Vector and Tensor Analysis 1950

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


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


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


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


[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


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


[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


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


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


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'ìq°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


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


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


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


[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


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.


18

[SEC. 9

apparent that r - ro is parallel to T so that r - ro = AT,

--
xo

y - yo -z - zo=A m

n

(30)

By allowing A to vary from - oo to + oo we generate every point on the line. (e) Equation of a sphere with center at Po(xo, yo, zo) and radius a.

In Fig. 21 obviously (r - ro) (r - ro) = a2, or (x - xo) 2 + (y - yo) 2 + (z - zo) 2 = a2 Problems

1. Add and subtract the vectors a = 2i - 3j + 5k,

b = -2i+2j+2k Show that the vectors are perpendicular.

2. Find the cosine of the angle between the two vectors

a = 2i - 3j + k and b = 3i - j - 2k. 3. If c is normal to a and b, show that c is normal to a + b,

a - b.

4. Let a and b be unit vectors in the x-y plane making angles

a and 6 with the x axis. Show that a = cos a i + sin a j, b = cos 9 i + sin 3 j, and prove that

cos (a - 6) = cos a cos S + sin a sin 8 5. Find the equation of the cone whose generators make an angle of 30° with the unit vector which makes equal angles with the x, y, and z axes. 6. The position vectors of the foci of an ellipse are c and - c, and the length of the major axis is 2a. Show that the equation of the ellipse is a4 - a2(r2 + c2) + (c r)2 = O. 7. Prove that the altitudes of a triangle are concurrent. 8. Find the shortest distance from the point A(1, 0, 1) to the

line through the points B(2, 3, 4) and C(-1, 1, -2).


SEC. 91

19

9. Let a = 2i - j + k, b =i-3j - 5k. Find a vector c so that a, b, c form the sides of a right triangle. 10. Let r be the position vector of a point P(x, y, z), and let a be a constant vector. Interpret the equation (r - a) a = 0.

11. Given a = 2i - 3j + k, b = 3j - 4k, find the projection of a along b.

12. Show that the line joining the end points of the vectors

a =2i -j - k, b = -i + 3j - k with common origin at O is parallel to the x-y plane, and find its length.

13. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

14. Let a = a, + a2 where a, b = 0 and a2 is parallel to b. Show that a2 = (iblb

b, al = a -

(a . b) b. JbJ2

15. Derive (25). 16. Verify (26).

17. Find a vector perpendicular to the vectors a = i - j + k,

b=2i+3j-k.

18. Let a = f(t)i + g(t)j + h(t)k, and define h'(t)k da = f'(t)i

+ g'(t)j +

dt

Show that

da dt

a=

lal

dal dt

19. Find the angle between the plane Ax + By + Cz + D = 0 and the plane ax + by + cz + d = 0. 20, a, b, c are coplanar. If a is not parallel to b, show that

aa ca ab cb Icb bbi c= aa ab ab bb

ca ab

b

21. For the an", Asa defined by (24) and (25), show that a, -Ap r = Sp-

y-1

where &0°= Iifa=l4,Se=Oif aPd f. 22. If B1, B2, B3 are the components of a vector B, that is, B = B'i + B2j + B3k (see Example 8 in regard to the super-


20

[SEC. 10

scripts), show that for a rotation of axes the components of the 3

vector B become (BI, B2, B3) where B° _ I ayaB#, a = 1, 2, 3, Ps1

and B = BIi' + B2j' + Mk'. Read Example 8 carefully. 23. Show that for a rotation of axes, BICI + B2C2 + B3C3 = BICI + B2C2 +,&3103

This shows the invariance of the scalar product for rotations of axes. The invariance here refers to both the numerical invariance of the scalar product and the formal invariance, 3

BaC' = Z BaC'a aa1

amt

24. Prove the statements made in Example 9. 25. Generalize the statements of Example 9 for n-tuples (a', a2

a")

10. Vector, or Cross, Product. Given any two nonparallel vectors a and b, we may construct a third vector c as follows: When translated so that they

have a common origin, the two vectors a, b form two sides of a parallelogram. We define c to be perpendicular to the plane of this parallelogram with magnitude equal

to the area of the parallelogram. We choose that normal obtained by the motion of a right-hand screw when a is rotated into b (angle of rotation less than 180°) (see Fig. 22). A cross is placed between the vectors a and b to denote the vector c = a x b. The vector c is called the cross, or vector, product of a and b and is given by F ia. 22.

c=axb=lallbl sin0E where JEl = 1.

The area of the parallelogram is

A = !alibi sin 0

(31)

TIIE ALGEBRA OF VECTORS

SEc. 121

21

The cross product will occur frequently in mechanics and electricity, but for the present Ave discuss its algebraic behavior. It

is obvious that a x b = -b x a, so that vector multiplication is not commutative. If a and b are parallel, a x b = 0. In particular, a x a = 0. 11. The Distributive Law for the Vector Product. We desire

to prove that a x (b + c) = a x b + a x c. Let

u= ax(b+c)-axb-axc and form the scalar product of this vector with an arbitrary vector v. We obtain

x(b+c)] -

x b)

In Sec. 13 we shall show that a (b x c) = (a x b) c. Hence

This implies either that u = 0 or that v is perpendicular to u. Since v is arbitrary, we can choose it not perpendicular to u. Hence u = 0 and

ax(b+c) =a xb+a xc

(32)

This proof is by Professor Morgan Ward of the California Institute of Technology. 12. Examples of the Vector Product Example 10

ixi=jxj=kxk=0 jxk=i, kxl=j

ixj=k,

For the vectors a = all + a2j + ask, b = b1i + b2j + b3k we obtain a x b = (a2b3 - a3b2)i + (a3b1 - alb3)j + (aib2 - a2b1)k by making use of the distributive law of Sec. 11. Symbolically we have

axb = a1

a2

k as

bl

b2

b3

i

j

(33)

where (33) is to be expanded by the ordinary method of determinants.


22

Example 11.

a = 2i - 3j + 5k, b = -i + 2j - 3k, so that k

i

j

2

-3

5

-1

2

-3

a xb = Example 12.

[SEC. 12

Sine law of trigonometry

c=b - a

c xc=cx(b-a)

0=cxb-cxa

or

c xa = c xb

Fia. 23.

However, if two vectors are equal, their magnitudes are equal so

that lellal sin I = lcilbl sin a and a

b

c

sin a

sin fl

sin ti

Example 13. Rotation of a Particle. Assume that a particle is rotating about a fixed line L with angular speed w. We assume that its distance from L remains constant. Let us define the angular velocity of the particle as the vector w, whose direction is along L and whose length is w. We choose the direction of w in the usual sense of a right-hand screw advance (see Fig. 24).


SEC. 13]

23

It is our aim to prove that the velocity vector v can be represented

by w x r, where r is the position vector of P from any origin taken on the line L. Let the reader show that v and w x r are parallel. Now lw x rl = wa = speed of L P, so that

a

13. The Triple Scalar Product. Let us consider the scalar a (b x c). This

OP

%

v

scalar represents the volume of the parallelepiped formed by the coterminous sides a, b, c, since

a- (b xc) =fallbllclsin0cosa = hA = volume

(35)

A being the area of the parallelogram with sides b and c, the altitude of the parallelepiped being denoted by h (see Fig. 25).

Fia. 25.

Now

a (b x c) = (a,i + a2j + ask) .

= ai(b2ca - bac2) + a

i

j

k

b,

b2

bs

c1

C2

C3

(baci - bic3) + as(bic2 - b2c1)

so that a1

a2

as

a (b x c) = b,

b2

bs

C1

C2

C3

(36)

24


[SEC. 14

Notice that a (b x c) = (a x b) c since both terms represent the volume of the parallelepiped. It is also very easy to show that the determinant of (36) represents (a x b) c. We usually write a (b x c) = (abc) since there can be no confusion as to where the dot and cross belong. We note that (abc) = (cab) = (bca)

and that (abc) _ - (bac) = - (cba) = - (acb). These results follow from elementary theorems on determinants. We are thus allowed to interchange the dot and the cross when working with the triple scalar product. This result was used to prove (32). If the three vectors a, b, c are coplanar, no volume exists, and we at once have (abc) = 0. In particular, if two of the three vectors are equal, the triple scalar product vanishes. 14. The Triple Vector Product. The triple vector product a x (b x c) plays an important role in the development of vector analysis and in its applications. The result is a vector since it is the vector product of a and (b x c). This vector is therefore perpendicular to b x c so that it lies in the plane of b and c. If b is not parallel to c, a x (b x c) = xb + yc, from Sec. 6. Now dot both sides with a and obtain x(a b) + y(a c) = 0, since a [a x (b x c)] = 0. Hence

(a X

where = (a b) = X,

X is a scalar, so that

a x (b x c) _ X[(a c)b - (a b)c]

(37)

In the special case when b = a, we can quickly prove that X = 1. We dot (37) with c and obtain c

[a x (a x c)] = X[(a c)2 - a2c2J

or

-(a x c)2 = X[(a c)2 - a2c2J by an interchange of dot and cross.

Hence

-a2c' sine O= A(a'c2 cost 0- a2c2) = -Xa'c2 sin2 0

so that A = 1. Hence a x (a x c) = (a c)a - (a a)c. From this it immediately follows that (a x b) x b = (a b)b - (b b)a. Now we prove that A = 1 for the general case. We dot (37) with b and obtain


SEC. 15]

25

x(b xc)]

-[(a xb) x

b

(a

x b) x implying X = 1.

(a b)(b c)

Thus

a x (b x c) = (a c)b - (a b)c

(38)

We leave it to the reader to show that

(a x b) x c = (a c)b - (b c)a Notice that a x (b x c) 5-1 (a x b) x c. If b is parallel to c, (38) reduces to the identity 0 = 0, so that (38) holds for any three vectors. The expansion (38) of a x (b x c) is often referred to

as the rule of the middle factor. More complicated products are simplified by use of the triple products. For example, we can expand (a x b) x (c x d) by considering (a x b) as a single vector and applying (38).

(a x b) x (c x d) = (a x b d)c - (a x b c)d = (abd)c - (abc)d

(39)

Also

(c x d) d)

(b c)

d)

(40)

15. Applications to the Spherical Trigonometry. Consider the spherical triangle ABC (sides are arcs of great circles) (see Fig. 26). Let the sphere be of radius 1. Now from (40) we see that

(a x b) (a x c) = (b

c) - (a

b) (a c)

The angle between a x b and a x c is the same as the dihedral angle A between the planes OAC and OAB, since a x b is perpendicular to the plane of OAB and since a x C is perpendicular to the plane of OAC. Hence sin y sin # cos A = cos a - cos y cos,6.


26

[SEC. 15

FIG. 26.

Problems

1. Show by two methods that the vectors a = 2i - 3j - k, b = -6i + 9j + 3k are parallel. 2. Find a unit vector perpendicular to the vectors

a=i-j+k

b=i+i - k 3. A particle has an angular speed of 2 radians per second, and its axis of rotation passes through the points P(0, 1, 2),

Q(1, 3, -2)

Find the velocity of the particle when it is located at the point R(3, 6, 4).

4. Find the equation of the plane passing through the end points of the vectors a = a1i + a j + ask, b = b1i + bd + bsk, c = c1i + cd + cak, all three vectors with origin at P(0, 0, 0).

5. Show that (a x b) x (c x d) = (acd)b - (bcd)a. 6. Prove that d x (a x b) (a x c) = (abc) (a d).

7. If a + b+ c= 0, prove that a x b= b x c= c x a, and interpret this result trigonometrically.


SEC. 15]

21

8. Four vectors have directions which are outward perpendiculars to the four faces of a tetrahedron, and their lengths are equal to the areas of the faces they represent. Show that the sum of these four vectors is the zero vector. 9. Prove that (a x b) (b x c) x (c x a) = (abc)2. 10. If a, b, c are not coplanar, show that

d = (dbc) a + (abc)

(adc) (abc)

b+

(abd) c (abc)

for any vector d. 11. If a, b, c are not coplanar, show that

d= (c(abc)d) a x b + - (a(abc)d) b x c -1-

(b d)

(abc)

cxa

for any vector d. 12. Prove that

a x(b xc) +b x(c x a)+ c x (a xb) =0 13. The four vectors a, b, c, d are coplanar. Show that (a x b) x (c x d) = 0. 14. By considering the expansion for (a x b) x (a x c), derive a spherical trigonometric identity (see Sec. 15). 15. Show that (a x b) (c x d) x (e x f) = (abd) (cef) - (abc) (def) = (abe) (fcd) - (abf) (ecd) = (cda) (bef) - (cdb) (aef) 16. Find an expression for the shortest distance from the end point of the vector r, to the plane passing through the end points of the vectors r2, r8, r4. All four vectors have their origin at P(0, 0, 0). 17. Consider the system of equations

a1x + b,y + c1z = d, a2x + b2y + c2z = d2

(41)

asx + bsy + c3z = da

Let a = aui + a2j + aak, etc. Write (41) as a single vector equation, and assuming (abc) F6 0, solve for x, y, z.

18. Find the shortest distance between two straight lines in space.

28


[Ssc. 15

19. Show directly that a x (b x c) _ (a c)b - (a b)c, where a, b, c take on the values i, j, k in all possible ways. Now show that, (38) is linear in a, b, c, that is,

a x [b x (ac + ad)] = as x (b x c) + #a x (b x d) etc. Since any vector is a linear combination of i, j, k, explain why (38) holds for all vectors. 20. If a and b lie in a plane normal to a plane containing c and

d, show that (a x b) (c x d) = 0. 21. If (a', a2, a3), (b', b2, b3) are the components of the vectors a, b, show that three of the nine numbers Ca9 obtained by considering cab = aab# - a8ba, a, a = 1, 2, 3, represent the components of a x b, that three others represent b x a, while the remaining three vanish. Represent c,--P as a matrix and show that cap = - G a

By considering a rotation of axes (see Example 8), show that the EaP in the new coordinate system are related to the caP in the old a

a

coordinate system by the equations &A = I I a aa,i9c°r, r-1 a=1 a, 6 = 1, 2, 3. Show that caP _ -Cfta. The numbers c' = a2b3 - a3b2, c2 = a3b' - a1b3, c3 = a'b2 - a2b'

a

are the components of a x b. Show that ca = 1 apa#S under a B=1

rotation of axes (see Prob. 22, Sec. 9). 22. We can construct a x b by three geometrical constructions. We first construct a vector normal to b lying in the plane of a

and b. We project a onto this vector, and finally we rotate this new vector through an angle of 90° about the axis parallel to b, magnifying this newly constructed vector by the factor Ibl The final result yields a x b. Use this to prove that

a x (b + c) = a x b + a x c

CHAPTER 2 DIFFERENTIAL VECTOR CALCULUS

16. Differentiation of Vectors. Let us consider the vector field

u = a(x,y,z,t)i+13(x,y,z,t)j+ '(x,y,z,t)k

(42)

At any point P(x, y, z) and at any time t, (42) defines a vector. If we keep P fixed, the vector u can still change because of the time dependence of its components a, S, and y. If we keep the time fixed, we note that the vector at the point P(x, y, z) will, in general, be different from that at the point Q(x + dx, y + dy, z + dz)

Now, in the calculus, the student has learned how to find the change in a single function of x, y, z, t. What difficulties do we encounter in the case of a vector? Actually none, since we easily

note that u will change if and only if its components change. Thus a change in a(x, y, z, t) produces a change in u in the x direction, and similarly changes in f3 and y produce changes in u

in the y and z directions, respectively. We are thus led to the following definition:

dai.+d,6i+dyk

du = du

as (-aa

ax dx

+

ay

(

dy +

as

as

az

dz + at

d) i

-dz +(axdx+aydy+a +

(43)

ax dx +

atdt)j dy -}-

az dz }- at dt J k

For example, let r = xi + yj + zk be the position vector of a moving particle P(x, y, z) in three-space.

Then

dr = dx i + dy j + dz k 29


30

[SEC. 16

and

v

dt

a=

dt 1

d 2r

+

2y

z

dt2 = dt2 I

+

dt

+

dt

,

k

(44)

2

j + dt2 k dt2

(45)

Equations (44) and (45) are, by definition, the velocity and acceleration of the particle. We have assumed that the vectors i, j, k remain fixed in space. If the vector u depends on a single variable t, we can define du dt

lim

u(t + At) - u(t)

(46)

At

At-,o

It is easy to verify that (46) is equivalent to (43). Example 14. Consider a particle P moving on a circle

(see Fig. 27).

C -A-.

:Q/

-44-1k

-4- -

angular speed w = - (Fig 28).

Fio. 27.

We note that

r=rcos0i+rsin0j

so that

v = dt = (-r sin 6i+rcos Bj) de and

a = dt

(dA2 (-r cos 0 i - r sin B j) dt/ dt2 = C

Therefore the acceleration is

a = -w2r

(47)

The point P has an acceleration toward the origin of constant magnitude w2r.

This acceleration is due to the fact that the

SEC. 16]

DIFFERENTIAL VECTOR CALCULUS

31

velocity vector is changing direction at a constant rate; it is called the centripetal acceleration. z

Y

Fra. 28.

Example 15.

FIG. 29.

Let P be any point on the space curve (Fig. 29)

x=x(s) y = y(s) z = z(s)

where s is are length measured from some fixed point Q.

r = x(s)i + y(s)j + z(s)k

Now (48)

so that dr _ dx

,

dy

,

dz

(49)

ds

and

dr dr

dx 2

dy z

(dz 2

ds ds = ds + \dsI + \ds) =dx2+dy2+dz2=1 ds2

from the calculus. Hence ds is a unit vector.

As As -> 0, the

position of Ar approaches the tangent line at P.

Hence (49)

represents the unit tangent vector to the space curve (48).


32

[SEC. 17

17. Differentiation Rules. Consider

'p(t) = u(t) v(t) 'P(t + At) - ,P(t) = u(t + At) v(t + At) - u(t) o(t) Now

u(t + At) = u(t) + Au v(t + At) = v(t) + AV (see Fig. 27), so that

c(t+ At

At

At

At

and passing to the limit, we obtain d (u v)

dt

_

- udtdv + dudt'v

(50)

Similarly

d(u x v) -

dt

d(fu)

dv

u

v

(51)

x dt + dt

-

du f

dt

du

dt

d_f

+

u

(52)

dt

Notice how these formulas conform to the rules of the calculus. Example 16. Let u(t) be a vector of constant magnitude. Therefore

u u = u2 = constant By differentiating we obtain du

u'dt+

du

u'dt=0


SEC. 17]

3m

Hence either dt = 0 or dl is perpendicular to u. This is an important result and should be fully understood by the student. The reader should give a geometric proof of this theorem. Example 17. In all cases y u u = u2 where u is the length of u. Differentiation yields 2u

' dt =

2u d and du

u'du

(53)

auat -

This result is not trivial, for

FIG. 30.

du. Example 18.

Idul

Now r = rR, where R is a

Motion in a Plane. unit vector (see Fig. 30). Hence

V-dt dtR Now

-I-r dR

R is perpendicular to R (see Example 16).

Also

IdRI

= de We can easily verify this by differ-

since R is a unit vector.

entiating R = cos B i + sin 9 j.

Hence v = Wt-

R + r d8 P, where

P is a unit vector perpendicular to R. Differentiating again we obtain

a

_ _

dv dt

d2r

dr dR

dt'R+dl

dt

dr d6

d20

+dtdt P+rdt2

dB dP P+rdt dt

or

d2r a_dt

since

R+2dtd6P+rd- P-r(d8)2R

P=

-d6R

(54)

[Sic. 17


34

Thus I d2r

- (dO)2]Rid(odO\p dt r dt dt

(55)

Problems

1. Prove (51) and (52). 2. Prove (54).

/

3. Differentiate l r - dr1 with respect to t. 4. Expand dt [p x (q x r)].

5. Show that

d

r x dt) = r x

d2r

6. Find the first and second derivatives of

dr d2r

(r dt dt2

7. r = a cos wt + b sin wt; a, b, w are constants. Prove that r

xdt = wa xb andd

+w2r = 0.

8. If r x dt = 0, show that r has a constant direction. 9. R is a unit vector in the direction r. Show that

RxdR= r x dr 7.2

10. If dt = w x a,

d = w x b, show that

dt(axb)=0x(axb) 11. If r = aew' + be-,", show that dt; -- w2r = 0. a, bare constart vectors.

12. Find a vector u which satisfies dtu = at + b.

stant vectors.

Is u unique?

13. Show that

d dt

()

r dt

r dt r.

a, b are con-


SEC. 17]

35

14. Let r° be the position vector to a fixed point P in space, and let r be the position vector to a variable point Q lying on a space curve r = r($). Show that if the distance Pte(- is a minimum, then

r - r° is perpendicular to the tangent at Q. Show also that z

l2

r ds2 + ds,

d1r r0

ds2

15. If u = a(x, y, z, t)i + 13(x, y, z, t)j + y(x, y, z, t)k, show _ thatdu _ au au dx au dy au dz dt

at

+

+

ax dt

ay dt

+ az dt

z

Fia. 31.

16. The transformation between rectangular coordinates and spherical coordinates is given by

x=rsin0cos0 y=rsin0sinrp z = r cos 0 where 0 is the colatitude, (p is the longitudinal or azimuthal angle,

and r is the magnitude of the position vector r from the origin to the particle in question. Find the components of the velocity and acceleration of the particle along the unit orthogonal vectors er, ee, e,, (see Fig. 31).

36


1SEc. 1b

17. Consider the differential equation (i)

du+2Adu+Bu=0

where A, B are constants. Assume a solution of the form

u(t) = e'°'C, where C is a constant vector, and show that u(t) = Clew,' + C2e'°2' is a solution of (i), w1, w2 being roots of w2 + 2Aw + B = 0. Consider the cases for which A2 - B < 0,

A 2 - B = 0, A2 - B > 0. 18. Find the vector u which satisfies

a

d3u

d2u

such that u = i,

du

2

te

dt3

d

du

0

= j, d = k for t = 0.

19. If u, is a solution of

d8u+Ad2u+Bdu+Cu=0 and if u2 is a solution of

d dtu+Adtu+Bdt+Cu=F(t)

show that u1 + U2 is a solution of (ii) provided A, B, C are independent of u. Why is this necessary? 20. A particle moving in the plane of (r, 8) has no transverse

acceleration, that is,

dt

t r2

df) = 0. Show that the radius

vector from the originrto the particle sweeps out equal areas in equal intervals of time. 18. The Gradient. Let p(x, y, z) be any continuous differentiable space function. From the calculus dV

=

a dx + a dy + a dz ax

ay

az

(56)


SEC. 181

37

Now let r be the position vector to the point P(x, y, z).

r = xi + yj + zk If we move to the point Q(x + dx, y + dy, z + dz) (Fig. 32),

dr = dxi + dyj + dzk Now notice that (56) contains the terms dx, dy, dz and the terms aP, aSP,

aP

ax 8y az

We define a new vector formed from gp by taking its

three partial derivatives. Let del rp =

gradient p be defined by

Vsp=axi+ yJ+aZk

(57)

We immediately see that d(p = dr VV

(58)

We shall now give a geometrical interpretation of IV-p. At the point P(xo, yo, zo), so has

the value gp(xo, yo, zo) so that go(x, y, z) = gp(xo, yo, zo)

represents a surface which obviously contains the point P(xo, yo, zo)

FIG. 32.

As long as we move along this surface, (p has the constant value ,p(xo, yo, zo) and d-r = 0. Consequently, from (58), (59)

Now Vgp is a vector which is at once completely determined after ,p has been differentiated, and Eq. (59) states that Vgp is perpendicular to dr as long as dr represents a change from P to Q, where Q remains on the surface v = constant. Thus V(p is normal to

all the possible tangents to the surface at P so that V(p must necessarily be normal to the surface (p(x, y, z) = constant (see

38


Fig. 33). Let us now return to dip = dr Vv.

[Sec. 18

The vector Vp is

fixed at any point P(x, y, z), so that dip (the change in p) will depend to a great extent on dr. Certainly dcp will be a maximum when dr is parallel to Vp, since dr VV = Idrl IV pl cos 0, and cos 8

is a maximum for 0 = 0°.

VO

Thus Vp is in the direction of maximum increase of p(x, y, z).

Let ldrl = ds so that d(P

ds

=u

V(p

(60)

where u is a unit vector in the direction dr. Hence the change of (p in any direction

the projection of Vp on the unit vector having this is

Fia. 33.

Example 19.

direction.

To find a unit vector normal to the surface

x2 + y2 - z = 1 at the point P(1, 1, 1).

Here

,P(x,y,z) =x2+y2-z

V(p=2xi+2yj-k

=2i-2j-katP(1,1,1)

Thus

N=2i-2j-k 3

Example 20. We find Vr if r = (x2 + y2 + Z2)1. The surface r = constant is a sphere. Hence Vr is normal to the sphere and so is parallel to the position vector r. Thus Vr = kr. Now

dr = dr Vr = k dr r = kr dr

from (53)

Therefore

k=1 r

and

Vr = r = R

r

Example 21

Vf(u) = f'(u) Vu,

u = u(x, y, z)

(61)

39


SEC. 18]

Proof:

Vf(u) =

i+ayj+azk

a i + f'(u) = r(u) ax au

au j ay

+

au k

f'(u) az

= f'(u) Vu Example 22 Vf(ul, u2,

. . .

, un)

=a i+ayj+azk _

of aua 1

n

=

,

au, ax

of aua

,

aua ay

a

If Vua aua

of au-aua az

) (62)

Consider the ellipse given by rl + r2 = constant (see Fig. 34). Now V(rl + r2) is normal to the ellipse. Let Example 23.

Y

Fio. 34.

T be a unit tangent to the ellipse. Thus V(rl + r2) T = 0, and W2 - T

(63)

But from Example 20, Vrj is a unit vector parallel to the vector AP, and Vr2 is a unit vector parallel to the vector BP.

Equation

40


[SEC. 19

(63) shows that AP and BP make equal angles with the tangent to the ellipse. 19. The Vector Operator V. We define

v=ia +iaay +kaz

(64)

Notice that V is an operator, just as dx is an operator in the differ-

ential calculus. Thus

v - (1-+jay +kaz)'p a vector operator because of its components a

a

a

ax ay az

It will help us in the future to keep in mind that V

acts both as a differential operator and as a vector. Example 24

v (uv) =

i a (uv) ax

q!!!) + k a (uv) + J ay az

C'ax+j

y

a+kc1v )u+Clax +jau+k az

V(uv) = U Vv + v Vu

y

)v

-az

(65)

This result is easily remembered if we keep in mind that V is a differential operator, so that we can apply the ordinary rules of calculus.

Problems

1. Find the equation of the tangent plane to the surface xy - z = 1 at the point (2, 1, 1).

Sac. 191


41

2. Show that V(a r) = a, where a is a constant vector and r is the position vector. 3. If r = (x2 + y2 + Z2)'1, find Vr" by explicit use of (57). 4. If So = (r x a) (r x b), show that

V,p = b x (r x a) + a x (r x b) when a and b are constant vectors. 5. Let gp = x2 + y2. Find IVpI and show that it is the maximum change of V.

6. Find the cosine of the angle between the surfaces x2y + z = 3

and x log z - y2 = -4 at the point of intersection P(- 1, 2, 1). 7. What is the value of Vv(x, y, z) at a point that makes So a maximum?

8. The surfaces g(x, y, z) = constant and #(x, y, z) = constant are normal along a curve of intersection. What is the value of V(p V4' along this curve?

9. What is the direction for the maximum change of the space function gp(z, y, z) = x sin z - y cos z at the origin? 10. Expand V(u/v) where u = u(x, y, z), v = v(x, y, z). 11. Let r and x be the distances from the focus and directrix to any point on a parabola. We know that r = x. Show that (R - i) T = 0, where T is a unit tangent vector to the parabola, and interpret this equation.

12. Show that the ellipse r, + rz = c, and the hyperbola r, - ra = c2 intersect at right angles when they have the same foci.

13. If Vv is always parallel to the position vector r, show that

rp=(p(r),r== x2+ysd-z2. 14. Find the change of g = xyz in the direction normal to the surface yx2 + xy2 + z'y = 3 at the point P(1, 1, 1). 15. If f = f(xl, x', xa) (see Example 8), and if x°t = x01(yl, y2, ya),

a = 1, 2, 3, show that of

of axe

aye

aya

a = 1, 2, 3


42

Using the fact that s

of

of ay a

axa

9-1 ay8 axa

axa

_

3

ax"

aye

1 ay° ax8

ax8

`

[SEC. 20

1 if« _ R , show that 0 if a 0 ,e

'

a = 1, 2, 3.

16. Apply the results of Prob. 15 above to the transformation

x = r cos 0 y = r sin 0 z=z 1 of , of are the components of Vf along the and show that af,

Or r ae az

three mutually orthogonal unit vectors e, es, e. which occur in cylindrical coordinates.

dr 17. If (_ v(x, y, z, t), show that d,P= ate +

V(p.

18. Ifr =xi+yj+zk,find xyevzi:'

+ log

-X

z

(x + z + y)

19. Ifu =u(x,y,z,t)showthat dt = au, +Cad V>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)


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 )


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,


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


[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


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


[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]


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


[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]


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


[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).


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)


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


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


[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


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.


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)-+


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


[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


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


(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


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?


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


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.


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


[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)


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


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-


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


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)


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)


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


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 àvl

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)


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


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


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


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


[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


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


[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)


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


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.

---


SEc. 401

Now

af au

Eic + Fv

ice

of au

f

CIE

au

+ 2uv

OF

au 2f

+ v2

87

aG

au


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


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
The set of numbers whose squares are less than 3 is certainly

bounded, for if x2 < 3 then obviously -2 < x < 2. However, the set of numbers whose cubes are less than 3 is unbounded, for x3 < 3 is at least satisfied by all the negative numbers. This set is, however, bounded above. By this we mean that there exists

a finite number N such that x < N if x3 < 3. Certainly N = 2 89

90


[SEc.41

does the trick. Specifically, a set of elements S is bounded above if a finite number N exists such that x < N for all x in S. Let the student frame a definition for sets bounded below. We shall, in the main, be concerned with sets that contain an infinite number of distinct points. The rational numbers in the

interval 0 < x < 1 form such a collection. Let the reader prove that between any two rationals there exists another rational.

Limit Point. A point P will be called a limit point of a set S if every open interval containing P contains an infinite number of distinct elements of S. For example, let S be the set of numbers (1/2, 1/3, 1/4, . . . , 1/n, . . .). It is easy to verify that any open interval containing the origin, 0, contains an infinite number of S. In this case the limit point 0 does not belong to S. It is at once apparent that a set S containing only a finite number of points cannot have a limit point. Neighborhood. A neighborhood of a point is any open interval containing that point.

Interior Point. A point P is said to be an interior point of a set S if it belongs to S and if a neighborhood N of P exists, every element of N belonging to S. If S is the set of points 0 < x 5 1, then 0 and 1 are not interior points of S since every neighborhood of 0 or 1 contains points that are not in S. However, all other points of S are interior points.

Boundary Point. A point P is a boundary point of a set S if every neighborhood of P contains points in S and points not in S. If S is the set 0 5 x <_ 1, then 0 and 1 are the only boundary

points. A boundary point need not belong to the set. 1 is a boundary point of the set S for which x > 1, but 1 is not in S since 1 > 1. Complement. The complement of a set S is the set of points not in S. The complement C(S) has a relative meaning, for it depends on the set T in which S is embedded.

If S, for example,

is the set of numbers -1 < x < 1, then the complement of S relative to the real axis is the set of points lxl > 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


[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


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


[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) <= s for all x on (a, b). Now consider the set s - e, s - e/2, . . . , s - e/n, ... , where e > 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


[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 <_ 1 into the intervals

(n+1) Sx
(I If \n/

f \n }

)I

2

=1 +3

+ 25

+ N2

We cannot bound this sum for all finite N since the series diverges

as N -> w. N was chosen as an odd integer.


100

[SEc.48

48. Arc Length. Consider the curve given by x = f(t), y = c(t), and assume no multiple points. Divide the parameter t in any manner into n parts, say a = t0 < 11 < t2 < '

'

' < to = ,8

and consider Sn =

i=1

1[x(ti)

- x(ti-1)]2 + ly(ti) - y(tti-1)]2J} This is the length of the straight-line segments joining

Y

the points x(ti), y(ti), i = 0, . . . , n (see Fig. 46). If the set of all such lengths, obtained by all finite meth-

1) 2,

ods of subdivision, is bounded,

we say that the curve is rectifiable and define the length of the curve as the supremum of these lengths.

An important theorem is

----x the following: A necessary

and sufficient condition that the curve described by

F1o. 46.

x = f(t),

y = 'P(t) be rectifiable is that f (t) and V(t) be of bounded variation.

Let

n

A=

{=1

[I f (ti) -

B = I IIf(t;) f-l

so that

f(4.-1)12

+ 14p(ti) - p(ti--1) J 2J1

f(4--1)I + Ip(k)

- v(4-1)I I

A==
Consequently if the curve is rectifiable, f (t) and '(t) are of bounded variation, and conversely, if f (t) and ap(t) are of bounded variation, A is bounded, and hence the curve is rectifiable. If f(t) and V'(t) are continuous, then from the law of the mean, I f(4) - f (ti--1) I

= I f (ii) (ti - ti-1) I < A I ti - ti-il, where A is the supremum of If(t)l, a 5 t =< # and 4_1 5 Ti 5 ti.

INTEGRA TION

SEC. 49] n

n

E If()

i-1

- f(t;-1)I < A 11(4i=1

101

- till = A(# - a)

Similarly, p(t) is of bounded variation, so that the curve is rectifiable. Under these conditions it can easily be shown that the arc length is given by

s - f8 [(dt)2 + (

p)2J dt

(154)

49. The Riemann Integral. We now develop the theory of the Riemann integral in connection with line integrals. We need a curve over which an integration can be performed and a

function to be integrated over this curve. Let r

aSt

x

y=¢(t)

be a rectifiable are. Let f (x, y) be a function continu-

ous at all points of the curve r. Subdivide the parameter t < t, = 6. Let the coordiinto n parts, a = to < ti < t2 < and let Osi be the length of are joining nates of Pi be Pi-1 to Pi. Since f (x, y) is continuous, it will take on both its minimum and maximum for each segment Pi_1P;. Multiply each are length by the maximum value of f (x, y) on this are, say f v, and form the sum n

J = A fii (xi, yi) A8i Let .7 be the infemum of all such sums. Similarly, let K be the supremum of all sums when the minimum value of f(x, y), say f,,, is used. If J = K, we say that f (z, y) is Riemann-integrable

over the curve r, and write

J = K = fr f(x, y) ds = LfI(t),

+ * )} dt

whenever So' and J' are continuous. That J = K for a continuous function defined over a rectifiable curve is not difficult to prove. For,

J-K=

n i-1

102

VECTOR AND TENSOR A V.I LYSIS

[SEC. 50

and from the uniform continuity of f(x, y) we can subdivide the curve r into arcs such that the difference between the maximum and minimum values of f (x, y) on any are is less than any given

e > 0 so that JJ

- Kl < e11lAsil < eL

(155)

We leave it as an exercise for the reader to prove that we can make

J as close to J as we please and similarly that the difference

R=R1+R2 (ii)

FIG. 47.

FIG. 48.

between K and K can be made arbitrarily small, for a sufficiently large number of subdivisions. Since the a of (155) is arbitrary,

we can make the difference between J and K as small as we please so that J = K since they are fixed numbers. 50. Connected and Simply Connected Regions. A region R is said to be connected if, given any two points of the region, we can join them by an arc, every point of the are belonging to the region R. In Fig. 47, (i) is connected; (ii) is not connected. In (i), R is the nonshaded region.

INTEGRATION

Sic. 51]

103

If every closed curve of a connected region R can be continuously shrunk to a point of R, we say that the region is simply connected. In Fig. 48, (i) is connected, but not simply connected; (ii) is simply connected. In (i) the curve r cannot be continuously shrunk to a point. An analytic expression for simple connectedness can be set up, but we shall omit this. 51. The Line Integral. Let

f = X(x, y, z)i + Y(x, y, z)j + Z(x, y, z)k

and consider the line integral f ( f ds) ds along a rectifiable space curve r given by r = r(s).

Since

dsds = dr = dx i + dy j + dz k we have

Jr (f ds/

ds = fr X dx + Y dy + Z dz

(156)

We use (156) as a means of evaluating the line integral. If the space curve is given by x = x(t), y = y(t), z = z(t), then (156) reduces to

ff

dr =

fo'

[x(t)

dt + Y(t) dt + Z(t) dtJ

dt

(157)

In general, the line integral will depend on the path joining the two end points of integration. If f f dr does not depend on the curve r joining the end points of integration, we say that f is a conservative vector field. If f is a force field, we define fB f - dr as the work done by f as the unit particle moves from A A to B. We shall now work out a few examples for the reader. Example 55. Let f = x2i + yaj, and let the path of integra-

tion be the parabola y = x2, the integration being performed We exhibit three methods of solution. (a) Let x = t, so that y = t2. Thus

from (0, 0) to (1, 1).

f = t2i + t6j,

r = ti + t2j,

dr = (i + 2tj) dt

104


[SEc. 52

and ft t of f

dr = f.' (t2 + 2t7) dt =

(b) Since y = x2 everywhere along r, f = x2i + x6j along r, and dr = dx i + dy j = (i + 2xj) dx, so that 1

Jr

JO

(x2 + 2x') dx = A

(c)

r (1, 1)

J(o, 0)

((1,

f dr =

1)

J(0, 0)

1+y'1- 7

xa

x2 dx + y$ dy - 3

0

4 (0

12

(c) shows that the integral from (0, 0) to (1, 1) is independent of the path since x2 dx + y3 dy is a perfect differential, that is, x2 dx + ys dy = d[(xs/3) + (y4/4)]. Example 56. f = yi - xj, and let the path of integration be y = x2 from (0, 0) to (1, 1). Then (1

f(1'1)ydx-xdy=

0,

ro, o)

, o)

f' x2dx-x(2xdx) _ -

Next we compute the integral by moving along the x axis from

x = 0 to x = 1 and then along the line x = 1 from y = 0 to y = 1. We have

f

f("

(0,0))

( o, o)

1, o)

0, o)

Along the first part of the path, f = -xj and dr = dx i, since y = dy = 0. Along the second part of the path, f = yi - j, dr = dy j since x = 1 and dx = 0. Thus

f

(111) 0)

x=1

v

f.--=o -

fv-o

f01-dy=-1

i

The line integral does depend on the path for the vector field

f = yi - xj. We say that f is a nonconservative field. 52. Line Integral (Continued). Let us assume that f can be written as the gradient of a scalar (p(x, y, z), that is, f = 0p. Then the line integral JAB f dr is independent of the path of

INTEGRATION

SEC. 521

105

integration from A to B since

JAfdrJAVdrLd(B)(A) (158) Our final result in (158) depends only on the value of (p(x, y, z) when evaluated at the points A(xo, yo, zo), B(xi, yi, z,) and in no way depends on the path of integration. If our path is closed,

then A = B and (p(B) = p(A), so that the line integral around any closed path vanishes if f = Vip. However, the region for which f = must be simply connected. Let us consider the following example. Example 57. Let

f

- yi

LL

xi

=x2+y2 -rx2+y2

Then f = VV where p = tan-' (y/x), and if we integrate f over the unit circle with center at the origin, we have d

(tan_ixJ)fo2T dO

2r

so that our line integral does not vanish. The region for which f = V(p is not simply connected since p is not defined at the origin.

We now prove that if f f dr is independent of the path, then f is the gradient of a scalar V. Let P(x, y, z) =

JP(z,y.z)

f

dr

o(xo, YO, zo)

JP(fdr)ds

(159)

Now (P(x + Ax, y, z) - (P(x, y, z) Ax

and since the line integral is independent of the path joining P to Q, we choose the straight line from P to Q as our path of integration, that is, dr = dx i. Thus lim AX-40

cP(x+Ox,y,z) --P(x,y,z) =av ax

Ax

iim AX-0

f

z

x+Ax

X (x, y, z) dx Ox

= X (x, y, z)

106


[SEC. 52

We have assumed that f is continuous.

from the calculus. similarly

-

= Y(x, y, z),

y

Now

az = Z(x, y, z)

so that

f=

a(P

ax

av i + a7

,

+

a,P

az

k = VV

(160)

If f has continuous derivatives, we can easily conclude whether f is the gradient of a scalar or not. Assume f = VV, or (1)

X=-"`P,

(3)Z=

(2)

az

Differentiating (1) with respect toy and (2) with respect to x, we see that ax ay

Similarly

ay ax

aYaZ az

ay

ax

az

(161)

azax This is the condition that V x f = 0. Conversely, assume V x f = 0. Let ,P(x,y,z) = f:oX(x,y,z)dx+f,0 Y(xo, y, z) dy

+

z,

Z(xo, yo, z) dz

Now a'P

ax = X (x, y, z) aq _ -L. f.. ay

f

dx+

Y(xo,y,z)

-0Y

ax dx + Y (x,), y, z)

= Y(x, y, z) - Y(xo, y, z) + Y(xo, y, z) = Y(x, y, z)

(162)

INTEGRATION

Sic. 53]

Similarly

107

Consequently

7.(x, y, z).

f

axi+ayj+azk=Drp

We have proved that a necessary and sufficient condition for f to be the gradient of a scalar is that.

V xf =0

(163)

If f = V or V x f = 0, f is said to be an irrotational vector. Example 58.

Let f = 2xyezi + x2ezj + x2yezk.

Vxf=

k

i

j

a ax

ay

a az

2xyez

x2ez

x2yeZ

a

Then

=0

and

=

JX

2xy9 dx +

02ez dy +

02. 0 . 9 dz

= x2yeZ

so that f= V (x2yeZ + constant). Problems

1. Given f = xyi - xj, evaluate f f dr along the curve y = x3 from the origin to the point P(1, 1). 2. Show that if the line integral around every closed path is zero, that is, if jFf dr = 0, then f = V p.

3. Show that /r dr = 0. 4. Show that the inverse-square force field f = -r/r3 is conservative.

The origin is excepted.

5. f = (y + sin z)i + xj + x cos z k. Show that f is conservative and find rp so that f = V p.

6. Evaluate f x dy - y dx around the unit circle with center at the origin.

7. If A is a constant vector, show that /A dr = 0,

jrAxdr=0 53. Stokes's Theorem. We begin by considering a surface of the type encountered in Chap. 3. Now consider a closed

108


[SEC. 53

rectifiable curve r that lies on the surface. As we move, along the curve r, keeping our head in the same direction as the normal ar car x ) we keep track of the area to our left. It is this surface au av u curve

that we shall keep in touch with,

and r

will

be the

boundary of this surface.

We neglect the rest of the surface r(u, v).

We now con-

sider a mesh of networks on the surface formed by a collection of parametric curves.

v curve Fia. 49.

Of course, the boundary r will not, in general, consist of arcs of these parametric curves (see Fig. 49). Consider the mesh A BCD. Let

the surface coordinates of A be (u, v), so that A(u, v), B(u + du, v),

C(u+du,v+dv) D(u, v + dv) are the coordinates of A, B, C, D. We also assume that the parametric curves are rectifiable. Now consider ABCD f dr

The value of fat A is f(u, v); at B it is f(u + du, v); at D it is f(u, v + dv). Now

f (u + du, v) = f (u, v) + dfu = f(u, v) + (dru V)f

= f(u, v) + (au du v) f except for infinitesimals of higher order. Similarly

f(u,v+dv)

=f(u,v)+(-dvv)f

INTEGRATION

Six. 53]

109

Hence, but for infinitesimals of higher order, -auc1r

f+ fi au} dude

av -

(164)

Now

(v x f)

for ar (au

x av

an

Or

= (° x f) x au

°/

C\au

J

av

av ^ L \av

Hencee fiABCD f

dr = (v x f) a x

a du dv

v)fI ar au

(165)

ar ar Now d x a du dv = area of sector ABCD in magnitude, and

au av its direction is along the normal. We define

dd =

Or

Or

aux& du dv

(166)

so that

IABCDfdr= (V We now sum over the entire network. Interior line integrals will cancel out in pairs leaving only fir f dr. Also

I

f f (V

(V

a

over surface 8

as the areas approach zero in size. We thus have Stokes's theorem :

fir f dr = f f (V x f) dd 8

(167)


110

[SEC. 54

Comments

1. The reader may well be aware that (165) does not hold for

a mesh that has r as part of its boundary. This is true, but fortunately we need not worry about the inequality. The line integrals cancel out no matter what subdivisions we use, and for

a fine network the contributions of those areas next to r contribute little to f f (V x f) dd. The limiting process takes care of this apparent negligence. 2. We have proved Stokes's theorem for a surface of the type r(u, v) discussed in Chap. 3. The theorem is easily seen to be true if we have a finite number of these surfaces connected continuously (edges).

3. Stokes's theorem is also true for conical points, where no dd can be defined. We just neglect to integrate over a small area covering this point. Since the area can be made arbitrarily small, it cannot affect the integral. 4. The reader is referred to the text of Kellogg, "Foundations of Potential Theory," for a much more rigorous proof of Stokes's theorem. 54. Examples of Stokes's Theorem Example 59. Let r be a closed Jordan curve in the x-y plane. Let f = -yi + xj. Applying Stokes's theorem, we have

ff ff S

ax

ay x

az

-y 0 =2ffdydx=2A S or

Area A =-

xdy - ydx

For the ellipse x = a cos t, y = b sin t, dx = -a sin t dt, dy = b cos t dt and

A=

f27

ab(cos2 t + sin 2 t) dt = grab

(168)

INTEGRATION

SEc. 541

Example 60.

111

If f has continuous derivatives, then a necessary

and sufficient condition that ff dr = 0 around every closed path is that V x f = 0. If V x f = 0, then Cf U. dr = f f (V x f) dd = 0. Cons

versely, assume f t dr = 0 for every closed path. If V x f p4 0, then V x f 0 at some point P. From continuity, V x f: 0 in some region about P. Choose a small plane surface S in this region, the normal to the plane being parallel to V x f. Then

f- dr = f f V x f- dd > 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


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


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


[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


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
If! [V2cc+ (Vcc)21 dr = ff `p vv - dd = f

0

s

R

and hence fff (VV) 2 dr = 0 so that V p - 0 inside R, and R

g = Vp - 0, so that f = f 1 inside R. We can also prove that f is uniquely determined if its divergence and curl are known throughout all of space, provided that f

tends to zero like 1/r2 as r--, ao. We duplicate the above proof and need Jim f p VV dd = 0. If cp tends to zero like 1/r, then V

zero like 1 /r as r oo. Example 71. Let f = f(x, y, z)a, where a is constant. Apply ing (178), we obtain


120

a-

[SEC. 55

f ffdo= f f fv.(fa)dr=a f f fVfdr s

V

V

Hence fffdd

s

fff

=

7fdr

(190)

V

We leave it to the reader to prove that

JJdo*f = Jff(V*f)dr s

(191)

V

Problems

1. Prove that f f dd x f = f f f (V x f) dr. y

s

2. Prove that ff dd = 0 over a closed surface S. s

c constants, show that = 4(a + b + c), where S is the surface of a unit

3. If f = axi + byj + czk, a,

Jff.do

b,

s

sphere.

fffdd

4. By defining grad f = Jim

As

gradf =

snow that

OT

4,-+O

afi+ afj+afk ax ay az

fff.dd 5. By defining div f = lim °3 A?--+O

div f =

r2 sin 0 Lar

GT

(r2 sin of,) +

, show that

e (r sin 9 fe) + a

for spherical coordinates.

6. Show that

f

fJ

JfJ (P 0fdr.

(rf,)]

INTEGRATION

SEC. 55]

121

7. If w = V x v, v = V x u, show that '9

R

v2dr = f f u xv.dd+ f f S R

8. Show that

f J f w Vu . Vv dr = f f uw Vv . dd - f if u V . (w Vv) dr 9. If v = Vp and V . v = 0, show that for a closed surface

f J f V2dr = Jf pv.dd. 10. Show that f f IrJ2r . da = 5 fJJ r2 dr.

11. If f = xi - yj + (z2 - 1)k, find the value of f f f . do over the closed surface bounded by the planes z = 0, z = 1 and the cylinder x2 + y2 = 1.

12. If f is directed along the normal at each point of the boundary of a region V, show that f f f (V x f) dr = 0. V

13. Show that f f r x da = 0 over a closed surface. S

14. Given f = (xyez + log (z + 1) - sin x)k, find the value of

ffv x f . da over the part of the sphere x2 + y2 + z2 = 1 above S

the x-y plane. 15. Show that (xi + yj)/(x2 + y2) is solenoidal. 16. If f 1 and f2 are irrotational, show that f 1 x f2 is solenoidal. 17. Find a vector A such that

f =- yzi - zxj + (x2+y2)k = V xA. 18. If r, 0, z are cylindrical coordinates, show that De and Find the vector potentials. 19. Let S, and S2 be the surface boundaries of two regions V1 and V2. Let r be the distance between two elementary

V log r are solenoidal vectors.


122

Show that

volumes d-r, and dT2 of V1 and V2.

fv, rn:-2 dT, fsI rm dd1 = -m(mn + 1) fV (IT2

.fs, dii2 and that

1s= dd2

show that V2X = for V2Y, V2Z.

fs, log r ddl = -

fv,

cIT2

f

dTl

V1

r

V xf ='1i+¢2j+4,ak,f = Xi + Yj + Zk,

20. IfV - f =

Vxf=zi.

[SEC. 56

49 O

-

- a3y -

z2' and find similar expressions

Find a vector f such that V f = 2x + y - 1,

56. Conjugate Functions. Let us consider the two-dimensional vector field w = u(x, y)i + v(x, y)j and an orthogonal vector field wi = v(x, y)i - u(x, y)j. Obviously w - wl = 0. What are the conditions on u(x, y), v(x, y) which will make w and w1 irrotational? From Stokes's theorem

x w dd = ff ffv s s { w1 dr = jJv = Jf

w dr =

av

au

ax

ay

au

ax-

dydx av

ay

(192)

dy dx

A necessary and sufficient condition that both w and w1 be irrotational is that 52).

av

ax

- au = 0 and - au -- av = 0 (see Sec. ax

ay

a4

This yields av ax

_

au ay

av

au

ay

ax

(193)

The reader who is familiar with complex-variable theory will immediately recognize (193) as the Cauchy-Riemann equations, which must be satisfied for the analyticity of the complex func. tion w = v(x, y) + iu(x, y), i =

INTEGRATION

SEC. 56]

123

On differentiating (193), we obtain

=

V 2U

a2u

axe

+

a2u 2

=

49Y

a2v V2v=+a axey = a2v

2

(194)

and

au av ax ax

+

au av ay ay

_

If functions u(x, y), v(x, y) satisfy Eqs. (193) we say that The importance of such func-

they are harmonic conjugates. y

v

P (x, yl

Ix

Fia. 51.

tions is due to the fact that they satisfy the two-dimensional Laplace's equation given by (194). If u satisfies V2u = 0, we say that u(x, y) is harmonic.

Let us now consider two rectangular cartesian coordinate systems, the x-y plane and the u-v plane (Fig. 51). Let

r=xi+yj. Now to every point P(x, y) there corresponds a point Q(u, v) given by the transformation u = u(x, y), v = v(x, y). Hence the vector w = u(x, y)i + v(x, y)j corresponds to the vector

r -- xi+yj. If now P(x, y) traverses a curve C in the x-y plane, Q(u, v) will trace out a corresponding curve r in the u-v plane. The curve u(x, y) = constant in the x-y plane transforms into

the straight line u = constant in the u-v plane. Similarly,

124


[Ssc. 56

v(x, y) = constant transforms into the straight line v = constant. The two straight lines are orthogonal. Do the curves

u(x, y) = constant v(x, y) = constant intersect orthogonally? The answer is "Yes"! The normal to the curve u(x, y) = constant is the vector ay i+u

=au

VU

a

and the normal to the curve v(x, y) = constant is Vv =

av

ax

av 49V

i+

yj

au av 0 from (194). ax ax + ay ay = Example 72. Consider the vector field w = 2xyi + (x2 - y2) j.

so that Vu Vv =

Ou av

Here u = 2xy, v = x2 - y2, and av

= au

ax

ay

av

_

= 2x

au _ _ 2

ay

ax

r

y

so that u and v are conjugate harmonics.

The curves

u = 2xy = constant and v(x, y) = x2 --- y2 = constant are orthogonal hyperbolas which transform into the straight lines u = constant, v = constant, in the u-v plane (Fig. 52). Example 73. Consider

w = ( tan-1

x)

i + J log (x2 + y2)j

Here u(x, y) = tan-1 (y/x), v = I log (x2 + y2), and au

av

y

ax

ay

x2 + y2

au

av

x

ay

ax

x2 + y2

so that u and v are conjugate harmonics.

SEc. 56]

INTEGRATION

125

Y

I

0'

Fia. 52.

(3)

(4)

------------

Fra. 53.

The circles x2 + y2 = constant transform into the straight lines v = J log c, while the straight lines y = mx transform into the straight lines u = tan-' m (Fig. 53). Example 74. If u(x, y) is given as harmonic, we can find its conjugate v(x, y). If v(x, y) does exist satisfying (193), then

126


dv=axdx+-dy=aydx Now consider the vector field f =

Vxf=-t

au ay

f&c. 56

au ax dy

i - au j. ax

We have that

x + a2 ) k = 0 by our assumption about u(x, y). y

Hence f is irrotational and so is the gradient of the scalar v, f = Vv, and

v = fa au a dx - jyOU dy + c 1!

8x

(195)

As an example, consider u = x2 - y2, which satisfies V2u = 0. Hence

v=

fox

- 2y dx - f." 2 . 0 dy + c = - 2xy + c Problems

1. Find the harmonic conjugate of xa - 3xy2, of ex coo y, of x/(x2 + y2). 2. Show that u(x, y) = sin x cosh y and v(x, y) = cos x sinh y are conjugate harmonics and that the curves u(x, y) = constant, v(x, y) = constant are orthogonal. What do the straight lines y = constant transform into? 3. If u(x, y), v(x, y) are conjugate harmonica, show that the angle between any two curves in the x-y plane remains invariant

under the transformation u = u(x, y), v = v(x, y), that is, the transformed curves have the same angle of intersection.

CHAPTER 5 STATIC AND DYNAMIC ELECTRICITY

57. Electrostatic Forces. We assume that the reader

is

familiar with the methods of generating electrostatic charges. It is found by experiment that the repulsion of two like point charges is inversely proportional to the square of the distance between the charges and directly proportional to the product of their charges. The forces act along the line joining the two charges. We define the electrostatic unit of charge (e.s.u.) as that charge which produces a force of one dyne on a like charge

situated one centimeter from it when both are placed in a vacuum. The electrostatic intensity at a point P is the force that would act on a unit charge placed at P as a result of the rest of the charges, provided that the unit test charge does not affect the original distribution of charges. For a single charge q placed at the origin of our coordinate system, the electric intensity, or field, is given by E = (q/r3)r. For many charges the field at P is given by

E=-

4'a s a=1 ra

(196)

ra

where ra represents the vector from P to the charge qa.

We have seen in Example 25 that V V. (r/r3) = 0. Consequently,

V.E=0

(197)

so that the divergence of the electrostatic-field vector is zero at any point in space where no charge exists. Hence E is solenoidal except where charges exist, for there E is discontinuous. If the coordinates of P are t, n, t and the coordinates of qi are xi, yi, zi,

then ri =

I(rS

l

V1

ri

ri

- xi)2 + (,l - y,)2 + a

(pJ

- z,)2]1, and

_

rr;3 (xi - )i -I- (yi- 7/)i + (zi -t)k

ot

r ri

an

127

r;


128

[SEC. 58

so that (196) reduces to

E = -Vp

(198)

n

where P =a-1 I qa/ra. We call w the electrostatic potential. For any closed path which does not pass through a point charge,

we have f E dr = - ^p dr = - f d(p = 0. Thus E is also irrotational, and

VxE=O

(199)

We also note that f E dr = cp(P) - p(-o) = tp(P), since ,p(oo) = 0. Hence the work done by the field in taking a unit charge from P to oo is equal to the potential at P. 58. Gauss's Law. Let S be an imaginary closed surface that

does not intersect any charges. In Example 66 we saw that J f (qr/r3) dd = 4Tq. This is true for each charge q; inside S. S

Hence

JS ! La a-1

qara dd = 41r raa a-1

qa

(200)

For a charge outside 8,

Jf.do=JfJv.()dr=0

(201)

since there is no discontinuity in qr/r3, r > 0. Adding (200) and (201), we obtain Gauss's law,

JJE.dd=4irQ

(202)

s

where Q is the total charge inside S. The theorem in words is that the total electric flux over any closed surface equals 4w times the total charge inside the surface. Example 75. We define a conductor as a body with no electric field in its interior, for otherwise the "free" electrons would move and the field would not be static. The charge on a conductor must reside on the surface, for consider any small volume contained in the conductor and apply Gauss's theorem.

SEC. 58]

STATIC AND DYNAMIC ELECTRICITY

129

f f E dd = 4xq and since E = 0, we must have q = 0. This is true for arbitrarily small volumes, so that no excess of positive charges over negative charges exists. Hence the total charge must exist on the surface of the conductor. If a body has the property that a charge placed on it continues to reside where placed in the absence of an external electric field,

Fic. 54.

we call the body an insulator. Actually there is no sharp line of demarcation between conductors and insulators. Every body possesses some ability in conducting electrons. At the surface of a conductor the field is normal to the surface, for any component of the field tangent to the surface would cause a flow of current in the conductor, this again being contrary to the assumption that the field is static (no large-scale motion of electrons occurring). Such a surface is called an equipotential surface. The field is everywhere normal to an equipotential

surface, for the vector E = -V(p is normal everywhere to the surface V(x, y, z) = constant. Example 76. Consider a uniformly charged hollow sphere E. We shall show that the field outside the sphere is the same as if

130


[SEC. 58

the total charge were concentrated at the center of the sphere

and that in the interior of the sphere there is no field.

Let P be any point outside the sphere with spherical coordinates r, 0, cp. Construct an imaginary sphere through P con(see centric with the sphere Fig. 54). From symmetry it is

obvious that the intensity at any point of the sphere is the same Fic. 55. as that at P. Moreover, the field is radial. Applying Gauss's law, we have f f E dd = 4,rQ, or

f fEdS=Ef fdS=4irr2E=4aQ so that

E= r

and

E=

r

(203)

Q

Q

We leave it to the reader to show that E = 0 inside I. Example 77. Field w thin a parallel-plate condenser.

Consider

two infinite parallel plates with surface densities a and -a.

Fic. 56.

From symmetry the field is normal to the plates. We apply Gauss's law to the surface in Fig. 55 with unit cross-sectional area.

f f E do = E = 4ara

(204)

so that the field is uniform. Example 78. We now determine the field in the neighborhood of a conductor. We consider the cylindrical pillbox of Fig. 56 and apply Gauss's law to obtain

SEC. 58]


131

EA = 4raA or

E = 47ro N

(205)

where a is the charge per unit area and N is the unit normal vector to the surface of the conductor. Example 79. Force on the surface of a conductor. We consider

a small area on the surface of the conductor. The field at a point outside this area is due to (1) charges distributed on the rest of the conductor (call this field E1), and (2) the field due to the charge resting on the area in question, say E2 (see Fig. 57). From Example 78, El + E2 = 41rv. Now the field inside the Ei+E2

P

Fia. 57.

conductor at the point P' situated symmetrically opposite P is zero from Example 75. The field at P' is E1 - E2 = 0. Thus El = 2o per unit charge. For an area dS the force is dE = (2av) (v dS) = 27rv2 dS

(206)

This force is normal to the surface. A charged soap film thus tends to expand. Problems 1. Two hollow concentric spheres have equal and opposite

charges Q and -Q. Find the work done in taking a unit test charge from the sphere of radius a to the sphere of radius b, b > 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.


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,


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
where M is the bound of p in the neighborhood of P.

I i l xf p

dT

_ fJf Rp V

Thus

I < M? (e2 + el )

where e' is the radius of the sphere R' surrounding P. The Cauchy criterion holds, so that the limit exists. In much the same way we can show that

E = f f f Ta dr

(208)

and that at a point P where a charge exists

E(P) = lim f uJ -t dr r +0

converges.

Now from Gauss's law

ff 8

f f f pdr v


134

[SEC. 59

In order to apply the divergence theorem to the surface integral,

we must be sure that V E is continuous at points where p is continuous. We assume this to be true, and the reader is referred

to Kellogg's "Foundations of Potential Theory" for the proof of this.

Thus

JJJ(v.E)dr = 47r f f f p dr

(209)

V

V

Since (209) is true for all volumes, it is easy to see that

V E = 4irp

(210)

Since E _ - Vs,, we have

provided V E and p are continuous. Poisson's equation V2,p = -4ap

(211)

and at places where no charges exist, p = 0, so that Laplace's equation, V2,p = 0, holds. Example 80. In cylindrical coordinates V 20

r L or

(r Or + e

r

t90/

+ 8z \r 8z/ J

Consider an infinite cylinder of radius a and charge q per unit length. At points where no charge exists, we have V2,p = 0. Moreover, from symmetry, p depends only on r. Thus

r dr = constant = A

vAlogr+B E_ -Vip= -A r,

r=xi -f- yl

Also 42rc = (Er)r_a = -A/a, so that q = 2,raa = -A/2, and

E=2r r2

(212)


SEC. 601

135

Example 81. To prove that the potential is constant inside a conductor. From Green's formula we have

Jfpvcc.ddJfJ(s7co)2dr+JJJrpv2codr Inside the conductor no charge exists so that V 2(p = 0.

Moreover,

for any surface inside the conductor, E = -Vgp = 0 so that Jff (vp)2 dr = 0 for all volumes V inside the conductor. V

Therefore (VV)2 = 0, and

app

ax

=

app

ay

=

app

az

_ 0, so that p = con-

stant inside the conductor. Problems

1. Solve Laplace's equation in spherical coordinates assuming

the potential V = V(r). 2. Find the field due to a two-dimensional infinite slab, of width 2a, uniformly charged. Here we have p = p(x) and must solve Laplace's equation and Poisson's equation separately for free space and for the slab, and we must satisfy the boundary condition for the potential at the edge of the slab. The space occupied by the slab is given by -a S x < a, - co < y < oo. 3. Solve Laplace's equation for two concentric spheres of radii a, b, with b > a, with charges q, Q, and find the field. 4. Solve Laplace's equation and find the field due to an infinite uniformly charged plane.

5. Prove that two-dimensional lines of force also satisfy Laplace's equation. 6. Show that rp = (A cos nx + B sin nx) (Cell + De-Av) satisfies

x+

a2 $

= 0.

49Y2

7. If Cpl and S02 satisfy Laplace's equation, show that cpl + 4o and Ipl -- rp2 satisfy Laplace's equation. Does cl92 satisfy Laplace's equation? 8. If (pi satisfies Laplace's equation and cp2 satisfies Poisson's equation, show that cpi + 4p2 satisfies Poisson's equation. 60. Dielectrics. If charges reside in a medium other than a vacuum, it is found that the inverse-square force needs readjustment. That this is reasonable can be seen from the following

136


[SEC. 61

considerations. We consider a parallel-plate condenser separated by glass (Fig. 59). Assuming that the molecular structure

/

of glass consists of positive and negative particles, the electrons being bound to the

.+++++++++++.....++++

0/0

---

Glass -----j/

nucleus, we see that the field due to the oppositely charged plates might well cause a dis-

Fm. 59

placement of the electrons away from the negative plate and toward the positive plate. This tends to weaken the field, so that E = 47rv/x, where x > 1. x is called the dielectric constant. It is found experimentally that E -_ (qq'/Kr9)r for charges in a dielectric. Applying this force, we see that Gauss's law is modified to read

fjE

.d

d=

4 Q

(213)

K

and if x is a constant,

f

f

41rQ

(214)

where D is defined as the displacement vector, D = xE _ -x V. Poisson's equation becomes V D = -V (K Vg) = 4rp, and for constantx 4rp

(215)

x

For p = 0 we still have Laplace's equation V2(p = 0. In the most general case, we have 3

D; = Z x;;E;, i=i

i = 1, 2, 3

where D = Dli + D2j + Dak, E = E,i + E2j + Eak, and x;; = K;,. 61. Energy of the Electrostatic Field. . Let us bring charges q,, q2, . . . , q,. from infinity to positions P1, P2, . . . , P,., and calculate the work done in bringing about this distribution. It takes no work to bring q, to P,, since there is no field. To bring q2 to P2, work must be done against the field set up by Q1. This amount of work is glg2/rl2, where r12 is the distance between Pl and P2. In bringing q3 to P3, we do work against the separate


SEC. 61]

137

fields due to q, and q2. This work is gig3/r13 and g2g3/r23. continue this process and obtain for the total work

=- -

We

n q;q'i

W

(216)

ri,

The J occurs because gig2/r12 occurs twice in the summation process, once as glg2/r12 and again as g2g1/r21. The quantity W n

is called the electrostatic energy of the field.

Since (pi =

n

we have W =

gipi.

i=1

;al

q;/rii,

For a continuous distribution of charge,

we replace the summation by an integral, so that

W= jfff pv dr

(217)

Now assume that all the charges are contained in some finite We have V D = 41rp so that

sphere. W

fff

fff

8- f f f Applying the divergence theorem, W

ff

87 S

ff v

Now p,is of the order of 1/r for large r, and D is of the order of 1/r2, while do is of the order of r2. We may take our volume of integration as large as we please, since p = 0 outside a fixed sphere.

Hence lim f f cpD dd = 0, so that s

w= 8A f f f (E D) dr

(218)

138


[SEC. 62

The energy density is w = (1/8ir)E D. For an isotropic medium, D= KE and W = (1/87r) Jff KE2 dr. Example 82. Let us compute the energy if our space contains a charge q distributed uniformly over the surface of a sphere of radius a. We have

=a D = E = qr,r>> r

and

D=E=O,r
The total energy is 22x x

W=s-

ffJ-sin O drdOdcp

q2

2a

62. Discontinuities of D and E at the Boundary of Two Dielectrics. Let S be the surface of discontinuity between two media with dielectric constant K1 and K2. We apply Gauss's law to a pillbox with a face in each medium (Fig. 60). Assuming no charges exist on the surface of K1 discontinuity, we have K2

so that n2=_nl

Si nce nl

D2 n2 = 0.

= - n2, we have DN, = DN,

(219)

FIG. 60.

We have taken the pillbox very flat so that the sides contribute a negligible amount to the flux. Equation (219) states that the normal component of the displacement vector D is continuous across a surface of discontinuity containing no charges. We next consider a closed curve r with sides parallel to the surface of discontinuity and ends negligible in size (Fig. 61). Since the field is conservative,

ft dr = 0

or

Er1 = Er1

(220)

SEC. 63]


139

In other words, the tangential component of the electric vector E is continuous across a surface of discontinuity. Combining (219) and (220), we have DN,

DN,

ET,

E,,

K1EN,

K2EN,

ET,

ET,

or

Fio. 62.

for isotropic media.

Hence

tan01_Kl tan 02

(221)

K2

which is the law of refraction (see Fig. 62). 63. Green's Reciprocity Theorem. Let us consider any distribution of volume and surface charges, the surfaces being conductors. Let p be the volume density and a the surface density. If p is the potential function for this distribution of charges, then

02p = -4up. We shall make use of the fact that E = -Vp and that at the surface E. = 4arv, or E . dd = -4uo dS. A new distribution of charges would yield a new potential function cp' such that V2sp' = -41ro'. Our problem is to find


140

1SEC. 63

a relationship between the fundamental quantities p, o, p of the old distribution and p', o', of the new distribution. To do so, we apply Green's formula

JJJ

(vV2V

- ip'V',p)

JJ(cV,' -

v v) dd

S

V

which reduces to

-4v ii (,Pp '-V 'p)dr=4a f f('-rp'o)dS s

V

or

f if rpp' dr +

ff

' ds = f f f v p dr +

jJ

dS

p

(222)

This is Green's reciprocity theorem. It states that the potential (p of a given distribution when multiplied by the corresponding charge (p', a,) in the new distribution and then summed over all of the space is equal to the sum of the products of the potentials (pp') in the new distribution by the charges (p, o) in the old distribution, that is, a reciprocal property prevails. Example 83. Let a sphere of radius a be grounded, that is, its potential is zero, and place a charge q at a point P, b units from

the center of the sphere, b > 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]


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


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


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


[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.


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.


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


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


148

[SEC. 67


[(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


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


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)


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


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
For r > b, V = (q/r) 0

and for r < b, V = (q/b) 2 (r/b)'Pn(cos 0). Moreover, we have 0


154

[SEC. 68

an induced potential of the form

V = q I (Anrn + Bnr n-1)Pn(cos 0)

(250)

0

which is due to the spheres, the An and Bn undetermined as yet. Hence

For r > 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


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


[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


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


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]


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


[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 <---- u (permeability) D = KE F----- B = µH

(267)

0

74. Solid Angle. Let r be the position vector from a point P N to a surface of area dS and unit normal N, that

is, do = N dS. We define the solid angle subtended at P by the surface dS to be (see Fig. 67)

dct =

r3

P Fia. 67.

The total solid angle of a surface is

J'r.do r S 3

Example 92.

(268)

Let S be a sphere and P the origin so that 12(P)=

4rdS=4u r

fJr

Example 93. The magnetic dipole is the exact analogue of the electric dipole. We consider a magnetic shell, that is, a thin sheet magnetized uniformly in a direction normal to its surface (Fig. 68). Let $ be the magnetic moment per unit area and


SEC. 75]

161

assume S = constant. The potential at P is given by

sff r,MdS=13 f f rras=On Now let P and Q be opposite points on

the negative and positive sides of the surface S.

We have

,p(Q) = -P(4ir - 11)

so that the work done in taking a unit positive pole from a point P on the negative side of the shell to a point Q on the positive side of the shell is given by

P Fur. 68.

W = f," H H. dr = - f,' Vp . dr = *(P) - F(Q)

=6U+$(4,r- 9) W = 4x-8

(269)

75. Moving Charges, or Currents. If two conductors at different potentials are joined together by a metal wire, it is found that certain phenomena occur (heating of the wire, magnetic field), so that one is led to believe that a flow of charge is taking place. Let v be the velocity of the charge and p the density of charge. We define current density by j = pv. The total charge passing through a surface per unit time is given by

ffpv.do_ Jfj.dd Now the total charge inside a closed surface S is Q = f f f p dr. R

If there are no sources or sinks inside S, then the loss of charge

per unit time is given by - aQ = -

ffpv.dd=-

f f at dr. R

dr

Thus

162


[SEC. 76

Applying the divergence theorem, we have 0

(270)

or

at

=

0

Equations (270) are the statement of conservation of electric charge.

We define a steady state as one for which p is independ-

ent of the time, a = 0, which implies V j = 0. It has been found by experiment that if E is the electric field, then

j = XE = -A VM

(271)

where X is the conductivity of the metal. This is Ohm's law. a

For the general case, j, = I Xa,gE#, and the simplest case, 69=1

X = constant, so that V2(p = 0 for the steady state.

We now compute the work done on a charge q as it moves The

from a point of potential p, to one of potential 02, Ip2 > 92. energy at (pi is qpl and at V2 is q(p2. The loss in energy is W = (1v1 - Iv2)q

This loss in electrical energy does not go into mechanical energy, since the flow is assumed steady. Hence the electrical energy

is converted into heat, Q = (,p1 - p2)q. The power loss is

P = dt

= (1P1 - (P2) dq,

and since 01 - 02 = RJ (another form

of Ohm's law, where R is resistance and J current), we have

P=RJ2

(272)

76. Magnetic Effect of Currents (Oersted). Experiments show that electric currents produce magnetic fields. The mathematical expression for the magnetic field is given by dH =

Jr x dr (273) cra


SEC. 76]

163

where r is the vector from the point P; P is the point at which we calculate the magnetic field dH due to the line current J in that

portion of the wire dr, and c is a constant, the ratio of the electrostatic to the electromagnetic unit of charge (see Fig. 69).

Q (tn'r)

Biot and Savart

established this law for straight-line currents. For a closed path

Jrxdr

H=

(274)

P (x,y, z)

cra

Fra. 69.

- x)2 + (n - y)2 + ( - z)2];, and V(1/r) = r/ra, + j a + k az. Hence where V = i clx Now r = [(E

y

H=1fiJVI c r

x(Jdr r

c

since V does not operate on dr and J is a constant. Thus H = V x A, where A = (1/c) J dr/r = (11c) f f f i dr/r is integrated over all space containing currents.

Now V A = fif V (j/r) dr

f f (j/r) s

dd, so that if all

currents lie within a given sphere, we may push the boundary of R to infinity, since nothing new will be added to the integral yielding A. But when S is expanded to a great distance, j = 0

on S, so that that V A = 0. Also V x H = -VIA. Now since A = (1/c)

f f f j dr/r, or 0

A. = (1 /C)

f f f is dr/r, A. = As

(1 /c)

f f f j dr/r,

is dr r fff

164


we have from Sec. 69 that V2A = -- (47r/c)j.

[SEC. 76

Thus

VxH=47j c

(275)

Example 94. The work done in taking a unit magnetic pole around a closed path r in a magnetic field due to electric currents is

ff

c

S

ff s

For an electric current J in a wire that loops r, we have (276)

Example 95. The magnetic field at a point P, r units away from an infinite straight-line wire carrying a current J, is obtained by use of Example 94.

H dr = H(2irr) =

4a c

J

so that

H=

27 cr

We compute the dimensions of c//. Now

Example 96.

f. = qaq,'/Kr2, and f. = qqm'/µr2 so that [M][L]

[g612

[q+x]2

[712

[K][L]2

[µ][L]2

and

[J]

[dq,/dt]

[c]

[c]

[g] [c][T]

[M14[L][K]}

[c][T]2

F rom (276) , Work

Unit pole

=

H

,

dr

c

so that [M][L]2 _ [J] [gm][T]2

= 4'rJ

[c]

SEc. 77]


165

and

[Ml}[L];

[M]}[Ll'[K]i

[7'][µl}

[c][TJ2

yielding

L/i

[TL]

We see that c// has the dimensions of speed. We shall soon see the significance of this.

(2)

(1)

Fca. 70.

77. Mutual Induction and Action of Two Circuits.

Consider

two closed circuits with currents J, and J2 (Fig. 70). The magnetic field at 0 due to J, is H, = V x A, where

A,=J, r dr c (1)r We define the mutual inductance of the two circuits as the mag-

netic flux through the surface B due to a unit current in (1). This is

IM=

f f L da= f f a

a

1(2) A1. dr = c 1(2)

(1(1)

drl)

dr2

Hence

M=

dr, dr,

1 C

f(2)J(1)

r

(277)

The current element J2 dr2 seta up a magnetic field, so that from Newton's third law of action and reaction, any magnetic

166


[SEC. 77

field will act on J2 dr2 with an equal and opposite force. Thus

df = J 2 drz x H, =

J,cJ2

(L) dr, x V r

x dr2

(278)

and integrating over (2) we obtain

f-J,J2f drzxf (2)

C

OZ

(1)

r

xdr,

Now dr2 x (TI 1 x

r

dr,) = V 1 (dr, dr2) - (dr2 "V 1) dr, r r

and J 2) [dr2 - v(1/r)] dr, =

f=

I2)

d(1/r) dr, = 0, so that

Jc 2

J (2)1(1)

(C'

1) (dr,

dr2)

This is the force of loop (1) on loop (2).

(279)

It is equal and

opposite to the force of loop (2) on loop (1), this being immediately deducible from (279) when we keep in mind that

v-=-V1

1

r2,

r12

In (279), r = r2,. Example 97. We find the force per unit length between two long straight parallel wires carrying currents J, and J2. We use (278) and the result of Example 95. We have H, = (2J,/cd)i at right angles to the plane containing the wires. Hence

df=Jzdr2x 2J, di.

=

2J,J2

dr2xi

and the force per unit length is F = 2J,J2/cd. If the currents are parallel, F is an attractive force; if the currents are opposite, F is a repulsive force. Problems

1. From (278) show that f = J2 f f dd2 x (V X H,). a 2. Find the force between an infinite straight-line wire carrying

a current J, and a square loop of side a with current J2, the

SEC. 791

167


extended plane of the loop containing the straight-line wire, and the shortest distance from the wire to the loop being d. 3. A current J flows around a circle of radius a, and a current J' flows in a very long straight wire in the same plane. Show

that the mutual attraction is 4irJJ'/c(sec a - 1), where 2a is the angle subtended by the circle at the nearest point of the straight wire.

4. Show that A = (J/c) f f dd x V (1 /r) for a current J in a S

closed loop bounding the area S.

For a small circular loop, show

that A = (M x r/r$), where r is very much larger than the radius of the loop and is the vector to the center of the circle, and where

M=c f f dd 78. Law of Induction (Faraday). It has been found by experiment that a changing magnetic field produces an electromotive force in a circuit. If B is the magnetic inductance, the flux through a surface S with boundary curve r is given by f f B dd. S

The law of induction states that

-ca f f Applying Stokes's theorem, we have

_ - =VxE C at

(280)

The time rate of change of magnetic inductance is proportional to the curl of the electric field. Equation (280) is a generalization of V x E = 0, which is true for the electrostatic case in which

B = 0 and for the steady state for which atB = 0.

79. Maxwell's Equations. Up to the present we have, for an electrostatic field, V x E = 0, V D = 4,rp and, for stationary currents, V x H = (41r/c)j, V- B = 0.


168

Now V x E_- 1CaB at

[SEC. 79

a generalization of V x E = 0.

is

Maxwell looked for a generalization of V x H = (41r/c)j. He decided to retain the two laws: (1) V D = 4rp as the definition

of charge, and (2) V j +

at

= 0 as the law of conservation of

charge.

Let us assume

VxH=4w(j+x)

(281)

C

We take the divergence

as a generalization of V x H = (47r/c) j. of (281) and obtain

(282)

so that

= - V j = at = Oar at (V . D)

V

aD 0

so that

We can choose Z =

VxH= w- (i

+--aD) (283)

C

We call -

aD t

the displacement current.

We rewrite Maxwell's equations

V D = 4irp (284)

V X

4c \1 + 41r aD1 t

SEC. 801


169

We have in addition the equation

f=p(E+1vxBJ

(v)

\\

(285)

C

where f is the force on a charge p with velocity v moving in an electric field E and magnetic inductance B. This result follows from Sec. 77.

Problems

1. Show that the equations of motion of a particle of mass m and charge e moving between the plates of a parallel-plate condenser producing a constant field E and subjected to a constant magnetic field H parallel to the plates are

md-

= Be -

dy

d

dx = He ML dt dtz

Given that

d

dt

= x = y = 0 when t = 0, show that

z = (E/wH) (1 - cos wt), y = (E/(X) (wt - sin wt), where He

w = -

m

2. From (iii) of (284) show that V - aB = 0.

3. From (i) and (iv) of (284) show that

D) -

at

4. Write down Maxwell's equations for a vacuum where

j=p=O,D=E,B=H.

80. Solution of Maxwell's Equations for "Electrically" Free

Space. We have p = j = 0 and c, u are constants. Equations (284) become


170

[SEC. 80

V E=0

(i)

VH=0 vxE=

(ii)

aH

(286)

K aE vxHcat

We take the curl of (iii) and obtain

V x (V x E) = V(V E) - V2E _ or

V2E =

µK a2E

C2 at2

(287)

by making use of (i) and (iv). Similarly

V2H -

JLK a2H C2

8t2

(287a)

Equation (287) represents a three-dimensional vector wave equation. To illustrate, consider a wave traveling down the x axis

with velocity V and possessing the wave profile y = f(x) at t = 0. At any time t it is easy to see that y = f(x - Vt). From y = f(x - Vt) we have

a2-y

= f"(x - Vt) and

at= = V2f"(x - V0 so that a2y ax2

1 a2y V2 at2

(288)

Equation (287) represents three such equations, and µK/c2 plays

the same role as 1/V2, so that c/V has the dimensions of a velocity (see Example 96).

c = 3 X 1010 by experiment. z

Example 98.

We solve the wave equation V2f = Y2 at2 in

spherical coordinates where f = f(r, t).


SEC. 80J

171

Vf = f'(r, t) Vr = f' r V2f

=

V.(rr/

3f+rf" = rf V - r+V(

)'r f,

Our wave equation is 2 of

19 2f

1 a2f V28t2

are+rar

(289)

Now let u(r, t) = rf(r, t); then

af_1au Or

r Or

u

02fi0 2u2au

r2

are

r are

r2 Or

2

+

r2

and substituting into (289), we obtain a2u V2 at2

82u

are

of which the most general solution is

u(r, t) = g(r - Vt) + h(r + Vt) and so

.f(r, t) = r [g(r - Vt) + h(r + Vt)]

(290)

is the most general solution of (289).

Let us now try to determine a solution of Maxwell's equations

for the case E = E(x, t), H = H(x, t).

Now V E = 0, so that which implies

aE=(x, t)

8E.(x, t) ax = 0.

ax

+

aEy(x, t) ay

+

8Eg(x, t) az

0,

We are not interested in a uniform

field in the x direction, so we choose E. = 0. Hence

E = EE(x, t)j + Es(x, t)k and similarly

H = Hy(x, t) j + H.(x, t)k

172


[SEC. 80

Now we use Eq. (iii) of (286), V x E _ - c axt , so that i

j

k

a

a

a

ax

ay

az

0

Ey

E.

AaH, i

LaH,

at c at

c

k

or

(i)

OE.

,.8H4

ax

c

at

aE,

aH,

ax

at

(291)

Similarly, on using (iv) of (286), we obtain (i)

aH,

K aE

8x

c at

aH,

K aE,

ax

c at

(292)

The four unknowns are E., E,, H,,, H,, which must satisfy (291) and (292). If we choose H = E, = 0, we see that (i) of (291) and (ii) of (292) are satisfied. Differentiating (ii) of (291) with respect to x and (i) of (292) with respect to t, we obtain a2Ey

axe

UK 61%

C2 at2

(293)

We leave it to the reader to show that 82H. ax2

Ax a2H, c2

at=

(294)

'These equations are of the type represented by (288). Hence a solution to Maxwell's equations is E = [E,,(') (x - Vt) + Eyes) (x + Vt)Jj H = [H,(')(x - Vt) + H,{2>(x + Vt)Jk

(295)

where V = c(AK)"}.

Both waves are transverse waves, that is, they travel down the x axis but have components perpendicular to the x axis.

SEC. 801


173

Also note that E H = 0, so that E and H are always at right angles to each other.

By letting H. = Er, = 0, we can obtain another solution, E = E.(x, t)k, H = H (x, t)j. These two solutions are called the two states of polarization, the electric vector being always oriented 90° with the magnetic vector. Example 99. We compute the energy density. 2

w,n=

_

_

We _

BH 2

and w = w, +

2 µH.2

2 µH2

2 = 2 =w.

=2

2w, = 2 ,. = KE 2, and for both waves

w, = K(E,,2 + E.2). We have here used the fact that

(see Prob. 1). Example 100. Maxwell's equations in a homogeneous conducting medium are

V.E=4p K

V.H = 0 OH

7E

c at

VxH=4c1uE+4v

a/

Assume a periodic solution of the form z)e-;'e

E = Eo(x, y, H = Ho(x, y,

z)e-'"e

Substituting into (iv), we obtain e--;tee V

x Ho =

4 1t oe.-;'eE0 -

iKW

or

VxHo

4a/ =-lo-Eo 2Kw

C

4r

(296)


174

[SEC. 81

This equation is the same as that which occurs for "electrically" free space with a complex dielectric coefficient. Problems

1. By letting E = f(x - Vt), Hz = F(x -- Vt), V = c/, show that H. = VKIIA E. 2. Derive (287a). a2y = 3. Letr=x- Vt, s =x+ Vt, and show thataX2

1 a2y V2 at2

2

reduces to -ay = 0. Integrate this equation and show that the Or as

general solution of (288) is y = f(x - Vt) + F(x + Vt), where f and F are arbitrary functions. 4. Prove that Maxwell's equations for insulators (a = 0) are

v x H=

KaE C at

V x E_- c a

and

(297)

5. Show that the solution of (297) can be expressed in terms of a single vector V, the Hertzian vector, where

aV 2

H=-vx at

and V satisfies V2V =at2 c. 6. Prove

thatE=-VxeH= at c

-V(V.W)+Kµa

2W

C2 at2

isa

192W

solution of (297), provided that W satisfies V2W = e

at2

7. Derive (294). 8. Look up a proof of the laws of reflection and refraction. 9. By considering (i) and (iv) of Example 100, show that

P=

Poe-4x'ì`

81. Poynting's Theorem. Our starting point is Maxwell's equations. Dot Eq. (iii) of (284) with H and Eq. (iv) with E and subtract, obtaining

c(H-V xE -

xH) = -H aB at

ID at

(298)


SEc. 821

Now from (218) aw, at

1

E

4ir

aD

aw,,

at

at

_

175

H aB 4r at 1

Let us write j = jo + jc where jo represents the galvanic current and j. = pv, the conduction current. Now dr

awmec,,,,a;c,,

awM

at

at

and from Sec. 75 it is easy to prove that E ja is Joule's power loss =

aQ

Moreover, H V x E - E V x H = V (E x H), so

that we rewrite (298) as

c V (E X H) = -41r

aw,

aw, at

+

at

aQl

awm

+

at

+

at

(299)

Integrating over a volume R and applying the divergence theorem, we obtain

f if aQ dr + 4r f f E x H- dd

f f at dr J

(300)

where w is the total energy density. We define s = (c/4x)E x H as Poynting's vector. Equation (300) states that to determine the time rate of energy loss in a

given volume V, we may find the flux through the boundary surface of the vector s = (c/4T)E x H and add to this the rate of generation of heat within the volume. It is natural to interpret Poynting's vector as the density of energy flow. Problems

1. Find the value of E and H on the surface of an infinite cylindrical wire carrying a current. Show that Poynting's vector represents a flow of energy into the wire, and show that this flow is just enough to supply the energy which appears as heat.

2. Find the Poynting vector around a uniformly charged sphere placed in a uniform magnetic field.

3. If E of Sec. 80 is sinusoidal, E = Eo sin co(x - Vt)k, find the energy density after finding the magnetic wave H.

82. Lorentz's Electron Theory. For charges moving with velocity v, j = pv, and Maxwell's equations become


176

[SEc. 82

4irp

(i) (ii)

(iii)

(301)

(iv)

V

xH

= 4c

D)

+

(pv

These equations are due to Lorentz. From (ii) we can write B = V x (Ao + Vx) = V x A. Substitute this value of B into C

(iii) and obtain V x E

v x Af or its equivalent

vx(E+catA/

0

I

Thus E + A is irrotational, so that E + 1

-vgp.

A

Let D= KE, B = pH, and substitute into (iv). We have I µ

V x (V xA) =

-

r Ipv+ 4a(

z at2A-vaC1 SO

t

c

and since V x (V x A) = --V2A + V(V A), we obtain z

v2A

cl-p

v + V j,

c2 at2

(302)

where

vA+c2 at Now

= -K V2(p -

K(--AKa2S0/ C

at

c

Kp, 12,p

4rp

_

14

c2 812

K

so that

_ vg

cat

at2

(303)

SEC. 821


177

Equations (302) and (303) would be very much simplified if

VA+

we could make

KA a

c- at

__ O.

This is called the equa-

tion of gauge invariance.

Let us see if this is possible. i aA° = Now B = V x A0 and E -v4p° so that + C at l aA

I aAo

cat - vp

E = -cat where A = A0 + V. Thus c1

(BA

aA0

at

at

1a

-cat °x =

v((P°

c°)

and

_

Iax _ C at 1 a2x C ate

(p + constant

asoo

aso

at

at

Now we desire

a' c2 at

ao C

z

at

C

at21

or 2

vzx

a°

-v A0

cz atz

cz

(304)

The right-hand side of (304) is a known function of x, y, x, t. This equation is called the inhomogeneous wave equation, and if the equation of gauge invariance is to hold, we must be able to solve it. If we can solve it, the Lorentz equations will reduce to four inhomogeneous wave equations and so will also be solvable. They are

v2A-c2

a2t2

=

v25p-=-= K a2c C2 at 2

4cµP,v,

(305)

4w K

P


178

(SEC. 83

Problems

1. For the Lorentz transformations (see Prob. 11, Sec. 24), show that a2,p

825,

1 a2p c2at2

a24,

a2tp

a2,p

a2(p

1 a2V

c2ai2

the D'Alembertian. We call 2. Consider the four-dimensional vector C; = (A1, A2, A3, -q'), 2

i = 1, 2, 3, 4, the A; satisfying V2A = C2 at2' while p satisfies 2

V 2v = -

with H = V x A, and E=

4,

-cat -

Let

X, = x) x2 = y, x3 = z, x' = ct, and show that

_ F"

Show that

0

_ aC;

8C; C axi

- H.

Hr

axf

-HY

0 Hx

E.

EY

H, - Es

- Hz - Es 0 E.

--E, 0

axr,s = 0, i = 1, 2, 3, 4, yields V x H =

-1

V E = 0, Fit = -F;;, i or j = 4, otherwise Fu = F;1.

at

and Also

show that

aFj-

aF.# axy

+

axp

3. If P;;

+

= °

ax.

aH

yields V x E _ 4

aF#,

a, P, y = 1,2,3,4

and V H = 0.

4

axa axs

c

formations P12 = -171 =

show that for the Lorentz trans-

Ht

ll -

(y2/C2)]l

Complete the ma-

trix P 1.

83. Retarded Potentials. Kirchhoff's Solution of

72

1

V2

2

av

- 47rF(x, y, z, t)


SEC. 83]

179

To find the solution (p(P, t) of the inhomogeneous wave equation at t = 0, we surround the point P by a small sphere of

Fia. 71.

radius e, and let S be the surface of a region R containing P (see Fig. 71). We apply Green's formula to this region.

fJ f

dT = f f

V2,p

R

Vp - V#) dd

B

+ f f (# V-p - p Vu')

dd

(306)

9

We choose for ¢ a solution of 'V24, - V2 a

= 0. We know

that 4, = f(r + Vt)/r is one such solution, where f is arbitrary. Equation (306) now becomes

V2f

If

(4,,,2,p

t_V

actd-r-4'rfff FOd-r= f f R

.. .

B

+ fJ a

.

(307)


180

[SEc. 83

Equation (307) is true for all values of t so that we may integrate

(307) with respect to t between limits t = tl and t = t2. We obtain

III

at

1:2l

at

j:2

dT - 4

dt j f f F4, dT

f dt (if .. rr

=

.+

Now on E, ¢ _ (1/E)f( + Vt) and

v o da = -

j

f

ar,._ = 1 [f(E + Vt) E

(308)

+ Vt)] dS

2

so that (308) reduces to 12

V2

f

If [f(r + Vt) &p at r

- 4w f if + 91

Vf'(r + Vt)1t2

r

fiFf(r+

t)dt dr = f,t: dt

Ef'(E + Vt)

dr

t,

f(E + Vt) R]

`

f

E

[f( Vt) 1

, da .+ f f

JJ

E2

...}

(309)

s

Let us now return to a consideration of f(r + Vt). Since f is arbitrary, let us choose f ° 0 for jr + VtI > 6, with the addias f(r + Vt) d(r + Vt) = 1, where 6 is tional restriction that arbitrary for the moment. Notice that f = 0 for Ir + Vti > a. Now let us choose t2 > 0 and tl negatively large, so that for all values of r in the region R, jr + Vt[ > a. Hence

f

[f(r + Vt) &p r at L

_

_

Vf'(r + Vt)

r

Itt,

since Ir + V121 > 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


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)


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


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)


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


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)


192

Now r = rR, where R is a unit vector.

SEC. 88

Hence R

v


\ 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


[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


[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


[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-


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


[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

Jò

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.


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


[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.


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.


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


[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


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


[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


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


[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
- dt8 sin So

(379)

d

dd

(d)2

do

d2

(dB 2

+

+Cdt

+ 2 cos 8

dcp d#

dt dt

(380)

For the fixed frame

a

w=-

wy

do

dcp

sin>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


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<'(C-A)/ALot

so that co. = a cos at wy = a sin at

(383)

where a = [(C - A)/A]wo and a is a constant of integration. Now w2 = ws2 + Wye + ws2 = a2 + wp2 = constant, so that the

magnitude of the angular velocity remains constant during the motion.

Moreover,

fixed space.

d

= 0, so that H is a constant vector in

We choose the z' axis for the direction of H. Now

H = Awzi + Bw,j + Cwk = Aa cos at i + Aa sin at j + Cwok

(384)

This shows that H rotates around the z axis (of the body) with constant angular speed a = [(C - A)/A]wo, and since H is fixed in space, it is the z axis of the body which is rotating about the fixed z' axis with constant angular speed -a = [(A - C)/A]wo.

Also H k = I HI cos 0 = Cwo, so that 9 is a constant since I HI = constant. We say that the top precesses about the z' axis. 106. The Top (Continued). We have assumed above that the weight of the top or gyroscope was negligible, or that the gyroscope was balanced, that is, suspended with its center of mass at the point of support, so that no torques were produced. We

MECHANICS

SEC. 106]

223

shall now assume that the center of mass, while still located on the axis of symmetry, is not at the point of support. We now have the following situation (Fig. 93) :

L = 1kx(.-Wk') = WI sin ON The three components of the torque are Lz = WI sin 0 cos 'v

L" = - Wl sin 0 sin p L. = 0 ZI

Fro. 93.

Euler's equations become

Wl sin 0 cos p = A - WI sin 8 sin (p = A

dw=

dt dwdt"

+ (C - A)w"ws

+ (A - C)wws

(385)

0=CdsforA=B Multiplying Eqs. (385) by wx, w", Co., respectively, and adding, we obtain Hence w: = coo.

2

d (Awx2 + Bw"2 + Cw,2) = Wl sin 0(wy cos

w" sin (p)

(386)

224


[SEC. 106

From (379) we have wz cos P - w sin cp = de, so that (386) becomes

1 d (A w=2 +

d8 B

2 dt

dt

and integrating Cwz2 = -2W1 cos 0 + k

Awx2 +

or, again using (379), (dO)2

(j)2

= a - a cos B

sin2 B +

(387)

a and a are constants.

Now since Le = L k' = 0) we have HZ- = constant. Also H = AwJ + Bw j + Cwsk, so that

9+Cwocos0

sin

= constant

Replacing wr and w by their equals from (379), we have A

sin2 9

sine,p + d sin (p sin B cos

+ di cos2 V sin2 0

- dte cos V sin (p sin 9) + Cwo cos 8 = constant

(388)

or A d sin2 9 + Cwo cos 9 = constant = He. Let 16 = H,,/A, b = Cwo/A, so that (388) becomes

d' -bcos9 dt - sin2 9

(389)

From (379) d4' w, = wo =

cos B + LIP

dt

(390)

dt

Using (389), (387) becomes ll2

(

sin9 s 9/

x

+(de = a - a cos 0

(391)

MECHANICS

SEC. 107]

Let z

cos 0, so that dt = - sin 6

225

-, and

2

( - bz)2 +

Cdt

= (a - az) (1 - z2)

Hence

t = fos [(a - az) (1 - z2) - (g - bz) 2]-} dz

(392)

This integral belongs to the class of elliptic integrals. If we can integrate and find z, then we shall know

9-bz

dip

dt - 1 - z2'

dt

d4,

= wa

d'y

- dt

The reader should look up a complete discussion of elliptic integrals in the literature.

Fm. 94.

107. Inertia Tensor. The moment of inertia of a rigid body about a line through the origin may be computed as follows. Let the line L be given by the unit vector ro = li + mj + nk, and let r be the vector from 0 to any point P in the body, r = xi + yj + zk

(see Fig. 94).

The shortest distance from P to L is given by

226


[SEC. 107

D2 = r2 -- (r r0)2 _ (x2 + y2 + z2) - (lx + my + nz)2 (12 + m2 + n2)(x2 + y2 + z2) - (lx + my + nz)2 _ 1l2(y2

=

+ z2) + m2(z2 + x2) + n2(x2 + y2) - 2mnyz

- 2lnzx - 2lmxy

Thus

I = f f fpD2dzdydx = Ale + Bm2 + Cn2 - 2mnD - 2n1E - 2lmF Let us replace 1, m, n by the variables x, y, z, and let us consider the surface

,p(x, y, z) = Axe + By2 + Cz2 --- 2Dyz - 2Ezx - 2Fxy = 1 (393)

A line L through the origin is given by the equation x = It, y = mt, z = nt. This line intersects the ellipsoid rp(x, y, z) = 1 for t satisfying

(Al2 + Bm2 + Cn2 - 2Dmn - 2n1E - 2lmF)t2 = 1

or t2 = 1/I. The distance from the origin to this point of intersection is given by d = (1212 + m2t2 + n2t2)} = t =

I-f

so that (394)

We know that a rotation of axes will keep I fixed, for the line and the body will be similarly situated after the rotation. We

now attempt to simplify the equation of the quadric surface ,p(x, y, z) = 1. First, let us find a point P on this surface at which the normal will be parallel to the radius vector to this point. The normal to the surface is given by Vip, so that we desire r parallel to Vv, which yields the equations

Ax - Ez - Fy

By - Dz - Fx

Cz - Dy - Ex

x

y

z

(395)

Any orthogonal transformation (Example 8) will preserve the form of (393) and (395) with x, y, z replaced by x', y', z' and A, B, . . . , F replaced by A', B', .. . , P. Now choose the

227

MECHANICS

SEC. 107]

z' axis through P so that x' = 0, y' = 0, z' _

satisfy (395).

This yields - E'/O = - D'/0 = C', which means that

E'=D'=0 and (393) reduces to

A'x'' + B'y'' + C'z'' - 2F'x'y' = 1 The rotation

(396)

x"=x'cos0-y'sin0

y"= x' sin0+y'cos0 z" = z'

with tan 20 = F'/(B' - A') reduces (396) to A"z" + D''y.,, + cf/z,F= = 1

(397)

This is the canonical form desired. We have thus proved the important theorem that a quadratic form of the type (393) can always be reduced to a sum of squares of the form (397) by a rotation of axes. In the proof we made the assumption that there was a point P such that r is parallel to V o, which yielded (395). We could have arrived at Eqs. (395) by asking at what point on the sphere x2 + y2 + z2 = 1 is (p(x, y, z) a maximum. Since p(x, y, z) is continuous on the compact set

x2+y2+z2 = 1 such a point always exists. Equations (395) are then easily deduced by Lagrange's method of multipliers. We can arrange the constants of inertia into a square matrix

-E

I = -F

B -D

-E -D

(398)

C

The elements of the matrix (an array of elements) are called the components of I. Under a proper rotation we have shown that we can write

A" 0

I= 0 0

B" .

0

0 0 C"

(399)

228


[SEC. 107

In general, under an orthogonal transformation, I will become

- E'

-F1

A'

I = -F'

-D'

B'

-D'

- E'

(400)

C'

and the components of I in (400) will be related to the components of I in (398) according to a certain law. We shall see in Chap. 8 that I is a tensor and so is called the inertia tensor. Referring back to (367), we may write

A -F -E H _ -F B -D

wZ

Hz

ws

H=

-E -D

wb

C

(401)

from the definition of multiplication of matrices, where

w = w1 + wyJ + w If we write (398) as

111

121

I31

1112

122

I32

113

123

I3

and

Ho' = H,i + H2J + H3k

w=w1i + w2j + w3k, then (367) may be written 3

H, = I, I,-way

j= 1, 2, 3

(402)

aa1

which is equivalent to the matrix form (401). Problems

1. Find the moments and products of inertia for a uniform cube, taking the cube edges as axes. 2. Show that the moment of inertia of a body about any line is equal to its moment of inertia about a parallel line through the center of mass, plus the product of the total mass and the square of the distance from the line to the center of mass.

3. Find the angular-momentum vector of a thin rectangular sheet rotating about one of its diagonals with constant angular speed we.

229

MECHANICS

SEC. 1071

4. If 3

Hp =

3

Ipawa,

Hp = I Ipacoa a-1

a=1 3

lip = I ap'Ha,

3

coo =

a-1

apawa,

3 = 1, 2, 3

a-1

for arbitrary wa, show that 3

3

1 Ipaaa = I Ia°apa, a-1

6, a = 1, 2, 3

amt

5. Let us consider the form

I = x2 + 9y2 + 18z2 - 2xy - 2xz + 18yz We may write

I = (x2 - 2xy - 2xz) + 9y2 + 18yz + 18z2

(x-y-z)2+8y2+16yz+1722

y - z)2 + 8(y + z)2 + 9z2 X2 +Y2 + Z2 (x

where X = x - y - z, Y = V"8- (y + z), Z = 3z, a set of linear transformations from x, y, z to X, Y, Z. This method may be employed to reduce any quadratic form to normal form. However, the linear transformations may not be a rotation of axes. Reduce I to normal form by a rotation of axes.

CHAPTER 7 HYDRODYNAMICS AND ELASTICITY

108. Pressure. The science of hydrodynamics deals with the motion of fluids. We shall be interested in liquids and gases, a liquid or gas being defined as a collection of molecules, which, when studied macroscopically, appear to be continuous in structure. A liquid differs from a z

solid in that the liquid will yield to any shearing stress, however small, if the stress is continued long enough. All liquids are compressible

to a slight extent, but for many purposes it is simpler to consider the liquid as being incompressible. We shall

also be highly interested in

FIG. 95.

These are

perfect fluids.

liquids which possess no shearing stresses. We now show that the pressure is the same in all directions for a perfect fluid. Let us consider the motion of the tetrahedron ORST (see Fig. 95). The face ORT has a force acting on it, since it is in contact with other parts of the liquid. Under the above assumption, this force acts normal to the face. Call it Af,,. If we divide Af" by the area of the face ORT, AA,,, we

obtain the pressure on this face, P,, =

f"

AAy

The limit of this

quotient is called the pressure in the direction normal to the face

ORT. The y component of the pressure on the face RST is P cos l4. Let f" be the y component of the external force per unit volume, and let p be the density of the fluid. of motion in the y direction is given by

The equation

+f"oT=d = 230

p A zr

y

dt°

(403)

HYDRODYNAMICS AND ELASTICITY

SEC. 109]

since

dt

(p Ar) =

dm dt

231

= 0. Now AA = AA,. cos 0, so that (403)

becomes

(Pv - Pn) + fv A

As A --- 0, we have

AA

oAv dt

Oz Cp

(404)

i

0, so that if we assume f,,, d

dt2

p

finite, we must have P = P.. Similarly, P = P. = P. = p. Since the normal n for the tetrahedron can be chosen arbitrarily, the pressure is the same in all directions and p is a point function, p = p(x, y, z, t). We leave it to the student to prove that at the boundary of two perfect fluids the pressure is continuous. 109. The Equation of Continuity. Consider a surface S bounding a simply connected region lying entirely inside the liquid. Let p be the density of the fluid, so that the total mass of the fluid inside S is given by p(x,y,z,t)d-r M=

ff R

Differentiating with respect to time and remembering that x, y, z are variables of integration, we obtain M dd =

f J fat dr

(405)

Now there are only three ways in which the mass of the fluid inside S can change: (1) fluid may be entering or leaving the surface.

The contribution due to this effect is JJ vp dd. s

(2) matter may be created (source), or (3) matter may be destroyed (sink). Let 4,(x, y, z, t) be the amount of matter created or destroyed per unit volume. For a source, 4, > 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


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<)zj+

2. Showthaty=

(x2 + y2)2

i+ (x2 + y2)2

x2 + y2 kiss

possible motion for an incompressible perfect fluid. Is this motion irrotational? 3. Prove that, if the normal velocity is zero at every point of the boundary of a liquid occupying a simply connected region, and moving irrotationally, rp is constant throughout the interior of that region.

Sic. 110]

233


4. Prove that if v is constant over the boundary of any simply connected region, then (p has the same constant value throughout the interior. 5. Express (407) in cylindrical coordinates, spherical coordinates, rectangular coordinates. 110. Equations of Motion for a Perfect Fluid. Let us consider the motion of a fluid inside a simply connected region of volume V and boundary S. The forces acting on this volume are

(1) external forces (gravity, etc.), say, f per unit mass; (2) pressure thrust on the surface, - p dd, since dd points outward. The total force acting on V is

F = fff pfdr - f f pdd

=

f f f (pf - Vp)dr

The linear momentum of V is

M = J117 pvdr and the time rate of change of linear momentum is

dMd =

fffpvdr

I tf f pdT + f f f v(pdr)

since the volume V changes with time. However, p dr is the mass of the volume dT, and this remains constant throughout the motion, so that

d

(p dT) = 0.

Since F =

dd , we obtain

fff(pf-vp)drfffpdr This equation is true for all V, so that

pf - VP=p dv d

I

or

dt

f - p VP

This is Euler's equation of motion.

(411)

234


From (76) we have that

dvav dt

at

SEC. 111

+ (v V)v, so that an alterna-

tive form of (411) is (412)

Vp

Also from Eq. (9) of Sec. 22, Vv2 = 2v x (V x v) + 2(v - V)v, so that (412) becomes av

at

1

+1VV2-Vx(VxV)=f-PVp 2

(413)

111. Equations of Motion for an Incompressible Fluid under the Action of a Conservative Field. If the external field is con-

servative, f = -Vx, so that f - (1/p) Vp = -V[x + (p/p)] if p = constan t. Hence (413) becomes (IV

\

-vx(Vxv)=-V(x+P+2v2)

(414)

-

We consider two special cases: (a) Irrotational motion. v = V\p and V x v = 0, so that (414) becomes

at

= -v l x + + 2 v2). \ P

(b) Steady motion. a = 0, so that (414) becomes

l 1 vX(Vxv)=VlX++1v2 P For this case we immediately have that v.[V(x+p+1v2)]=0

p

2

Hence V[x + (p/p) + +v2] is normal everywhere to the velocity field v. Thus v is parallel to the surface X + (p/p) + j v2 = constant. The curve drawn in the fluid so that its tangents are parallel to the velocity vectors at corresponding points is called a streamline. We have proved that for an incompressible

perfect fluid, which moves under the action of conservative

SEC. 111]


235

forces and whose motion is steady, the expression X + (p/P) + -v2 rernains constant along a streamline. This is the general form of Bernoulli's theorem. If X remains essentially constant, then an increase of velocity demands a decrease of pressure, and conversely.

1.

Problems If the motion of a perfect incompressible fluid is both steady

and irrotational, show that x + (p/p) + V2 = constant.

2. If the fluid is at rest, dt = 0. Show that V x (pf) = 0, and hence that f V x f = 0. This is a necessary condition for equilibrium of a fluid. Why must pf be the gradient of a scalar if equilibrium is to be possible?

3. If a liquid rotates like a rigid body with constant angular velocity w = wk and if gravity is the only external force, prove that p/p = 1w2r2 - gz + constant, where r is the distance from the z axis.

4. Write (411) in rectangular, cylindrical, and spherical coordinates.

5. A liquid is in equilibrium under the action of an external

force f = (y + z)i + (z + x)j + (x + y)k. Find the surfaces of equal pressure.

6. If the motion of the fluid is referred to a moving frame of reference which rotates with angular velocity w and has translational velocity u, show that the equation of motion is dw Dr D2r du f--Vp=dt+dt Xr+wx (wxr)+2wxdt+ dt2 1

P

and that the equation of continuity is at

(P Rdt-) r

For a simply connected region R with boundary S, the kinetic energy of R is 7. The energy equation.

T=ifffpv2dr

R

Let the surface S move so that it always contains all the original mass of R. Show that

236


dT

= ffJ R

dt

-dv2

dt

= If! =

[SEc. 112

v.f!Vpdr

f f f v.fdr- f f

f f f pdt(dr)

S

R

(415)

R

Analyze each term of (415). 8. For irrotational flow show that

at = p(p),

- (x + P + 2 v2) + C(t), z

f-D(t).

-f- 2 + x + t 112. The General Motion of a Fluid. Let us consider the

and if p =

velocities of the particles occupying an element of volume of a

Let P be a point of the volume or region, and let vp represent the velocity of the fluid.

fluid at P (Fig. 96). The veloc-

ity at a nearby point Q is vQ = vp + dvp = Vp + (dr . V)vp from (75).

(416)

By (dr V)vp we

mean that after differentiation, the partial derivatives of v are

calculated at P. We now replace dr by r for convenience, so that r = xi .+ y} + zk if we consider P as the origin and x,

Fio. 96.

y, z large in comparison with x2, y2, z2, zy, etc.

now becomes vQ = vp + (r V)vp. Now

Equation (416)

r x (V x w) +

(417)

from (9), (10), (12) of Sec. 22. Now let

w = (r V)vp = x av + y av + z avl

al p

ay p

azlp

(418)

SEC:. 1121

and hence


-- P+yayay ax av ax ax

ay av

237

az avI P

+z azazP

=w We did not differentiate the

(' ayP axP I

IV P

' since they have

been evaluated at P and so are constants for the moment. using (417), we obtain

Thus,

w = j V(r w) + J(V x w) x r Moreover, v (418)], and

(419)

=Vp+w, so thatV xv =V xw = (Vxv)p [see VQ = VP+ -(V x v)P

(420)

It is easy to verify that r w is a quadratic form, that is,

Axe+Bye+Cz2+2Dyz+2Ezx+2Fxy and so by a rotation (Sec. 107), we can write and

IV(r w) = axi + byj + czk We may now write (420) as

VQ = VP + w x r + (axi + byj + czk)

(421)

where w = J(V x v)p. Let us analyze (421), which states that the velocity of Q is the sum of three parts: 1. The velocity vp of P, which corresponds to a translation of the element. 2. w x r represents the velocity due to a rotation about a line through P with angular velocity J(V x v) p.

3. axi + byj + czk represents a velocity relative to P with components ax, by, cz, respectively, along the x, y, z axes. The first two are rigid-body motions; they could still take place if the fluid were a solid. The third term shows that particles at

238


(SEc. 113

different distances from P move at different rates relative to P. If we consider a sphere surrounding P, the spherical element is trans-

lated, rotated, and stretched in the directions of the principal axes by amounts proportional to a, b, c. Hence the sphere is deformed into an ellipsoid. This third motion is called a pure strain and takes place only when a substance is deformable. Each point of the fluid will have the three principal directions associated with it. Unfortunately, these directions are not the same at all points, so that no single coordinate system will suffice for the complete fluid. The most general motion of a fluid is that described above and

is independent of the coordinate system used to describe the motion. It is therefore an intrinsic property of the fluid. 113. Vortex Motion. If at each point of a curve the tangent

vector is parallel to the vector w = J(V x v), we say that the dx

dy

wz

wy

curve is a vortex line. This implies that dx =

=

dz where w,

dx, dy, dz are the components of the tangent vector and

w=wi+wyJ+w L The integration of this system of differential equations yields the vortex lines. The vortex lines may change as time goes on, since, in general, w will depend on the time. Let us now calculate the circulation around any closed curve in the fluid.

C=

ff (V xv) dd

(422)

s r If V x v = 0, then C = 0. This is true while we keep the curve r fixed in space. Let us now find out how the circulation

changes with time if we let the particles which comprise P move according to the motion of the fluid. As time goes on, assuming continuity of flow, the closed curve will remain closed. Now

e=

(423}

r r' where s is are length along the particular curve v, at some time t. At an instant later the curve t' has moved to a new position given by the curve r". The velocity of the particles over this path is

239


SFC. 1141

slightly different from that over r, and, moreover, the unit tangents ds have changed.

The parameter s is still a variable of

integration and has nothing to do with the time. Therefore dt

dt

ds

ds + $6 v

Er- ds

= T dt . ds

+

v.

d

dt Pdsl

ds \dt/

ds

ds

(424)

Euler's equation of motion (411) for a conservative field,

f = -VX is

dv dt

1

= -VX - Vp = -VV, where V = x + f dp/p.

There-

1

fore

2I

2

dt

_

--0d(V-J2) =0

(425)

We have arrived at a theorem by Lord Kelvin that the circulation around a closed curve composed of a given set of particles

remains constant if the field is conservative, provided that the density p is a function only of the pressure p. If we now consider a closed curve

lying on a tube made up of vortex lines, but not encircling the tube (see Fig. 97), then

C=

ff 8

s i nce dd i s normal to V x v.

Fr om Fio. 97. Kelvin's theorem, C = 0 for all time, so that the curve r always lies on the vortex tube. 114. Applications Example 109. Let us consider the steady irrotational motion of an incompressible fluid when a sphere moves through the fluid


240

ISEC. 114

with constant velocity. Let the center of the sphere travel along the z axis with velocity vo. We choose the center of the sphere as the origin of our coordinate system. From Sec. 110, Prob. 6, we have

f - pop=

at

and

0

Hence

Dr

=

Now at points on the surface

so that V2p = 0.

of the sphere we must have CYT)radWLY = 0, so that

(a?Y-a = 0. Or

We look for a solution of Laplace's equation satisfying this boundary condition, so that we try 'P = (see Sec. 67).

(Ar +

! r/J

cos 8

(426)

We need B -CA-

a3

cosB=0

so that B = a2A/2. Moreover, at infinity we expect the velocity of the fluid to be zero, so that the velocity relative to the sphere should be -vo. Hence aVs az z

.= A =

- vo

and s

P = -vo (r + 2r2

cos 0

(427)

The velocity of the fluid relative to the sphere is given by v = Vsp and the velocity of the fluid is v = Vp + vok. Example 110. Let us consider a fluid resting on a horizontal surface (x-y plane) and take z vertical. Let us assume a transverse wave traveling in the x direction. For an incompressible

Sec. 114]


241

fluid z+az?

°2`P-ax

0

2

(428)

We assume a solution of the form p = A (z)e

aLT(X-'i)t.

Sub-

stituting into (428), we obtain 2w

e

tZ_voi f

41x2

- 2 A (z) +

d2A dx2

I=0

so that d2A

_

41r2

(429)

\2 A

dz2

The solution to (429) is A = Ape
(x - vt)]

(Aoe(2r/X)z + Boe-(zrn*)z) cos

(430)

L

The fluid has no vertical velocity at the bottom of the plane on

which it rests, so that v. =

= 0 at z = 0. This yields

L(P

az

A 0 = B0, so that = Ao(e(2r')= + ec2r/l.>=) cos

I2r

(x - Vi) J

= 20 cosh (-- z)

cos

[

xr (x - vt)]

(431)

From Prob. 8, Sec. 110, we have

-(x++2v2)+C(t)

at

and for a gravitational potential, x = gz, so that at

=

- (gz +

p

+

2

v2)

+ C(t)

(432)

We now assume that the waves are restricted to small amplitudes and velocities, so that we neglect Jv2. Moreover, at the


242

[SEC. 114

surface, p, the atmospheric pressure, is essentially constant, so

that dp = 0. Differentiating (432), we obtain a2(

_

dC

az .9 at

ate

+

(433)

dl

and again at the surface at = vZ = az , so that (433) becomes dC

a_p

a2V

(434)

g az + dt

ate

Substituting (431) into (434), we obtain 2

zcosLA (x - Vt)

-v2 a2 A0 cosh

g

2w-

z cos

I2r

(x - vt) ] + dt

(435)

In order for C to be dependent only on t, we must have the coefficient of cos [(27r/A) (x - vt) ]identically zero in (435). Ao

(- v

227r2

cosh

27r

z+

g

sink

27r

z=0

Hence (435a)

or

v2 =

g tanh - z

In deep water z/X is large so that tank

z

1, and the

velocity of the wave is v = (Ag/2,r)*.

Problems

1. Show that for steady motion of an incompressible fluid under the action of conservative forces, (v O)w - (w V)v = 0, wherew = D x v. 2. Show that dt (Pl = \P of v for a conservative system.

SEc. 1151


243

3. If C is the circulation around any closed circuit moving

with the fluid, prove that

dC

=

p d (1) if the field is con-

servative and if the pressure depends only on the density.

4. Show that v = 2axyi + a(x2 - y2)j is a possible velocity of an incompressible fluid.

5. Verify that the velocity potential (p = A[r + (a2/r)] cos 0 represents a stream motion past a fixed circular cylinder. 115. Small Displacements. Strain Tensor. In the absence of external forces, a solid body remains in equilibrium and the forces between the various particles of the solid are in equilibrium because of the configuration of the particles. If external forces

are added, the particles (atoms, molecules) tend to redistribute themselves so that equilibrium will occur again. Here we are interested in the kinematic relationship between the old positions

r

p°

of equilibrium and the new. We shall assume that the deformations are small and continuous. We expect, from Sec. 112, that

so po

r

p

Is p

Fia. 98 .

in the neighborhood of a given point Po, the remaining points will

be rotated about Po and will suffer a pure strain relative to Po. Let r be the position vector of P relative to Po, and let s be the displacement vector suffered by P, and so the displacement suffered by Po (Fig. 98). Then

S = so + ds = so + (r

V)so

(436)

Let s = u(x, y, z, t)i + v(x, y, z, t)j + w(x, y, z, t)k. Since we will be dealing with static conditions,

s = u(x, y, z)i + v(x, y, z)j + w(x, y, z)k From (420),

S = so+J(V xs)P,

(437)

where

w=x-asax

as

PO

as

+yay Pc +zoz

PO

since s = v At.

We are interested in the position of P after the deformation (now P') relative to the new position of Po (now Po ). This is

244


[SEC. 115

the vector r' = r `}- s - so, or

r' = r+4(D xs)po x r +

(438)

Since J(V x s)P, x r represents a rigid-body rotation about P0, we ignore this nondeformation term and so are interested in r + 'I'p(r w). Now

auj aw a-l +2 D Cx tax +xyax +xz at av au awl I r + yz D xy 1

Po

+2

y Po

P.

+

y2

ay PO

0

ay Po

aw

+ 2D

zx azlPo

+ zy a P.

az Po

and

r+

1

aul x Ci

x

+

(au

avl x (aw au ax! + 2 \ax + az J i

au

_y

+ 2 Cay +

ax av

av

_z (aw

2y+

+yCl+ay x (aw

y aw

au

av

az)+2Cay+az

8v

aw 1 az

Jk

(439)

The partial derivatives are evaluated at the point Po. Let us now consider the matrix

au i + ax

lIsr'II =

au

av

2 ay

+ Ox

1

+ + ax/ i aw aul i law 2 Cay

2 C 8x

+

y

azl 2 ày +

(440)

av az

The nine components of this matrix form the strain tensor. If

we write r = x'i + x2j + xak and r' = y'i + y2j + yak

(see

245


SEC. 115)

Example 8), then r' = r + I V(r w) may be written i = 1, 2,3

yi = sfix' + S2 'X2 + 83ix3, or 3

y' =

(441)

Sa'xa

a-1

We shall see in Chap. 8 that since r and r' are vectors, then, of

necessity, the s/ are the components of a tensor. Notice that Sii = s; , so that the tensor is symmetric. The ellipsoid which has the equation

(

aul

avl

(

1 + ax)xr+ 1 +ay y2+ aw

(awl 1

ay +aavlxy

+ az)z2+ au

aw

av

+ (ay + az

(au

yz + (ax

+

az

zx = 1

(442)

is called the strain ellipsoid. From Sec. 107 we know that we can reduce the ellipsoid to the form

Ax" + By'2 + Cz'z = 1

by a proper rotation. The strain tensor becomes entirely diagonal, A

0

0

0 B 0 0 C 0 In the directions of the new x', y', z' axes, the deformation is a pure translation, and these directions are called the principal directions of the strain ellipsoid. Let us now compute the change in the unit vectors, neglecting

the rotation term. The unit vector i has the components (1, 0, 0), so that from (438) and (439) i

(

au

1

au

av

- r1 = \1 + ax) 1 + 2 (ay +

1

aw

ax 1 + 2

ax

(LU)

By neglecting higher terms such as

_ Iril

I

1+

au ax

21

+

az/ k

u-

y

I

.

a_ul

Similarly j --> r2, and lr2l = 1

we have

+

avl

ay

; k - rs,

246


and fr31 = 1 +

awl

[SEC. 116

The angle between r1 and r2 is given by

. I

cos B =

au +r,jjr2+

av

v ay + ax'

.

The terms of the strain tensor are now fully understood. The volume of the parallelepiped formed by r1, r2, r3 is

au

av

r2xrs = 1+ ax + ay +

V

so that V

_

V

_ au

aw az

aw

av

(443)

ax+8y +az

V

The left-hand side of (443) is independent of the coordinate system, so that V s is an invariant. Finally, we see that the deformation tensor due to the tensor - V (r w) has the components

1

2\ay+8x/ 2(ax+ az)

ax au

8v'\

2 (ay aw 1

+ ax/

2 \ ax

av

ay

au

1 (au

+

az

2 \ay + az

+

axi/1I

av

aui 2 (axl

au

av

2 (ay

+ az

1

aw az (444)

where

u1 =u, u2=v, u3=w,

x1=x x2=y x3=z

116. The Stress Tensor. Corresponding to any strain in the body must be an impressed force which produces this strain. Let us consider a cube with faces perpendicular to the coordinate axes. In Sec. 108 we assumed no shearing stresses, but now we consider all forces possible between two neighboring surfaces.

SF:C. 116]

247


Let us consider the face ABCD (Fig. 99). It is in immediate contact with other particles of the body. As a consequence, the resultant force tx on the face ABCD can he decomposed into three forces: txx, tyx, tzx, where txx is the component of tx in the x direc-

tion, t, is the component of tx in the y direction, and tzz is the ,

z

y

Fia. 99.

component of tt in the z direction. We have similar results for the other two faces and so obtain the matrix tzz

tzy

tzz

tyz

tyy

tyz

tzz

tzy

G.

(445)

These are the components of the stress tensor. By considering a tetrahedron as in Sec. 108, we immediately see that if dd is the vectoral area of the slant face, then the components of the force f on this face are f, = tzx dsz + tzy d s + tzz dsz fy = tyx dsz + tyy dsy + tyz dsz fz = tzz ds, + tzy dsy + tzz dsz

where dsz = i dd, dsy = j dd, dsz = k dd.

(446)

248


[SEC. 117

We immediately see that tyx =

Of.,

fx,

tx

asv

as., 3

and that fi = I tia dsa, where f, = f,, f2 = f,,, f3 = f., t12 = t,,, a=1

.

.

.

.

We shall see later that this explains why the ti; are called the components of a dv tensor.

Let us now consider the resultant force acting on a volume V with boundary 8

z

(see Fig. 100). (446)

Wehavefrom

fz = txx dsx + txv ds + tzx dst

so that Y

fz _

F. =

JJtxzdSx

+ t,, ds, + tx, dss

=

!s J t.dd

Fio. 100.

where t = tz, + Applying the divergence theorem, we obtain Fz

- JIf Jr V

acz=

acZy

ax

+

4-9t"

ay

+

az

dT

t,, k.

(447)

with similar expressions for F,,, F. By letting V --p 0, we have that the x component of the force

per unit volume must be Vax- +

qty"

y

+

tz

117. Relationship between the Strain and Stress Tensors. In the neighborhood of a point P in our region, let us choose the three principal directions of the stress tensor for the axes of our cartesian coordinate system. If we assume that the region is isotropic (only contractions and extensions exist), a cube with faces normal to the principal directions will suffer distortions only along the principal axes. Hence the principal directions of the

249


SEC. 1171

strain ellipsoid will coincide with those of the stress ellipsoid. this coordinate system el ie17JJ =

0

0

0

It,

0 0

0

0 0 t2

In

0 0

(448)

0

0

Our fundamental postulate relating the shear components with those of the stress will be Hooke's law, which states that every tension produces an extension in the direction of the tension and is proportional to it. We let E (Young's modulus) be the factor of proportionality. Experiments also show that extensions in fibers produce transverse contractions. The constant for this phenomenon is called Poisson's ratio a. We thus obtain for the relative elongations of the cube in the three principal directions the following: 1 -SET a el =E;ti --(t2+t3) _ E E 1

0,

or

t2 -

e2 1

e3 =

E

E(t1+t2+t3)

(t3 + tl) =

1 r

t2 -

E (t1 + t2 + t3)

t3 "-

(t1 + t2 + t3)

(449)

1-r-a-

a 3 - - (11 + t2)

The formulas for el, e2, e3 apply only in the immediate neighbor-

hood of a point P. Since points far removed from P will have different stress ellipsoids, the principal directions will vary from point to point. Hence no single coordinate system will exist that would enable the stress and strain components to be related by the simple law of (449). Let us therefore transform the components of the stress and strain tensors so that they may be referred to a single coordinate system. The reader should read Chap. 8 to understand what follows. If he desires not to break the continuity of the present paragraph, he may take formula (456) with a grain of salt, at least for the present. Example 8, Probs. 21 and 22 of Sec. 11, and Prob. 21 of Sec. 15 will aid the reader in what follows. If x1, x2, x3 are the coordinates above, and if we change to a new coordinate system 11'. T2, 23 where 3

x' = I a'xx, a-1

i = 1, 2, 3

(450)


250

(SEC. 117

then the transformation (450) is said to be linear. Notice that the origin (0, 0, 0) remains invariant.. If, furthermore, we desire distance to be preserved, we must have 3

3

(xi)

(.Ti), =

In Chap. 8 we shall easily show that this requires

=0ifij

3

a=1

= 1if ij

a.,aaia = S=i

(451)

Equation (451) is the requirement that (450) be a rotation of axes. Moreover, since we are dealing with tensors, we shall see that the components of the strain tensor in the x'-x2-x3 coordinate system

are related to the components in the x'-a 2-x3 system by the following rule: 3

3

e;i = I I a,aa1 ea#

(452)

'6=1 a=1

If we now let i = j and sum on i, we obtain `3

3

3

3

L, ei, =i=1I I0=1 Z a=1 aiaMeag

:=1

3

3

3

Saisea# = I eaa

a-1

0-1 a-1

so that e11 + 922+933 = e11 + e22 + e33 = el + e2 + e3

(453)

This is an invariant obtained from the strain tensor [see (443)]. A similar expression is obtained for the stress tensor; namely,

that 111+122+133 = t11 +t22+133 = t1+t2+t3 Equations (449) may now be written as

el -1+0 E t1+ e2=1

(454)

e3=1Eor

ts+

HYDRODYN.4.1IICS AND ELASTICITY

SEC. 117]

251

where 4, is the invariant E (tl + t2 + t3)

-

(III + 122 + 133)

From Eq. (452) we have 3

eij =

3

E aiaa,8eas B=1 a-1

and since eap = 0 unless a = 6 [see (448)], we obtain 3

3

1 aiaaJaeaa = I aiaajaea a=1 a=1 3

a

-l

aiaaja (1

1+ a F,

E

to

+)

3

3

I aiaajala + Y' 1 aiaaja a-1

a=1

and 1 + o

eij = E 3

3

of + 4,5ij

(455)

3

since 1j = I I aiaajalas = I aiaaala. 6=1 a -l

a=1

Equation (455) is the relationship between the components of the strain and stress tensors when referred to a single coordinate

system. We have Q111

ell =

- E (tll + 122 + 133)

1

1

922 =

E +Q_ E

122 -

oT

-

(111 + t22 + 138)

O e33 =

1

1 912 =

+ 133 - E, (111 + E22 + 133) +v_ E

112

1+0. 923 =

E

l23

1 +Q_ 981 =

E

131

(456)

252


[SFC. 11

Solving Eels. (456) for the t;; and removing the bars, we obtain t11 =

1+

[ell +

01

1

(ell + e22 + e33)

2au

E

o

aw

av

(au

=i+o, axl-2vax+ay+az

= tss

=

E rav

au

v

J

av awt22

1+aLay1-2v (ax+ay+az)] E taw v 1+aaz+1-2QV*s E

t12=t21= t23 = t32 =

=

t31 - t13 =

E

(457) (au

av

2(1+o) ay+ax aw E av 2(1 + v) \az + ay E (aw au

2(1+Q)ax +az

Equation (447) now becomes

E

F=

[d?u

-}- v

ax2

+

(a2u 1 - 20 \49x2

E

+ _

1

2(1+o)

a2w

+ ax ay +

ax az

a2u

2(1 + u) aye

E -2(1+a)IV

a2v

.

a (au

a2w

a2v

a2u

ay ax

+ az ax + az2

av

aw

1

zu+1 -2aaxax+ay+ az J

[vu+ 1 1

The forces per unit volume in the y and z directions are Fy

2(1+a)[Dzv+1 12vay(V - s)I E

{V2W+

a

1 I-2oaz

(0 s)

so that

f=

+ 2(1

[V2S r)

V(V s)] + 1 1 2cr

(458)

If we let R = RJ + R j + R2k be the external body force per unit volume, p the density of the medium, then Newton's second

253


SEc. 117]

law of motion yields

R+ 2(1 +o,)[V2S+-Lv(V.S)]

= P

For the case R = 0, (459) reduces to E 2(1 + a) [v2s

+

1

- 2a °(°

2

s)

=p

at2

(460)

In Sec. 70 we saw that a vector could be written as the sum of a solenoidal and an irrotational vector. Let s= s1 + s2j where V s, = 0 and V x s2 = 0. Since (460) is linear in s, we can consider it as satisfied by s, and s2. This yields E V2s1_- p a2S, 2(1 + a) ate and

(461) 2(1

E +a,)

1

[V2s2

+ 1 -2a °(° . s2)

a2s2

= p ate

However,

V X (V X 62) = V(V S2) - V 'S2 = 0

so that E(1 T a)

(1 +a)(1 - 2a)

V2S2

= pa2s2 at2

(462)

In Sec. 80, we saw that (461) leads to a transverse wave moving with speed Vt = E/2(1 + a)p. Equation (462) is also a wave equation, but the wave is not transverse. Let us assume that the wave is traveling along the x axis. Then

s2=s2(X-Vt) = u(x - Vt)i + v(x - Vt)j + w(x - Vt)k

=0

V X S2 =

axJ+-k=0

clx

254


[SEC. 117

so that w and v are independent of x and therefore are independent

of x - Vt. We are not interested in constant displacements, so that s2 = u(x - Vt)i, and the di,,placement of s-2 is parallel to the direction of propagation of the wave. longitudinal. The speed of the wave is

The wave is therefore

L'(1 - o.)

V,=

P(1 + v)(1

2v)

In general, both types of waves are produced, this result being useful in the study of earthquakes. Problems 1. Derive (451). 2. If P, f--, f3 are the components of a vector for a cartesian

coordinate system, prove that the components P, P, j'3 of this vector in a new cartesian coordinate system are related to the old 3

components by the rule J' =

i = 1, 2, 3, using the

n=i

coordinate transformation (450). 3. If the body forces are negligible and if the medium is in a

state of equilibrium, show that V2S +

1 -1

V(V s) = 0.

4. If the strain of Prob. 3 is radial, that is, if s = s(r)r, find the differential equation satisfied by s(r).

5. Assuming a = 0 for a long thin bar, find the velocity of propagation of the longitudinal waves.

6. If µ = E/2(1 + v) (modulus of rigidity) and Ea

A=

(1 + a) (1 - 2Q)

show that Eq. (459) becomes 49

R+AVss+ x

7. Why do we use ats instead of

's

Pats x

in in (459)?

8. A coaxial cable is made by filling the space between a solid

core of radius a and a concentric cylindrical shell of internal radius b with rubber. If the core is displaced a small distance

Ssc. 118]


255

axially, find the displacement in the rubber. Assume that end effects, gravity, and the distortion of the metal can be neglected. 118. Navier-Stokes Equation. We are now in a position to derive the equations of motion of a viscous fluid. In the case of nonviscous fluids, we assumed no friction between adjacent layers of fluid. As a result of friction (viscosity), rapidly moving layers tend to drag along the slower layers of fluid, and, conversely, the

slower layers tend to retard the motion of the faster layers. It is found by experiment that the force of viscosity is directly proportional to the common area A of the two layers and to the gradient of the velocity normal to the flow. If the fluid is moving in the x-y plane with speed v, then the viscous force is

since av is the gradient of the speed normal to the direction of flow. 17 is called the coefficient of viscosity.

We shall let Pt; be the stress tensor and u;; the strain tensor for the fluid analogous to t; and e;; of the previous paragraphs. We have

_

E au 2(1 + v) ay

P12

Ov

+

ax

where u, v, w are the components of the velocity vector v (see Sec. 112). For a fluid moving in the y direction with a gradient

in the x direction, we have u = 0 and au = 0, so that ay

_

P12

Hence the term

E 2(1

E av _ av 2(1 + v) ax = '' ax must be replaced by q.

or)

In addition to the stress components due to viscosity, we must add the stress components due to the pressure field, which we assume to be

-p I

0 0

0

0

-p

0

0

-p


256

[SEC. 118

The equations of (457) become (463)

P;i = 2rio;i + X(911 + 022 + r33)aii - pa+i

where X is undetermined as yet. We see that

Now let i = j and sum on j.

P11 + P22 + P33 = (2rj + 3X)(ail + 022 + 033) - 3p

We know that P11 + P22 + P33 is an invariant and that for the static case P11 + P22 + P33 = -3p. Consequently we choose 2, + 3X = 0, so that (463) becomes p;, = P;, = 2+10;1 - I V(0-11 + 022 + 033) - pa+i

(464)

Moreover, the velocity vector is given by

for small velocities.

v = u11 + 1627 + u3k

and 0;i =

1

au;

2 C axi

auiax:

+

' so that div v = V v = 011 + 022 + 033-

To obtain the equations of motion, we note that from (447) ap11 fl -

ax, 3 c3

= jG =1

49p12

+

axe

x1 = x where x2 = y

49p13

+ axi

x3=z

apli ax

3

and in general f;

I apt'

Hence

-i

du;

pF; + f1 = p dt becomes 3

pF. +

ap;i axi

du:

(465)

P dt

j =1

where F; is the external force per unit mass.

From (464)

a0;i 2 a(diy v) ap ap;i 17 = 2,i axi - 3 axi aì - axi a+i axi _ 'q a au; au) 2 a(V v) +1 axi 3 = axi axi + ax'

-

-

ap a; - axi

257


SEC. 1181

and 3

api;

3

a

au;

au;

2

+ axi/

1-1 ax?

3'

v)

a (V

ap

ax,

The equations of motion (465) are dui

p dt - = pFi + v

I 3

J

1

a aui ax' Cax'

ap 2 a(V v) + au; -7 ax' axi axi 3

(466)

or

pat

= pf +,q V2v - - V(V v) - Vp

(467)

For an incompressible fluid V v = 0, and

pd Along with (467) we have the equation of continuity

at

=0

Problems 1. Derive (467) from (466). 2. Consider the steady flow of an incompressible fluid through a small cylindrical tube of radius a in a nonexternal field. Let

v = vk and show that p = p(z) and +1 V2v =

ap

Show that the

boundary conditions are v = 0 when r = a, and v = v(r), r2 = x2 + y2, and that = r dr (r dr) Hence show that v = (A/4i7)(r2 - a2), where A is a constant and LP = A. 3. Consider a sphere moving with constant velocity vok (along the z axis) in an infinite mass of incompressible fluid. Choose the center of the sphere as the origin of our coordinate system.

Show that the equation of motion is p aE = q V2v - Vp and that

the boundary conditions are v = 0 for r = a, v = -vok at r = oo

258


[SEC. 118

and that for steady motion a = 0 for any quantity 4, associated with the motion. Moreover V v = 0. We shall assume that v and the partial derivatives of v are small. Show that this implies (i)

Vp = I V2v

Hence prove that V2p = 0.

Now let v = -V(p + wlk and show

that awi

=v4o z

49Z

Assuming p = -,1 V2p and V2w1 = 0, show that (i) and (ii) will be satisfied. Let w, = 3voa/2r, rp = coz - (voa3z/4r3) + (3voaz/4r),

P=

3,gvoaz 2r3

and show that V2w1 = 0,

V2p = 0,

p = -n V2rp

v= -V(p+wik=0forr=a v = -vok for r = co

4. Solve for the steady motion of an incompressible viscous fluid between two parallel plates, one of the plates fixed, the other moving at a constant velocity, the distance between the plates remaining constant. 5. Find the steady motion of an incompressible, viscous fluid surrounding a sphere rotating about a diameter with constant angular velocity.

No external forces exist.

CHAPTER 8 TENSOR ANALYSIS AND RIEMANNIAN GEOMETRY

We shall be interested in sums

119. Summation Notation. of the type

S = alx, + a2x2 + .

. + anxn

.

(468)

We can shorten the writing of (468) and write n

S = Z aixi

(469)

i=i

Now it will be much more convenient to replace the subscripts of the quantities x:, x2,

.

. .

, x by superscripts, x', x2,

. . . , X11.

The superscripts do not stand for powers but are labels that allow us to distinguish between the various x's. Our sum S now becomes n

S=

(470)

aixi

We can get rid of the summation sign and write (471)

S = aixi

where the repeated index i is to be summed from I to n. This notation is due to Einstein. Whenever a letter appears once as a subscript and once as a superscript, we shall mean that a summation is to occur on this letter. If we are dealing with n dimensions, we shall sum from 1 to n. The index of summation is a dummy index since the final result is independent of the letter used. We can write

S = a;xi = a;x' = a,,xa = as0. Example 111. If f = f(x', x2, calculus 259

,

xn), we have from the

0,60


. + -CIx"dx"

df = ax dxl + ax dx2 +

_ = of

[SEC. 120

of dx' ax'

dxa

axa

;.nd

df afdJxa

dt - axa dt

The index a occurs both as Hence we first sum on a, say from

Example 112. Let S = gasxax#.

a subscript and superscript. 1 to 3. This yields

S = 9iax'x8 + 92ax'x' + g38xaxe

Now each term of S has the repeated index # summed, say, from I to 3. Hence S = g11x'x1 + 912x'x2 + 913x'x3 + 921x2x' + g22x2x2 + 923X2x3 + 931x3x' + g32x3x2 + 933x8x3

and S = gaoxaxa represents the double sum 3

3

S = Z I g-Oxaxs

0-1 a-1 We also notice that the gap can be thought of as elements of a square matrix gii 912 913 921

922

923

931

932

933

120. The Kronecker Deltas. We define the Kronecker o to be equal to zero if i - j and to equal one if i = j:

a;=0,ipj 1,i=j We notice that Si = 52 _ 1

1

b_ = 03 =

= SA = 1, =

r

+1 = u

(472)

TENSOR ANALYSIS

SEC. 120)

If xI, x2,

261

... , x" are n independent variables, then

ax' axl

= S,

i

for if

axe = 1, and if i

j, there is no change in the vari-

able x' if we change x1 since they are independent variables, so

that

ax!

ax1

= 0.

Example 113. Let S = aaxa. as -=

ax"

Then

as ga"

cis

= as ax , and

ax

Now Sµ = 0 except when a = µ, so that on summing on a we a(axa) obtain = a". ax"

Let S = apxax# = 0 for all values of the vari. . , x^. We show that a,, + a;; = 0. First differentiate S with respect to x' and obtain

Example 114. ables x', x2, .

T

axa axp as = aa$xa xa = 0 + a, ax' ax' axi = aaflxaaf + aa$67x8 = 0

= aaixa + a;se = 0 Now differentiate with respect to xi so that 028 axe

ax'

- a.v + aiob = 0

and

a,,+a;,

0

We define the generalized Kronecker delta as follows: The superscripts and subscripts can have any value from 1 to n.

If at least two superscripts or at least two subscripts have the same value, or if the subscripts are not the same set of numbers as the superscripts, then we define the generalized Kronecker delta to be zero. If all the superscripts and subscripts are separately distinct, and the subscripts are the same set of numbers Qs the superscripts, the delta has the value of + 1, or --1, accord-

262


[SEC. 120

ing to whether it requires an even or odd number of permutations to arrange the superscripts in the same order as the subscripts. For example, 6x23 = 1, 6213 - - 1, 61231 23 = 1, 6163 = 0, a1438 - -11 a123

=

0221

= 6312 323

0,

61213

=0

It is convenient to define eilh ... i.. =

(473)

and Eiji, ... i. =

Problems

1. Write in full aaxa=bi,a,i= 1, 2, 3. 2. If aae7xax8xY = 0, show that ai;k + aki; + a; + aJik + akii + aiki = 0.

3i,ti:'412 ...n i.,i,...i. 6;,i:...;.. 3. Show that 612..,, i. = 4. If yi = aaxa, zi = by, show that zi = bas;x#. 5. Prove that bas = are - a8r baBratYaapi = arst + airs

+ astr - sari - aisr - arts

6. Show that the determinant I at at can be written 2

a1

a11

a2'

atz

a2

2

a2

= Etiiaia? = etiia1

2

and that at

a2

2

z

a1 8

a1

1

a3 at z

a2

a3

8

a$$

a2

7. Prove that ei1i, ... ie =

8. Show that aij

= Eiika;a;ak = eiikai4a8 e'li2 ' ' i*.

ax axi 9. If yi = yi(xl, x2) ... , xe), i = 1, 2, . . . , n, show that ayi ax" ai assuming the existence of the derivatives. Also a; = axa y

i

show that ayi

-

axe

e

yi

as = a?. Show that a2yi

axa ax¢

axa 8x8 ayi

8yk

ayi

492xa

+ Oxa ayi ayk - 0

TENSOR ANALYSIS

SEC. 1211

10. If

= cp(x', x 2 .

. .

.

. .

n),

xi = xi(y', y2'

i = 1,2,

.

yn)

,

= xi(y)

n, and if

show that

263

.

.

,

.

y=

yn)

= p[xl(y), x2(y),

a

Given 9,, = aoi

arpi

ayayi -

(ax# _

C) p

acs

.

.

.

,

xn(y)]

, show that

axe, axil

axa) ay, ayl

121. Determinants. We define the determinant lal by the equation a1 at 2

a1

a2 at

at an 2

2

a2

an

i,a1a2 n

a1

n

n

a2

l all

an

an

=

Erik ... i"ai,a?2

.

.

.

(474)

a"n

The reader should note that this definition agrees with the definition for the special case of second- and third-order determinants which he has encountered in elementary algebra. The definition of a determinant as given by (474) shows that it consists of a sum of terms, nn in number. Of these, n! are, in general, different from zero. Each term consists of a product of elements, one element from each row and column. The sign . of the term a a'a a depends on whether it takes an even or odd permutation to regroup i1i2 in into 12 . . n. Since i1 and i2 are dummy indices, we can interchange them so that fail = Ei,i,... ina,'a2

.

.

.

ann = ei,i, ... inal=a2

.

. , ann

An interchange of the subscripts 1 and 2, however, will mean that an extra permutation will be needed on i2iii3 in. This changes the sign of the determinant. Hence interchanging two columns (or rows) changes the sign of the determinant. As an


264

[SEc. 121

immediate corollary, we have Ia;I = 0 if two rows (or columns) are identical. Let us now examine the sum bill:... i.

= Ei,i, ...

a.

(475)

If jl, j2, . . . , jn take on the values 1, 2, . . . , n, respectively, we know that (475) reduces to (474). If the ji, j2, . , j,. take on the values 1, 2, . . . , n, but not respectively, then we have interchanged the rows. An even permutation reduces (475) to Ia;I, and an odd permutation of the i's reduces (475) to -Ia 1. If two of the j's have the same value, (475) is zero, since if two rows of a determinant are alike it has zero value. Hence

..

Es, ...,.a`;<1

. .. a;-

(475a)

i=

and t,44 ...

i

ai.

(475b)

ai Ehj, ... i.

Example 115. We now derive the law of multiplication for two determinants of the same order. I ail

IasIEC,;,...

Ibil

bri

i.bi'b2

Ei,i, ... i.a is =

= Ei,i, ...

We have

.

. .

s,

a.,b

(ai,bi,)

i.(a1j1,bi')

s, 2

1

b2

.

.

.

{

,. x )

. b%

I ciI

where abi

(476)

We now derive an expansion of a determinant in terms of the cofactors of the elements. We have Example 116.

I41 = Ei,i,...i.aia8 a; (Ei,i,... j.a2

. an

.

.

.

.

.

a'*)

= aIAQ

AQ is called the cofactor of a,.

where Ao = In general,

a'0-b5-Pa'-#-K +,. .

= apAP.

. an)

(f4 not summed)

Hence aiAa = IaIB;

(477)

TENSOR ANALYSIS

SEC. 121]

265

Also

aaA7 = IaI

(477a)

Example 117. Let us consider the n linear equations

a, i = 1, 2,

yi = aQxa,

.. , n

(478)

Multiplying by Af, we obtain

A°y' =a' Asxa so that summing on i, we have from (477)

A"y' = Ialaxa = Ialxx

If IaI 0 0, then A°y'

_ y'(cofactor of a' in IaI)

IaI -

Example 118.

(479)

IaI

Let yi = y'(x', x2,

. ,

x"), i = 1, 2,

. . .

, n.

i

In the calculus it is shown that if ayi) 96 0 at a point P and if the

partial derivatives are continuous, we can solve for the x' in terms of the y's, that is, xi = xi(y', y2, ... , y") in a neighborhood of P. Now we have identically ayi

8i

ayi

ay' axa axa ayi

Forming the determinant of both sides, I O;I

from (476).

ayi axa

ay

- lax- ayi

ax

Hence axi ayl

The determinant I I

is called the Jacobian of the y's with respect

to the x's. We have shown that


266

J \xl,y2, ', x2,

. . .

xn ,,yry/Jfy',X2, 1, y2, .

.

.

y xn) -

[SEc. 121 1

(480)

Example 119. If the elements of the (leterminant lal are functions of the variables x', x22, . . . , x", we leave it to the student

to prove that

_ Aa0 aao

alai

(481)

axµ

axp

As a special case, suppose a =

ay

i

, where y' = y'(x', x2,

, xn).

49x1

a yi axa Let us consider j to be fixed for the moment. ax- ay, yt a , Xi = ax -, so that Yi = aQXa. If Let Yi = 491, ax y

Now a`t =

0

l al

ax11

then

Ya(cotactor of a#" in lal) lal

from (479), or a

8x8

S; cofactor of aye in

ay ax

cix1 1

(cofactor of Lyi in

ay 8x

ax

and so 1

(cofactor of

a?

in

Applying this to (481), we have

ay

ay 8x8

ax

TX 8y1

267

7'EXSOR ANALYSIS

Sic. 121

a,y

a

ay axa

8x

a2ya

ax, ays ax), axa

axµ or

(482)

la

a log

axa

laxI

ft"

a2ya

aya axµ

axa

We shall make use of this result later.

Example 120. Let yi = y'(xl, x2,

,

ay

xn),

We

3-1 0.

ax

wish to prove that a2xµ

ayk aye

axa ax'6 axA

-

a2yi

ayi ayk ay` axa axa

ayi axa

Now S' = ax- -,a so yi that upon differentiating with respect to yk, we obtain ayi

0 - axa

axa

a2xa

ayk aye

a2yi

axa

(483)

+ ay; axa axa ayk 1+

Multiplying both sides of (483) by ayi and summing on i, we obtain axa axa axµ

a2xµ

o = ayk ayay' + As a special case, if y = f (x),

d 2x dy 2

ayk

ayi

&2yi

Q.E.D.

axa axa

= - (dx3d2y dy dx2

Problems i

1. What is the cofactor of each term of

z

i

al

a2

a3

2

2

2

a1

a2

a3

asl

a32

a33

2. Prove (481), (477). 3. If JAI is the derminant of the cofactors of the determinant IaI, show that I:4I = IaJ"-'.


268

[SEC. 122

4. If a = a, show that A; = A. Is laI = jail in all cases? 5. Find an expression for

aaxa

ayi ayi ayk

6. If z' = z'(y1, y2, ... , yn), y' = yt(x1, that J(z/y)J(y/x) = J(z/x). 7. If 9i; = gap _

a ia2i-, show that I#I = IgI a2{

x2,

.

. .

, x"), show

2 I

_I

axa

_

8. If u' = ua axa, Vi = V. clt. -, show that uaVU = uaVa, where xi = x{(x1 x2

xn).

ax{ ax i 9. If u{ = ua , show that u{ = ua ax" axa a ax- 00 10. If gi; = gaa , show that gi; = 9a$ a2i axi axi a { , ui = ua axi azi , show that u{ = ua 11. If u{ = ua axa axa axa 82i

12. Apply (476) for the product of two third-order determinants. 13. If A; = B,-s ayi ax , show that A{ = B, and that IAI = IBI, A;A; = B )$B,!.

14. If X is a root of the equation Iai; -- Xbifl = 0, show that X

is also a root of Id;; - Ail = 0 provided that B;; = aaa

!ii =

baa

axa axa ax

a axi

a

axa axa

aji

0.

122. Arithmetic, or Vector, n-space. In the vector analysis studied in the previous chapters, we set up a coordinate system with three independent variables x, y, z. We chose three mutually perpendicular vectors i, j, k, and all other vectors could be written as a linear combination of these three vectors. Any vector could have been represented by the number triple (x, y, z), where we imply that (x, y, z) -= xi + yj + zk. The unit vectors could have been represented by (1, 0, 0), (0, 1, 0), and (0, 0, 1).

A system of mathematics could have been derived solely by defining relationships and operations for these triplets, and we need never have introduced a geometrical picture of a vector. For example, two triplets (a, b, c), (a, #, y) are defined to be equal

TENSOR ANALYSIS

SEC. 1221

269

if and only if a = a, b = P, c = y. We define the scalar product as (a, b, c) (a, f4, -y) = as + bfi + cy. The vector product, differentiation, etc., can easily be defined. Addition of triples is

defined by (a,b,c)+(a,#,y)=(a+a,b+$,c+y). If A is a real number, then A(a, b, c) is defined as (Aa, Ab, Ac). The set of all triples obeying the rules (i) (ii)

(a, b, c)+(a,0,y) _ (a+a,b+i,c+y)

(484)

A(a, b, c) = (Aa, Ab, Ac)

is called a three-dimensional vector space, or the arithmetic space of three dimensions. It is easy to generalize all this to obtain the arithmetic n-space. Elements of this space are of the form (x', x2, . . . , x"), the x'

taken as real. In particular, the unit or basic vectors are (0, 0, . . . (1,0,0, . . . ) 0), (0,1,0, . . . ,0),

... ,

We shall designate V. as the arithmetic n-space. By a space of n dimensions we mean any set of objects which

can be put in one-to-one reciprocal correspondence with the arithmetic n-space. We call the correspondence a coordinate system. The one-to-one correspondence between the elements or points of the n-space and the arithmetic n-space can be chosen in many ways, and, in general, the choice depends on the nature

of the physical problem. Let the point P correspond to the n-tuple (x', x2, We now consider the n equations y' = y'(x', x2,

... , x"),

i = 1, 2,

.

.

.

.

. ,

x^).

.

,n

.

, n (486)

(485)

and assume that we can solve for the x', so that x' = x'(y', y2,

... , y"), i = 1, 2,

.

We assume (485) and (486) are single-valued. It is at once

obvious that the point P can be put into correspondence with the n-tuple (y', y2, y"). The n-space of which P is an element is also in one-to-one correspondence with the set of (y', y2, . .. , y"), so that we have a new coordinate system. The point P has not changed, but we have a new method for attaching numbers to the points. We call (485) a transforma-

... ,

tion of coordinates.

270


[SEC. 123

123. Contravariant Vectors. We consider the arithmetic n-space and define a space curve in this V by

= 1, 2, ..

x` = x`(t),

at<_/i

n (487)

Note the immediate generalization from the space curve x = x(t),

y = y(t), z = z(t): In our new notation x = x', y = x2, z = x3. We We remember that

dx d dz , are the components of a tangent dt dt' dt

vector to this curve. Generalizing, we define a tangent vector to the space curve (487) as having the components d.ri

'j = 1, 2,

dt

. .

.

,n

(488)

Now let us consider an allowable (one-to-one and single-valued) coordinate transformation, of the type (485). We immediately

have that y' = y'(x1, x2,

. . . ,

x") = y`tx'(t), x2(t),

. .

.

, x"(t)1 = y'(t)

as the equation of our space curve for observers using the y The components of a tangent vector to the same space curve (remember the points of the curve have not changed; only the labels attached to these points have changed) are given by coordinate system.

dyi

i = 1, 2,

. .

.

,n

(489)

dt

Certainly the x coordinate system is no more important than

the y coordinate system. We cannot say that dt is the tangent s

vector any more than we can say dt is the tangent vector. If we considered all allowable coordinate transformations, we would obtain the whole class of tangent elements, each element claiming

to be the tangent vector for that particular coordinate system. It is the abstract collection of all these elements that is said to be the tangent vector. We now ask what relationship exists between the components of the tangent vector in the x coordinate system and the components of the tangent vector in the y coordi-

TENSOI? ANALYSIS

SF(,. 123]

nate system.

271

We can easily answer this question, for

We also notice that

dys

0y1 dxa

dt

axa dt

d = ay- dydls dx1

(490)

We leave it as an exercise

that this result follows from (490) as well as from (486). We now make the following generalization: Any set of numbers . , n, which transform accordAi(xe, x2, . . . , xn), i = 1, 2, ing to the law

..

_

AL(P, 22,

,

xn) = Aa(xl, x2,

. .

,

.

x')

axa

(491)

axa

under the coordinate transformation x= = 21(x', x2, . . . , xn), are said to be the components of a contravariant vector. The vector is not just the set of components in one coordinate system but is rather the abstract quantity which is represented in each coordinate system x by the set of components Ai(x).

We immediately see that the law of transformation for a contravariant vector is transitive. Let A'=Aaa2i,

OX*

.Ai=Aai a2a

Then

A' _ A8 ax a = Aa ax8 a28

ax _Aaax

axa ate

axa

which proves our statement. If the components of a contravariant vector are known in one coordinate system, then the components are known in all other

allowable coordinate systems by (491). A coordinate transformation does not give a new vector; it merely changes the components of the same vector. We thus say that a contra variant vector is an invariant under a coordinate transformation. An object of any sort which is not changed by transformations of coordinates is called an invariant. Example 121. Let X, Y, Z be the components of a contravariant vector in a Euclidean space, for an orthogonal coordinate


272

[SEC. 123

system, and let ds2 = dx2 + dy2 + dz2. The components of this vector in a polar coordinate system -,re

R=X-ar+Yaary 0

ae ae X ax + Y ay

ar

+Zaz= cos 0X+ sin 0Y

CIO

_

+ Z az^

-sine X + cos e Y

Z=Xa +Yaz+Zaz=Z

r

r

y

where r = (x2 + y2)4, 0 = tan-' (y/x), z = z. The components R, 0, Z are not the projections of the vector A = Xi + Yj + Zk on the r, 9, z directions. However, if the 0-component is given the dimensions of a length by multiplying by r, we obtain Or = - sin 0 X + cos 9 Y, which is the projection of A in the 9-direction. We multiplied by r because r d8 is arc length along the 0-curve. R, 0, Z are the vector components of the vector A in the r-e-z coordinate system, whereas R, re, Z are the physical components of the same vector. Problems

1. If A'(x), B'(x) are components of two contravariant vectors, show that C'"(x) = A'(x)B'(x) transforms according to the

ax' ar

law C*' = Cap axa ate, where C" =

2. Show that if the components of a contravariant vector vanish in one coordinate system, they vanish in all coordinate systems. What can be said of two contravariant vectors whose components are equal in one coordinate system?

3. Show that the sum and difference of two contravariant vectors of order n is another contravariant vector. 4. If X, Y, Z are the components of a contravariant vector in an orthogonal coordinate system, find the components'in a spherical coordinate system. By what must the 0 and Sp components be multiplied so that we can obtain the projections of the vector on the 0- and (p-directions?

5. If A' = Aa a ' show that A' = Aa

' a.

TENSOR ANALYSIS

SEC. 125]

6. Referring to Prob. 1, show that Cs, N

7. If Ai = ax1 Aa

273

Caa

ax` ax' axa al$

N

i

axa, show that Ati =

i l-A-

I

az

124. Covariant Vectors. We consider the scalar point function So = co(x', x2, . . . , x"), and form the n-tuple (492)

Now under a coordinate transformation av ay'

a(p

axa

(493)

axa ay'

so that the elements of the n-tuple l

1 y2

related to the elements of (492) by (493).

,

y"J are

We say that the axi

are the components of a covariant vector, called the gradient of gyp. More generally, if

J i=Aaa' L'

(494)

the A; are said to be the components of a covariant vector. The remarks of Sec. 123 apply here. What is the difference between a contravariant and a covariant vector? It is the law of transformation! The reader is asked to compare (494) with (491). We might ask why it was that no such distinction was made in

the elementary vector analysis. We shall answer this question in a later paragraph. 125. Scalar Product of Two Vectors. Let A'(x) and &(x) be the components of a contravariant and a covariant vector. We form the scalar AaBa. What is the form of AaBa if we make a coordinate transformation? Now

ax a

Aa=A8axe'

Baaxa c12'


274

[SEc. 126

so that AaBa = AFB,

AaBa = A"B,

axa axQ axp axa

a2s

= Asfl = AaBa

Hence AaBa is a scalar invariant under a coordinate transformation. The product (AaBa) is called the scalar, or dot, product, or inner product, of the two vectors.

Problems 1. If Ai and Bi are components of two covariant vectors, show that Ci; = A;B, transforms according to the law Oil = Cap

axa axe axa ax?

2. Show that C = AB; transforms according to the law axp C"i

- C8 V axa

. a

3. If Ai = A. axi show that A, = Aa

axi

4. If p and J, are scalar invariants, show that grad (vO) = c grad P + 0 grad cp grad [F(op)] = F'(,p) grad rp

5. If A'Bi is a scalar invariant for all contravariant vectors A', show that Bi is a covariant vector.

126. Tensors. The contravariant and covariant vectors defined above are special cases of differential invariants called tensors.

The components of the tensor are of the form Ta4b .. b;,

where the indices a,, a2, the integers 1, 2,

.

. .

, a,

b1, b2,

. . . ,

b, run through

... , n, and the components transform accord-

ing to the rule ax N

1 b,b,...

ax

alas

TaW*A:...,q,

anal

. .

axa* axle,

. aXa

alb,

. .

. axig a-.

(495)

TENSOR ANALYSIS

SEC. 126]

We call the exponent N of the Jacobian tensor field.

275

1092X1

the weight of the

If N = 0, we say that the tensor field is absolute;

otherwise the tensor field is relative of weight N. A tensor density occurs for N = 1. The vectors of Sees. 123 and 124 are absolute tensors of order 1. The tensor of (495) is said to be contravariant of order r and covariant of order s. If s = 0, the tensor is purely contravariant, and if r = 0, purely covariant; otherwise it is called a mixed tensor. Two tensors are said to be of the same kind if the tensors have the same number of covariant indices and the same number of contravariant indices and are of the same weight. We can construct further tensors as follows: (a) The sum of two tensors of the same kind is a tensor of this kind. The proof is obvious, for if a...b _ axN

Tc ...

d - ax

b= A7c.

d

a

... 8xrOxa axd axa

To ... r age

ax°

laXlN S-..

""T a

axr axa

.

. .

.

alb ax"

.

.

atd axa

.

a ax0

then

Ua...b c...d

Fa...b)

c...d

c...d

_ Ua... c...z

ax N

ax'

axb

a2

(b) The product of two tensors is a tensor. We show this for a special case. Let Tba =

a axQ a2"

alt T s axlaxa Ox 3

19i =

a2

02i

Sc or,

so that

= ax s (T bs')

,ax

ax8 a2a all (TWO)

a26 ax* ax"

The new tensor is of weight N + N' = 3 + 2 = 5. (c) Contraction.

Consider the absolute tensor Atk

-

a ax0 axy axi AOY

a2' axk axa

276


[SEC. 126

Replace k by i and sum. We obtain A,i - Aay = Aay a = Apa

43X'9 axy axi

axi axi axa

M ax,

y

13x#

a = Aay axi sa ax axa

axa

axi

so that Ali are the components of an absolute contravariant vector. In general, we equate a certain covariant index to a contravariant index, sum on the repeated indices, and obtain a new tensor. We call this process a contraction. (d) Quotient law. We illustrate the quotient law as follows: Assume that AiB,k is a tensor for all contravariant vectors Ai. We prove that B,k is a tensor, for AiE1 = AQBay

= A'Bay

-

CIO axy 492i

x6

axi ax; axa

axa axy

ax axk

or

l

axa axy

Ai ` $,k - Bay axp avk

=0

Since A' is arbitrary, we must have f k = Bay result. Example 122. tensor, for

a2 ax

the desired

The Kronecker delta, S, is a mixed absolute =

axi axa

' aga axi' Example 123.

axa axk

axi axa = 80 a

axa axs

= asa

If Ai and Bi are the components of a contra-

variant and a covariant vector, then C = AiB, are the components of a mixed tensor, for s

Ai = Aa ax , axa

axa B' _ B a ali

-

277

TENSOR ANALYSIS

SEC. 1261

CO that AaB#

C

axa axe =

Cs

axa Ox'

Let gij be the components of a covariant tensor

Example 124.

ax-,9x#

Taking determinants and applying

so that gi, = gap - =

axt ox' Example 115 twice, we obtain lax

ax

I#I} = IgIl

or

Igl =1g1 l ax

a2

Now if Ai are the components of an absolute contravariant vector, then A{ = Aa ax Y so that ax

B, = IgI}A' =

B.

ax

aax

IgI'Aa

Of

axa

Ba2i

ax-

are the components of a vector density. This method affords a means of changing absolute tensors into relative tensors. Example 125. Assume gap dxa dxO an invariant, that is,

so that Bi °

IgI#Ai

9aP d7a d2O = gp dxa dxfi a

Now d2a = axu dxµ, so that gap

axa aaxu ax,

dxu dx = g,,, dxu dx'

or (9aO

a0

°1° ^ 9u axu ax

dxu dx' = 0

(496)

If we assume gap = gaa, then since (496) is identically zero for arbitrary dxi, we must have (see Example 114)

278


ISEC. 126

axa axs 9µY = 9a$

(496a)

axµ axY

or the gµ, are components of a covariant tensor of rank 2. Example 126. If the components of a tensor are zero in one

coordinate system, it follows from the law of transformation (495) that the components are zero in all coordinate systems. This is an important result. Example 127. Outer product of two vectors. Let A; and B; be the components of two covariant vectors, so that C;; = A;B; = AaEp

axa ax#

axti axe =

C°

axa a28

axti axe

The C:; are the components of a covariant tensor of second order,

the outer product of Ai and B. Example 128. By the same reasoning as in Example 127, we have that Cc; = A;B; - A;B, are the components of a covariant tensor of the second order. Notice that C,, is skew-symmetric, for C;; = -C;;. For a three-dimensional space

11C1;II =

0

A1B8 - AaBI A2B3 - A3B2

-(A2B3 - A3B2)

0

A1B2 - A2B1

0

-(A1B2 - A2B1) -(A1B8 - A3B1)

The nonvanishing terms are similar to the components of the vector cross product.

Problems

1. If A = Aao

axa ax0

ai

axe, show that A;; = Aaa

axa axa

ax' axe

2. Show that A,, can be written as the sum of a symmetric and a skew-symmetric component. 3. If A are the components of an absolute mixed tensor, show that A; is a scalar invariant. 4. If Aa# are the components of an absolute covariant tensor, and if Aa#A$,, = 8 , show that the Aal are the components of an absolute contravariant tensor. The two tensors are said to be reciprocal.

SEC. 1271

TENSOR ANALYSIS

279

5. If Al, and A1, are reciprocal symmetric tensors, and if ui are components of a covariant vector, show that A,;uiul = Aiiu,u;, where ui = Ai°u,,. 6. Let Ail and Bit be symmetric tensors and let ui, vi be components of contravariant vectors satisfying

(A,,-KBi;)ui=0 i,j= 1, 2, (A,, -K'Bii)v' = 0,

.

. .

n

KX K

AiiutuWhy i

is K Prove that Ai,uiv' = Bi,uiv' = 0, and that K = BiJuiui an invariant? 7. From the relative tensor A of weight N, derive a relative scalar of weight N. 8. If Ami is a mixed tensor of weight N, show that Amw is a mixed tensor of weight N. 9. Show that the cofactors of the determinant lair, are the components of a relative tensor of weight 2 if ail is an absolute covariant tensor.

10. If Ai are the components of an absolute contravariant vector, show that axe are not the components of a mixed tensor.

127. The Line Element. In the Euclidean space of three dimensions we have assumed that ds2 = dx2 + dye + dz2

In the Euclidean n-space we have

ds2 = (dx')2 + (dx2)2 + . .. + (dxn)2 = bo dx° dxa If we apply a transformation of coordinates xi = xi(21, 22,

.

.

.

,

2' )

we have that dx' = axi d2a, so that (497) takes the form ds2 = 6,0

ax° axe dxµ dx axµ aV

We may write ds2 = vr d2" dz', where 9vr = a°a

axa 8x#

a_

n

ax° ax° ax+' any

(497)


280

[SEc. 127

Thus the most general form for the line element (ds)2 for a Euclidean space is the quadratic form ds2 = gas dxa dx5

(498)

The gap are the components of the metric tensor (see Example 125).

The quadratic differential form (498) is called a Rie-

mannian metric. Any space characterized by such a metric is called a Riemannian space. It does not follow that there exists

a coordinate transformation which reduces (498) to a sum of squares.

If there is a coordinate transformation xi = xi(y', y2,

.

.

.

,

y")

such that ds2 = Say dya dy5, we say that the Riemannian space is Euclidean. The y's will be called the components of a Euclidean coordinate system. Notice that gas = Sap. Any coordinate system for which the g;; are constants is called a Cartesian coordinate system. We can choose the metric tensor symmetric, for

gii = Ij(gv + gig) + (gii - gii) and the terms $(gt, - gii) dx' dx' contribute nothing to the sum ds2. The terms J(g,i + gii) are symmetric in i and j. Example 129. In a three-dimensional Euclidean space ds2 = tax 1)2 + (dx2)2 + (dx3)2 for an orthogonal coordinate sys-

tem, so that 1

0 0

0 0 1

0

0

1

Let

x' = r sin 0 cos p = y' sin y2 cos y$ x2 = r sin 0 sin rp = y' sin y2 sin y2

x3=rcos9=y'cosy2 Now axa 8x5 9ti(r, g, w) = gas ay; ayi

ax' ax'

axe axe

OxI 49x8

- ay: ayi + ay: ayi + ay' ayi

TENSOR ANALYSIS

SEC. 127]

281

Hence

gll = (sin y2 cos y3)2 + (sin y2 sin y3)2 + (cos y2)2 Similarly 933 = (y')2(sin y2)2,

922 = (y') 2,

gi,=0 for i5j

so that ds2 = (dy')2 + (y')2(dy2)2 + (y' sin y2)2(dy3)2 = dr2 + r2 do2 + r2 sin2 0

is the line element in spherical coordinates. Since the g's are not constants, a spherical coordinate system is not a Cartesian coordinate system. Example 130. We define gii as the reciprocal tensor to gi,, that is, gi.2 gai = d,f (see Prob. 4, Sec. 126).

The gii are the signed

minors of the g;i divided by the determinant of the gi;. For spherical coordinates in a Euclidean space 1

0

0

r2

0 0

0

0

r2 sin2 B

1

0

0

0

2 r2

0

0

0

1

r2 sin2 9

Example 131. We define the length L of a vector Ai in a Riemannian space by the quadratic form L2 = g pA°A#

(499)

The associated vector of Ai is the covariant vector i = gia

It is easily seen that Ai = gipAp, so that L2 = gapg'

g"'A"A.

We see that a vector and its associate have the same length. If L2 = 1, the vector is a unit vector. Example 132. Angle between two vectors.

unit vectors.

Let Ai and B, be

We define the cosine of the angle between these


282

[SEC. 127

two vectors by

cos 0 = A'Bi = Aig;;Bi = gi,AiB' = giiA,B1 = gi'A;B,

(500)

If the vectors are not unit vectors, gi,AiB; (500 a)

cos 0 =

If gi;A'B' = 0, the vectors are orthogonal. We must show that Assuming a posiI cos 01 < 1. Consider the vector AA' + tive definite form, that is, gaaza# > 0 unless z' = 0, we have 1Bi.

gas(XAa + µB")(XAa + kB") > 0 or

y = A2(gaAaAa) + 2Aµ(gasA"B1) +µ2(gaaBaBa) > 0

This is a quadratic form in A2/µ2, so that the discriminant must be negative, for if it were nonnegative, y would vanish for some value of A/µ or µ/A. Hence gaaAaBP <

(gaaAaA8)4(gaaBaBa)}

or ,cos 01 < 1. Moreover, if Ai = kBi, it is easy to see that cos 0 = ± 1. Hence I cos el < 1. Example 133. A hypersurface in a Riemannian space is given by x' = x'(ul, u2). If we keep ul fixed, ul = uo, we obtain the space curve x' = x'(uo, u2), called the u2 curve. Similarly, x' = x'(ul, uo) represents a ul curve on the surface. These curves

are called the coordinate curves of the surface. once that on the surface 2

We have at

axa axa

ds2 = gad dxa dxa = I gap - -dui du'

_

au' aui ca-1 axa axa g°a aui au

ds2 = h;; du' du' where hi, = gaa

axa axa

au'aui

dui du' (501)

TENSOR ANALYSIS

SEC. 127]

Example 134.

283

Let us consider

The special theory of relativity.

the one-parameter group of transformations

x = ,B(x - Vt)

y=y

(502)

z = 2

xJ

where,6 = [1 - (V2/c2)1-} and V is the parameter. c is the speed of light. These are the Einstein-Lorentz transformations (see Prob. 11, Sec. 24). The transformations form a group because (1) if we set V = 0, we obtain the identity transformations x = 2, y = y, z = 2, t = t; (2) the inverse transformation exists since 2 _ #(x + Vt), y, t = z, t = O[t + (V/c')x], the inverse transformation obtained by replacing the parameter V by - V; (3) the result of applying two such transformations

yields a new Lorentz transformation, for if

where

_ [1 - (W'/c')]-;, then

x = (2- Ut) y=?! z

t=fi -Ux c where

U`

V-I-W

- 1 + (VW/c):'

^ r1 _ `\

U2)-1 c

I

We now assume that (x, y, z, t) represents an event in space and time as observed by S and that (x, 9, 2, 1) represents the same event observed by S (see Fig. 101).


284

The origin b has :f = y = 2 = 0' so that from (502)

[SEC. 127

dx dt

= - V'

showing that S moves with a constant speed - V relative to S. Similarly S moves with speed + V relative to S. y

.7, 1,q 'P1:Mf,7 14

-V

S

S

X

X

z

z

Fia. 101.

From (502) we see that 0 and 0 coincide at t = 0. At this instant assume that an event is the sending forth of a light wave. The results of Prob. 11, Sec. 24 show that x2 + y2 + z2 t2

- xa _'

_2

12

+

22

-- 4 C

so that the speed of light is the same for both observers. This is one of the postulates of the special theory of relativity. Starting with this postulate and desiring the group property, we could have shown that the transformations (502) are the only transformations which keep dx2 + d y2 + dz2 - cz dt2 = 0 an invariant. Let us now consider a clock fixed in the S frame. We have x = constant, so that dx = 0, and from (502) dt = 0 dt. Hence a unit of time as observed by S is not a unit of time as observed by S because of the factor 0 5-6 1. 8 remarks that S's clock is running slowly.

The same is true for clocks fixed in the S frame.

TENSOR ANALYSIS

SEC. 1271

285

We choose for the interval of our four-dimensional space the invariant ds2 = c2 dt2 - dx2 - d y2 -dz2 = (dx') 2 - (dx') 2 - (dx2) 2

- (dxa) 2

where x' = x, x2 = y, x$ = z, x' = ct. The interval ds2 yields two types of measurements, length and time, but takes care to distinguish between them. If we keep a clock fixed in the S frame, then dx = dy = dz = 0, so that ds2 = c2 dt2, and the measurement of interval ds is real and proportional to the time dt.

Now if we keep t fixed, dt = 0, and ds' = - (dx2 + dye + dz2)

so that ds is a pure imaginary, its absolute value denoting length as measured by meter sticks in a Euclidean space. We shall describe the laws of physics by tensor equations, the components of the tensors subject to the transformations (502). This will guarantee the invariance of our laws of physics. The momentum of a particle of mass mo will be defined by pa = mo d8 . If the speed of the particle is u,

()2 ()2 u2

+

+ \dt/2

as measured by S, then

d82 = c2 dt2 - (dx2 + dye + dz2) = (c2 - u2) dt2 so that mo dxa I mo dxa pa = (C2 - u2)} dt = C [1 - (u2/c2)]} dt

and

p = [1

mo - (u2/C2)]}

-- m

We define the Minkowski force by the equations

_

f a - C2

d(

dxa

ds m0 da )

I

d (m

[1 - (u2/C2)]; dt

a

dt

'

a = 1, 2, 3, 4

286


ISEC. 127

The Minkowski force differs from the Newtonian force by the factor [1 - (u2/C2))-1. The work done by the Newtonian force

Fa = d- (m ddt

for a displacement dxa is d

dE _

(mxa) dxa

a-1

I (mxa dxa + dd -4- dxa j a1 mou du [1 - (u2/C2)11

and integrating, E = [1 - (u2/c2))-lmoc2 - moc2 = (m - mo)c2, with E = 0 for u = 0. Expanding [1 - (u2/c2)1-1 in a Maclaurin series, we have E Jmou2 for (u2/c2) << 1. The reader is referred to Probs. 1, 2, and 3 of Sec. 82 for the application of special relativity theory to electromagnetic theory. Let the reader derive (285) by use of this theory, choosing the frame S so that at a particular instant the charge p is fixed in this frame. The force on the charge as measured by S is given by (285).

Problems 1. For paraboloidal coordinates x1 = yly2 cos ys x2 = y1y2 sin ya x8 = 3[(yl)2 - (y2)2] show that ds2 = [(y')2 + (y2)21[(dyl)2 + (dy2)21 + (yly2)2(dy$)2.

2. Show that for a hypersurface in a three-dimensional 8

Euclidean space, h1 = a-1

axa axa

t9lis au?

3. Show that the unit vectors tangent to the ul and u2 curves are given by

1

ax

18x

1

and

-V//-22aU2

Vh11aul 4. If w is the angle between the coordinate curves, show that

cos w = his/.

TENSOR ANALYSIS

SEc. 127J

287

. . . , x") = constant determines a hypersurface If dx` is any infinitesimal displacement on the hyper-

5. (p(x', x2,

of a V,,.

Why does this show that the ax

surface, we have a dxa = 0. ax

are the components of a covariant vector that is normal to the hypersurface? 6. If (p(x', x2,

. . . ,

x") = constant, show that a unit vector

a aP

gra avp

normal to the surface is given by (g°8 ax, axo axa 7. Consider the vector with components (dx', 0, 0, .

. . ,

0).

Under a coordinate transformation the components become 1

"

2

Cax' dx', ax' dx', ax' dx' . Consider the new components for the vectors with components (0, dx2, . . . , 0), . . . , (0, 0, . .. ) 0, dx"), and interpret . . . ,

d2' dxa ... dx"

=

1a1 dx' dx2 .

.

Using the result of Prob. 7, Sec. 121, show that

is an invariant.

. dx" gl dx' dx2

dx"

We define the volume by

V=ff.

. .

f

dx' dxa .

.

. dx"

8. Show that ds is a unit vector for a V".

9. The surfaces xti = constant, i - 1, 2, ... , n, are called the coordinate surfaces of Riemannian space. On these surfaces all variables but one are allowed to vary. This determines subspaces of dimensions (n - 1). If we let only x' vary, we obtain a coordinate curve. Show that the unit vectors to the coordinate curves are given by a; = 1//, i = 1, 2, . . . , n, and that the angle of intersection between two coordinate curves is given by r

cosm;; _

gc,

9g

10. Show that a length observed by S appears to be longer as observed by S. How does S compare lengths with S?

288


[SEC. 128

11. Let jx, jj,, jz, p be the components of a vector as measured What are the components of the same vector as measured by S? by S.

12. Let

d2 x

d 29 dz2 72

die' die' die

be the components of the acceleration of a 2z

2y

dz

particle as measured by S. Find dt2, dt2, dt2 from (502). 128. Geodesics in a Riemannian Space.

If a space curve in a

Riemannian space is given by xi = xi(t), we can compute the distance between two points of the curve by the formula s

dxa dxO 4 r ( - ` i \g°a dt dt dt

(503)

To find the geodesics we extremalize (503) (see Sec. 40). The differential equations of the geodesics are [see Eq. (146)] d (of of

of

ax' -

dt

where f = (go '.#)} =

(504)

Now

Wt

of ax'

0

1 (aga9 zaxs

2f \ ax'

and

d (af

d (ga;xa + gist

dt \az'

dt \ 2 ds/dt Fz

2 ds/dt

/

(g.a + g+sio + agxps xaxs + x° ve 2

2(ds/dt)2 dt2

(ga:xa +

d

2

If we choose s for the parameter t, s = t, dt = 1, dtQ = 0, and use the fact that gii = gii, (504) reduces to a

1 /agai

g'ax + 2

610

ague

+

axa

agoal/ xa x o =0 - -axti

(505)

TENSOR ANALYSIS

SEC. 1281

280

Multiplying (505) by gri and summing on i, we obtain gri

xr

+

`(89-i

.99io

+

2

axi /

axa

or d2xr

dxa dxP

,

(506)

+ r°a ds ds -

dS2

where 9r r° (.9g.,

+

2 \ ax$

ra$

a9oa

ago

axa

axQ

(507)

The functions r" are called the Christoffel symbols of the second kind. Equations (506) are the differential equations of the geodesics or paths. Example 135. For a Euclidean space using orthogonal coordinates, we have ds2 = (dxl) 2 + + (dx") 2, so that gap = a.# and

ft"

d1

Hence the geodesics are given by ds = 0 or

0.

xr = ars + br, a linear path. Example 136. Assume that we live in a space for which dal = (dx')2 + [(x1)2 + c2](dx2)2, the surface of a right helicoid immersed in a Euclidean three-space. We have 0

1

{1gi?il =

0

0

1

1

0

(x1)2 + c2

(x')2 .+ C2

Thus we have

rill =0, r21 = r14 = 0, 1

x1

r212

(x') 2 + C2

1 -- -x1, r22

r2

so that the differential equations of the geodesics on the surface are d2x' ds2 d2x2

d82

2x'

dx2 2 x

1

T8

- 0

dx' dx2

+ (x')2 + C2 d8 ds =

0

290


(SEC. 129

Problems 1. Derive the r;k of Example 136. 2. Find the differential equations of the geodesics for the line element ds2 = (dxl)2 + (sin x')2(dx2)2.

3. Show that rqn = r'a. 4. For a Euclidean space using a cartesian coordinate system,

show that r, = 0. 5. Obtain the Christoffel symbols and the equations of the geodesics for the surface

xl = ul cos u2 x2 = u' sin 412

x3 = 0

This surface is the plane x3 = 0, and the coordinates are polar coordinates.

6. From (507) show that axµ = g,sroals + g,arp,,. 7. Obtain the Christoffel symbols for a Euclidean space using cylindrical coordinates. Set up the equations of the geodesics. Do the same for spherical coordinates. 8. Write out the explicit form for the Christoffel symbols of the first kind : { i, jk I = g;,rlk. 9. Let ds2 = E due + 2F du dv + G dv2. Calculate IgJ, g'',

i, j = 1, 2. Write out the rk. axk axe + a2x° axe 10. If r = r;7 axs a. ax ax ax ax ax show that r;,y - rry

are the components of a tensor.

129. Law of Transformation for the Christoffel Symbols. Let the equations of the geodesics be given by d2xi

ds2

d i J dxk

_

+ r "k ds ds ^ 0

(508)

and

AV ds2

+

P 2k

d.V d2k

ds ds

_-

°

(5 0 9)

for the two coordinate systems xi, xi in a Riemannian space. We now find the relationship between the r k and r; . Now

TENSOR ANALYSIS

SEc. 129J

dxi ds

__

a2' dxa

axa ds

d2x`

and

_

a2xi

291 dxa dxa

axa axa ds ds

ds2

d2x°

9.

+ axa ds2

Substituting into (509), we obtain &V d2xa

axa ds'

dxa dxa

a22 i

+ ax# axa ds A +

rik axi axk dxa dxa

axe, ax" ds as

0

(510)

°

We multiply (510) by axi and sum on i to obtain d2x°

dx2

axe axk + a22i 49x°l dxa dxa + (rijk axa axa axgr a2i axa axa a-V ds ds -

0

Comparing with (508), we see that (using the fact that rlk = rk;) rik

axa axy axi

a2x°

Oxi

aY axe axk axa + axi axk a2°

This is the law of transformation for the rk.

(511)

We note that the

rk are not the components of a tensor, so that the rk may be zero in one coordinate system but not in all coordinate systems. Example 137. From (481) we have algl 8

= Iglgae ax'

and from Prob. 6, Sec. 128, ax#

= garap + g°arPA

so that = gas9°ar,, + gang°aFBp

= a;rap + a1r00

=rap+r;p=2rµ

or

C log axp

_

ra p

(512)

292


[SEC. 129

Example 138. We may arrive at the Christoffel symbols and their law of transformation by another method. Differentiating the law of transformation axe, ax8

g'' = ga$ a2i a2; with respect to xk, we have ag;i

agar axY axa ax8

axk - axY a2k

gay

ax0 aaxa

492X#

axa

49.t- __-

j a.Ti ak a +a2k I02i x2a2

(513)

If we now subtract (513) from the two equations obtained from it by cyclic permutations of the indices i, j, k, we obtain axs axY axi

ti

a

r" =

2 g°' (agk, axj

a2xa

axi

r - rdy afj a2k axa + 02i axk axa

511a)

where agi,

+ axle

_

agile

ax"

Example 139. Let us consider a Euclidean space for which ds2 = (dx 1) 2 + (dx2) 2 + .

-

. + (dx") 2

In this case the r (x) = 0. In any other coordinate system, we have a2x° a2i TV a2k ax°

If the new coordinate system is also Cartesian (the g,, = constants), then r k = 0, or 82x°

aP a2k

=0

x° = alga + b°

(514)

where a', b° are constants of integration. Hence the coordinate transformation between two cartesian coordinate systems is linear. If, furthermore, we desire the distance between two points to be an invariant, we must have n

n

d2° d2- _

dx° dx° _ ff-1

n

W-1

a:a- dxa dx8 C-1

TENSOR ANALYSIS

SEC. 129]

293

so that n

I a,-,a; = Sas

(515)

V-1

A linear transformation such that (515) holds is called an orthogonal transformation. For orthogonal transformations,

'

=g"° ax. axe

a' axe

axa axe a:' CIO

reduces to S;; =

We multiply both sides by

amt

ax),

and

sum on i, so that azf axe

axp

8x'

8x8

is a = #0 axe - iii

= aaa

(516)

Now let us compare the laws of transformation for covariant and contravariant vectors. We have

A' = Aa Replacing

621

a2' _, axa

A; =

Aa

axa 82'

(517)

by a from (516), we see that axa

n

a-1

Aaa

(518)

so that orthogonal transformations affect contravariant vectors in exactly the same way that covariant vectors are affected [compare (517) and (518)]. This is why there was no distinction

made between covariant and contravariant vectors in the elementary treatment of vectors. Problems

1. From (511) show that

_ r'k

82xC, of all To ax- + afj axk axe

a ax8 axy a2'

=ray


294

[SEC. 129

2. By differentiating the identity gaga; = S , show that ag{k

axi =

_gkxr h4 - 9htrh;

3. Derive (513) by performing the permutations.

s0

2 aQ

4. If az.

xk

ax = 0, show that

0. az axk =

5. If 090 My axi

r;k

a21°

axi

= rPr ax; 021; axa + axi axk at°

show that r;k =

a axA 8x' axi

6. If xa = xa(ul, u2, if h;; = gas

axa axs

Taui

r;, a.C axk . . . ,

axa

+

a2x' axi axi axk ax°

u*), a = 1, 2,

.

.

.

, n, r < n, and

and if 1

i {rik)h = 2 h.,

(ah auk

ahk,

ah,kl

+ aui _ au°

show that hair k)k = gap(ra,)o

axa axs axy axa a2xs + gas aui aui auk aut aui auk

7. Define gs(x) by the equation gas(x) = µ(x)gas(x). We see that the metric tensor gas(x) is determined only up to a factor of multiplication u(x). In this space (conformal) we do not compare lengths at two different points, or, in other words, the unit of length changes from point to point. Show that T;r(x) = r;, (x} + rPYaa + PSaY - ga°gstimo

1alogµ

where 'Po = 2 axe

, and ra sy are defined by (507) using

I'sa

gas and gas.

8. Prove that a geodesic of zero length (minimal geodesic) [that is, xa(s) satisfies

(506)

and

dxa dxs

gas ds A

= 0] remains a

minimal geodesic under a conformal transformation.

295

TENSOR ANALYSIS

SEC. 130J

130. Covariant Differentiation. Let us differentiate the absolute covariant vector given by the transformation

At-Aaaxa axi

We obtain aA;

aAa axa axa

axe

axa ax= ax' + Aa axi axi

It is at once apparent that

aA; axe

a2xa

(519)

are not the components of a

tensor. However, we can construct a tensor by the following device: From (511a) (see page 292) ax' axr axa ax; axe axo

82x,7

+

axa

axi axi ax"

(520)

Multiplying (520) by A. and subtracting from (519), we obtain

-A &V

aA;

a 1'

_

( axa

-Ar O

aa

Ox- axa ax' axe

(521 )

so that if we define

A;;=-Aa1 8x1 49A;

(522)

we have that A..s = Aa.a

axa axa

axi axi

and A;,, is a covariant tensor of rank 2. The tensor is called the covariant derivative of A; with respect to xi. The comma will denote covariant differentiation. For a cartesian coordinate sysaA;

tem, r,,E = 0, so that A;,1 = axe our ordinary derivative.

For a scalar of weight N we have A=I-

821

A

296


(SEC. 130

so that axlN aA axa ax 8xa 02'

+N

ax N-1

a

ax

lax

aax'

_ laxe a2xs

ax

and from (482),

Hence

ax ax i ax

aA

axi -

ax N aA axa latl axa axi

Multiplying r* = n.,

+

ax N axa

N

H

82x0

axs &V C120

a x, axa

A

(523)

2

axa

+ axi

axa Hi axQ

by NA and subtracting

from (523), we have OA

ax

- NAr; _

N (claAx a

ax

clx

- NA r-.1, i a

(524)

Hence A,, -= ax - NAr`,, is a relative covariant vector of weight N. A.

It is called the covariant derivative of the relative scalar

For a cartesian coordinate system it reduces to the ordinary derivative. In general, it can be proved that if Tp,s,::: p; is a relative tensor of weight N, then

+

,,p

axe

+

TW-a,ry" -A-

a v, Ta,a,...a,

Ain

(525)

is a relative tensor of weight N, of covariant order one greater than Ts,,:::a;, and it is called the covariant derivative of TB,A :::8. Example 140. We have

ag;;

gii,k = axk - gwl'.k - g+Nr k

so that from Prob. 6, Sec. 128, g i.k = 0

(526)

TENSOR ANALYSIS

SEC. 130J

297

Example 141. If p is an absolute scalar,

_ gyp, we call

-p,; = ax the gradient of V. Example 142.

Curl of a vector. aA; ant vector. We have A;,; = axj

aaxA;

- Aar .

axi

aA1

- ax` is a covariant tensor of rank 2.

It is called the curl of the vector Ai.

of the gradient of a scalar, Ai = curl A; ==

Similarly,

"' - Ar,i

A,.; Hence Ai,5 - A;,; =

Let A; be an absolute covari-

If the A; are the components

a-P,

axi

then

- ax' ax' = o ax1 ax' 621P

a2ip

so that the curl of a gradient is zero. It can be shown that the converse holds. If the curl is identically zero, the covariant vector is the gradient of a scalar. Example 143.

Intrinsic derivatives.

tensors, we know that Ai.;

dxi

ds

.

is a covariant vector.

the intrinsic derivative of A;. A;.,

dxi ds

8A; as

Since Ai,1 and

dx

are

We call it

We have

aAi dxi

a dxf

a axe ds - A°`r`' dA;

ds -

ds

dx1 (527)

òar"ds

and write the intrinsic derivative of A; as

BA;

as

Example 144. The divergence of an absolute contravariant vector is defined as the contraction of its covariant derivative.

298


(SEC. 130

Hence

div A' = A' = ax Now rQ; as =

a logx

IgI

div A

-4- AarQ,

from (512), so that a

1

_

ax-

t

div As =

+ 1

Aa

ax.

N/191

aa a (V f gj Aa)

(528)

In spherical coordinates, we have 1

`'9=

0 0

0 r2 0 0 r2 sin2 0 0

i = r2 sin 0

so that

div A' =

i

a

r2 sin 8 ar

(r2 sin 0 A,)+

a. d9

(r2 sin 0 AB)

+ app (r2 sin 0 Ac)

and changing At and A" into physical components having the dimensions of A (see Example 121), we have

div A' =

1

Car (r2 sin 0 A*) + a9 (r sin 0 A°) + a (rAc) J

r2 sin o Example 145. The Laplacian of a scalar invariant. If p is a

scalar invariant, -r,; is the gradient of jp, and the div (p,,) is called the Laplacian of .p.

a

Lapcp=VZ(p=div(g,)=diva' Thus v 2(p

1,91

aaa (v' Fg19°°

)

TX

(529)

TENSOR ANALYSIS

Sec. 1301

into a contravariant vector so that we could

We changed apply (528).

299

aThe associate of

81P

is g«i LIP.

ax'

axe

In spherical coordinates i

1

0

0

0

r2

0

0

0

r2 sin2 B

0

0

t g' l =

7

I

0

0

r2 sin 9

so that

VF =

a

1 r2 sin B r2 sin 0 ar \\\ 1

{

Example 146.

aF ar

-1-

a (sin B aF' + a / 1 aF' 00 \ 00a l\sin 9 app/)

In Example 144 we defined the divergence of

the vector A' as div A' = A

Ma

+ Aar' - For a Euclidean Cti

space using cartesian coordinates, the rk = 0, so that 0A2 OA' div A'=axl+axe+ ... + aA^ axn

The quantity A* is a scalar invariant. If we let Ni be the components of the unit normal vector to the surface do, then AaNa is also an invariant. In cartesian coordinates the divergence theorem is

JJ

divAdr = ffA.dd = JJA.N&r

In tensor form it becomes

fJfAT = JJA"Nad r

(530)

We can obtain Green's formula by considering the covariant vectors 4O,; and y ,;. Now let A, = kp,i - soO,c

300


{SEC. 130

The associated vector of A; is A° = g'=A; = g°.i(4,v,s - 4,i).

We easily see that A' = g°i(4y,; -

Now g°'co,,Q is an

invariant and in cartesian coordinates reduces to a29

+ (ax1)2

a2_

+ (ax2)2

a24

_ Lap (p

(ax3)2

Hence, using (530), we obtain

f if (t' Lap so - v Lap ¢) dr =

f f g°iA;NQ do S

=

ff

Svf.,)N; dv

S

Let us consider the covariant vector Fa. We multiply it by the contravariant vector d--° and sum on a to Example 147.

obtain the invariant F. dxa = F.

ds

ds, which reduces to f dr

in cartesian coordinates. In Example 142 we constructed the curl of a vector, which turned out to be a tensor of rank 2. We now construct a vector whose components will also be those of

the curl of a vector. We know that F.,s is a tensor. Now define 4ah = 1

w

1

if a, ft, y is an even permutation of 1, 2, 3;

'

is an odd permutation of 1, 2, 3;

4°Ar = 0 otherwise. Thus 41st

-, 1

211 -

-L

I

4132 =

1

411! = 0

191

We obtain a new invariant G" = - 4"0''Fa,F

In cartesian coordinates Fa,P = ate, and 1 = -EarilFR .a

= -j E2111F2 s + 021F3,2

= aFs

aF2 axe - ax=

TENSOR ANALYSIS

SEC. 130]

Hence Stokes's theorem in tensor

and similarly for G2 and G3. form reads a

f F.

301

ds ds =

f

do

(531)

Problems

1. By starting with Ai = All A.1

a2i Oza

, show that

axi + Aara1

is a mixed tensor. 2. Prove that (g{aAa),, = g{aA'.

3. Prove that (A-B.).1 = AaBa,, + A Ba. 4. Prove that 14i = 0. 5. Use (529) to find the Laplacian of F in cylindrical coordinates.

6. Prove that a;

7. Prove that

0. 1

axa (vflg g`-) + rQgas = 0.

8. As in Example 143, show that the intrinsic derivative

as

ds

+

Aarocp

ds

is a contravariant vector.

9. Show that the intrinsic derivative of a scalar of weight N

is b = ds - NAr;, ds,

so that if A is an absolute constant,

10. Show that (g,#AaAO) j = gapA ;Ap + 9apAaA,'

11. Show that A; , = aA' + ax°

mixed tensor A. 12. Show that DZ(vo)

13. Show that Aaa =

r;,A' for an absolute

V;k -I- 2 0v - DO -}- J, pz9. 1 0 (VTg_j A") - Abr. gl exa

302


[Sac. 130

14. If A; = A;(x, t) is a covariant vector, show that oA; at

_

8A; at

+ Ai,J

dx' dt

av;

_

and hence that the acceleration fc °

at

8vi

at + v ,,v'.

15. Let X. be an arbitrary vector whose covariant derivative Consider Jf Ta,' XaN,6 da, and apply s the divergence theorem to the vector TOXa. Hence show that vanishes; that is, Xa,$ = 0.

fJ T aaNN da = ff7 T dT. S

16. Let s; be the displacement vector of any particle from its position of equilibrium (see Sec. 115). We know that s;, is a covariant tensor. The relative displacements of the particles are given by as; = s,,J dx

_ 4(s,,J + sJ,:) dx' +

s,,.) dx'

Show that the term 4(8;,J - 8J,;) dxJ represents a rotation. We define the symmetric strain tensor E;J by the equation Ei; = +(s.,J + ss,;)

The stress tensor T;, is defined by the equations AF; = TJNJ Ao-

where AF; is the force acting on the element of area Aa with normal vector Ni (see Sec. 116). Let f, be the acceleration of the volume dT and F, be the force per unit mass acting on the mass in question. Show that

f j f pF, dT + f f TJNJ da =

f f f pf, dT

or using contravariant components,

fffa pF'dr+ f sf T*JNJda= fffR pf'dT

TENSOR ANALYSIS

SEC. 1311

303

Now deduce the equations of motion

pFr + T = pfr If Tn = pgri, show that pFr - p.igr' = pfr or

[see (411)]

pFr --- p.r = pfr

The equations of the geodesics

131. Geodesic Coordinates. are given by

dxi dxk

d2xi + r,k

as2

= 0

ds as

where the Ik transform according to the law ask

-

rasY

49x8 ax" axi

ax axk axa

192xa

axi

+ 491i axk axa

(532)

We ask ourselves the following question : If the I k are different

from zero at a point x` = qi, can we find a coordinate system such that rk 0 at the corresponding point? The answer is "Yes"! Let

(xi - q') + a and

so that

I

(533)

= 1, and moreover the transforma-

ax1i

a

tion (533) is nonsingular.

q«)(xd - 4s)

The point xi = q' corresponds to the

point x' = 0. Now differentiating (533) with respect to x', we obtain axii

a=

+ (ri OP )

because of the symmetry of ri,8.

Q(xa - qa) ate? I

Hence 6z! = 6f. Q

Differentiating (534) with respect to xk, we obtain a2x8 axa axs a2x' + (rae) a + (r10) a(x0 q°`) axk a21 axk ax' axk axi

(534)


304

[SEC. 131

so that ax° axe

a2x'

axk axj

= - (ray) 4 ask Q

Q

axe

-(raa)Q6-50 _ -k Substituting into (532), we obtain

(rij k)o = (rfY)Qaj kaa -

(rk)Qaa

= (r k), - (r;k)Q = o,

Q.E.D.

Any system of coordinates for which (r7k)P = 0 at a point P is called a geodesic coordinate system. In such a system, the covariant derivative, when evaluated at the origin, becomes the ordinary derivative evaluated at the origin. For example, \(Mi (As')O _

axi )0

+ (ra;)o(Aa)o = ax, )0

since rq; = 0 at the origin. The covariant derivative of a sum or product of tensors must obey the same rules that hold for ordinary derivatives of the calculus, for at any point we can choose geodesic coordinates so

that A'1

OA'

aBi

ax;

+axi

_ a(Ai + Bi) ax'

(A'+B');

and A', + B`; - (A' + B'),, is a zero tensor for geodesic coordinates. Hence A + Bì - (Ai + B'),1 is zero in all coordinate systems, so that A` + B`; _ (A' + B').; We leave it as an exercise for the reader to prove that (A'B;).k = A'kB7 + A'B1,k

Equation (533) yields one geodesic coordinate system.

There

are infinitely many such systems, since we could have added c,fiy(x) (x° - q') (xa - qP) (xr - q7) to the right-hand side of (533) and still have obtained (r k)o = 0.

A special type of geodesic coordinate is the following: Let x' = x'(s) be a geodesic passing through the point P, x' = xo, and

TENSOR ANALYSIS

SEc. 131]

let

xi r

305

Define

X' = Vs

(535)

where s is arc length along the geodesic. Each V determines a geodesic through P, and s determines a point on this geodesic.

Hence every point in the neighborhood of P has the definite coordinate xi attached to it. The equations of the geodesics in this coordinate system are d2z1 dS2 2

d21 dxk 1 + r'k ds ds =

0

i

But ds = V and ds = 0, so that rikl:'k = 0

(536)

Since this equation holds at the point P for all directions t', we

must have r + r 1 = 21 k = 0, so that the x= are geodesic coordinates. The X' = `s are called Riemannian coordinates. Example 148. If ' is a unit vector, we have

gaa°

=1

The intrinsic derivative is a gas s e + gaak°

since (g.p),1 = 0 (see Example 140). adx.

gQs°

0. Usp + -r" $) =

dt1

Q

as

=

0

Hence gaat"

as

0, and

We see that the vector dx°

is normal to the vector t'. Problems

1. Show that

dx°, dxgab

ds ds

remains constant along a geodesic.

2. Show that for normal coordinates 2, r;2' k = 0.


306

[SEC. 132

3. If s is are length of the curve C, show that the intrinsic i

derivative of the unit tangent

ds

in the direction of the curve

has the components pi -

d2xs

+

ds2

.

dx1 dxk

r'k

A

TS

What are the components for a geodesic? a

4. Prove that bt (X"Ya) = sat Ya + Xa sat.

132. The Curvature Tensor. Let us consider the absolute contravariant vector Vi. Its covariant derivative yields the mixed tensor

v`; =

a'

ax; + V ar°`'

On again differentiating covariantly, we obtain

avt.

v',k = axk' + v rak

= a 2V

s

axk 49x1

- viark

+ aV a ri + Va arm, + axk

'

49xk

(a a

+ VOr;, rak

Interchanging k and j and subtracting, we have Vt;k - Vsk; = V"Ba;k

(537)

where

Since V';k - Vik; and Vi are tensors, Va;k must be the components

of a tensor, from the quotient law (Sec. 126). It is called the curvature tensor. We can obtain two new tensors of the second order by contraction.

TENSOR ANALYSIS

SEC. 133]

307

Let Ri'

Ba;

ara

axa

ara

axa + r ara; - r%raa

(539)

This tensor is called the Ricci tensor and plays an important role in the theory of relativity. We obtain another tensor by defining Si; = Be..

ara aa i

ara a;

ax1

axi

(540)

Evidently Si; = -8;i. and if we use the fact that a log "ICI

- ra

1aµ.

ax"

we have that a2 log

a2 log -VIgl

axi ax1

ax1

=

0

axi

Now Ri; - R;; = Si; = 0, so that the Ricci tensor is symmetric in its indices. We could have deduced this fact by examining (539) directly. The invariant R = gi"Ri; is called the scalar curvature. 133. Riemann-Christoffel Tensor. The tensor Rhitk = gpaBk

(541)

is called the Riemann-Christoffel, or covariant curvature, tensor.

Let us note the following important result: Assume that the Riemannian space is Euclidean and that we are dealing with a cartesian coordinate system. Since r;k(x) = 0, we have from (538)

B';ki = 0

(542)

in this coordinate system. But if B';k= = 0 in one coordinate system, the components are zero in all coordinate systems. Hence if a space is Euclidean, the curvature tensor must vanish. We shall show later that if B1,-k1 = 0, the space is Euclidean.

308


[Sac. 134

If we differentiate (538) and evaluate at the origin of a geodesic coordinate system, we obtain Bajk a

a2raj

_

- ax° axk

a2rak 8x° axj

Permuting j, k, a and adding, we have the Bianchi identity (543)

Bakj.o + Ba0j.k + Baka,1 = 0

134. Euclidean Space. We have seen that if a space is Euclidean, of necessity Bike = 0. We shall now prove that if the Bki = 0, the space is Euclidean. Now axfi axy 00 ayj ayk axa

rik(y) = rsY

a2xa ayi

+ ayj ayk aXa

If there is a coordinate system (x', x2, r;,.(x) = 0, then a2xa

. .

.

,

xn) for which

ayi

r;k(y) = ayj ayk axa

(544)

and conversely, if (544) holds, the r;y(x) = 0. Now let us investigate under what conditions (544) may result. We write (544) as a2xa

axe

ayj ayk = ayi rk(y1

(545)

which represents a system of second-order differential equations. Let us define axa

u7-aycc

(546)

so that (545) becomes auk = u7Nkly) ayj

(547)

For each a we have the first-order system of differential equations

given by (546) and (547), which are special cases of the more general system

TENSOR ANALYSIS

SEC. 1341

azk

ay'

= f (z', z2,

. . .

z", z"+1, y`,

,

yn)

309

k=1, 2,

..,n+I

j=1,2,...,n

(548)

If we let z' = x°, z2 = 211, z3 = 112i and (547) reduce to (548).

.

.

2

2

We certainly must have aye ayj = ayj afk 8y1

, Zn+' = un, Eqs. (546)

.

yi, and this implies

_ aft af, av ay' - ay' + aZM ay'

afk az"

+

az

(549)

or k

afk

afk

ay'

+ azi, fill

+ fk u

ay

az"

fj

If the f; are analytic, it can be shown that the integrability conditions (549) are also sufficient that (548) have a solution satisfying the initial conditions z1; = zo at yi = yo. The reader is referred to advanced texts on differential equations and especially to the elegant proof found in Gaston Darboux, "Lecons sysOmes orthogonaux et les coordonnees curvilignes," pp. 325-336, Gauthier-Villars, Paris, 1910. The integrability conditions (549), when referred to the system (546), (547), become 1

jkua

nklua

aj"U'.=0

(550)

The first equation of (550) is satisfied from the symmetry of the I';k, and the second is satisfied if 19j"n = 0. Hence, if f"kj = 0, we can solve (545) for z' = x° in terms of y', y2, . . . , y". For the coordinate system (x', x2, . . , x"), we have T k(x) = 0.

Problems

1. Show that Rhijk = - Rihik = - Rhikj and that Riijk = Rhikk = 0

2. Show that Rhijk + Rhkij + Rhiki = 0-

3. If Rij = kgii, show that R = nk.

310


[SEC. 134

4. For a two-dimensional space for which g12 = 921 = 0, show that R12 = 0, 811922 = R22911 = R1221 and that

R_

R1221 911922

R:; = JRg.;. 5. Show that B k1 76 0 for a space whose line element is given by ds2 = (dx')2 + (sin xl)2(dx2)2. 6. Derive (550) from (546), (547), (549). 1 49R

7. If R = gtiaR,, show that (R'),; = 2 ax;.

CHAPTER 9 FURTHER APPLICATIONS OF TENSOR ANALYSIS i

135. Frenet-Serret Formulas. Let Ai =

ds

be the unit tangent

vector to the space curve xi = xi(t), i = 1, 2, 3, in a Riemannian space. In Example 148 we saw that the contravariant vector t

as

is normal to Xi.

Let us define the curvature as K =

0} g°0 88 aas °

and the principal unit normal µi by aai

as

_ =

Kµi

(551) i

Since pi is a unit vector, we know that aS is normal to pi.

Now

g°0A°,us = 0, so that the intrinsic derivative yields Sµa

bA°

as

since g°0,; = 0 or a3s

µB = 0

+ g°0

g°'eA°

bs

Hence

0. aµ0

g°a A°

+ Kg°aa°i8 = 0 as

or

(B68 µs

+ KAS

g°aa°

=0

(552)

since gp°µs = g°aX 'X 8 = 1. i

Equation (552) shows that Ss + KAi is normal to V, and since 5Fdi

as

and Ai are normal to µi,

aµi

as

+ KAi is also normal to µi.

311

We


312

[SEC. 135

define the binormal vi by the equation

-

v'

Sµ + KAt

1

(553)

(68

T

or aIAi

as

-= -#c? - TY'

(554)

(i)

where ,r is called the torsion and is the magnitude of

Since v` is normal to both X and µ', we have gaPvax$

=0 (555)

g«avO'µP = 0

By differentiating (555) and using (551), (554), (555), we leave it to the reader to show that gaaµa

/ - ava as

(T/!a

=0

(556)

The vector rµe - as is thus normal to all three vectors V, Af, Pi.

Since we are dealing in three-space, this is possible only if av TEIZ

as

= 0, or avi

=

as

(557)

Writing (551), (554), (557) in full, we have the Frenet-Serret formulas

&

dxP

ds

ds

KAi

i

d+

µa s = - (Ki + Tv`) a

ds

+

Pa

ds

= Tµ

(558)

SEC. 1361 FURTHER APPLICATIONS OF TENSOR ANALYSIS 313

= 0,

For a Euclidean space using cartesian coordinates, the r and (558) reduces to the formulas encountered in Sec. 24.

Problems 1. Derive (556). 2. Using cylindrical coordinates, ds2 = (dxl)2 + (x')2(dx2)2 + (dxa)2

and for a circle xl = a, x2 = t, x3 = 0. Expand (558) for this case, and show that K = 1/a, r = 0. 2i -K2X + U' - KTY'. 3. Show that Ss = as 4. Since (558) is true for a Euclidean space using cartesian coordinates, why would (558) hold for all other coordinate systems in this Euclidean space? .

2

5. Since 'Xi = d3 , show that

dx

d

a

+ ra' ds

Q

dare the

com

ponents of a contravariant vector. 136. Parallel Displacement of Vectors. Consider an absolute contravariant vector Ai(xl, x2, . . . , xn) in a cartesian coordinate system. Let us assume that the components Ai are con-

stants. Now A' = Aa

a2'

A' = Aa

axa

axi a2a

so that dA: - _ Aa

822'

axfi

8xs axa (921y

dxy

since dA' = 0. We thus obtain

dA = Ao

axe axa d,y ax' axa a2y 492' a22i

From (511a) (see page 292)

-

a2i r~"° = a2y a2° axa a2xa

a2xi

since r, = 0

axO axa

ax8 axa axy axe

;

from (483)

so that dAi = - A°1;y d2y

(559)


314

[SEC. 136

In general, a Riemannian space is not Euclidean. We generalize (559) and define parallelism of a vector field Ai with respect

to a curve C given by x' = x'(s) as follows: We say that Ai is parallelly displaced with respect to the Riemannian V,, along the curve C, if dAi ds

dx*

°T ds

or

aas=

=0

d-si+r,,A°ds

(560)

We say that the vector Ai suffers a parallel displacement along the curve. Notice that the intrinsic derivative of A' along the curve xi(s) vanishes. d2xi + dxa dxe = 0, In particular, for a geodesic we have ri

°s ds ds

ds2

so that the unit tangent vector

dxi

d suffers a parallel displacement

along the geodesic. Example 149. Let us consider two unit vectors Ai, Bi, which

undergo parallel displacements along a curve. We have cos 0 = gapAaB$ and

a(cos 0) as

_ gas

SAa

as

B$

aBQ

+

gaaAa

as

=0

so that 0 = constant. Hence, if two vectors of constant magnitudes undergo parallel displacements along a given curve, they are inclined at a constant angle. Two vectors at a point are said to be parallel if their corresponding components are proportional. If A' is a vector of constant magnitude, the vector B' = WA', cp = scalar, is parallel to A'. If A' is also parallel with respect to the V. along a curve t

xi = xi(s), we have Ss = 0. Now aBi

as -

0A' as

dc*

+ ds

`4s

d A'

ds

1 dip

- p ds

r

d(log y) d8

SEC. 137] FURTHER APPLICATIONS OF TENSOR ANALYSIS 315

We desire B' to be parallel with respect to the V along the curve, so that a vector B' of variable magnitude must satisfy an equation of the type a13'

B =

f(s)Bb

(561)

if it is to be parallelly displaced along the curve. Problems

1. Show that if the vector At of constant magnitude is parallelly displaced along a geodesic, it makes a constant angle with the geodesic.

2. If a vector A' satisfies (560), show that it is of constant magnitude.

3. If a vector Bi satisfies (561) along a curve r, by letting At = #Bi show that it is possible to find ,y so that At suffers a parallel displacement along IF.

4. Let xi(t), 0 < t < 1, be an infinitesimal closed path. The change in the components of a contravariant vector on being parallelly displaced along this closed path is 1 Ai = -jrra,Aa dxs, from (560). Expand Aa(x), ri (x) in Taylor series about xo = x4(0), and neglecting infinitesimals of higher order, show that

AA' = }RO,,A'

xy dx# - x0 dxy

where Rte,, is the curvature tensor (see Sees. 132, 133, 134).

137. Parallelism in a Subspace. We start with the RieV,,, ds2 = gab dxs dxs. If we consider the

mannian space, transformation

xa = xa(ul, u2,

.

. . ,

u"),

m
(562)

we see that a point with coordinates ul, u2, . . . , u"' is a point of V. and also a point of V,,. The converse is not true, for given the point with coordinates x', x2, . . . , x", there may not exist U', u2, . . . , U. which satisfy xa = xa(ul, u2, . . , ur), since

m < n. Now axa 8xs sub 8u1

ds2 = 94 dxa dxs = gab -

= h;; du' du'

- du' du;

316


[SEC. 137

so that the fundamental metric tensor in the subspace, V,,,, is given by jE;j =gap

axa M aui au1

a

Now dxa = aua dui, so that if due are the components of a contra-

variant vector in the Vm, dxa are the components of the same In general, if ai(u',

vector in the V,,.

.

.

.

,

u»1), i = 1, 2,

. . . ,

m, are the components of a contravariant vector in the V., we say that

a = 1, 2, = aui ai

Aa

. .. , n

(563)

are the components of the same vector in the V,,. Ma We now find a relationship between and as, where s is

arc length along the curve ui = ui(s) or the space curve xa = xa[ui(s)]

Differentiating (563), we have dAa ds

axa dai

au' A +

a2xa duy i a aui aui Ts

and

6Aa as

dAa + (r *-y ay ds

dxy ds

_

a2xa du' as aui ds + aui aui ds

ax,- dai

ax8 axy du1

+ (rayaauui aui ds Hence 9'a

ax' Ma auk Ss dai

_ ;k ds + ai

du' ds

ax, 9oa

auk SSsa

h`k dst

CIO axy ax' 9a«(ray>°

+ ai

ds

aui au' auk h`k(r'i)h,

+ 9Qa

a2xa ax°l aui aui auk

(see Prob. 6, See. 129)

Sr;c. 1371 FURTHER APPLICATIONS OF TENSOR ANALYSIS 317

Hence ax" SAa 9aa

_

auk as

hlk

I dal ds

dull

+

(r'i)ha ds y

J

and

ax" SAa 9°a

auk as

Sat hck

(564)

as

From (564) we see that if ai is parallelly displaced along xa[ui(s)], that is, if

as

i

= 0, then Ss = 0. Thus the theorem:

If a curve C lies in a subspace V. of V,,, and a vector field in V. is parallel along C with respect to V,,, then it is also parallel along C with respect to V.Problems 1. Prove that if a curve is a geodesic in a

it is a geodesic in

any subspace V. of V,,. Consider the unit tangents to the geodesics.

2. By considering k fixed, show that auk is a contravariant vector of V,, tangent to the uk curve, obtained by considering ul, u2,

.

uk-1, uk+',

.

.

. . .

u'" fixed in the equations

xa = xa(u'

um)

3. If ai is parallel along C with respect to the Vm, show that Ma is normal to the space V, that is, normal to the ui curves, as

i = it 2,

.

.

.

, M.

4. Under a coordinate transformation ui = ui(u', . .. , um), i = 1, 2, . . . , m, the xa remain invariant. Hence show that axa

xaa =

au',

agaa

9ad.i - ax° x.f

where the covariant derivatives are performed relative to the aai

metric hi;, that is, a1, = aa + ak(r k au'

318


5. Show that ga$(x kxa + x {x k) + xaxIxk

[SFC. 138 a

0, where

ax, = covariant differentiation is with respect to ui and h;;. 6. Show that A ; = x;ai + x°,as, for each a. 138. Generalized Covariant Differentiation. The quantities aui are contravariant vectors if we consider i fixed; for if xa = xa(ul

and ya = ya(xl,

. .

u'")

xn), a = 1, 2,

. ,

aya aui

_

.

.

.

, n, then

aya axa CIO aui

a

showing that the aui transform like a contravariant vector. axa

However, if we consider a as fixed, the -- , i = 1, 2, aui

. .. , m,

are covariant vectors in the V.; for if u' = ui(ul, .. . , u'n), axa axa and we have - = - - We propose to consider tensors of this

aui our au' type, Latin indices indicating tensors of the V., and Greek indices indicating tensors of the V.. Let us consider the tensor A7. We wish to derive a new tensor which will be a tensor in the V. for Greek indices and a tensor in

V. for Latin indices. We consider a curve C in V. given by u` = ui(s) and by xa = xa(s) in V,,. Let bi be the components of a vector field in V. parallel along C with respect to and let ca be the components of a vector field in V. parallel along C with

respect to V. We have dbi

ds dca

rrkb,

+

dd k

s

=0 (565)

dx# dsds=0

We now consider the product bicaA;. In V. this product is an invariant (scalar product) for each i, and in V. it is a scalar invariant and is a function of are length s along C. Its derivative is

SEc. 1381 FURTHER APPLICATIONS OF TENSOR ANALYSIS 319 i ds

(b°c«A;)

bica

dsi +

biAi"

ds

°

since

ds

c°A; k\

d

= bica d' + A; rag making use of (565).

+ d3

- Air,k

ds

)

Since b1 and ca are arbitrary vectors, and

d (b'c«A;) is a scalar invariant, it follows from the quotient ds

law that

_j + `4'r ,,

dA°

dx#

ds

ds

dull

(566)

- A`r' k ds

is a tensor of the same type as A7. We call it the intrinsic derivative of A; with respect to s. We may write (566) as aA°

axe

(auk + A; r«a auk

- Air;)\ duk ds k

and since this is a tensor for all directions ds (the directions

k

of C are arbitrary), it follows from the quotient law that aAuk°

Aff:k =a

axe + A;r auk - Air?k

(567)

is the generalized covariant derivative of A7 with respect to the V,,,.

Problems

a contravariant vector in V,i? 1. Why is A ", ill 2. Show that aA" Aa;:, =

- r#rA r,A,; ax* au,

ax?

aut

- rAflk

is a mixed tensor, by considering the scalar invariant b8c'daA';. a

3. Show that x = x`; = aus, and that


320

_

(SEC. 139

a2xa

44 = au' aui - r .xh + r;yx x = xai + FRxx4.

Show that

ga$xQ4 k = 0, and show by cyclic

permutations that gaaxa;x = 0. 5. The x°i of Prob. 4 arek normal to the vectors 4k, the tangent vectors to the surface. Hence the x"i are components of a vector normal to the subspace Vm. If Ni are the components of a unit normal to Vm, we must have x = b;)Na. We call B = b;i du' dui the second fundamental form. Show that b;i = gapx" NP. If

the V is a Euclidean V3, gas = as, show that b11 = e, b12 = f, b22 = g (see Sec. 35) for the subspace r = r(u, v). 139. Riemannian Curvature. Schur's Theorem. Let us consider a point P of a Riemannian space. We associate with P two independent vectors X,, A2. These vectors determine a pencil of directions at P, given by a1Al + a2A2 = aiX;

Every pair of numbers at, a2 determines a direction Ea. Since the geodesics are second-order differential equations, the point P and the direction Sa at P determine a unique geodesic. The locus of all geodesics determined in this manner will yield a surface. In a Euclidean space the surface will be a plane, since the geodesics are straight lines and two vectors determine a plane.

We now introduce normal coordinates ya with origin at P. The equations of the geodesics take the form ya = has, where ka

dya

= C ds p, and the geodesic surface is given by ya = a'8A1 + a2sX2

= ut?4 + u2A2 = uiX7

(568)

j summed from 1 to 2 and ul = a's, U2 = a2s. The element of distance on the surface is given by ds2 = h;i du' dui

(569)

and if ds2 = gab dya dyfi for the V, then aya ayo

hti

= gap au' aui = gasX A

(569a)


where the gas represent the components of the fundamental metric tensor in the system of normal coordinates. Now let R;;kc be the components of the curvature tensor for the surface S with coordinates ul, u2. Let us note the following: The gas of a Riemannian space completely determine the Christoffel symbols r;,, which in turn specify completely the RiemannChristoffel tensor Ras,.a. Once the metric of a surface embedded in a V. is determined, we can determine the F;k for this surface, and the R;;kc can then be determined. We need not make any reference to the embedding space, V,,, to determine the R:;k1 it is apparent that the ht; can be determined without leaving the surface, so that all results and formulas derived from the h;; are intrinsic properties of the surface. All we are trying to say is

that ds2 = h21 dus du' is the fundamental metric tensor for a Riemannian space which happens to be a surface embedded in a Riemannian V,,. We shall use Latin indices for the space determined by the metric h;; and Greek letters for the V,,. The indices of R;;k1 take on the values 1 or 2, and from Prob. 1, Sec. 134, we have that R1212 = R2121 = -R1221 = -R2112 = R2222 81111 = 1?1122 = R1122 = . . = 81121 = . .

(570)

=0 If we make an analytic transformation, uti = uti(ul, u2), i = 1, 2, then R;;k1(u) = Rabcd(u)

aua aub au` and anti au' auk aul

and

au!, aub auc and aul au2 aa1 au2 au1 au2 2 au1 au2 = R1212ka-lvau2

81212 = Rabcd

au2aul

(571)

by making use of (570). Thus R 1212 =81212IJ1,LL

(572)

aUa

aub

Moreover, ds2 = h;; du= du', and h;; = hab -- --- so that aut alai

322


IF = I hIJ 2.

[SEC. 139

We rewrite (572) in the form R1212

R1212

Ih

IhI

(573)

Equation (573) shows that K is an invariant, and it is called the Gaussian curvature. It is an intrinsic property of the surface. We now determine an alternative form for K in terms of the directions Ai and X and the curvature tensor for the V at the point P. The coordinate transformation between the Christoffel symbols is given by (see Prob. 6, Sec. 129) h;1rik(u) = gaprs.,(y)

ay- 8y8 ayT

au' auj auk

+ gaa

ay° 8u1 auk aui a2yP

which reduces to har3k = g.pr,,'X AgAk aya

since

aui -

''

(574)

a2ya

0, from (568).

auk auj

At the point P, rP1(y) = 0 (see Sec. 131), so that h;irik = o or himhiir;k = r, = 0. Hence the curvature tensor can be written R1212(P) = h11Rg12(P)

= h1i

[ar1(P) au2

from (538) and (541). From (574), h11r`21(u) = of Riemannian coordinates

or, hit

since

ar'22(P)i au1 J

(575)

XRX , so that at the origin

= gaN

y*

(576)

ag-0 = g,,911.,. + g,ar p = 0 at the origin (Prob. 6, See. 128), y

and from (569a) auk = X'XOXk agy = 0. Y h1i

12 = gap

ayr

"

Similarly (577)

Sec. 139] FURTHER APPLICATIONS OF TENSOR ANALYSIS 323

Using (538), (541), (576), (577), it is easy to show that R1212 = A72A1t2Rapyr

Finally, Ihl _

hll

h12

h2l

h22

hllh22 - hit

and aya c3y9

hi, = h12 h22

gap

c3u1 c3u1

= ga9Xx1

= gaAX2

so that Ihi

=

x7x2xip21'(gaagp. - ga,gpa)

(579)

Thus 7

K

aps,X aj X 2'a1 2

(580)

We are now in a position to prove Schur's theorem. If at each point of a Riemannian space, K is independent of the orientation (X "j, as), K is constant throughout the space. It follows at once if K is independent of X,, X7i that Rap,,, = K(gaagp, - ga,gpe)

and so Rapa=, = K,N(gaegp, - ga,gp,,) = K,,(gaagp, -- gaogpa) Rap,, Raaµ.e = K,,(ga,gpµ - gw.gp,)

Adding and using Bianchi's identity, (543), we have K.p(ga,gp - ga,gpa) + K.,(ga,,gsa -- ga,,gp,M) + K,n(ga,gp,, - gawgpr)

=0

Multiplying the above equation by ga° and summing, we have K, ,(ngp, - gp.) + K.,(gp - ngp,,) + K,,gp,, - K,,,gp, = 0 or

(n - 2)gp,K,,, = (n - 2)K,,gpp and gp,K,,, = gp K,,, or 5,'.K,,, =

is no arbitrary orientation.

K,,, if n > 2. For n = 2 there


324

[SEC. 140

If we choose o = r X u, K, = 0. This is true for all µ since IL can be chosen arbitrarily from I to n. Hence K = constant throughout all of space. Such a space is said to be of constant curvature. Problems 1. Derive (571). 2. Derive (575). 3. Derive (578).

4. For a V3 for which gii = 0, i 54 j, show that if h, i, j are unequal, 1

Rii = - Riaai 9hh 1

1

R, = - Raica + - Rhiih gii

s

gii

1

R=

Riiii

9ii9ii

5. If R; = gaiRii, show that R"a =

l aR 2 axi

6. If Rii = kgii (an Einstein space), show that R = giiRii = nk,

or Rii = (R/n)gii 7. Show that a space of constant curvature K is an Einstein space and that R = Kn(1 - n). 140. Lagrange's Equations. Let L be any scalar invariant function of the coordinates q', q2, . .. , qn, their time derivatives 41, 42, ... , 4n, and the time t: L = L(qi, 4i, t) = L(qi, qi, t) If we perform a transformation of coordinates, qa

a = 1, 2, the

. . .

= ga(gl, q2,

, n, then qa =

aga aqa

.

.

.

, qn)

so that 4a is a function of

Now aL 84a aL aL aqa aqi = aqa aqi + a4a aqi aL aqa aL a2ga = aqa aqi + a4a a4i

aq0

41

(581)


where we consider q' and q' as independent variables in L. Now also

aL _ aL a4 aqa aq; aq'

aL aqa aqa aqr

so that d

(582)

dt \agi1 - ala a \aq"I + aq q8 aqt Subtracting (581) from (582), we obtain d (IL) dt /

d 8L,) L dt \aqa

aqaq-

aq'

which shows that the covariant vector.

d

- (aqa

aqa

c

qla

are the components of a

For a system of particles, let L = T - V, where T is the 1 1 (ds;12 _

kinetic energy; T

:=1 2

I/

ma

2 s V (1 xI x1 x1

1

i1

2

s

2

x2, x2f x2f

s

7

xn)

is the potential function, - aV = (F*),. Then e

aL dt ax; _d

_

aL ax;

-d(magma) _ dt

n

1

8ga0 x,a2

2

ax *

_ m;

'

+ aV -.

ax;

=max;-(F,)a, ifgaQ=aae and m,x; - (FT), = 0 for la Euclidean space and Newtonian

ax vanishes in all coordinate W \-x systems. We replace the x; by any system of coordinates q', q2, . . . , qn which completely specify the configuration of the mechanics.

Hence

system of particles, and Lagrange's equations of motion are

(;) _ aL

d aL

aq=

0, r= 1 ' 2'

'n

(583)

Example 150. In spherical coordinates, a particle has the square of the velocity v2 = t2 + r262 + r2 sin2 0 rp2, so that


326

[SEC. 140

L = T - V = 2 (t2 + r262 + r2 sin2 0 02) - V

-

= m(r92 + r sing 0 #2) - aV _d

aL

dt

ar

= mr

and one of Lagrange's equations of motion is

mr - m(r62 + r sin 2 9 02) + aV = 0 c1r

Since - aV represents the radial force, the quantity r - (r62 + r sing 0 02) must be the radial acceleration. If no potential function exists, we can modify Lagrange's equa-

tions as follows: We know that T = (m; 2)gapxaO is a scalar invariant, so that

d (BT) aT Qr = dt

aT axr

(584)

are the components of a covariant vector. In cartesian coordinates, the Q, are the components of the Newtonian force, so that Q, is the generalized force vector. If f, are the components of the force vector in a y'-yL -y" coordinate system, then -

aya

Qr - fa axf and

Qr dx r =

f

a

49y

a

a

dxr = fa dya

is a scalar invariant. The reader will immediately realize that fa dya represents the differential of work, dW, so that aW Q+

= ax'

(585)

SEC. 1401 FURTHER APPLICATIONS OF TENSOR ANALYSIS 327 , xi-', We obtain Qi by allowing xi to vary, keeping x', x2, . . , x% fixed, calculate the work A Wi done by the forces, and compute

xi+',

.

d Wi Qi =..L-lim -, O AV

i not summed

Example 151. A particle slides in a frictionless tube which rotates in a horizontal plane with constant angular speed co. The only horizontal force is the reaction R of the tube on the particle.

We have T = (m/2) (j62 + r2A2), so that (584) becomes

mr" - mr02 = Q, dt (mr2o) = Qe

with Q, = 0, Qe =

R(r de) do

(586)

= rR. The solution to (586) is

r = Aew' + Be", gR = 2mw dt, since 6 = w. Problems

1. A particle slides in a frictionless tube which rotates in a vertical plane with constant angular speed w. Set up the equations of motion. 2. For a rigid body with one point fixed,

T =JA(w;+wy)+jCwa Using Eulerian angles, show that if Q, =

0, then

C(O + cos 0 +') = R

A¢sin20+Rcos0=S

A#-A,2sinocoso+R,1 sino =Qe where R, S are constants of integration.

3. If T =

q")gags, show that 2T = oL(q, aq , q^, p,,

4. Define p, = q* = gr(gl)

.

.

aT aq-

q°°.

and assuming we can solve for and show that the Hamiltonian

328


[SEC. 141

II defined by H = pages - L satisfies

II = T + V = It (a constant) where V = V (q',

.

.

,

q"),

T = aas(q',

.

show that aH

.

. ,

gn)gags.

Also

pr

a' all = g. ap'.

These are Hamilton's equations of motion; the pr are called the generalized momentum coordinates. Show that they are the components of a covariant vector. 5. By extremalizing the integral f " L(x1, x', t) dl, show that Lagrange's equations result. 6. If the action integral

e [(h

A=

f`

dxs - V )gas '-6 a aA ]

1

dX

is extremalized, show that the result yields d dt

aT

IT

aV - ax' = - ax,

Cair

where T + V = h, the constant of energy. The path of the particle is the same as the geodesic of a space having the metric ds2 = 2M(h - V)gas dxa dxs

141. Einstein's Law of Gravitation.

We look for a law of

motion, which will be independent of the coordinate system used,

describing the gravitational field of a single particle. In the special theory of relativity, the line element for the space-time coordinates is given by

ds2 = -dx2 - dy2 - dz2 + c2 dt2 -dr2 - r2 do2 - r2 sin2 0 dp2 + c2 d12

(587)

In the space of x, y, z, t, the gas are constants and the space is

SEc. 1411 FURTHER APPLICATIONS OF TENSOR ANALYSIS 329

flat (Euclidean), so that BJk: = 0. For a gravitating particle we postulate that the Ricci tensor Ri, vanish (see Probs. 5 and 6 of this section).

Since Ri; = R;;, a four-dimensional space yields

n(n + 1)/2 = 10 equations involving the gi; and their deriva1 aR , where From Prob. 7, Sec. 134, we have R ,i =

tives.

2 ax;

R = gi°R°;, R = g°pRay, and for j = 1, 2, 3, 4 the 10 equations are essentially reduced to 6 equations. We assume the line element (due to Schwarzchild) to be of the form ds2 = -e"''' dr2 - r2 d82 - r2 sin2 0 dp2 + e'(') dt2 (588)

so that our space is non-Euclidean. We do not include terms of the form dr do, etc., because we expect our space to be homogeneous and isotropic. We have 911 =

-ea,

-r2,

g22

and

g11 = -e,

g33 = -r2 sin2 0, i j gi, = 0,

-r-2'

g22

agke

ago;

(axk

+

ax'

(589)

g33 = - (r2 sin2 6)-1

gii = 0,

g44 = e-',

Now r,k - 21' 9

944 = e'

ag;k - ax°

iPd j

and since g'° = 0 for

i o e, we have rik

n

+ ax's - agi)

Caxk

i not summed

If i, j, k are different, then rk = 0. We also see that rii

_

1 agii 2 g ,i axi

9ii

rik =

gii kk

2

1

agkk

rkk = - 2 9ii axi Applying (590), we have

(590)


330

1_

r11 1

r22

,tag, _ 1da

1

2g 1

= _

_ =

or

agaa

1 g11 1

raas=

1

ag44

1

dv

or

2

dr

z

1

s

_= -r sine 0 e-A

Or

--g - _

112 = 2

gs2 ag zz

r1aa=

1

1

a9aa

22

29 Or

1

933

2

rt = -219 4

44

4

- sin 0 cos 0

00 =

an..

gsa2

(591)

r

Or

_

-re a

=

g11 a92z

2

1441

2 dr

Or

2

1

[SEC. 141

agsa

ae

r = cot 0

8g 44

1 dv

Or

2 dr

and all other r;k vanish. From (539)

ar:

R;; _

ar8,

axe - axa + r

- r°raa

so that R11 -

a ar1R

+ art, +

4 ar14

ar

Or

Or

+

rilril + risrs1 + ri3r 1

+ ri4ril - r1i(ri1 + rig + ria + r44) 1(dv12

1

1

r2

r2+2dr2+r2+r2+4`drl

1

d2D

1

1

l da 2r dr

1 da 2r dr l da dv 4 dr dr

(592)

by making use of (591). Hence Einstein's law R;; = 0 yields

_ R11

1 (_f)2

1 d2v

2 dr2

+

4 Cdr

1 A dv 4 dr dr

1A

r dr -

0


Similarly R22 = e-a C 1 -

R33 = sine 8

2 C

1= 0 dr) I-

r (dr

(ddrv

1

1+

L

da

dr I

1 (dv 2

e.-a r_ 1 dzv R44 =

_

2r

2 dr2

4

- sin2 0 = 0

1 dX dv

+ 4 dr dr

dr

(593)

1 dvl _

r drJ - 0

Dividing R44 by e'-x and adding to R11, we obtain

d +dr =o or

X + v = constant = co

We desire the form of (588), as r --f oo, to approach that of (587). co..

ev

C1

This requires that X and v approach zero as r approaches

Hence \X + v = 0 or X

P.

+ r dr) = 1. Let y =

From R22 = 0 we have

so that dr = 1 dy and 'Y r

yCl+ydr) or

dy

dr

1-y

r

and

2m

(594)

r where 2m is a constant of integration. The equations of the geodesics are d2x'

a82 +

r;k

dx' dxk

d ds = 0

which yield d20

2

CdLp)2

dr d_9

Z-2 + 2T E ds A

+

2

r33 Vs

J

=0

332


or

d20

+

d,82

a dB ds

If 0 = 2,

2 dr db

- sin 0 cos 8

r ds ds

a 2

= 0 initially, then 0 =

boundary conditions. We also obtain

(595)

0

ds

satisfies (595) and the

()2

1'11(ds/2

ds2+

(j2

[SEC. 141

ds+ r44(d- 2 =0

or

dt ds2

1 da

+ 2 dr

(dr)2 ds

-

-x

re a (thp)2

1 e v-'

+2

dv (dt)2

dr ds/

=0

(596)

=0

(597)

making use of 0 = 7r/2. Also d29

ds2

dr dcp

a 2I'19

ds ds

dt dr

d21

ds2

=

+ 2x14 ds ds -

0

d2

or

ds2

0

2drdip

+r

d2t

or

ds2

+

dsds dv dt dr dr ds ds

- 0 (598)

Integrating (597) and (598), we obtain r2

log

dt d

d(p

A

= h

(599) dt

+ v = log c

or

c

ds = y

(600)

where h and c are constants of integration. Equation (588) becomes

ds2 = - 1 dr2 - r2 d(P 2 + y dt2 y

or 1

-

1

y \ds2 - r2

2

\ds2 + y

or

(l r4 + y (r d(p)2 r2 y 1

1

z

SEc. 1411 FURTHER APPLICATIONS OF TENSOR ANALYSIS 333 or

Ch drl 2

2m

h.2

- c2 - 1 r2 dip/ + r2

h2 + 2m r

r

r2

and writing it = 1/r, we obtain

()2

au + u2 =

C

z -

h2

1

+ h2 u + 2mu8

and differentiating, we finally obtain

d + u = h2 + 3mu2 d2

(601)

2U

We obtain an approximate solution of (601) in the following manner: We first neglect the small term 3mu2 = 3m/r2, for large r.

ad z

The solution of dz + u = h2 is

T =u=-[l+eeos((p-w)j This is Newton's solu-

where e, w are constants of integration.

tion of planetary motion. We substitute this value of u in the term 3mu2, and we obtain m 3m3 6m3 W,p2+u =h2+h; + h4 ecos(cp-w) d2u

33

+ h4 [1 + 23

cos 2((p - co)]

We now neglect certain terms which yield little to our solution and obtain dz 6m'3 d2U + u = h2 + e Cos (,p - w) From the theory of differential equations the solution of our new equation is z

U = h2

+ e cos ((p - w) + h2

= h2 [1 + e cos (rp - co - e)J, where e = (3m2,'h2)cp and e2 is neglected.

sin ((p - w) approximately


334

(SEC. 141

When the planet moves through one revolution, the advance of the perihelion is given by 8(w + e) = (3m2/h2) 3 = 6irm2/h2.

When numerical results are given to the constants, it is found that the discrepancy between observed and calculated results on the advance of the perihelion of Mercury is removed.

Problems 1. Derive (591). 2. Derive (593).

3. For motion with the speed of light, ds = 0, so that from (599), h = oo, and (601) becomes s

d2u + u = 3mu2

(602)

2

Integrate d Y + u = 0, replace this value of u in 3mu2, and P

obtain an approximate solution of (602) in the form

u = co`p+ 2(coal jp+2sin2(p) where R is a constant of integration.

Since u = 1/r, x = r coa rp,

y = r sin tp, show that

m x2 +

2y2

x = R + R(x2+y2)1 The term (m/R) (x2 + 2y2)/(x2 + y2)} is the small deviation of the path of a light ray from the straight line x = R. The asymptotes

are found by taking y large compared with x. Show that they

are x = R + (m/R) (± 2y) and that the angle (in radians) between the asymptotic lines is approximately 4m/R. This is twice the predicted value, on the basis of the Newtonian theory, for the deflection of light as it passes the sun and has been verified during the total eclipse of the sun.

4. If R;; = ag;; is taken for the Einstein law, show that if y = el, then y = 1 - (2m/r) - }are and d2U

d2 + U = h2 + 3mu2 - 3 h2 us 5. Assume the following: ds2 = gap dxa dxi, gap = 0 for a p& agaa gae

1, 8x4

= 0,

dxa ds

'

0, a = 1, 2, 3,

dx4 %:W

ds

1,

SEC. 142] FURTHER APPLICATIONS OF TENSOR ANALYSIS 335 41

_ J944+ constant

xl = x, x2 = y, xa = z, x4 = ct. Show that the equations of the

dzi geodesics reduce to Newton's law of motion die + a ' = 0, i = 1, 2, 3.

6. With the assumptions of Prob. 5 show that R44 = 0 yields Laplace's equation V24 = 0. Tensors. 142. Two-point The tensors that we have studied have been functions of one point. Let us now consider the functions ga,s(xl, x2) which depend on the coordinates of two points. We now allow independent coordinate transformations at the two points M1(x;, x;, . . . , x;), M2(x4i x2, xs, . . . , If in the new coordinate systems xl, 22 we have ga.0(

1, 22) = g, (xl, x2)

8xi 8xg2

8-i -

(603)

then the go.,A are the components of a two-point tensor, a covariant

vector relative to Ml and a covariant vector relative to M2. Indices preceding the comma refer to the point M1, indices If we keep the coordinates of M2 fixed, that is, if 2_ = x', then (603) reduces to following the comma refer to M2.

xs) = gw.a(xl, x2)

dxi

8x

(604)

1

so that relative to Ml, g,,s behaves like a covariant vector. A similar remark applies at the point M2. We leave it to the reader to consider the most general type of two-point tensor fields. We could, indeed, consider a multiple tensor field depending on a finite number of points. What difficulties would one encounter for tensors depending on a countable collection of points? We may consider a two-point tensor field as special one-point tensors of a 2n-dimensional space subject to a special group of coordinate transformations. The scalar invariant ds2 = ga,s(xl, x2) dxi dxs

(605)

is an immediate generalization of the Riemann line element. Indeed, when x, and x2 coincide, we obtain the Riemann line


336

element.

sdt-

[SEC. 142

Assuming ds2 > 0 for a < t < 0, we can extremalize

fd

fB(ga'odx" dx, dt

dt

(606)

dt

and obtain a system of differential equations. z:

1

+

ra+.o,x-°ze1 1

x$ + r:

+

C..'6x1 axe2= 0 a,

xax2 + CO,

(607)

10=0

where

rye, =g ''° a x ' a9°.e

ax`

r,"# = g°'' axe

_ a9°.,) C`

axiz

=

9"'

;

a9µ..

_

ax1°

ag°. (608) axµ IF

dx;

9"'`9µ.i = a;7r

x1 = ds

The unique solutions of (606), xi(s), x4(s), subject to the initial conditions x1(8o) = ao, X '(so) _ Oo, 1 '(so) = a', xs(so) = o;, are called dyodesice, or dyopaths.

Problems 1. Derive (607). 2. Show that the CQ,#(x1, X2) are the components of a two-point

tensor, a mixed tensor relative to M1, and a covariant vector relative to M2. 3. Show that the law of transformation for the linear connec-

tion ri, is (2 1, x2) = r°";,(xl, x2)

4. Show that rip, = r

,

axi ax; '9X' azi awl ax;

a2xi

axi

+ ax; awl axi

if and only if g.,p =

axa,

where

1

,p,g(x x2) is a scalar relative to M1 and a covariant vector relative to M2. If also r: = r' , show that of necessity

a2 9°`'0

= axi axQ

where 4, is a scalar relative to both M1 and M2.

SEc. 142] FURTHER APPLICATIONS OF TENSOR ANALYSIS 337

5. If ds2 = -el, dr, dr2 - r,r2eM dpi

e=

-

[1mM2 (m + L

el=(1+rl ex

=

er dt, dt2

- (mm2M ' + r' M)2 r2

1 1 11

M)2 rl J

C1+m/

e2µ-r

show that the two-point tensors

ax ar:'

1

a

2

+

ro,

a,c

9x°

ax-1

T.a . a,T

2

vanish identically (m, M are constants). Show that the dyodesics satisfy rir2 ds1 = he-,,

= he-a

rir2 dt1

ds

__

Cie_

dt2

ds = C2e d2v d,P2

+

Mm

v L

1 + [1

+ (m/M))'hi

M 11 + (m/M)]2h2l

+3Mv2r1+(mMM)J 2

L

provided that Mr2 = mr1, v = 1/r1, hi = (M/m)h. For m << M, d M Mm/hi << 1, we have +v + 3Mv2, the Einstein solution for the motion of an infinitesimal particle moving in the field of a point gravitational mass M.

REFERENCES Brand, L. "Vectorial Mechanics," John Wiley & Sons, Inc., New York. 1930. Brillouin, L. "Les Tenseurs," Dover Publications, New York 1946.

Graustein, W. C. "Differential Geometry," The Macmillan Company, New York. 1935. Houston, W. V. "Principles of Mathematical Physics," McGraw-Hill Book Company, New York. 1934. Joos, G. "Theoretical Physics," G. E. Stechert & Company, New York. 1934. Kellogg, O. D. "Foundations of Potential Theory," John Murray, London. 1929.

McConnell, A. J. "Applications of the Absolute Differential Calculus," Blackie & Son Ltd., Glasgow. 1931. Michal, A. D. "Matrix and Tensor Calculus," John Wiley & Sons, Inc., New York. 1947. Milne-Thomson, L. M. "Theoretical Hydrodynamics," The Macmillan Company, New York. 1938.

Page, L. "Introduction to Theoretical Physics," D. Van Nostrand Company, New York. 1935. Phillips, H. B. "Vector Analysis," John Wiley & Sons, Inc., New York.

1933.

Smythe, W. R. "Static and Dynamic Electricity," McGrawHill Book Company, New York. 1939. Thomas, T. Y. "Differential Invariants of Generalized Spaces," Cambridge University Press, London. 1934. Tolman, R. C. "Relativity Thermodynamics and Cosmology," Oxford University Press, New York. 1934. Veblen, O. "Invariants of Quadratic Differential Forms," Cambridge University Press, London. 1933.

Weatherburn, C. E. "Elementary Vector Analysis," George Bell & Sons, Ltd., London. 1921. "Advanced Vector Analysis," George Bell & Sons, Ltd., London, 1944. 339


3.10

"Differential Geometry," Cambridge University Press, London.

1927.

"Riemannian Geometry," Cambridge University Press, London. 1942. Wilson, W. "Theoretical Physics," vols. I, II, III, Methuen & Co., Ltd, London. 1931, 1933, 1940.

INDEX A

Acceleration, angular, 186 centripetal, 30-31, 184 Coriolis, 210-211 linear, 30, 184, 210-211 Action integral, 328 Addition, of tensors, 275 of vectors, 2

Angular momentum, 196-200 Angular velocity, 22 Are length, 71, 100 Archimedean ordering postulate, 92 Arcs, 98 rectifiable, 98, 100 regular, 98 Arithmetic n-space, 268-269 Associated vector, 281 Associative law of vector addition, 3 Asymptotic directions, 81 Asymptotic lines, 81 Average curvature, 78 B

Bernoulli's theorem, 234-235 Bianchi's identity, 308 Binormal, 58, 312 Biot-Savart law, 163 Boundary point, 90 Boundary of a set, 91 Bounded set, 89 Bounded variation, 99 C

Calculus of variations, 83-85

Cartesian coordinate system, 280, 292

Cauchy-Riemann equations, 122 341

Cauchy's criterion for sequences, 97 Cauchy's inequality, 13 Center of mass, 194 Centripetal acceleration, 30-31, 184 Ceva's theorem, 8 Characteristic curves, 69 Charges, 127 moving, 146 Christoffel symbols, 289-293 law of transformation of, 290-291 Circulation, 238 Closed interval, 89 Closed set, 91 Commutative law of vector addition, 3 Complement of a set, 90 Components of a tensor, 274 Components of a vector, 8, 270-271 Conductivity, 162 Conductor, 128 field in neighborhood of, 130-131 force on the surface of, 131 Conformal space, 294 Conjugate directions, 80 Conjugate functions, 122, 143 Connected region, 102-103 Conservation of electric charge, 162 Conservative field, 103 Continuity, 95 equation of, 231-232 uniform, 95 Contraction of a tensor, 275-276 Contravariant tensor, 274-275 Contravariant vector, 270-272 Coordinate curves, 52 Coordinate system, 9, 269 Coordinates, geodesic, 303-305 Riemannian, 305 transformation of, 269 Coriolis acceleration, 210-211

342


Couple, 197 Covariant, differentiation, 295-296 generalized differentiation of, 318319

Direction cosines, 13 Directional derivative, 38, 297

Discontinuities of D and E, 138139

Covariant tensor, 274-275 Covariant vector, 273 Curl, 45, 55, 297, 300 of a gradient, 46, 297 Currents, displacement, 168 electric, 161-162 Curvature, average, 78 of a curve, 58, 311 Gaussian, 78, 322 lines of, 78 Riemannian, 307, 320-323 tensor, 306-307 Curve (see Space curve) Curvilinear coordinates, 50, 70 curl, divergence, gradient, Laplacian in, 54-55 D D'Alembertian, 178 Del (v), 40 Deflection of light, 334 Deformation tensor, 246 Desargues's theorem, 7 Derivative, covariant, 295-296 intrinsic, 297-298 of a vector, 29 Determinants, 263-267 cofactors of, 264 derivative of, 266 multiplication of, 264 Developable surfaces, 70 Diameter of a set, 93 Dielectrics, 135-136 Differentiation, covariant, 295-296 generalized covariant, 318-319

rules, 32 of vectors, 29 Dipole, 157-158 energy of, 158 field of, 158 magnetic, 160-161 moment of, 157 potential of, 157

Displacement current, 168 Displacement vector, 136 Distributive law, 3, 11, 21 Divergence, 42, 54, 120, 297-298 of a curl, 46 of a gradient, 44 Divergence theorem of Gauss, 114120, 299

Dot, or scalar, product, 10 Dynamics of a particle, 189 Dynamics of a system of particles, 194

Dyodesics, 336

E Edge of regression, 69-70 Einstein, Albert, law of gravitation, 328-329 space, 324

special theory of relativity, 283286

summation notation, 259

Einstein-Lorentz transformations, 283

Electric field, 127 discontinuity of, 138 polarization of, 158-159 Electromagnetic wave equations, 170-173 Electrostatic dipoles, 157-158 Electrostatic energy, 136-138 Electrostatic field, 127 Electrostatic flux, 128 Electrostatic forces, 127 Electrostatic intensity, 127 Electrostatic polarization, 158 Electrostatic potential, 128 Electrostatic unit of charge, 127 Electrostatics, Gauss's law of, 128

Green's reciprocity theorem of, 139-140 Ellipsoid of inertia, 226 of strain, 245

INDEX Energy, equation for a fluid, 235-236 of electromagnetic field, 175 of electrostatic field, 136-138 kinetic, 201 Envelopes, 69 Equation, of continuity, 231-232 of gauge invariance, 177

of motion for a fluid, 233-236, 302-303 Equipotential surfaces, 129 Euclidean space, 8, 279, 308-309 Euler's angular coordinates, 219-221 equation of motion, for a fluid, 233, 302-303

for a rigid body, 216-217 Euler-Lagrange equation, 85 Evolutes, 66

343

Gauss, divergence theorem of, 114120

electrostatic law of, 128 Generalized force vector, 326 momentum, 328 Geodesic coordinates, 303-305 Geodesics, 83, 288-289 minimal, 294 Gradient, 36, 120, 273, 297 Gravitation, Einstein's law of, 328329

Newton's law of, 190 Green's formula, 118, 299-300

Green's reciprocity theorem, 139140

Gyroscope, motion of, 222-225

H

F Faraday's law of induction, 167 Field, 9 conservative vector, 103 nonconservative vector, 104 solenoidal vector, 117 steady, 9 uniform, 10 Fluid, 230 general motion of, 236-238 Force moment, 196-200 Foucault pendulum, 213-215 Frenet-Serret formulas, 60, 311-312 Functions of bounded variation, 99100

conjugate, 122 continuous, 95 properties of, 96 harmonic, 123 Fundamental forms, first, 71 second, 74-75 Fundamental planes, 62-63 normal, 62 osculating, 62 rectifying, 63 G

Gauge invariance, 177 Gauss, curvature, 78, 322

Hamilton's equations of motion, 328 Harmonic conjugates, 123 functions, 123, 143 Heine-Borel theorem, 94 Helix, 60 Hooke's law, 249 Hypersurfaces, 282 I

Images, method of, 141-143 Induction, law of, 167 Inertia, moment of, 216 product of, 216 tensor, 225-228 Inertial frame, 211 Infemum, 92 Inhomogeneous wave equation, 177 solution of, 178-182 Insulator, 129 Integral, line, 101, 103-105 Riemann, 101 Integrating factor, 111

Integration, of Laplace's equation, 145

of Poisson's equation, 155 Interior point, 90 Interval, closed, 89 open, 89


344

Intrinsic equations of a curve, 63 Invariant, 271 Involutes, 64 Irrotational motion, 232 Irrotational vectors, 107, 111

J Jacobian, 49, 265 Jordan curves, 98

K Kelvin's theorem, 239

Kepler's laws of planetary motion, 191-193

Kinematics, of a particle, 184 of a rigid body, 204-207 Kinetic energy, 201 Kirchhoff's solution of the inhomogeneous wave equation, 178-182 Kronecker delta, 260-262 L

Lagrange's equations, 324-327 Laplace's equation, 123 integration of, 145 solution in spherical coordinates, 146-149

uniqueness theorem, 119 Laplacian, 45, 298-299 in cylindrical coordinates, 55 in spherical coordinates, 56, 299 Law of induction, 167 of refraction, 139 Legendre polynomials, 148-149 Legendre's equation, 148 Limit point, 90 Line, of curvature, 78 element, 279 of Schwarzchild, 329 of force, 132 integral, 101, 103-105 Linear function, 3 set, 89

Liquids, general motion of, 233-234 Lorentz's electron theory, 175-177 transformations, 61, 283

M

Magnetic dipole, 160-161 effect of currents, 162-164 Magnetostatics, 160 Mass of a particle, 189, 285 Maxwell's equations, 167-169

for a homogeneous conducting medium, 173 solution of, 169-173 Menelaus' theorem, 8 Meusnier's theorem, 75 Minkowski force, 285 Moment of inertia, 216 Momentum, 196 angular, 196-200 generalized, 328 relative angular, 199-200 Motion, in a plane, 33 irrotational, 234 relative, 187-188 steady, 234 vortex, 238-239 Moving charges, 161-162

Mutual induction of two circuits, 165-166

N Navier-Stokes equation, 255-257 Neighborhood, 90 Newton's law of gravitation, 190 Newton's law of motion, 189, 211 Nonconservative field, 104 Normal acceleration, 184 plane, 62 to a space curve, 58, 311 to a surface, 73, 109 Number triples, 15, 268-269 O

Oersted, magnetic effect of currents, 162-165

Ohm's law, 162 Open interval, 89 Open set, 90 Orthogonal transformation 292-293 Osculating plane, 62

345

INDEX

P Parallel displacement, 313-315 Parallelism in a subspace, 315-317 Parametric lines or curves, 71 Particle, acceleration of, 30, 210 angular momentum of, 196 dynamics of, 189 kinematics of, 184 mass of, 189, 285 momentum of, 196 Newton's laws of motion for, 189, 211

rotation of, 22 velocity of, 30, 184, 209 Particles, system of, 194 Perihelion of Mercury, 334 Permeability, 160 Planetary motion, 190-193 Point, 89 boundary, 90 interior, 90 limit, 90 neighborhood of, 90 set theory, 89 Poisson's equation, 132-134 integration of, 155 Poisson's ratio, 249 Polarization, 158-159 Potential, of a dipole, 157 electrostatic, 128 vector, 117 velocity, 232 Power, 162 Poynting's theorem, 174-175 Poynting's vector, 175 Pressure, 230 Principal directions, 77-78 Q

Quadratic differential form, 280 Quotient law of tensors, 276

R Radius of curvature, 24

Recapitulation of differentiation formulas, 48 Reciprocal tensors, 281

Rectifying plane, 63 Refraction, law of, 139 Regions, connected, 102 simply connected, 102-103 Regular arcs, 98 Relative motion, 187-188 time rate of change of vectors, 208 Resistance, electric, 162 Retarded potentials, 178-182 Ricci tensor, 307 Riemann integral, 101 Riemannian, coordinates, 305 curvature, 307, 320-323 metric, 280 space, 280

geodesics in, 288-289 hypersurface in, 282 Riemann-Christoffel tensor, 307-308 Rigid bodies, 203 motion of, 215-225 S

Scalar, 1 curvature, 307 gradient of, 36, 273, 297 Laplacian of, 45, 298-299

product of vectors, 10, 273-274 Schur's theorem, 323-324 Schwarzchild line element, 329 Second fundamental form, 74-75 geometrical significance, 75 Sequence, 97 Cauchy criterion for convergence of, 97 Set, 89 bounded, 89 closed, 91 complement of, 90 countable, 93 diameter of, 93 infemum of, 92 limit point of, 90 linear, 89 open, 90 supremum of, 91 theorem of nested, 93 Simply connected region, 102

346


Sink, 231 Solenoidal vector, 117 Solid angle, 160 Source, 231 Space, conformal, 294 Space curve, 31

Tensor, components of, 274 contraction of, 275-276 contravariant, 274-275 covariant, 274-275 curvature, 306-307 deformation, 246 inertia, 225-228 mixed, 275 Ricci, 307 Riemann-Christoffel, 307-308 strain, 243-246 stress, 246-248 two-point, 335 weight of, 275 Tensors, 274-278 absolute, 275 addition of, 275 cross product of, 278 outer product of, 278 product of, 275 quotient law of, 276 reciprocal, 281 relative, 275 Theorem, of Ceva, 8 of Desargue, 7 of Menelaus, 8 Top, motion of, 222-225 Torque, 196-200 Torsion of a space curve, 59, 312 Transformation of coordinates, 269 Trihedral, 59 Triple scalar product, 23 Triple vector product, 24 Two-point tensors, 335

arc length of, 100-101 curvature of, 58, 311 intrinsic equations of, 63 Jordan, 98 on a surface, 72 radius of curvature of, 58 tangent to, 31, 58, 311 torsion of, 59, 312 unit binormal of, 59, 312 unit principal normal of, 58, 311 Space of n-dimensions, 268-269 Special theory of relativity, 283-286 Spherical coordinates, 35, 50 indicatrix, 68 Steady field, 9 Stokes's theorem, 107-112, 300-301 Strain, ellipsoid, 245 tensor, 243-246 Streamline, 234 Stress tensor, 246-248 Subtraction of vectors, 3 Summation convention, 259 Superscripts, 14, 259 Supremum, 91-92 Surface, 70 are length on, 71 asymptotic curves on, 81 average curvature of, 78 conjugate directions on, 80 U curves on, 72 developable, 70 Uniform continuity, 95 first fundamental form of, 71 Uniform vector field, 10 Gauss curvature of, 78 Uniqueness theorems, 119 geodescis of, 83 Unit charge, 127 normal to, 73, 109 principal directions on, 77 V second fundamental form of, 74 Vector, associated, 281 T basis, 3, 8 center of mass, 194 Tangent to a space curve, 31, 58, 311 components of, 8-9 Tangential acceleration, 184 conservative field, 108

INDEX Vector, contravariant, 270-272 covariant, 273 curl of, 45, 55, 297, 300 definition of, 1 differentiation of, 29 displacement, 136 divergence of, 42, 54, 120, 297-298 field, 9

irrotational, 107, 111 length of, 1, 281 operator del (v), 40 physical components of, 272 potential, 117 solenoidal, 117 space, 268-269 sum of a solenoidal and an irrotational vector, 156-157 unit, 1, 281 zero, 1 Vectors, addition of, 2, 275 angle between two, 10, 281-282 differentiation of, 29 equality of, 1 fundamental unit, 8 linear combination of, 3

347

Vectors, parallel, 2, 314 parallel displacement of, 313-315 scalar, or dot, product of, 10, 273274

subtraction of, 3 triple scalar product of, 23-24 triple vector product of, 24-25 vector, or cross, product of, 20-23 Velocity, angular, 22-23 linear, 30, 184, 209 potential, 232 Vortex motion, 238-239 W

Waves, equation of, 170 inhomogeneous equation of, 177 longitudinal, 253-254 transverse, 172, 253 IN'eierstrass-Bolzano theorem, 92 Work, 103, 202

Y Young's modulus, 249

Vector and Tensor Analysis 1950

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