. If P* and Q* are the mean centers of p points Pi and q points Q respectively, prove that Mean (PA)2 = Mean (P*Pi)2 + Mean (Q*Q1)2 + (P*Q*)2.
CHAPTER II MOTOR ALGEBRA
29. Dual Vectors. The vector f, bound to the line whose equation, referred to the origin 0, is
rxf = fo,
(1)
is completely determined by the two vectors f, fo, its Plucker coordinates. These obviously satisfy the relation, f fo = 0.
(2)
The vector f does not depend upon 0; but fo, the moment of f about 0, becomes (3)
fp=fo+POxf,
when the origin is shifted from 0 to P. We now amalgamate f and fo into a dual vector.
F=f+efo
(4)
where a is an algebraic unit having the property e2 = 0. If f = 1, F is called a unit line vector. The unit line vectors,
A=a+eao, stand in onetoone correspondence with the c0 4 lines of space. A line vector F of length X always may be written F = XA, where A is a unit line vector. Unit line vectors depend upon four independent scalars, general line vectors upon five. Finally, for applications in mechanics, we consider dual vectors F without the restriction (2). The vectors f, fo then involve six independent scalars. The dual vector (or the entity it represents) then is called a motor, provided its resultant vector f is independent of the choice of 0, while its moment vector fo changes in accordance with (3) when the origin is shifted to P. Line vectors are thus special motors for which f fo = 0; for unit line vectors also f f = 1. 63
MOTOR ALGEBRA
64
§ 30
30. Dual Numbers. In analogy with the complex numbers x + ix', W. K. Clifford introduced dual numbers x + ex', in which x, x' are real and e is a unit with the property e2 = 0.
In x + ex', x is called the real part and x' the dual part. We write
x{ex'=y+ey' when x=y,x'=y'; x+ex'=0 when x = 0, x' = 0.
Addition and multiplication of dual numbers are defined by the equations : (1) (2)
(x+ex')+(y+ey') =x+y+e(x'+y'), (x + ex') (y + ey') = xy + e(xy' + x'y)
Observe that (2) may be obtained by distributing the product on the left and putting e2 = 0. From these definitions we see that addition and multiplication are commutative and associative and that multiplication is distributive with respect to addition. In fact, the formal operations are precisely those of ordinary algebra followed by setting e2 = e3 = . . . = 0. The negative of x + ex' is defined as x  ex'.
If A = a + ea', B = b + eb', the difference X = A  B and quotient Y = A/B satisfy, by definition, the equations B + X = A, BY = A. We find that (3)
A  B = a  b + e(a'  b');
and that BY = A has the unique solution, a'b  ab' A a
B=b+ e
(4)
b2
when and only when b F6 0. Division by a pure dual number eb' is not defined. The quotient (4) may be remembered by means of
the device (a + ea') (b  eb')/ (b + eb') (b  eb') used in complex algebra.
A dual number a + ea' in which a 5=1 0 is said to be proper; the product and quotient of proper dual numbers are also proper. If the dual product (2) is zero, there are three alternatives:
x=x'=0;
y=y'=0;
x=y=0.
The last shows that a dual product can vanish when neither factor is zero; for any two pure dual numbers have zero as their product.
MOTORS
§ 31
65
If the function f(x) has the derivative f'(x), we define its value for the dual argument X = x + Ex' by writing down its formal = 0; thus Taylor expansion and setting E2 = E3 = Ax + Ex') = AX) + Ex'f' (x)
(5)
In particular, (6)
sin (x + Ex') = sin x + Ex' cos x,
(7)
cos (x + ex') = cos x  ex' sin x.
When x = 0, we have sin Ex' = Ex', cos ex' = 1; consequently (6) and (7) have the form of the usual addition theorems of the sine and cosine. Note also that sine X + cost X = 1. 31. Motors. We now can characterize a motor as a dual multiple of a unit line vector. Thus, on multiplying the unit line vector
A = a + Eao (a a = 1, a ao = 0) by the dual number X + EX', we obtain the dual vector, M = m + Emo = (X + EX') (a + Ea.o).
(1)
On equating the real and dual parts of both members, we obtain m = Xa,
(2)
mo = Xao + A'a.
To show that M is a motor we need only verify that mo transforms in accordance with (29.3):
MP =mo+POxm.
(3)
In view of (2), this result follows from
mp = Nap + A'a = X (ao + PO a) + X'a _ (Xao + X'a) + PO x (Xa).
From (3), m mp = m mo; hence the scalars m m,and m mo are invariants in the sense that they are not altered by a shift of origin.
From (2),
(4)
m
0, we call the invariant, X'
(5)
mmo
mm
MOTOR ALGEBRA
66
the pitch of the motor.
§31
Choosing A > 0, we have the unique
solution,
A=ImI, Aa=m,
A'=,zim1, Aao=moµm.
When m 0 0, M has the dual length, A + EX, = I m { (1 + eµ) (6) and its axis is along the line vector, (7)
ImIA=m+e(moµm).
The equation of the axis is therefore
rxm =mo gym=
(8)
(mxmo)Xm
;
this shows that the axis passes through the point Q given by f mxmo
r=OQ=
(9)
;
and, from (3),
mQ=mo 
(10)
(mxmo)Xm
=µm.
At all points on its axis M has the same moment µm, a fact also apparent from (7). Only for points P on the axis is the moment mp parallel to m.
Motors for which m
0 (A 0) are called proper. Proper motors are screws if m mo 0 0 (A, A' 0 0) ; line vectors if m mo = 0 (A 0 0, A' = 0). Only proper motors have a definite axis.
If m = 0, mo 0 0 (A = 0, A' 0), M = emo is pure dual; then mo is not altered by a change of origin. In this case M is called a couple of moment mo. A couple may be regarded as a screw of infinite pitch with an axis of given direction but arbitrary position in space.
Finally if m = 0, mo = 0, the motor M = 0, then A = A' = 0 from (2). Hence M = 0 only when its dual length A + eA' = 0. Our classification of motors is therefore as follows: Screw
0;1
} Proper
Line vector Couple Zero
(m=0, (m=0,
m0 =0):A=0,A' =0.
MOTOR SUM
§ 32
67
If two proper motors M, N are connected by a linear relation with proper coefficients,
(a + ia')M + ($ + ia')N = 0,
as 34 0,
either may be expressed as a dual multiple of the other; hence both M and N are multiples of the same unit line vector and are therefore coaxial. Conversely, if M and N are coaxial, they satisfy a linear relation with proper coefficients.
32. Motor Sum. The sum of the motors M = m + em0, N = n + eno is defined as the motor,
M + N = m + n + e(mo + no).
(1)
That M + N is a motor follows from (29.3) :
mp+nr = mo+no+POx(m+n) THEOREM 1.
The sum of two line vectors is a line vector only when
their axes are coplanar and their vectors have anonzero sum. Proof.
If M and N are line vectors, M + N is a line vector
when and only when
m+n
0,
and (m + n) (mo + no) = 0.
Since m mo = 0, n no = 0, the latter condition reduces to (2)
which is precisely the condition (26.14) that the axes be coplanar. THEOREM 2. If two line vectors intersect, their sum is a line vector
through the point of intersection. Proof. If r1 is the position vector of the point Pl in which the
line vectors M and N intersect, we may take mo = rl x m, no = r1 x n. Since the axis of M + N has the equation,
r (m + n) = mo + no = r1 x (m + n), it passes through the point P1. THEOREM 3.
If the line vectors M and N are parallel and n =
Xm (X F6 1), the axis of their sum divides any segment from M to N in the ratio X/1.
68
MOTOR ALGEBRA
§ 33
Proof. If Pl, P2 are points on the axes of M and N, we may
take mo = rl x m, no = r2 x n. The line vector,
M + N = (1+A)m+E(r,+Ar2)xm; and the equation of its axis, r x (1 + A)m = (r1 + Xr2) x m,
is satisfied by the point r = (r1 + Ar2)/(1 + X) which divides PiP2 in the ratio A/1. The division is internal when m and n have the same direction (X > 0), external when they have opposite directions (A < 0). THEOREM 4. couple.
The sum of two couples, if net zero, is another
Proof. If M and N are pure dual, M + N, if not zero, is also pure dual.
If M and N are line vectors such that n = m and P1, P2 are points on their respective axes, M + N is a couple of THEOREM 5.
moment (r1  r2) x m. Proof.
Since M = m + er1 x m, N = m + ere x (m), M + N = e(rl  r2) x m.
33. Scalar Product. The scalar product of the motors M = m + emo, N = n + eno is defined as the result of distributing the product,. (m + emo) + (n + eno), namely, (1)
This dual number is independent of the choice of origin; for on computing mp, np from (29.3), we find that (2)
The definition (1) shows that (3)
The definition (1) for M M. N differs from that given by R. von Mises ("Motorrechnung, ein neues Hilfsmittel der Mechanik," Z. angew. Math. Mech., vol. 4, 1924, p. 163) who defines M N as the invariant real scalar (2). The definition (2) is suggested by applications in mechanics; its consequences are (a) the familiar rules of vector algebra do not all carry over into motor algebra, and (b) the elegant generalization of the scalar product of unit vectors, given in (5), is lost.
SCALAR PRODUCT
§ 33
69
In order to compute the scalar product of two unit line vectors A = a + ea0i B = b +  ebo, we shall suppose first that they are not parallel and write axb = esin (P(u) = t + P.
Let equations (1) define a regular analytic space curve. The position vector of the point P(t) is (2)
r = i x(t) + j y(t) + k z(t) = r(t);
and, if we write t, z for dr/dt, dx/dt, (3)
t = it(t) + j y(t) + k i(t) F6 0.
If Po(t = to) is a fixed point on the curve, the length of the arc s from Po to P is defined as (4)
s=
fV
dt,
to
an analytic function of t which is positive or negative, according as t > to or t < to; moreover (5)
ds/dt = VTt > 0,
The inverse function t = 0. Putting t = 
+ w X (AS  AR).
 
VECTOR FUNCTIONS OF ONE VARIABLE
122
>
>
du
du
§ 56
But, since AS  AR = RS = PQ = u, d =
(3)
+ 0).U; dt
in particular du
(4)
dt
=wxu
if u is fixed in a'.
When u = w, the angular velocity of the frame a', we have, from (3), dw
d'w
dt
dt
(5)
The vector dw/dt is denoted by a and called the angular acceleration vector of a. Example. Kinematic Interpretation of Frenet's Formulas. At any point P of a space curve, the trihedral TNB may be used as a frame of reference.
If P moves along the curve with unit speed, ds/dt = 1 and s = t if t = 0 at the origin of arcs. Then the arc s may be interpreted as the time, and dr/ds = T is the velocity of P. If w is the angular velocity of TNB referred to a "fixed" frame a, we have, from (4), dT
ds =
dN
, x T,
ds = p1 x
N,
dB
ds = "
x B.
A comparison with Frenet's Formulas (44.6) shows that 0) = S, the Darboux vector. The Darboux vector of a space curve is the angularvelocity vector of its moving trihedral TNB. Since the vertex P of the trihedral has the velocity T, its motion is represented completely by the velocity motor,
V =S+eT=rT+KB +Er. From (54.7) and (54.8), we see that the axis of this motor passes through the point Q given by >
PQ =
8xT S
8
K
ST
= ,,2+,2N and vQ = SS S =
r K2 +
'28.
In general, the trihedral TNB has, at every instant, a screw motion: a combina
tion of the angular velocity S about an axis through Q and the velocity vQ along this axis. The velocity of translation vanishes when and only when the curve is plane (r = 0); then S = KB, PQ = pN. For plane curves TNB has, at
every instant, a pure rotation of angular velocity KB about the center of curvature.
THEOREM OF CORIOLIS
§ 57
123
57. Theorem of Coriolis. Let v and a denote the velocity and
acceleration of a particle P, relative to a frame , while v' and a' denote these vectors relative to a frame ' in motion with respect to . Then if 0 and A are origins fixed in and a', we have
r= OP,
dr
a= dv;
v=  ,
dt
dt
r' = AP
v'=
d'r' dt
d'v'
a
dt
,
If w is the angular velocity of ' with respect to , we have, on differentiating,
 3
r = OA + r', twice with respect to the time, and, on making use of (56.3), dr'
V =VA+
dt
= VA +wxr'+v', dr' dv' a=aA+axr'+wx+ dt dt
/
(a
dw
dt
J
a A+ a x r' + w x ((o x r' + V) + w x v'+ a'
a4+axr'+wx(wxr')+2wxv'+a'. The velocity and acceleration of the point Q of the frame j' with which P momentarily coincides are called the transfer velocity and transfer acceleration of P. To find them, put v' = 0, a' = 0 in the foregoing equations; thus VQ = VA + wxr',
aQ = aA + ax r' +wx Our equations now read (1)
V=vQ+V',
(2)
a=aQ+2wxv'+a'.
Equation (1) restates the theorem on the composition of velocities already proved in § 55. Equation (2) shows that an analogous theorem for the composition of accelerations is not in general true;
VECTOR FUNCTIONS OF ONE VARIABLE
124
§ 57
we have, in fact, the additional term 2w x v', known as the Coriolis acceleration.
If we regard the frame
as "fixed" and rates of
change referred to it as "absolute," while the corresponding rates
referred to ' are "relative," we may state (2) as the The absolute acceleration of a particle is equal to the sum of its transfer acceleration, Coriolis acceleration, and relative acceleration. THEOREM OF CORIOLIS.
The Coriolis acceleration, 2w x v', vanishes in three cases only: (a) co = 0; the motion of ' relative to is a translation.
(b) v' = 0; the particle is at rest relative to '. (c)
v' is parallel to co.
A particle of mass m, acted on by forces of vector sum F, has the equation of motion F = ma. When the motion is referred to a rotating frame the equation of motion becomes N U
P(Q)
(3)
ma' = F  maQ  2mw x v'.
If we regard maQ and 2mw x v' as fictitious forces, ma' equals a sum of forces just as in the case of a fixed frame. The term
V O(A)
Fio. 57a
2mw x v' is called the Coriolis force. When ' has the constant angular velocity w about
a fixed axis through A, a = 0, aA = 0, and aQ = w x (w x r'). Taking 0 at A and the zaxis along w (Fig. 57a), we have
maQ = mw2 (r'  k r' k) = mw2 (OP  ON) = mw2 NP; this vector perpendicular to the axis and directed outward is called the centrifugal force on the particle. Example. Particle Falling from Rest. Refer the particle, originally at 0, to the revolving frame Oxyz attached to the earth: Oz points to the zenith (along a plumb line), Ox to the south, Oy to the east (Fig. 57b). The earth
revolves from west to east about the axis SN and its angular velocity at north latitude a is _ 2r (k sin X  i cos a) radians/see. 24 X 602
When the particle is at rest relative to the earth, the force acting upon it is its local weight mg; hence in (3) F  maQ = mg. Therefore the equation of motion becomes (i)
a'=g2o,xv',
§ 57
THEOREM OF CORIOLIS
125
where a' and v' are the acceleration and velocity relative to the earth. We shall integrate (i) by successive approximations under the initial conditions = 0, v' = 0 when t = 0 and with g regarded as constant. 1. If we neglect the term in w,
a' = g,
v' = gt,
=zgt2.
The displacement r' is along the plumb line.
Fla. 57b
2. With v'  gt, (i) gives a' = g  2tw x g,
v'=gt12wxg, ' = zr
gt2  3 taw x g.
Since g = gk, the second term gives an eastward deflection,  3 taw x g = 3 t3wg cos Xj.
3. With the last value of v', a' = g  2tw x g + 2t2w x (w x g), v' = gt  t2w x g + 3 taw x (w x g), = 2gt2  3t3w x g + Gt4w x (w x
g).
VECTOR FUNCTIONS OF ONE VARIABLE
126
The last term is a deflection in the meridian or xzplane.
§ 58
Since w x g = wg
cos X j, wng
w x (w X g) =
r=
1 s tow" 9 sin X cos X i +
cos x (i sin X + k cos X),
1 taw 3 9
cos X J' 1 t2  114w2 9 cos' X) k. (29 6
The first and second terms give deflections to the south and east; the third shows that the particle in t seconds falls through a distance z = 2gt2  I t4w2g cost X.
58. Derivative of a Motor. Let the motor M= m + EmA be referred to the frame a' in which A is a fixed point; and let ' itself have the motion V= o + EV A with respect to the frame . If m and mA are functions of the real variable t, the derivative of
M relative to a' is
dm
d'M (1)
dt
dt
d'mA
+
e
dt
In order to compute dM/dt relative to a we must remember that mA depends not only upon t but also upon the point A, which is in motion relative to a. If A moves to B in the interval At, Om.4
mB(t + At)  m4(t)
At
'At
MA (t + At) + AB X m(t + At)  m.4 (t) At
mA (t + At)  mA (t) At
+ OB  At OA X m(t + At),
and hence, if rA ° OA, dmA
AMA
c l.o
dt 
At
drA
+
xm.
dt
Relative to the frame a, we therefore define dM
dm
dmA
dt
dt
dt
(2)
drA
+
dt
m)
To verify that dM/dt is a motor, we must show that dmP dt
drp
+
dt x m =
dmA
+ dt 
dr,4
dt
) dm
x m + PA X
dt
(31.3)
DERIVATIVE OF A MOTOR
§ 58
127
This equation, in fact, follows from
mp = mA + (rA  rp) x m on differentiation. If t denotes time, drA
VA; dt =
and
dm
wxm+
dt =
d'm
d'mA
dmA
dt =
dt '
dt
'
Substituting these results in (2) gives
from (56.3).
dM (3)
dt
d'M
= (0M+ E((axmA + VA xm)
+
dt
or, in view of (34.1),
d'M
dM (4)
V X M +
dt
dt
This is the motor analogue of (56.3). Making use of the definition (2), we may verify that the following rules of differentiation are valid : dM
d ( 5)
dt ( M
+ N)
dt
d ( 7)
dt d
( 8)
dt
+
dt
dt '
dM
d (6)
dN
(XM) = X
(M N) = M (M
x
N)
=
dt
dA
+
dt
M, dM
dN '
dt
+N
dt
N dM
dN Mx
dt '
+
.
dt
Example 1. If the motion of a rigid body is given by the velocity motor V = w + EVA (A a fixed point of the body), its acceleration motor is dV
= a+e(aA +VAXW).
The axis of this motor is called the instantaneous axis of acceleration; its equation may be written from (31.8).
VECTOR FUNCTIONS OF ONE VARIABLE
128
§ 59
Example 2. Let F = f + EfA be a force acting on a rigid body whose velocity motor is V = w + EvA. Then dF
(i)
dt + E ( dt + VA x
f)
.
If F acts at the point P of the body, fp = 0 and dF
(ii)
= df + Evp x f, referred to P.
Hence, referred to A,
d
(iii)
do+E(vPxf+APxdt/
let the reader show that (i) and (iii) are consistent. Example 3. The moving trihedral TNB at the point P of a space curve r = r(s) may be regarded as a rigid body having the velocity motor,
V=S+Evp=8+ET, as the point P traverses the curve with unit speed. The three line vectors T = T, N = N, B = B through P are fixed in the trihedral; hence, from (4),
ds =VxT,
dN =VxN,
B
=VxB.
These are Frenet's Formulas in motor form. Written out in full they become
T = KN,
dN = 7T + KB + EB,
d _ rN  EN;
for example dB
 =(S+ET)xB=dB EN=TNEN. 59. Summary: Vector Derivatives. The derivative du/dt of a vector function u(t) is defined as the limit of Au/At as At approaches zero. If u = OP is drawn from a fixed origin and P describes the
curve C as t varies, du/dt is a vector tangent to C at P in the direction of increasing t. If I u I is constant, C is a circle and du/dt is perpendicular to u. The derivative of a constant vector is zero. The derivatives of
the sum u + v and the products fu, u v, u x v are found by the familiar rules of calculus; but for u x v the order of the factors must be preserved.
§ 59
SUMMARY: VECTOR DERIVATIVES
129
If s is the are along a curve r = r(s), dr/ds = T, a unit vector tangent to the curve the direction of increasing s. The unit prin
cipal normal ft to the curve has the direction of dT/ds; and the unit binormal B = T X N; then [TNB] = 1. The vectors of the moving trihedral TNB change conformably to Frenet's Formulas: dT
= SXT,
Kft
ds dN
KT
+ TB = 6 X ft,
ds
dB
rft
= SXB;
ds
K is the curvature, r the torsion of the curve; and the Darboux vector
o = TT + KB is the angular velocity of the moving trihedral as its vertex traverses the curve with unit speed (ds/dt = 1). For plane curves r = 0. If t denotes time, a particle P traveling along the curve r = r(t) has the velocity and acceleration, dr
v
dt '
a

dv
d2r
dt
dt2
If v = ds/dt is the speed, and p = 1/K the radius of curvature of the path, V = VT,
dv
v2
dt
p
a =  T+  ft.
If A is any point of a free rigid body, the velocities of its points are given by the velocity motor, V = w + EvA ;
here co, the angular velocity vector, is the same for any choice of A.
Since V is a motor, for any point P of the body, Vp = VA + PA Xw = vA +(0 XAP.
If, at any instant, co 0, the body has a screw motion about the axis of V, the instantaneous axis of velocity; this reduces to a pure rotation if V is a line vector (co VA = 0). If co = 0 the motion is an instantaneous translation of velocity VA.
130
VECTOR FUNCTIONS OF ONE VARIABLE
If the frame a' has the angular velocity relative to , the rates of change of a vector u relative to these frames are connected by du
d'u
dt
dt
+
w .U.
A particle P has the velocity and acceleration v', a' relative to frame a', v, a relative to the frame a; then, if the motion of a' relative to a is given by the motor V = co + EvQ, where Q is the point of ' coinciding at the instant with P,
a = aQ +2wxv' + a'.
v = VQ +v',
The term 2w X v' is the acceleration of Coriolis. When a' is in trans
lation relative to , the velocity equation may be written v = vl, + V'. The derivative of the motor M = m(t) + emA(t) is defined as dM
dm
dmA
at = at +
at_
drA
+
Xm
at
If t denotes time, and d'M/dt refers to a frame a' having the motion V = co + evA, relative to a frame , then d'M
dM dt
V X M +
dt
PROBLEMS
1. If r and x are the distances of a point on a parabola from the focus and
directrix, r  x = 0. Show that (R  i)  T = 0, and interpret the equation. 2. Prove that the tangent to a hyperbola bisects the angle between the
focal radii to the point of tangency. [Cf. § 44, ex. 2.1 3. An equiangular spiral cuts all vectors from its pole 0 at the same angle a (R T = cos a). If (r, 0) are the polar coordinates of a variable point P on the spiral, show that (a) (b)
ds/dr = sec a, 1 dr
rd0
= cot a,
(c)
s  so = (r  ro) sec a; log r = (9  Bo) cot a; ro
p = ds/dB = r/sin a;
and that the center of curvature is the point where the perpendicular to OP
at 0 cuts the normal at P.
PROBLEMS
13 1
4. If r = r(s) is a plane curve. r, show that (a) r1 = r + eN (c cont.) is a parallel curve r1 (§ 45, ex.l at it tance c from r; (b) Si = s  c, provided Si = s = 0, when ,y = 0;
ii: l dis
(C) P1 = P  C.
5. A curve r = r(s) has the property that the locus r1 = r + rT (e const.) is a straight line. Prove that the curve is plane, in fact, a tractrix. [Cf. § 50, ex. 5.)
6. An involute of a curve r, r = r(s), is a curve r1 which cuts the tangents of r at right angles. Prove that r has the oneparameter family of involutes,
r1 = r + (c  s)T
(c = const.);
and, if we take T1 = N, ds1/ds = (c  s)K. 7. Show that the involute of the circle,
y = a sin t,
x = a cos t,
obtained by unwrapping a string from the point t = 0 is
y1 = a(sint  tonst);
x1 = a(cost +tsint),
and that s1 = Zat2 gives the arc along the involute. 8. The cylinders x2 + y2 = a2, y2 + z2 = a2 intersect in two ellipses, one of which is x = a cos t, y = a sin t, z = a cos t.
Show that its radius of curvature is
P = a(l + sine t)I/''2. 9. The equation of a cycloid is
y = all  cos t).
x = a(t  sin t), Prove that
t
1
2
2J
(a)
T = [sin , cos
(b)
>y = (i, T) = 2  2 ,
,
ds/dt = 2a sin
2
P = 4a sin 2
(44.3);
(c) The equation of its evolute is x1 = a(t + sin t),
y1 = a(1  cost).
10. Find the vectors T, N, B and the curvature and torsion of the twisted
cubic,
x = 3t, y = 312,
z = 20,
at the points where t = 0 and t = 1. Write the equations for the normal and osculating planes to the curve at the point t = 1. 11. Show that curvature and torsion of the curve, x = a(3t  t3),
y = 3at2,
z = a(3t + t3),
are
K = r = 1/3a(1 +
t2)2.
132
VECTOR FUNCTIONS OF ONE VARIABLE
12. Find the envelope of the family of straight lines in the xyplane, r = p(O)R + XP,
where R and P = k x R are the unit vectors of § 44, 0 = angle (i, R), and p is the perpendicular distance from the origin to the line. Show that the curve,
r, = pR + p'P is the envelope, and that
dsl/de = p + p".
Ti = T,
13. Find the envelope of the family of lines for which the segment included between the xaxis and yaxis is of constant length c. (In Problem 11 put p = c sins cos 0.1 Show that the envelope has the parametric equations, x = c sin3 0,
y = C C083 0;
and that the entire length of the curve is 6c. 14. Show that the curvature and torsion of the curve, are
/2 t,
x = et, y = et,
z=
Sc = T
+ at)2.
15. Verify that the curve, z = a cos t,
x = a sin2 t, y = a sin t cos t,
lies on a sphere. Show that the curve has a double point at (a, 0, 0) for the parameter values t = ±r/2, and that the tangents to the curve at this point are perpendicular.
16. If a curve r = r(s) lies on a sphere (r  c) (r  c) = a2, show that dp r  C = pN  TIB. ds
Hence, prove that
d (ldP)
pr+ ds
r ds
0
is a necessary and sufficient condition that a twisted curve (r d 0) lie on a sphere. 17. The points on two curves r and r, are in onetoone correspondence.
If T = T, at corresponding points, prove that N = N1,
B = B1,
dsl/ds = K/Kt = r/T,.
When both curves cut the rays from 0 at the same angle, show that r, = cr; and that s, = as, if both arcs are measured from the same ray. 18. A curve r is called a Bertrand curve if its principal normals are principal normals of another curve r1; then r1 is also a Bertrand curve, and
rl = r + XN, Ni = eN, where e = 1 or 1.
PROBLEMS
133
Show, in turn, that
Ti d = (1  cK)T + cTB;
X = c (const.),
(a)
11
(b) If
= angle (B, Bl), taken positive relative to N, B1 = B cos (P + T sin p,
ET1 = T cos rp  B sin rp;
= const., and
(c)
ET1
ds1
ds =
K sin p  r COS gyp, ds1
Eds =
(d)
COS cp + T sin gyp;
ds
1  CK Cr _ cos p sin
;
K1 = E(Ti cot (p + 1/c);
C2TT1 = Sin2 (P,
(e)
ds1
Kl
since K1 = 0, the last equation determines E. '(f) As to the Darboux vectors, Si ds1/ds = S. 19. The characteristics of a oneparameter family of planes whose homogeneous coordinates (§ 26) are (a(t), ao(t)), are the limiting lines of inter
section of the planes corresponding to the parameter values t and t + h as h  0. The line of intersection of the two planes has the homogeneous coordinates (26.10),
ao(t + h)a(t)  ao(t)a(t + h) [;
{a(t) x a(t + h),
and, if we divide both by h and pass to the limit h + 0, show that we obtain
(a x a', apa  aoa') as the coordinates of the characteristics. 20. From the result of Problem 19 show that the three families of planes
associated with the twisted (r 0 0) curve r = r(s), namely, (T, r T),
normal planes,
(N, r N),
rectifying planes,
(B, r B),
osculating planes,
have as characteristics the respective families of lines: [B, (r + pN) x B],
parallel to B through the centers of curvature;
(8, r x 8),
parallel to S through points of the curve;
(T, r x T),
tangent to the curve (§ 48).
21. If a particle P of mass m is subject to a central force F = mf (r)R, where
r = OP and R is the unit vector along OP, its equation of motion is z
m dt2 = F or
dv
= f (r)R.
Prove that r x v = h, a vector constant; hence, show that the motion is plane
and that r = OP sweeps out area at a constant rate (Law of Areas).
134
VECTOR FUNCTIONS OF ONE VARIABLE
22. When mf(r) = ymM/r2 in Problem 21 the particle P is attracted towards a mass M at 0 according to the law of inverse squares: dv dt
(a)
Show, in turn, that
=  k R
(k
rxv = h
(b)
h = r2 R x dR
(c)
,
=kdR,
(d)
dt xh
(e)
v x h = k(R + e),
where a is a constant vector.
From (b) and (e),
r x v h = h2,
r vxh = kr(1 +acose)
where 0 = angle (e, r); thus obtain the equation of the orbit, hz/k
(f)
1 + e Cos 0
a conic section of eccentricity a referred to a focus as pole. When e < 1, prove that the orbital ellipse is described in the periodic time,
7,  area of ellipse _ 2a
k a'
h/2 and, hence, (g)
T2/a3 = 4,r2/yM
(Kepler's Third Law).
(See Brand's Vectorial Mechanics, § 177.)
23. If the plane Ax + By + Cz = 1, fixed in space, is referred to rectangular axes rotating with the angular velocity w = [wl, w2, w3], show that
rdA dB dC] l dt
'
dt '
dt
=
[A, B, C] x [WI, (02Y w3].
24. A particle P has the cylindrical coordinates p, p, z (cf. § 89, ex. 1). If P moves in the pzplane so that dp/dt and dz/dt are constant while the plane itself revolves about the zaxis with the constant angular speed dip/dt = w, find the acceleration of P, (a) By direct calculation. (b) By use of the Theorem of Coriolis. 25. A real, everywhere convex, closed plane curve with a unique tangent at each point is called an oval. It can be shown that an oval has just two tangents parallel to every direction in the plane. The distance between these tangents is the width of the curve in the direction of the perpendicular. Prove that the perimeter of an oval of constant width b is 7rb (Barbier's Theorem).
CHAPTER IV LINEAR VECTOR FUNCTIONS
60. Vector Functions of a Vector. A vector v is said to be a function of a vector r if v is determined when r is given; and we write v = f(r). Since r is determined by its components, f(r) is a function of two or three scalar variables according as r varies in a plane or in space. A vector function may be given by a formula, as f(r) _ (a X r) x r; or it may be defined geometrically. Thus if
r = OP varies over the points P of a given surface, and v = OQ is the vector perpendicular on the tangent plane to the surface at
P, v = f(r). A vector function f(r) is said to be continuous for r = ro if (1)
lim f(r) = f(ro). r  ro
This means that, when the components of r approach those of ro in any manner, the components of f (r) approach those of f (ro). A vector function is said to be linear when (2)
(3)
f(r + s) = f(r) + f(s),
f (Xr) = Af (r),
for arbitrary r, s, X. For example, linear vector functions are defined by the formulas kr, a X r, a b r, in which k, a, b are constant. It can be shown that, when a continuous vector function satisfies the relation (2), it also satisfies (3) and is therefore linear.
Since we assume that (2) holds when r = s = 0, f(0) = 2f(0); hence, for any linear vector function, (4)
f(0) = 0.
A linear vector function is completely determined when f(a,), f(a2), f(a3) are given for any three noncoplanar vectors al, a2, a3. For, if
we express r in terms of al, a2, a3 as a basis,
r = xlal + x2a2 + x3a3, 135
136
LINEAR VECTOR FUNCTIONS
§ 61
we have, from (2) and (3), (5)
f(r) = x'f(a,) + x2f(a2) + x3f(a3)
Let a', a2, a3 denote the set reciprocal to a,, a2, a3 (§ 23); then, if we write f(ai) = b1, xt = r at,
f(r) may be written in either of the forms: (6)
f(r) = r
(7)
f(r) = (b,al + b2a2 + b3a3) r.
(albs + a2b2 + a3b3),
These formulas represent the most general linear vector function.
61. Dyadics. In linear vector functions of the form,
f(r) = alb, r+a2b2.r+...+anbn.r, we now regard f (r) as the scalar product of r and the operator,
= alb, + a2b2 + ... + anbn.
(1)
Assuming the distributive law for such products, we now write
f (r) = ' r.
(2)
Following Willard Gibbs, we call the operator 4) a dyadic and each of its terms aib1 a dyad. The vectors ai are called antecedents, the vectors bi consequents. While a b is a scalar and a x b a vector, the dyad ab represents a new mathematical entity. Gibbs regarded ab as a new species of product, the "indeterminate product." We shall find indeed that this product conforms to the distributive and associative laws, but
is not commutative (in general ab 0 ba). Since r follows. in (2), we call r a postfactor. If we use r as a prefactor, we get, in general, a different linear vector function:
g(r) = r
4).
We proceed to develop an algebra for dyadics, laying down for this purpose definitions for equality, addition, and multiplication of dyadics. Definition of Equality. We write 4) = NY when (3)
4>
r = 'Y r for every vector r.
DYADICS
§ 61
137
If s is an arbitrary vector, we have from (3) s  (4)  r) = s 
or
r
Since s 1 and s ' are two vectors that yield the same scalar product with every vector r, these vectors must be equal; thus s iD = s  T for every vector s.
(4)
Conversely, from (4) we may deduce (3). Therefore 4) = T when either (3) or (4) is fulfilled. The Zero Dyadic. We write = 0 when (5)
4)
r = 0 for every r.
The preceding argument shows that 4) = 0 also when s S. 4) = 0
(6)
for every s.
Definition of Addition. The sum 4) +' of two dyadics is defined by the property,
(4) +'Y) r=
(7)
If
for every r.
is given by (1), and ' = c1d1 + C2d2 + ... + Cmd,n,
4) +' = a1b1 + ... + anbn + c1d1 + ... + c..d,,,..
In the sense of this definition, 1 (or ') is the sum of its dyad terms, thus justifying our notation. The order of these dyads is immaterial; but the order of the vectors in each dyad must not be altered, for, in general, ab r ba r and hence ab 54 ba. The Distributive Laws,
(8), (9)
a(b + c) = ab + ac,
(a + b)c = ac + bc,
are valid. The proof follows at once from the definition of equality; thus from a(b + c) r = (ab + ac) r,
we deduce (8). We may now perform expansions as in ordinary algebra, if the order of the vectors is not altered; for example:
(a+b)(c+d) =ac+ad+bc+bd.
LINEAR VECTOR FUNCTIONS
138
§ 62
If X is any scalar, (Xa)b = a(Xb),
(10)
and we shall write simply Xab for either member.
From (60.6) or (60.7) we conclude that any dyadic D can be reduced to the sum of three dyads. To effect this reduction on 4) as given by (1), express each antecedent ai in terms of the basis ei, e2, e3, ai = ale, + aie2 + a;e3,
expand by the distributive law, and collect the terms which have the same antecedent: thus 4) _
(a1 ei
+ a2e2 + a3e3)bi = elf, + e2f2 + e3f3,
where f;
We also may reduce 4) to the sum of three dyads by expressing each consequent bi in terms of the basis el, e2, e3, and then expanding and collecting terms which have the same consequent; then 4) assumes the form, 4) = giei + 92e2 + 93e3
Thus it is always possible to express any dyadic so that its antecedents or its consequents are any three noncoplanar vectors chosen at pleasure.
If a,, a2, a3 are noncoplanar, a dyadic 4) is completely determined by giving their transforms (§ 60). If
4)  a, =b,,
4>  a2 = b2,
(D  a3= b3,
and the set a', a2, a3 is reciprocal to al, a2, a3, we have explicitly (11)
4) = bla' + b2a2 + b3a3.
For a physical example of a dyadic the reader may turn to § 116, where the stress dyadic is introduced. The name tensor, now used in a much more general sense, originally was ajTplied to this dyadic. 62. Affine Point Transformation. If we draw the position vectors,
r=OP,
r'=4 r=OQ,
from a common origin 0, the dyadic 4) defines a certain transformation of the points of space: to each point P corresponds a defi
COMPLETE AND SINGULAR DYADICS
§_63
139
nite point Q. If, when P ranges over all space, Q does likewise, this transformation is called affine; the dyadic 4) then is called complete.
Important properties of an affine point transformation follow at once from the equations,
4) (a+b) = which characterize a linear vector function. Since  0 = 0, the transformation leaves the origin invariant. Lines and planes are transformed into lines and planes. Thus, for variable x, the Line
r = a + xb  Line r' = a' + xb';
and, for variable x, y, the Plane r = a + xb + yc  Plane r' = a' + xb' + yc'. The transformed equations always represent lines and planes when 4) is complete; for we shall show in § 70 that b 54 0 implies b' F6 0, and b x c 5xI 0 implies b' x c' 54 0.
63. Complete and Singular Dyadics. Given an arbitrary basis, e1, e2, e3, we can express any dyadic in the form, (1)
4) = glee + 92e2 + g3e3
If we express r in terms of the reciprocal basis,
r = xle' + x2e2 + x3e3, then, by virtue of the equations, et e' = St, 4)
r = x1g1 + x282 + x383
When g1, g2, g3 are noncoplanar, r' = 4) r assumes all possible
vector values as r ranges over the whole of space. If we put 3 a r = OP, r' = OQ, the dyadic defines an affine transformation r' _ ( r of space into itself; 3dimensional Pspace goes into 3dimensional Qspace. A dyadic having this property is said to be complete. A complete dyadic cannot be reduced to a sum of less than three dyads; if, for example, we could reduce 4) to the sum of two dyads, ab + cd, all vectors r would transform into .vectors r' = a b r + c d r parallel to the plane of a and c.
LINEAR VECTOR FUNCTIONS
140
§ 63
If, however, g1, g2, g3, are coplanar, but not collinear, we can express each gi in terms of two nonparallel vectors f1, f2, and reduce
c to the sum of two dyads: 4, = f1h1 + f2h2.
(2)
This dyadic transforms all vectors r into vectors r' = 1  r in the plane of f1 and f2; 3dimensional Pspace goes into 2dimensional
Qspace. A dyadic having this property is said to be planar. A planar dyadic cannot be reduced to a single dyad ab; for then all vectors r would transform into vectors a b . r parallel to a. If 91, g2, g3 are collinear, we can replace each gi by a multiple of a single vector f and reduce 4' to a single dyad:
4 = fh.
(3)
This dyadic transforms all vectors r into vectors r' = fi  r parallel to f; 3dimensional Pspace goes into 1dimensional Qspace. Such a dyadic is called linear. Finally, if g1, g2, g3 are all zero, 4' = 0. Planar, linear, and zero dyadics collectively are called singular. The point transformation, OQ = (D OP,
corresponding to a singular dyadic 4' reduces 3dimensional Pspace to a 2, 1, or 0dimensional Qspace.
This discussion shows that, when  is reduced to the form (1) in which the consequents are noncoplanar, then 4, is complete,
planar, linear, or zero, according as the antecedents are noncoplanar, coplanar but not collinear, collinear, or zero. In particular, we have the
THEOREM. A necessary and sufficient condition that a dyadic al + bm + cn be complete is that the antecedents a, b, c and consequents 1, m, n be two sets of noncoplanar vectors.
As a corollary, (4)
4)  r = 0 implies r = 0 when  is complete.
For, if  = al + bm + cn,
alr+bmr+cnr=0, and hence r = 0.
1  r = m  r = n  r = 0,
CONJUGATE DYADICS
§ 64
64. Conjugate Dyadics.
141
The dyadics,
4) = alb, + a2b2 + ... + anon, + bn.an,
,D, = b1a1 + b2a2 +
are said to be conjugates of each other. In general 4) and (D, are different dyadics; but evidently
4) r=r
(1)
define the same linear vector function. If two dyadics are equal,
their conjugates are equal; for 4, =' implies that 4)
by definition.
r, and hence 4), A dyadic is called : Symmetric
if
Antisymmetric if
4), = 4),
4 =  4).
The importance of these special types of dyadics is due to the THEOREM. Every dyadic can be expressed in just one way as the sum of a symmetric and an antisymmetric dyadic. Proof.
For any dyadic 4), Ave have the identity, =
(1)
4)+4), 2
+
4)
2
=
+ 52;
since
= *I
4rc =
= Q,
St, =
2
2
41 and Sl are, respectively, symmetric and antisymmetric. Moreover 4) can be so expressed in only one way. For, if 41
+Q_N,,+9,
gave two such decompositions, we have, on taking conjugates, q/  SZ =
and hence
_ V, Sl = S2'.
LINEAR VECTOR FUNCTIONS
142
§ 65
65. Product of Dyadics. If the transformations corresponding to two linear vector functions,
v=4)  r, are applied in succession, their resultant,
is a third linear vector function; for w1 + w2 corresponds to r1 + r2, and Xw to Xr.
This function is written
w= and 'I' ( is called the product of the dyadics ' and 4), taken in 4) is therethis order. The defining equation for the product fore (1)
(*
for every r.
From (1) we find that the distributive and associative laws hold for the products of dyadics: (2)
4) ('I' + S2)
{ 4
(3)
((F +')
+
S2,
(4)
Proofs.
For every vector r,
(I' + l)] r = 4 [( + 0)
r]
r+l r] r)+ (St
r+ [( +'F)
St]
r = ((F +'I')
r; r)
(St
r)+
r=
[(`I'
r)
' ) ' r] r)
r)
PRODUCT OF DYADICS
§ 65
143
Equations t2), (3), (4) follow from these results by the definition of equality. In the proofs of (2) and (3), definition (1) justifies the first and last steps; in the proof of (3), (1) is applied in every step.
explicitly, we first find the In order to compute a product product of two dyads. By definition,
for every r, and hence (ab) (cd) = (b c) ad.
(5)
The product of ab and cd, in this order, is the scalar b c times the dyad ad.
Similarly,
(ab) = (d a) cb,
(cd)
(6)
which in general differs from (5). Making use of the distributive law, we now may form the product of any two dyadics, m
n
' = E c;d;,
4) = E aibi, i=1
i=1
by expanding into nm dyads. n
(7)
m
=E i=1 j=1
(bi c,)aidr
The conjugate of 4) ' is m
n
71 (c; (I' ) = Ej=1 i=1
(8)
bi)d;ai
The conjugate of the product of two dyadics is the product of their conjugates taken in reverse order.
Making use of (8), we now find that
r
(4)
 *) = (4)
'P ) ,
r = ( * ,( 1 ) , )r
hence
r ((D 'Y) = (r (D) 4, for every r, an associative law analogous to (1) but with r as prefactor. If el, e2, e3 form a basis and e', e2, e3 the reciprocal basis, we can express any two dyadics in the form, (9)
4) = fie, + f2e2 + f3e3,
' = elg1 + e2g2 + e3g3
LINEAR VECTOR FUNCTIONS
144
§ 66
In the product 4  ', six dyads vanish, and we find (10)
4)
`I' = figs + f2g2 + f393
If 4) and' are complete, fl, f2, f3 and g1, g2, g3 are noncoplanar sets; then ( 4, is also complete. But if (F or 4, is singular, one of these sets must be coplanar, and 4) 'Y is likewise singular. Therefore the product of two dyadics is complete when and only when both dyadic factors are complete.
If 4) is complete, it may be "canceled" from equations such as
4.4,=O,4, (F=0, to giveT =O. Thus, if are noncoplanar in (10), and hence g1 = g2 = g3 = 0. We also = 4) St; for this is equivalent to (4,  9) = 0, and hence 4,  SZ = 0. 66. Idemfactor and Reciprocal. The unit dyadic or idemfactor I is defined by the equation,
may "cancel"  in (F
I
(1)
r=r
for every r.
The idemfactor is unique (§ 60). For any basis e1, e2, e3, we have I e; = e I e' = e'; hence, from (60.7), (2)
I = e1e' + e2e2 + e3e3 = e'e1 + e2e2 + e3e3;
in particular the selfreciprocal basis i, j, k gives (3)
I = ii + jj + kk.
Evidently I is symmetric and complete. THEOREM. In order that a, b, c and u, v, w form reciprocal sets, it is necessary and sufficient that (4)
au+bv+cw=I.
When the sets are reciprocal, we have just proved (4). Conversely, if (4) holds, a, b, c are noncoplanar, since I is complete; let a', b', c' denote the reciprocal set. Using these vectors as prefactors on (4) gives u = a', v = b', w = c'. Multiplying a dyadic by I leaves it unaltered; for, from
r (I.4)) _ we conclude that (5)
4I=I
(F = 4.
IDEMFACTOR AND RECIPROCAL
§ 66
145
If 4 I = I, 4) and ' are both complete, since I is complete From
(§ 65).
T
or
(4,
4) = I. Two dyadics 4), 4, are said to be reciprocals of each other when
4) 41 _I (D =I,
(6)
and we write 4, = 4)1, 4) = T1 A complete dyadic 4) has a unique reciprocal (D1. If (7)
(D1 = flee + f2e2 + f3e3,
4) = elf, + e2f2 + e3f3,
for their products in either order give I: 4) .
4>I
= e,e' + e2e2 + e3e3,
4)1
4) = f If, + f'f2 + f3f3.
Since dyadic multiplication is associative 4,I) _ 4) ('F 1) 4,1 = 4) (4) F) (T1
(D1 = I;
(4) . *)I = 'FI .,I>,.
(8)
Making use of (8), we have (4) . * . Q) 1 = 21 , ggeneral,
and, in
(4)
q') 1 =
the reciprocal of the product of n dyadics is the
product of their reciprocals taken in reverse order. For positive integral n we define 4,n
= (P . 4)n = (4)i)n = (4)n)I;
4)°=I. With these definitions, (12)
4)m . 4,n = Pm+n,
(4)m) n.
_
4)mn,
for all integral exponents. But owing to the noncommutativity of the factors in a dyadic product, (4>. ')n is not in general equal ,0 cpn . *n.
LINEAR VECTOR FUNCTIONS
146
§ 67
By means of reciprocal dyadics, we readily may solve certain vector and dyadic equations. Thus, if (F is complete, the equations, (F
r=v,
r 4 ) =v,
4 )4 1=2,
2,
(F
have the respective unique solutions:
r=4)1 v, If (F = Maibi, we define the dyadics:
67. The Dyadic (1) x v.
 x v = Ea1bi x v,
(1)
v x 4) = Tv x aibi.
From these definitions we have the relations: (2)
(4) x v)
r = 4) (vxr),
r (v x (F) = (rxv)
(F;
(r(P)v,
(3)
When 4, = I, these give (4)
r = vxr.
(Ixv) = Since these equations hold for any r, (6)
Ixv = vxI,
(7)
(I x v), _ Ixv.
Thus I x v is antisymmetric and planar; it transforms all vectors f into vectors perpendicular to v. Since
aibi (I x v) = aibi x v,
(I x v)
aibi = v x aibi,
the definitions (1) give (8)
(P
(Ixv) _
xv,
(Ixv)  4) = vx(F.
Consequently, the operations v x and x v on vectors or dyadics may be replaced by dyadic products (I x v) and (I x v). For any vectors u, v, r we have [I x (u x v)]
r = (u x v) x r = (vu  uv)  r,
and hence (9)
1 x (u x v) = vu  uv.
FIRST SCALAR AND VECTOR IN\'AItIANT
§ 68
147
From this identity we arrive at the general form of any antisymmetric dyadic. THEOREM.
If the dyadic f = 2;aibi is antisymmetric, and the
vector co = tai x bi,
l = 21xw.
(10)
Proof.
The conjugate of f is St = 2;biai; hence
2SZ = 2; (biai  aibi) = EI x (ai x bi) = I x w. Every antisymmetric dyadic is planar. 68. First Scalar and Vector Invariant. We may express a dyadic (P = 2;aibi in various forms by substituting vector sums for ai, bi, and expanding and collecting terms by applying the distributive law. For all these forms there are certain functions of the vectors, in terms of which is expressed, which remain the same. These functions, which may be scalar, vector, or dyadic, are called invariants of the dyadic. Each step of the process in changing = 2;aibi from one form to another may be paralleled by the same step in transforming the scalar and the vector, Corollary.
(1)
Cpl = tai bi,
(2)
4) = 1dai x bi,
obtained by placing a dot or cross between the vectors of each dyad of (D. Each application of the distributive law in the transformation of 4) is also valid in the corresponding transformations of 01
and 4; and, just as 1 is not altered by these changes, the same is true of the scalar cpj and vector 4. These quantities are therefore invariants of 4 with respect to the transformations in question. They are called, respectively, the scalar (or first scalar invariant) and vector of the dyadic.
For example, if P = ij + jk + kk, c1
1,
4)=ixj+jxk+kxk=k+i.
For the idemfactor I = ii + fl + kk, the scalar is 3 and the vector 0; and we obtain these same values if I is expressed in terms of an arbitrary basis: I = ele' + e2e2 + e3e3. Again, for the dyadic,
S2 = IxV = iixv+jjxv+kkxv,
LINEAR VECTOR FUNCTIONS
148
§ 69
we have co, = 0, and
w=ix(ixv)+Jx(Jxv)+kx(kxv)
2v;
thus S2 = ZI x w, as in (67.10). The scalar or vector of the sum of two dyadics is the sum of their scalars or vectors: thus, if (3)
'Pi=4i+Wi, 4=4+w.
4) =`I'+0;
It is principally to this property that these invariants owe their importance. In (64.1), we have expressed any dyadic 4) as the sum of a symmetric and antisymmetric dyadic: (4)
The antisymmetric part S2 =
z
(4)  4) has the vector,
From the theorem of § 67 we now have
SZ = 2Ix4.
(5)
Hence, from (4), (6)
24=4) +F,,Ix4, +Ix
From (6), we have the THEOREM. A necessary and sufficient condition that a dyadic be symmetric is that its vector invariant vanish.
69. Further Invariants. We may obtain further invariants of the dyadic c = 1aibi by processes that are distributive with respect to addition. The most important of these are the dyadic, (1)
4)2 = 2;aixajbixbj, i.l
called by Gibbs the second of I ; its scalar invariant, (2)
(P2 = 2 (ai x a3) (bi x bj); i,)
and the scalar, (3)
(P3 = * i.i.k
(aixaj .ak)(bixbj .bk).
FURTHER INVARIANTS
§ 69
149
In (1) the summation is taken over all permutations i, j. When
i = j, the dyad vanishes; when j  i, the permutations i, j and j, i give the same dyad, so that each dyad occurs twice in the final sum; this doubling is avoided by the factor 2.
In (3) the summation is taken over all permutations i, j, k. When two subscripts are the same, the term vanishes; when i, j, k all differ, the 3! = 6 permutations of these subscripts give the same term, so that each term occurs six times in the final sum; i this multiplication of terms is avoided by the factor s Let CF be reduced to the threeterm form:
CF = al+bm+cn.
(4)
The invariants considered thus far are now (5)
(7)
4 = axl+bxm+cxn, 42 = bxcmxn+cxanxl+axblxm,
(8)
IP2 = (b x c)
(9)
(p3 = (a x b
(6)
(m x n) + (c x a)
(n x 1) + (a x b)
(l x m),
c) (1 x m n).
The numbers 01, P2, 03 often are called the first, second, and third scalars of 4).
If CF is singular, the antecedents a, b, c or the consequents = 0, 1, m, n in (4) will be coplanar, and p3 = 0. Conversely, if a, b, c or 1, m, n are coplanar sets and 4) is singular. Therefore a dyadic is singular when and only when its third scalar is zero. If V3 = 0, 4) must be planar, linear, or zero. * When CF is linear it can be reduced to the form al and cF2 = 0. Conversely, if 4'2 = 0, 4) will be linear or zero; for, if we choose a noncoplanar set a, b, c as antecedents in (4), b x c, c x a, a x b are also noncoplanar, and `1'2 = 0 implies that
mxn = nxl = lxm = 0; then 1, m, n are parallel or zero. Therefore we may state the THEOREM. Necessary and sufficient conditions that a dyadic CF be Complete
'P3 3' 0,
Planar
are that has the characteristic numbers X1, X2, 1\3 for the directions r1r r2, r3, prove that 4>2 has the characteristic numbers X2X3, X3X1, X1X2 corresponding to the directions r2 x r3i r3 x r1, r1 x r2. [Cf. (70.2).]
5. If 4, = 4,, prove that *2 = 4>2,,, + = 4 and h = cir >k2 = v2, 4,3 = v3 6. Prove that the scalar invariants of 4' = 4>2 are 01 = 'c1  2'P2, 2 = 'P2  2'vl'P3, '3 = 'P3 [Use Prob. 3 and (71.4).]
PROBLEMS
7. Given the invariants c2, +, iants for (a) k4),
17 5
, v3 of 4), find the corresponding invar
,
(c) 41.
(b) P x u,
8. Compute (P2, +, and the six scalar invariants of (69.13), namely,
+ ' +, + ' 4) ' +, + ' p2  +,
'Pl, V2, 4'3;
for the dyadic 1
3
2
2
0
4
1
2
3
=
9. If D = I x e and e is a unit vector, prove that 4)2
=  (I  ee), 4'3 = 4), D4 = I  ee, 4)5 = (P.
10. Show that the symmetric part of (P, namely, 4' = 1(4) +4,), has the scalar invariants
h = c1, 42 = V2  4+'
11
4
413 = V3
11. Prove that ('t  ')2 = `1'2 *212. Prove that the first three scalar invariants of 4
1
and 4 4) are the
same. m
13. Ifs
n.
a;b,, 4' 1=1
c;d1,
J=1
we define the doubledot product 'F:4' as the scalar m
n
E(a`.c1)(b`d,). 1I j_1 Prove that
++4':
(a) (b)
(uv) :4' = u
4)
v,
4 ) :1
(c) If 4) is given in the nonion form (77.1), 'P11 + V12 +'P13 +'P2I + P22 + V23 +
'P21
+'P32
+P2
Hence 4, = 0 when and only when 4):4' = 0. 14. Any central quadric surface with its center at the origin has an equation of the form, (1)
r 'F r = 1,
where 4, is a symmetric dyadic; for, if 4) is reduced to the standard form (72.4), we have
axe + $y2 + yz2 = 1.
This represents an ellipsoid, an hyperboloid of one sheet, or an hyperboloid of two sheets, according as a, j3. y include no, one, or two negative constants.
176
LINEAR VECTOR FUNCTIONS
If p 5*1 0, the equation (1) associates with every point (1, p) the plane p, 1), its polar plane. Prove that: (a) If the point (1, p) lies on the quadric surface, its polar plane (4' p, 1)
(4'
is tangent to the quadric at (1, p). [Find the points where the line r = p + Xe cuts the quadric when e 4' p = 0.]
(b) If the polar plane of (1, p) passes through (1, q), the polar plane of (1, q) passes through (1, p).
15. The diametral plane of any point (1, p) on the quadric surface (1) is the locus of the midpoints of all chords parallel to p. Show that the diametral plane of (1, p) is (4). p, 0). 16. If three points (1, u), (1, v), (1, w) on the ellipsoid r 4) r = 1, satisfy the equations, U. the position vectors u, v, w are said to form a conjugate set. Show that: (a) The vectors u, v, w and 4'  u, 4) v, 4) w form reciprocal sets.
(b) 4'1 = uu + vv + ww. (c) For any conjugate set u, v, w, of the ellipsoid r (D r = 1, the sum u u + v v + w w and the product u x v w are constant. 17. Verify by direct computation that the dyadic 4' in Problem 8 satisfies its HamiltonCayley Equation. 18. A rigid body with one point 0 fixed has the inertia dyadic K relative to 0 (§ 72, ex. 2). If the angular velocity of the body at any instant is w, show that its moment of momentum H (defined as f r x v dm) and kinetic energy T (defined as
if v
v dm) are given by
19. If the forces acting on the body of Problem 18 have the moment sum M about 0, the equation of motion is dH/dt = M. Show, from (56.3), that dH dt
=K dw dt
+wxK
w.
Let K = Aii + Bjj + Ckk (72.10) when referred to the principal axes of inertia (fixed in the body). Then if w = [WI, w2i w3], M = [Ml, M2, M3] referred to these axes, deduce Euler's Equations of Motion: Awl  (B  C)w2w3 = M1, Bm2  (C  A )w3w1 = M2, (A  B)w1w2 = M3
20. In Problem 19 show that d7'/dt = M w (the energy equation). 21. In Problem 18, suppose that the only forces acting on the body are its weight W and the reaction R at the support 0; then if the center of mass is at 0,
W passes through 0 and M = 0. (a) H is a constant vector. (b) T is a constant scalar.
Prove in turn that
PROBLEMS
17 7
(C) If w = OP, the locus of P in space is the invariable plane w H = 2T; :uul the locus of P in the body is the energy ellipsoid w K w = 27'. (d) The energy ellipsoid is always tangent to the invariable plane at P. (c) The body moves so that its energy ellipsoid rolls without slipping on
the invariable plane (Poinsot's Theorem). [See Brand's Vectorial Mechanics, § 219.]
22. The vectors of a dyadic 'F are fixed in a rigid body having the instantaneous angular velocity w, relative to a "fixed" frame. Show that, relative to this frame, d'F/dt=ca x4) 4) xw. 23. A rigid body revolving about its fixed point 0 has the angular velocity w = [wl, W2, w3] referred to fixed rectangular axes through O. If I1, 12, I3 are
the moments of inertia about the fixed axes x, y, z and 123, I31, 112 are the products of inertia for yz, zx, xy, show that
dIl/dt = 2(I12w3  1132), ..
,
dI23/dt = (13  12)wl + 113W3  1122, ... ,
[If K is the inertia dyadic of the body relative to 0, Il = i K i, 123 = j K K. k, etc. See § 72, ex. 2.] 24. The axis of a homogeneous solid of revolution has the direction of the unit vector e. If its principal moments of inertia at the mass center G are A, A, C, show that the inertia dyadic at G is
KG =AI+(CA)ee. Hence find the moments and products of inertia with respect to fixed rectangular axes x, y, z through G if e = [1, in, n].
25. If G is the mass center of a rigid body of mass in and r* = OG, show that the inertia dyadic at 0 is Ko = m(r* r* I  r*r*) + KG. Hence compare moments and products of inertia for parallel axes at 0 and G. 26. If X is arbitrary parameter < a2 but b2 or c2, the dyadic,
x211 + b2 J1
c2kk {
(a > b > c),
defines a oneparameter family of confocal quadric surfaces r % r = 1. These are ellipsoids if X < c2, hyperboloids of one sheet if c2 < \ < b2, hyperbolas of two sheets if b2 < X < a2. Prove that
(a) The central quadrics r T r = 1 and r O r = 1 are confocal when and only when I'  01 = kI. (b) Two confocal central quadrics of different species intersect at right angles.
[O  4, = ke *.]
CHAPTER V DIFFERENTIAL INVARIANTS
82. Gradient of a Scalar. The points P of a certain region may be specified by giving their position vectors r = OP; and we shall on occasion refer to P as the "point r." A scalar, vector, or dyadic which is uniquely defined at every point P of a certain region is called a point function in this region and will be denoted by f (r). For example, the temperature and velocity of a fluid at the points of a threedimensional region are scalar and vector point functions respectively.
A scalar point function f (r) is said to be continuous at a point P1 if to each positive number e, arbitrarily small, there corresponds a positive number S such that
If(r)  f(r1)
0. Hence Vf has the direction of n, and its magnitude is the value of df,/ds in this direction. Writing this normal derivative df/dn, we have (7)
For example, if f = r, the distance OP from the origin, the level surfaces are spheres about 0 as center, and n = R, the unit radial vector; the normal derivative dr/dn = 1, and
Vr=R.
(8)
When f is a function f (x, y, z) of rectangular coordinates, Vf is given by (4). In particular, (9)
Vx=i,
Vy = j,
Vz = k.
If f is a function f(u, v, w) of variables which themselves are functions of x, y, z, we have df
of du
of dv
of dw
ds
au ds
av ds
aw ds
or, in view of (5),
\
Vu Of
au
+Vvaf +Vwaf1 aw
av
As this equation holds for every e, (10)
Vf = Vu
of
+ Vv
Ou
af
av
+ Vw
Of .
aw
When f = f (u, v) or f = f (u), Of
Vf = Vu of + Vv , av au
Vf = Du
Of
au
,
§ 83
C1t.ADIENT OF A VECTOR
181
respectively; for example,
Vu' = nu"1 Vu,
V(uv) = v Vu + it Vv,
1
V log u =  Vu. It
When f is constant, Vf = 0; conversely, Vf = 0 implies of/ax = of/ay = of/az = 0, and f is constant. If f (r) is differentiable in a region R, Vf is defined at all points of R. If moreover Vf is continuous in R, we say that f (r) is continuously differentiable in R. Example. Gradients in a Plane. The gradient of a point function f(x, y) in the xyplane is
vf=afi+Of j. ay ax At any point P, Vf is normal to the level curve f(x, y) = c through P. For a function f(r, 0) of plane polar coordinates, Cf
of = or R
of
+00r'
in the notation of § 44; for Vr = R and V0 = P/r (from (44.5) dB/ds = 1/r in the direction perpendicular to R). For a function f(rl, r2) of bipolar coordinates,
Vf =OrlR1+_R2, where R1, R2 are unit radial vectors from 01, 02 directed to the point in question.
The families of curves u = c, v = c cut at right angles when Cu Cv = 0. For example, the ellipses rl + r2 = c and hyperbolas r,  r2 = c cut orthogonally since (R1 + R2) (R1  R2) = 0
83. Gradient of a Vector. Let f(r) denote a vector point function in a certain region. It is said to be continuous at a point P1
if to each e > 0 there corresponds a S > 0 such that I f (r)  f(r1) I < e when
Ir
 r1 I< S.
If f(r) is given by its rectangular components,
f(r) = f1i + f2j + f3k,
f(r) is continuous at P1 when the three scalar point functions fi, f2, f3 are continuous there.
DIFFERENTIAL INVARIANTS
182
§ 83
Let P1P be a ray drawn from P1 in the direction of the unit vector,
e = icosa+jcos$+kcosy. Along the ray r = r1 + so (s > 0) and f(r) is a function of s. We now define df
 = lim ds
f(rl + se)  f(r1)
s.o
s
as the directional derivative of f (r) at P1 in the direction e. If this limit exists on all rays issuing from P1, f(r) is said to be differentiable at Pl. Evidently f (r) is differentiable if its rectangular components f, are differentiable. If f(r) is given as a function f(x, y, z) of rectangular coordinates, we have. just as in § 82, (2)
of dz
of dy
df
of dx
ds
axds+ayds+azds of
=
cosa +
of ay
of

cosa+
cosy.
On replacing the cosines by e i, e j, e k, df (3)
of
ds =
of
of
e1+ja +kaz y
,
a formula entirely analogous to (82.3). The dyadic in parenthesis is called the gradient of f(r) and is written grad f or (4)
of
of Vf = ofi+ j  +k 
ax
ay
az
Vf is a dyadic point function; thus, when (3) is written (5)
df
the direction enters only through the prefactor e. The gradient Vf at P effects a synthesis of all the directional derivatives of f at P. In effect, the dyadic Vf replaces the infinity of vectors df/ds. Since df/ds, as defined by (1), does not depend upon any specific
choice of coordinates, (5) shows that Vf has the same property.
DIVERGENCE AND ROTATION
§ 84
In fact Vf is determined by giving the directional derivatives df/dsi in three noncoplanar directions ei: df
+ e2 df + e3 df
of _ e'
(6)
ds3
ds2
ds1
For, if we put df/dsi = ei Vf, the right member becomes (e'ei + e2e2 + e3e3) Vf = I I. Vf = Vf. When f is function of rectangular coordinates f (x, y, z), Vf is given by (4). More generally, for f(u, v, w) we have df ds
of du au ds
of dv
of dw
av ds
aw d s
of
C
au
+ VV
of
of
av
aw
for any e; and, from the definition of dyadic equality, of of of Vf = Vu+Vv+Vw
(7)
au
aw
av
84. Divergence and Rotation. The first scalar invariant of the gradient Vf of a vector f is called the divergence of f and is written
The vector invariant of Vf is called the rotation or curl of f and is written V x f, rot f, or curl f.
In terms of rectangular coordinates, we therefore have the defining equations: (1)
Vf=gradf=iaxof  +j ayof +k azof
,
= i ax of + j
,
(2)
V f = divf
3)
xf = rotf
of
of
ay
az
of of of = ix+ jx+kx__. ax ay az
If f is resolved into rectangular components,
f=ifl+jf2+kf3,
DIFFERENTIAL INVARIANTS
184
§ 84
we obtain from (1) its gradient of in nonion form with the matrix:
afl/ax af2/ax af3/ax of = afl/ay af2/ay af31ay
(4)
af2/az
8fl/az
af3/az
The first scalar and vector invariants of of now become
v
(5) (6)
axl+ a19h
+
f=
az
,
vxf=i  fj(azax }k(ax ay)/ Gay f3
afl
af2
af3
af2
afl
az
The last expression is easily remembered when written in determinant form: j k (7)
Vxf =
a
a
a
ax
ay
az
fl
f2
f3
These expressions for divergence and rotation may be written down
at once, if we regard V  f and V x f as products of the vector operator del (or nabla), (8)
a a V =axai+j+k, ay az
with the vector f = if, + jf2 + kf3. For the position vector r = xi + yj + zk, we have Or yr = i+ j Or+k Or= ii+ jj +kk,
ax
ay
az
the idemfactor; hence (9)
When f =
Vr = I,
rot r = 0.
div r = 3,
the gradient of a scalar, (9 (P
fl=ax,
f2=ay
a4'
,
f3=az
DIVERGENCE AND ROTATION
§ 84
185
Using these components, (5) and (6) gives a2
( 10 )
+
ax2
.
(11)
a2V
(P
v v'P
a2,p
+
ay 2
az2
V X VV = 0.
The differential operator V V or div grad is called the Laplacian and often is written a2
a2
__
v2
(12)
ax2
a2
+ ay2 +az2
When f = V x g, the rotation of a vector g, A
C193
092
ay
az
_a91
a93
ax '
az
f2
f3
a92
C191
ax
ay
In this case (5) and (6) give (13)
0,
(14)
Vx(Vxg) _ V(V.g)  O29.
Proof of (14) : From (6),

r a292
ay ax
x291

a291
aye
az2
= i j a tag, + a92 + lax
=i
ax
ay
a
ax
(O. g)  V291
a293
+
a931
az J
az ax Jt
+ .. .
 v291 I + .. .
+...
= V(V . g)  V2g.
Note that, if we regard V as an actual vector and expand V x (V x g), but keep the V's to the left of g, we obtain V(V g) (V V)g, the correct result. The proofs of (11), (13), and (14) depend upon changing the order of differentiations in mixed second derivatives; this is always valid when the derivatives in question are continuous.
DIFFERENTIAL INVARIANTS
186
§ 85
The foregoing differential relations of the second order also may be written (10)
div grad p = V2
position vector r = OP changes by the increment Ar = PP. Then, if f(r) is any differentiable tensor point function, the total differential of f (r) is defined by the equation : (1)
In particular, when f = r, Vf = I, and dr = Or. (2) The differential of the position vector is the same as the increment The defining equation (1) therefore may be written (3)
df=drVf.
DIFFERENTIAL INVARIANTS
198
§91
If f is a function f (u, v, w) of curvilinear coordinates,
Vf = Vufu+VvfV+Vwfw, and, from (3), since dr Vu = du, etc., df = fu du + f dv + fw dw. (4)
91. Irrotational Vectors. A vector function f (r) is said to be irrotational in a region R when rot f = 0 in R. If V(r) is a scalar function with continuous second derivatives, its gradient VV is irrotational; for, from (84.10),
rot grad ,p = 0.
(1)
Conversely, if rot f = 0, and f is continuously differentiable in R, we shall show that f may be expressed as the gradient of a scalar V(r). Using rectangular coordinates, let f = f1i + f2j + f3k; then, if rot f = 0, the three determinants of the matrix, aax a/ay
(2)
f2
(fl,
a/az\ f3
all vanish (84.7). Under this condition we shall determine a scalar function ,p(r), so that Vv = f, that is, (3)
= f1(x, y, z),
ax
ay
= f2(x, y, z),
az
= f3 (x, y, z)
Let (xo, yo, zo) be an arbitrary point of R. On integrating app/ax = f1 with respect to x and regarding y and z as constant parameters, we have x
P=
(4)
J
f1(x, y, z) dx + «(y, z),
o
_ dx+=
where «(x, y) is a function of x and y as yet undetermined. Hence x afl a« /' af2 a« app ay
,,
_J 49Z
1o
ay
ay
dx+, or ay
Ixo ax
f
a
f2(x, y, z) = f2(x, y, z)  f2(xo, y, z) + xafl
a«
 dx +  _ az az
xaf3

 dx + ax az
,
ay
or
a«
f3 (x, y, z) = f3 (x, y, z)  f3 (xo, y, z) +

IRROTATIONAL VECTORS
§ 91
199
Instead of three equations (3) for gp(x, y, z), we now have two equations, a«
a« (5)
ay
= f2(xo, y, z),
az
= f3(xo, y, z)
for «(x, y), with the condition aft/az = af3/ay. The problem has been reduced from three to two dimensions. On integrating a«/ay = f2 with respect to y, we have
f f v
a=
(6)
f2(xo, y, z) dy + #(z),
bo
where $(z) is a function of z to be determined.
a«fY af2 dy az
vo
az
d(3
+
dz
Y a af3

y
dy +

,
dz
Hence
or
f3(xo, y, z) = f3(xo, y, z)  f3(xo, yo, z) +
d
da (7)
dz
= f3 (xo, y0, z)
We now have one equation (7) to determine $(z); and J f3 (xo, yo, z) dz.
(8)
o
f
On collecting the results (4), (6), and (8), we have finally (9)
'=
ffi
(x, y, z) dx +
z
Y
f2 (xo, y, z) dy + f f3 (xo, yo, z) dz.
Yo
D irect substitution shows that Vip = f; and, if
is a second func
tion for which V4, = f, V(¢  cp) = 0, and 4  (p is a constant. Thus (9) gives the solution of equations (3), determined uniquely except for an additive constant. There are evidently five other forms for ,p which may be obtained from (9) by permuting 1, 2, 3 and making the corresponding permutation on x, y, z; for example, (10)
c = ff2(x
z
y , z) dy + ffs(x, yo, z) dz + ffi(x, yo, zo) dx.
Y
Moreover, xo, yo, zo may be given any values that do not make the integrands infinite.
DIFFERENTIAL INVARIANTS
200
§91
In mathematical physics, it is customary to express an irrotational vector f as the negative of a gradient. Thus, if 4, we have
f = V ';
(11)
¢ is then called the scalar potential of f. Example 1. When
f = 2xzi + 2yz2j + (x2 + 2y2z  1)k,
we find that rot f = 0; hence f = V. With xo = yo = zo = 0, we have, from (9),
=f X2xz dx +f 2yz2 dy 0
0
 f Zdz = x2z + y2z2  z. 0
The differential equation,
Example 2. Exact Equation. (i)
is said to be exact when f dr = dp, the differential of a scalar. If (i) is exact,
we have, from (90.3), dr f = dr vv; and, since dr is arbitrary, f = When f is continuously differentiable, f = V implies rot f = 0, and conversely.
Therefore we may state the
THEOREM. If f is a continuously differentiable vector, in order that f dr = 0 be exact it is necessary and sufficient that rot f  0. Thus, in view of ex. 1, the equation,
2xz dx + 2yz2 dy + (x2 + 2y2z  1) dz = 0,
is exact and may be put in the form dp = 0. Its general solution is that is,
= c,
x2z + y2z2  z = C. If (i) is not exact, a scalar X which makes Xf dr = 0 exact is called an integrating factor of (i). The preceding theorem shows that X must satisfy the equation, (ii)
rotaf = 0 or VX x f + X rot f = 0.
On multiplying (ii) by f , we have
rot f = 0. Hence, when f dr = 0 admits an integrating factor, f rot f = 0. (iii)
f
Con
versely, when f is continuously differentiable, the condition (iii) ensures the existence of an integrating factor. We shall prove this in § 105. For this reason f rot f = 0 is called the integrability condition for f dr = 0. When f dr = 0 is integrable, ?f = V (p. Then the vector field f(r), being parallel to Dip, is everywhere normal to the level surfaces rp = c. The condition (iii) therefore implies the existence of a oneparameter family of surfaces everywhere normal to f. Thus the geometrical content of the condition f rot f = 0 is that the vector field f (r) is surfacenormal.
SOLENOIDAL VECTORS
§ 92
201
92. Solenoidal Vectors. A vector function f(r) is said to be solenoidal in a region R when div f = 0 in R. If g(r) is a vector function whose components have continuous second derivatives, rot g is solenoidal, for, from (84.13), div rot g = 0.
(1)
Conversely, if div f = 0, we shall show that f may be expressed as the rotation of a vector g. Using rectangular coordinates, let f = f1i + f2j + f3k, and (84.5) af3 aft div f aftax= ++= 0. az ay
(2)
We now shall determine a vector g = g1i + g2j + g3k, so that i (3)
f = rot g = a/ax 91
j
k
c1/ay
a/az
92
93
We first find a particular solution of (3) for which g3 = 0; then (3) is equivalent to the scalar equations: (4)
a91
4992
fl = 
az
f2 =
,
az
f3 =
,
4992
ax
_
1991
ay
The first two equations of (4) are satisfied when z
(5)
92 =  f f i (x, y, z) dz,
91 = f f2 (x; y, z) dz + 49(x, y) ;
w z
Zo
in these integrations x and y are regarded as constant parameters, and a(x, y) is a function as yet undetermined. In order that these functions satisfy the third equation of (4),
 Zaf + J dz  as = f3, ay (ax ay/ .
aJ2
Zp
or, in view of (2),
I.
Zafs az
as
dz   = f3 (x, y, z). ay
When we perform the integration, this reduces to f3 (x, y, z0) 
as ay
= 0,
DIFFERENTIAL IN VARIANTS
202
§ 92
an equation which is satisfied by taking v
a (X, y) = 
f3 (x, y, zo) dy. Yo
Hence the vector g, whose components are z
(6)
v
gi = ff2(x, y, z) dz  ff3(x, y, zo) dy, a
z
92 =  f fi (x, y, z) dz,
93 = 0,
zo
is a particular solution of our problem. If G is any other solution, rot G = rot g = f, and hence
rot (G  g) = 0.
(7)
But any irrotational vector may be expressed as the gradient of a scalar p (§ 91); hence the general solution of (7) is G  g = o,p, where ,p(r) is an arbitrary twicedifferentiable scalar. When div f = 0, the general solution of rot g = f is therefore g + Vip; its rectangular components are obtained by adding c /cx, 8,p/cy, 8,p/8z to the components of g given in (6). In mathematical physics, the solenoidal vector f = rot g is said to'be derived from the vector potential g. Example 1. When
f = x(z  y)i + y(x  z)j + z(y  x)k, we find that div f = 0. Therefore f = rot g; if we take zo  yo = 0, the particular solution (6) is 91 = f
Zy(x  z) dz =xyz 
Zyz2,
0
z
g2
f x(z  y) dz =
 Zxz2 + xyz,
ga = 0.
0
Example 2. If u and v are continuously differentiable scalars, the vector Vu x Vv is solenoidal, for, from (85.3), Vu x Vv = rot (uVv).
Conversely, we shall show in § 104 that a continuously differentiable solenoidal vector always can be expressed in the form Vu x Vv.
SURFACES
§ 93
203
93. Surfaces. A surface is represented in parametric form by the equations: x = x(u, v),
(1)
y = y(u, v),
z = z(u, v).
We assume that the three functions of the surface coordinates u, v
are continuous and have continuous first partial derivatives, a requirement briefly expressed by saying that the functions are continuously differentiable. In order that equations (1) represent a proper surface, we must exclude the two cases: (i) the functions x.(u, v), y(u, v), z(u, v) are constants: equations
(1) then represent a point; (ii) these functions are expressible as functions of a single variable t = t(u, v); equations (1) then represent a curve. In case (i) all the elements of the Jacobian matrix,
(2)
vanish. In case (ii), the rows of the matrix are dx/dt, dy/dt, dz/dt
multiplied by at/au and at/av, respectively, and all of its tworowed determinants vanish identically. We exclude these cases by requiring that the matrix (2) be, in general, of rank two; then, at least one of its tworowed determinants, (3)
A =
yu
zu
yv
zv
B =
zu
xu
Zv
xv
C
_
I xu
Yu
xv
,
yv
is not identically zero.
Even when the matrix in general is of rank two, the three determinants A, B, C, all may vanish for certain points u, v. Such points are called singular, in contrast to the regular points, where at least one determinant is not zero. If we introduce the position vector to the surface, (4)
r = xi + yj + zk = r(u, v),
(5)
and the condition for a regular point may be written (6)
ru x r, / O.
DIFFERENTIAL INVARIANTS
204
§94
If we introduce new parameters u, v by means of the equations,
u = u(u, v),
(7)
v = v(u, v),
we shall require that the Jacobian of this transformation, a (u, v)
J =
(8)
FK 0.
a(u, i)
Then equations (7) may be solved for u and v, yielding
u = u(u, v),
(9)
v = 17(u, v),
and the correspondence between u, v and u, v will be one to one. Since
(lp)
_
X
rU
au au
av) x (ru au av au
ru   rU 
av\
r  J = J ru
X
ry,
aU
the requirement J 0 0 makes ru x rv 0 a consequence of (6). Thus a point which is regular with respect to the parameters u, v is also regular with respect to u, D.
94. First Fundamental Form. A curve on the surface r(u, v) may be obtained by setting u and v equal to functions of a single variable t:
u = u(t),
(1)
v = v(t).
A tangent vector along the curve (1) is given by dr dt
and (2)
The are s along the curve is defined as in (43.4) :
fro'
t t dt;
and
ds
dt
=
Since du = ft dt, dv = v dt by definition, on multiplying (2) by dt2, we have (3)
ds2 = ru ru du2 + 2ru r du dv + r r, dv2.
This first fundamental quadratic form usually is written (4)
ds2 = E du2 + 2F du dv + Gdv2,
FIRST FUNDAMENTAL FORM
§ 94
where
E = ru
(5)
F = ru
ru,
G = r r,,.
r,,,
Moreover, from (20.1), (ru x rv)
(ru x rv) _
ru  ru
ru rv
hence
ru x rv 12 = EG  F2
is positive at every regular point. The curves v = const (ucurves) and u = const (vcurves) are called the parametric curves on the surface. For these curves dv = 0 and du = 0, respectively, and the corresponding elements of arc are (7)
dsl = 1/E du,
ds2 = N/G dv.
Since the vectors ru, rv are tangent to the ucurves and vcurves, respectively, the parametric curves will cut at right angles when and only when .(8)
The vector ru x rv is normal to the surface. The parallelogram formed by the vectors ru du, rv dv, tangent to the parametric curves and of length dsl, ds2i has the vector area (§ 16), (9)
dS = ru x rv du dv.
We shall call this the vector element of area; the scalar element of area is (10)
dS = I ru x rv I du dv =
E
du dv.
The unit normal n to the surface will be chosen as
n=
ruxrt E
=
ruxrv
H
then dS = n dS. At every regular point the vectors ru, r, n form a dextral set; for (12)
[rurvn] = H = VEG
> 0.
DIFFERENTIAL INVAItIANTS
206
§ 95
95. Surface Gradients. Let f(u, v) be a differentiable function,
scalar, vector, or dyadic, which is defined at the points of the surface r = r(u, v). We shall compute the derivative of f(u, v) with respect to the are s along a surface curve u = u(t), v = v(t). Along this curve, df
du
ds
fu ds
(1)
dv
+ fv ds '
where du/ds = is/s, dv/ds = v/s, and, from (93.4),

ds
ds = V L ic2 + 2Ficv + Gv2. dt
(2)
If we apply (1) to the position vector r(u, v), we obtain the unit tangent vector e to the curve: du dv e=r,,,+rv ds ds
(3)
Let a, b, c denote the set reciprocal to ru, r,,, n; then, since [rur ,n] = H,
_ rvxn H
b
_ nxru II
,
c =
ruxrv
= n.
H Now from (3) a e = du/ds, b e = dv/ds; hence (1) may be (4)
a
,
written df
ds
We now define afu + bf as the surface gradient of f and denote it by V8f t or Grad f: (5)
V8f = Grad f = afu + b f,,.
Grad f has the characteristic properties: (6)
e Grad f =
(7)
ru Grad f =
df
ds
,
of au ,
n Grad f = 0;
r Grad f =

of av
t In surface geometry we shall write of for the surface gradient.
96
SURFACE DIVERGENCE AND ROTATION
207
If f (r) is a tensor of valence v, Grad f is of valence v + 1. At any point (u, v) of the surface, Grad f is in effect a synthesis of all the values of df/ds for surface curves through this point. Since Grad f depends only on the point.(u, v), df/ds is the same for all
surface curves having the unit tangent vector e at this point. At any point (u, v) of the surface Grad f is completely determined when df/dsi is given for two directions ei in the tangent plane at (u, v). If the set reciprocal to el, e2, n is e', e2, n,*
Grad f = el
(8)
df + e2 df dsi
ds2
for, by virtue of equations (6), the right member of (8) may be written
(elel + e2e2 + nn) Grad f = I Grad f where I is the idemfactor (§ 66). In particular if el and e2 = n x el
are perpendicular unit vectors, el = el, e2 = e2. In view of (8), Grad f is independent of the coordinates x, y, z and of the surface parameters u, v.
96. Surface Divergence and Rotation. If f(r) is a tensor point function defined over the surface r = r(u, v), its surface gradient,
V8f=afu+bf,,,
(1)
has the invariants, (2)
V,f = axfu +bxf,,.
(3)
We recall that the set a, b, n is reciprocal to ru, rv, n. When f (r) is a vector, V8f is a planar dyadic; then the scalar and vector invariants of oaf are called the surface divergence and surface rotation and are written (4)
V8
f = Div f,
V8 x f = Rot f.
The second of V8f is the linear dyadic, 1
(V3f)2 = axbfuxfv = Hnfuxfv; the second scalar of V8f is therefore n f,, x f v/H. * Since el x e2 = An, e3 = n, we have e3 = Xn/X = n.
DIFFERENTIAL INVAItIANTS
208
196
For the position vector r to the surface, we have V8r = ar.u + br v =
(6)
Div r = 2,
(7)
I  nn,
Rot r = 0.
For the unit normal n,
Rot n = 0.
(8)
To prove this, put f = n, a = r x n/H, b = n x ru/H in (3); then H Rot n = (rz, x n) x nu + (n x ru) x n,,.
= Now, from n n = 1, we have, on differentiation with respect to u and v,
n nu = n nv = 0; and, from ru n = r n = 0, ru nv = rv nu = n ruv; (10) (9)
hence H Rot n = 0. From the defining equations (1), (2), (3), we may derive various expansion formulas. Thus, if X is a scalar, f a tensor, V3(af) = (V3X)f + XV3f,
(11)
(Af) _ (V8A) f + XV8 f,
(12)
V.
(13)
V8 x (Xf) _ (V8X) x f + XV3 x f.
If g and f are vectors, (14)
Div (g x f) _ (Rot g)
f  g Rot f;
and, in particular,
Div (n x f) = n Rot f.
(15)
Proof of (14) :
Div(gxf) (Rot g)
f  g Rot f.
SPATIAL AND SURFACE INVARIANTS
§ 97
209
97. Spatial and Surface Invariants. If f(r) is a tensor function of valence v defined over a 3dimensional region including the
surface r = r(u, v), its spatial gradient at a point (u, v) of the surface may be computed from (86.4). If el, e2 are vectors in the
tangent plane at (u, v) and e3 = n, then, at all points of the surface,
df df df df Vf = e'+e2+n= V8f +n ds2 do do dsl
(1)
;
here the set e', e2, n is reciprocal to el, e2, n and df/dn denotes the derivative of f in the direction of n. Moreover, if v > 0, df
df
df
df
ds.l
ds2
do
do
(2)
V
(3)
df df df df Vxf = elx+e2x+nx= V8xf +nx. ds2 do dsl do
From (1), we see that the tensors of valence v + 1,
nx Vf = nx V j,
(4)
are the same over the surface; therefore both may be written Thus n x Vf may be computed solely from the values of f on the surface. The same is true of the invariants obtained from n x Vf by dotting and crossing in the first position. Since
n x Vf.
nxVf = nxafu+nxbft,,
(5)
from (96.1), this process yields (n x a) fu
+
(nxb) f = n
(a x fu
+ b x f,) = n V. x f,
n208f nV8 f. Here n V8f means that n is dotted into the second vector from the left in each term of V8f. These invariants remain the same when computed from the corresponding spatial quantities. Thus we verify at once, from (3), (6)
and, from (1) and (2), (7)
nVfnV.f=n2V8fnV8f.
DIFFERENTIAL INVARIANTS
210
§ 97
When f is a vector, these invariants become (6)'
n rot f = n Rot f,
(7)'
(grad f) n  n div f = (Grad f) n  n Div f.
We now express n x Vf and its invariants (6) and (7) in terms of the surface coordinates u, v. Since a, b, n and ru, r,,, n are reciprocal sets, we have, from (5), 1

nx Vf
(8)
1
(r,fu  ruf
II
{(rvf)u  (ruf),, },
since ruv = r,,,, if these derivatives are continuous. Dotting and crossing now yields
n Vxf
(9)
1
II II{(rvxf)u(ruxf)v}.
(10)
Since ru, rv are tangent to the surface, (9) shows that: If a vector f is everywhere normal to the surface, n rot f 0. With f = V3g in (9) we obtain the important identity,
=0;
(11) for then
ru f = gu f = gv, When f = pq, a dyad, we have, from (9),
(95.7).
rv
1
n V x (pq) =
II
Pq) u  (ru pq)
(r,, {
1
1
=
H
P)u  (ru
{ (r,,
P) v } q +
H
p (rvqu  rugv),
that is,
n V x (pq) = (n rot p)q + p n x Vq. For future use we compute the invariant (10) when f = ng and
(12)
g is an arbitrary tensor: n2 Vf
nVf=
H
1(rvxng)u  (ruxng)v} 1
= agu + bg,, +
H
{(rvxn)u  (ruxn)v} g.
§ 98
SUMMARY: DIFFERENTIAL INVARIANTS
211
Now
Grad g = agu + bg n Grad n = (an. +
(95.5),
2
n=0
(96.9) ;
hence, on putting f = n in (10), we have
n Div n =
1
H
{ (r x n)u  (ru x n) v
}
.
Thus, with f = ng, (13)
n? Vf  n V f = Grad g  (Div n) ng.
98. Summary : Differential Invariants. Let f (r) be a tensor point function of valence v; its derivative at the point P and in the direction of the unit vector e is defined as df
 = lim ds
f (r + se)  f (r) (s > 0).
s
s  o
If df/dsi denote the directional derivatives corresponding to three noncoplanar unit vectors ei, the gradient of f, namely,
grad f = Vf = e'
ds 1
+ e2
df
+ e3
2
has the property,
,
63
df
 =e V.
ds
Vf, a tensor of valence v + 1, thus gives a synthesis of all the directional derivatives of f at point r. If the vectors ei are i, j, k,
Vf = if., + jfv + kfz
(fz = of/ax, etc.).
From Vf (valence v + 1) we derive the invariants V f (valence v  1) and V x f (valence v) by dotting and crossing in the first position. When f is a vector (v = 1),
V f = div f, the divergence of f, V x f = rot f,
the rotation of f.
When r is a function r(u, v, w) of curvilinear coordinates Vf = Vu fu + Vv fv + Vw f.W,
Vr = Vu ru + Vv r + Vw rw = I.
DIFFERENTIAL INVARIANTS
212
The sets Vu, Vv, Vw and ru,
r',,,
ru, are reciprocal.
§ 98
With J =
we have
J Vf = (r x ru, f)u + cycl,
J V f = (r, x ru, f)u + cycl,
ru
J V x f = ((r x ru,) x f)u + cycl =
r
ru,
a/au a/av a/aw
ru, ru, are mutually orthogonal, the coordinates u, v, w are called orthogonal. Then if a, b, c denote a dextral set of or
thogonal unit vectors, ru = Ua, r = Vb, ru, = Wc, and J = UVW in the preceding formulas. For any tensor f (r),
V Vf = V2f,
V (V f) = 0,
V. Vf = 0, Vx (Vxf) = V(V _ f) 
V2f.
The operator V V. V is called the Laplacian.
A vector f is called irrotational if rot f = 0, solenoidal if div f = 0. An irrotational vector f can be expressed as the gradient of a scalar (f = Dip) ; a solenoidal vector f can be expressed as the rotation of a vector (f = rot g). For the surface r = r(u, v), the fundamental quadratic form is ds2 = E due + 2F du dv + G dv2, where
At a regular point,
E>0,
and the unit surface normal is defined as n = ru x The derivative of a tensor f (r) along any surface curve tangent to the unit vector e at the point (u, v) is df
ds
the set a, b, n is reciprocal to ru, r, n. The surface gradient,
V8f = Grad f = a fu + b f,,,
PROBLEMS
213
has the properties,
e. Grad f =
df
n Grad f = 0;
,
ds
df
Grad f =  ,
r.u
r, Grad f =
au
df av
From V3f we derive the invariants,
Oaxf = axfu+bxfv,
O3 f =
by dotting and crossing in the first position. When f is a vector, V8
f = Div f, the surface divergence of f,
Og x f = Rot f,
the surface rotation of f.
The tensor n x of and its dot and cross invariants,
n?of are not altered when V is replaced by V8. They may be computed
solely from the values f(r) assumes on the surface.
Thus, if
H= n x of =
1
H
{
(ru.f) v } ;
and the invariants follow by dotting and crossing between r,, r and f. PROBLEMS
1. If R = r/r is the unit radial vector, prove that div R = 2/r, rot R = 0. 2. For any scalar function f (r) of r alone prove that
V2f(r) = frr + 2fr/r. If v2f (r) = 0, show that f = A/r + B. S. If a is a constant vector, prove that
grad (a. r) = a, div (a x r) = 0, rot (a x r) = 2a. 4. If a is a constant vector, prove that grad (a  f)
div (a x f) = a rot f, rot(axf) Specialize these results when f = r.
DIFFERENTIAL INVARIANTS
214
5. For any vector point function f prove that 6. If Sp and ¢ are scalar point functions, prove that
div (Vp x v) = 0 7. If e is a unit vector, prove that div (e r)e = 1,
rot (e r)e = 0;
div (e x r) x e = 2,
rot (exr) xe = 0.
8. If a is a constant vector, prove that V(r x a) = I x a. 9. Show that Laplace's Equation V2f = 0 in cylindrical and spherical coordinates is and
1 a2f
1 of
a2f
a2f
+ p (9p + p2
a22
a/ af\ r{  sin0J+
sine 0 5;i
apt
a2(rf) are
1
1
sin 0 a0
a0
a2f
= 0,
=0.
10. Prove that 1
1
 (3RR  I)
vC T =
where R is a unit radial vector. 11. If f is a vector point function, prove that V
(Vf), = grad div f;
V x (Of), _ (V rot f),.
12. If Vf is antisymmetric, prove that rot f is constant and that the dyadic Vf itself is constant. 13. If X is a scalar point function, prove that
V(,I) = VXI,
V V. (XI) = VX,
V x (XI) = VX x I.
14. If f is a vector point function, prove that V
v x (I x f) = (Vf),  I div f.
(I x f) = rot f,
15. For the dyad fg prove that V V. (fg) = (div f)g + f
V x (fg) _ (rot f)g  f x Vg.
Vg,
In particular, if r is the position vector, V V. (rr) = 4r,
16. If
V x (rr) _ I x r.
is a constant dyadic, prove that div 4 . r = 91;
rot e r =
+ (§ 69).
17. If the scalar function f(p, gyp) in plane polar coordinates is harmonic, prove that f(p', tip) is also harmonic. 18. If the scalar function f(r, 0, Sp) in spherical coordinates is harmonic, prove that r 'f (r ', 0, gyp) is harmonic.
PROBLEMS
215
19. If f, g, h are scalar point functions, prove that V2(fg) = g02f + 2Vf Vg + fV2g;
V2(fgh) = gh V2f + hf V2g + fg v2h + 2fvg vh + 2gvh of + 2hvf vg. 20. If u, v, w are orthogona' coordinates and f(u), g(v), h(w) are scalar functions of a single variable, show that
V2(fgh) = fgh { f + egg + hh 21. If a is a constant vector and f = ar", prove that
of = nrs2ra,
e'f = n(n + 1)ri2a.
22. If f = r"r, prove that
of = r"I + nrnerr,
V2f = n(n + 3)rn2r.
23. If f = r"r, find a scalar function p such that f = vcp. 24. Prove that rot f = 0, and find V so that f = vp: (a)
f = yzi + zxj + xyk;
(b)
f = 2xyi + (x2 + log z)j + y/z k;
f = c r, cD a constant symmetric dyadic.
(c)
25. If a is a constant vector and f = r" a x r, prove that
div f = 0,
rot f = (n + 2)rna  nrn2(a r)r.
26. Find a vector g such that rot g = a x r (cf. § 92). 27. Prove that
a r
=  rot 
.
r'
28. The position vectors from 0 to the fixed points P1, P2 are r1, r2.
(a) If f = (r  r1) x (r  r2), show that
divf =0, (b) Prove that
rotf =2(r2 r1).
V(rrl) (rr2) =2rr1r2.
29. If f(u, v) is a vector function defined over the surface r = r(u, 1'), prove
that
H Div f =n (ruxf r, xfu).
30. If the vector f(u, v) is always normal to the surface r = r(u, v), prove that Rot f is tangent to the surface or zero. 31. If u, v, w are curvilinear coordinates, prove the operational identities: a a a 'le = ruxrw+ru,xru+ruxrv; av aw au [rurvru7
a
a (ruxr,,)xV =rv  ruav; au
a ru.V =_.
au
CHAPTER VI INTEGRAL TRANSFORMATIONS
99. Green's Theorem in the Plane. Let R be a region of the xjplane bounded by a simple closed curve C which consists of a finite number of smooth arcs. Then, if P(x, y), Q(x, y) are continuous functions with continuous first partial derivatives, (1)
J
I1
CIQ
x
 y) dx dy =
f
(P dx F Q dy),
where the circuit integral on the right is taken in the positive sense; then a person making the circuit C will always have the region R to his left.
Fra. 99a
Consider first a region R in which boundary C is cut by every line parallel to the x or yaxis in two points at most (Fig. 99a). Then, if a horizontal line cuts C in the points (x1, y), (x2, y),
f f aQ dx dy =
f
a
{Q(x2, Y)  Q(xi, y)) dy
a fQ(X2,
= =
y) dy + fQ(xiy) dy
fQ(x, y) dy. 216
GREEN'S THEOREM IN THE PLANE
§ 99
217
And, if a vertical line cuts C in the points (x, yl), (x, y2), b
If
ay
dy dx = f I P(x, Y2)  P(x, Y1)) dx
fb(
yl) dx 
J
P(x, y2) dx
6
=  fP(x, y) dx. On subtracting this equation from the preceding, we get (1). We now can extend this formula to more general regions that may be divided into a finite number of subregions which have the property that a line parallel to the x or yaxis cuts their boundary
Frc. 99b
Fia. 99c
in at most two points. For in each subregion formula (1) is valid, and, when these equations are added for all the subregions, the surface integrals add up to the integral over the entire region; but the line integrals over the internal boundaries cancel, since each is traversed twice, but in opposite directions, .leaving only the line integral over the external bounding traversed in a positive direction (Fig. 99b). The boundary of R may even consist of two or more closed curves: thus in Fig. 99c the region R is interior to Cl but exterior to C2; we may now make a cut between Cl and C2 and traverse the entire boundary of R in the positive sense as shown by the arrows.
The line integrals over the cut cancel, for it is traversed twice and in opposite directions; and the resultant line integral over C consists of a counterclockwise circuit of Cl and a clockwise circuit of Cam.
INTEGRAL TRANSFORMATIONS
218
§ 100
Although Green's Theorem is commonly stated for scalar func
tions P, Q, it is evidently true when P and Q are tensors with continuous first partial derivatives; for, when P and Q are referred to a constant basis, the theorem holds for each scalar component. 100. Reduction of Surface to Line Integrals. A surface is said to be bilateral if it is possible to distinguish one of its sides from
the other. Not all surfaces are bilateral. The simplest unilateral surface is the Mobius strip; this b may be materialized by taking a a,
a
b'
rectangular strip of paper, giving it one twist, and pasting the ends ab and a'b' together (Fig. 100a). This surface has but one side; if we move a point P along its median line and make a complete FIG. 100a
circuit of the strip it will arrive at the point P directly underneath. Since we can travel on a continuous path from one side of the Mobius Strip to the opposite, these sides cannot be distinguished. This is not the case with a spherical surface, which has an inside and an outside, or any portion S of the surface botinded by a simple closed curve C; for we cannot pass from one side of S to the other without crossing C. Let S be a portion of a bilateral surface r = r(u, v) bounded by' a simple closed curve C which consists of a finite number of smooth arcs. The surface itself is assumed to consist of a finite number
of parts over which the unit normal n is continuous. The positive sense on C is such that a person, erect in the direction of n, will have S to the left on making the circuit C. If such an oriented surface is continuously deformed into a portion of the xyplane bounded by a curve C', n becomes k (the unit vector in the direction +z) and the positive circuit on C' forms with n = k a righthanded screw.
Let f (r) be a tensor point function (scalar, vector, dyadic, etc.) whose scalar components have continuous derivatives over S. Such a tensor is said to be continuously differentiable. When the foregoing conditions on S and C are fulfilled, we then have the BASic THEOREM I. If f(r) is a continuously differentiable tensor point function over S, the surface integral of n x Of over S is equal to the integral of Tf taken about its boundary C in the positive sense: (1)
fn
Vf dS =
JTf ds.
§ 100
REDUCTION OF SURFACE TO LINE INTEGRALS
Proof.
219
Since
n x Vf =
1
II
dS = H du dv,
(ref),},
{
fn x Vf dS =
{ (r ,f).u 
du dv,
IS, s'
where S' is the region of the uvplane which contains all parameter
values u, v corresponding to points of S.
We now may apply
U
FIG. 100b
Green's Theorem in the plane to the last integral, letting u and v play the roles of x and y; thus
fs, { (rtf)u 
du dv =
f(rtf du +
dv),
where the circuit integral is taken about the curve C', which forms the boundary of S', in the positive sense, that is, in the direction of a rotation of the positive uaxis into the positive vaxis. If we
regard u and v as the functions of the are s on the curve C, it becomes
I
/
du dv ruf+rf)ds =ic Tfds, ds
ds
\
ru
du ds
+
ry
dv
dr
ds
A
= T,
the positive unit tangent along C. Formula (1) thus is established.
INTEGRAL TRANSFORMATIONS
220
§100
If we introduce the vector elements of surface and of arc,
dS = n dS,
dr = T ds,
the basic theorem for transforming surface to line integrals becomes
fdS X Vf =fdr f,
(2)
in which dS and dr must be written as prefactors.
If f is a vector, dyadic, etc., we may obtain the other integral transformations from (1) by placing a dot or a cross between the first two vectors in each term (dyad, triad, etc.) of the integrands; for both of these operations are distributive with respect to addition and therefore may be carried out under the integral sign. This process applied to the tensor n x Vf gives n V X f and n z Vf  nV f; we thus obtain the important formulas: (3)
ffl.VxfdsfT.fds,
(4)
f (n z Vf  n V f) dS = if T X f ds.
When f is a vector function, (3) becomes Stokes' Theorem:
fn.rotfds=fT.fds
(3)'
,
a result of first importance in differential geometry, hydrodynamics and electrodynamics.
If S is a closed surface which is divided into two parts S1, S2 by the curve C, we may choose n as the unit external normal; then, from (1),
fnxVfds = fTlf ds, c
fn,< Vf dS = fT2f ds,
where T2 =  T1, since the positive sense on C regarded as bounding S1 is reversed when regarded as bounding S2. On adding these integrals, we find that the integral of n X Vf over the entire surface is zero. We indicate this by the notation, (5)
fnx Of dS = 0,
§ 101
ALTERNATIVE FORM OF TRANSFORMATION
321
which denotes an integral over a closed surface. On taking dot and cross invariants in (5), we have also (6)
fn.VxfdS=0,
(7)
f(n'4Vf_nV.f)dS=0.
Example 1. Put f = r in (4) and (7); then, since grad r = I, div r = 3,
J'ndS = 2irxdr,
fnds = 0.
The last result, which states that the vector area of a closed surface is zero, generalizes the polyhedron theorem of § 17.
Example 2. When div f = 0, we have seen that f = rot g (§ 92); hence
f Thus, if f = Vu x Vv, div f = 0, and we may take g as uVv or vVu; hence
Jn.
Vu x Vv dS =
it dv = 
Jv du = 2 f(u dv  v du).
In particular, when u = x, v = y, n = k, this formula expresses an area in the
xyplane as a circuit integral 2
J(x dy  y dx) over its boundary.
101. Alternative Form of Transformation. If we replace f by ng in (100.4), where g is any continuously differentiable tensor, we have, from (97.13), (1)
f { Grad g  (Div n) ng } dS =
fT x n g is.
This integral transformation is quite as general as that given by basic theorem I. For, if we replace g by of and then cross in the first position, the integrand on the left becomes a x (nuf + n f.u) + b X
n fv) _ n x Grad f,
since Rot n = 0; whereas, on the right,
(Txn) Xnf = Tf. We thus retrieve the basic transformation (100.1).
INTEGRAL TRANSFORMATIONS
222
§ 102
We now shall express (1) in slightly different notation. On the right the vector T x n = m is the unit normal to C tangent to S and pointing outward. If we write f instead of g and J =  Div n (J is not the Jacobian of § 88), (1) now becomes
f
(2)
(vsf + J nf) dS =
fm f ds.
On dotting and crossing in the first position of the tensor integrands, we obtain the additional formulas:
f(vs.f+Jn.f)ds =
(3)
f
f
f f= Div n is called the mean curvature of the surface (§ 131).
Surfaces for which J = 0 are called minimal surfaces; for such surfaces the integrands on the left reduce to the first term. On the plane, n = k, a constant vector, and Div n = 0; then J = 0 in (2), (3), and (4). When f is a vector, these equations become (if we write dA for dS) : (5)
fGrad f dA =
(6)
fDivfdA
(7)
JRotfdA = fm f ds, 
i(m f ds,
where m is the external unit normal to the closed plane curve forming the boundary. We may deduce (6) from (7) by replacing
fbykxf;for
rf
m x (k x f) = km f. 102. Line Integrals. When f(r) is a vector, consider the line Rot (k x f) = k Div f,
integral, (1)
J'f.dr= o
to
dr dt
dt,
ta ken over a curve C: r = r(t) from the point P0 to P. The value of this integral depends on the curve C and the end points Po and
LINE INTEGRALS
§ 102
223
P, but not on the parameter t. For, if we make the change of parameter t = t(T), t
Jcu
f dr dt= dt
pT TO
fdrf  dr. dt dr ar r
dr dt
dr
,,
Let us consider under what conditions the line integral (1) is independent of the path Cthat is, when its value is the same for all paths from Po to P. We shall call a closed curve reducible in a region R if it can be
shrunk continuously to a point without passing outside of R. Thus in the region between two concentric spheres all closed curves are reducible. However in the region composed of the points within a torus, all curves that encircle the axis of the torus are irreducible. A region in which all closed curves are reducible is called simply connected. Thus the region between two concentric spheres is simply connected, whereas the interior of a torus is not. If f is continuously differentiable, and rot f = 0 in a region R, the line integral of f around any reducible closed curve in R is zero;
for we can span a surface S over R that lies entirely within R (shrinking the curve to a point generates such a surface), and then, by Stokes' Theorem,
ff . dr = fn. rotfdS=0. We may now prove the THEOREM.
If f is a continuously differentiable vector, and rot f
= 0 in a simply connected region, the line integral
J'f
dr between
any two points of the region is the same for all paths in the region joining these points. Proof. If C1 and C2 are any two curves POAP, POBP joining PO to P, the line integral of f around the circuit POAPBPO is zero; for all closed curves in a simply connected region are reducible. Hence
J
f dr=0.
fPBP, oA P
P
Po
INTEGRAL TRANSFORMATIONS
224
§ 103
Under the conditions of this theorem, the line integral (1) from a fixed point Po to a variable point P of the region depends only upon P; in other words the integral defines a scalar point function,
p(r) = Jf dr.
(2)
o
Let us compute d 0), the force F is upward if the flow is horizontal and to
the left (v =  Vi); we then have a lift of pyVj per unit length of cylinder. The counterclock circulation diminishes the flow velocity below the cylinder and increases it above; by Bernoulli's Theorem this results in an excess of pressure below with a consequent upward lift. In the absence of circulation about the cylinder (y = 0), F = 0, and there can be no lift. This is the case in § 127, ex. 2, where
w = V(z + a2/z),
v = dw/dz =  V(1  a2/z2),
and a1 = 0 in the Laurent series.
HYDRODYNAMICS
278
§ 129
129. Summary: Hydrodynamics. For a perfect (nonviscous)
fluid the stress dyadic is pI; at any point within a fluid the pressure is the same in all directions and exerted normal to a surface element. If f (t, r) is any tensor function associated with a fluid particle moving with the velocity v, its substantial rate of change is df dt
of
Vf, at + V
where of/8t is the local rate of change. In the Eulerian method of dealing with fluid motion, the aim is to compute v, p, and p as functions of the independent variables r, t. The (kinematic) equation of continuity is 8p
at
+div(pv) =
dp
dt
+pdivv=0;
and, for an incompressible fluid (dp/dt = 0), becomes div v = 0. The Eulerian Equation of Motion is
P=f dp,
dt=V(Q+P),
(E)
when the body forces have a potential Q and the density a function of p only. From this we can deduce the Differential Equation of Helmholtz for the vorticity co = 1 rot v:
dt \p/ =
P
Vv.
In the Lagrangian method the aim is to follow the motion of the fluid particles from their initial positions ro (t = 0) to their positions r after a time t. The independent variables are now ro and
t; the position r of a particle at time t is a function of ro and t. We may take gradients relative to ro(V0) or r(V); and these conform to the relations, Vof = Vor Of,
Vf = Vro Vof;
Vor
Vro = I.
If J is the third scalar of the dyadic Vro, the equation of continuity becomes
pJ = po,
or dJ/dt = J div v.
SUMMARY: HYDRODYNAMICS
§ 129
279
The Lagrangian Equation of Motion corresponding to the Eulerian Equation (E) is (L)
Vor
dv
dt
=  Vo(Q + P).
From this we deduce Cauchy's integral of Helmholtz's Equation, 0)0 _=.Vor;
(0
P
Po
f
and this in turn shows that vortexlines (co x dr = 0) move with
the fluid. We find, moreover, that the circulation
v dr over
any closed curve moving with the fluid does not alter with the time.
When the motion is irrotational (rot v = 0), v =  VV, where p is the velocity potential, and
iv
dr = 0 over any reducible
curve. When the body forces are conservative and p is a function of p alone,
Q+P+
Zv2
 a = f (t), an arbitrary function of t.
For a steadymotion equation (E) gives
vxroty=V(Q+P+Zv2); whence Bernoulli's Theorem: Q + P + Zv2 is constant along a
streamline or vortexline, and this constant is absolute if the steady motion is also irrotational. For an incompressible fluid in plane motion, the integral independent of the path,
¢(r) =
x k dr (k I. plane of motion),
defines the stream function. The curves 4, = const are the streamlines (v x dr = 0). If the motion is also irrotational, both stream function and velocity potential are harmonic (V2# = 0, V2V = 0), and V4, = k x VV. This is the vector equivalent of the Cauchy
Riemann Equations which guarantee that w = p + iii is an analytic function of the complex variable z = x + iy in the plane of motion. The complex potential w therefore has a unique derivative dw/dz; and the negative of its conjugate gives the complex velocity
vector: v = w'.
HYDRODYNAMICS
280
PROBLEMS 1. A mass of liquid is revolving about a vertical axis with the angular speed
f(r), where r is the perpendicular distance from the axis. With cylindrical coordinates r, 0, z (we replace p, p in § 89, ex. 1 by r, 0 to avoid conflict with
the notation of Chapter VII), let a, b, c be unit vectors in the directions of r,., rg, rZ (Fig. 89a). If the angular velocity w = f(r)c, prove that
v = rf(r)b,
rot v = r dr [ref (r)]c.
[Cf. (89.7).]
2. If the motion in Problem 1 is irrotational, show that
a
PC,
and that the velocity potential v =
a v=rb,  ao is not single valued (a, $ are
constants). For a liquid of constant density p under the action of gravity alone, show that the pressure is given by
gz+p+const. 3. If a fluid is bounded by a fixed surface F(r) = 0, show that the fluid must
satisfy the boundary condition v VF = 0. More generally, if the bounding surface F(r, t) = 0 varies with the time, show that the fluid satisfies the boundary condition
F + v 7F = 0
[Cf. (120.3).]
4. A sphere of radius a is moving in a fluid with the constant velocity u. Show that at the surface of the sphere the velocity of the fluid satisfies the condition
!vu) (r  ut) = 0.
5. A fluid flows through a thin tube of variable cross section A. Show that for a tube PoP of length s a
J8PA ds + pAv
0;
hence deduce the equation of continuity at (pA) + as (pAv) = 0.
6. From the equation of fluid equilibrium (118.1) show that when p is a function of p alone, rot R = 0 and that the potential of R is P. If p is not a function of p alone, show that equilibrium is only possible when R R. rot R = 0. [Cf. (121.7) for P.]
PROBLEMS
281
7. A gas flows from a reservoir in which the pressure and density are po, po into a space where the pressure is p. If the expansion takes place adiabatically,
p/py = const. (y is the ratio of specific heats), show that
P=
y
p
y1p
Neglecting body forces and the velocity of the gas in the reservoir, show that the velocity v of efflux when the motion becomes steady is given by v2 =
2y
y=1
1  (p/po) 7 } .
7?0
y  1 PO
The velocity of sound in a gas is c = 1'yp/p; hence show that 2
v2 _
 1
(c0  c2).
y
8. If a body of liquid rotates as a whole from r = 0 to r = a with the constant angular velocity coo, and rotates irrotationally with the angular velocity to = woa2/r2 when r = a, we have the socalled "combined vortex" of Rankine. If z = za at the free surface when r = a, show that the free surface is given by
Z = za.+(r2a2) a2`
Z = Z. +w2a2  1! 1  2J , 2 \ r
r 5 a, r
a.
Prove that the bottom of the vortex (r = 0) is a distance w2a2/g below the general level (r = oc) of the liquid. 9. If the vorticity is constant throughout an incompressible fluid, prove that V2v = 0. [Cf. (84.14).] 10. When the motion of an incompressible fluid is steady, deduce from Helmholtz's Equation (122.6) that w Vv = v Vw. 11. If an incompressible liquid in irrotational motion occupies a simply connected region, show that
fv2dv
=
f
dodS'
where p is the velocity potential and the normal derivative dp/dn is in the direction of the external normal to the bounding surface. [Cf. (108.1).] 12. Prove Kelvin's theorem: The irrotational motion (v) of an incompressible fluid occupying a simply connected region S with finite boundaries has
less kinetic energy (T =
p
f
z the same boundary conditions.
v v dV) than any other motion (vl) satisfying
[Put vi = v + v'; then div v' = 0 and n v' = 0, where n is the unit normal vector to the boundary S. Show that Ti = T + T'.]
HYDRODYNAMICS
282
13. Under the conditions of Problem 11, show that if (a) v = const. over the boundary, or (b) dip/dn = 0 over the boundary, then V is constant throughout the region and v = 0. Hence show that the irrotational motion of a liquid occupying a simply connected region is uniquely determined when the value of either w or d'p/dn is specified at each point of the boundary. 14. If the body forces have a potential Q, the integrals, T
f a pv2 dV,
U= f pQ dV,
represent the kinetic and potential energy of the fluid within the region of
integration.
Show that for an incompressible fluid
d(T[U)_15. Assuming that the earth is a sphere of incompressible fluid of constant density p and without rotation, show that the pressure at a distance r from its center is p = 2gpa(1  r2/a2), where a is the radius of the earth.
[If y is the constant of gravitation, the attraction on a unit mass at the distance r is 47rtipr3/r2
= gr/a
where g is the attraction when r = a; hence the body force grR/a has the potential 'gr2/a.] Compute the pressure at the center if p is taken equal to the mean density of the earth, (pg = 5.525 X 62.4 lb./ft.3). 16. In example 1 of § 127, consider the motion when n = r/a(0 < a < r). Since the lines 0 = 0, 0 = a are parts of the same streamline = 0, we have the steady irrotational motion of a liquid within two walls at an angle a. Find the radial and transverse components of v at any point (r, 0).
17. Discuss the plane, irrotational motion when the complex potential w = rp + i¢ (§ 127) is given by z = cosh w. Show that the equipotential lines and streamlines are the families of confocal ellipses and hyperbolas: y2
xz
c2 cosh' v
+
x2 C2 COS2 '
c2 sinh2 ,p 
1,
2
c2 sin2
 1'
with foci at (fc, 0). Show that the streamlines 4, = nor, where n is any positive integer, correspond to the part of the xaxis from x = fc to x = f co. If we regard this as a wall; we have the case of a liquid streaming through a slit of breadth 2c in an infinite plane.
CHAPTER VIII GEOMETRY ON A SURFACE
130. Curvature of Surface Curves. Let C be any curve on the surface r = r(u, v) with unit normal n (94.11). As the point P traverses C with unit speed, the Darboux vector, 6 = TT+KB,
(1)
gives the angular velocity of the moving trihedral TNB. Consider now the motion of the dextral trihedral Tnp (p = T x n) associated with the curve (Fig. 130). Since the trihedrals Tnp and
T
T points upward from paper
FIG. 130
TNB have T in common, the motion of Tnp relative to TNB is a rotation about T. If p = angle (N, n), taken positive in the sense of T, the angular velocity of Tnp relative to TRB is (d a/2, is called a tractrix. The equation u = a cos yG shows that the segment of any tangent between the curve and zaxis has the constant length a. This property is characteristic of the tractrix (ef. § 50, ex. 5).
140. Bonnet's Integral Formula. The differential formula for the total curvature, (1)
K = n rot R,
has an important integral equivalent. If we integrate K over a portion of a surface S bounded by a simple closed curve which consists of a finite number of smooth arcs, we have, by Stokes'
BONNET'S INTEGRAL FORMULA
§ 140
309
Theorem,
fKdS= from (134.12); hence (2)
fKds =
f f do 
f(. 
ds,
y ds.
If the bounding curve has a continuously turning tangent,
fdo
= 2ir and (2) becomes (3)
fKdS
= 2ir
fds.
f
This very important formula was discovered by the French geom
eter, Bonnet, in 1848. The integral
K dS over S is called the
integral curvature of S. Let us first apply (2) to the figure bounded by two geodesic arcs
APB and AQB meeting at A and B. Then denoting the interior angles at the corners by A and B (Fig. 140a), we have (4)
fK dS = fdo = 2ir  (ir  A)  (ir  B) =A+ B.
Since A + B > 0 this equation is impossible when K = 0; hence, on a surface of negative or zero total curvature, two geodesic arcs
Fu}. 140a
FIG. 140b
GEOMETRY ON A SURFACE
310
3 14t)
cannot meet in two points so as to enclose a simply connected area. We may state (4) as follows: THEOREM 1. The integral curvature of a geodesic lune is equal to the sum of its interior angles.
We next apply (2) to a geodesic triangle ABC (Fig. 140b) : that is, the figure enclosed by three geodesic arcs. Again denoting the interior angles at the corners by A, B, C, we have
fKdS = fdO=27r (7rA)  (irB)  (rrC), (5)
fKds=A+B+C_.
THEOREM 2 (Gauss).
The integral curvature of a geodesic triangle
is equal to its "angular excess," that is, the excess of the sum of its angles over 7r.
When K is constant, the integral curvature
f
K dS is the prod
uct of K by the area S. If K is identically zero, as on a plane, the sum of the angles of a geodesic triangle is 7r. If K 34 0, we have THEOREM 3 (Gauss). On a surface of constant nonzero total curva
ture, the area of a geodesic triangle is equal to the quotient of its angular excess by the total curvature. Example 1. A sphere of radius r has the constant total curvature K = I /r2. For all normal sections at a point are great circles of curvature 1/r; all points are umbilical points and k1k2 = 1/r2. Formula (5) now states that: The area of a spherical triangle is r2 times its angular excess in radians. If e denotes the angular excess in right angles, we therefore have the formula,
Area = r2 2 e =
4,rr2
8
e,
that is, the area of a spherical triangle equals the area of a spherical octant times the angular excess in right angles. Example 2. Let us decompose a closed bilateral surface S, consisting entirely of regular points (§ 93), into a number f of curvilinear "polygons" Si whose sides are analytic arcs. These sides meet in vertices at which at least three arcs come together. Since S is regular and bilateral, the surface has a continuous unit external normal n which defines a positive sense of circuit on each polygon by the rule of the righthanded screw. When positive circuits
BONNET'S INTEGRAL FORMULA
§ 140
311
are made about two adjoining polygons, their common side is traversed in opposite directions. Now for each Si we have, by Bonnet's Theorem (2),
J(i KdS=27r
(7rai;)Js
where ail are the interior angles at the vertices of Si. If these equations for all of the f polygons Si are added, we have
fKdS
i
j
(a 
2irf 
for the integrals f y ds over adjoining sides cancel in pairs since y changes sign with change of direction. In each polygon the number of interior angles equals the number of sides; hence 2;2:7r = 2eir, where e is the total number of sides or edges. Moreover the sum of the interior angles about any vertex is
27r; hence TdIai; = 2v7r, where v is the total number of vertices. We thus obtain
fe+v=1
(6)
KdS.
is,
From the right member of (6) we see that the number on the left is independent of the manner in which S is subdivided into polygons. From the left member
we conclude that cb K dS is not altered when S is deformed into another completely regular surface. If S can be continuously deformed into a sphere, we have
fK dS =
(7)
47rr2 = 4a
for a sphere, and (8)
f  e + v = 2.
Evidently f  e + v is not altered by any continuous deformation of S, even though the resulting surface is not completely regular. Thus, if S is transformed into a polyhedron (a closed solid with plane polygons for faces), the relation (8) still holds good and constitutes the famous Polyhedron Formula that Euler discovered in 1752: Faces + vertices  edges = 2. Next suppose that S can be continuously deformed into a torus. From (138.6), for a torus, (9)
f"KdS=ffKb(a+bcoso)dedv
2
= 0
f
2"
cos e do dv, = 0;
0
hence f  e + v = 0 for any surface continuously deformable into a torus. This relation holds, for example, for any ring surface with polyhedral faces.
GEOMETRY ON A SURFACE
312
§140
Example 3. Parallel Displacement. A vector f, that always remains tangent to a surface S, is said to undergo a parallel displacement along a surface curve C when the component of df/ds tangential to the surface is zero; then df
df nx=0 or
(10)
ds=>`n.
Since
the length of f must remain constant during a parallel displacement. Now the trihedral Tnp associated with the curve C has the angular velocity,
w = tT + yn + kp
(130.3').
If the angle 0 = (T, f) is reckoned positive in the sense of n, the trihedral fng (g = f x n) has the angular velocity n dB/ds relative to Tnp. Hence, as f moves along C with unit speed, the trihedral fng revolves with the angular velocity,
SZ = to + n  = tT + (y + dB/ds)n + kp, hence df/ds = St x f, and
dfxn = (12 xf) xn = (n  fl)f =
+do> f. C'Y
Hence f will undergo a parallel displacement along the curve C when and only when de
y+  =0.
(11)
If f is tangent to C (or, more generally, when 0 remains constant), f will undergo a parallel displacement along C only when C is a geodesic. THEOREM. If a tangential surface vector is given a parallel displacement about a smooth closed curve C, it will revolve through an angle, (12)
27r  if y ds = f K dS.
Proof. From (11) we see that f revolves through the angle,
if
ds

fy ds
relative to T, the unit tangent vector of C. Since T itself revolves through 2r
in passing around C, the total rotation of f is 27r Bonnet's Integral Formula (3).
fy
ds,
orJK dS by JJJJJ
NORMAL SYSTEMS
§ 141
313
141. Normal Systems. At every point of the surface S, r = r(u, v), a straight line is defined by the unit vector m(u, v). The points of this twoparameter family of lines are given by
rl = r + Am.
(1)
Under what circumstances do these lines admit an orthogonal surface? That is, when do they form the system of normals to a surface?
When A is a definite continuous function A(u, v), r1 is the position vector of a surface S1. If V denotes a surface gradient relative to S, we have, from (1), Vr1 = Or + (VX)m + AOm,
+VA,
since m m = 1. In Vr1 = arlu + brl,,, the postfactors are tangent to S1; hence, in order that m be normal to S1, it is necessary and sufficient that (Or1) m = 0, that is,
 VA = (Or) m = m  (m n)n
(2)
(132.5).
But (2) implies that
n rot m = 0;
(3)
conversely, from § 103, theorem 2, (3) implies that (4)
mg = (Or) m = VA where X= 
f m dr, ro
the integral being taken over any path from ro to r(u, v) on S. We have thus proved THEOREM 1.
In order that a twoparameter family of lines
r(u, v) + Xm(u, v) form a normal system, it is necessary and suffi
cient that n rot m = 0, where n is the unit normal to the surface r = r(u, v). When m fulfils condition (3), we can find a oneparameter family
of surfaces normal to the lines r + Am. We need only determine A(u, v) from (4) with some definite choice of ro. Since ro may be chosen at pleasure, all possible values of A are then given by
314
GEOMETRY ON A SURFACE
§ 142
X + C, where C is an arbitrary constant. We thus obtain a oneparameter family of surfaces, r1 = r + (X + C)m, normal to the lines. Consider now a geodesic field over S with the unit tangent vector e. Since y = n rot e = 0 at every point of S, we have The tangents to the geodesics of a field form a normal system of lines. THEOREM 2 (Bertrand).
We conclude with an important theorem in geometrical optics. THEOREM 3 (Malus and Dupin). If a normal system of lines is reflected or refracted at any surface, it still remains a normal system. n
FIG. 141
Let el denote the unit vectors along the incident rays and e2 the unit vectors along the refracted (or reflected) ray (Fig. 141). If µi denotes an absolute refractive index, µ1 sin 01 = Proof.
µ2 sin 02 (Snell's Law) and e1, e2, n are coplanar; hence,
(µ2e2  µ1e1) x n = 0 and nrot (µ2e2  µ1e1) = 0, by the theorem following (97.10). For the reflected ray (§ 75),
e2 = (I  2nn)
e1
and n  rot (e2  e1) = 0.
In either case: n rot e1 = 0 implies n  rot e2 = 0142. Developable Surfaces. A developable surface (§ 138) is the envelope of oneparameter family of planes. The limiting line of intersection of two neighboring tangent planes is a line, or ruling, on the surface. A developable is therefore a ruled surface. The equation of a ruled surface may be written (1)
r = p(u) + ve(u),
DEVELOPABLE SURFACES
§ 142
315
where p = p(u) is a curve C crossing the rulings and e(u) is a unit vector along the rulings. The surface normal is parallel to ru x rv = (Pu + veu) x e.
When the normal maintains the same direction along a ruling, the surface is developable. For the same plane is tangent to the sur
face along an entire ruling; and, as each value of u corresponds to a ruling and hence to a tangent plane, the surface is enveloped by a oneparameter family of planes. Consequently the surface (1) is developable when and only when pu x e and eu x e are parallel; that is, when pu, e and eu are coplanar:
pu e x eu = 0.
(2)
Case 1: e x eu = 0. Since e e = 1, we have e . eu = 0 and hence eu = 0. Therefore e is constant, and (1) represents a general cylinder. Case 2: e x eu 34 0. We then may write (§ 5)
Pu = a(u)e + 5(u)eu.
(3)
With a function A(u), as yet undetermined, let us write (1) as (4)
r= (P+Xe)+(vX)e=q+(vX)e,
where (5)
q(u) = p(u) + X(u)e(u).
Then
qu = pu + Xue + Xeu = (a + Xu)e + (S + x)eu; and, if we choose X = ,6, (6)
qu = (a  {3u)e.
If a = l3u along C, qu = 0 and q is constant. Then (4) represents a cone of vertex q if the vectors e(u) are not coplanar, a plane if they are coplanar.
In general, however, a 0 $u; then (6) shows that q(u) traces a curve having e as tangent vector. The surface (4) then is generated by the tangents of the curve q = q(u); it is the tangent surface of this curve. Developable surfaces are planes, cylinders, cones or tangent surfaces.
GEOMETRY ON A SURFACE
316
§143
143. Minimal Surfaces. In § 131 the mean curvature of a sur
face was defined as J =  Div n = kl + k2. With f = 1, the integral theorem (101.2) becomes
fJnds=Jmde.
(1)
Next put f = r in (101.4) ; then, since Or = I  nn, Rot r = 0, and we have (2)
fr
x Jn dS
=Jrxmds.
These equations admit of a simple mechanical interpretation: THEOREM 1. The normal pressures Jn (per unit of area) over any simply connected portion of surface S bounded by a closed curve C are statically equivalent to a system of unit forces (per unit of length) along the external normals to C and tangent to S.
Consider, now, a soap film with a constant surface tension q per
unit of length and subjected to an unbalanced normal force pn per unit of area. Since any portion of the film bounded by a closed curve C is in equilibrium, we must have
fpn dS + 2q Jmds= f (p + 2Jq)n dS = 0; the surface tension q is doubled, because it is exerted on both sides
of the film. We therefore conclude that (3)
p = 2Jq.
Thus, for a soap bubble of radius r, the normal curvature is everywhere
k = «x  n = rc cos Tr = 11r
and J = 2/r;
thus the pressure inside of a soap bubble exceeds the pressure out
side by an amount p = 4q/r. In particular, if the film is exposed to the same pressure on both sides, p = 0; then J = 0 at all points of the film. A surface whose mean curvature J is everywhere zero is called a minimal surface. Thus a soap film spanned over a wire loop of any shape and with atmospheric pressure on both sides materializes a minimal surface. Let So be a minimal surface bounded by a closed curve C. Its existence is guaranteed by its physical counterpart, the soap film
MINIMAL SURFACES
§ 143
317
spanned over C. Imagine now that So is embedded in a field of minimal surfaces, that is, a oneparameter family of surfaces in a certain region R enclosing So, such that through every point of R there passes one and only one surface of the family. Such a field may be generated, for example, by the parallel translation of So.
At each point of R, the unit normal no to the minimal surfaces may be so chosen that no is a continuous vectorpoint function; and, from (97.2),
div no = Div no + no
dno
do
= J + 0 = 0.
Now let S be any other surface in R spanning the curve C and enclosing with So a volume V of unit external normal n. From the divergence theorem,
n=
f
n no
over
n no
cos
dS <
fdS.
We have thus proved THEOREM 2. A minimal surface spanning a closed curve and belonging to a field of minimal surfaces has a smaller area than any other surface spanning the same curve and lying entirely in the field. Example 1. In order to find a surface of revolution which is also minimal, we set J = 0 in (138.5). The resulting differential equation,
uz" + z'(1 + z'2) = 0 may be written z'z" z72
z'z"
1
+z'2+u=0
and has the first integral uz'/1/1 + z'2 = C. Solving for z', we have z' _ C/1/u2  C2; hence
z + k =Ccoshr C ,
u
=Ccoshz Ck.
This represents a catenary in the uzplane; when revolved about the zaxis it generates a surface known as the catenoid. Catenoids are the only minimal surfaces of revolution.
GEOMETRY ON A SURFACE
318
§ 143
Example 2. The right helicoid is a surface generated by a line which always cuts a fixed axis at right angles while revolving about and sliding along the axis at uniform rates. In other words, the right helicoid is a spiral ramp. A right helicoid about the zaxis has the parametric equations,
z = as,
y = u sin v,
x = u cos v,
(4)
where u, v are plane polar coordinates p, p in the xyplane. The lines v = const are the horizontal rulings on the surface; the lines u = const are circular helices on the cylinder x2 + y2 = u2. Equations (4) give the vector equation,
r = u R(v) + av k,
(5)
where R is a unit radial vector (R . k = 0).
Now
r = uP + ak; ruxr uk  aP
ru = R,
n
_ I ru x rv I
=
a2
a
nu = (u2 + a2)2 rv,
a nv = (u2 + a2)2 ru.
Since ru r, = 0, the parametric lines are orthogonal; and, since ru . nu = 0,
r n = 0, they are also asymptotic lines. From nu x n = K ru x r (132.20), we have
K=
 a2 (u2 + a2)2
and, since nu x r + ru x n = 0, J = 0 (132.19). The right helicoid is a minimal surface. Its negative total curvature is constant along any helix u  const.
Along the lines of curvature,
(rudu +rvdv) x (nudu
ruxnudu2 +rvxnndv2 = 0,
or du2  (u2 + a2) dv2 = 0. The two families of these lines therefore satisfy the differential equations:
du/
u2 + a2 = fdv.
On integration, these give u sinh1 = c f v or u = a sinh (c f v) a
as the finite equations of the lines of curvature. Example 3. On a minimal surface, the asymptotic lines form an orthogonal system (§ 131, ex. 2). If we choose them as our field curves, k1 = k2 = 0, and the Codazzi Equations (137.13), (137.14) become 27211 +
i = 0,
2y1t1 

ds2
= 0.
SUMMARY: SURFACE GEOMETRY
144
319
Since K = ti, dK dsl
dt1 = 2t1_ 472K,
dK ds2
ds1
dt1 = 2t1= 4y1K; ds2
dK dK VK = el  + e2  = 4K(y1e2  72e1) = 4K n x R dsl ds2
(134.3).
144. Summary: Surface Geometry. On a surface r = r(u, v) of
unit normal n, the angular velocity of the trihedral Tnp (p = T x n) along a curve is w = tT + yn + kp. If (p = angle (N, n), positive in the sense of T,
t = r+ dv/ds,
y= K sin cp,
k = K cos V
are the geodesic torsion, geodesic curvature, and normal curvature, respectively. From do (1)
ds
=wxn= kT  tnxT,
we conclude that k and t are the same for all surface curves having
a common tangent at a point. Important surface curves and their differential equations: Lines of curvature (t = 0) : Asymptotic lines (k = 0) : Geodesics (y = 0) :
d2r nx
ds2
dr do x= 0; ds ds dr do dsds= 0; = 0 or Vr
d2r ds2
= 0.
The equations for geodesics follow from dT
xn=KNxn=yT. ds
If e and e' = n x e are perpendicular directions in the tangent plane,
Vn = k ee + k' e'e' + t ee' + t' e'e. Since Rot n = 0, Vn is symmetric, and t + t' = 0. The first and second scalar invariants of  Vn,
J =  Div n = k + k',
the mean curvature;
K = kk' + tt' = kk'  t2, the total curvature,
GEOMETRY ON A SURFACE
320
§ 144
have the same value for any pair of perpendicular directions. If the point is not an umbilic (k constant for all directions), k attains its extreme values in the orthogonal principal directions e1, e2 for which t1 = 0, t2 = 0. In terms of the principal curvatures k1, k2,
J = k1+'k2iK=k1k2. If a, b, n are reciprocal to ru, r,,, n, the first and second fundamental forms are
Or = E as + F(ab + ba) + G bb,
Vn = Las+M(ab+ba) }Nbb; and the fundamental quantities are given by
L = ru nu,
M = ru nv = rv nu,
N = rv nv.
If e1i e2 = n x e1 are the unit tangent vectors along a field of surface curves and their orthogonal trajectories, we have
Vet = Re2  Pn,
Ve2 =Re1  Qn,
Vn = Pet + Qe2,
where
P = do/ds1,
Q = dn/ds2,
R = y1e1 + y2e2.
Codazzi Equations: n V x (Vn) = 0; or
Gauss Equation: n V x (De1) = 0; or
n rot R = K. For any field of curves of tangent vector e,
y = n rot e. The last two equations show that y and K may be expressed in terms of E, F, G and their partial derivatives. When embedded in a geodesic field, the arc of a geodesic is shorter than any other curve lying within the field and having the same end points.
If a field of curves of tangent vector e cuts the field e1 at the angle 0 = (eli e), positive in sense of n, R = y1e1 + y2e2 = ye + y'e'  V0.
PROBLEMS
321
Bonnet's Integral Formula,
fK d S= fdo 
ds,
is the integral ("Stokian") equivalent of the Gauss Equation, n  Rot R = K. If the simple closed curve C has a continuously turning tangent
fdo = 27r, but if the path has angles (in
terior angles ai),
fdo = 2ir If C is a geodesic triangle,
 as).
fK dS = al + a2 + a3  ir; and, on
a surface of constant K 31 0, the area of a geodesic triangle is
(al + a2 + a3  ir)/K. A sphere of radius a has the constant positive curvature 1/a2; a tractrix of revolution for which the tangential distance to the asymptote is a has the constant negative
curvature 1/a2. A minimal surface (J = 0) embedded in a field of minimal surfaces and spanning a closed curve C has a smaller area than any other surface lying within the field and spanning C. Soap films spanned over wire framework materialize minimal surfaces. The right helicoid is a ruled minimal surface. The catenoid is the only minimal surface of revolution. PROBLEMS
1. If u, v, are plane polar coordinates r, 0, prove that the surface obtained by revolving the curve z = f(x) about the zaxis has the parametric equations:
x = u cos v,
y = u sin v, z = z(u).
2. A straight line, which always cuts the zaxis at right angles, is revolved about and moved along this axis. The surface thus generated is called a conoid.
If u and v are plane polar coordinates in the xyplane, show that the conoid has the parametric equations: x = u cos v,
z = z(v).
y = u sin v,
In particular when dz/dv is constant the conoid is a right helicoid: x = U cos v, y = u sin v, z = av. 3. The central quadric surface, x2
y2
a2 ± b2 ±
z2
_
1,
GEOMETRY ON A SURFACE
322
is an ellipsoid, a hyperboloid of one sheet, or a hyperboloid of two sheets according
as the terms on the left have the signs (+, +, +), (+, +, ), (+, , ). Show that the corresponding parametric equations are x = a sin u cos v, y = b sin u sin v,
z = c cos u;
x = a cosh u cos v, y = b cosh u sin v, z = c sinh u; y = b cosh u sinh v,
x = a cosh u cosh v, 4. The paraboloid,
x2 a2
z = c sinh u.
t =, y2
2z
b2
c
is elliptic or hyperbolic, according as the terms on the left have the signs (+, +) or (+, ). Show that the corresponding parametric equations are
x = au cos v, y = bu sin v, z = Zcu2; x = au cosh v, y = bu sinh v, z =.
4cu2.
5. The position vector,
r = f(u) + g(v), traces a surface of translation. Show that any curve u = a (const.) may be obtained by giving the curve rl = g(v) a translation f(a); and that any curve v = b may be obtained by giving the curve r2 = f(u) a translation g(b). What are the surfaces,
r=f(u)+bv,
r=au+bv+c?
6. Show that a right circular cylinder of radius a has the constant mean curvature J = 1/a. 7. For the surface xyz = a3 show that 3a6
Jr = 2a3(x2 + y2 + z2)
(X2y2 + y2z2 + 22x2)2 '
(x2y2 + y2z2 + 22x2)7
[Cf. § 132, ex. 2.]
8. For the elliptic paraboloid, x2/a2 + y2/b2 = 2z/c
prove that the total curvature K 5 c2/a2b2. 9. For the surface of revolution about the zaxis,
r = iu cos v + ju sin v + kz(u), show that
L=
E=1+z'2 F=O, G=u2; 2'2)1, M = 0, N = uz'/(1 + 2'2)1.
[Cf. § 132.1
From (132.21) and (132.22) deduce the values of K and J and from (132.25) find the principal curvatures. Check with (138.4) and (138.5).
PROBLEMS
323
10. For the conoid,
r = iu cos v + ju sin v + kz(v), show that
E=1, F=0, G=u2+z'2=H2; L = 0, M = z'/H, N = uz"/H; K = z'2/H4, J = uz"/H3.
[Cf. § 132.1
When z = av, show that the resulting right helicoid is minimal (§ 143); and that its principal curvatures are fa/(u2 + a2). 11. Show that the curvatures J and K have the dimensions of (length) 1 and (length) 22. Verify this in Problems 6, 7, 8, 9, 10.
12. The position vector of a twisted curve r is given as a function ro(s) of the The surface generated by its tangents is
are.
r=rc(v)+uT(v) where v = s and u is the distance along a tangent measured from r. Show that
this tangent surface has the curvatures K = 0, J = z/i u Jx. What are the principal curvatures?
13. If a parabola is revolved about its directrix, show that the principal curvatures of the surface of revolution satisfy the relation 2k1 + k2 = 0.
[With the notation of § 138, ex., the parabola has the equation z2 _ 4a(u  a) when the directrix is the zaxis and u = 2a at the focus.] 14. The vectors a, b, n and ru, r,,, n form reciprocal sets of vectors over the surface r = r(u, v) [Cf. § 95]. Hence show that any vector f(u, v), defined over the surface, may be written f 15. Prove that
H2a = Gr  Fr,,,
H2b = Fru + Er,,.
[Use Prob. 14 and (132.18).]
16. Prove that
FM  GL H2
II,
a
'b=
'
H2
FM  EN
FN  GM H2
FL  EM
I
H2
[Use Prob. 15.]
17. Prove the derivative formulas of Weingarten:
HZnu = (FM  GL)ru + (FL  EM)r,,,
H2n _ (FN  GM)ru + (FM  EN)r,,. [Use Prob. 14 and Prob. 16.]
;
GEOMETRY ON A SURFACE
324
18. Show that the asymptotic lines of the surface r = r(u, v) have the differential equation,
(ru du + r dv) (nu du + n dv) = 0 [Cf. (130.10)]; or, in terms of fundamental quantities,
Ldu2+2Mdudv+Ndv2 = 0. 19. Prove that the asymptotic lines on a right helicoid (Prob. 10) are the parametric curves and form an orthogonal net. 20. For the ruled surface,
r = p(u) + ve(u), prove that M = pu e x eu/H, N = 0.
(142.1).
21. Prove that a ruled surface is developable when, and only when, M = 0. [Cf. (142.2).] 22. Prove that the asymptotic lines of a developable surface, not a plane, are the generating lines (counted twice). [Differential equation: du 2 = 0.]
23. Prove that, along a curved asymptotic line, (a) The osculating plane is tangent to the surface. (b) The geodesic torsion equals the torsion: t = r. (c) The geodesic curvature is numerically equal to the curvature: y = ±K. (d) The square of the torsion is equal to the negative of the total curvature:
2 = K. 24. Prove that
(a) If an asymptotic line is a plane curve other than a straight line, it is also a line of curvature. (b) An asymptotic line of curvature is plane. 25. Prove that the following conditions are necessary and sufficient in order that the surface be
(i) A plane: L = M = N = 0. (ii) A sphere: E/L = F/M = GIN. 26. Show that the lines of curvature of the surface r = r(u, v) have the differential equation, n (r, du + r dv) x (nu du + n,, dv) = 0
[Cf. (138.1)]; or, in terms of fundamental quantities, I
du2
E
du dv F
L
M
N
dv2
I
G
= 0.
27. Show that the parametric curves are lines of curvature when, and only
when, F = 0, M = 0. 28. In the notation of § 132, ex. 2, prove that the lines of curvature on the surface z = z(x, y) have the differential equation, dy2
dx dy
dx2
1 + p2
pq
1 + q2
= 0.
PROBLEMS
325
29. Show that the hyperbolic paraboloid of Prob. 4 has the parametric equations:
x = 2a(u + v), y = 2b(u  v), z = cuv. Prove that the differential equations of its asymptotic lines and lines of curvature are, respectively, dvl2
du dv = 0,
\du /
a2 + b2 + c2y2 a2 + b2 + c2u2
and find the uvequations of the asymptotic lines and lines of curvature. 30. If the parametric curves on a surface are its lines of curvature, show that
nu = kiru,
nv = k2rv.
31. Two surfaces that have the same normal lines are called parallel. Surface points on the same normal are said to correspond; thus
r(u, v) = r(u, v) + An(u, v) determines corresponding points on the parallel surfaces S and S. Prove that the distance X between the parallel surfaces is constant. [If we take the lines of curvature of S as parametric curves,
ru = (1  Aki)ru + Xun,
rv = (1  Xk2)rv + Xn
(Prob. 30); and, on cross multiplication,
ffn = (1  Xki)(1  Ak2)Hn + Xu(1  Xk2)n x rv + X ,(l  Xkl)ru x n.
Now ii = En where e = ±1; and as n, ru, r are mutually orthogonal,
H = E(1  Xki)(1  Xk2)H,
Xu = Xv = 0.1
32. Prove that the lines of curvature on parallel surfaces correspond and that the corresponding principal curvatures satisfy the equations, kl
_
Eke
Ekl
1  Xkl '
T2 =
1  xk2
Hence the mean and total curvature of S are given by E(J  2XK) J _ 1rJ+X2K' 
K
_
K
1xJ+x2K
33. Show that surfaces parallel to a surface of revolution are also surfaces of revolution.
34. Prove the "Theorema egregium" of Gauss by establishing Baltzer's formula for the total curvature K:
Fuv  UGuu Evv H4K=I Fv2Gu
2Eu
Fu  2Ev
E
F
F
G
0 ZEv
'Gu
U Ei,
Gu
E F
F G
GEOMETRY ON A SURFACE
326
From (132.10) and (132.22) show that
Method.
H4K = [ruurx][rv,;r r,;]  [ruvrurv]2. Using (24.14), the right member becomes ruu
ruu ru
rvv
ruu
rv
ruv
ruv
ru
ruv
ruv
ru rvv
E
F
ruv
ru
E
F
rv
F
G
ruv
rv
F
G
rvv
rv
In both determinants the upper left element has the cofactor EG  F2; hence we may replace these elements by ruu rvv  ruv ruv and 0, respectively. Now, by differentiating we may show that rut,,
ru
rvv
rv =
i
2F!u, zGv,
ruu rvv  ruv
ruv
ruv
ru = 2iEv,
rv =
ruv  Fuv
2Gu>
ru = F,,
rvv
ruu
r,,  Fu 
i2Gu,
2Ev;
 2!Evv. vv
1Guu
35. On the surface r = r(u, v) let us write u1 = u, u2 = v; el = ru,
e2 = rv;
D1 =
el = a,
aau
,
D2 = ava ;
e2 = b;
where the Greek indices have the range 1, 2.
then e"
Moreover let
gap = ea ep, 9ap = ea e$, g = det gap =
hap = n DaDpr = hpa;
ep = SQ,
911
912
921
922
h = det ha p.
Prove that
(a) 9u= E,g12=F,922=G;g=EGF2. (b) 911 =
922
g12

9
912 922 = 9
922
9
that is, g"p is the reduced cofactor of gap in det gap.
(c) h11=L,h12=M,h22=N;h=LNM2;
(d) Daep  Dpea;
(e) Dagpy + Dogy.  D.rgap = 2ey Daep.
(f) Daep = ey(Dagp , + Dpgya  Dygap) + nhap (summed over y = 1, 2). (g) e" = g"pep (summed over (3 = 1, 2). [Cf. § 145.] 36. The Christoffel symbols Papa, g, y = 1, 2) for a surface r = r(ul, u2) having gap du" dup as first fundamental form (§ 132) are computed from the equations: Pap = ey . Daep.
PROBLEMS
327
With reference to Prob. 35 prove that
(a) ra0 = r'a (b) rP# =
Diggya  Dyga#) (summed over p = 1, 2).
(c) 29r1i = 922 D1911  2g12 D1912 + 912 D29uu, 29ri2 = 922 D2911  912 D1922,
29r22 = 912 D2922 + 2g22 D2912  922 D1922
[Use the formula of part b; for example, 21'1ii
= 911(D1911 + D1g11  D1911) + 91Y(D1912 + D1921  D2911),
1
29t1 = 922 Dig,,  912(2D1912 
(d) From the formulas for r,# obtain the formulas for rap. (e) From Prob. 35 (f) prove the derivative formulas of Gauss: X Daep = raryea + heron (summed over a = 1, 2).
37. From n efl = 0 show that eo Dan = han. Hence prove the derivative formulas of Weingarten;
Dan = hafieO (summed over S = 1, 2). Check these results with Prob. 17. 38. With the notations of Prob. 35 show that equations (132.21) and (132.22) may be written J = g0h,,#, K = h/g (a, 0 summed over 1, 2). 39. Show that the tangential projection of the vector curvature dT/ds = KNi of a surface curve is d2r
dua duDr
= ey
ds2 + raA ds ds (
summed over a, 0, y = 1, 2. Hence prove that the equations of a geodesic on a surface are y
d3
B
dua
+ rad
ds
ds
=0 a
1 T = ep ds ,
= ep ds2 + Daeg d
(Y = 1, 2). 8
,
Vr = eyes. Cf. Prob. 36. ]
CHAPTER IX TENSOR ANALYSIS
145. The Summation Convention. Expressions which consist of a sum of similar terms may be condensed greatly, without any essential loss of clarity, by indicating summation by means of repeated indices. This usage, originally due to Einstein, is stated precisely in the following SUMMATION CONVENTION. Any term, in which the same index
(subscript or superscript) appears twice, shall stand for the sum of all such terms obtained by giving this index its complete range of values. This range of values, if not understood, must be specified in advance. By way of illustration, we repeat some of the formulas of § 24, using the summation convention. The index range is 1, 2, 3. The two forms of a vector (24.1) and (24.2) become (1)
u = u;e'.
U = uiei,
In these equations i and j are summation (or dummy) indices. But any other letter will do as well; thus u = carer = ule1+ u2e2 + u3e3.
In order to compute u  v, we first recall the defining equation for reciprocal sets, (2)
ei e' = Si
(23.3).
Now (3)
u . v = (uiei) . (v,e') = uiv; 'Vi' = uiei;
we here use different indices in expressing u and v in order to get the 32 = 9 terms in the expanded product (six are zero). Similarly, (4)
u v = (uiei) . (vie;) = uiv' S = u1vi. 328
DETERMINANTS
§ 146
329
Note that the summation indices in the preceding examples appear once as subscript and once as superscript. The significance of this arrangement (not required by the summation convention) soon will be apparent. 146. Determinants. A permutation of the first n natural numbers is said to be even or odd, according as it can be formed from 123 n by an even or odd number of interchanges of adjacent numbers. The total number of permutations of n different numbers is n!; one half of these are even and one half odd. For example, when n = 3, 123, 231, 312 are even permutations, 213, 132, 321 are odd. Consider now a permutation ijk r of the numbers 123 n.
We then define the permutation symbols ijk...r and as equal to 1 or 1, according as ijlc . r is an even or an odd pereijk""r
mutation of 123 . . . n; but, if any index is repeated, the epsilon is zero. For example, when n = 3, E123 = 231 = 312 = 1
E213 = 132 = 321 =  1
112 = 122 = 222 =
0.
The epsilons just defined are useful in dealing with determinants. To be specific, we shall take n = 3; but all the formulas apply without change of form for any value of n. From the definition of a determinant : 1
(1)
a = det a _
2
3
a1
a1
a1
2
as
a2
3
1
2
a3
a3
a2 3
a3
ei jka 12a3
e'k ai1 aj2ak3
The implied summations on i, j, k produce 33 = 27 terms; of these,
21 involve repeated indices and therefore give a zero epsilon, whereas 6 involve permutations of 123 and give precisely the 3! = 6 terms of the determinant. (Write them out!) The theorems relative to an interchange of rows and columns
i
are given by (2)
Proof.
eijkarasat
aer8t =
'
ijkarL atk a8
When r, s, t are 1, 2, 3, equations (2) reduce to (1).
Since rst changes sign when two adjacent indices are interchanged,
TENSOR ANALYSIS
330
§ 146
(2) will beestablished when the right members are shown to have
this same property. Now Eijkarasat = eijkasarat =  ejikasa r aki
and, if we interchange the summation indices i, j, the last expression becomes  eijkasajat . If we multiply the first equation of (2) by e" and sum on rst, we obtain 3! a =
(3)
eijkerstarasae.
On the left er8tergt, summed over the 3! permutations of 123, equals 3!; on the right, the summation extends over all six indices and produces 36 = 729 terms. Many with zero epsilons vanish; and each nonvanishing term such as a1la22a3 3 appears six times, corresponding to the 3! permutations of its factors. Equation (3) thus is not useful for computation; its importance rests on the information it gives about the determinant when its elements are transformed.
The cofactor of an element az in the determinant (1) is defined as its coefficient in the expansion of the determinant. The cofactor of a; is denoted by A'. To find A',* strike out the row and
column in which ai stands; then A equals the resulting minor taken with the sign (1) i+j. If the elements of any row or column are multiplied by their respective cofactors and added, we obtain the determinantits Laplace expansion; but, if the elements of any row (column) are multiplied by the corresponding cofactors of another row (column) and added, the sum is zero. These important properties both are included in aiA; = a'Ai = a S .
(4)
Here r is the summation index. If in (4) we put j = i, both r and i are summation indices, and we get a7AT = a; A, = a Si,
(5)
where S = Si + S2 + S3 = 3 (n = 3).
But, if we put j = i in
(4) and suspend summation on i, we obtain the Laplace expansions, (6)
a7,Ar = arAi = a (i fixed),
for S = 1 for a fixed i.
DETERMINANTS
§ 146
331
A cofactor divided by the value of the determinant is called a reduced cofactor. The reduced cofactor of a2 is therefore
a; = A,/a
(7)
(a F6 0).
In terms of reduced cofactors, equations (4) become aTc = 8a.
(8)
When the elements of a determinant are written aid, its definition becomes a = E'.'kalia2Ja3k = E1Jkailaj2ak3
(9)
The reduced cofactor of ai3 then is written ai
and equations (8)
become aira'r = aria'' = SL.
(10)
If the elements ai; are functions of a variable x, we have, from (9), da
_ ilk (dali dal; da3k\ a2ja3k + a1i a3k + aiia2i = x x dx ) E
dx
=
dalAli+da2jA21+da3kA3k dx
dx
dx
'
where A 'j denotes the cofactor of aij. Summing on two indices, we may write (11)
da= daii Ai' dx
(i, j = 1, 2, 3).
d.r,
The derivative of a determinant is the sum of the products formed by differentiating each clement and multiplying by the cofactor of the element.
The properties (8) of reduced cofactors enable us to solve a system of linear equations when the determinant of the coefficients is not zero. Consider, for example, the equations: (12)
ax' = yt
(i, j = 1, 2, 3; a
0).
To solve these for x', multiply (12) by the reduced cofactor ak and sum on i : the left member becomes S; x' = xk, and we obtain xk = a;yi, or, on replacing k by j, (13)
x' = aiyi
(i, j = 1, 2, 3).
TENSOR ANALYSIS
332
5 147
To find the product of the determinants a = det a, b = det b;, we have, from (1) and (2), ab = EiJkaia2a3b
(14)
s t ia2 ja3 kEr8tbsrbjbk
= a1
= Erst(albi)(aabj)(a3kbk)
r
=
8
ErstC1C2Ct3,
where c% = aabj = aib1 F a2b2 f a2b33.
(15)
From (15), we see that the element in the ith row and jth column of the product ab is given by the sum of the products of the corresponding terms in the ith row of a and the jth column of bthe socalled "rowcolumn" rule. Since the value of a determinant is not altered by an interchange
of rows and columns, we also can compute ab by "rowrow," "columnrow," and "columncolumn" rules. For example, the product 1
a1
(16)
2
al
3
al
1
a1
2
1
1
a2
a3
1
0
0
0
1
0
0
0
1
a2
a2
a2
ai
a2
a3
1
z
3
3
3
3
a3
a3
a3
a1
a2
a3
= 1.
by use of the rowrow rule. THEOREM. If a determinant a 0, the determinant formed by replacing each element by its reduced cofactor is 1/a.
If we solve equations (13) for yi by using the reduced cofactors in det a, we will obtain (12); in other words, the reduced cofactor
of a in its determinant is a. The two determinants in (16) thus are reciprocally related: each is formed from the reduced cofactors of the other. .'. 147. Contragredient Transformations. Let us now introduce a new basis e1i e2, e3 by means of the linear transformation, 61 = cie1 + ciez + Cie3, (1)
e2 = c2e1 + c2e2 + 4e3,
e3 = c3e1 + c3e2 + c3e3,
CONTRAGREDIENT TRANSFORMATIONS
§ 147
333
where the coefficients ci are real constants whose determinant 0.
c
In brief, ei = c'ej,
(1)
c = det ca / 0.
The condition c 0 ensures that e1, e2, e3 are linearly independent. For, if there were three constants At such that Aiei = 0,
Aicej = 0;
then
the linear independence of the vectors ej requires that c'Ai = 0, and, since c P 0, these equations only admit the solution Ai = 0. This argument applies without change to space of n dimensions, in which a basis consists of n linearly independent vectors.
In the present case (n = 3), we also may argue as follows. From (1), we have 161
. e2 x e3
=
cic2c3 ei ej x ek = C'iC2iC3Eijk e1 ' e2 x e3;
hence, on writing E = el e2 x e3, we have
E _ (det c')E = cE.
(2)
Since the vectors ei form a basis E 96 0, hence c E
0 implies
0 and the linear independence of the vectors e1. For any vector u, u = 'uiei = ujej;
(3)
hence, on substitution from (1), ujej,
or, since the vectors ej are linearly independent, u.l = Ci,u
(4)
If y is the reduced cofactor of c in the determinant c, we have on solving equations (4),
ut =
(5)
or, written out in full, ul = riu1 + 'Y2u2 + Y3u3, (5)
u2
= 'Ylui + Y2u2 + Y3 u3,
u3 = 'iu1 + Y32 u2 +
'Y3u3.
3
TENSOR ANALYSIS
334
§ 148
The linear transformations (1) and (5) are said to be contragredient.
Their matrices,
C=
Cl
C1
c1
1
C2
2 C2
3 C2
C31
2 C3
3 C3
,
71
72
r = 71
2
72
3
3
72
71
73 2
73 3
73
are so related that any element of r is the reduced cofactor of the corresponding element of C in its determinant c; and any element of C is the reduced cofactor of the corresponding element of r in its determinant y.
In order to state this definition analytically, we remind the reader of the following definitions from matrix algebra. The product AB of two square matrices A and B is the matrix whose element
in the ith row and jth column is the sum of the corresponding products of the elements in the ith row of A and jth column of (rowcolumn rule). The transpose of a matrix A is a matrix A' obtained from A by interchanging rows and columns. The unit matrix I has the elements Si (ones in the principal diagonal, zeros elsewhere).
If we compute Cr' or rC' and make use of the equations, (6)
cz7r = S,
7rc1 = bj,
we find that
Cr'=rC'=I.
(7)
These equations characterize contragredient transformations.
Two matrices whose product is I are said to be reciprocal. Thus C and r' or C' and r are reciprocal matrices. From (7), we see that contragredient matrices are so related that the transpose of either is the reciprocal of the other.
148. Covariance and Contravariance. When new base vectors are introduced by means of the transformation, (1)
ei = c e5,
det c
0.
the components of any vector u are subjected to the contragredient transformation, (2)
ui = 7;u'.
In view of (24.3), this may be written
uei=7jue';
§ 148
COVARIANCE AND CONTRAVARIANCE
335
and, as this holds for every vector u, (3)
The basis ei, reciprocal to ei, thus is transformed in the same way as ui. To find the transformation for the components ui, multiply (1) by u ; then, from (24.4),
ui = cu
(4)
a transformation cogredient with (1). Thus quantities written with subscripts transform in the same way (cogrediently); and quantities written with superscripts also transform in the same waybut the latter transformations are contragredient to the former.
Thus the position of the index indicates the character of the transformation.
A vector u is said to have the contravariant components ui, the covariant components ui. These terms suggest variation unlike and like that of the base vectors ei; in other words, the transformation
(1) is regarded as a standard. Thus [u', u2, u3] and [U1, u2i u3] are two ways of representing the same vector u; in the first it is referred to the basis ei, in the second to the reciprocal basis e' [u1, u2, u3J often is called a contravariant vector, [u1, u2, u3] a co'
Actually, the vector u is neither contravariant or covariant, but invariant. variant vector.
By using the properties of the reduced cofactors, eir'Yr i
r
i
= Yi = ai, we may solve equations (1) to (4) for the original base vectors and components (§ 146). Thus the equations, (5)
(6)
ei = c,ei,
ui = ciuj,
et = 'Yie',
ui = Yiui;
e' = c ,%J,
uz
 yiu',
have the solutions, (7)
ei = ?j6j,
ut = c ui.
Note that the matrices c, y in (7) are the matrices cz, y' of (6) transposed. To be quite clear on this point the reader should write out equations ei = (.'e; and et = c;ei in full. From (1) and (3), we find that ei ' e' = cier yie8 = ciYaSr = c 'Yr = 31j,
which shows that the new bases have the fundamental property of reciprocal sets.
TENSOR ANALYSIS
336
§ 149
An expression, such as uiv i, uie i, ei e i, summed over the same upper and lower indices, maintains its form under the transformation. Take, for example, the scalar product uivi: making use of (2) and (4), we have uiv2 = Ciur78vs = bryurv8 = u'1T.
The index notation thus automatically indicates quantities that are invariant to affine transformations of the base vectors. 149. Orthogonal Transformations. When the basis ei and the new basis ei both consist of mutually orthogonal triples of unit vectors, the transformation, ei = ciei,
(1)
is called orthogonal. Since both bases are selfreciprocal (§ 23), the corresponding transformation (148.3) between the reciprocal bases, ei = 7e', now becomes ei = 7e;.
Since this transformation must be the same as (1), C11
(2)
C=
c2
ci c2
c3
C3
3 C3
1
C3
ci
7i 72
=
2
71
72
3
3
71
72
73
7sq
=r
3
73
The matrix of an orthogonal transformation is called an orthogonal matrix. From (2), we have the THEOREM. Every element of an orthogonal matrix is its own reduced cofactor in the determinant of the matrix.
In view of the properties of reduced cofactors, the coefficients of an orthogonal transformation satisfy the equations: (3)
crc; = Cic; = 8L.
The matric equation (147.7) characterizing contragredient transformations becomes (4)
CC' = I
for orthogonal transformations. Thus a real matrix is orthogonal when its transpose equals its reciprocal.
If c denotes the determinant of C, we have, from (4), (5)
c2 = 1,
c = f 1.
QUADRATIC FORMS
§ 150
337
If the bases ei and ei are both dextral or both sinistral, we can bring the trihedral e1e2e3 into coincidence with e1e2e3 by continu
ous motionin fact, by a rotation about an axis through the origin. At any stage of the motion, the base vectors are related
by equations of the form (1); and, as the motion progresses, the coefficients c change continuously from their initial to their final values S , and c = det ci becomes det S = 1. But, since this determinant equals =E1 at all stages of the motion and must change continuously if at all, c = 1 when the bases have the same orientation. If one basis is dextral, the other sinistral, e1e2e3 and ele2(e3) have the same orientation. Hence the determinant of the transformation, el = c1er,
e3 = caer, namely, c, must equal 1; thus c = 1 when the bases have dife2 = C3er,
ferent orientations. 150. Quadratic Forms. A real quadratic polynomial, (1)
A(x, x) = aijxixj
(i, j = 1, 2,
,
n),
for which aij = aji is called a real quadratic form in the variables The symmetry requirement aij = aji entails no x1, x2, , x"`. loss in generality; for, if aij 0 aji, the form is not altered if we replace aij and aji by (aij + aji). The determinant of2 the coefficients, a = det aij, is called the discriminant of the form. The form is said to be singular if det aij = 0, nonsingular if det aij 0 0. Associated with the quadratic form A (x, x) is the bilinear form, (2)
A(x, y) = aijx'y' = A(y, x),
known as its polar form.
The expansion of aij(xi + Xyi) (x' + Xyj) shows that a quadratic form and its polar are related by the identity, (3)
A (x + X y, x + Xy) = A (x, x) + 2XA (x, y) + X2A (y, y).
If we make the linear transformation, (4)
xi=c yr,
c=detc;F 0,
the form (1) becomes (5)
B(y, y) = br4yry8 where
bra = aijcTc'.
TENSOR ANALYSIS
338
§150
From the rule for multiplying determinants, det bra = det (ai;cT) det c8 = det ai; det cr det c8;
the discriminant of B(y, y) is therefore (6)
b = det bra = c2a.
Linear transformations of nonzero determinant do not alter the singe lar or nonsingular character of a quadratic form.
If the form A (x, x) is singular, it can be expressed in terms of fewer than n variables which are linear functions of x. For since det ai, = 0, the system of n linear equations ai;xi = 0 has a solution (xa, xo, , xo) which does not consist entirely of zeros; thus we may assume that xo 54 0. Now A(xo, y) = ai;xoy' = 0, and, from (3), A (x + Axo, x + Axo) = A (x, x),
irrespective of the value of A.
If we now choose A = xl/xo
and write y i = x, i + Axp = x i  x4x1 /x0,
we have A (y, y) = A (x, x). Since y' = 0, A (y, y) is expressed in
, y". terms of the n  1 variables y2, y3, Henceforth we shall consider only nonsingular forms. A nonsingular form is said to be definite when it vanishes only for
xl = x2 =
= x" = 0. For all other sets of values, the sign of
a definite quadratic form is always the same. To prove this, sup, x" and A(y, y) < 0 pose that A(x, x) > 0, for the set x1, x2, Then, from (3), for the set y', y2, , y".
(7)
A (x + Ay, x + Ay) = 0
is a quadratic equation in A having two distinct real roots A1i A2; for
{A(x, y) 12  A(x, x)A(y, y) > 0.
Consequently, the form vanishes for two different sets of values, xi + Alyi and xi + A2yi and therefore cannot be definite. For a definite quadratic form, the quadratic equation (7) in A must have a pair of either complex roots or equal roots; hence, for a definite form, (8)
JA (x, y) 12  A(x, x)A(y, y) < 0,
the equal sign corresponding to the case xi + Ayi = 0, (i = 1. , n), in which the sets xi and yi are proportional.
§ 152
RELATIONS BETWEEN RECIPROCAL BASES
339
A definite quadratic form is called positive definite or negative definite according as its sign (for nonzero sets) is positive or negative. A nonsingular form which is not definite is called indefinite: such a form may vanish for values x` other than zero. When A (x, x) is positive definite, it is well known that we can find a real linear transformation (4) which will reduce A (x, x) to a sum of squares: (9)
B(x, x) = Siyiyj = ylyl + y2y2 + ... + ynyn
The discriminant of a positive definite form is therefore positive;
for, from (6), b = 1 = c2a, and a = 1/c2 > 0. 151. The Metric.
Using the notation,
9i; = ei ej = gji,
(1)
for the nine scalar products of the base vectors, we have (2)
u . v = (uiei) . (viej) = giptivj,
u u = gi;uiu'.
Thus u u is a real quadratic form, giiU'u1 + g22u2u2 + 933u3u3 + 2g12u'u2 +
2923u2u2
+ 2931u3u1r
in the variables u1, u2, u3; and, since u u > 0 when u 0, this form is positive definite. Moreover, u v is given by the associated bilinear (polar) form in ui, vj. From (2), we have (3) (4)
I
gijl(itljr
UI=
Cos (u, V) =
gijuw' UV _ VgijuiujVVgijvivi lUl VI
Thus, when the six constants gij are known, we can find the lengths of vectors and the angles between them when a fixed unit of length is adopted. The quantities gjj, since they make measurements possible, are said to determine the metric of our 3space; and gijuiuj is called the metric (or fundamental) quadratic form. 152. Relations between Reciprocal Bases. The discriminant c' the metric form is (1)
9=
911
912
913
921
922
923
931
932
933
:" Bother, Introduction to Higher Algebra, New York, 1907, p. 150.
TENSOR ANALYSIS
340
§ 153
The reduced cofactor of gij in this determinant is denoted by gij; and we have, from (146.10), the relations:
= 9rigr' = Since any vector may be written girg.r
(2)
u=
Si.
ei,
ei = gije' We may solve these equations for ej by multiplying by gik and (3)
summing on i; thus
k j = e k, 9ikei = 9ikgijej = 3je or, if we interchange i and k, ei = 9kiek
(4)
From (4), we have also ei ej = gkiek ej = gkiak = gji (5)
gij =
ez
e' = 9'i
Making use of (3) and (4), we now have, for any vector u, (6)
(7)
ui = ei
u = gijuj
The equations enable us to convert contravariant components of a vector to covariant and vice versa. Finally, from (24.14), we have (8)
(9)
g = det gij = det (e1 ej) = [eie2e312 = E2,
det gij = det (ei ej) = [ele2e312 = E2 = 1 9
153. The Affine Group. When an origin 0 is given, the constant basis el, e2, e3 defines a Cartesian coordinate system x1, x2, x3; for any point P is determined by the components of its position vector :
OP = x'el + x2e2 + x3e3. The components xi of OP are called the Cartesian coordinates of P relative to the basis ei.
THE AFFINE GROUP
§ 153
341
When the base vectors ei are subjected to the transformation,
et=ce;,
(1)
the invariance of OP, namely, i''es = xZei,
(2)
induces a contragredient transformation on the coordinates. For,
if we substitute from (1) in (2), we find that x' = ci"xi; that is, xi = c;x'.
(3)
The matrix c? in (1) is transposed in (3); and, while (1) expresses
the new base vectors in terms of the old, (3) expresses the old coordinates in terms of the new. We now can solve (3) for the new coordinates by multiplying by the reduced cofactors ti; and summing; thus we find xiy, = xk, or (4)
x' = y2xi,
y = det yj = 1/c.
The transformation (3) is called affine, or, more specifically, centered affine, since the origin 0 has not been altered. The centeredaffine transformations with nonzero determinant form a group, that is,
(a) the set includes the identity transformation: ri = S ; a ; ' = x`,
det 6 = 1;
(b) each transformation of the set has an inverse; (c) the succession of two transformations of the set is equivalent to a single transformation of the set. Thus the transformations, a x',
x' = bj;xk, give xa = Ckx1,
where ck = a bk and
det ck = (det a) (det bk) F4 0.
Our present point of view is that a transformation of the base vectors induces a definite transformation of coordinates. But the reverse point of view is adopted when transformation groups more general than the affine are under consideration : the coordinates are transformed, and we then inquire as to corresponding transformation of the base vectors.
TENSOR ANALYSIS
342
a 151
In the case of the affine group the transformation (3) of coordi
nates obviously entails the transformation (1) of base vector.,. For, if we put x' = x'cy in (2), we have
xi(czej  ei) = 0 for all xi, that is, (1) must hold. 154. Dyadics. Under the affine transformation, (1)
ei = c ei,
xi = y;xi
ei = y;ei,
ti = e xi
(c
0),
we have also (2)
(148.6).
Contravariant and covariant components of a vector, ut and ui, transform like x i and xi, respectively. The sets of components u i and ui often are called contravariant and covariant vectors. Actually they are different representations of the same invariant vector,
U = uiei = uiei.
(3)
Nine (3 X 3) numbers Tii which transform like the nine products of vector components uivi, namely,
Tii =
(4)
yryBT'8
are called contravariant components of a dyadic. Similarly, nine numbers Ti; which transform like uiei, namely, T,i = ('icjTra,
(5)
are called covariant components of a dyadic. Finally, nine num
bers Ti; or T;i which transform like u ivi or uivi, respectively, namely, (6)
7,'i =
yrc;TB,
T:' =
are called mixed components of a dyadic. Note that in Tti the upper index comes first, in Tii the lower index. Just as ui and ui are two representations of a single invariant vector u, the components Tii, Tii, Ti, T;i are four representations of one and the same dyadic, (7)
T = Tiieiei = Tiieiei = Ttieiei = Ti eiei.
Note that the indices on components and base vectors always are placed in the same order from left to right. All forms of T given
ABSOLUTE TENSORS
§ 155
343
in (7) are invariant; for example, on using (1) and (4), we have T'16x.e
i ckek = 3,h6kTr8e1,ek = Thkehe b, = YrYsjTrsc'eh i r
from the properties of reduced cofactors (146.8). The coefficients gij of the metric quadratic form are tensor components; for, from (151.1), (8)
9ij = ei ' ej = (tier) ' (cjes) = c' jug,.
in agreement with (5). The same is true of gij; for, from (152.5), (9)
9i7
= e ' e' = (Yrer)  (Yses) = Y:Y39r8, b, S transforms
in agreement with (4). Again, if we define as a mixed tensor,
ei ej = (Yrer)  (c;es) = Yrc;a8.
(10)
Indeed, gij, gij and b are all components of the same tensor, (11)
gijeiej = gijeiej = beiej = 8jeiej = eiei = eiei,
namely, the idemfactor (§ 66). In fact, if we use the formulas,
ei =
(12)
girer,
ei = girer
(§ 152),
and remember the properties of gij, g" as reduced cofactors, any two members of (11) can be shown to be equal; for example, gijeie' = 9ij9irere' = b;erei = eje',
gije ej = gc,e gjrer = Sie er = e ei. i
i
r
ti
i
Since eiei and eiei represent the same tensor, the order of the indices in its mixed components, the Kronecker deltas, is immaterial. Any component of T may be expressed in terms of components of another type by means of the metric form. For example, we have, from (7) and (12),
Tit = ei,T.ej = 9trer'T'ej = 9irT,J, = 9'r ei'T er = girgjs er  T  e8 = girgj8Trs
155. Absolute Tensors. We next define absolute tensors with respect to the centeredaffine group of transformations: xi = cxj
TENSOR ANALYSIS
344
Scalars p(x', x2, x3) which have one component in each coor(linate system given by
gxl, x2, x3) = P(x', x2, x3)
(1)
are said to be absolute tensors of valence zero.
Vectors have three components, and these may be of two types u,, ui. The laws of transformation, 2li = ciuj,
(2)
ui = Yu
characterize absolute tensors of valence one. The vector itself,
u = uiei = uiei, is invariant to the transformation: u = u. Dyadics have 32 = 9 components, and these may be of 22 = 4 types: Ti;, Tij, T'j, Tij. If these components transform, respectively, like the products ujv;, uivj, uiv;, uivj of components of two absolute vectors, they are said to form absolute tensors of valence two. The dyadic itself,
T = Tijeie' = Ti'eiej = TZjeie' = T'jetej, is invariant to the transformation: T = T. In general, an absolute tensor of valence m is a set of 3' components that transform like the product of m absolutevector components; and, since each of these may be covariant or contravariant, the components are of 2' types. One of these types is purely contravariant (T1 .k), another purely covariant (Tij...k); the remaining types are mixed and have both upper and lower indices.
Consider, for example, a tensor T of valence three (a triadic); its 33 = 27 components may be of 23 = 8 types. Its covariant components Ti;k will transform like uiv;wk, namely, Tijk = c2cjckT,.t,
(3)
and the mixed components Tijk like uivjwk, namely,
Ti
(4)
jk
rjk
= ea'Ya t Tr
at
The tensor itself,
T = T.;keiejek = Tijkeiejek =
,
RELATIVE TENSORS
§ 156
345
is invariant to the transformation; for example,
T = Tijkeiejek (CiciCkTrst) (7ea) (' beb) (} cec) SaSbS,'Trsteaebec
= Trstereset = T. Note that the indices on the base vectors in T have the same order as the indices on the component but occupy opposed positions. We may solve the 27 equations (3) for the original components Trst by multiplying by yayby, and summing; we thus find
j k (5)
r s t
yaybycTijk = SaSbScTrst = Tabc
In similar fashion we find, from (4), (6)
yaCbCkijk
= brbbb 'r = Tab`.
156. Relative Tensors. When the coordinates are subjected to an affine transformation, (1)
xi = Cr,
;xti
(c = det c s 0),
= yrxr
the transformation equations, for all tensor components thus far considered contain only the coefficients c, y. More generally, we may have equations of transformation, such as (2)
T.jk = CNCr j i
t
r ,
which involve the Nth power of the determinant c. In this case the quantities in question are said to be components of a relative tensor of weight N. When N = 0, equations (2) reduce to (155.4) for an absolute tensor. As a consequence of (2), the relative tensor T = Trseere8et becomes (3)
T = cNT
when the transformation (1) is effected. The law of transformation of a tensor component is determined by (i) its valence: the number of its indices; (ii) its type: the position of its indices from left to right; (iii) its weight: the power of c that enters into the transformation equations.
TENSOR ANALYSIS
346
§ 157
If the weight of a tensor is not specified, it is assumed to be zero; the tensor is then absolute. A relative scalar so of weight N (valence zero) is transformed according to gxl, x2, x3) = CNc,(x1) x2, x3).
(4)
The box product of the base vectors, E = el e2 X e3, is a relative scalar of weight 1; for, from (147.2),
E = cE.
(5)
The box product of the reciprocal base vectors, el . e2. e3 = E1, is a relative scalar of weight  1; for E1 = c1E1, from (5). The determinant,
g = det gij = E2
(6)
(152.8),
is a relative scalar of weight 2; for E2 = c2E2. THEOREM. The permutation symbols Eijk and eijk, regarded as the same set of numbers in all coordinate systems, are components of relative tensors of weight  1 and 1, respectively.
Proof. If the formulas (146.2) are applied to the determinants,
y = det ?j = 1/c ,
c = det c2j
we have CEijk =
iCjCkerat'
YEzlk = yr"Ys7
Brat.
Since Eijk = Eijk, Eijk = Eijk by definition, these equations assume the form, e
= C8Ctk jarat, z
Eijk
j kerat _ Y iryayt
required by relative tensors of weight 1 and 1. 157. General Transformations. Three equations, (1)
x' = f z(xls x2
x3) f
in which the functions f i are singlevalued for all points of a region
R and which can be solved reciprocally to give the three equations, (2)
xt = gi(xl 22,
3)
in which the functions gi also are single valued, determine a onetoone correspondence between the sets of numbers xi and xi. If we regard both sets of numbers as coordinates of the same point,
GENERAL TRANSFORMATIONS
§ 157
347
(1) defines a transformation of coordinates. The affine transforma
tion is the particular case of (1) in which the functions f i are linear and homogeneous in x', x2, x3. Consider now the totality of such transformations in which the functions f i(x', x2, x3) are analytic functions having a nonvanishing Jacobian in R: ax
azi
=det 3,6 0.
(3)
ax' The implicitfunction theorem ensures the existence of the solutions (2) of equations (1) in a sufficiently restricted neighborhood of any point. In order to have a transformation of coordinates OX
in R, we must assume that the solution (2) exists and is single valued throughout R. The coordinate transformations (1) thus defined form a group; for
(a) they include the identity transformation xi = xi whose Jacobian is 1; (b) each transformation (1) has an inverse (2) whose Jacobian ax/ax I is the reciprocal of (3); (c) the succession of two transformations of the set is equivalent to another transformation of the set. As to (c), the two transformations, xL = fi(xl x2 x3)
xz = h'(x' x2 x3)
are equivalent to
xz = h'[f`(x),f2(x),f3(x)] = ji(xl, x2, x3), for which the Jacobian, /0xi axr axi
detax' = det\axr(ax'/

/
axi
\
axr
= (det
/ \
( det )
L
0.
ax'/ ax Moreover, since f i and hi are analytic functions, the functions ji are likewise analytic. In the affine transformation, (4)
xi
= yrxt, i
xi
ir , = crx
we have axi (5)
axr
yr
=r axr axi
z;
TENSOR ANALYSIS
348
§ 157
the Jacobians, axi
(6)
det  = y, axr
axi
det  = c,
and yc = 1.
axr
A tensor of weight N, with respect to the affine group, transforms according to the pattern, (7)
TU'.k =
CNyTy3CtkTrst.
In view of (5) and (6), this equation may be written (8)
axi ax' axt
ax
Ti'k at
IN
axr axs axk
Now, by definition, a set of 33 functions Tr. $t is said to form a tensor
of valence 3 and weight N with respect to the general transforma
tions (1) provided the components transform according to the pattern (8). This equation becomes (7) when the transformation is affine and constitutes a natural generalization of (7). The corresponding equation for a tensor of any valence or type is now obvious. In particular, for contravariant and covariant components of an absolute vector, axi
(9), (10)
uy =axr ur,
_
axr
ii = ax ur. i
Since the new tensor components are linear and homogeneous in terms of the old components, we have the important result: If the components of a tensor vanish in one coordinate system, they vanish in all coordinate systems.
More generally, tensor equations maintain their form in all coIf any geometrical or physical property is expressed by means of an equation between tensors, this equation in an arbitrary coordinate system expresses the same property. Although the coordinates themselves are not vector components in the general group (1), their differentials dxi are the components of a contravariant vector; for, from (1), axi axi axi axi dxI + (11) dxi = dx2 +  dx3 =  dxr. ax' axi axr ax2 We regard the differentials dxr as the prototype of contravariant vectors; and the rule of total differentiation gives the correct pattern for their transformation. ordinate systems.
 
GENERAL TRANSFORMATIONS
§ 157
349
The partial derivatives app/axi of an absolute scalar p(x', x2, x3) are the components of a covariant vector; for, since
0(.'
x2, x3) =
P(xl, x2,
x3),
when we replace xi on the right by the values (2), a P ax2
(12)
ax
ax
1
ax i
_ axr aV ax ax r
a,p ax3 ax3 at i
+ ax2 ax i
We regard the derivatives ap/axr as the prototype of covariant vectors; and the chain rule for partial differentiation gives the correct pattern for their transformation. In the affine transformation, the position vector r = xiei, and
ei = ar/axi.
(13)
We now adopt (13) as the definition of ei in any coordinate system
x'; then
ar ar axr ei===er. axr
(14)
axraxt
axi
axz
The base vectors ei thus transform after the pattern in (10) and therefore merit a subscript.
The base vectors ei must transform after the pattern in (9), namely, a. t
ey =  er;
(15)
axr
for, since er e8 = S8, ei
axr axj
axr axj
axj
49V ax8
a.2'i ax'
axi
and the sets ei, e' are also reciprocal. We now can show that (16)
gij =
gij =
S
transform as tensor components; we need only make the replacements indicated by (5) in the proofs of § 154. In fact gij, gi', S; are all components of the idemfactor ere' = erer; thus S; sometimes is written gg. (17)
Moreover,
ei = ei ere' = gire',
ei = ei. erer = girer.
TENSOR ANALYSIS
350
158. Permutation Tensor. The box product E = el
e2 x e3 is
a relative scalar of weight 1: ax
E
ax
for the proof of (147.2) applies when we replace c; by axe/a.V. Moreover, E' = e' e2 x e3 is a relative scalar of weight 1. The permutation tensor is defined by (2)
e1  ej x ek eiejek = ei ej x ek eiejek.
These triadics are equal absolute tensors.
ej =
ei = gireT)
For, if we put ek = gkter,
9ise3,
the left member becomes er es x e` ereset. Moreover, (157.14) and (157.15) show that (3)
= ei ej x ek eiejek.
C
From (2), we see that the covariant and contravariant components of the permutation tensor are (4)
e1  ej x ek = EijkE,
ei ej x ek =
EijkE1
Since E and E' are relative scalars of weight 1 and 1, Eijk and Eijk are components of relative tensors of weight  1 and 1. This is the theorem of § 156; the former proof still holds when obvious changes are made.
159. Operations with Tensors. The three basic operations on tensors are addition, multiplication, and contraction. 1. Addition of tensor components of the same valence, weight, and
type generates a tensor component of precisely the same characters. Example.
If Pij, Qij are both of weight N,
Fij + Qij _
ax N axr ax" ax
(Pre + Qr.)a.ri azj
Hence Tit = Pij + Qij transforms in exactly the same manner as Pij and Qij.
2. Multiplication of tensor components of valence m1i m2, of weight N1, N2, and of arbitrary type generates a tensor component of valence m1 + m2i of weight N1 + N2, and of a type which is defined by the position of the indices in the factors.
OPERATIONS WITH TF''\SORS
§ 159
Example.
351
Let Pi' and Uk have the weights 2 and 1; then
ax ' ati a' P'uk = I ai ly axr axe Pr8
at'.
ax
Ox
axt axkut
az
=
as
3
az' axt
ax'' ax8 D.
PT8 ut.
Hence T'jk = P''uk transforms as a tensor of valence 3, of weight 3, and of the type indicated by its indices. Even though P''uk = ukPi', the product tensors,
uP = ukPz'ekeie
Pu = Pi'ukeie?ek, are not the same.
3. Contraction. In any mixed tensor component an upper and lower index may be set equal and summed over the index range this generates a tensor component of the same weight and of valence two less. Its type is determined by the remaining indices, not involved in the summation. Example 1.
In the absolute component T'!.k set j = k; then
u=_VV jT1.11'+'V.22+T133 is a contravariant vector, for Tt =
a ax ax T3 = az ax T ..t ..  ax"" ax8 all Of i
i
i
t
t
ax8
all =38T'st = ax,*
axr
T
Trt al'
"` = ax"
ur.
A tensor component may also be contracted with respect to two indices on the same level; we need only raise or lower one index and then contract as previously. Example 2. To contract the absolute component V% on the indices i, j we first lower the index j 4 !r T.jk = T..k9i*;
setting i = j now gives the covariant vector
! Vk = T ::..kgir = T ..kgij. We may also obtain this vector by lowering the index i and setting i = j: Tick =
Tt"k = V" k9ri
Contraction on a pair of indices is equivalent to the scalar multiplication of the corresponding base vectors in the complete tensor.
In the preceding examples, T = T!.ke,e ek,
u = Ti'.keiej ek = r'.ks;ei = T .,ei, v = T:'.kei ejek = Tt'.kgi,ek
352
TENSOR ANALYSIS
§ 160
The proof that contraction produces tensors follows from the equation : T'.'.keie;ek = TY?.keie,ek.
When the new base vectors on the left are expressed in terms of the old, this becomes an identity; and, since scalar multiplication is distributive with respect to addition, it remains an identity when base vectors in the same position are multiplied on both sides. Thus, in ex. 2, v. V = T'.kgiiek = As long as a contracted tensor has two or more indices, the foregoing process may be repeated, each contraction reducing the valence by two but leaving the weight unaltered. Thus the tensor T!.kh may be contracted twice to yield two different scalars 7".'.i; and T'!.;i, each consisting of nine terms. These are obtained from
the invariant tensor TT'.kheiejekeh by the formation of ei  ek, e; eh and ei  eh, e; ek, respectively.
Contraction often is combined with multiplication. For example, if we multiply the vectors u, v.; and then contract, we obtain their scalar product u'vi. Again the product A B of two dyadics defined in § 65 is equivalent to tensor multiplication followed by contraction on the two inner indices. Thus, if A = Ati,eV , B = Bkhekeh, A  B = Ai;Bkhe'e' . ekeh = Aj;B'he'eh. The product of the idemfactor I = ere? and a tensor T of valence In general IT differs from TI; but the contracted products reproduce T:
m is a tensor IT of valence m + 2.
160. Symmetry and Antisymmetry. A tensor is said to be symmetric in two indices of the same type (both covariant or both contravariant) if the value of any component is not changed by permuting them. It is antisymmetric or alternating in two indices
of the same type if permuting them in any component merely changes its sign. Thus, if Tabc = Tbac,
Tabc = Tcba,
the tensor T is symmetric in a, b, alternating in a, c.
KRONECKER DELTAS
§ 161
353
Symmetry or antisymmetry are properties that subsist after a general transformation of coordinates. Thus, for the preceding example, we have Tijk =
axa axb axe a
ax ax axk
axb axa a.T.e
Tabc =
ax ax £ axk
Tbac = Tjik
axe axb axa
axk axj axi
(Tcba)
A tensor cannot be symmetric or alternating in two indices of different types; for a property such as Ti j = Vi i does not subsist after a transformation of coordinates. A tensor is said to be symmetric in any set of upper or lower indices if its components are not altered in value by any permutation of the set. The subsistence of this property in one coordinate system ensures it in all. A tensor is said to be alternating (or antisymmetric) in any set of upper or lower indices if its components are not altered in value by any even permutation of the indices and are merely changed in sign by an odd permutation of these indices. A tensor Tij or Tijk (T ij, Tijk) which is alternating in all indices is called a bivector or trivector, respectively. In such alternating tensors, all components having two equal indices are zero. In a trivector Tijk, the nonzero components can have but two values, ±T123; moreover, the contracted product, EijkTijk = 3!T123
In this connection we remind the reader that a given permutation of n indices from some standard order can be accomplished by a succession of transpositions of two adjacent indices and that the total number of transpositions required to bring about a definite permutation is always even or always odd. The permutation in question is said to be even or odd in the respective cases. 161. Kronecker Deltas. We now generalize the simple Kronecker delta S by introducing two others, defined as follows: aiajbkc
(1)
 Eabr E
(2)
aab = EabrE
ij
ijk ijr ijr = aabr
Since the epsilons have weights of 1 and 1, ba"bk, is an absolute tensor of valence six; hence b b, formed by contracting S, is an absolute tensor of valence four.
TENSOR ANALYSIS
354
§ 162
Those definitions show that S b and bay can only assume the values 0, 1, 1; they are evidently alternating in both upper and lower indices. Their precise values in any case are as follows: If both upper and lower indices of a generalized delta consist of the same distinct numbers chosen from 1, 2, 3, the delta is 1 or 1 according as the upper indices form an even or odd permutation of the lower; in all other cases the delta is zero. This rule is a direct consequence of the properties of the epsilons. We have, for example, 32
012
12 =
023   1,
1,
123
6213
231
x123 = 0123  1,
23
13
011 = 021
o. ;
v321
8123 = 0
= 0123
162. Vector Algebra in Index Notation. The three operations on tensors enable us to give a succinct account of vector algebra. Vectors are to be regarded as absolute unless stated to be relative; and we shall often speak of components as vectors.
If w = u + v, we have
vi =ui+vi, or wi=ui+vi; obviously vector addition is commutative and associative. The tensor product of two vectors u, v is the dyad uv. On contraction, uv yields the scalar product, (1)
u . v = uivi = uivi = gijuivj = giju1vj.
From (1), we have
U. (v+w) = The antisymmetric dyadic (bivector) P = uv  vu is called the outer product of u and v; its covariant components are Pjk = UjVk  UkVj = 8jkuavb.
(2)
The dual of Pjk (cf. § 170) is the contravariant vector of weight 1: p =
(3)
2
e ijk Pik = E''ikujvk;
Z
its components, U2V3  u02,
U3V1  u1V3,
u1V2  u2v1Y
are the same as the nonzero components Pik. Since E1 is a scalar of weight 1, the components E'p' are absolute. The corresponding vector,
§ 162
VECTOR ALGEBRA IN INDEX NOTATION
(4) Elpiei = E1EijkeiujVk = E1
355
= uXv (24.9). v1
V2
V3
Similarly, from the outer product,
P'k = bab26avb = u'vk  ukvJ, we obtain as dual the covariant vector of weight  1: (2)'
(3)'
i pi =_ZEijkPJk = eijklljvk.
The absolute components Epi again give u X v : (4)'
el
e2
e3
Epiei = Eeijkeiujvk = E ul
u2
u3
vl
v2
v3
= uxv
(24.10).
From (4) or (4)', we have uXv =  V X u,
U (v + w) . u x v + U X W.
The components of the triple product u x (v X w) are (u X (v X w))' = E1EijkUj(V X w)k
(4)
(4)'
= EhE'fkujEEkabvawb
=
EijkEabkujvawb
(161.2)
= SabUjva'Uib
= uj(viwj  vjwi) = (u w)vi  (u  V)wi; hence
u X (v X w) = (u w)v  (u v)w. The box product u x v w is given by either of the absolute scalars, (5)
(5)'
(U X V)'wi = E'Cijk,ujvkwi = E1
(u X v)zw = Eeijku'vkw'
=E
ul
u2
U3
Vi
V2
V3
W1
W2
W3
U1
u2
u3
V1
V
w1
in agreement with (24.12).
2
w2
V3
w3
TENSOR ANALYSIS
356
§ m3
163. The Afne Connection. Any vector can be expressed as a linear combination of the base vectors. When the 32 derivatives of the base vectors ae;/axi = Die; are thus expressed, (1)
Die; = ref + r e2 + r 'e3 = ri,er,
the 33 coefficients r, are called components of the affine connection. If we multiply (1) by ek , the righthand member becomes rzJj r = r ; hence
r = ek
(2)
Die;.
The law of transformation for r is given by (3)
r = ek
axk ec
Die; =
ax c
\ /  DQ I I axb eb ), /
l ax i
/ \ax
'
where (4)
a axb Di =axia  _axa _ Da; axi axi axa
The differential operator Di transforms like a covariant vector (hence
the subscript). In (3), Da acts on both scalar and vector factors following; hence, from (1), axk
ax`
e
axa axb ( a2xb eb+raber
axiax'
axi axi
Since e° eb = bb, ec er = 6'r, this gives a2xb axk
axa axb axk +    rab axiaxj axb axi ax' ax`
(5)
for the desired law of transformation. Therefore r is a tensor component when and only when a2xb
49Pi axi
= 0,
axb
 = CJ (const), ax'
and hence xb = c;x', if the coordinates xi and xi have a common origin.
The components r of the affine connection (between the deriva
tives of the base vectors and the vectors themselves) are tensor components only with respect to affine transformations.
THE AFFINE CONNECTION
§ 163
357
Although the Irk are not tensor components for general transformations, we shall see that the index notation still serves a useful purpose. j Since e1 = D;r (157.13),
P = ek DiD;r = ek D;Dir = Pk;
(6)
and, from (5), we see that the symmetry of P in the subscripts persists after a transformation of coordinates. If we transform coordinates from x to IF, we have, from (3), (7)
P4  \axk k/
\ax1'
\4924
or, on making the replacements, axk
k
Di =  Da,
_r _ PPQ

e;
axi
ax,
(8)
axb
(9x,
e =  es,
axa
axr
ax'
eb,
(axb
l
l
ax, ec/ \a2p Da/ \ai7 eb/
a relation of the same form as (3). Consequently, the succession of transformations r p I` > f produces a transformation r  r of the same form as r + 1. We express this property by saying that the transformation (3), or its equivalent (5), is transitive. If, in particular, x i = x', we may delete all the tildas () in (7) ; this equation then gives, on expansion, (9)
a2xJ P,,
air
axP axq ax'
a.
a. axr
k
I'2J.
axP a:rq axk
These equations constitute the transformation inverse to (5) and also may be derived by solving these equations for P'ab. This solution may be effected by the multiplication of (5) by axi ax' axr axp ax4 axk
t The coordinates xi present a similar situation. For afflne transformations they are components of a vector; for general transformations this is not the case, but the indices still serve to indicate that their differentials dxi are vector components.
T1E;NSOR ANALYSIS
358
§163
On writing y = x, the equation of transformation (5) becomes:
ax, a2' +   rab lay, ayj ayi ay' i
(10)
axb
ayk
1
axc
If we now make the change of coordinates,
xr = xo + yT 
(11)
2(rPq)oypy4,
where the gammas are computed for x = xo, the point x" = xo corresponds to yr = 0. Since ay'r/ayi = ax,
ayi =
Si

a2xr
ayi a y.i
  (r)o;
hence, at the point x' = xo (yr = 0), the brace in (10) becomes
(r )0 +aj(rab)o = 0. Consequently, all the gammas I' = 0 vanish at the origin y'' = 0 of the new coordinates. Such a system of coordinates is termed geodesic. Since the gammas are not tensors, the equations riI; = 0
(y = 0) do not imply r = 0 (x = x0). Example 1.
For cylindrical coordinates p, p, z, we have (§ 89, ex. 1) r = [x, y, z] = [P cos ip, P sin ,p, z].
If we put x1 = p, x2 ='P,x3 =z, el = Dlr = [cos p, sin p, 0], e2 = D2r = p[ sin p, cos {p, 0],
e3=D3r=[0,0,1]; Die, = 0,
D2ei = [ sin gyp, cos (a, 0] = 1 e2,
D3ei = 0,
P
D2e2 = p[cos rp, sin p, 0] = pei,
D3e2 = 0,
D3e3 = 0.
Hence all gammas are zero except
rig = r2, = 1/P,
r22 = P
Example 2. For spherical coordinates r, sp, B, we have (§ 89, ex. 2)
r = [x, y, z] = r[sin 0 cos p, sin 0 sin p, cos 0].
KINEMATICS OF A PARTICLE
§ 164
359
If we putxi =r,x2= V,x3=0,* ei = Djr = [sin 0 cos gyp, sin 0 sin gyp, cos 01,
e2 = D2r = r sin 01  sin gyp, cos p, 0], e3 = D3r = r[cos 0 cos gyp, cos 6 sin gyp,  sin 0]; D2e1
D1e1 = 0,
=
1 1
D3e1
e2,
r D2e2 = r sin2 6 e1  sin 0 cos 0 e3,
=
1
I e3,
r
D2e3 = cot 0 e2,
D3e3 = rel. The nonzero gammas are therefore
r12=r21=1/r,
r13=r31= 1/r,
r22 =  sin 0 cos 0,
r23
=
1132
= Cot 0,
r22= r sin2 B, r33 = r.
164. Kinematics of a Particle. We now may find velocity and acceleration of a particle in general coordinates:
a = dv/dt = akek. In rectangular coordinates x1 = x, x2 = y, x3 = z we have vk = V = vkek,
(1)
dxk/dt (52.7). Hence, in general coordinates 2i, these components become 8xk dxi
dxk
vk =   _ axi dt dt The time derivatives of the coordinates, which we write ±k, thus transform as contravariant vector components. For the acceleration we have, from (1), (2)
a=
dvk
dt
ek + vk
aek dxi
ax i
dt =
vkek +
vtvkr:kei;
and, on interchanging the summation indices j, k in the last term, (3)
a = (vk + viv'r ,)ek
Therefore, in any coordinate system the velocity and acceleration components are vk = xk ak = xk + x°'x'I'). (4) Since the base vectors ei are not, in general, unit vectors, we must distinguish between the components vk and ak and the numerical values of the terms in vkek, akek. Thus the terms v'e1, a'e1 have the numerical values vllell, a'lell. * The order r, p, 0 is sinistral and [ele2e3] < 0; cf. p. 197.
TENSOR ANALYSIS
360
Example 1.
Cylindrical Coordinates p, ,p, z.
§ 1&
From § 163, ex.
1, el, e2, e3
have the lengths 1, p, 1; hence the values of the velocity terms in (1) are a, pp, z
From the nonzero gammas r22 = p, rig = 1/p, we have
a1 = a  Psp2, a2 = sp + 2Pcp/P, a3 = z; and the numerical values of the acceleration terms are
p' + 2p',
P  Pcp2,
z.
Example 2. Spherical Coordinates r, p, 0. From § 163, ex. 2, e1, e2, e3 have
the lengths 1, r sin 0, r; hence the velocity terms in (1) have the values, r, rcc sin B, r9.
With the nonzero gammas r22 = r sin 2 0, r33 = r we have, from (4), al = r  p2r22 + O2r33 = r  rrp2 sin2 B  r92;
with r12 = 1/r, r23 = cot 6,
a2 = + 2r,pr12 + 2,pOr23 = + 2r(p/r + 24 cot 0; and, with r13 = 1/r, r22 =  sin 6 cos 0, a3 = 9 + 2rOri3 + ,,2r22 = 9 + 2r9/r  ,p2 sin B cos 0. Hence the values of the acceleration terms are r,p sin 0 + 2i ,p sin 0 + 2r09 cos 0,
1'  r,p2 sin2 0  r92,
r9 + 2r9  r02 sin B cos 0.
165. Derivatives of e i and E. From the relation e' er = S*, we have, on differentiation, (Die')
er =  e' Deer =  rsr,
and hence
Die'
(1)
rrer
From Die; = r jer and (1), we have
Di(e,e') = rz,ere'  rreer = 0,
(2)
on interchanging the summation indices r, j in the second term. Since the product E = el e2 x e3 is distributive with respect to addition, its partial derivatives may be found by the familiar rule for differentiating a product, DiE = (Diet) . e2 x e3 + el (Die2) x e3 + el e2 x (D:e3) = r,r'1er e 2 x e3 + ri2e1 er x e3 + ri3e1 1
2
3
_ (ri1 + ri2 + ri3)e1 . e2 x e3,
e2 er
§ 166
AFFINE CONNECTION AND METRIC TENSOR
361
or, if we apply the summation convention,
DiE = rirE. The derivative of Ei = el e2 e3 is therefore (3)
(4)
DiEi =
E2DiE
rirEi
From (3), we have
DjDiE = (Djrir + rarr .,)E; and, since DjDiE = DiDjE, (5)
Dj rir = Dir;r
166. Relation between Affine Connection and Metric Tensor. On differentiating gij = ei ej, we have
Dk9ij = rkier  ej + rkjei  er = rkjJrj + rkjgir Since Dkgij and F are both symmetric in the indices ij, there are 3 X 6 = 18 quantities in each set. Equations (1), 18 in number and linear in the 18 gammas, may be solved for the latter. We first introduce the notation, (1)
(2)
F9rk = rij,k,
just as if r k were a tensor whose upper index was lowered. Then we have also (3)
rij,rgrk = r ,
for the left member equals
ri,gar9rk = I't S = r We note that Fij, k is also symmetric in the indices i, j. Moreover (4)
Diej = F jer = rij,se9,
if we put e,. = g,.ses. We may now write (1) in the form: (5)
rki,j + rkj,i = Dkgij.
If we permute i, j, k cyclically in (5), we obtain the equations,
rij,k + rik,j = Di9jk,
rjk,i + rji,k = Dj9ki;
TENSOR ANALYSIS
362
§ 167
and, upon subtracting (5) from their sum, we obtain ri.i.k =
(6)
Z
(Digjk + D;gki  Dkgi1)
We may now compute r from (3). In the older literature, the components of the affine connection are denoted by [ } = r., [ii,k] = rij,k, .V
Z)
These Christoffel symbols of the first and second kind, respectively, therefore are defined by the equations: J
l
7
} = gkr[ij, r]. [ [ij, k] = (Digik. + D7gki  Dkgii), z 167. Covariant Derivative. The gradient of an absolute tensor T is defined as aT (7)
vT=eh
(1) Since T = T,
eh
aT axh
axh
aT aT ax8 8 r=axh_ ar__ =e = Sre axr
ax8 axh
8
aT axr ;
hence VT is a tensor of valence one greater than T. If the components of T are T ;.'.'.k (the order of the indices is not specified), the components of VT are written (2)
Vh
Tab
i;...k,
the index h on V corresponding to the differentiation a/axh. This is a covariant index, for the operator Dk = a/axh transforms like a covariant vector: (3)
axr axr a Dh==Dr. axh axr a
axh
axh
For this reason the components (2) are called covariant derivatives of Tab
If T is a relative tensor of weight N and valence m, ENT is an absolute tensor, whose gradient, ehDh(ENT),
is an absolute tensor of valence m + 1. If this is multiplied by EN, we again obtain a relative tensor of weight N; this tensor is defined to be the gradient of T and written (4)
VT = ENehDh(ENT).
COVARIANT DERIVATIVE
§ 167
363
When N = 0, (4) reduces to (1). The components of VT, denoted by prefixing Vh to the components of T as in (2), are called covariant derivatives.
From (165.3), we have DhE_N
and hence
= NE'V1DhE = NrhrEN,
Dh(E'YT) = Ev(DhT  NrhrT),
VT = eh(Dh  Nrhr)T. Thus the operator, (5)
V
= eh(Dh  Nrhr), r
applied to any invariant tensor T (with its complement of base vectors) generates another tensor VT of the same weight and valence one greater. We next compute the components of VT explicitly, where
T = Tq '...ceaeb ... e,eiej ... ek. If T is a relative tensor of weight N, VT contains the term, ' e k. k eheaab e,e iej N rrhr It remains to compute the part of VT due to the operator ehDh. (6)

Tab'
ij...
Now Dh acts on the "product" of T ;.'.'.k and a series of base vectors. Since this product is distributive with respect to addition, DhT may be computed by the usual rule for a product; hence c ab DhT = (DhTi;...k)eaeb . . . eye iaj
+
. .
. ek
eceiej
ab
In the second line put Dhea = rhaer, ...
... ek + .. .
i j ec(Dhe )e ... ek +
,
Dhe, =
and in the successive terms interchange the summation indices In the third line put r and a, r and b, , r and c. i r , ... , k r Dhe kDhei  rhre   rhre , and in the successive terms interchange the summation indices r and i, r and j, , r and k. When this is done, each term of VT contains the same complex of base vectors, eheaeb ... eceiej ... ek,
TENSOR ANALYSIS
364
§ 167
and the component VhTt6.'.k of VT equals the sum of their scalar coefficients :
v Tab e = D
17
k
Tab . C q k
+ Tij.k rchr
a Tij...krhr + Trj...k rhi 
 Tij...r rhk
r

r
Nrr rTab c hr
This is the general formula for the covariant derivative of any tensor component, absolute or relative. The last term is absent when T is absolute (N = 0). For every upper index
hik
v Ta*:
contains a term
T°::.r:: t k r* hr
y
and, for every lower index *, OhTa' *:: °k
contains a term
 T"
'
a rr.
We consider now some important special cases. If p is a relative scalar of weight N, vhcp = Dhp  Nrhr(c
(8)
Since E and E1 are scalars of weight 1 and 1, we have, from (165.3) and (165.4), (9) (10)
vhE = DhE  rhrE = 0,
vhE1 = DhE' + rhrE1 = 0. Again, since g = det gij = E2 is a relative scalar of weight 2, (11) Vhg = DhE2  2r;,rE2 = 0. If v = vie1 = vie' is an absolute vector, (12)
VhV = Dhvti + vrrhr,
r vhvi = Dhvi  vrrhi. These expressions are the mixed and covariant components of one and the same dyadic Vv. For an absolute dyadic T, the components of VT may have the (13)
forms :
(14)
VhT2 = DhT'' + Tr1rhr + Tirrhr,
(15)
VhT`j = DhVj + T'jrhr  Tirrhj, VhTij = DhTij  Trjrhi  T,rrhj.
(16)
§ 168
RULES OF COVARIANT DIFFERENTIATION
365
For the metric tensor, (17)
G = gijeiej = gijeiej = eiei = I,
we have, from (165.2), (18)
VG = e'DhG = ehDh(eiei) = 0.
The components of VG therefore vanish: (19)
Chgij = 0,
Vhgii = 0,
Vhbi = 0.
Note that (166.1) is equivalent to Vkgij = 0. Example. We can write
VT = eh(Dh  Nrhr)T = ehvhT,
i20)
if we regard Vh as an operator that acts only on scalars. Covariant differentiation then is defined by this operational equation. With this convention, we have also
VVT = ei(Di  Nrir)ej(Dj  Nrn)T = eiejVjVjT. The second member may be written eiej(Di
 Nrit) (Dj  Nrjr)T  eirise'(Dj  Nrr )T = eiej ( (Di  Nrit) (Dj  Nrir)  rij(D.
Nr )1T
when indices j and s are interchanged; hence
viv jT = (Di  N rir) (Dj  Nrrr,)T  rijV,T, ViViT = (D,  Nrjr) (Di  Nrsr)T  r! v.T, and, on subtraction, (21)
(CiV1  Vjvi)T = (D1Dj  DjDi)T,
in view of (165.5). On the left the operators Vr act only upon the scalar components of T. We note that (21) applies to relative as well as absolute tensors.
168. Rules of Covariant Differentiation. 1. The covariant derivative of the sum or product of two tensors may be computed by the rules for ordinary differentiation.
If we introduce a system of geodesic coordinates yi 1163), the corresponding gammas 1 , will all vanish at the point yi = 0; hence, from (167.7), OTab...r=DTab...r (1) h ij...k h ij...k,
at the origin of geodesic coordinates, which moreover can be chosen at pleasure. For example, we have the tensor equation, Vh(Tt'uk) = (VhTz3)uk + T''Vhuk,
TENSOR ANALYSIS
366
§ 169
in geodesic coordinates and therefore in any coordinates (§ 157). Consequently, the sum and product rules of ordinary differentiation also apply in covariant differentiation. 2. The covariant derivatives of the epsilons and Kronecker deltas are zero.
Since these tensors have constant components, their covariant derivatives vanish at the origin of a system of geodesic coordinates; hence they vanish in all coordinate systems. 3. The operation of contraction is commutative with covariant differentiation. For example, if we contract T=jk on the indices i, j to form T?ik = SiT jk,
we have OhVik = BiZVO
jk.
4. The components of the metric tensor (gij, gij, S) may be treated as constants in covariant differentiation. This follows at once from (167.19). For example, Civj = Vi(g7rvr) = gjiViv'.
Thus we may find Vivj by lowering the index j in Divj; in other words, Divj and Vivj are components of one and the same tensor: Vv = V vieiej = V1vjeiei. Evidently the operation of raising or lowering an index is commutative with covariant differentiation. 5. The relative scalars E, E1 and g may be treated as constants in covariant differentiation.
Since E, E1 and g = E2 are relative scalars of weight 1, 1, 2, respectively, we have, from (167.4), V Vg = 0; VE = 0, for in each case Dh(ENT) = DO = 0169. Riemannian Geometry. Any set of objects which can be placed in onetoone correspondence with the totality of ordered . sets of real numbers (.r.1, x2, , x") satisfying certain inequalities, I xi  ai I < ki (a1 and ki > 0 are constants), is said to form a region of space of n dimensions.$ We speak of
'=0,
(x1) x2,
. ,
x") as a point; but the actual objects may be very
$ Veblen, Invariants of Quadratic Differential Forms, Cambridge, 1933, p. In some applications the numbers xi may be complex.
13.
RIEMANNIAN GEOMETRY
§ 169
diverse.
367
Thus an event in the spacetime of relativity may be
pictured as a point in fourspace; and the configurations of a dynamical system, determined by n generalized coordinates, often are regarded as points in nspace. If we associate the space (x', x2, , x") with an arbitrary nonsingular quadratic form, 9ii dxi dx',
g = det gij 0), (9ij = 9ii, we have a Riemannian space with a definite system of measure(1)
ment prescribed by this form (§ 151) ; and the geometry of this metric space is called Riemannian geometry. We assume that the coefficients gij are continuous twicedifferentiable functions of x. The base vectors ei are not specified; but their lengths and the angles between them may be found from the relations: 9ij = ei e,. The reciprocal base vectors now are given by
(2)
ei =9irer,
(3)
where gii is the reduced cofactor of gij in det gij; for (4) ei e; = girgir = d
(152.2).
Moreover (5)
ei e' = girS* = gij
If we now transform coordinates from xi to xi, we assume that the new base vectors are given by (6)
axr
ei =  er,
axi
ei =  es;
axb
ax8
then ei and e' still form reciprocal sets, for axr a'V
8r = o.
a:C i ax8
In view of (2) and (5), equations (6) show that gjj and gi' transform like absolute dyadics. These tensors often are called the fundamental covariant and contravariant tensors of Riemannian geometry. By use of the equations, (7) e = g,rer, ei = they permit us to raise and lower indices (§ 154) and thus represent any tensor of valence m by 2' types of components. girer,
TENSOR ANALYSIS
368
§ lti'9
At a given point, gij gives the orientation of the base vectors ei relative to each other. In order to determine the relative orientation of sets of base vectors at different points, we must know the components r of the affine connection, defined by
Diej = rer.
(8)
Then, just as in § 165, we have also
Die'
(9)
Tier
(165.1).
We now assume that the affine connection is symmetric (r = r ). The gammas then are determined by the metric tensor (§ 166) : (10)
rkV = 129kr(Digjr + Djgri  Drgij) I
The epsilons in nspace, defined in § 146, have n subscripts or superscripts. The equations, ax .. (9.t xi axj axk Eb... (11)

. .
ax ax
axa axb IEij ...k =
axa axb
.ax`
'
ax°
...axk Eab...ej,
ax ax` 49V t We shall call the base vectors constant if Die, = 0 (i, j = 1, 2,. , n); then raf = 0 and the functions gij = ei e; are also constant. Conversely, (12)
when gii are constants, (10) shows that r!J = 0, and hence Die, = 0. In Riemannian space it is not, in general, possible to introduce coordinates xi for which the base vectors ei are constant. Only flat space (§ 175) is compatible with such Cartesian coordinates; then each point has the position vector
r = xiei and ei = ar/axi. Moreover, for any coordinates 2i in flat space, we have (cf. § 163, ex. 1, 2) (i)
ei =
ax' 021 e'
=
or ax'ar ax' &V
=
.
at i
The geometry on a surface with the metric tensor gi; is Riemannian (n = 2). Unless the surface is flat (a plane, for example), constant base vectors cannot be introduced. If we regard the surface as immersed in Euclidean 3space each surface point has the position vector r = xi + yj + zk, where
x = x(u, v), y = y(u, v), z = z(u, v) are the Cartesian equations of the surface. If we write xl = u, x2 = v, the
base vectors on the surface may be taken as ei = ar/axi; for equation (i) shows that these vectors transform in the manner prescribed in (6). Here r is a position vector in Euclidean 3space; but, in general, the surface points have no position vector in the Riemannian 2space they define. Any Riemannian space of n dimensions may be regarded as immersed in a Euclidean space (§ 178) of n(n + 1)/2 dimensions. This theorem has not as yet been rigorously proved; see Veblen, op. cit., p. 69.
RIEMANNIAN GEOMETRY
§ 169
369
generalized from (146.2), show that eij"'k and eij...k are relative tensors of weight 1 and 1; for these are the powers of the Jacobian I ax/ax I when it is transferred to the righthand side. There are n types of Kronecker deltas in nspace: S,, S b, S bc, up to Sa'bc'...f with 2n indices. As in § 161, they are defined in terms of the epsilons. In the case n = 4, for example, i 1 ibcd Sa =  eabcde
1
Sab   eabcde
ijk
1
abc
ijcd ,
2!
3!
eabcde
ijkd
,
ijkh
aabcd = eabcde
ijkh
The rule given in § 161 for the value (0, 1, or 1) of a generalized delta still applies. Moreover the preceding definitions show that all the deltas are absolute tensors, alternating in both upper and lower indices. See Prob. 42. The nrowed determinant, (13)
g = 1t etii...ke,s...tgir9js ... gkt n!
[cf. (146.3)],
generalized from (152.1), is the contracted product of two epsilons of weight 1 and n absolute dyadics. Hence the discriminant g of the fundamental quadratic form is a relative scalar of weight 2. The cofactor of gij in g is ggij; hence, from (146.11) and (166.5), we have
Dhg = gg"Dhgij = ggi (rhi, j + rhj, or, in view of (166.3), (14)
Dhg = grhi + gr,. = 2grhr
From (14), we have also (15)
DhV_ \'gr,.,,,
a result of the same form as (165.3) with E replaced by /g. In defining the gradient of a tensor, we replace E by \/g in (167.4); thus, in Riemannian nspace, N
N
VT = g2ehDh(g 2T). The components of VT again are given by (167.7). Hence this formula for the covariant derivative is still valid in Riemannian (16)
geometry.
TENSOR ANALYSIS
370
§ 170
Since the metric tensor G = gjkejek = ekek still has the property DhG = 0, VG and its components vanish as before: Vhb = 0. Moreover, from (16), vq = gehDh1 = 0, and Ohg J = 0,
Ohgij = 0,
(17)
Ohg = 0.
(18)
170. Dual of a Tensor. An mvector is a tensor of valence m which is alternating in all indices (cf. § 160). For convenience in
wording, we also regard scalars (m = 0) and vectors (m = 1) as mvectors. In nspace we can associate with any mvector (0 < m < n) an (n  m)vector, its dual, defined as follows: If and Qij...k are mvectors of weight N, their duals are the (n  m)vectors, 1
(1)
ab...c ij...k
m.
li
gab...c 
(2)
Pj...k,
Qij...kEij...k
m.
ab...c,
of weights N + 1 and N  1, respectively. Note that the epsilons have n indices in nspace; and, in the contracted products, the contravariant tensor is written first, and the summation indices are adjacent.
A scalar Sp has two duals, Eab .htp, Eab . hV, according as we use (1) or (2); they have the same numerical components but different weights. The dual of an nvector Tijk...h is the scalar: ijk...h
1 E
n!
Tijk...h = T123...n.
m < n), dual dual T = T. (3) Proof. Write T = P, dual T = p. Using (2) and (1), we have THEOREM.
If T is an mvector (0
1
(dual
pab
c
(n  m) !
=
1
Eab...c
(n  m) !m! 1
i;...k
_ m!brs tPi;...k = Pra...t)
rs.
ra...t
[Cf. Prob. 42.]
DIVERGENCE
§ 171
371
since Pip .k is alternating in all indices. Hence dual p = P; and, similarly, dual q = Q. When T is an nvector, say Tijk...h, (1) gives the dual T123 . . n now (2) gives dual dual T = Eijk...hT123...n =
Thus the theorem holds in this case also, provided both dualizing equations (1), (2) are used. 171. Divergence. The gradient of v = viei is Vv = vhviehei.
The first scalar of this dyadic is V11) div v = %ib = If vi is an absolute vector, the divergence is the absolute scalar Viv'; this definition applies in nspace and for any coordinate (1)
system. When vi is a relative vector of weight 1, VhV'
= Dhvi + 17rrhr  vtrhr
(167.7).
On contracting with h = i, the second and third terms cancel; for vrrsr = virir, since r and i are summation indices. Hence (2) div v = Divi (wt. v = 1). vi has the weight 1 imparted by the scalar
If vi is absolute,
/; hence
viva = vi(9i91vi) = 9lvd(91v=) = 91Di(91vi), in view of (2). Thus (3)
div v = 9
Di( / vi)
(wt. v = 0).
The Laplacian V2p of the scalar p is defined as div Vp. If rp is absolute, v = V p is an absolute vector; then Vr = Drop,
Vi
= 9irDrcp,
and, from (3), (4)
v2(p
Di(V J 9irDrcp)
The divergence of any tensor T is defined as the gradient VT contracted on the first and last indices. Thus if T has the com
TENSOR ANALYSIS
372
1 172
porients T 'j kh of valence m and weight N, (divT)ij...k
(5)
VhTij...kh
=
is a tensor of valence m  1 and weight N. THEOREM. When T i' ., kh is an mvector of weight 1, div T is the (m  1)vector, (div T) ij...k = DhTij...kh, (6) where V in the defining equation is replaced by D. Moreover,
div div T = 0
(7)
From (167.7), VhTij...kh = DhTij...kh +
(m > 1).
Proof.
+ rhrT
... + rk
1,hrTrj...kh +
Tij...rh
ij" . kr  rhrT ij... kh.
The two final terms cancel, as we see on" interchanging the summation indices h, r in the last term. The remaining terms containing 1'hr vanish separately on summing over h and r; for rh,. is
symmetric and T antisymmetric in these indices. Thus (6) is proved.
From (6) and the alternating character of T,
(div div T) 'j... =
DkDhT'j... kh = 0.
172. Stokes Tensor. The gradient of the covariant vector vk is the dyadic Vhvk. From this we form the antisymmetric dyadic, Pij = Sf V hvk,
(1)
known as the Stokes tensor. t Ohvk =
and, since az4r (2)
When Vk is albsolute, hVk 
rrhkvr, .
= 0 owing to the symmetry of rr , we have
Pij = SfDhvk = Divj  Djvi
(wt. v = 0).
In 2space we can form from Pij the relative scalar of weight 1, (3)
2eijPij = P12 = Dlv2  D2v1
t Veblen, op. cit., p. 64.
(wt. V = 0),
STOKES TENSOR
§ 172
373
and from this the absolute scalar, 1
(4)
Vg
(Dlv2  D2v1)
(wt. V = 0).
This is the absolute invariant on the surface with the fundamental form gig dxi dx', written n rot v or n V X v $ in § 97. In 3space we can form from Pii the relative vector of weight 1,
2
wi = 1EijkP7k = Ei'kD7'v k,
(5)
having the components, D2v3  D3v2i
D3v1  D1v3,
D1v2  D2v2.
The absolute vector,
rot v =
(6)
1
wiei =
1
1/9
Ei'keiDjvk,
may be written as a symbolic determinant (cf. § 146): el
rot v =
(7)
1
V9
D1 V1
e2
e3
D2 D3 V2
V3
Comparing this with (88.19) now shows that rot v is in fact the rotation of v previously defined.*
In § 91 we proved that a vector v is the gradient of a scalar in 3space when and only when rot v = 0. In nspace we have the corresponding R.
THEOREM. Let v be a continuously differentiable vector in a region Then, in order that v be a gradient vector,
(8)
V = OAP,
vi = Dip,
it is necessary and sufficient that the Stokes tensor vanish in R:
Pi,=DivaD;vi=0.
(9)
Proof. The condition is necessary; for, if vi = Dip, Pi; vanishes identically, owing to the continuity of the second derivatives of cp.
In (97.9), H=v19, u=x1, v=x2, ru=el, rv=e2,
r,.f
= f2 in our present notation.
* In (88.19), J notation.
u = x1, ru = el, ru f = fl, etc., in our present
TENSOR ANALYSIS
374
§ 173
The condition is sufficient. For the method of § 91, extended to case of n coordinates, leads to the function,
f
z1
(10)
(P =
vj (xl,
al
I.
x2
+
x2)
... , xn) dxl
v2 (a', x2, ... xn) dx2
a2
" f"
f
3
v3 (ale a21 x3,
xn) dx3 ...
(ai a2 ..
an1 xn) dxn
zn
n
v
where all a2, , an are the coordinates of a fixed but arbitrary point. In the ith integral xi is the variable of integration while xi+1' ... xn are regarded as constant parameters. We now can , show from (9) and (10) that Dig = vi. Let us compute, for example, D3V. Only the first three integrals in (10) contain x3; their derivatives with respect to x3 are, respectively, v3(xl, x2) ... , xn) _ v3(al) x2, ... , x"), v3(a1 ,
x2, ... , xn)
 v3(al, a2, ... 'X n),
v3(al, a2, ... , xn),
when we make use of D3vi = Dlv3, D3v2 = D2v3 in the first and second; hence D34P = v3(xl, x2,
,
xn).
173. Curl. We define the curl of a covariant tensor Tb...d of valence m (m < n) as the (m + 1)vector: (1)
(curl T)hij...k =
1
m.
bhij
.k VaTbc. d
When m = 0, T = P (a scalar), 0! = 1, and curl cp = ahV aco = V MP
is the gradient of gyp.
When m = 1, T = v (a vector), curl v is the Stokes tensor (172.1).
In general, we have the THEOREM. When Tab.. d is an absolute tensor of valence m < n, (2)
(curl T)hij ... k =
Shaijc b ......kD..Tbc...d,
M.
CURL
§ 173
375
where V in the defining equation is replaced by D. Moreover,
(m < n  1).
curl curl T = 0
(3)
Proof.
From (167.7), we have
VaTbc...d = DaTbc...d  rabTrc...d
1'adTbc...r.
Hence in the right member of (1) there are m terms of the type, Sabc...dl,r 7,
1
ab
rc...d,
m!
these all vanish separately when we sum over the subscripts of I'a*. Thus (2) is proved. From (2) we have Sahi kD (1(,'_ Sab d D T 1 = (curl curl T) 1
1)'
(m + 1
m!
a
g
.
.
Sfet...a DgDaTb...d
which vanishes when we sum over g and a. When Tbc...d is an mvector, the summation in (1) or (2) may
be carried out in m + 1 stages by setting a = h, i, j,
k in turn and summing over the other m indices. Thus, from (1), we ,
have (4)
(curl T)hij...k = VhTij...k + (1)mViTj...kh + ... ,
taking the m + 1 cyclical permutations of the subscripts hij .. k in order and placing ( 1)' before the second, fourth, terms. In the cases m = 1, 2, 3, we thus obtain
(curl T)ij = VjTj  ojTi, (curl T)ijk = ViTjk + VJTki + VkTij, (curl T)hijk = VhTijk  ViTjkh + VVTkhi  VkThij
When Tbc...d is an absolute mvector, we obtain, from (2), an equation of the form (4) with V replaced by D.
When S is an absolute tensor of valence m  1, T = curl S is an absolute mvector and curl T = 0. Then (4) becomes (8)
0=
(1)'DiTjk...h + .
TENSOR ANALYSIS
376
§ 175
Thus if vi is absolute, T i j = S Davb is an absolute bivector, and
DiTjk + DjTki + DkTij = 0.
(9)
174. Relation between Divergence and Curl. For alternating tensors, we have the THEOREM.
If T be d is an n2vector (n2 < n),
dual curl T = div dual T,
(1)
provided dual T is taken contravariant when T is a scalar. Proof.
From § 170, 1
(dual curl T)pq"'r
ii ... k(CUrlT) ij...k Epq...r ij...kaab ..kdOaTb...d
(m + 1)!m! pq... r ab... dVaT
b...d
M! Va
(1. E pq "r
Tb ...d
= Va(dual T)P . ra
= (div dual T)pq*'*.
On taking duals of both members of (1), we have
curl T = dual div dual T.
(2)
Moreover, if we replace T in (1) by dual T, we have also
div T = dual curl dual T.
(3)
175. Parallel Displacement. A vector p is said to undergo a parallel displacement along a curve xi = pi(t) when dp/dt = 0 along this curve. dp
dt
If p = pkek, we have
dxi dxi = dpk dpkek + pk aek  ek + pk rikej, axi dt dt dt
dt
or, on interchanging summation indices j, k in the last term, dp dt
(+pr)ek. dt
PARALLEL DISPLACEMENT
§ 175
377
Hence, if dp/dt = 0, the components pk satisfy the differential equations: dpk
(1)
dt
pj
+
dx'
dt
(k = 1, 2, ... , ii).
0
r". =
A solution pk(t) of this system, satisfying the arbitrary initial conditions pk(0) = ak, defines a vector at each point t of the curve. The vector ak at the point PO (t = 0) is said to undergo a parallel displacement along the curve into the vector pk(t) at the point P. In (1) dxi/dt depends upon the functions pi(t) defining the curve; hence, in general, the solutions pk(t) will change when the curve is altered. In other words, the vector pk obtained by a parallel displacement of ak from PO to P depends, in general, upon the path connecting these points. The length of a vector p and the angle between two vectors p, q remain constant during a parallel displacement; for, if dp/dt and dql dt vanish along a curve, we have also d
d
dt(P  q) = 0.
dt(P.P) =0,
We shall say that a vector remains constant during a parallel displacement.
If s is the are along the curve, ds2
= gi; dxi dx1 = g;xix' dt2.
If we choose the are as parameter (t = s), we have gjjziz' = 1, an equation which states that the tangent vector dxi/ds to the curve is of unit length. If a curve has the property that its unit tangent vectors dxi/ds are parallel with respect to the curve, it is said to be a path curve for the parallel displacement. With t = s, pk = dxk/ds, (1) gives the differential equations of the path curves, d2xk
dxi dx'
ds2
t' ds ds 
0.
(2)
The path curves are the "straightest" curves in our Riemannian spacethe analogues of straight lines in Euclidean geometry. When equations (1) can be satisfied by functions pk(x', , x") of the coordinates alone, the parallel displacement is independent of the path, and the space is said to be flat. Then dpk
apk dxi
dt
axe dt
TENSOR ANALYSIS
378
§ 175
and the ordinary differential equations (1) are replaced by the partialdifferential equations: ap''
axi + p'1' = V pk = 0.
(3)
Since Vipk are the components of Vp = e'Dip (§ 167), we see that a flat space contains vectors p(x', , X'), such that Vp = 0, or
(i = 1, 2, ... , n). Dip = 0 Since pk = p ek, we see that (4) is equivalent to (4)
D,Pk = p Dzek. For any fixed value of k, the n functions p D,ek are components of a gradient vector, and for this it is necessary and sufficient that (5)
Di(p D,ek)  D,(p Diek) = 0
(§ 172, theorem) ;
or, since Dip = 0,
p (DiD,  D,Di)ek = 0. These equations must hold independently of the initial conditions imposed upon p and are therefore equivalent to
(DiD;  D,Di)ek = 0.
(6)
Since Di transforms like a covariant vector (167.3), the operator
Di, = DiD;  D,Di = a bDaDb,
(7)
transforms like a covariant dyadic; for
DiD' 
(at.,
a2xb
axb
ax,
Da) \ax1 Dbl
Di, _ DiD,  D,Di =
axa axb
axiaxI Db + azs a
D0Db,
ax, axb
. Dabat, ax'
Moreover, axa axb
Di;ek
\1
axa axb axc
(axc
Dabec,
Caxi at, Dab/ \axk ec/
at  4921 axk
since D,,Dkxc = 0; hence eh
axa axb axc axh
DiJek =  axk at t at,
axd
ed Dabec
§ 175
PARALLEL DISPLACEMENT
379
This equation shows that e' Di;ek is an absolute mixed tensor of valence four, say (8)
Rijkh = eh Di;ek.
The components of this curvature tensor R are thus the coefficients in the equation, (9)
Rijkheh;
Dijek =
the condition (6), necessary for flat space, now assumes the tensor form, (10) h Rtijk = 0.
If it holds in one coordinate system, it holds in all. When the space is flat, we can choose n linearly independent vectors ai at a point P and, by giving them parallel displacements to neighboring points obtain a set of constant base vectors ei = ai in a region about P. For these base vectors, we have
gi; = ai a; = const,
I'k = 0
(166.6),
and the corresponding coordinate system x is called Cartesian. To determine the Cartesian coordinates y = x corresponding to the base vectors ai, we have ayr
ek =axk a,., aek
c12yr
OX,
ax' axk
ar,
and, on dotmultiplying (11), member for member, by ay8
 er = a8 , axr
ays (12)
r
ax, rA
a2y3
t
axi axk
From (11), we have the necessary conditions for the integrability of equations (12): D;ek = Dke;,
Di,jek = 0;
t This equation also follows from (163.9) with Tj = 0.
TENSOR, ANALYSIS
380
that is,
r r rjk = rkj,
Rijk
h
§ 176
= 0.
These conditions are also sufficient for the complete integrability of equations (12).$ When these conditions are fulfilled, (12) ad, x") which with ay/axi take on arbimits solutions y(x', x2, trary values at a given point. If we place the origin of the Car
tesian coordinate system at the point x0, we have y = 0 when x = x0; and, if we choose n linearly independent sets of initial values,
axi  pi,
a
= p2,
... ,
ax"
(i = 1, 2, ... , n),
= pn
when x = x0, we obtain n corresponding solutions y'(x) whose Jacobian I ay/ax I = det p 0 when x = x0. In the region about x = x0 for which I ay/ax 19 0, the n functions yj(x) thus obtained define a Cartesian coordinate system. In brief, we have the important THEOREM. A necessary and sufficient condition that a Riemann space, with symmetric afne connection, be flat is that its curvature tensor vanish identically.
We may readily compute the components Rijkh from (8):
*.k'; eh
(DiDjek  D;Diek)
= eh {Di(r;ker)  Dj(riker)I eh
{(Dirjk)er
 (Djrik)er} +
and, on putting eh e, (13)
eh
{rjkrires  rikrjres},
eh e8 = 887 we have
Ri;kh = D2r'  Djr k + r; r r 
176. Curvature Tensor. From the defining equation for the curvature tensor, Dijek = ji rer, we obtain the covariant components, (1)
eh Dijek = Rijk r9rh = Rijkh
$ Cf. Veblen, op. cit., p. 701.
CURVATURE TENSOR
§ 176
:381
We now can express the covariant curvature tensor Rijkh in terms of the gammas :
Rijkh = eh (DIDjek  D;Diek) (166.4) = eh' {Di(rjk,re')  Dj(rik,rer)} Sh{Dirjk,r  Djrik,rI  0h{ rjk,rria  rik,rrja};
(2)
Rijkh = Dirjk,h  Djrik,h  rjk,rrik + rik,rrjh
Since
rjk,rrih = grerikrih = rjkrih,s, we also may write (2) in the form: (3)
Rijkh = Dirjk,h  Djrik,h  rikrih,r + rr,krjh,r.
Rijkh has the following types of symmetry:
Rijkh + Rijkh = 0; Rijkh + Rijhk = 0; Rijkh + Rjkkh + Rkiih = 0;
Rijkh  Rkhij = 0. Proofs.
(I) follows from Dij = Dji. Since the scalars ek
eh
= gkh have continuous second derivatives (§ 169),
gives (II). Thus Rijkh is alternating in its first two and last two indices.
Since the affine connection is symmetric (§ 169), we have (4)
Die; = r er = r;ier = Djei.
Hence, on adding the identities,
Dk(D1ej  Djei) = 0,
Di(Djek  Dkej) = 0,
Dj(Dkei  Diek) = 0, we obtain
Dijek + Djkei + Dkjej = 0, which, on multiplication by e h , gives (III).
TENSOR ANALYSIS
382
§ 176
Now (IV) is a consequence of (I), (II), and (III). From (III), we have
Rijkh + Rikih + Rkiih = 0, Rjkhi + Rkhji + Rhjki = 0,
Rkhij  Rhhkj  Rikhj = 0,  Rhijk  Rijkk  Rjhik = 0. When we add these equations and make use of (I) and (II), only the underlined terms survive, namely, 2Rijkh  2Rkhij, and we obtain (IV). The symmetry relations (I) to (IV) reduce the number of independent components of Rijkh to i22n2(n2  1). Proof. Rijkh = 0 when i = j or k = h (I, II); hence, in general, the number of nonzero components is (nC2)2 = n2(n  1)2/4. But, when ij ; kh, these are paired, because Rijkh = Rkhij (III); hence, if we add the number nC2 of unpaired components Rjjij to the preceding total, we obtain double the number of components with distinct values. The number of components with distinct values thus is reduced to 1n2(n
4
n(n  1)
1)2
+
2
1=*n(n1)(n2n+2).
These are still further reduced by the ,,C4 relations (III); for i, j, k, h must all be different in order to get a new relation. If, for example, i = j, Riikh + Rikih + Rkiih = Rikih + Rkiih = 0
is already included in (I). The number of linearly independent components is therefore
n(n  1)(n2  n + 2) 
n(n  1)(n  2)(n  3) 24
=
n2(n2  1).
For n = 2, 3, 4 this gives 1, 6, 20 linearly independent components Rijkh, respectively. Example 1. When n = 2, the contracted product, eijEkhR,,kh
= 4R1212,
CURVATURE TENSOR
§ 176
383
is a relative scalar of weight 2 (§ 169); hence R1212/g is an absolute scalar. Now, from (2), R1212 = D1r21,2  D2r11,2  r21,.ri2 + rll,rr22
(5)
We shall compute this expression when the base vectors are orthogonal: g12 = 0.
From (166.6),
rll,1 = !Digll, r12,1
=
r11,2 = iD2g11, r12,2 = 2D1g22,
ZD2911,
r22,, = 2D1922,
r22,2 = 3Dzg22.
Moreover, since g = 911922, 911 = 1/911, g2 = 1/g22; hence
r21,rr12 = r21,,ri2 + r21,2r12 = r21,,r,2,1 911 + r21,2r,2,2 g22

(D2g11)2
+
(D1922 )2
4911
4922
r11,rr22 = rll,lr22 + r11,2r22 = r11.1r22,1 911 + rll,2r22,2 g22 (Dig,,) (D1922) 491,
(D2911) (D2g22)
4922
Substituting these results in (5) gives 81212  DiD1922 + 2 D2D2911
Dlgll + D1922 D19zz + Dw11 + D2922 D2g11 l j
St l\ 911
922
922
911
J
2
\
The absolute scalar, (6)
K
_81212 9
(
1
219
Dl
1
(.\/g
D19zz +
J
\
D2
D2 911 J
,
is precisely the total curvature of the surface whose fundamental differential form is ds2
= gll dx1 dx1 + 922 dx2 dx2;
for, if we put x1 = u, x2 = v, 911 = E, g22 = G, H =, (6) agrees with (139.6).
Example 2. We may contract the tensor, (7)
R41k
in essentially two different ways.
h
= ghrRijkr,
TENSOR ANALYSIS
384
§ 177
With h = k, we have (8)
Rt1x
k
= gkrRiikr = 0,
since gkr and Rijkr are, respectively, symmetric and antisymmetric in k, r. From (175.13), we see that (8) is equivalent to the identity:
Dirjr = Djr;,.
(9)
With h = i, we obtain the Ricci tensor, (10)
Rik = R0 ' = 9ihRijkhi this is a symmetric dyadic; for, from (IV), (11)
Rkj = 9''`Rikih = 9'Rihik = ghiRhiki = Rik.
The first scalar of this dyadic, R = gikRik = gikgihRijkh,
(12)
is an absolute invariant. In the case n = 2, g12 = 0 considered in ex. 1, we have (13)
R = 2g11g22R2112
= 2R1212/9 = 2K.
177. Identities of Ricci and Bianchi. Ricci Identity. In analogy with
Dij = D,Dj  DjDj, we also write Vij = ViVj  V, Vi. With this notation, (167.21) becomes
VijT = DijT
(1)
for any tensor, absolute or relative, with its base vectors. On the left Vii acts only on scalars (cf. § 167, ex.); on the right Dij acts only on base vectors, for Dij
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