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International Series in Pure and Applied Mathematics William Ted Martin, CONSULTING EDITOR
VECTOR AND TENSOR ANALYSIS
a International Series in Pure and Applied 'Mathematics WILI.LA11 TED MARTIN, Consulting Editor
Complex Analysis BELLMAN Stability Theory of Differential Equations AIILFORS
Bucx Advanced Calculus CODDnNUTON AND LEVINSON Theory of Ordinary Differential Equations G01.011 13 AND SHANKS Elements of Ordinary Differential Equations of Real Variables GRAVES The Theory of
GRIFFIN Elementary Theory of Numbers IIILDEBRAND rodurl ion to Numerical Analysis Principles of Nunurriral Analysis LAS b;leuu'nts of Pure aunt Applied Mathematics LASS Vector and Tensor Analysis LEIGIITON An Introduction to the Theory of Differential Equations NEHAIU Conformal Mapping NEWELL Vector Analysis ROSSER Logic for 'Mathematicians RUDIN Principles of Mathematical Analysis SNEDDON Elciuents of Partial I)iiTerential Equations Pourier Transforms SNEDDON STOLL Linear Algebra and 'Matrix Theory WEINSTOCK Calculus of Variations
VECTOR AND TENSOR ANALYSIS BY
HARRY LASS JET PROPULSION LABORATORY
Calif. Institute of Tech. 4800 Oak Grove Drive Pasadena, California
New York Toronto London McGRAW-HILL BOOK COMPANY, INC. 1950
VECTOR AND TENSOR. ANALYSIS Copyright, 1950, by the '.11e(_;ratt-Bill Book Company, Inc. Printed in the 1 nited States of America. All rights reserved. 't'his hook, or parts thereof, may not be reproduced in any form without, perrnis.sion of the publishers.
To MY MOTHER AND FATHER
PREFACE This text can be used in a variety of ways. The student totally unfamiliar with vector analysis can peruse Chapters 1, 2, and 4 to gain familiarity with the algebra and calculus of vectors. These chapters cover the ordinary one-semester course in vector analysis. Numerous examples in the fields of differential geometry, electricity, mechanics, hydrodynamics, and elasticity can be found in Chapters 3, 5, 6, and 7, respectively. Those already acquainted with vector analysis who feel that they would like to become better acquainted with the applications of vectors
can read the above-mentioned chapters with little difficulty: only a most rudimentary knowledge of these fields is necessary in order that the reader be capable of following their contents,
which are fairly complete from an elementary viewpoint. A knowledge of these chapters should enable the reader to further digest the more comprehensive treatises dealing with these subjects, some of which are listed in the reference section. It is hoped that these chapters will give the mathematician a brief introduction to elementary theoretical physics. Finally, the author feels that Chapters 8 and 9 deal sufficiently with tensor analysis and Riemannian geometry to enable the reader to study the theory of relativity with a minimum of effort as far as the mathematics involved is concerned. In order to cover such a wide range of topics the treatment has necessarily been brief. It is hoped, however, that nothing has been sacrificed in the way of clearness of ideas. The author has attempted to be as rigorous as is possible in a work of this nature.
Numerous examples have been worked out fully in the text. The teacher who plans on using this book as a text can surely arrange the topics to suit his needs for a one-, two-, or even threesemester course.
If the book is successful, it is due in no small measure to the composite efforts of those men who have invented and who have vii
viii
PREFACE
applied the vector and tensor analysis. The excellent works listed in the reference section have been of great aid. Finally, I wish to thank Professor Charles de Prima of the California Institute of Technology for his kind interest in the development of this text. HARRY LASS URBANA, ILL.
February, 1950
CONTENTS PREFACE .
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vii
CHAPTER 1 THE ALGEBRA OF VECTORS .
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1
1. Definition of a vector 2. Equality of vectors 3. Multipli-
cation by a scalar 4. Addition of vectors 5. Subtraction of 6. Linear functions 7. Coordinate systems 8. Scalar, or dot, product 9. Applications of the scalar product to space geometry 10. Vector, or cross, product 11. The distributive law for the vector product 12. Examples of the vector product 13. The triple scalar product 14. The triple vector product 15. vectors
Applications to spherical trigonometry
CHAPTER 2 DIFFERENTIAL VECTOR CALCULUS . . . . . . . . . . . . . 29 16. Differentiation of vectors 17. Differentiation rules 18. The gradient 19. The vector operator del, V 20. The divergence of a vector 21. The curl of a vector 22. Recapitulation 23. Curvilinear coordinates
CHAPTER 3 DIFFERENTIAL GEOMETRY . . . . . . . . . . . . . . . . . 58 24. Frenet-Serret formulas 25. Fundamental planes 26. Intrinsic equations of a curve 27. Involutes 28. Evolutes 29. Spherical indicatrices 30. Envelopes 31. Surfaces and curvilinear coordinates 32. Length of arc on a surface 33. Surface curves 34. Normal to a surface 35. The second fundamental form 36. Geometrical significance of the second fundamental form 37. Principal directions 38. Conjugate directions 39. Asymptotic lines 40. Geodesics
CHAPTER 4
INTEGRATION . . . . . . . . . . . . . . . . . . . . . . . . 89 41. Point-set theory 42. Uniform continuity 43. Some properties of continuous functions 44. Cauchy criterion for sequences 45. Regular area in the plane 46. Jordan curves 47. Functions of bounded variation 48. Arc length 49. The Riemann integral ix
CONTENTS
x
50. Connected and simply connected regions 51. The line inte53. Stokes's theorem 54.
gral 52. Line integral (continued)
Examples of Stokes's theorem 55. The divergence theorem (Gauss)
56. Conjugate functions
CHAPTER 5
STATIC AND DYNAMIC ELECTRICITY . . . . . . . . . . . 127 57. Electrostatic forces 58. Gauss's law 59. Poisson's formula 60. Dielectrics 61. Energy of the electrostatic field 62. Discontinuities of D and E 63. Green's reciprocity theorem 64. Method of images 65. Conjugate harmonic functions 66. Integration of Laplace's equation 67. Solution of Laplace's equation in spherical coordinates 68. Applications 69. Integration of Poisson's equation 70. Decomposition of a vector into a sum of solenoidal and irrotational vectors 71. Dipoles 72. Electric polarization 73. Magnetostatics 74. Solid angle 75. Moving charges, or currents 76. Magnetic effect of currents (Oersted) 77. Mutual induction and action of two circuits 78. Law of induction (Faraday) 79. Maxwell's equations 80. Solution of Maxwell's equations for electrically free space 81. Poynting's theorem 82. Lorentz's electron theory 83. Retarded potentials CHAPTER 6 MECHANICS . . . . . . . . . . . . . . . . . . . . . . . . . 184 84. Kinematics of a particle 85. Motion about a fixed axis 86. Relative motion 87. Dynamics of a particle 88. Equations of motion for a particle 89. System of particles 90. Momentum and angular momentum 91. Torque, or force, moment 92. A theorem relating angular momentum with torque 93. Moment of momentum (continued) 94. Moment of relative momentum 95. Kinetic energy 96. Work 97. Rigid bodies 98. Kinematics of
a rigid body 99. Relative time rate of change of vectors 100. Velocity 101. Acceleration 102. Motion of a rigid body with one point fixed 103. Applications 104. Euler's angular coordinates 105. Motion of a free top about a fixed point 106. The top (continued) 107. Inertia tensor CHAPTER 7 HYDRODYNAMICS AND ELASTICITY . . . . . . . . . . . . 230 108. Pressure 109. The equation of continuity 110. Equations of motion for a perfect fluid 111. Equations of motion for an incompressible fluid under the action of a conservative field 112. The general motion of a fluid 113. Vortex motion 114. Applications 115. Small displacements. Strain tensor 116. The stress tensor 117. Relationship between the strain and stress tensors 118. Navier-Stokes equation
CONTENTS
xi
CHAPTER 8 TENSOR ANALYSIS AND RIEMANNIAN GEOMETRY. . . . . 259 119. Summation notation 120. The Kronecker deltas 121. Determinants 122. Arithmetic, or vector, n-space 123. Contravariant vectors 124. Covariant vectors 125. Scalar product of two vectors 126. Tensors 127. The line element 128. Geodesics in a Riemannian space 129. Law of transformation for the Christoffel symbols 130. Covariant differentiation 131. Geodesic coordinates 132. The curvature tensor 133. RiemannChristoffel tensor 134. Euclidean space
CHAPTER 9 FURTHER APPLICATIONS OF TENSOR ANALYSIS . . . . . . 311 135. Frenet-Serret formulas 136. Parallel displacement of vectors 137. Parallelism in a subspace 138. Generalized covariant differentiation 139. Riemannian curvature. Schur's theorem 140. Lagrange's equations 141. Einstein's law of gravitation 142. Two-point tensors
REFERENCES .
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INDEX .
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339
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341
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CHAPTER 1 THE ALGEBRA OF VECTORS 1. Definition of a Vector. Our starting point for the definition of a vector will be the intuitive one encountered in elementary physics. Any directed line segment will be called a vector. The length of the vector will be denoted by the word magnitude. Any physical element that has magnitude and direction, and hence can be represented by a vector, will also be designated as a vector. In Chap. 8 we will give a more mathematically rigorous definition of a vector. Elementary examples of vectors are displacements, velocities,
forces, accelerations, etc. Physical concepts, such as speed, temperature, distance, and specific gravity, and arithmetic numbers, such as 2, i, etc., are called scalars to distinguish them from vectors. We note that no direction is associated with a scalar. We shall represent vectors by arrows and use boldface type to Ca-, a] a, a, a, indicate that we are speaking of a vector. In order to distina guish between scalars and vec-
tors, the student will have to
Fia. 1.
adopt some notation for describing a vector in writing. The student may choose his mode of representing a vector from Fig. 1 or may adopt his own notation. To every vector will be associated a real nonnegative number equal to the length of the vector. This number will depend, of course, on the unit chosen to represent a given class of vectors. A vector of length one will be called a unit vector. If a represents the length of the vector a, we shall write a = jal. If jal = 0, we define a as the zero vector.
2. Equality of Vectors. Two vectors will be defined to be equal if, and only if, they are parallel, have the same sense of direction, and the same magnitude. The starting points of the vectors are immaterial. It is the direction and magnitude which 1
VECTOR AND TENSOR ANALYSIS
2
[SEC. 3
are important. Equal vectors, however, may produce different physical effects, as will be seen later. We write a = b if the vectors are equal (see Fig. 2).
Fia. 2.
3. Multiplication by a Scalar.
If we multiply a vector a by a
real number x, we define the product xa to be a new vector parallel to a whose magnitude has been multiplied by the factor x.
Thus 2a will be a vector which is twice as long as the vector a
Z-a
Fm. 4.
FIG. 3.
and which has the same direction as a (see Fig. 3). We define -a as the vector obtained from a by reversing its direction (see Fig. 4).
We note that x(ya) = (xy)a = xya
(x + y)a = xa + ya Oa = 0
(zero vector)
It is immediately seen that two vectors are parallel if, and only if, one of them can be written as a scalar multiple of the other. 4. Addition of Vectors. Let us suppose we have two vectors given, say a and b. We form a third vector by constructing a triangle with a and b forming two sides of the triangle, b adjoined to a (see Fig. 5). The vector starting from the origin of a and ending at the arrow of b is defined as the vector sum a + b.
We see that a + 0 = a, and if a = b, c = d, then
a+c=b+d
THE ALGEBRA OF VECTORS
SEc. 6]
3
From Euclidean geometry we note that
a+b=b+a
(1) (2) (3)
(a+b)+c=a+(b+c) x(a+b) =xa+xb
(1) is called the commutative law of vector addition; (2) is called the associative law of vector addition; (3) is the distributive law
for multiplication by a scalar. The reader should have no trouble proving these three results geometrically.
a+b+c Fio. 5.
5. Subtraction of Vectors. Given the two vectors a and b, we can ask ourselves the following question: What vector c must be added to b to give a? The vector c is defined to be the vector
a - b. We can obtain the desired result by two methods. First, construct -b and then add this vector to a, or second, let b and a have a common origin and construct the third side of the triangle. The two possible directions will give a - b and b - a (see Fig. 6). Thus a - b = a + (-b).
a-b
a-b Flo. 6.
6. Linear Functions. Let us consider all vectors in the twodimensional Euclidean plane. We choose a basis for this system of vectors by considering any two nonparallel, nonzero vectors. Call them a and b. Any third vector c can be written as a linear
VECTOR AND TENSOR ANALYSIS
4
[SEC. 6
combination or function of a and b,
c = xa + yb
(4)
The proof of (4) is by construction (see Fig. 7). Let us now consider the following problem: Let a and b have a common origin, 0, and let c be any vector starting from 0 whose end point lies on the line joining the ends of a and b (see Fig. 8). B
a
FIG. 8.
Let C divide BA in the ratio x: y where x + y = 1. In particular, if C is the mid-point of BA, then x = y = fr. Now
c=OB+BC
= b + x(a - b)
=xa+(1 - x)b so that c = xa + yb
Now conversely, assume c = xa + yb, x + y = 1.
(5)
Then
c=xa+(1 -x)b =x(a-b)+b We now note that c is a vector that is obtained by adding to b the
vector x(a - b), this latter vector being parallel to the vector a - b. This immediately implies that the end point of c lies on the line joining A to B. We can rewrite (5) as
c-xa-yb=0
1-x-y=0
(6)
SEC. 61
THE ALGEBRA OF VECTORS
5
We have proved our first important theorem. A necessary and sufficient condition that the end points of any three vectors with common origin be on a straight line is that real constants it m, n exist such that
la+mb+nc=0
l+m+n=O
(7)
with l2+m2+n2p,, 0. We shall, however, find (5) more useful for solving problems. Example 1. Let us prove that the medians of a triangle meet at a point P which divides each median in the ratio 1:2. C
0 FIG. 9.
Let ABC be the given triangle and let A', B', C' be the midChoose 0 anywhere in space and construct the vectors from 0 to A, B, C, A', B', C', calling them a, b, c, a', b', c' (see Fig. 9). From (5) we have points.
a'=4b+ic b' = Ja + 4c
(8)
Now P (the intersection of two of the medians) lies on the line joining A and A' and on the line joining B and B'. We shall thus find it expedient to find a relationship between the four vectors a, b, a', b' associated with A, B, A', B'. From (8) we eliminate the vector c and obtain 2a' + a = 2b' + b or
*a' + '}a = *b' + b
(9)
6
VECTOR AND TENSOR ANALYSIS
[SFC. 6
But from (5), -sa' + -ia represents a vector whose origin is at 0 and whose end point lies on the line joining A to A. Similarly, ,'jb' + *b represents a vector whose origin is at 0 and whose end point lies on the line joining B to B'. There can only be one
y
vector having both these properties, and this is the vector p = OP. Hence p = tea' + is = -b' -fib. Note that P divides AA' and
BB' in the ratios 2: 1. Had we considered the median CC' in connection with AA', we would have obtained that p + J e, and this completes the proof of the theorem. Example 2. To prove that the diagonals of a parallelogram bisect each other. Let ABCD be the parallelogram and 0 any C
FIG. 10.
point in space (see Fig.
10).
The equation d - a = c - b
implies that ABCD is a parallelogram.
Hence
Ja+4c=lb+Id =p so that P bisects AC and BD. Problems
1. Interpret I a a 2. Give a geometric proof of (3). 3. a, b, c are consecutive vectors forming a triangle. What is the vector sum a + b + c? Generalize this result for any closed polygon.
4. Vectors are drawn from the center of a regular polygon to its vertices. From symmetry considerations show that the vector sum is zero.
SEc. 61
THE ALGEBRA OF VECTORS
7
5. a and bare consecutive vectors of a parallelogram. Express the diagonal vectors in terms of a and b. 6. a, b, c, d are consecutive vector sides of a quadrilateral. Show that a necessary and sufficient condition that the figure be
a parallelogram is that a + c = 0 and show that this implies
b + d = 0. 7. Show graphically that lal + lbl >_ la + bl. From this show that la - bl >_ lal - lbi. 8. a, b, c, d are vectors from 0 to A, B, C, D.
If
b-a=2(d-c) show that the intersection point of the two lines joining A and D and B and C trisects these lines. 9. a, b, c, d are four vectors with a common origin. Find a necessary and sufficient condition that their end points lie in a plane.
10. What is the vector condition that the end points of the vectors of Prob. 9 form the vertices of a parallelogram? 11. Show that the mid-points of the lines which join the midpoints of the opposite sides of a quadrilateral coincide. The four sides of the quadrilateral are not necessarily coplanar. 12. Show that the line which joins one vertex of a parallelogram to the mid-point of an opposite side trisects the diagonal. 13. A line from a vertex of a triangle trisects the opposite side. It intersects a similar line issuing from another vertex. In what ratio do these lines intersect one another? 14. A line from a vertex of a triangle bisects the opposite side. It is trisected by a similar line issuing from another vertex. How does this latter line intersect the opposite side? 15. Show that the bisectors of a triangle meet in a point. 16. Show that if two triangles in space are so situated that the three points of intersection of corresponding sides lie on a line, then the lines joining the corresponding vertices pass through a common point, and conversely. This is Desargues's theorem.
17. b = (sin t)a is a variable vector which always remains parallel to the fixed vector a. What is rically the meaning of
A.
fib?
Explain geomet-
8
VECTOR AND TENSOR ANALYSIS
[SEC. 7
18. Let v, be the velocity of A relative to B and let v2 be the velocity of B relative to C. What is the velocity of A relative to C? Of C relative to A? Are these results obvious? 19. Let a, b be constant vectors and let c be defined by the equation c = (cos t)a + (sin t)b When is c parallel to a? Parallel to b? Can c ever be parallel do d2c If a and b to a + b? Perpendicular to a + b? Find 9
dt dt2
are unit orthogonal vectors with common origin, describe the positions of c and show that c is perpendicular to
dc
20. If a and b are not parallel, show that ma + nb = ka + 3b implies m = k, n = j. 21. Theorem of Ceva. A necessary and sufficient condition that the lines which join three points, one on each side of a triangle,
to the opposite vertices be concurrent is that the product of the algebraic ratios in which the three points divide the sides be -1. 22. Theorem of Menelaus. Three points, one on each side of a
triangle ABC, are collinear if and only if the product of the algebraic ratios in which they divide the sides BC, CA, AB is unity. 7. Coordinate Systems. For a considerable portion of the text we shall deal with the Euclidean space of three dimensions. This is the ordinary space encountered by students of analytic geometry and the calculus. We choose a right-handed coordinate system. If we rotate the x axis into the y axis, a right-hand screw will advance along the positive z axis. We let i, j, k be the three unit vectors along the positive x, y, and z axes, respectively. The vectors i, j, k form a very simple and elegant basis for our three-dimensional Euclidean space. From Fig. 11 we observe that
r=xi+yj+zk
(10)
The numbers x, y, z are called the components of the vector r. Note that they represent the projections of the vector r on the x, y, and z axes. r is called the position vector of the point P
SEC. 71
THE ALGEBRA OF VECTORS
9
and will be used quite frequently in what follows. The most general space-time vector that we shall encounter will be of the form
u =u(x,y,z,t) =a(x,y,z,t)1+#(x,y,z,t)j + y(x, y, z, t)k
(11)
It is of the utmost importance that the student understand the meaning of (11). To be more specific, let us consider a fluid in motion. At any time t the particle which happens to be at the z
Fm. 11.
point P(x, y, z) will have a velocity which depends on the coordinates x, y, z and on the time t. As time goes on, various particles arrive at P(x, y, z) and have the velocities u(x, y, z, t) with components along the x, y, z axes given by a(x, y, z, t), Xx, y) z, t), y(x) y, z, t).
Whenever we have a vector of the type (11), we say that we An elementary example would be the vector u = yi - xj. This vector field is time-independent and so is have a vector field.
called a steady field.
At the point P(1, -2, 3) it has the value
- 2i - j. Another example would be u = 3xzeq - xyztj + 5xk. We shall have more to say about this type of vector in later
VECTOR AND TENSOR ANALYSIS
10
[SEC. 8
chapters and will, for the present, be interested only in constant vectors (uniform fields).
A moment's reflection shows that if a = ali + a2j + ask, b = b1i + ba + b3k, then a + b = (ai + b1)i + (a2 + b2)j + (as + b3)k (12) xa + yb = (xai + yb1)i + (xaz + ybz)j + (xaa + yb3)k 8. Scalar, or Dot, Product. We define the scalar or dot product of two vectors by the identity a b = JaJJbJ cos 0
(13)
where 0 is the angle between the two vectors when drawn from a common origin. It makes no difference whether we choose 0 or - 0 since cos 0 = cos (- 0). This definition of the scalar product
arose in physics and will play a dominant role in the development of the text.
Fm. 12.
From (13) we at once verify that
a a
ab
then
the
b
b=
a a
a
b
b
ab
= (proj a)b (bI = (proj b)a (al
(17)
SEC. 8J
THE ALGEBRA OF VECTORS
11
With this in mind we proceed to prove the distributive law, which
states that (18)
From Fig. 13 it is apparent that
[proj (b+c)]aIaF f
I
= (proj b)a Iai + (proj C)a Ial
Fm. 13.
since the projection of the sum is the sum of the projections. Let the reader now prove that
Example 3. To prove that the median to the base of an isosceles triangle is perpendicular to the base (see Fig. 14). From (5) we see that
m=1
1
so that
0 FIG. 14.
which proves that OM is perpendicular to AB. Example 4. To prove that an angle inscribed in a semicircle is a right angle (see Fig. 15).
_j.k:g=o
1'1
_
C2
co$B
2ab
-b2+a2_.
6
Example
that
$o
.]a
PIG.
cb
_a)
(b
a -a Ab
b2
_
C
C.
.
T'`iq°n°metr
of
Lau,
Cone
&
xample
E
18.
I°
j'i
c,a
c
z a
0
angle
aright
4$CAis
that
so
0
_
AC
BC
-,
$
+c
s
[SEc.
BC AND
c a
'
yAC
YECr?,oR
12
THE ALGEBRA OF VECTORS
SEC. 8]
13
Hence if a = a1i + a2j + ask, b = b1i + b2j + b3k, then
a b = albs + a2b2 + a,b3 Formula (19) is of the utmost importance. Example 7. Cauchy's Inequality
(19)
Notice that 0 a = 0.
(a - b)(a - b) _ 1a121b12 cost 0 5 1a12Ib12
so that from (19) (albs + aab2 + aab3)2 S (a12 + a22 + as2)(b12 + b22 + b 32)
In general n
n
n
I aaba s (Z aa2)1(I ba2)1
a-1
a-1
a-1
(20)
Let i' be a unit vector making angles a, fl, 7 with the x, y, z axes. The projections of i' on the x, y, z axes are cos a, cos ft, cos -y, so that Example 8.
i' = cosai+cos0 j+cos7k
=pli+qlj+rlk
(21)
pi, q1, ri are called the direction cosines of the vector Y. Similarly, let j' and k' be unit vectors with direction cosines p2, qa, r2 and pa, qa, r,. Thus Notice that p12 + q12 + r12 = 1.
j' = p2i + q2j + r2k
k' = pai+qsj+rak
()
We also impose the condition that i', j', k' be mutually orthogonal, so that the x', y', z' axes form a coordinate system similar to the x-y-z coordinate system with common origin 0 (see Fig. 17).
We have r = r' so that xi + yj + zk = x'i' + y'j' + z'k', where x, y, z are the coordinates of a point P as measured in the x-y-z coordinate system and x', y', z' are the coordinates of the
same point P as measured in the x'-y'-z' coordinate system. Making use of (21) and (22) and equating components, we find
that
x = p1x' + ply' + pgz' y = q1x' + qty' + qaz' z = r1x' + r2y' + raz'
(23)
14
VECTOR AND TENSOR ANALYSIS
1Src. 8
We now find it more convenient to rename the x-y-z coordinate system. Let x = x1, y = x2, z = x3, where the superscripts do not designate powers but are just labels which enable us to differentiate between the various axes. Similarly, let x' = x1, y' = x2'
y
FIG. 17.
Z' = 28.
Now let a,a represent the cosine of the angle between the xa and V axes. We can write (23) as 3
xa = I a0a.V,
a = 1, 2, 3
(24)
a-1
By making use of the fact that i' j' = j' k' = k' i' = 0, we can prove that 3
xa = I Asax#, 0-1
a = 1, 2, 3
(25)
where Asa = a). We leave this as an exercise for the reader. Let us notice that differentiating (24) yields axa
- = a,a,
of, or = 1, 2, 3
(26)
SEC. 91
THE ALGEBRA OF VECTORS
15
Example 9. The vector a = a'i + a2j + a'k may be represented by the number triple (a', a2, a'). Hence, without appealing to geometry we could develop an algebraic theory of vectors.
If b = (b', b2, b3), then a + b is defined by the number triple (a' + b', a2 + b2, a3 + b3), and xa = x(al, a2, a') is defined by the number triple (xa', xa2, xa3). From this the reader can prove
that (a', a2, a3) = a'(1, 0, 0) + a2(0, 1, 0) + a3(0, 0, 1)
The triples (1, 0, 0), (0, 1, 0), (0, 0, 1) form a basis for our linear vector space, that is, the space of number triples. We note that the determinant formed from these triples, namely, 1
0
0
1
00
0
0 =1 1
does not vanish. Any three triples whose determinant does not vanish can be used to form a basis. Let the reader prove this result. We can define the scalar product (inner product) of two triples by the law (a - b) = a'b' + a2b2 + a3b3. 9. Applications of the Scalar Product to Space Geometry (a) We define a plane as the locus of lines passing through a
fixed point perpendicular to a fixed direction. Let the fixed point be Po(xo, yo, zo) and let the fixed direction be given by the vector N = Ai + Bj + Ck. Let r be the position vector to any -4
point P(x, y, z) on the plane (Fig. 18). Now POP = r - ro is perpendicular to N so that or
[(x - xo)i + (y - yo)j + (z - zo)kl - (Ai + Bj + Ck) = 0 and
A(x - xo) + B(y - yo) + C(z - zo) = 0
(27)
This is the equation of the plane. The point Po(xo, yo, zo) obviously lies in the plane since its coordinates satisfy (27). Equation (27) is linear in x, y, Z. (b) Consider the surface Ax + By + Cz + D = 0. Let P(xo, yo, zo) be any point on the surface. Of necessity,
Axo+Byo+Czo+D=0.
16
VECTOR AND TENSOR ANALYSIS
[SEC.9
Subtracting we have
A(x - xo) + B(y - yo) + C(z - zo) = 0
(28)
Now consider the two vectors Ai + Bj + Ck and
(x - xo)i + (y - yo)j + (z - zo)k
Fia. 18.
Equation (28) shows that these two vectors are perpendicular. Hence the constant vector Ai + Bj + Ck is normal to the surface at every point so that the surface is a plane.
Fia. 19.
(c) Distance from a point to a plane. Let the equation of the plane be Ax + By + Cz + D = 0, and let P(E, % r) be any point in space. We wish to determine the shortest distance from P
THE ALGEBRA OF VECTORS
SEC. 91
17
to the plane. Choose any point Po lying in the plane. It is apparent that the shortest distance will be the projection of
PoP on N, where N is a unit vector normal to the plane (see Fig. 19).
Now
d = JPoP NJ =
IAZ+B, +Cr+DI
(29)
(A2 + B2 + C2)f
where use has been made of the
fact that
Axo+Byo+Czo+D=O.
FIG. 20.
(d) Equation of a straight line through the point Po(xo, yo, zo)
parallel to the vector T = li + mj + A. From Fig. 20 it is z
Fia. 21.
VECTOR AND TENSOR ANALYSIS
18
[SEC. 9
apparent that r - ro is parallel to T so that r - ro = AT,
-- u. 20. If u(tx, ty, tz) = t"u(x, y, z), show that (r V)u = nu. 20. The Divergence of a Vector. Let us consider the motion of a fluid of density p(x, y, z). We assume that the velocity field is given by f = u(x, y, z)i + v(x, y, z)j + w(x, y, z)k. This type of motion is called steady motion because of the explicit independence of p and f on the time. We now concentrate on the flow through a small parallelepiped ABCDEFGH (Fig. 35), of dimensions dx, dy, dz.
Let us first calculate the amount of fluid passing through the face ABCD per unit time. The x and z components of the velocity f contribute nothing to the flow through ABCD. The mass of fluid entering ABCD per unit time is given by pv dx dz. The mass of fluid leaving the face EFGH per unit, time is
r
pv +
a(pv) 1 dy dx dz ay
SEC. 20)
DIFFERENTIAL VECTOR CALCULUS
43
The loss of mass per unit time is thus seen to be equal to a(pv)
ay
dx dy dz
If we also take into consideration the other two faces, we find that the total loss of mass per unit time is
[a(Pu)+o(Pv)+a(Pw)]dddz z
A
dy
E
ry
Fia. 35.
so tha t a(Pu) ax
+
a(pv)
ay
+
(66)
a(Pw) az
represents the loss of mass per unit time per unit volume. This quantity is called the divergence of the vector pf. We see at once that VV
. (pf )
= di v (pf) =
a(Pu) ax
a(Pv)
+
ay
+
a(Pw) az
= 1 dM V dt
(67 )
VECTOR AND TENSOR ANALYSIS
44
[SEc. 20
since i, j, k are constant vectors. M and V are the mass and volume of the fluid.
The divergence of any vector f is defined as V f. We now calculate the divergence of p(x, y, z)f. V
a(q.u)
+ a(te) +
ax
ay
_
(ax
au
av
x
+ay+
aw az
a(cyw) az
arp
a'p
sip
+ uax +v ay +u'aa
(68)
We remember this result easily enough if we consider V as a vector differential operator. Thus, when operating on spf, we first keep p fixed and let V operate on f, and then we keep f fixed and let V operate on V(V V is nonsense), and since f and Vp are
vectors we complete their multiplication by taking their dot product. Example 25.
Compute V f if f = r/r' (inverse-square force).
V (r-'r) = r-3V r+r Vr-'
V (r- 3r) = 0
(69)
This is an important result. The divergence of an inverse-square force is zero. We note that y
Example 26.
What is the divergence of a gradient?
k/
axe+ayz+az,
DIFFERENTIAL VECTOR CALCULUS
SEc.21]
45
This important quantity is called the Laplacian of 0.
= v2 = axe
v
(70)
aye
az2
21. The Curl of a Vector. We postpone the physical meaning of the curl and define
curl f=vxf=
_ -J
Caw
vxf = i
ay
+
j
a
a
ax
ay
az
u
v
w
_ awl
1
az
i
a
+k
ax
az
_ au
1
ax
(71)
ay,
Example 27
=0
Vxr= Example 28 i
j
k
a
a
a
ax
ay
az
cpu
vv cv
1
v x (wf) =
[a(te) _ a()1 ay
az
JJ
[a(sou) az
a(caw)
ax
+ k [a(;)
a(,Pu)]
ay
ax v x (cOf) = cO
aw
[i (ay
av
az
+j
au
Ow
az
ax
U
Vx('vf)=(Pvxf+Via xf
v
w
(72)
46
VECTOR AND TENSOR ANALYSIS
[SEC. 21
This result is easily obtained by considering V as a vector differential operator. Example 29. To show that the curl of a gradient is zero. i a V X (VP) = ax
j
k
a ay
a az
i
av av av ax
ay
- az ay ay az a2
a2IP
az 02,p
a2,p
+ i
\ az ax
ax az/ + k \ax ay a2,
C1 2,p
ay ax/
Hence
Vxvio =0
(73)
provided (p has continuous second derivatives. Example 30. To show that the divergence of a curl is zero.
a au_aw'
a aw_av
V (v x f) = ax ay
az
a2u
a2ul
ay az
az ay
+ ay az a2V
ax /
a(av_au) + az \ax a2w
a2v
+ az ax
ax az
+
ay/ a2w
ax ay - ay ax
Thus (74)
Example 31. V.
What does (u V)v mean? We first dot u with
This yields the scalar differential operator a
a
a
ux ax + ua ay + U. az Then we operate on v obtaining uxax+Uyay+u$az
Thus
df=axdx+af dy+azdz y
=dxa+dya+dzaz y
DIFFERENTIAL VECTOR CALCULUS
SFC. 211
47
and
df = (dr V)f
(75)
since dr = dx i + dy j + dz k. If f = f(x, y, z, t),
df = (dr V)f + a dt
(76)
Example 32
+vsa = vJ + vyj + vZk (v - V)r = v
(77)
where r is the position vector xi + yj + zk. Example 33. Let us expand V(u v). Now u x (V x v) = Vti(u
v) - (u V)v
Here we have applied the rule of the middle factor, noting also that V operates only on v. Vn(u v) means that we keep the components of u fixed and differentiate only the components of v. v) - (v V)u. Adding, we Similarly, v x (V x u) = obtain
Vu(u v) + Vn(u v) = u x (V x v) + v
x u)
+ (u V)v + (v V)u and
u x(V xv) + v x(V xu) +
(78)
Example 34
V x (u x v) = (v V)u - v(V u) + u(V v) - (u
V)v
(79)
48
VECTOR AND TENSOR ANALYSIS
[SEc. 22
Example 35
V. (u xv) = Vu (u xv) +Vv (u xv) (V
(80)
(V
Example 36
V X (V x v) = V(V v) - V2v Let A = V x ((pi) where V2,p = 0. pute A V x A. Since A = V(p x i, we obtain Example 37.
V x A = V x V xi =
V
iV2
(81)
We now com-
a cix
from (7), Sec. 22. Thus
AVxA=
i
j
k
app
app
app
a2
ax
ay
az
ax2+j ay ax+kazax
1
0
0
app a2 = az ay ax
a2y
()2
-P
app a2g
ft azax
If also sp = X(x)Y(y)Z(z), we can immediately conclude that 22. Recapitulation. We relist the above results: 1. V (UV) = u Vv + v Vu 2. V
V
xv+V(pxv
4. V x (V(p) = 0
5. V. (Vxv)=0 6. V. (u x v) _ (V x u) - v - (V x v) - u
7. V x (u x v) _ (v V)u - v(V u) + u(V v) - (u V)v 8.
V
Vxr
=0 13. df = (dr V)f + a dt
Sxc. 22]
DIFFERENTIAL VECTOR CALCULUS
14. dco = dr VV +
49
LP dt
15. V- (r 3r) =0 Problems
1. Show that V2(1/r) = 0 where r = (x2 + y2 + z2)}. 2. Compute V2 r, V2r2, V2(1/r2) where r = (x2 + y2 + z2)3. 3. Expand V(uvw).
4. Find the divergence and curl of (xi - yj)/(x + y); of
xcoszi+ylogxj -z2k.
5. If a = axi + fiyj + yzk, show that V(a r) = 2a. 6. Show that V x V(r)r] = 0 when r = (x2 + y2 + Z2) i and
r = xi + yj + zk. 7. Let u = u(x, y, z), v = v(x, y, z). Suppose u and v satisfy an equation of the form f(u, v) = 0. Show that Vu x Vv = 0. 8. Assume Vu x Vv = 0 and assume that we move on the surface u (x, y, z) = constant. Show that v remains constant and hence v = f(u) or F(u, v) = 0. 9. Prove that a necessary and sufficient condition that u, v, w satisfy an equation f(u, v, w) = 0 is that Vu Vv x Vw = 0, or au
au
au
ax
ay
az
av
ft
av
ax
ay
az
0
aw aw aw ax
ay
az
This determinant is called the Jacobian of (u, v, w) with respect to (x, y, z), written J[(u, v, w)/(x, y, z)].
10. If w is a constant vector, prove that V x (w x r) = 2w,
where r=xi+yJ+zk.
11. If pf = Vp, prove that f V x f = 0. 12. Prove that (v V)v = 4 Vv2 - v x (V x v). 13. If A is a constant unit vector, show that V x (v xA)] = 14. If fl, f2, f3 are the components of the vector f in one set of
rectangular axes and 11, /, / are the components of f after a
50
VECTOR AND TENSOR ANALYSIS
[SEC. 23
rotation of axes (see Example 8), show that a fa
a=1
xa
aa_
=
a=1 x
so that V f is a scalar invariant under a rotation of axes.
Also
see Prob. 21, Sec. 9. 15. Prove (79), (80), (81). 16. Let f = f1i + f2j + f3k and consider nine quantities afi axi
9is
of, axis
i, j = 1, 2, 3
Show that gi; = -gi and that three of the nine quantities yield the three components of V x f. Use this result to show that
V x((pf) =('V xf+Vcpxf. 23. Curvilinear Coordinates. Often the mathematician, physicist, or engineer finds it convenient to use a coordinate system other than the familiar rectangular cartesian coordinate system. If he is dealing with spheres, he will probably find it expedient to describe the position of a point in space by the spherical coordinates r, 8, sp (see Fig. 31). Let us note the following: The sphere x2 + y2 + z2 = r2, the cone z/(x2 + y2 + z2)'1 = cos 8, and the plane y/x = tan sp pass through the point P(r, 0, Sp). We may consider the transformations
r = (x2+y2+Z2)} a = cos- ' Sp =
z
(x2 + y2 + z2)
tan-1 y
X
as a change of coordinates from the x-y-z coordinate system to the r-B-#p coordinate system. The surfaces r = (x2 + y2 + z2)} = cl 8 = cos-' [z/(x2 + y2 + z2)}1 = c2, O = tan-' y/x = c3 are respect tively, a sphere, cone, and plane. Through any point P in space, except the origin, there will pass exactly one surface of each type, the coordinates of the point P determining the constants c1, c2, c3. The intersection of the sphere and the cone is a circle, the circle of latitude, having e, as its unit tangent vector at P. This circle is called the (p-curve since r and 0 remain constant on this curve
SEC. 23]
DIFFERENTIAL VECTOR CALCULUS
51
so that only the coordinate p changes as we move along this curve. The intersection of the sphere and the plane yields the 0-curve, the circle of longitude, while the intersection of the cone and plane yields the straight line from the origin through. P, the r curve. ee and er are the unit tangent vectors to the 0- and r curves, respectively. The three unit vectors at P, er, ee, e, are mutually perpendicular to each other and can be considered as forming a basis for a coordinate system in the neighborhood of P. Unlike i, j, k, they are not fixed, for as we move from point to point their directions change. Thus we may expect to find more
complicated formulas for the gradient, divergence, curl, and Laplacian when dealing with spherical coordinates. Since Vv is perpendicular to the plane cp = constant, we must have V(p parallel to e, Hence e9, = h3 Vsp, where hs is the scalar factor of proportionality between e,, and Vp. If drs is a vector tangent to the p-curve, of length dss = jdr3j, we have from (58)
dip = drs VP = dr3
e,p
hs
=
ds3
hs
so that ds3 = hs dip.
Hence h3 is
that quantity which must be multiplied into the differential change of coordinate gyp, namely, d-p, to yield arc length along the p-curve.
Thus e,, = r sin 0 V(p, while similarly er = Vr and ee = r VB. We note that er = ee x e,, = r2 sin 0 VO X VV,
eB=e,xe,.=rsin 0V(pxVr
e,=erxee=rVrxVB
Any vector at P may be represented as f = flex + flee + f3ec. The scalars f1, f2, fs can be functions of r, 0, p. We may also
represent f as f = f1 Vr + f2r VO + far sin 0 VV and also by f = f1r2 sin O VO x V(p + f2r sin BV(p x Vr + far Vr x V8.
We also
note that the triple scalar product Vr V9 x Vtp is equal to (r2 sin 0)-1 and that dV = dsl ds2 ds3 = r2 sin 0 dr d9 dp. Spherical coordinates are special cases of orthogonal curvilinear coordinate systems so that we will proceed to discuss these more general coordinate systems in order to obtain expressions for the gradient, divergence, curl, and Laplacian. Let us make a change of coordinates from the x-y-z system to a u1-u2-u3 system as given by the equations Ul = u1(x, y, z) U2 = u2(x, y, z) ug = u3(x, y, z)
(82)
52
VECTOR AND TENSOR ANALYSIS
[SEC. 23
We assume that the Jacobian J[(ul, u2, u3)/(x, y, z)] ;P,-, 0 so that
the transformation (82) is one to one in the neighborhood of a point. A point in space is determined when x, y, z are known and hence when ui, u2, u3 are known. By considering ui(x, y, z) = ci
u2(x, y, z) = C2, u3(x, y, z) = es, we obtain a family of surfaces. Through a point P(xo, yo, zo) will pass the three surfaces ui(x, y, z) = ui(xo, yo, zo) z
y
FIG. 36.
u2(x, y) z) = u2(xo, yo, Zo), and u3(x, y, z) = u8(xo, yo, zo). Let us
assume that the three surfaces intersect one another orthogonally. The surfaces will intersect in pairs, yielding three curves which intersect orthogonally at the point P(xo, yo, zo). The curve of intersection of the surfaces uI = ci and u2 = C2 we shall call the us curve, since along this curve only the variable us is allowed to change. Let u1, u2, us be three unit vectors issuing from P tangent to the ui, us, us curves, respectively (see Fig. 36).
DIFFERENTIAL VECTOR CALCULUS
SEc. 231
53
Now Vu3 is perpendicular to the surface u3(x, y, z) = u3(xo, yo, zo)
so that Vu3 is parallel to the unit vector u3. Hence u3 = h3 Vu3 where h3 is the scalar factor of proportionality between us and Vu3.
Now let dr3 be a tangent vector along the us curve,
dr3' = dss.
Obviously dr3 us = dss, and
so that from (58) dss = h3 du3
(83)
We see that hs is that quantity which must be multiplied into the differential coordinate dus so that are length will result. For example, in polar coordinates ds = r do if we move on the 0-curve,
so that r = h2. Similarly, ul = h1 Vu1j u2 = h2 Du2, so that u1 = U2 X u3 = h2h3 Vu2 X Vu3 u2 = us x u1 = h3h1 Vu3 X Vu1 113 = u1 x u2 = h1h2 Vu1 x out
(84)
and
Du1 Du2 X Vu3 =
u1 U2
h1 h g
x
Us s
= (h1h2ha)-1
(85)
Note that the differential of volume is dV = ds1 ds2 ds3 = h1h2h3 du1 due du3
and making use of (85) as well as Prob. 9, Sec. 22,
dV = j (_x, y, z ) du1 due du3 u1, U2, u3
Example 38.
In cylindrical coordinates ds2 = dr2 + r2 d02 + dz2
so that h1 = 1, h2 = r, h3 = 1.
(86)
VECTOR AND TENSOR ANALYSIS
54
[SEC. 23
If f = f (U1, u2, u3), then from Example 21,
Example 39.
of
yf =
vu3 Vul + au2 ` f vu2 + of au3
au1
of
of
of
Vf = h1 au1 u1+----u2+-''U3 h3 au3 h2 49u2 1
1
1
(87)
In cylindrical coordinates
of
ar
R+p+k r a9
az
Our next attempt is to obtain an expression for the divergence
of a vector when its components are known in an orthogonal curvilinear coordinate system. Now f = flul + f2u2 + f3u3
= f lh2ha Vu2 x vu3 + f 2h3h1 Vu3 x Vul + fahlh2 Vu1 x vu2
from (84).
Consequently
V f = V (f 1h2h3) vu2 x Vu3 + f 1h2h3 V (Vu2 X Vu3) + V(f 2h3h1) Dua x Vu1 + f 2h3hly (Vu3 X Vu1) + V(f3h1h2) Vu1 X Vu2 + f3h1h2V (Vul x vu2)
Now V(f1h2h3) VU2 X Vu3 =
(88)
a(flh2ha ) Vu1 Vu2 x Vu3, and out
V (vu2 x vua) = 0 ao that (88) reduces to [o(h2h3fi)
1
h1h2h3
49U,
+
a(h3h1f2) au2
h,h2f3)
+
(89
49u3
)
If we apply (89) to the vector VV as given by (87), we obtain
VV2
1
a
h1h2h3 au1
h2h3 a
{
h1 49U,
a (hAi a +au2
h2 au2
a (h,h2 a
+ aus
h3
au] (90)
55
DIFFERENTIAL VECTOR CALCULUS
SEC. 23]
This is the Laplacian in any orthogonal curvilinear coordinate system. Example 40. V2V
In cylindrical coordinates
l r
a
ar
r
a
ar
-I-
a
1a_
aB
r aB
+az- r aza
a
(91)
Example 41. Solve V2V = 0 assuming V = V(r),
r = (x2 + y2)i From (91)
1 d (LV r J =0 dr r dr
r
or
ddV = c1
and
V =cllogr+C2 Finally we obtain the curl of f. f = f1u1 + f2u2 + faun = f 1h1 Vu1 + f2h2 Due + fshs Vua and
V x f = V(f1hi) x Vul + V(f2h2) X Vu2 + V(faha) X Vus
since V x (Vul) = V X (Vu2) = V x (Vua) = 0. Now V(f1h1) X VU, =
a(f1h1)
49U,
Vul X Vul +
a(f1h1)
Vu2 X Vu1
49U2
+ a(flhl aus
Vus X Vul
us
Replacing Vu2 x Vul by -- hlh2, etc., we obtain
Vxf=
ul f a(hsf3) h2h3 L
au2
u2 ra(hlfl) ^ a(hsfa) )1 - a(h2f2 aul aua J + hsh1 L aua ua ` a(h1f1) (92) La(h2f2)
+
1h2
49U1
49U2
56
VECTOR AND TENSOR ANALYSIS
[Sec. 23
Problems 1. For spherical coordinates, ds2 = dr2 + r2 d92 + r2 sine 9 d(p2 where B is the colatitude and (p the azimuthal angle. Show that V Z V=
a I r2 sin B a r2 sin 9 ar \\\ ar 1
I
a
}-
(sin a9 `
6
aV aB/J
+ a (sin 9 2. Solve V2V = 0 in spherical coordinates if V = V(r). 3. Express V - f and V x f in cylindrical coordinates. 4. Express V f and V x f in spherical coordinates by letting a, b, c be unit vectors in the r, 9, (p directions, respectively. 5. Write Eq. (92) in terms of a determinant. 6. Show that V x [(r V9)/sin 0) = V(p where r, 0, (p are spherical coordinates. 7. If a, b, c are the vectors of Prob. 4, show that as
as
ar=0'
a9= b,
ab
=0
ab
as =sin9c
= -a
ar
a9
ac
ac
ar=0'
a9=0'
ab
= cos 9 c
a(P
_
ac
= - sin0a - cos0b
8. If x = r sin 9 cos (p, y = r sin 9 sin (p, z = r cos 9, then the form ds2 = dx2 + dy2 + dz2 becomes ds2 = dr2 + r2 d92 + r2 sin2 9 d(p2 3
Prove this. If, in general, ds2 = I (dxa) 2, and if a-1
xa = xa(y', y2, y3)
a = 1, 2, 3, show that axa axa
ds2 = a,
..ydyOdy
_ I go, dys X-f
Y
dyo dyr dy''
SEC. 231
DIFFERENTIAL VECTOR CALCULUS
57
where 3
90Y = aI
ax" axa
a" ay"
Check this result for the transformation to cylindrical coordinates:
x = r cos 0 y = r sin 0 z=z and obtain ds2 = dr2 + r2 d02 + dz2.
9. By making use of V2V = V(V V) - V x (V x V), find V2V for V = v(r)e,, V being purely radial (spherical coordinates). Find V2V for V = f(r)e, + (p(z)e, in cylindrical coordinates. 10. Find V IV if V = w(r)k x r. 11. Consider the equations a2S
(X -f 'U)V(V
X, µ, p constants.
s) + u V2s = p ate
Assume s = eiP'sl, p constant, and show that
(X + p)V(V s1) + (A + pp2)sl = 0
Next show that [V2 + (µ + pp2)/(X + µ)](V s1) = 0, X+µ
0.
12. If A = V x (¢r), V2¢ = 0, show that 1
a a2Y'
sin 0 appaOar
a¢ a2Y' a0acpar
so that A V x A = 0 if, moreover, ¢ = 13. Show that Cpi = Ae9 + Bey + Ce° satisfies V2sp1 = ci, and show that if 402 satisfies V°02 = 0, then rp = Cpl + 4p2 also satisfies V2ip = gyp. Find a solution of V2Sp = -(p.
CHAPTER 3 DIFFERENTIAL GEOMETRY
24. Frenet-Serret Formulas. A three-dimensional curve in a Euclidean space can be represented by the locus of the end point of the position vector given by
r(t) = x(t)i + y(t)j + z(t)k
(93)
where t is a parameter ranging over a set of values to < t < ti. We assume that x(t), y(t), z(t) have continuous derivatives of all orders and that they can be expanded in a Taylor series in the neighborhood of any point of the curve.
We have seen in Chap. 2, Sec. 16, that ds is the unit tangent vector to the curve. Let t =
ds-
Now t is a unit vector so that
its derivative is perpendicular to t. Moreover, this derivative, dse
tells us how fast the unit tangent vector is changing direction
as we move along the curve. The principal normal to the curve is consequently defined by the equation dt
= Kn
(94)
ds
where K is the magnitude of ds and is called the curvature.
The
reciprocal of the curvature, p =I 1K, is called the radius of curvature. It is important to note that (94) defines both K and n,
K being the length of ds while n is the unit vector parallel to dt At any point P of our curve we now have two vectors t, n at ds right angles to each other (see Fig. 37). This enables us to set up 58
DIFFERENTIAL GEOMETRY
SEC. 24]
59
a local coordinate system at P by defining a third vector at right angles to t and n. We define as the binormal the vector
b = t xn All vectors associated with the curve at the point P can be written as a linear combination of the three fundamental vectors t, n, b, which form a trihedral at P. z
0
Y
x Fia. 37.
Let us now evaluate - -- and
dn
Since b is a unit vector, its
derivative is perpendicular to b and so lies in the plane of t and n. Moreover, b t = 0 so that on differentiating we obtain
t=
0.
Hence `
must be parallel to n.
is also perpendicular to t so that
dd_b
Consequently, ds = rn, where r by defini-
tion is the magnitude of -. r is called the torsion of the curve.
VECTOR AND TENSOR ANALYSIS
60
[SEC. 24
Finally, to obtain dn, we note that n = b x t so that b
ds
b x ds +
x t = b x Kn + rn x t =- Kt- rb
The famous Frenet-Serret formulas are dt
Kn ds
do
- (Kt + Tb)
ds
95)
db rn ds
Successive derivatives are functions of t, n, b and the derivatives of K and r. Example 42.
The circular helix is given by
r = a cos ti+asintj+btk t =ds = (-a sin ti+acostj + bk) st and
(dl (a2sin2t+a2cos2t+b2) 2
1=
(.)2 (a2 + b2)
Hence
t = (-a sin ti+acostj+bk)(a2+b2) 4
Now
Kn = d = (-a cos t i - a sin t j)(a2 + b2)-t so that K = a(a2 +
b2)-i
Also
i
j
b = t x n = -a sin t a cost -cos t -sin t
k b
(a2 + b2)-1
0
= (b sin t i - b cos t j + ak) (a2 +
b2)-+
DIFFERENTIAL GEOMETRY
SEC. 241
61
and
db ds
= rn = (b cos t i + b sin t j)(a2 + b2)-1
so that T = b(a2 + b2)-1
Problems
1. Show that the radius of curvature of the twisted curve x = log cos 0, y = log sin 0, z = V2 0 is p = csc 20. 2. Show that r = 0 is a necessary and sufficient condition that a curve be a plane curve.
(r'r"r"').
3. Prove that T = K
4. For the curve xz = a(3t - te), y = 3at2, z = a(3t + t3), show that K = T = 1/3a(1 + t2)2.
5. Prove that
AA _= ds . ds
Kr,
do db d8 - ds
=
0,
dt dn _ d8 . ds
= 0.
6. Prove that r"' = -K2t + K'n - rKb, where the primes mean differentiation with respect to are length.
7. Prove that the shortest distance between the principal normals at consecutive points at a distance ds apart (s measured along the arc) is ds p(p2 + 8. Find the curvature and torsion of the curve r`2)_;
y = all - cos u),
x = a(u - sin u),
z = bu 9. For a plane curve given by r = x(t)i + y(t)j, show that
x,y - y,x
[(x')2 + (y1)2]1
10. Prove that (t't"t"') = K5
ds (K)
11. Show that the line element ds2 = dx2 + dy2 + dz2 - c2 dt2 remains invariant in form under the Lorentz transformation
x=
- yt [1
- (V2/c2)1I
y=I
z=2 t=
- (V/c2)x [1 - (V2/c2)J*
62
VECTOR AND TENSOR ANALYSIS
[SEC. 25
V, c are constants. The transformation ct = iT, i leads to the four-dimensional Euclidean line element
ds2=dx2+dy2+dz2+dr2 12. If xa = xa(s), a = 1, 2, . . . , n, represents a curve in an n-dimensional Euclidean space for which d82
= (d_-1)2 + (dx2) 2 +
.
. + (dxn) 2
define the unit tangent vector to this curve, this definition being a generalization of the definition of the tangent vector for the case
n = 3. Show that the vector
d2xa
ds2
) a = 1, 2,
.
.
.
, n, is normal
to the tangent vector, and define the unit principal normal n, and curvature K, by the equations d2xa
dta
d82 - ds =
Klnla,
a = 1, 2,
..,n
a
Show that to
a-1
ds
a = 1, 2,
dnla = - K1. !Is
. . .
Define the second curvature K2 and unit
normal n2 by th e equati ons d
..
, n, is normal to nl and that
d
la
= -Klt a + K2n2 a, a =
1, 2,
i
. , n, and show that n2a is normal to to and nla if K2 ;P'- 0. Continue in this manner and obtain the generalization of the
Frenet-Serret formulas. 25. Fundamental Planes. The plane containing the tangent and principal normal is called the osculating plane. Let s be a variable vector to any point in this plane and let r be the vector to the point P on the curve. s - r lies in the plane and is conse-
quently perpendicular to the binormal. The equation of the osculating plane is
(s - r) b = 0
(96)
The normal plane to the curve at P is defined as the plane through P perpendicular to the tangent vector. Its equation is
DIFFERENTIAL GEOMETRY
SEC. 26]
63
easily seen to be
(s - r) t = 0
(97)
The third fundamental plane is the rectifying plane through P perpendicular to the normal n. Its equation is
(s - r) n = 0
(98)
Problems
1. Find the equations of the three fundamental planes for the curve
x = at,
y=bt2,
z=cta
2. Show that the limiting position of the line of intersection of two adjacent normal planes is given by (s - r) n = p where s is the vector to any point on the line. 26. Intrinsic Equations of a Curve. The curvature and torsion of a curve depend on the point P of the curve and consequently on the are parameter s. Let is = f(s), r = F(s). These two equations are called the intrinsic equations of the curve. They owe their name to the fact that two curves with the same intrinsic equations are identical except possibly for orientation in space. Assume two curves with the same intrinsic equations. Let the trihedrals at a corresponding point P coincide; this can be done by a rigid motion. Now ds
(t1. t2) = t1
,cn2 + xni t2
T (nl n2) = n1 (-Kt - rb2) + n2 (--Kt, -- rbi)
d Adding, we obtain
s
0
(99)
64
VECTOR AND TENSOR ANALYSIS
(SEC. 27
so that
constant = 3 since at P
tl=t2,
n1=n2,
(100)
b1=b2
Since (100) always maintains its maximum value, we must have dr, _ dr2 tl = t2, n1 n2, bi = b2 so that or r1 = r2 locally. ds
ds
Hence the two curves are identical in a small neighborhood of P. Since we have assumed analyticity of the curves, they are identical everywhere.
Problems
1. Show that the intrinsic equations of x = a(9 - sin 8), y = a(l - cos 8), z = 0 are p2 + s2 = 16a2, 7- = 0, where s is measured from the top of the are of the cycloid. 2. Show that the intrinsic equation for the catenary y=a'(ex/a+e-(sla)) 2
is ap = s2 + a2, where 8 is measured from the vertex of the catenary.
Fla. 38.
27. Involutes. Let us consider the space curve r. We construct the tangents to every point of r and define an involute
DIFFERENTIAL GEOMETRY
SEc. 271
65
as any curve which is normal to every tangent of r (see Fig. 38). From Fig. 39, it is evident that
r,=r+ut
(101)
is the equation of the involute, u unknown. Differentiating (101), we obtain
dr,
ds,^lt
A
Cdr
d_u \ ds
ds+ u ds+dst
ds,
where s is are length along r and s, is arc length along r'. (95), (102) becomes r
ti =
(t+uua+dut) d ds ds,
(102)
Using
(103)
Now t t, = 0 from the definition of the involute so that du 1 +=0
and
u=c - s Fta. 39.
(104)
Therefore r, = r + (c - s)t, and there exists an infinite family of involutes, one involute for each constant c. The distance between corresponding involutes remains a constant. An invo-
r
Fia. 40.
lute can be generated by unrolling a taut string of length c which has been wrapped along the curve. The end point of the string generates the involute (see Fig. 40). What are some properties of the involute?
VECTOR AND TENSOR ANALYSIS
66
[SEc. 28
r1=r+(c-s)t ti = dr,
dr / dt Ids + lC S) ds ds, -
(C - s)
t]
ds dsl
ds K - n ds1
Hence the tangent to the involute is parallel to the corresponding normal of the curve. Since ti and n are unit vectors, we must have (c - s) K d = 1. The curvature of the involute is l
do ds
dt1
obtained from - = Kin, ds= ds1 --= dsl
(-Kt -7-b)
Hence
(c - s)K
+ r2
r K12
K2
K2(C - 8)2
(105)
28. Evolutes. The curve t' whose tangents are perpendicular to a given curve is called the evolute of the curve. The tan-
gent to r' must lie in the plane of b and n of r since it is perpendicular to t. Consequently
Fla. 41.
rl=r+un+vb is the equation of the evolute.
tl
=
dr1
= dr
dsl -
{ds
Differentiating, we obtain
d_n db d_u dv A + u ds + v ds + ds n + d8 bl ds,
= It + u(-Kt -rb) +vTn+ds n +d8b]dsl Now t t1 = 0, which implies I - uK = 0 or u = dv) tl=L(-ru+as , b+( +ds)n
1
= p. Thus
K
d8
Also t1 is parallel to r1 - r = un + vb (see Fig. 41). Therefore (dv/ds) - UT
(du/ds) + Pr
u
V
DIFFERENTIAL GEOMETRY
SEC. 28]
67
or
T=
uv' - vu' u2+v2
= dsd- tan-' -vu)
Therefore
=
ra
o
Tds=tan--'U -C -V
and v = p tan (,p - c) since u = p. Therefore r1 = r + pn + p tan (,p - c)b
(106)
and again we have a one-parameter family of evolutes to the curve T.
Problems
1. Show that the unit binormal to the involute is Kb - Tt
b1 = (C - S)KK1
2. Show that the torsion of an involute has the value = T1
T-K
dS
] [K(K2 + r2)(C - 8)I-1
3. Show that the principal `normal to the evolute is parallel to the tangent of the curve 1'. 4. Show that the ratio of the torsion of the evolute to its curva-
ture is tan (,p - c). 5. Show that if the principal normals of a curve are binormals (equal vectors not necessarily coincident) of another curve, then c(K2 +,r2) = K where c is a constant. 6. On the binormal of a curve of constant torsion T, a point Q is taken at a constant distance c from the curve. Show that the binormal to the locus of Q is inclined to the binormal of the given curve at an angle
tan-'
CT2
K(C2r2 + 1)}
7. Consider two curves which have the same principal normals (equal vectors not necessarily coincident). Show that the tangents to the two curves are inclined at a constant angle.
VECTOR AND TENSOR ANALYSIS
68
[SEC. 29
29. Spherical Indicatrices
(a) When dealing with a family of unit vectors, it is often convenient to give them a common origin and then to consider
the locus of their end points. This locus obviously lies on a unit sphere. Let us now consider the spherical indicatrix of the tan-
gent vectors to a curve r = r(s).
The unit tangent vectors are t(s) = ds- Let r, = t. Then =
t'
dr,
= dt ds
ds1
ds dal -
ds
ds,
Thus the tangent to the spherical indicatrix P is parallel to the normal of the curve at the corresponding point. Moreover, Fla. 42.
1 = K d1' t, = n. Let us now find the curvature K, of the indicatrix.
We obtain do ds =K,n,='--=-(-Kt-rn) ds ds, as,
dt,
1
K
and K2 + r2 K12
2
K
(b) The spherical indicatrix of the binormal, r1 = b. entiating, dr, ds A ds = = re
t,=--- ds ds1
ds,
Therefore
rds=1
ds1
t,=n
and
ds,
Differentiating,
dt, dS1
= x,n1 =
do ds A dS;
_
1
r
and K12 =
K2 + r2 T2
`-Kt - rn)
Differ-
SEC. 30J
DIFFERENTIAL. GEOMETRY
69
Problems
1. Show that the torsion of the tangent indicatrix is T(dK/ds) - K(dr/ds) T2)
Ti
K(K2 +
2. Show that the torsion of the binormal indicatrix is Ti
T(dK/ds) - K(dT/ds) T(K2 r2)
+
3. Find the curvature of the spherical indicatrix of the principal normal of a given curve. 30. Envelopes. Consider the one-parameter family of surfaces F(x, y, z, c) = 0. Two neighboring surfaces are
F(x, y, z, c) = 0 and F(x, y, z, c + Ac) = 0. These two surfaces will, in general, intersect in a curve. But these equations are equivalent to the equations F(x, y, z, c) = 0 and F(x, y, z, c + Ac) - F(x, y, z, c) Ac
=0
where Ac # 0. As Ac --> 0, the curve of intersection approaches a limiting position, called the characteristic curve, given by
F(x, y, z, c) = 0 aF(x, y, z, c) ac
-0
(107)
Each c determines a characteristic curve. The locus of all these curves [obtained by eliminating c from (107)] gives us a surface called the envelope of the one-parameter family. Now consider two neighboring characteristics
F(x,y,z,c) = 0
aF(x,,z,c) ac
=0
and
(108)
F(x, y, z, c + Ac) = 0
aF(x, y, z, c + Ac) = 0
which, in general, intersect at a point.
ac
The locus of these points
is the envelope of the characteristics and is called the edge of
70
VECTOR AND TENSOR ANALYSIS
[SEc.31
The edge of regression is given by the three simultaneous equations F(x, y, z, c) = 0 regression.
aF(x, y, z, c) = 0 ac
(109)
a2F(x,y,z,c) = 0 ace
Example 43. Let us consider the osculating plane at a point P.
From (96) we have [s - r(s)] b(s) = 0. If we let P vary, we obtain the one-parameter family of osculating planes given by
f(x, y, z, s) = [s - r(s)] b(s) = 0 where s is the parameter and s = xi + yj + A. Now
of = as
dr ds
A = (s - r) b + (s -r).ds
of = 0, we obtain (s - r) n = 0. as plane.
This locus is the rectifying
The intersection of f = 0 and as = 0 obviously yields
the tangent lines which are the characteristics. a If
as2
in, and setting
Now
= -t n + (s - r) (-Kt - A) + (s - r) n da
It is easy Y to verify Y that s = r satisfies f =
a
as
=
az 4982
-
= 0, so that
the edge of regression is the original curve r = r(s). A developable surface, by definition, is the envelope of a oneparameter family of planes. The characteristics are straight lines, called generators. We have seen that the envelope of the
osculating planes is the locus of the tangent line to the space curve P.
In general, a developable surface is the tangent surface
of a twisted curve. A contradiction to this is the case of a cyiinder or cone.
31. Surfaces and Curvilinear Coordinates. the equations x = x(u, v) y = y(u, v) z = z(u, v)
Let us consider (110)
DIFFERENTIAL GEOMETRY
SEc. 32]
71
where u and v are parameters ranging over a certain set of values. If we keep v fixed, the locus of (110) is a space curve. For each v, one such space curve exists, and if we let v vary, we shall obtain
a locus of space curves which collectively form a surface. We shall consider those surfaces (110) for which x, y, z have continuous second-order derivatives. Equation (110) may be written
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k
(111)
where the end point of r generates the surface. The curves obtained by setting v = constant are called the u curves, and similarly the v curves are obtained by setting u = constant. The parameters u and v are called curvilinear coordinates, and the two curves are called the parametric curves. 32. Length of Arc on a Surface. If we move from the point r to the point r + dr on the surface, the distance ds is given by
(arudu+-dvN 2
ds2 =
or
2
C_ J
due + 2
9r
o9r
au . av
du dv + (av)2 dv2
or
ds2 = E du2 + 2F du dv + G dv2
(112)
where
E
\12
au/ '
F
ao9r .
u av'
2
G
(Ov)
Equation (112) is called the first fundamental form for the surface
r = r(u, v).
In particular, along the u curve, dv = 0, so that
(ds) = 1'E du and similarly
(113)
(ds), = VG_ dv Now
Or
and av are tangent vectors to the u and v curves, so
that the parametric curves form an orthogonal system if and only
Or
Or
au av
72
VECTOR AND TENSOR ANALYSIS
Example 44.
1Szc. 33
Consider the surface given by
r = r sin 8 cos v i+ r sin 8 sin V j+ r cos 8 k,
r = constant
Differentiating,
ar = r cos 0 cos sP i + r cos 0 sin V j - r sin 8 k d0 ar
a st
= -r sin6sinpi+rsin8coscpj
and is
E
a8/ =
c1r
r2,
F=-e
c1r
2
= 0,
G=
r2 Sin2 0
so that ds2 = r2 d82 + r2 sin2 0 dcp2 and the 0-curves are orthogonal to the 9-curves. Of course the surface is a sphere.
33. Surface Curves. By letting u and v be functions of a single variable t, we obtain
r = r[u(t), v(t)]
(114)
which represents a curve on the surface (111). Along this curve, (arau ar dvl dr = ` du + Wt dt. dr is completely determined when du dt av 1 and dv are specified, so that we will use the notation (du, dv) to specify a given direction on the surface. Now consider another
curve such that ar =
su au + av av, where su and av are the differential changes of u(t) and v(t) for this new curve. Now
dr or = E du au + F(du av + dv au) + G dv av
(115)
so that two curves are orthogonal if and only if
Edu su +F(duav +dv &u) +Gdv av = 0 or
dv E+FCa6Vu+au/+Gd -=0
(116)
If we have a system of curves on the surface given by the differential equation P(u, v) su + Q(u, v) av = 0, the differential equa-
DIFFERENTIAL GEOMETRY
SEC. 341
73
tion for the orthogonal trajectories is given by
E+F,(_P+dv1 - GPdv Q
``
by
P
bu
Q
since -
dull
Q du
=
0
(117)
Problems 1. Find the envelope and edge of regression of the one-param-
eter family of planes x sin c - y cos c + z tan 8 = c, where c is the parameter and 0 is a constant. 2. Show that any two v curves on the surface r = u cos v i + u sin v j + (v + log cos u)k cut equal segments from all the u curves. 3. Find the envelope and edge of regression of the family of x2
ellipsoids c2 (a2
+
y
z2
+ c2 = 1 where c is the parameter.
b2
4. If 8 is the angle between the two directions given by P due + Q du dv + R dv2 = 0
show that tan 0 = H(Q2 - 4PR)}/(ER - FQ + GP), where H
ar
8 avl.
8u
tvhl
5. Prove at the differential equations of the curves which bisect the angles between the parametric curves are
VEdu-VGdv=0 and 1/E du + 1i dv - 0. 6. Given the curves uv = constant on the surface r = ui + vj, find the orthogonal trajectories. 7. Show that the area of a surface is given by f f (EG
- F2)1 du dv
34. Normal to a Surface. The vectors
or
and av are tangent
to the surface r(u, v) along the u and v curves, respectively. Consequently,
ar
8r
au x 49V
,
is a vector normal to the surface. Note
VECTOR AND TENSOR ANALYSIS
74
[SEC. 35
that au need not be a unit tangent vector to the u curve since the
parameter u may not represent are length. Since
(ds),, = 1/E du a necessary and sufficient condition for u to be arc length is that E = 1. We define the unit normal to the surface as n _
(Or/au) x (ar/av) (118)
(ar/au) x (ar/av) 35. The Second Fundamental Form.
Consider all the planes
through a point P of the surface r = r(u, v) which contain the normal n. These planes intersect the surface in a family of curves, the normals to the curves being parallel to n. We now compute the curvature of any one of these curves in the direction (du, dv). Let ds be length of are along this curve. Now
dr Or du Or dv t=ds=auds+avds Therefore
d2r
dt
ds2
ds
_
492r
du 2
au2 (ds)
K"n
2
a2r d_u dv
a2r dv 2
+ au av A ds +
av2 (ds)
Or d2u
Or d2v
+
au ds2
+
av ds2
119)
and
(du)2
n)
since n
-
Or
au
(n au2)
n
- = 0. av
Kn =
ds+ 2 (n au2av) ds ds )z
Therefore
edue+2fdudv+gdv2 ds2
Kn
edue+2fdudv+gdv2 Edue+2Fdudv+Gdv2
(120)
DIFFERENTIAL GEOMETRY
SEC. 361
75
where we define a2r
e=n-au2'
a2r
a2r _n. auav' f
(121)
The quantity e due + 2f du dv + g dv2 is called the second fundamental form. Now consider any curve t on the surface and let its normal be n, at a point P, the direction of r being (du, dv) at P. Let r' be
Fia. 43.
the normal curve in the same direction (du, dv) with normal n at P (Fig. 43). We have
r" K
=n r"K
since n rl" = r" n for two curves with the same (du, dv) [see (119)].
Therefore COS0=
-
Kn K
so that K = K,, sec 0
(122)
This is Meusnier's theorem.
36. Geometrical Significance of the Second Fundamental Form. We construct a tangent plane to the surface at the point r(uo, vo). What is the distance D of a neighboring point r(uo + Du, vo + Av)
VECTOR AND TENSOR ANALYSIS
76
on the surface, to the plane? It is D = Ar n. r(uo + Au, vo + AV) = r(uo, vo) + 1
+
2!
from the calculus.
Au2 + 2
Now
ar
+ av
a2r
AV
a2r
+ av2 Av2 + au av Au AV
Consequently
1/
D=
-
(a2r
ar Au au
[SEC. 36
z
z
Av2
except for infinitesimals of higher order.
Thus
2D = e du2 + 2f du dv + g dv2
(123)
Problems 1. For the paraboloid of revolution
r=ucosvi+usinvj+u2k show that E = 1 + 4u2, F = 0, G = u2, e = 2(1 + 4u)`}, f = 0, g = 2u2(1 + 4u2)-}, and find the normals to the surface and the normal curvature for the direction (du, dv).
2. What are the normal curvatures for directions along the parametric curves? 3. Find the second fundamental form for the sphere
r=rsin0coscpi+rsin6sincpj+rcosOk r = constant.
4. Show that the curvature K at any point P of the curve of intersection of two surfaces is given by K2 Sln2 0 = K12 + K22 - 2K1K2 COS 0
where 9j, K2 are the normal curvatures of the surfaces in the direc-
tion of the curve at P, and 0 is the angle between their normals. 5. Let us make a change of variable u = u(u, v), v = v(u, 1). Show that E, F, G transform according to the law
E
lz Caul
+ 2F
au au
+G(av \au
2
DIFFERENTIAL GEOMETRY
SEC. 37]
F
au au
au av
77
av au
av av
=Eaudo +F (aft do +audo )+Gag do UP-
(avl 2
au av
au\2
E
Ii + 2F avav
+ G `avl
and that E due + 2F du dv + G dv2 = E due + 2F du do + G W. Also show that
e=±
(
e
au + (au)2
au au au av g
-±
[e
(au)2
av
au av
(OV\2 1
au au + g au J au av av au au av au av au av
+ 2f avav+ g
37. Principal Directions.
(K .E - e) due +
Zf
av av
au CIO]
(av12 av
J
From (120) we have
f) du dv +
g) dv2 = 0
(124)
or
A due + 2B du dv + C dv2 = 0
This quadratic equation has two directions (du, dv), (Su, Sv), which give the same value for x,,. These two directions will coincide if the quadratic equation (124) has a double root. is true if and only if
B2-AC= (K.F'-f)2-
This
0
or ,
2(F2 - EG) +
Moreover, we have
so that
du
gE - 2fF) + (f2 - eg) = 0
(125)
- B and d = - C if B2 - AC = 0, A
(K. E - e) du + (x F - f) dv = 0 f) du + (xnG - g) dv = 0
6126)
The solutions of (124) give the two directions for a given x,,. When x is eliminated between (124) and (125), the two directions coincide and satisfy
(Ef - Fe) due + (Eg - Ge) du dv + (Fg - Gf) dv2 = 0 (127)
VECTOR AND TENSOR ANALYSIS
78
[SEc. 37
The two directions, solutions of (127), are called principal directions and are the only ones with a unique normal curvature, that is, no other direction can have the same curvature. The normal curvatures in these two directions are called the principal curvatures at the point. The average of the two principal curvatures is
HeG+gE-2fF
(128)
2(EG - F2)
which is obtained by taking one-half of the sum of the roots of (125). The Gaussian curvature K is defined as the product of the curvatures, that is,
f2eg K= F2EG
(129)
A line of curvature is a curve whose tangent at any point has a
direction coinciding with a principal direction at that point. The lines of curvature are obtained by solving the differential equation (126). The curvature of a line of curvature is not a principal curvature since the line of curvature need not be a normal curve. Example 45. Let us consider the right helicoid
r= ucospi+usine'j+cook We have Or =cosSPi+sinrpj,
au 2
zr
a2G2-0'
8ua
= -u sins
= -sin+Cos(pj, a2r
Q= -ucosipi - usinrpj
Hence
(ar)2
au-
1,
F=
(Or\2 au - am
0'
G=
am
= u2 + c2
DIFFERENTIAL GEOMETRY
SEC. 371
79
Also
n = (ar/au) x (,9r/app) = (c sin p i - c cos v j + uk) (C2 +
u2)-}
(ar/au) x (ar/ap)j 82r
a2r
f=n
au ago
= -c(c2 +
u2)-#
2
Equation (125) yields - (u2 + C2)K,, + C2(C2 + u2)-1 = 0
whence C
u2+C2 The average curvature is H = 2
the Gaussian curvature is K =
(u2
2
C2)2'
+ C2) =
0, and
The differential
equation (126) for the lines of curvature becomes u2)-}
-c(c2 +
dug + c(c2 + U2)Id tp2 = 0
so that d(p = ±
du
= ± log (u + \/u2 + c2) + a
and
(C2 + u2) b
and the lines of curvature are given by
r = u cos [± log (u +
u2 + c2) + aji + u sin -p j + cspk.
Referring to (126) for the two principal directions, we have dv
du
av
Eg - Ge
+
Fg - Gf _ Ef - Fe du au Fg - Gf au dv av
(130)
Substituting (130) into (116) we obtain
E - F(Fg-Gf/ +G \F'g -Gf/ so that the principal directions are orthogonal.
0
VECTOR AND TENSOR ANALYSIS
80
[SEC. 38
Now let us choose the principal curves as the parametric lines.
Thus u = constant, v constant are to represent the principal curves. These two curves satisfy the equation du dv = 0, so that from (127) we must have
Ef - Fe = 0
Fg-Gf=0
Eg - Ge
0
From these equations we conclude that
f(Eg - Ge) = gfE - feG=Feg - eFg =0 and F(Eg - Ge) = 0, so that f = F = 0. We have shown that a necessary and sufficient condition that the lines of curvature be parametric curves is that
f=F=0
(131)
Problems 1. Find the lines of curvature on the surface
x=a(u+v),
y=b(u-v),
z=uv
2. Show that the principal radii of curvature of the right conoid x = u cos v, y = u sin v, z = {f (v) are given by the roots of f'=K2
- of"(u2 + fF )iK - (u2 + ft )2 = 0
3. The surface generated by the binormals of the curve r = r(s)
is given by R = r + ub. Show that the Gauss curvature is K = -.r2/(1 + r2u2)2. Also show that the differential equation of the lines of curvature is
-T2 du2 -
(K + Kr2u2 + d3 u)
du ds + (1 + r2u2)T M = 0
38. Conjugate Directions. Let P and Q be neighboring points on a surface. The tangent planes at P and Q will intersect in a straight line 1. Now let Q approach P along some fixed direction. The line 1 will approach a limiting position 1'. The directions PQ and 1' are called conjugate directions. We now compute the analytical expression for two directions to be conjugate. Let n be the normal at P and n + do the
DIFFERENTIAL GEOMETRY
SFc. 38]
normal at Q, where dr = PQ = tion of 1' be given by Sr =
Or
au
Su
Or
an
du + ar
Sv.
+ av
Or
-
81
dv.
Let the direc-
Since Sr lies in both
planes, we must have Sr n = 0 and Sr (n + dn) = 0. These two equations imply Sr do = 0, or (au
Su
+
av
Sv)
Can du +
a dv)
=0
(132)
Expanding, we obtain far
(_au
-) du bu + I (-av -) Sv du + (auau Or
anav
Or an aU) au dvJ
anau
ar an
+
av
av
Sv dv = 0
(133)
Now n au = 0, so that by differenuiating we see that ar
an
a2r
which implies
an ar
au au
a2r -n- au, - = -e
an
Or
an Or
av au
au av
Similarly
-f
an Or 9
so that (133) becomes e du Su + f (du Sv + dv Au) + g Sv dv = 0
(134)
If the direction (du, dv) is given, there is only one corresponding conjugate direction (Su, Sv), obtained by solving (134). Now consider the lines of curvature taken as parametric curves. Their directions are (du, 0), (0, Sv). Equation (134) is satisfied by these directions since f = 0 for lines of curvature, so that the lines of curvature are conjugate directions.
82
VECTOR AND TENSOR ANALYSIS
[SFc. 39
39. Asymptotic Lines. The directions which are self-conjugate are called asymptotic directions. Those curves whose tangents are asymptotic directions are called asymptotic lines. If a direcdv
tion is self-conjugate,
=
du
Sv,
Su
so that (134) becomes
e due + 2f du dv + g dv2 = 0
(135)
We see that the asymptotic directions are those for which the second fundamental form vanishes. ture rc, vanishes for this direction.
If e = g = 0, f
Moreover, the normal curva-
0, the solution of (135) is u = constant,
v = constant, so that the parametric curves are asymptotic lines if and only if e = g = 0,f3PK 0. Example 46. Let us find the lines of curvature and asymptotic lines of the surface of revolution z = x2 + y2. Let x = u cos v, y = u sin v, z = u2, and
r=ucosvi+usinvj+u2k We obtain Or
=cosvi-{-sinvj+2uk,
ar= -u sinvi+ucosvj av
au azr
8u2
a2r
=2k' a2r av2
au av
= -sinvi+cosvj
- -ucosvi - usinvj
n= (ar/au) x (ar/av)
(ar/au) x (ar/av)I = (-2u2 cos v i - 2u2 sin v j + uk)u-1(1 + 4u2)'I
Therefore z
=2(1+4u)-},
au2 z
g=n
f=n aua2rav =0
= 2u2(1 +4 U2)-f 2
DIFFERENTIAL GEOMETRY
SEC. 40] ar
83
ar
Also F = -- - = 0, so that f = F = 0, and from (131) the au av parametric curves are the lines of curvature. The asymptotic lines are given by due + u2 dv2 = 0. These are imaginary, so that the surface possesses no asymptotic lines. Problems 1. Show that the asymptotic lines of the hyperboloid
r=acos0seeipi+bsin0sec4, j+ctan 4k are given by 0 ± ¢ = constant. 2. The parametric equations of the helicoid are
x=ucosv,
y=usinv,
z=cv
Show that the asymptotic lines are the parametric curves, and
that the lines of curvature are u + V 'u2+ c2 c2 = Ae}°. Show that the principal radii of curvature are ± (u2 + c2)c-1. 3. Prove that, at any point of a surface, the sum of the normal curvature in conjugate directions is constant. 4. Find the asymptotic lines on the surface z = y sin x. 40. Geodesics. The distance between two points on a surface (we are allowed to move only on the surface) is given by
8 = I.1 [E (dt/2 + 2F dt dt + G (i)]1 dt
(136)
Among the many curves on the surface that join the two fixed points will be those that make (136) an extremal. Such curves are called geodesics. We wish now to determine the geodesics. To do this, we require the use of the calculus of variations, and so we say a few words about this important method. Let us first consider the integral fQ(xt,yi) (1 (x,, VS)
+ y'dx
(137)
We might ask what must be the function y = y(x) joining the two points P and Q which will make (137) a minimum. The reader might be tempted to say, y' = 0 or y = constant, since the integrand is then a minimum. But we find that y = constant will
84
VECTOR AND TENSOR ANALYSIS
[SEC. 40
not, in general, pass through the two fixed points. Hence the solution to this problem is not trivial. We now formulate a more general problem: to find
y
y = y(x) such that
j'f(x, y, y') dx is an extremal. ry
Ar
}' M
The function
f(x, y,)y ' is given. It is y(x) and so also y' (x) that are unknown . Let y = y (x) be that function which makes
I.-, B
,
(138)
It
(138) an extremal Now 1 t Y(x, a) = y(x) + arp(x), where
a is arbitrary and independ-
P.x
b
a
ent of x and jp(x) is any function with continuous first
Fzo. 44.
derivative having the property that jp(a) = p(b) = 0 (see Fig. 44). Under our assumption,
J(a) = j,'f(x, Y, ?') dx is an extremal for a = 0. dJ
=
da a-0
(139)
dJ
da0 = 0 or
Consequently
Ib\ay,+a-
(140)
since
of as
of 81' of 81" aY as + aY' as
of
if
+
of C7F
'p,
and for a=0, of
a> r
of
of
of ay'
Tay ,
e
We now integrate the right-hand term of (140) by parts and obtain P] b
fb ay
dx
+
Lay
-
Jab
dx ay'/ `p
dx = 0
(141)
DIFFERENTIAL GEOMETRY
SEc. 40]
85
Now (p(a) = p(b) = 0 by construction of v(x), so that (142)
Now let us assume that
a (-'
ay
ay
dx
(ay')
is continuous.
If
is not identically zero on the interval (a, b), it
will be positive or negative at some point. If it is positive at x = c, it will be positive in a neighborhood of x = c from continuity (see Sees. 42 and 43). We can construct (p to be positive on this interval and zero elsewhere. Then
Jb[afd(of)]
ay`p dx > 0
dx
y
so that we have a contradiction to (142). Consequently, the function of y(x) must satisfy the Euler-Lagrange differential equation of
d of dx Cay'
ay
0
(143)
If f = f(y, y'), we can immediately arrive at an integral of (143).
Let us consider
f dx (
y yl
ay
1J, +
a y
y [Of
_
lay
f-y
aof y,
yy
of ay
d dx ay'
(t\1 = 0
= constant
d (af 1 y/ dx TO from (143)
(144)
is an integral of (143) if f = f(y, y'). Example 47. To extremalize (137), we have f = (1 + y'2)1, which is independent of x. From (144), f - y
of
1
= constant =
a
VECTOR AND TENSOR ANALYSIS
86
[SEC. 40
so that
(1 +y') - Y'(1
J
a
y'2),
and
y'= and finally
y= ±
av-1x+ 0
(145)
The constants a and fi are determined by noting that the straight line (145) passes through the two given fixed points. Example 48.
If f = f Cx',x 2, ... , x",
dt2) ...
dtl,
dtn?
t);
fhfdt is an extremal when the xa(t) satisfy
then
d (ftaf dt a
for a = 1, 2,
. .. , n with is =
of axa = o X.
(146)
The superscripts are not
powers but labels that enable us to distinguish between the various variables. The formulas (146) are a consequence of the fact
that
fo f dt must be an extremal when x`(1) is allowed to vary
while we keep all other x1 fixed, j = 1, 2, ... , i - 1, i + 1, ,n. Let us now try to find the differential equations that u(t) and v(t) must satisfy to make (136) an extremal. We write
s= f"
(E,42 + 2Fuv + Gv2)} dt
and apply (146), where x' = u and x2 = v. We thus obtain
dt \a4/
au - 0
off`
of
dt avl
av
d
o
(147)
(148)
where
f = (E,42+ 2Fuv + Gv2)i = dt'
E = E(u, v), etc.
---
DIFFERENTIAL GEOMETRY
SEc. 401
Now
af au
Eic + Fv
ice
of au
f
CIE
au
+ 2uv
OF
au 2f
+ v2
87
aG
au
so that (147) becomes
d Eu + Fvl _ 2t2 + 2"(W/au) + v2(aG/au) dt
ds/dt /
2 ds/dt
(149)
and if we choose for the parameter t the are length s, then t = a
and dt = 1, so that (149) reduces to d ds
(Eic + Fv) = 2( u2 aE + 2uv
OF
+ 62
-n/A opu
while similarly (148) yields
d (Fu+Gv)
\
(150)
_2(u2av +24vav +v2av/
In Chap. 8 we shall derive by tensor methods a slightly different system of differential equations. Example 49. Consider the sphere given by
r = a sin 0cos(pi+a sin 0 sin cpj+acosOk where ds2 = a2 do2 + a2 sin2 0 dp2 so that E = a, F = 0, G = a2 Sin2 0, and OE
80
_
aEE
ac
_
OF
_
OF
a9
av
_
aG
_
ap
8G
0'
aB
_
- a sin 29 2
Hence (150) reduces to 2
ds a ds/
2 ds
sin 2B
d (a sin2 0 ds) = 0 Integrating (151) we have sine 0 LIP = constant.
d
(151)
We can choose
our coordinate system so that the coordinates of the fixed points
VECTOR AND TENSOR ANALYSIS
88
are a, 0, 0 and a, 0, 0. Hence sin' 8
ds
= 0, and
[SEC. 40
ds
= 0, so that
0. Hence the geodesic is the are of the great circle joining the two fixed points. Example 50. Let us find y(x) which extremalizes
f y(1 + y")} dx
Since f = y(1 + y")# = f(y, y'), we can apply (144) to obtain a first integral. We obtain y(1 + y'')t -- y''y(1 + y'')-l = a-1, and simplifying this expression yields y' = ± (a2y2 - 1)I. A further integration yields ay = cosh (0 ± ax). These are the curves (catenoids) which have minimum surfaces of revolution.
Problems 1. Find the geodesics on the ellipsoid of revolution 2 x2+ z2 b2 = 1 + a2
Hint: Let x = u cos v, z = u sin v. 2. Show that the differential equation of the geodesics for the right helicoid x = u cos v, y = u sin v, z = cv is du TV
+
1
[(u2 + C2)(u2 + C2
h2)1],
h = constant
-h
3. Prove that the geodesics on a right circular cylinder are helices.
4. Show that the perpendicular from the vertex of a right circular cone to the tangents of a given geodesic is of constant length. 5. Find y(x) which extremalizes
f'[(l + y')/y)]} dx.
CHAPTER 4 INTEGRATION
41. Point-set Theory. In geometry and analysis the student has frequently made use of the concept of a point and of the notion of a set or collection of elements (objects, points, numbers, etc.). We shall not define these concepts, but shall take their notion as intuitive. We may be interested in the points subject
to the condition x2 + y2 < r2. These will be the points interior to the circle of radius r with center at the origin. We might also be interested in the rational points of the one-dimensional continuum. For convenience, we shall consider only points of the real-number axis in what follows. Any set of real numbers will be called a linear set. The integers form such a set, as do the
rationals and irrationals. All the definitions and theorems proved for linear sets can easily be extended to any finitedimensional space. Closed Interval. The set of points { x j satisfying a < x S b will be called a closed interval. If we omit the end points, that
is, consider those x that satisfy a < x < b, we say that the interval is open (open at both ends). For example, 0 S x 5 1 is a closed interval, while 0 < x < 1 is an open interval. Bounded Set. A linear set of points will be said to be bounded if there exists an open interval containing the set. It must be emphasized that the ends of the interval are to be finite numbers, which thus excludes - oo, + oo. An alternative definition would be the following: A set of numbers S is bounded if there exists a finite number N such that
-N 1. But the complement of -1 < x < 1 relative to the set -1 S x S 1 is the null set (no elements). The complement of the set of nationals relative to the reals is the set of irrationals, and conversely. Open. Set. A set of points S is said to be an open set (not to be
confused with open interval) if every point of S is an interior
SEC. 411
IXTBGRATION
91
point of S. For example, the set S consisting of points which satisfy either 0 < x < 1 or 6 < x < 8 is open.
Closed Set. A set containing all its limit points is called a closed set. For example, the set (0, 1/2, 1/3, 1/4, . . . , 1/n, .) is closed, since its only limit point is 0, which it contains. Problems
1. What are the limit points of the set 0 < x < 1? Is the Open? What are the boundary points? 2. Repeat Prob. 1 with the point x = removed. 3. Show that the set of all boundary points (the boundary)
set closed?
of a set S is closed. 4. Prove that the set of all limit points of a set S is closed.
5. Prove that the complement of an open set is closed, and conversely.
6. Why is every finite set closed? 7. Prove that the set of points common to two closed sets is closed. The set of points belonging to both Si and S2 is called the intersection of Sl and S2, written SI n828. Prove that the set of points which belong to either S, or 82 is open if S, and S2 are open. This set is called the union of S, and S2, written S, U S29. An infinite union of closed sets is not necessarily closed. Give an example which verifies this. 10. An infinite intersection of open sets is not necessarily open. Give an example which verifies this. Supremum. A number s is said to be the supremum of a set of points S if 1. x in S implies x 5 s
2. t < s implies an x in S such that x > t Example 51. Let S be the set of rationale less than 1. Then 1 is the supremum of S, for (1) obviously holds from the definition of S, and if t < 1, it is possible to find a rational r < 1 such
that t < r, so that (2) holds. We give a proof of this statement in a later paragraph. Example 52. Let S be the set of rationals whose squares are to less than 3, that is, SI x2 < 3. Certainly we expect the be the supremum of this set. However, we cannot prove this without postulating the existence of irrationals. We overcome
92
VECTOR AND TENSOR ANALYSIS
[SEC. 41
this by postulating Every nonempty set of points has a supremum
(152)
Hence the rationals whose squares are less than 3 have a supreIt is obvious that we should define this supremum as the square root of 3. The supremum of a set may be + oo as in the case of the set of all integers, or it may be - oo as in the case of the null set. The infemum of a set S is the number s such that 1. x in S implies s < x mum.
2. t > s implies an x in S such that x < t Example 53. Let a > 0 and consider the sequence a, 2a, 3a, ... , na, ... . If this set is bounded, there exists a finite supremum s. Hence an integer r exists such that ra > s - (a/2) so that (r + 1)a > s + (a/2) > s, a contradiction, since na 5 s for all n. Hence the sequence Ina} is unbounded. This is the Archimedean ordering postulate. Example 54. To prove that a rational exists between any two numbers a, b. Assumed: a > b > 0, so that a - b > 0. From
example 53, an integer q exists such that q(a - b) > 1, or qa > qb + 1. Also an integer p exists such that p 1 = p > qb. Choose the smallest p.
Thus p > qb ? p - 1. Hence
qa>gb+1zp>qb, and a > p/q > b. Q.E.D. With the aid of (152) we are in a position to prove the wellknown Weierstrass-Bolzano theorem. "Every infinite bounded set of points S has a limit point."
The proof proceeds as follows: We construct a new set T. Into T we place all points which are less than an infinite number of S. T is
not empty since S is bounded below. From (152), T has a supremum; call it s. We now show that s is a limit point of S. Consider any neighborhood N of s. The points in N which are less than s are less than an infinite number of points of S, whereas
those points in N which are greater than s are less than a finite number of points of S. Hence N contains an infinite number of
S, so that the theorem is proved. We have at the same time proved that s is the greatest limit point of S. A limit point may or may not belong to the set.
SEC. 41)
INTEGRATION
93
Problems 1, 2, 3, . . . is 1. The set 1/2, 1/3,. . . , 1/n, . . unbounded. Does it have any limit points? Does this violate the Weierstrass-Bolzano theorem? 2. Prove that every bounded monotonic (either decreasing or increasing) sequence has a unique limit. 3. Prove that lim r" = 0 if Irl < 1. Hint: The sequence r, r2, . . . , r", . . . is bounded and monotonic decreasing for r > 0, and r"+1 = rr" 4. Show that if P is a limit point of a set S, we can pick out a subsequence of 8 which converges to P. 5. Show that (152) implies that every set has an infemum. 6. Show that removing a finite number of elements from a set cannot affect the limit points. 7. Prove the Weierstrass-Bolzano theorem for a bounded set of points lying in a two-dimensional plane. 8. Let the sequence of numbers sl, 82, .. . , sn, . . . satisfy the following criterion: for any e > 0 there exists an integer N such that Is,,+F - snI < e for n ? N, p > 0. Prove that a unique limit point exists for the sequence.
.,
9. A set of numbers is said to be countable if they can be written as a sequence, that is, if the set can be put into one-to-one correspondence with the positive integers. Show that a countable collection of countable sets is countable. Prove that the rationals are countable.
10. Show that the set S consisting of x satisfying 0 5 x 5 1 is uncountable by assuming that the set S is countable, the numbers x being written in decimal form. Theorem of Nested Sets. Consider an infinite sequence of nonempty closed and bounded sets S1, S2, . . . , Sn, . such There that Sn contains that is, S1 M S2 D Sa M .
exists a point P which belongs to every Si, i = 1, 2, 3, . . . . The proof is easy. Let P1 be any point of S1, P2 any point of P,, S2j etc. Now consider the sequence of points P1, P2, . . . . This infinite set belongs to S1 and has a limit point P which belongs to 8, since Sl is closed. But P is also a limit of P, Pn+1, ... , so that P belongs to Sn. Hence P is in all Sn, n = 1, 2, . . . .
... ,
Diameter of a Set. The diameter of a set S is the supremum of all distances between points of the set. For example, if S is the
VECTOR AND TENSOR ANALYSIS
94
[SEc. 41
set of numbers x which satisfy I < x < 1, 3 < x < 7, the diameter of S is 7 - = 61. There are pairs of points in S whose distance apart can be made as close to 6,91 as we please.
Problems 1.
If a set is closed and bounded, the diameter is actually
attained by the set. 2.
Prove this.
If, in the theorem of nested sets, the diameters of the S
approach zero, then P is unique. Prove this. The Heine-Borel Theorem. Let S be any closed and bounded
set, and let T be any collection of open intervals having the property that if x is any element of S, then there exists an open interval 7' of the collection T such that x is contained in T.
The theorem states that there exists a subcollection T' of T which is finite in number and such that every element x of S is contained in one of the finite collection of open intervals that comprise T'. Before proceeding to the proof we point out that (1) both the set S and the collection T are given beforehand, since it is no great feat to pick out a single open interval which completely covers a bounded set S; (2) S must be closed, for consider the set S(1, 1/2, 1/3, . . . , 1/n, . . .) and let T consist of the following set of open intervals: TIIx such that Ix
- 11 < 22
T21x such that Ix
- _f < 32
1
T,,I x such that Ix - 1 n < (n + 1)2
It is very easy to see that we cannot reduce the covering of S by eliminating any of the given T,,, for there is no overlapping of these open intervals. Each T,, is required to cover the point 1/n of S contained in it.
Src. 421
INTEGRATION
95
Proof of the theorem: Let S be contained in the interval
-N < x < N.
This is possible since S is bounded. Now divide this closed interval into two equal intervals (1) -N < x _:5- 0 and
(2) 0 < x < N. Any element x of S will belong to either (1) or (2).
Now if the theorem is false, it will not be possible to cover
the points of S in both (1) and (2) by a finite number of the given collection T, so that the points of S in either (1) or (2) require an infinite covering. Assume that the elements of S in (1) still require an infinite covering. We subdivide this interval
into two equal parts and repeat the above argument. In this way we construct a sequence of sets S,
S2
S3
,
such
that each Si is closed and bounded and such that the diameters of the Si -* 0. From the theorem of nested sets there exists a unique point P which is contained in each Si. Since P is in S, one of the open intervals of T, say T, will cover P. rt'his Tp has a finite nonzero diameter so that eventually one of the Si will be contained in TD, since the diameters of the Si --> 0. But
by assumption all the elements of this Si require an infinite number of the { T } to cover them.
This is a direct contradiction
to the fact that a single Tp covers them. Hence our original assumption is wrong, and the theorem is proved. 42. Uniform Continuity. A real, single-valued function f (x)
is said to be continuous at a point x = c if, given any positive number e > 0, there exists a positive number a > 0 such that If(x) - f(c)I < e whenever Ix - cI < S. The a may well depend on the a and the point x = c. The function f(x) is said to be continuous over a set of points S if it is continuous at every point of S.
We now prove that if f (x) is continuous on a closed and bounded interval, it is uniformly continuous. We define uniform continuity as follows: If, for any e > 0 there exists a S > 0 such that If(XI) - f(X2)1 < e whenever Ix1 - x21 < S, then f(x) is said to be uniformly continuous. It is important to notice
that 8 is independent of any x on the interval. We make use of the Heine-Borel theorem to prove uniform continuity.
Choose
an e/2 > 0. Then at every point c of our closed and bounded set there exists a S(c, a/2) such that If(s) - f (c)I < e/2 for c - a < x < c + S. Hence every point of S is covered by a 26-neighborhood, and so also by a a-neighborhood. By the Heine-Borel theorem we can pick out a finite number of these
96
VECTOR AND TENSOR ANALYSIS
[SEc. 43
Let S, be the diameter of the smallest of this finite collection of 6-neighborhoods. Now consider any two points x, and x2 of S whose distance apart is less neighborhoods which will cover S.
than bl. x1.
Let xo be the center of the 6-neighborhood which covers
From continuity lf(xi) - f(xo)I < e/2. But also x2 differs
from xo by less than 26, so that I f (x2) - f (xo) I < e/2. Consequently If(xl) - f(x2)1 < e. Q.E.D. 43. Some Properties of Continuous Functions (a) Assume f(x) continuous on the closed and bounded interval a < x < b. As a consequence of uniform continuity, we can prove that f (x) is bounded. Choose any e > 0 and consider the corresponding S > 0 such that If(XI) - f(X2)1 < e whenever
-
x21 < S. Now subdivide the interval (a, b) into a finite number of S-intervals, say n of them. It is easily seen that Ix1
max If(x) I < If (a) I + ne. (b) If f (a) < 0 and f (b) > 0, there exists a c such that f (c) = 0,
a < c < b. Let {x} be the set of all points on (a, b) for which f (x) < 0. The set is bounded and nonvacuous since a belongs to {x}, for f(a) < 0. The set {x} will have a supremum; call it c. Assume f (c) > 0. Choose e = f (c) /2. From continuity, a 8 > 0 exists such that If(x) - f (c) I < f(c)12 if Ix cI < 6. Hence f(x) ? Jf(c) for all x in some neighborhood of x = c. Hence c is not the supremum of f x 1. Similarly f (c) < 0 is impossible, so that f (c) = 0. (c) We prove that f (x) attains its maximum. In (a) we showed that f (x) was bounded. Let s be the supremum of f (x),
-
a 5 x S b. As a consequence, f (x) 0. Since s is a supremum, there exist x1, x2,
x,,, ... such that f (xl) > s - e, f (x2) > s - e/2, ... ,
... ,
Let { xn' } be a subsequence f xd } will have a limit point c. of f x,,) which converges to c. Then lim f (x.') z s, since -s'-+c
e/n -* 0 as n
oo.
But from continuity lim f(x,') = .f(c). 4C
Hence f (c) = s. Q.E.D.
SEC. 44]
INTEGRATION
97
Problems
1. In the proof of (c) we exclude f(c) > s. Why? 2. If f (x) is continuous on (a, b), show that the set of values {f(x) }
is closed.
3. If f (x) has a derivative at every point of (a, b), show that f (x) is continuous on (a, b). 4. If f (x) has a derivative at every point of (a, b), show that a
c exists such that f(c) = 0, a < c < b, when f (a) = f (b) = 0. 5. If f (x) has a derivative at every point of (a, b), show that
a c exists such that f(b) - f(a) _ (b - a)f'(c), a < c < b. 6. Show that if two continuous functions f(x), g(x) exist such that f (x) = g(x) for the rationals on (a, b), then f (x) = g(x) on (a,
7. Given the function f (x) = 0 when x is irrational, f (x) = 1/q when x is rational and equal to p/q (p, q integers and relatively prime), prove that f (x) is continuous at the irrational points of
(0, 1) and that f(x) is discontinuous at the rational points of this interval. 44. Cauchy Criterion for Sequences. Let x1, x2, ... ,
xn, ... be a sequence of real numbers. We say that L is the limit of this sequence, or that the sequence converges to L, if, given any e > 0, there exists an integer N depending on a such that IL - xnl < e whenever n > N(e). However, in most cases we do not know L, so that we need the Cauchy convergence criterion. This states that a necessary and sufficient condition that a sequence converge to a limit L is that given any e > 0, there exists an integer N such that Ixn - xml < e for n >_ N, m N. That the condition is necessary is obvious, for IL - xnl < e/2, xn,l < e/2 for m, n > N implies Ix,, - xml < e for m, n >_ N. The proof of the converse is not as trivial. Choose any e/2 > 0. IL
Then we assume an N exists such that Ix,, -- xml < e/2 for
N, so that the
m, n ? N. Hence Ixn+ < I xNl + (e/2), n sequence is bounded. We ignore xi, X2,
.
.
.
, xN_, since a finite
number of elements cannot affect a limit point. From the Weierstrass-Bolzano theorem this infinite bounded set has at least one limit point L.
Hence, given an a/2, there exists an x.,
with n > N, such that IL - xAi < e/2. But we also have
98
VECTOR AND TENSOR ANALYSIS
[SEC. 45
I xm - x,,l < e/2 for all m, n > N. Hence IL - xml < e for all m > N. Q.E.D. Problems
1. Show how the convergence of a series can be transformed into a problem involving the convergence of a sequence.
2. Show that the Cauchy criterion implies that the nth term of a convergent series must approach zero as n -- oo. 3. Show that the sequence 1, 1/2, 1/3, . . . , 1/n, . . . converges by applying the Cauchy test.
45. Regular Arcs in the Plane. Consider the set of points in the two-dimensional plane such that the set can be repreP2 sented
in
some coordinate
system by x = f (t), y = sp(t), a S t < 0, where f(t) and ap(t) are continuous and have continuous first derivatives.
Such
curves are called regular arcs.
A regular curve is a set of points consisting of a finite number of regular arcs joined
one after the other (see Fig. P4
Fia. 45.
45).
PoPI, P1P2, P2t 3, P3P4 are
the regular arcs joined at P1, P2, P3. Notice that there are at most a finite number of discontinuities of the first derivatives. In Fig. 45 the derivatives are discontinuous at P1j P2, P3.
46. Jordan Curves. The locus i x = f (t) a < t 5 fl, will be y = P(t), called a Jordan curve provided that f (t) and (p(t) are continuous
l curve correspond to two and that two distinct points on the distinct values of the parameter t (no multiple points). A closed Jordan curve is a continuous curve having f(a) ,o(a) = 9(fl) but otherwise no multiple points.
From this we see that a Jordan curve is always "oriented," that is, it is always clear which part of the curve lies between two points on the arc, and which points precede a given point. We
shall be interested in those curves which are rectifiable, or, in
IN TkGRA TION
Sr:c. 47]
99
other words, we shall attempt to assign a definite length to a given Jordan are or curve. 47. Functions of Bounded Variation. Let f(x) be defined on
the interval a < x < b. Subdivide this interval into a finite number of parts, say a = xo, x,, . . . x;, . . . , x,-1) x = b. Now consider the sum
- f(xo)I + I f(x2) - f(xi)1 + .
I f(xi)
. + l f(xn) - f(xn - 1)I n
_ i_1 ± I f(xi) - f (xt-1) If the sums of this type for all possible finite subdivisions are bounded, that is, if n
s=x
I f(xi) - f(xc-1)I < A < -o
(153)
we say that f(x) is of bounded variation on (a, b).
A finite, monotonic nondecreasing function is always of bounded variation since n
n
i=1
If(x=) - f(x=-,)1 = I If(x=) - f(xi-1)) = f(b) - f(a) = A t=1
An example of a continuous function that is not of bounded variation is the following: f(x) = x sin 2x,
0< x 5 1
f (O) = 0
Let us subdivide 0 5 x 0, a contradiction s Example 61. We see that an irrotational field is characterized by any one of the three conditions:
f = VP Vxf=0 f dr = 0
(i) (ii)
(iii)
(169)
for every closed path
Any of these conditions implies the other two. Example 62. Assume f not irrotational. Perhaps a scalar p(x, y, z) exists such that pf is irrotational, that is, V x (pf) = 0. If this is so,
pV xf+Vp xf = 0
(170)
and dotting (170) with f, we have, since f VA x f = 0, the equation
f- (V xf) = 0
(171)
If f = Xi + Yj + Zk, (171) may be written
XYZ a
a
a
ax
ay
az
=0
(172)
XYZ
In texts on differential equations it is shown that (172) is also sufficient for p(x, y, z) to exist. We call p(x, y, z) an integrating factor. Example 63. Let f = f(x, y, z)a, where a is any constant vector. Applying Stokes's theorem, we have
VECTOR AND TENSOR ANALYSIS
112
ff
[SEC. 54
x a) dd
s
ff s
a
f
f f= a
dd x
a
we have S
ff [a(V s
g) - (a V)gl dd
=a- f f (V.g)dd-a f f dd (summed).
V
There-
fore
xdr = a- f f (ddxv) xg and dr x g = f f (dd x V) x g S
(174)
We notice that in all cases
0 dr * f=
Jf(doxv)*f
(175)
S
The star (*) can denote dot or cross or ordinary multiplication. In the latter case, f becomes a scalar f.
INTEGRATION
Sec. 541
113
Problems
1. Prove that fdr = 0 from (175). 2. Show that L, 'dr x ri taken around a curve in thex-y plane is twice the area enclosed by the curve.
3. If f = cos y i + x(1 + sin y)j, find the value of ff - dr around a circle of radius r in the x-y plane.
4. Prove that fr dr = 0. 5. Prove that ff dd x r =
fir 2 dr.
s
6. Prove that f u Vv dr = - ,f v V u dr. 7. Prove that u Vv - dr = JJvu x Vv dd. s
8. If a vector is normal to a surface at each point, show that its curl either is zero or is tangent to the surface at each point. 9. If a vector is zero at each point of a surface, show that its curl either is zero or is tangent to the surface.
10. Show that fi a x r dr = 2a. Jf dd, if a is constant. S
11. If fi E dr = that V x E =
B dd for all closed curves, show -cat Jf s
---
C at 12. By Stokes's theorem prove that V x (V(p) = 0.
13. Show that fi dr/r = f f (r/r3) x dd where r = Id. 8
14. Find the vector f such that xy = f(O,(X'o, Y,) f dr. o)
15. If f = r/r3, show that V x f = 0 and find the potential yo such that f = Vsp.
16. Show that the vector f = (- yi + xj)/(x2 + y2) is irrotational and that (f r f dr = 2w, where r is a circle containing the origin.
Does this contradict Stokes's theorem? Explain.
17. Show that ff v x f dd = 0, where S is a closed surface. s
18. Let C1 and C2 be two closed curves bounding the surfaces Show that
S1 and S2.
114
VECTOR AND TENSOR ANALYSIS
k. ki r122 dr1
[SEC. 55
dr2 = -4 fJ dal f f dd2 St
s2
f, fc, r122 dr1 x dr2 = - 2 f f dot x ff dd2 Si
s2
where r12 is the distance between points on the two curves.
55. The Divergence Theorem (Gauss). Let us consider a region V over which f and V f are continuous. We shall assume that V is bounded by a finite number of surfaces such that at each surface there is a well-defined continuous normal. We shall also assume that f can be integrated over the total surface bounding V. Now, no matter what physical significance f has, if any,
we can always imagine f to be the product of the density and velocity of some fluid. We have seen previously (Sec. 20) that the net loss of fluid per unit volume per unit time is given by V f. Consequently, the total loss per unit time is given by
JfJ(v.f)dr
(176)
V
Now since f and V f are continuous, there cannot be any point in the region V at which fluid is being manufactured or destroyed;
that is, no sources or sinks appear. Consequently, the total loss of fluid must be due to the flow of fluid through the boundary
S of the region V. We might station a great many observers on the boundary S, let each observer measure the outward flow, and then sum up each observer's recorded data. At a point on the surface with normal vector area dd, the component of the velocity perpendicular to the surface is V N, where N is the unit outward normal vector. It is at once apparent that pV dd = f dd represents the outward flow of mass per unit time. Hence the total loss of mass per unit time is given by
Jff.dd
(177)
Equating (176) and (177), we have the divergence theorem:
JJf.do=fJf(v.f)d.r 8
V
(178)
INTEGRATION
SEC. 55]
115
For a more detailed and rigorous proof, see Kellogg, "Foundations of Potential Theory." We now derive Gauss's theorem by a different method. Let f be a differentiable vector inside a connected region R with rectifiable surface S. Surround any point P of R by a small element of volume dr having a surface area AS. Form the surface integral JJf. dd and consider the limit, As
lira
f
As
Ar-.0
DT
If this limit exists independent of the approach of Ar to zero, we define
Jff.dd div f = lim
As
AT-40
(179)
AT
We can write
tardivf= JJ
(180)
where e -40 as Ar -* 0. If we now subdivide our region into many elementary volumes, we obtain formulas of the type (180) for all of these regions. Summing up (180) for all volumes and then passing to the limit, we have
fffdivfdr__
(181)
Jffdd
Inthe derivation of (181) use has been made of the fact that for
each internal dd there is a -dd, so that all interior surface integrals cancel in pairs, leaving only the boundary surface S as a contributing factor. The sum of the ei Ar; vanishes in the limit, for I Zet Aril 5 ZIel... OT S IEIo,.=V, and if div f is continu-
ous, IeI,. - 0 as 2r --i 0. By choosing rectangular parallelepipeds and using the method of Sec. 20, we can show that
fff.do div f = slim
AS
AT
clu
c3v
aw
ax
+ ay +
49Z
(182)
116
VECTOR AND TENSOR ANALYSIS
[SFC. 55
for f = ui + vj + wk, which corresponds to our original definition of the divergence. Example 65. Consider a sphere with center at 0 and radius a.
Take f=r=xi+yj+zk, dd =
(r) dS = a
(ai)
(
xi+yj+zk)dS
Now V f =3, f dd = (1/a)(x2 + y2 + z2) dS = a dS on the sphere.
Applying (178), f _f f 3 dr = f f a dS, or 3V = aS, where
S is the surface area of the sphere. If V is known to be g 47a8, then S = 4aa2.
Example 6 6. Let f = qr/rs and let V be a region surround-
ing the origin and let S be its We cannot apply the divergence theorem to this resurface.
gion since f is discontinuous at
r = 0. We overcome this difficulty by surrounding the origin by a small sphere E of radius e (see Fig. 50). The divergence theorem can be applied to the connected region V'. The region V' has two boundaries, S and Fzo. 50.
E.
Applying (178),
ffJ V. () dr
f
8
r ad+J1 E
r"d
(183)
In Example 25 we saw that V (r/r3) = 0. This implies that (183) reduces to
ff.!.do=
(184)
Now for the sphere 1, dd = -r dS/e, since the outward normal to the region V' is directed toward the origin and is parallel to the radius vector. Hence
JJ.dd= fff.dd= ffds=4fq
INTEGRATION
SFe. 551
117
The integral f f f dd is called the flux of
since f f dS =
s
the vector field f over the surface S. square force f = qr/r3,
fJ f S
We have, for an inverse-
dd = 4irq
(185)
Example 67. A vector field f whose flux over every closed surface vanishes is called a solenoidal vector field.
From (178)
it is easy to verify that V f = 0 for such fields. Hence a solenoidal vector is characterized by either /f dd = 0 or 0.
Now assume f = V x g, which implies V f = V (V x g) = 0. Is the converse true? If V f = 0, can we write f as the curl of some vector g? The answer is "Yes"! We call g the vector potential of f. This theorem is of importance in electricity theory, as we shall see later. Notice that g is not uniquely determined since V x (g + Vp) = V x g. We now show the existence of g. Let f = Xi + Yj + Zk and assume
g=ai+ij+yk We wish to find a, $, y such that f = V x g, and hence
x
=---
Y= Z
ay
a#
ay
az
as
ay
az
ax
as
as
ax
ay
(186)
=---
Now assume a = 0. Then X =
ay ay
- 8-, Y = - aye Z = as. 8z
ax
Consequently if there is a solution with a = 0, of necessity $6=
fXzdx+o.(y,z)
y=Now V f = 0 or
ax
o xos
Ydx+r(y,z)
aY y
+ az) by assumption.
ax
VECTOR AND TENSOR ANALYSIS
118
f
Hence
ay
aY
JZo
az
dx
,"
+
+ + ar
- as
ay
az
ay
fx ax ax
aZ az /
ar ay
= X (x, y, z) - X (xo, y, z) +
_
[SEC. 55
aa
az
or
ao
ay -
az
so that (186) is satisfied by a = 0, S = I o Z dx + or (y, z), jX
y
Y dx by choosing r = 0, o(y, z) = - JZX(xo,y,z) dz.
Hence f =Vxgwhere
g=[jz
Z dx + c(y, z)] j - faVdxk.
In general
g=[ faZdx+v(y,z)]j- foYdxk+Vp
(187)
For example, if f = V(1/r), then V f = 0. Now
X=--
Y=-y,
s,
Z=-raz
where r2 = x2 + y2 + z2. Applying (187)
g=
x -z dx y dx (x2+y2+z2), j+1 (x2+y2+z2)$k+V
,
v arbitrary
and
g = (x2 + y2 + z2)#(y2 + Example 68. Green's theorem. and f2 = v Vu and obtain
z2)
We apply (178) to fl = u Vv
f f f V. (u Vv)dr= f f f R
(-zj + yk) + VSP
ff
R
S
(188)
f f f V V. (v Vu) dr = f f f (v V2u + VV. VU) d7 = Jf v Vu dd R
S
R
Subtracting, we obtain
f f f (u V2v - v V2u) dr = f f (u Vv - v Vu) dd (189) R
S
INTEGRATION
SEC. 55]
119
Example 69. Uniqueness theorem. Assume two functions which satisfy Laplace's equation everywhere inside a region and
which take on the same values over the boundary surface S. The functions are identical. Let V2(P = 1724, = 0 inside R and (p ='p on S.
Now from (188) we have
JJJov2odr+JfJvo.vodr=JJove.do R
R
Define 0 - (p - ¢ so that 1720 = 0 over R and 0=0 on S. Hence f f f (V0)2 dT = 0, which implies V0 = 0 inside R. R
Hence 0 = constant = p -'p, and since sc
on S, we must
have cp ='G. We have assumed the existence of Vp, V¢ on S. Example 70. Another uniqueness theorem. Let f be a vector
whose curl and divergence are known in a simply connected region R, and whose normal components are given on the surface S which bounds R. We now prove that f is unique. Let f1 be
another vector such that V f = V f1, V x f = V x f1, and f dd = f, dd on S. We now construct the vector g = f - f1.
We immediately have that V g = V x g = g dd = 0. In Sec. 52 we saw that if V x g = 0, then g is the gradient of a scalar Consequently, V g = V2ip = 0. Applying (188) cp, g = V a. The outer sphere is negatively charged. 2. Find the field due to any infinite uniformly charged cylinder. 3. Solve Prob. 1 for two infinite concentric cylinders.
VECTOR AND TENSOR ANALYSIS
132
[Sec. 59
4. Let ql, q2, . . . , q be a set of collinear electric charges residing on the line L. Let C be a circle whose plane is normal
to L and whose center lies on L. Show that the electric flux n
through this circle is N = 121rga(1 - cos Sa), where Na is the
a-I
angle between L and any line from qa to the circumference of C. 5. Let the line L of Prob. 4 be the x axis, and rotate a line of
force r in the x-y plane about the x axis (see Fig. 58). If no Y
xa q3
0
x2
xl
qa
qi
x
I
II
`
I
I
1
1
/
Fia. 58.
charges exist between the planes x = A, x = B, show that the equation of a line of force is n
a-1
qa(x - xa)[(x - xa)z + y2]i = constant
6. Point charges +q, -q are placed at the points A, B. The line of force that leaves A making an angle a with AB meets the plane that bisects AB at right angles in P. Show that
sin 2 = 59. Poisson's Formula.
E_
sin (+
PA B)
In Sec. 57, we saw that
-v
a-1 r,
STATIC AND DYNAMIC ELECTRICITY
SEC. 591
133
For a continuous distribution of charge density p, we postulate that the potential is =
fff
pdr
(207)
where the integration exists over all of space. At any point P where no charges exist, r > 0, and we need not worry about the convergence of the integral. Now let us consider what happens at a point P where charges exist, that is, r = 0. Let us surround the point P by a small sphere R of radius e. The integral
f f f (p dr/r) exists if p is continuous. We define 9 at P Y-R
as lim f f f (p dr/r). This limit exists, for using spherical y-R
coordinates,
If f f "d*T
=I
fo2r fo=
0 pr sin a dr d9 d> r
and
D=E=O,r a. The charge q will induce a charge Q on the sphere. We desire to find Q. We construct a new distribution as follows: Place a unit charge on the sphere, and
assume no other charges in space. The potential due to this charged sphere is p' = 1/r. For the sphere we have initially = 0, Q = ?, and afterward, cp' = 1/a, q' = 1. For the point P we have initially pp = ?, q = q, and afterward, V = 1/b, q' = 0. Applying the reciprocity theorem, we have
0.1+vp-0Q'+q a
b
so that Q = - (alb)q. This is the total charge induced on the sphere when it is grounded. Note that this method does not tell us the surface distribution of the induced charge.
.
Problems 1. A conducting sphere of radius a is embedded in the center of a sphere of radius b and dielectric constant K. The conductor
SFc.64]
STATIC AND DYNAMIC ELECTRICITY
141
is grounded, and a point charge q is placed at a distance r from
its center, r > b > a. Show that the charge induced on the sphere is Q = -Kabq{r[b + (K - 1)a])-1. 2. A pair of concentric conductors of radii a and b are connected by a wire. A point charge q is detached from the inner one and moved radially with uniform speed V to the outer one. Show that the rate of transfer of the induced charge (due to q) from the inner to the outer sphere is dQ
dt
= -gab(b - a)-'V(a + Vt)-2
3. A spherical condenser with inner radius a and outer radius b is filled with two spherical layers of dielectrics Kl and K2, the boundary between being given by r = J(a + b). If, when both
shells are earthed, a point charge on the dielectric boundary induces equal charges on the inner and outer shells, show that K1/K2 = b/a.
4. A conductor has a charge e, and V1, V2 are the potentials
of two equipotential surfaces which completely surround it (V1 > V2). The space between these two surfaces is now filled with a dielectric of inductive capacity K. Show that the change in the energy of the system is - e(V l - V2) (K - 1 )K 1. 5. The inner sphere of a spherical condenser (radii a, b) has a constant charge E, and the outer conductor is at zero potential.
Under the internal forces, the outer conductor contracts from radius b to radius b1. Prove that the work done by the electric forces is E2(b - b1)b-'b1-1.
64. Method of Images. We consider a charge q placed at a point P(b, 0, 0) and ask if it is possible to find a point Q(z, 0, 0) such that a certain charge q' at Q will cause the potential over the sphere x2 + y2 + z2 = a2, a < b, to vanish. The answer is "Yes"! We proceed as follows : From Fig. 63 we have 82 = z2 + a2 - 2az cos B
t== b2+a2-2abcos6 We choose z so that zb = a2, and call Q(a2/b, 0, 0) the image point of P(b, 0, 0) with respect to the sphere. Thus a2
s2 = a (a2 + b2 - 2ab cos 0) =
a2 b2
0
VECTOR AND TENSOR ANALYSIS
142
[SEC. 64
and
The potential at S due to charges q and q' at P and Q is
P=
s,-I- q
t
=-t Cbq'+qJ
and p = 0 if we choose q' _ - (alb)q.
FIG. 63.
The potential at any point R with spherical coordinates r, 0, (p is
_
(a/b)q
q
(r2 + b2 - 2rb cos 0)#
[r2
+ (a4/b2) - (2a2/b)r cos 0]1 (223)
with = 0 on S and V24 = 0 where no charges exist. Now let us consider the sphere of Example 83. The function of (223) satisfies Laplace's equation and is zero on the sphere. From the uniqueness theorem of Example 69, F of (223) is the potential function for the problem of Example 83. The radial field is given by Er
8
q(r-bcos9)
or
(r2 + b2 - 2rb cos B)' (a/b)q[r - (a2/b) cos 0] [r2 + (a4/b2) - (2a2/b)r cos B];
SEC. 651
STATIC AND DYNAMIC ELECTRICITY
143
and the surface distribution is given by
_ (Er)rs _
b2 - a2
q
4ir a(a2 + b2 - 2ab cos 9)'
47r
Problems 1. A charge q is placed at a distance a from an infinite grounded plane. Find the image point, the field, and the induced surface density. 2. Two semiinfinite grounded planes intersect at right angles. A charge q is placed on the bisector of the planes. What distribution of charges is equivalent to this system? Find the field and the surface distribution induced on the planes.
3. An infinite plate with a hemispherical boss of radius a is at zero potential under the influence of a point charge q on the axis of the boss at a distance f from the plate. Find the surface density at any point of the plate, and show that the charge is attracted toward the plate with a force q2 4f2
4g2a3 fa
+
(f' - a')2
65. Conjugate Harmonic Functions. If we are dealing with a two-dimensional problem in electrostatics, we look for a solution of Laplace's equation V2V = 0. The curves V (x, y) = con-
stant represent the equipotential lines. We know that these curves are orthogonal to the lines of force, so that the conjugate function U(x, y) (see Sec. 56) will represent the lines of force. We know that V2V = V2U = 0. We now give an example of the use of conjugate harmonic functions. In Example 73 we saw that
U(x, y) = A tan-' y x
(224)
V (X, y) = 2 log (x2 + y2)
are conjugate functions satisfying Laplace's equation. If we take V (x, y) as the potential function, then the equipotentials are the circles (A/2) log (x2 + y2) = C, or x2 + y2 = e2cie.
144
VECTOR AND TENSOR ANALYSIS
[SEC. 65
Hence the potential due to an infinite charged conducting cylinder is
log r2 = A log r, r2 = x2 + y2
V (X, y) = 2 log (x2 + y2) =
since A log r satisfies Laplace's2equation and satisfies the bound-
ary condition that V = constant for r = a, the radius of the charged cylinder. If q is the charge per unit length, then
_ aV q
(Er)r..a
ar
24ra
41r
41r
r-a
A 4ara
so that A = -2q and V = -2q log r.
1e=Jr)
0
U=U1
U=U0 (0=0)
Frs. 64.
If we choose U(x, y) = A tan-1 (y/x) as our potential function, then the equipotentials U = constant are the straight lines A tan-1 (y/x) = C, or y = x tan (C/A). As a special case we may take the straight lines 0 = 0, 0 = v as conducting planes raised to different potentials (see Fig. 64). The lines of force are the circles (A/2) log (x2 + y2) = V. The theory of conjugate functions belongs properly to the theory of functions of a complex variable. With the aid of the Schwarz transformation it is possible to find the conjugate functions associated with more difficult problems involving the twodimensional Laplace equation.
Problems
1. By considering Example 72, find the potential function and lines of force for two semiinfinite planes intersecting at right angles.
STATIC AND DYNAMIC ELECTRICITY
SEC. 66J
145
2. What physical problems can be solved by the transformation
x = a cosh U cos V, y = a sinh U sin V? Show that V2U = V2V = O
66. Integration of Laplace's Equation. Let S be the surface of a region R for which V24p = 0. Let P be any point of R, and let r be the distance from P to any point of the surface S. We make use of Green's formula
fJf (,p 02' - ¢ V2V) dr = f f (9 Vi'
DAP)
dd
R
We choose ¢ = 1/r, and this produces a discontinuity inside R, namely, at P, where r = 0. In order to overcome this difficulty, we proceed as in Example 66. Surround P by a sphere Z of radius e. Using the fact that V2,P = V24, = 0 inside R' (R minus the Z sphere), we obtain
0=
f8 f (9 V r
r
V-r) - dd +
f f (p V 1r - r Vp) - dd
(225)
//
B
//
Now V(1/r) = -r/r', and on the sphere X,
r
r8
r
e$
and (1/r) VV dd is of the order eIV(pf, so that by letting e --j 0, (225) reduces to
v(p) = 4
ffir
VV - sa V
r)
dd
(226)
This remarkable formula states that the value of Sp at any point P is determined by the value of (p and V(p on the surface S.
Problems
f
1. If P is any point outside the closed surface S, show that f V(1/r)] dd = 0, where V2cp = 0 inside S and
r is the distance from P to any point of S.
VECTOR AND TENSOR ANALYSIS
146
a2
2. Let rp satisfy V2cp =
aX2 +
[SEC. 67
2
2 = 0. Let r be the closed
1P
y°
boundary of a simply connected region in the x-y plane. If P is an interior point of t, show that
a[log (1/r)]
1) app
1
,P(P) - 2x
`0 an
1(`log r an
ds
where use is made of the fact that
ff
(u an -van) da `
(u V 2V - v V 2U) dA
A
n being the normal to the curve. 3. Let (p be harmonic outside the closed surface S and assume that ip --> 0 and rI V pl -> 0 as r -' oo. If P is a point outside 8, show that
P(P) = 4x Ifs (1r V p - ip v
T
dd
where the normal dd is inward on S. 4. Let so be harmonic and regular inside sphere 7. Show that the value of p at the center of is the average of its values over the surface of the sphere. Use (226). 67. Solution of Laplace's Equation in Spherical Coordinates. From Sec. 23, Prob. 1,
+ aB (sin
+
49(p
o
(sin
0
(227)
B
To solve (227), we assume a solution of the form (228)
V(r, 0, gyp) = R(r)6(0),P(,p)
Substituting (228) into (227) and dividing by V, we obtain sin 0
d8) d( r2 dR) - J +-A (sin O -+
R dr
1
dr
0 d9
d8
1
d2 -- =
4) sin 0 d(p2
0
STATIC AND DYNAMIC ELECTRICITY
SEc. 67j
147
Consequently
_-
_1 d dR\ r R dr dr 2
d
1
0 sin 0 dB
sin
0
d9 d9
1
d4
- sin 2 0 4, dp2
(229)
The left-hand side of (229) depends only on r, while the righthand side of (229) depends on 0 and gyp. This is possible only if both quantities are constant, for on differentiating (229) with
respect to r, we obtain
d
d (r2 LR
i
J
= 0. We choose as
the constant of integration c = -n(n + 1), so that R dr (r2 dR)
n(n + 1)
or
r2
4+24+n(n+1)R=0
(230)
It is easy to integrate (230), and we leave it to the reader to show
that R = Ar" + Br-n-1 is the most general solution of (230). Returning to (229), we have
1d
'e
= n(n -{- 1) sing 0 --
sin
(sin 0 de)
(231)
d0
41)
Since we have again separated the variables, both sides of (231) are constant. We choose the constant to be negative, -m2, m being an integer. This choice guarantees that the solution of d24 d(p2
-I-mq =0
is single-valued when p is increased by tar.
(232)
The solution of
(232) is 4) = A cos mtp + B sin mcp.
Finally, we obtain that 0(0) satisfies sin 0 d0 (sino)+[n(n+1)sin2o_m2Je=o (233) We make a change of variable by letting µ = cos 0,
dµ= -sin0d0
VECTOR AND TENSOR ANALYSIS
148
[SEC. 67
so that (233) becomes
[(1 - µ2)
(1 - µ2)
+ [(1
µ2)n(n + 1) - m2]0
dµ
= 0
(234)
If we assume that V is independent of p (symmetry about the z axis), we have m = 0, so that (234) becomes (235)
d
dIA
This is Legendre's differential equation. By the method of series solution, it can be shown that 0 = P"(14)
1
d"(µ2 - 1)n
2 -n!
dµn
satisfies (235); the P (µ) are called Legendre polynomials.
Two important properties of Legendre polynomials are
the
following: f 11 P,n(;&)Pn(p) dµ
=0
if m : n
J 11 P.'(µ) dµ = 2n + 1
(236) (237)
We give a proof of (236). P. and P. satisfy
[(1 dµ
- p!) dµn1 + n(n + 1)Pn = 0
(238) (239)
Multiplying (238) by P. and (239) by P. and subtracting, we obtain
P. d- [(1 - 2) a ,, - Pn d [(1
_ u2) d µ"'
+[n(n+1) -m(m+l)PPm=0
SFC. 671
STATIC AND DYNAMIC ELECTRICITY
149
or
-M2)(Pmddn-PndPm)]
µ[(1 + [n(n + 1) - m(m + 1)]P,aPm = 0 (240; Integrating between the limits -1 and + 1, we obtain
[n(n + 1) - m(m + 1)J f i P,nPn dµ = 0 and ifm -d n, 1
PmPn dµ = 0
A particular solution of (227) which is independent of gyp, that
is, aV = 0, is given by V(r, 0) = (Anrn + Bnr-' 1)Pn(cos 0). aip
Now it is easy to show that any sum of solutions of (227) is also a solution, since (227) is linear in V. Consequently a more general solution is
V=
n-0
(Anrn + Bn7rn-1)Pn(cos 0)
(241)
provided that the series converges. If we wish to solve a problem involving V2V = 0 with spherical
boundaries, we try (241) as our solution. If we can find the constants An, B. so that the boundary conditions are fulfilled, then (241) will represent the only solution, from our previous uniqueness theorems involving Laplace's equation. We list a few Legendre polynomials:
Po(µ) = 1 Pi(µ) = A P2(p) = . (3µs - 1) P,(µ) = 4(5Aa - 3µ)
P,(µ) = 9(35."4 - 30u$ + 3) P.(0) = 0, n odd P-(0) = (_1)n/x 1-3-5 ... (n - 1)
2.4.6 ... n
Pn(1) = 1 Pn(-µ) _ (-1)"Pn(µ)
(242)
uneven
VECTOR AND TENSOR ANALYSIS
150
[SEC. 68
Problems
1. Prove (237). 2. Solve V2V = 0 for rectangular coordinates by the method of Sec. 67, assuming V = X(x)Y(y)Z(z). 3. Investigate the solution of VI V = 0 in cylindrical coordinates. 68. Applications Example 84. A dielectric sphere of radius a is placed in a uni-
form field Eo = Eok. We calculate the field inside the sphere. The potential due to the uniform field is p = -Eoz = -Eor cos 0. There will be an additional potential due to the presence of the dielectric sphere. Assume it to be of the form ArPI = Ar cos 0 inside the sphere and Br-2PI ° Br-2 cos 0 outside the sphere. We cannot have a term of the type Cr-2 cos 0 inside the sphere, for at the origin we would have an infinite field caused by the presence of the dielectric. Similarly, if a term of the type Dr cos a occurred outside the sphere, we would have an infinite field at infinity due to the presence of the sphere. If we let VI be the potential inside and V11 the potential outside the sphere, we have
VI = --Eor cos 0 + Ar cos 0 (243)
Vu = --Eor cos B + B cos 0 r2
Notice that VI and Vn are special cases of (241). We have two unknown constants, A, B, and two boundary conditions,
VI= Vu at r=a or
DN, = D.Y.
(see Sec. 62).
a I
K
=
II
a
at r = a
(244)
From (243) and (244) we obtain
A=K-'Eo,
x+2
B_aaK-1Eo K+ 2
so that
VI = I
-'V+2 Eor cos 0 = -
2 Eoz K
+
(245)
STATIC AND DYNAMIC ELECTRICITY
SEC. 681
151
We see that the field inside the dielectric sphere is
E_ -vVI=K+2Eo and E is uniform of intensity less than E0 since K > 1. the sphere
VII = -Eor cos B +
K - 1 _ Eo cos 8 K + 2 r2
Outside (246)
The radial field outside the sphere is given by
aVII=Eocos0+2K +
E,
ocos9
2aa
For a given r the maximum E, is found at 0 = 0.
Example 85. A conducting sphere of radius a and charge Q is surrounded by a spherical dielectric layer up
to r = b (Fig. 65). Let us calculate the potential distribution.
From
spherical
symmetry V = V (r), so that we try
FlG. 65.
The boundary conditions are
VI= VIIatr=b
(i)
aVI
_
aVII
at r = b Or = Or J2rJT (a D dd = K
(11)
(iii)
Q= 41r
ff s
47
T a2 sin 0 d9 dp
J
From (i) Alb = (B/b) + C; from (ii) -A/b2 = -KB/b2; from (iii) Q = (K/4x) (B/a2) f f dS = KB.
Hence
VECTOR AND TENSOR ANALYSIS
152
VI = Q'
VII =
Q +90C
1
(SEC. 68
(247)
K
Example 86. A conducting sphere of radius a and charge Q is placed in a uniform field. We calculate the potential and the distribution of charge on the sphere. We assume a solution
V = -Eor cos 0 +
B
r
The boundary condition is
Q = 4v
Jf - avlr-a dS
so that Q
=
1
4
,
f2v
f-
(E cos 8 + a a2 sin 8 d8 dsp = B
and
V= -Eorcos8+Qr For the charge distribution aV
a-
cIr
4"
4I(E0cos8+Q
Example 87. Consider a charge q placed at A (b, 0, 0). Let us compute the potential at any point P(r, 8, gyp) (see Fig. 66). The potential at P is 4
V=
= q(r2 + b$ - 2rb cos
8)_;
There are two cases to consider:
(a) r < b. Let µ = cos 8, x = r/b, so that (1 - 2µx
V= b
+2)-}
Now (1 - 2µx + x=)-} can be expanded in a Maclaurin series in powers of x, yielding
SEC. 68]
STATIC AND DYNAMIC ELECTRICITY _
m
V=b
n= b
n0 I
153
m ()fl
n0
P.(--)
( 248)
The proof is omitted here that the Pn(p) are actually the Legendre polynomials. However, we might expect this, since V satisfies Laplace's equation and P,,(µ)rn is a solution of VI V = 0. z
Fia. 66.
(b) r > b. In this case
V = q I Pn6u) rn=0
/b n
(249)
r
Notice that each term is of the form Pn(µ)r-n-1, which satisfies Laplace's equation.
Example 88. A point charge +q is placed at a distance b from the center of two concentric, earthed, conducting spheres of radii a and c, a < b < c. We find the potential at a point P
for a b: Vl = q I [Anrn + (Bn + bn)r-n-1]Pn(cos 6) 0
(251)
For r < b: [(A,, +
V2 = q
b-n-I)rn +
Bnr-n-1]Pn(cos 0
0
The boundary conditions are
V1=0atr=c
(i) (ii)
(252)
V2 = O at r = a
These yield the equations
Ancn + (Bu + bn)c-n-1 = 0 (An + b-n-1)an + Bna n-1 = 0
(i) (11)
so that a2n+1 (C2n+l
Bn T bn+l (a2n+l
-- b2n+l)'
C2n+1)'
A. + b-(n+1) _
b-n-1(C2n+1
-
b2n+1)
a2n+1 - C2n+l
Hence
j m
V2(P) = q
n
-0
b2n+1 - C2n+1 bn+l(a2n+l - C2n+1)
(rn
a2n+1
- rn+1 Pn(cos 0)
(253)
Problems 1. Show that the force acting on the sphere of Example 86 is F = kQEok.
2. A charge q is placed at a distance c from the center of a spherical hollow of radius a in an infinite dielectric of constant K. Show that the force acting on the charge is
STATIC AND DYNAMIC ELECTRICITY
SEC. 691
155
n(n + 1) 2n+1 ,410n+K(n+ 1)(a/
(K - 1)g2'' c2
3. A point charge q is placed a distance c from the center of an
earthed conducting sphere of radius a, on which a dielectric layer of outer radius b and constant K exists. Show that the potential of this layer is m
V
=
(2n + 1 )b2n+1(rn - a2n+lr-n-1) q -c n c cn { [(K +1)n + 1]b 2n+1 + (n +1)(K -1)a 2n+1 } Pn(cos e) 0
4. Show that the potential inside a dielectric shell of internal and external radii a and b, placed in a uniform field of strength E, is
V=
9KE
9K - 2(1 - K2)[(b/a)3 - 1]
r cos 0
5. The walls of an earthed rectangular conducting tube of infinite length are given by x = 0, x = a, y = 0, y = b. A point charge is placed at x = xo, y = yo, z = zo inside the tube. Show that the potential is given by V = 8q
I (m2a2 + nil m-1
n2b2)-e-a-'b-'(mta'+n$t)i*(s-ss)
to
sin
n1rx0
a sin
nrx a
sin
mTryo
b
sin
miry b
69. Integration of Poisson's Equation. Instead of assuming that p is harmonic, let us consider that 1p satisfies V2co = -47rp. By applying Green's formula as in Sec. 66, we immediately obtain
,P(P) = Jff rdr+Jf (rVp
- jP 0
(254)
If we make the further assumption that rp is of the order of 1/r for large r and that jVjpj - 1/r2, we see that by pushing S out to infinity the surface integral will tend to zero. Our assumption is valid, for if we assume the charge distribution to be bounded by some sphere, then at large distances the potential will be of the order of I /r, since we may consider all the charges as essentially
156
VECTOR AND TENSOR ANALYSIS
[SEc. 70
concentrated at a point. Thus
V(P) = JJf e dr
(255)
70. Decomposition of a Vector into the Sum of Solenoidal and Irrotational Vectors. In Example 70, we saw that if Ifl tends to
zero like 1/r2 as r --> oo, then f as uniquely determined by its curl and divergence. We now proceed to write f as the sum of irrotational and solenoidal vectors. Let f dr
W(P) = JfJ
(256)
where r is the distance from P to the element of integration dr. If we write f = fli + f j + f k, W = WA + W,,j + Wek, then
f
ffrldr
W2 = f f f
f2
dr
(257)
M
W3=fffr$dr
00
We assume that the components of f are such that the integrals of (257) converge and that I Wl - 1/r, I V W,,l '' 1/r2, n = 1, 2, 3. From Sec. 69, V2W. = -4xfn, so that
V2W = -4,rf
(258)
From (256)
(259)
VxW= JJJf xVdr 40
Now
V x(V xW) =
--V2W
SEc. 711
STATIC AND DYNAMIC ELECTRICITY
157
so that f
=4-V
-4
X
X
and hence
f = V x A + Vcp
260)
where
A=4-VxW, = Problems 1. Show that (256) is a special case of (254). 2. Find an expression for (p(P) if V2(p = -41rp inside S and if P is on the surface S.
3. f = yzi + xzj + (xy - xz)k. Express f as the sum of an irrotational and a solenoidal vector. 71. Dipoles. Let us consider two neighboring charges -q and +q situated at P(x, y, z) and Q(x + dx, y, z). The potential at
the origin 0(0, 0, 0) due to -q is -q/r, and that due to +q is q/(r + dr), where r = (x2 + y2 + Z2)I and
r+dr= [(x + dx)2 + y2 + Z2]1 The potential at 0(0, 0, 0) due to both charges is q
r + dr
q
r
g2dr r2
Now dr = x dx/r, so that rp - qx dx/r3. If we now let q -p co and dx -+ 0 in such a way that q dx remains finite, we have formed what is known as a dipole. Let r be the position vector from the origin to the dipole, and let M = q dr, where dr is the vector from
the negative charge to the positive charge; !drl = dx. We have (M r)/r8 = (qr dr)/ra = (qx dx)/r3, so that
M is called the strength or moment of the dipole.
than one dipole, the potential at a point P is given by
For more
VECTOR AND TENSOR ANALYSIS
158
i= i
[SEC. 72
(Mi.ri) ri3
where r; is the vector from P to the dipole having strength M. Example 89. The field strength due to a dipole is E _ --Vv so that r (262) E = V rs ) rs ra
-
-
Example 90.
Potential energy of a dipole in a field of potential V.
Let (p, be the potential at the charge q and c2 the potential at the charge -q. The energy of the dipole is
L
app
ds = M W = spiq + IP2(-q) = q(spl - IP2) = g as ao where ds is the distance between the charges. Now
d,p = ds V p
so that W = M as VV = M V(p. 72. Electric Polarization. Let us consider a volume filled with dipoles. The potential due to any single dipole is given by (261). If we let P be the dipole moment per unit volume, that is, P = lim (AM/or), then the total potential due to the dipoles is A"o
= I f f r ra- dr
(263)
Now V V. (P/r) = (1/r)V P - [(P r)/r3]. The reason that we have taken V(1/r) t= r/r3 instead of -r/r3 is that
r = [( - x)2 + (n - y)2 + (J
- z)2]'
and V performs the differentiations with respect to x, y, z, the coordinates of the point P at which V is being evaluated. The coordinates t, ot, t belong to the region R and are the variables of
integration, and r = (E - x)i + (n - y)j + (r - z)k. (263) becomes `P=fffv.(!)dr R
fff R
Hence
SEC. 72]
STATIC AND DYNAMIC ELECTRICITY
159
and
(p -
ff!.dd_ fff_!d
r
(264)
by applying the divergence theorem. Example 91. Let us find the electric intensity at the center of a uniformly polarized sphere. Here P = Pok, so that V P = 0 inside R. Hence (264) becomes sp(x, y, z) =
Pok dd
ff
x)2 + (71 - y)2 + ( - z) 2]i
S
and
E=
fj
Po(k
dd)[(E - x)i + (n
- y)j + (1' - z)k]
[(E-x)2+(7] -y)2+(J -z)2]$
S
E(0,0,0)
f
f
ni + 'k)Po(k dd)
(265)
Now for points on the sphere, 52 +' 72+-2 = a2,
and letting t = a sin B cos -o, n = a sin 0 sin tp, is easily seen that (265) reduces to
E(0, 0, 0) = -frPok
a cos 0, it (266)
E is independent of the radius of the sphere. By superimposing (concentrically) a sphere with an equal but negative polarization, we see that the field at the center of a uniformly polarized shell is zero. Problems 1. Prove (262). 2. Prove (266). 3. If M1 and M2 are the vector moments of two dipoles at A and B, and if r is the vector from A to B, show that the energy - 3(M1 r)(M2 r)r-6. of the system is W = M1 4. The dipole-moment density is given by P = r over a sphere of radius a. Calculate the field at the center of the sphere. Mgr-a
160
VECTOR AND TENSOR ANALYSIS
[SEc. 73
73. Magnetostatics. The same laws that have held for electrostatics are true for magnetostatics with the exception that OZcp.. = 0 always, since we cannot isolate a magnetic charge. We make the following correspondences, since all the laws of electrostatics were derived on the assumption of the inversesquare force law, which applies equally well for stationary magnets.
EH
Magnetostatics
Electrostatics q
qm
D ---) B (magnetic induction) K a, I r + V4J > S. Moreover
ft
F f(r
+ Vt) dt = 1 fttip f(r + Vt) d(r + Vt)
=Vj
da F
f (x) dx
(310)
for a fixed r. Now if S is chosen very small, the value of (310) reduces to approximately
STATIC AND DYNAMIC ELECTRICITY
SBc. 831
1
r z=o (F)
V
IL f(x)
dx =
181
1
V
(r
t_-r/v
Hence the left-hand side of (3' 09) reduces to 4V
(311)
()t_-r/vdr
Now considering the right-hand side of (309), we see that
lim e-.0
ff
E
rf(E + Vt) Dtp L
E
(pf'(E + Vt)
+
R] dd = 0
E
since dd is of the order E2, and f, f , tp, Dtp are bounded for a fixed a.
We also have that
lim -
t: jI
dt f f
since f f dS =
'pE, f(E + Vt) . dS = -
1
v 47rcp(P)
(312)
and for small 6,
E
Finally,
fs f i:'
dtlf(r+ Vt)D
(rf'
Dr] dd
2
r2
L
ii fl'
= s
dt{f(r+Vt)Dp-Vr].dd
r
r
f
- f f j,t dt r f'(Dr . dd) = ff f .t dt
[f(r + Vt) r
+ -f Dr] . dd + f f j i s rV f at dt (Dr dd)
(313)
s
on integrating by parts and noticing that f} = 0. Finally the right-hand side of (313) becomes equivalent to
V sf f(rD t - -r/V -,p t--r/V
1
r
1
t3(o
rV at t- -r/V (314)
VECTOR AND TENSOR ANALYSIS
182
[SEC. 83
Combining (311), (312), and (314), we obtain V(P)
` 1 JR J
F=
dr
-r/ v 1
- i fJ
(v , ~
S
V r + rV at
V
r)
dd
(315)
t= -r/'V
Now let S recede to infinity and assume that (p,
when
evaluated at t = -r/V on the surface S, have the value zero until a definite time T. For large r, t = -r/V is negative and so is always less than T. Hence the surface element vanishes, and V
(P)=
JJJ
t- -r/V
di
-
(316)
The solutions to (305) are thus seen to be A (P,
t) =
f fJ W
4
pv
Ie
r t-(r/V)
,v(P,t)'JLI x t - (r/V) whe re V = Finally,
dr 317)
dr
B=VxA l aA E=-'cat-V
(318)
The physical interpretation of these results is simple. The values of the magnetic and electric intensities at any particular point P at any instant t are, in general, determined not by the state of the rest of the field (p, v) at that particular instant, but by its previous history. The effects at P, due to elements at a distance r from P, depend on the state of the element at a previous time t -- (r/V). This is just the difference in time required for the waves to travel from the element to P with the velocity V = c/ V AK, hence the name retarded potential. Had we considered the function f(r - Vt), we should have obtained a solution depending on the advanced potentials. Physically this is impossible, since future events cannot affect past eventsl
SEC. 831
STATIC AND DYNAMIC ELECTRICITY
183
Problem
1. A short length of wire carries an alternating current, j = pv = Io (sin wt)k, -1/2 5 z 5 1/2. (a) At distances far removed from the wire, show that
A=
j-o1
cr
sin w
(t
- rc//J k
and that in spherical coordinates A,.
=jot sin wit - r1 cos 0
\ c/l / Isin0 Ae= - Iotsinwltc// cr cr
A, =O (b) Show that H, = He = 0, and that Hw =
cr sin 0
cos w Ct - c) + r sin w IC
t - -) 1
(c) Find p from the equation of gauge invariance, and then
E,, E8, E from E + c A - V o.
CHAPTER 6 MECHANICS
84. Kinematics of a Particle. We shall describe the motion of a particle relative to a cartesian coordinate system. The motion of any particle is known when r = x(t)i + y(t)j + z(t)k is known, where t is the time. We have seen that the velocity and acceleration, relative to this frame of reference, will be given by
v' dt i +dt a=
d2y
d2x dtz 1
+
dt
j
dtz
k
d zz
+
dt2
k
The velocity may also be given by v = vt, where v is the speed
and t is the unit tangent vector to the curve r = r(t). Differentiating, we obtain
a=dt
dtt+vdsdt
by making use of (95).
=dtt+Kv2n
(319)
Analyzing (319), we see that the accelera-
tion of the particle can be resolved into two components: a tangential acceleration of magnitude dt, and a normal accelera-
This latter acceleration is called centripetal acceleration and is due to the fact that the velocity vector is changing direction, and so we expect the curvature to play a role here. For a particle moving in a plane, we have seen in Sec. 17, Example 18, that the acceleration may be given by tion of magnitude v2rc = v2/p.
2r
2l
- r ` de]R -}- r d (2 de P = [ dt r
a
184
MECHANICS
SEC. 84J
Example 101.
185
Let us assume that a particle moves in a plane
and that its a cceleration is only radial. In this case we must have r d [ r2 de1 = 0, and integrating, +r2 dte = h = constant.
From the calculus we know that the sectoral area is given by dA = }r2 dB (see Fig. 72). Thus dA = constant, so that
equal areas are swept out in equal intervals of time. Example 102. For a particle
Fie. 72.
moving around a circle r = b with constant angular speed coo = d
we have dt = 0 and
(r2wo) = 0, so that a = -bwo2R.
Example 103. To find the tangential and normal components of the acceleration if the velocity and acceleration are known.
v=vt, and
a v = vat Also
so that
as =
v
axv=va,nxt= -vanb
and
an = I$)V,i°I
Problems 1. A particle moves in a plane with no radial acceleration and constant angular speed wo. Show that r = Ae"o' + Be at. 2. A particle moves according to the law
r = cos t i + sin t j + t2k Find the tangential and normal components of the acceleration.
VECTOR AND TENSOR ANALYST °
186
[SEc. 85
3. A particle describes the circle r = a cos 0 with constant Show that the acceleration is constant in magnitude and directed toward the center of the circle. 4. A particle P moves in a plane with constant angular speed w about 0. If the rate of increase of its acceleration is parallel speed.
d2r
to OP, prove that ate = 4rw2 5. If the tangential and normal components of the acceleration
of a particle moving in a plane are constant, show that the particle describes a spiral. 85. Motion about a Fixed Axis. In Sec. 10, Example 12, we saw that the velocity is given by v = w x r. Differentiating, we obtain dw dr xr a=wxa+dt
a=wxv+axr
(320)
Since v = w x r, we where a is the angular acceleration --' at have also
wl
a=wx(wxr)+axr
_ (w - r)w - w2r + a x r
If we take the origin on the line of w in the plane of the motion, then w is perpendicular
tororor = 0, so that
a= -w2r+axr a x r is the tangential acceleraFm. 73.
tion, and w x (w x r) is the
centripetal acceleration. If we assume that a particle P is rotating about two intersecting lines simultaneously, with angular velocities cal, w2 (Fig. 73), we can choose our origin at the point of intersection so that
vi = wl x r, and the total velocity is
v2 = 632 x r
V = V1 + V2 = (wi + (02) x r
MECHANICS
SFC. 86]
187
A particle on a spinning top that is also precessing experiences such motion. 86. Relative Motion. Let A and B be two particles traversing curves r, and r2 (Fig. 74). r, and r2 are the vectors from a point 0 to A and B, respectively.
r2=r+rl
(321)
Definition: dt is the relative velocity of B with respect to A, written V4(B).
Fia. 74.
Differentiating (321), we have dr2
dr
dt - dt
dri
+
dt
or
Vo(B) = VA(B) + Vo(A)
(322)
More generally, we have
Vo(A) = V4,(A) + VA,(A1) + VA.(A,) + ..
.
+ Vo(A.)
It is important to note that V4(B) _ -VB(A). Example 104. A man walks eastward at 3 miles per hour, and the wind appears to come from the north. He then decreases his speed to 1 mile per hour and notices that the wind comes from the northwest (Fig. 75). What is the velocity of the wind? We have V0(W) = VM(W) + V0(M) G(ground)
VECTOR AND TENSOR ANALYSIS
188
ISEc. 86
In the first case
VM(W) = -kj,
V0(M) = 3i
so that
VG(JV) = -kj + 3i In the second case,
VM(W) = h(i - j),
VG(M) = i
so that
V0(W) = h(i - j) + i = (h + 1)i - hj, and
3=h+1,
-k= -h,
VG(W) =3i-2j
miles per hour, and its direction The speed of the wind is makes an angle of tan-' I with the south line.
N
x.
E
S Fia. 75.
Fzo. 76.
Example 105. To find the relative motion of two particles moving with the same speed v, one of which describes a circle of radius a while the other moves along the diameter (Fig. 76). We have
P=acos0i-}-asin0j,
adze=v
Q = (a - vt)i This assumes that both particles started together.
T -dQ=(-a sin0dO+v)i+acos0doi VQ(P) = v(1 - sin 0)i + v cos 6 j
MECHANICS
Sec. 87]
189
The relative speed is IVQ(P)I = [v2(1
- sin
0)2 + v2 cos2 B]} = 2}v(1 - sin B)}
Maximum IVQ(P)I occurs at 0 = 3x/2, minimum at 0 = x/2. Problems 1. A man traveling east at 8 miles per hour finds that the wind seems to blow from the north. On doubling his speed, he finds that it appears to come from the northeast. Find the velocity of the wind. 2. A, B, C are on a straight line, B midway between A and C. It then takes A 4 minutes to catch C, and B catches C in 6 minutes. How long does it take A to catch up to B?
3. An airplane has a true course west and an air speed of 200 miles per hour. The wind speed is 50 miles per hour from 1300. Find the heading and ground speed of the plane. 87. Dynamics of a Particle. Up to the present, nothing has
been said of the forces that produce or cause the motion of a particle. Experiment shows that for a particle to acquire an acceleration relative to certain types of reference frames, there must be a force acting on the particle. The types of forces encountered most frequently are (1) mechanical (push, pull), (2) gravitational, (3) electrical, (4) magnetic, (5) electromagnetic. We shall be chiefly concerned with forces of the types (1) and (2). For the present we shall assume Newton's laws of motion hold for motion relative to the earth. Afterward we shall modify this. Newton's laws are: (a) A particle free from the action of forces will remain fixed or will continue to move in a straight line with constant speed. (b) Force is proportional to time rate of change of momentum,
that is, f = dt (mv). In general, m = constant, so that
The factor m is found by experiment to be an invariant for a given particle and is called the mass of the particle. In the theory of
relativity, m is not a constant. my is called the momentum. (c) If A exerts a force on B, then B exerts an equal and opposite force on A. This is the law of action and reaction: fAB = -fha
VECTOR AND TENSOR ANALYSIS
190
[SEC. 88
By a particle we mean a finite mass occupying a point in our Euclidean space. This is a purely mathematical concept, and physically we mean a mass occupying negligible volume as compared to the distance between masses. For example, the earth
and sun may be thought of as particles in comparison to their distance apart, to a first approximation. 88. Equations of Motion for a Particle.
Newton's second law
may be written f = m dt = ma. We postulate that the forces acting on a particle behave as vectors. This is an experimental fact. Hence if fl, f2j . . . , f act on m, its acceleration is given by a
1
(fl+f2+....+fn)_
m
1I
mti_1
We may also write f = m d
dt2
,
f,_ mff
where r is the position vector from
the origin of our coordinate system to the particle. If the particle d
is at rest or is moving with constant velocity, then
dt2
= 0, and
so f = 0, and conversely. Hence a necessary and sufficient condition that a particle be in static equilibrium is that the vector sum of the forces acting on it be zero.
A standard body is taken as the unit mass (pound mass). A poundal is the force required to accelerate a one-pound mass one foot per second per second. The mass of any other body can be compared with the unit mass by comparing the weights (force of f12/ m2 gravit y at mean sea leve l ) o f th e two objects. This assumes the equivalence of gravitational mass and inertial mass. Example 106. Newton's law of gravi-
tation for two particles is that every pair of particles in the universe exerts Fzo. 77.
a mutual attraction with a force directed along the line joining the particles, the magnitude of the force being inversely
proportional to the square of the distance between them and directly proportional to the product of their masses. f12 = (Gmim2/r2)R (see Fig. 77). G is a universal constant.
Let
191
MECHANICS
SEC. 88]
the mass of the sun be M and that of the earth be m. We shall assume that the sun is fixed at the origin of a given coordinate system (Fig. 78). The force act-
ing on the earth due to the sun is f = - (Gm.M/r3)r From the second law GmM rs r
dv dt
d2r m dt2
so that dv
GM
dt
r3
M
(323)
Fia. 78.
Now d v) dt (r x
dv
= r x dt
and hence d
dt(rxv)=rx(-GMr) =0
This implies
r x v = h = constant vector or
rxa = dr
(324)
h
Since Ir x drl = twice sectoral area, we have 2 dA = Ihl, or equal
areas are swept out in equal intervals of time. This is Kepler's first law of planetary motion. Moreover, r I r x dt ] = r h = 0,
so that r remains perpendicular to the fixed vector h, and the motion is planar. Now
-GMrxh= -GMrx(rxv)
a xh from (324), and
d
d
(v x h) = d x h, so that (v x h)
GM
r x (r x v)
(325)
VECTOR AND TENSOR ANALYSIS
192
Now r = rR, where R is a unit vector.
SEC. 88
Hence R
v
so that (325) becomes
\ d(vxh)=-GMrx (rxrR) / =
-GMRx(RxdR) -GM K R .
ddR
t/
R - R2 dRJ
= GM ddR
(326)
since R is a unit vector. Integrating (326), we obtain
vxh=GMR+k and
h' = GMr + rk cos (R, k)
(327)
Thus h2/GM
r
(328)
1 + (k/GM) cos (r, k)
We choose the direction of the constant vector k as the polar axis, so that
r
h'/GM 1 + (k/GM) cos 8
(329)
This is the polar equation of a conic section. For the planets these conic sections are closed curves, so that we obtain Kepler's second law, which states that the orbits of the planets are ellipses with the sun at one of the foci. Let us now write the ellipse in the form S
r
1+ e cos 9
where e=
GM' p
k
MECHANICS
SEC. 88]
193
The curve ci osses the polar axis at 0 = 0, 0 = it so that the length of the major axis is
ep + ep 2, - 1+e 1-e
2p
2h2
1-e2
GM(1-e2
For an ellipse, b2 = a2 - c2 = a2 - e2a2, orb = a(l - ,2)1. area of the ellipse is A = Tab = ,ra2(1 - e2)1, and since
The
dA = Jh, dt
the period for one complete revolution is
T
2A
2ara2(1 - e2)i Zral - e2)1 = G1Mi
- h - a1GiM1(1
Thus
_
T2
47r2
0
GM =
constant, for all planets
(330)
This is Kepler's third law, which states that the squares of the periods of revolution of the planets are proportional to the cubes of the mean distances from the sun. Problems
1. A particle of mass m is attracted toward the origin with the force f = - (k2m/r6)r. If it starts from the point (a, 0) with the speed vo = k/21a2 perpendicular to the x axis, show that the path is given by r = a cos 0. 2. A bead of mass m slides along a smooth rod which is rotating
with constant angular speed w, the rod always lying in a horizontal plane. Find the reaction between bead and rod. 3. A particle of mass m is attracted toward the origin with a force - (mk2/r3)R._ If it starts from the point (a, 0) with velocity vo > k/a perpendicular to the x axis, show that the equation of the path is (OW k2)i r = a sec C 0
-
avo
I
4. In a uniform gravitational field (earth), a 16-pound shot leaves the putter's fingers 7 feet from the ground. At what angle should the shot leave to attain a maximum horizontal distance?
194
VECTOR AND TENSOR ANALYSIS
[SEC. 89
5. Assume a comet starts from infinity at rest and is attracted toward the sun. Let ro be its least distance to the sun. Show that the motion of the comet is given by r = 2ro/(1 + cos 0). 89. System of Particles. Let us consider a system consisting of a finite number of particles moving under the action of various forces. A given particle will be under the influence of two types of forces: (1) internal forces, that is, forces due to the interaction of the particle with the other particles of the system, and (2) all
other forces acting on the particle, said forces being called external forces.
If r; is the position vector to the particle of mass m,, then we shall designate f,(e) as the sum of the external forces acting on the jth particle, and f,(i) as the sum of the internal forces acting on this particle. Newton's second law becomes for this particle
f1( + f1( = m; d2r'
(331)
Unfortunately, we do not know, in general, f;('), so that we shall not try to find the motion of each particle but shall look rather for the motion of the system as a whole. Since Eq. (331) is true for each j, we can sum up j for all the particles. This yields n
J=1
f'c6> +
d 2r,
f'(i) ;=1
1-1
m' dt2
From Newton's third law we know that for every internal force n
there is an equal and opposite reaction, so that This leaves
A
f ®= 0.
n
n
l m,
f,
dt2
(332)
We now define a new vector, called the center-of-mass vector, by the equation n
Im,r,
n
ra =1' =
(333) n
;11
7-1
MECHANICS
SEC. 89)
195
The end point of r. is called the center of mass of the system. It is a geometric property and depends only on the position of the particles. Differentiating (333) twice with respect to time, we obtain Md2r` _ dt2
n
md?r, ' dt2
;=1
so that (332) becomes n
f
f.(e)=M
2_1°
(334)
z
)ml
Equation (334) states that the center of mass of the system accelerates as if the total mass were concentrated there and all the external forces acted at that point.
Fia. 79.
Example 107. If our system is composed of two particles in free space and if they are originally at rest, then the center of
mass will always remain at rest, since f = 0 so that d2° = 0, and r. = constant satisfies the equation of motion and the initial condition
0.
For the earth and sun we may choose the center
of mass as the origin of our coordinate system (Fig. 79). The equations of motion for earth and sun are m
d2r1 dt2
r2)2f = - (rlGmMR
M d2r2 dt2
_ (rlGmMR - 12)2
Since r. = 0, we have mr, + Mr2 = 0, and d2r1 d12
_
-GM
rl
[1 + (m/M))2 rig
196
VECTOR AND TENSOR ANALYSIS
[SEc. 91
This shows that m is attracted toward the center of mass by an inverse-square force.
The results of Example 106 hold by replac-
ing M by M[1 + (m/M)1-2. Problems
1. Show that the center of mass is independent of the origin of our coordinate system. 2. Particles of masses 1, 2, 3, 4, 5, 6, 7, 8 are placed at the corners of a unit cube. Find the center of mass. 3. Find the center of mass of a uniform hemisphere. 4. Find the force of attraction of a hemisphere on another hemisphere, the two hemispheres forming a full sphere. 90. Momentum and Angular Momentum. The momentum of a particle of mass m and velocity v is defined as M = mv. The total momentum of a system of particles is given by M = j m;v1. j-1
We have at once that dM dt
n
I mj j-1
dv1
dt =
n
I f; (e) = f
(335)
j-1
We emphasize again that the mass of each particle is assumed constant throughout the motion. The vector quantity r x my is defined as the angular momentum, or moment of momentum, of the particle about the origin 0. f The total angular momentum is given by n
H = E r; x mjvj
(336)
j-1 91. Torque, or Force Moment. Let Fla. 80.
f be a force acting in a given direction and let r be any vector from the origin
whose end point lies on the line of action of the force (see Fig. 80). The vector quantity r x f is defined as the force moment, or torque, of f about 0. For a system of forces, n
L = I r, x f, j-1
(337)
MECHANICS
SEC. 921
197
We immediately ask if the torque is different if we use a differ-
ent vector ri to the line of action of f.
The answer is in the
negative, for
(r, - r) x f = 0 since ri - r is parallel to f. Hence rl x f = r x f. What of the torque due to two
equal and opposite forces both acting along the same line? It is zero, for
r1xf+r1x(-f)
=r1x(f-f)=0
Two equal and opposite forces
with different lines of action constitute a couple (see Fig. 81).
Let ri be a vector to f and r2 a vector to -f. The torque due
Fio. 81.
to this couple is
L = r1xf+r2x(-f) = (r1 - r2) x f
The couple depends only on f and on any vector from the line of action of -f to the line of action of f. Problems 1. Show that if the resultant of a system of forces is zero, the total torque about one point is the same as that about any other point. 2. Show that the torques about two different points are equal, provided that the resultant of the forces is parallel to the vector joining the two origins.
3. Show that any set of forces acting on a body can be replaced by a single force, acting at an arbitrary point, plus a Prove this first for a single force. 4. Prove that the torque due to internal forces vanishes.
suitable couple.
92. A Theorem Relating Angular Momentum with Torque. We are now in a position to prove that the time rate of change of angular momentum is equal to the sum of the external torques for a system of particles.
198
VECTOR AND TENSOR ANALYSIS
[SEC. 93
Since n
H = Ir3xmjv
r, x
j=1
j=1
dri
m'dt
we have on differentiating
dH _ dt
n
Z r, x m,
dt2
n
I
j-1 r' x
aii = dt
n
d2r,
dri
dri
+ j=1 dt x m' dt
(f1" + f, (') ) n
I r; x
j=1
f;(e)
=L
(338)
93. Moment of Momentum (Continued). It is occasionally more useful to choose a moving point Q as the origin of our
FIG. 82.
coordinate system. space. We define
Let 0 be a fixed point and Q any point in n
dri
HQa = I (r, -- rQ) x m; di j-1
(339)
The superscript a stands for absolute momentum, that is, the velocity of m; is taken relative to 0, whereas the subscript Q stands for the fact that the lever arm is measured from Q to the particle m; (Fig. 82).
Differentiating (339), we obtain
MECHANICS
SEc. 941
dHQa
_
dt
drQl
(dri _ drQ
dl
dr,
dt /
+ L (r. -
dt
dr1
n
j-1
m;r,, so that M dt =
m1
dr,
-, and
l1
j=1
j-1
dt2
(r1 - rQ) x (f.(e) + f .(i))
xLm'dt+I
n
Now Mr =
rQ) x "ni
j=1
n
j=1
d2r;
nn`
X 7n,
(dt
j=1
199
n f,(i)=0,
j=1
r1 x f;(') = 0 from Sec. 91, so that dHQa
M
dt =
drQ
I
dF
n
(r, - rQ) dt x dt + j-1
x fj(e)
or dtQa
LQ (e)
-M
Qx dt
dtc
(340)
We can simplify (340) under three conditions:
1. Q at rest, so that
drQ
dt
=0
2. Center of mass at rest, dt` = 0
3. Velocity of Q is parallel to velocity of center of mass, drQ
dre
dt
dt
0
In all three cases
dHe
= LQ(e)
dt
(341)
In particular, if LQ(e) = 0, then He - constant, and this is the law of conservation of angular momentum. 94. Moment of Relative Momentum about Q. In Sec. 93 we assumed that the absolute velocity of each particle was known.
It is often more convenient to calculate the velocity of each particle relative to Q.
This is
dr, dt
- drQdt
We now define rela-
VECTOR AND TENSOR ANALYSIS
200
[SEC. 94
tive moment of momentum about Q as n
HQr =
j-1
(rj - rQ) x mj
(rj - rQ)
dl
(342)
Differentiating, dHQ*
dt
j-1 =
(d2rj
(r' - rQ) x m'
(r7 - rQ) x
dt2
dl2
d2rQ
`
(fj(ei
/
d2rQ
f3(i))
n
x
dt2
I m, (rj - rQ) j-1
We see that dtiQr
a,
LQ(' +
'rQ d12
n
x I mj (rj - rQ)
(343)
js1
Under what conditions does dd Q' = Lo d2rQ
dt2
?
We need
n
x I mj(rj - rQ) = 0
or
M
d2rQ
dt2
x (r, - rQ) = 0
(344)
Now (344) holds if 1. rQ = rQ, or Q is at the center of mass.
2. Q moves with constant velocity,
dQ l = 0.
2
3. r. - rQ is parallel to
2 dt'
Problems n
1. Show that
jet
a rQ
m4(r1 - rQ) x dt2 = M(r. - rQ) x
d2rQ dt2
2. A system of particles lies in a plane, and each particle remains at a fixed distance from a point 0 in this plane, each
MECHANICS
SEC. 95]
201
particle rotating about 0 with angular velocity w. n
Show that
Ho = Iw, where I = I m;r;2, and show that Lo = I at j=1
3. A hoop rolls down an inclined plane. What point can be
taken as Q so that the equation of motion (343) would be simplified?
95. Kinetic Energy. We define the kinetic energy of a particle of mass m and velocity v as T = jmv v. C M For a system of particles, r.-r n
T
n
1
2 mv,2 f=1
fdrs\ 2
1
2 m'
.1-1
(345)
dt 1
P;
`r
r
Now let r, be the vector to the center of 0 mass C (Fig. 83). It is obvious that
Fia. 83.
r;=rr+(r,-r-) so that dr; dt dr; dr; _ (dro)2 di dt dt
_ dr, dt
+
d dt
(r; - r-)
(r, - rc) -{+2 drd dt dt
I d. L dt
(r; - rc) ]
2
Hence 1 (dr12 dr" T=2M dtJ +dt
d
n
2
+ jn
Now Mr. =
1
2 m'
n
(346)
n
r,, so that
m;r, j-1
r`)] [dt (r' -
j-1
m1 alt (r; - r-) = 0, ;-1
and (346) reduces to
T=IM
()2
-}-
-1 2
m; [-d
dt
(r; - r.)]2 a)(347) 1
This proves that the kinetic energy of a system of particles is equal to the kinetic energy of a particle having the total mass
202
VECTOR AND TENSOR ANALYSIS
[SEC. 96
of the system and moving with the center of mass, plus the kinetic energy of the particles in their motion relative to the center of mass. 96. Work. If a particle moves along a curve r with velocity v
under the action of a force f, we define the work done by this force as
W=
fr f t
fr f
f acts at right angles to the path, no work is done. If the field is conservative, f = -V(p, the work done in taking
the particle from a point A to a point B is independent of the path (see Sec. 52). Now dvi = fj(a) + fi(.) dt dvi = fi(e) . V1. + mivi . dt m,
Vi
and integrating and summing over all particles, n
J`o
mivi
dt' dt =
La
i:1
fi(a)
. vi dt +
fee:' fi('i
. v; dt
or n
i=1
lmi[vi2(ti) - vi2(to)l = W(e)
W°
(349)
This is the principle of work and energy. The change in the kinetic energy of a system of particles is equal to the total work done by both the external and internal forces.
If the particles always remain at a constant distance apart, (ri - rk)2 = constant, the internal forces do no work. Let r, and r2 be the position vectors of two particles whose distance apart remains constant, and let f and -f be the internal forces of one particle on the other and conversely. Now
(r, - r2) (r, - r2) = constant
MECHANICS
Sec. 97]
203
so that
(r, - r2)
dr, di
- dr2dt/
0
(350)
Also
W(o =
f ff
f is parallel to r, - r2, Ave have f = a(r, - r2) and
f (v, - v2) = a(r, - r2) (v, - v2) = 0
from (350). Thus W(° = 0. Problems 1. A system of particles has an angular velocity w. n
T=
i-i
Show that
Jm;lw x r,J2.
2. If to of Prob. 1 has a constant direction, show that T = }Iw2, n
where I =
md;2, d; being the shortest distance from m; to
line of w.
3. Show that dT = w L, by using the fact that T =
4-mv;2
and that v, = to x r;. 4. Show that the kinetic energy of a system of rotating particles is constant if the system is subjected to no torques. What if L is perpendicular to w?
5. A particle falls from infinity to the earth. Show that it strikes the earth with a speed of approximately 7.0 miles per Use the principle of work and energy. 97. Rigid Bodies. By a rigid body we mean a system of particles such that the relative distances between pairs of points remain constant during the discussion of our problem. Actually no such systems exist, but for practical purposes there do exist such rigid bodies, at least to a first approximation. Moreover, the rigid body may not consist of a finite number of particles, but rather will have a continuous distribution, at least to the unaided eye. We postulate that we can subdivide the body into a great many small parts so that we can apply our laws of motion for particles to this system, this postulate implying that we can use second.
VECTOR AND TENSOR ANALYSIS
204
the integral calculus. the following form:
[SEC. 98
Our laws of motion as derived above take
T = f f f -pv2 dr,
p = density
R
ff pr dr f f f p dr R
dr,
( 351)
2
J
fR f f(e)
dt2
H = f f f prxvdr R
-it ff r x f(e) dr where f(.) is the external force per unit volume. 98. Kinematics of a Rigid Body. Let 0 be a point of a rigid body for which 0 happens to be fixed.
It is easy to prove that the velocity f V (P1 = yr
P
of any other point P of the body must be perpendicular to the line joining 0
to P, for if r is the position vector from 0 to P, we have r - r = constant throughout the motion so that 0.
Q.E.D.
r dt = We next prove that if two points of a rigid body are fixed, then all other particles of the body are rotating around the line joining these two points.
Let A and B be the fixed
points and P any other point of the body.
From above we have
so that P is always moving perpendicular to the plane ABP. Moreover, since the body is rigid, the shortest distance from P to the line A B remains constant, so that P moves in a circle
Sec. 981
MECHANICS
around AB (Fig. 84). could be written
205
We saw in Sec. 10 that the velocity of P
VP =wxrp Is w the same for all particles? Yes! Assume Q is rotating about AB with angular velocity w,, so that vQ = to, x rQ. Now (rP - rQ)2 = constant, so that (rP - rQ) (vp - vQ) = 0 or
(rP - rQ) ((a x rp - w, x rQ) = 0 Thus
x rP -
xrQ = 0
and rQ x rp
(w, - w) = 0
We leave it to the reader to conclude that w, = w. VA
B
1
FIG 85.'
If one point of a rigid body is fixed, we cannot, in general, hope to find a fixed line about which the body is rotating. However, there does exist a moving line passing through the fixed point so that at any instant the body is actually rotating around this line. The proof proceeds as follows: Let 0 be the fixed point of our rigid body and let r` be the position vector to a point A. From above we know that the velocity of A, VA, is perpendicular to rA. Construct the plane through 0 and A perpendicular to VA (Fig. 85). Now choose a point B not in the plane. We also have that vB rB = 0, so that we can construct the plane through 0 and B perpendicular to vB. Both planes pass through 0, so
206
VECTOR AND TENSOR ANALYSIS
[SEC. 98
that their line of intersection, 1, passes through 0. Now consider any point C on this line. We have vc rc = 0. Moreover, (rc - rA) - (rc - rA) ° constant, so that
(re-rA).(vc-VA) =0 and (rc - rA) VC = 0, since vA is perpendicular to (rc - rA). Similarly (re - rB) vc = 0. Hence the projections of vc in three directions which are nonplanar are zero.
This means that
FIG. 86.
Vc - 0, so that we have two fixed points at this particular instant. Hence from the previous paragraph the motion is that of a rotation about the line 1. If w is the angular-velocity vector, then v; = w x r;, where r; is the vector from 0 to the jth particle. Now let us consider the most general type of motion of a rigid -r represent a fixed coordinate system in space, body. Let and let 0-x-y-z represent a coordinate system fixed in the rigid body (see Fig. 86). Let ,o; and r; represent the vectors from 0'
and 0 to the jth particle, and let a be the vector from 0' to 0. We have ei = a + r,, and differentiating, dLD;
dt
- da dt
+
dr, dt
MECHANICS
SEC. 98]
207
Now l represents the velocity of Pi relative to 0.
This means
0 is fixed as far as Pi is concerned, and from above we know that dri
dt - w
Thus
x ri.
_dei =
da
dt
dt
v'
(352)
+wxr;
that is, the most general type of motion of a rigid body is that of
a translation
A dt
plus a rotation w x r;.
We next ask the following question: If we change our origin
from 0 to, say, 0" does w change? (Fig. 87.) The answer is
"No"! Let b be the vector
0
from 0' to 0". Then v;
rf
=t=' + w,xr;
But
and
db
da
dt
dt
ri = (a
+w x (b - a) -- b) + ri
Fio. 87.
Thus
vi = d +w x(b - a)+wi x (a -b)+w, xri
(353)
Subtracting (352) from (353), we obtain
(w - wi) x (b-a)+(w,-w) xr;=0 or
(w - wl) x (b-a-r,)=
xr;"=0
0 and not parallel to the vector w - wl, at any particular instant. Hence w, ° w. We can certainly choose an r;"
Problems
1. Show that if r, and r2 are two position vectors from the origin of the moving system of coordinates to two points in the rigid body, then r, dt2 + r2 - dtl = 0.
VECTOR AND TENSOR ANALYSIS
208
[SEC. 99
2. A plane body is moving in its own plane. Find the point in the body which is instantaneously at rest.
3. Show that the most general motion of a rigid body is
a translation plus a rotation about a line parallel to the translation.
99. Relative Time Rate of Change of Vectors. Let S be an y vector measured in
the
moving system of coordinates (Fig. 88).
S = S i + Sj + S,k
Fin. 88.
(354)
To find out how S changes with time as measured by an observer at 0', we differentiate (354), dS
dS=.
dt = dt 1+
dS
.
dt'+
dk dS, di _ dj dt k+S= dt+Sydt+S'dt
(355)
We do not keep i, j, k fixed since i, j, k suffer motions relative to
0'. But we do know that dt is the velocity of a point one unit along the x axis, relative to 0. Hence dk = w x k.
at dS
d
= w x is dt = w x it
Hence (355) becomes
dSs
dt - dt
i -}
dS dt
dS,
j + d k + w x (3j +
S1t)
and dS
dt
DS
S dt + w x
(356)
where DS represents the time rate of change of S relative to the moving frame, for S= is measured in the moving frame and so d _ t
is the time rate of change of S. as measured by an observer in the moving frame.
MECHANICS
SEC. 1001
209
Intuitively, we expected the result of (356), for not only does S change relative to 0, but to this change we must add the change in S because of the rotating frame. The reader might well ask, What of the motion of 0 itself? Will not this motion have to be considered? The answer is "No," for a translation of 0 only
pulls S along, that is, S does not change length or direction if 0 is translated. It is the motion of S relative to the frame O-x-y-z and the rotation about 0 that produce changes in S. Problems
1. Show that d dl 2. For a pure translation show that
dS
DS
dt =
dt
3. From (356) show that di = w x i.
Let P be any point in space and let 9 and r be the position vectors to P 100. Velocity.
from 0' and 0, respectively (see Fig. 89). Obviously a + r, so that _ dp _ A dr v
dt
dt
+
dt
Now r is a vector measured in
the O-x-y-z system, so that (356) applies to r. This yields dr _ Dr + to x r and
7t
dt
v
dt
This result is expected.
t
-dt
FIG. 89. xr+Dr
(357)
A is the drag velocity of P, co x r is
the velocity due to the rotation of the 0-x-y-z frame, and
Dr
is
the velocity of P relative to the 0-x-y-z frame. The vector sum is the velocity of P relative to the frame 0'--i-r.
VECTOR AND TENSOR ANALYSIS
210
(SEC. J 01
101. Acceleration. In Sec. 100 we saw that
v=ac
A dt+wxr+ Dr dt
To find the acceleration, we differentiate (357) and obtain
dv_d2p_d2a dt
dt2
dt2
d
+ dt (wxr) +
d(Drl dt
dt
(358)
We apply (356) to w x r and obtain
d
(wxr) = w x (wxr) + D (w x r)
Similarly d
dldtl d2p
dt 2
_
d2a dt2
+ w x (wxr) +
D xdt+d`dtl
Dr D2r do x r + 2w x dt dt + dt2
(359)
If P were fixed relative Dr = D2r to the moving frame, we would have 0 and consedt2 = dt quently P would still suffer the acceleration Let us analyze each term of (359).
d2a dt2
&
+wx(wxr)+dt xr
This vector sum is appropriately called the drag acceleration of the particle. Now let us analyze each term of the drag acceleration. If the moving frame were not rotating, we would have 2
0, and the drag acceleration reduces to the single term dta This is the translational acceleration of 0 relative to 0'. Now in Sec. 84 we saw that w x (w x r) represented the centripetal accel-
eration due to rotation and d x r represented the tangential component of acceleration due to the angular-acceleration vector
MECHANICS
SEC. 1011
d We easily explain the term relative to the O-x-y-z frame.
D2r
211
as the acceleration of P
What, then, of the term 2w x dr?
This term is called the Coriolis acceleration, named after its discoverer. We do not try to give a geometrical or physical reason for its existence. Suffice to say, it occurs in Eq. (359) and must be considered when we discuss the motion of bodies moving over the earth's surface. Notice that the term disappears for par-
ticles at rest relative to the moving frame, for then dt = 0.
It
also does not exist for nonrotating frames.
Now Newton's second law states that force is proportional to the acceleration when the mass of the particle remains constant. It is found that the frame of reference for which this law holds best is that of the so-called "fixed stars." We call such a frame of reference an inertial frame. Any other coordinate system moving relative to an inertial frame with constant velocity D d2
is also an inertial frame, since from (359) we have d e = because w = Of
66
1
0,
dt2r
da d1 constant, dta = 0. dt =
Let us now consider the motion of a particle relative to the earth. If f is the vector sum of the external forces (real forces, 2
that is, gravitation, push, pull, etc.), then d p = m, and (359) becomes D2r
d2a
Dr
d4,3
f
dt2 - w x (w x r) - d x r - 2w x dt + m (360)
dt2
This is the differential equation of motion for a particle of mass m with external force f applied to it. Example 108. Let us consider the earth as our rotating frame. The quantity w x ((a x r) is small, since jwl 27r/86,164 rad/sec,
and for a particle near the earth's surface, lrl , (4,000)(5,280) feet.
Also
d2
dt
- 0 over a short time; da ,= 0 over a short time;
212
VECTOR AND TENSOR ANALYSIS
[SEC. 101
so that (360) becomes D2r
-2w x dt +
dt2
m
(361)
Now consider a freely falling body starting from a point P at rest relative to the earth. Let the z axis be taken as the line joining the center of the earth
to P, and let the x axis be taken perpendicular to the z axis in the eastward direction.
We shall denote the latitude of the place by A, assuming A > 0. The equation of motion in the eastward direction is given by
yd1x
_
t2
S
(fm/, = 0.
tion it is
(see Fig. 90).
dj dt
, + Cfm
/
has no component eastward, so
We do not know
-gtk +
x
Now f (force of attraction)
Fio. 90.
that
-2 (w
at
but to a first approxima-
Moreover, w = w sin A k + w cos A j
i.
Hence (w x
at)
d2x
dt' =
0
= -wgt cos A, and
2wgt cos A
(362)
If the particle remains in the vicinity of latitude A, we can keep A constant, so that on integrating (362), we obtain dx dt
X=
=wgt2cosA 3ts cos A
(363)
(363) is to a first approximation the eastward deflection of a shot if it is dropped in the Northern Hemisphere. If h is the
MECHANICS
SEc. 1011
213
distance the shot falls, then h = Jgt2 approximately, so that 2
x = 3 wh
cos X
2gh1}
( J
Problems
1. Show that the winds in the Northern Hemisphere have a horizontal deflecting Coriolis acceleration 2wv sin X at right angles to v. 2. A body is thrown vertically upward. Show that it strikes the ground 'jwh cos X (2h/g)1 to the west. 3. Choose the x axis east, y axis south, z axis along the plumb line, and show that the equations of motion for a freely falling body are d2x
W
+2wsinUdt z- 2wcos0dty=0 dt +2w cos0dy =0
dtz-g-2w sin0dx =0 2
where 0 is the colatitude. CO
Fio. 91.
4. Using the coordinate system of Prob. 3, let us consider the motion of the Foucault pendulum (see Fig. 91). Let ii, i2, i$ be the unit tangent vectors to the spherical curves r, 8, p. We leave it to the reader to show that the acceleration
VECTOR AND TENSOR ANALYSIS
214
[SEC. 101
along the is vector is 2 cos e .06 + sin 0 0 when the string is of unit length.
The two external forces are mgk along the z axis and
the tension in the string, T = - Tr = - Tit. We wish to find the component of these forces along the i3 direction. T has no component in the is direction. Now k is = 0, so that mgk has no component along the is direction. Finally, we must compute Dr The velocity vector is the is component of -2w x Dr dt
_ Bit + sin 6 rpis
Also w = w(- cos A j - sin X k), so that we must find the relationship between i i2, is and i, j, k. Now
r = i, = sin0cosci+sin Bsin cpj+cos0k ail i2
cos 0 cos v i + cos 0 sin
= aB =
_
ail
1
sin B ai
1s
sin 0 k
- sin sp i + coo sp j
Thus
i = (i il)i, + (i i2)i2 + (i
i3)is
= sin 0 cos Tp it + cos 0 cos jP is -- sin rp is
j = sin 0 sin Sp i, + cos B sin 'P i2 + C0803 k = cos 0 it - sin B i2 11
is
i2
-2wx-=2w cos A sin 0 sin P + sin A cos 0 0
-sinAsin0 6
sine c
and
(-
r
2w x -D dt
= #(sin A sin 0 sin ip + sin A cos 0)
Equation (361) yields 2 cos 006+ sin 8;p = 2w(B sin A sin B sin rp + # sin A cos 0) (364)
MECHANICS
SEC. 102]
For small oscillations, sin 0
215
0, and (364) reduces to
, = w sin X
(365)
Hence the pendulum rotates about the vertical in the clockwise sense when viewed from the point of suspension with an angular speed w sin X. At latitude 30° the time for one complete oscillation is 48 hours.
5. Find the equation of motion by considering the i2 components of (361) for the Foucault pendulum. 102. Motion of a Rigid Body with One Point Fixed. The motion of a rigid body with one point fixed will depend on the forces acting on the body. Let O-x-y-z be a coordinate system fixed in the moving body, and let O-- be the coordinate system fixed in space. 0 is the fixed point of the body. In Sec. 94 we
saw that
-Or
= Lo.
Now Ho* = f if r x p dt dr. We can
replace dt by to x r (w unknown). Thus
H0? = f f f pr x (w x r) dr R
= fJf p[r2w - (r
w)r] dr
(366)
R
Let
w=w=i+wyj+w
r=xi+yj+zk
so that
r2w - (r w)r = (x2 + y2 + z2) (w i + wyj + W k) + (xwx + ywy + zws) (xi + yj + zk) = [(y2 + z2)wz - xywy - xzws]i
+ [-xyw. + (22 + x2)wy - yzw=lj + [-xzwz - yzwy -- (x2 + y2)w]k We thus obtain H0? = i[w= f f f p(y2 + z2) dr - w f f f pxy dr - wz f f f pxz dr]
+ j[ - w=f f f pxy dr + wyf f f p(z2 + x2) dr - w: f f f pyz dr] + k[ -w=f f f pxz dr - wyf f f pyz dr + w=f f f p(x2 + y2) dr] (367)
216
VECTOR AND TENSOR ANALYSIS
[SEC. 102
The quantities A = f f f p(y2 + Z2) dr
B=fffp(z2+x2)dr C = fffp(x2 + y2) dr D = f f fpyzdr E = f f fpzxdr
(368)
F = f f f pxy dr
are independent of the motion and are constants of the body. That they are independent of the motion is seen from the fact that for a particle with coordinates x, y, z, the scalars x, y, z remain invariant because the O-x-y-z frame is fixed in the body.
The quantities A, B, C are the moments of inertia about the x, y, z axes, and D, E, F are called the products of inertia. We assume the student has studied these integrals in the integral calculus.
Now from Sec. 99 we have
Lo =
dHor dt
_
DHor + w x Hor so that dt
DRO, dt + to x Hor
Hence
L=i+Lvj+LA=i(A
+
ds-F
+j(-F
w
+ k \-E
dt
dtE
dts/ +Bdty-Ddt;/
- D dt + C dta/
i
k
wy
wt
Aws - Fwv - Ewz,
WV
-Fws+Bwv - Dw=,
-Ews - Dwv+Ccos (369)
In the special case when the axes are so chosen that the products of inertia vanish (see Sec. 107), we have Euler's celebrated equations of motion :
MECHANICS
SEc. 1031
217.
L. = A d[ + (C - B)wYwz
L = B
dwY
(370)
+ (A - C)w,wZ dt
L. = C
dw, dt
+ (B
A)wzwy
103. Applications. If no torques are applied to the body of Sec. 102, Euler's equations reduce to A
(i)
dwz
dt-
+ (C
B)wYw, = 0
B dt +(A-C)w,wx=0
(371)
C d,+(B-A}wZwY=0
Multiplying (i), (ii), (iii) by
w,, respectively, and adding,
we obtain Aws
dws
dt
+ Bw dwY + Cw, dwa = 0 dt dt
Integrating yields Aws2 + Bw,,2 + Cw,2 = constant
(372)
This is one of the integrals of the motion. We obtain another integral by multiplying (i), (ii), (iii) by Aw,z, Bw,,, Cw,, and adding. This yields AZ w,
dws
dt
+
BZwy'dw'
dt
+
C2w,
dw, dt
=
0
so that A 2w12 + B2wV2 + C2w,2 =constant
(373)
If originally the motion was that of a rotation of angular velocity w about a principal axis (x axis), then initially
VECTOR AND TENSOR ANALYSIS
218
[SEc. 103
wz(0) = coo
0
(374)
ws(0) = 0
and we notice that (371) and the boundary condition (374) are satisfied by wt(t) = -coo
0
wZ(t) = 0
so that the motion continues to be one of constant angular velocity about the x axis. Here we have used a theorem on the uniqueness of solutions for a system of differential equations.
Now suppose the body to be rotating this way and then slightly disturbed, so that now the body has acquired the very We can neglect ww: as compared Euler's equations now become
small angular velocities w,,, wt.
to
and wswo.
B dty + (A - C)wswo = 0
Cdt
+(B0
(375)
wo = constant
Differentiating the first equation of (375) with respect to time and eliminating d s' we obtain B
z
dtzy +
(- C)((A - B) w02w = 0
(376)
If A is greater than B and C or smaller than B and C, then (A - C) (A - B) a2 =
C
> 0, and the solution to (376) is
wy = L cos (at + a)
Also ws =
aBL sin (at + a) by replacing w,, in (375). wo(A - C)
SEC. 1041
MECHANICS
219
Problems 1. Solve the free body with A = B for wzj co,, co,.
2. A disk (B = C) rotates about its x axis (perpendicular to the plane of the disk) with constant angular speed wo. A constant torque Lo is applied constantly in the y direction. Find w and W..
3. Show that a necessary and sufficient condition that a rigid
body be in static equilibrium is that the sum of the external forces and external torques vanish. 4. A sphere rotates about its fixed center. If the only forces acting on the sphere are applied at the center, show that the initial motion continues.
5. In Prob. 2 a constant torque Lo is also applied in the z direction.
Find w,, and ws.
104. Euler's Angular Coordinates. More complicated problems can be solved by use of Euler's angular coordinates. Let O-x'-y'-z' be a cartesian coordinate system fixed in space, and let O-x-y-z be fixed in the moving body (Fig. 92). The x-y plane will intersect the x'-y' plane in a line, called the nodal line N. Let 0 be the angle between the z and z' axes, L' the angle between the x' and N axes, and (p the angle between the nodal line and the x axis. The positive directions of these angles are indicated in the figure. The three angles &, 0, rp completely specify the configuration
of the body. Now d represents the rotation of the O-z'-N-T' frame relative to the O-x'-y'-z' frame; de represents the rotation of the O-z-N-T frame relative to the O-z'-N-T' frame, and finally, d(P
represents the rotation of the O-x-y-z frame relative to the
O-z-N-T frame. Therefore
+ dO + dt gives us the angular
velocity of the O-x-y-z framed* relative to the fixed O-x'-y'-z' frame, and
d1 + dO + d
(377)
220
VECTOR AND TENSOR ANALYSIS
[SEC. 104
The three angular velocities are not mutually perpendicular. We now define i, j, k, i', j', k', N, T', T as unit vectors along the x, y, z, x', y', z', N, T', T axes, respectively. Thus w =
d
k
wzi + w,j + wLk wz'i' + wy j' + wZ k'
Fia. 92.
Now it is easy to verify that i = cos rp N + sin cp T
j = - sinpN+cosjpT i' = cos 4, N - sin ' T'
(378)
MECHANICS
SEC. 1051
221
sin 4, N+cos#T' sin psin B cosspsin 0
sin
so that
ki
ws at
at
at
sin p sin 0 d + cos s dB day
do
at dp
at
w = cos (p sin 8 --- - sin (p day
W: = cos B d +
dt
Rewriting this, we have
dt
wy =
d4,
sin 0 sin (P + sin B cos
wI = d cos 8 +Also
wi=wss+w2+(O2=
cos Gt - cos0sin8dt
(381)
d +cos6d 105. Motion of a Free Top about a Fixed Point. Let us assume that no torques exist and that the top is symmetric (A = B). Euler's equations become
VECTOR AND TENSOR ANALYSIS
222
(i)
A
(ii)
A d' + (A - C)cvxwZ = 0
[SEC. 106
(382)
Cdts=O
(iii)
Integrating (iii), we obtain wz = w, = constant. and add to (i). We obtain by i =
Multiply
A dt (w= + iwv) + (C - A)wo(wy - iwz) = 0 or
A
d (co., + iwy) = iwo(C - A)(wx + i
)
Integrating, wz + 2Wy =
ae 0, and for a sink, 0 < 0. The net gain of fluid is therefore
ffJ1dT_ jfpv.dd
(406)
Equating (405) and (406) and applying the divergence theorem, we obtain
232
VECTOR AND TENSOR ANALYSIS
ap + V (Pv) = #(x, y, z, t)
[SEC. 109
(407)
This is the equation of continuity. For no source and sink, (407) reduces to 0 at + V - (Pv) =
(408)
If furthermore the liquid is incompressible, p = constant, aP
at
= 0, and (408) becomes
Vv=0
(409)
If the motion is irrotational, that is, if f v dr = 0, then V=
so that the equation of continuity for an incompressible
fluid possessing no sources and sinks and having irrotational motion is given by V2V = 0
(410)
We call p the velocity potential. We solve Laplace's equation for rp, then compute the velocity from v = Vrp. Problems 1. If the velocity of a fluid is radial, u = u(r, t), show that the equation of continuity is ap
at
LP
+ u ar +
P a
r2 ar
(r2u)
Solve this equation for an incompressible fluid, if '(r, t) = 1/r2. z _2xyz x z - 1
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