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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341
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CHAPTER 1 THE ALGEBRA OF VECTORS 1. Definition of a Vector. Our starting point for the definition of a vector will be the intuitive one encountered in elementary physics. Any directed line segment will be called a vector. The length of the vector will be denoted by the word magnitude. Any physical element that has magnitude and direction, and hence can be represented by a vector, will also be designated as a vector. In Chap. 8 we will give a more mathematically rigorous definition of a vector. Elementary examples of vectors are displacements, velocities,
forces, accelerations, etc. Physical concepts, such as speed, temperature, distance, and specific gravity, and arithmetic numbers, such as 2, i, etc., are called scalars to distinguish them from vectors. We note that no direction is associated with a scalar. We shall represent vectors by arrows and use boldface type to Ca-, a] a, a, a, indicate that we are speaking of a vector. In order to distina guish between scalars and vec-
tors, the student will have to
Fia. 1.
adopt some notation for describing a vector in writing. The student may choose his mode of representing a vector from Fig. 1 or may adopt his own notation. To every vector will be associated a real nonnegative number equal to the length of the vector. This number will depend, of course, on the unit chosen to represent a given class of vectors. A vector of length one will be called a unit vector. If a represents the length of the vector a, we shall write a = jal. If jal = 0, we define a as the zero vector.
2. Equality of Vectors. Two vectors will be defined to be equal if, and only if, they are parallel, have the same sense of direction, and the same magnitude. The starting points of the vectors are immaterial. It is the direction and magnitude which 1
VECTOR AND TENSOR ANALYSIS
2
[SEC. 3
are important. Equal vectors, however, may produce different physical effects, as will be seen later. We write a = b if the vectors are equal (see Fig. 2).
Fia. 2.
3. Multiplication by a Scalar.
If we multiply a vector a by a
real number x, we define the product xa to be a new vector parallel to a whose magnitude has been multiplied by the factor x.
Thus 2a will be a vector which is twice as long as the vector a
Z-a
Fm. 4.
FIG. 3.
and which has the same direction as a (see Fig. 3). We define -a as the vector obtained from a by reversing its direction (see Fig. 4).
We note that x(ya) = (xy)a = xya
(x + y)a = xa + ya Oa = 0
(zero vector)
It is immediately seen that two vectors are parallel if, and only if, one of them can be written as a scalar multiple of the other. 4. Addition of Vectors. Let us suppose we have two vectors given, say a and b. We form a third vector by constructing a triangle with a and b forming two sides of the triangle, b adjoined to a (see Fig. 5). The vector starting from the origin of a and ending at the arrow of b is defined as the vector sum a + b.
We see that a + 0 = a, and if a = b, c = d, then
a+c=b+d
THE ALGEBRA OF VECTORS
SEc. 6]
3
From Euclidean geometry we note that
a+b=b+a
(1) (2) (3)
(a+b)+c=a+(b+c) x(a+b) =xa+xb
(1) is called the commutative law of vector addition; (2) is called the associative law of vector addition; (3) is the distributive law
for multiplication by a scalar. The reader should have no trouble proving these three results geometrically.
a+b+c Fio. 5.
5. Subtraction of Vectors. Given the two vectors a and b, we can ask ourselves the following question: What vector c must be added to b to give a? The vector c is defined to be the vector
a - b. We can obtain the desired result by two methods. First, construct -b and then add this vector to a, or second, let b and a have a common origin and construct the third side of the triangle. The two possible directions will give a - b and b - a (see Fig. 6). Thus a - b = a + (-b).
a-b
a-b Flo. 6.
6. Linear Functions. Let us consider all vectors in the twodimensional Euclidean plane. We choose a basis for this system of vectors by considering any two nonparallel, nonzero vectors. Call them a and b. Any third vector c can be written as a linear
VECTOR AND TENSOR ANALYSIS
4
[SEC. 6
combination or function of a and b,
c = xa + yb
(4)
The proof of (4) is by construction (see Fig. 7). Let us now consider the following problem: Let a and b have a common origin, 0, and let c be any vector starting from 0 whose end point lies on the line joining the ends of a and b (see Fig. 8). B
a
FIG. 8.
Let C divide BA in the ratio x: y where x + y = 1. In particular, if C is the mid-point of BA, then x = y = fr. Now
c=OB+BC
= b + x(a - b)
=xa+(1 - x)b so that c = xa + yb
Now conversely, assume c = xa + yb, x + y = 1.
(5)
Then
c=xa+(1 -x)b =x(a-b)+b We now note that c is a vector that is obtained by adding to b the
vector x(a - b), this latter vector being parallel to the vector a - b. This immediately implies that the end point of c lies on the line joining A to B. We can rewrite (5) as
c-xa-yb=0
1-x-y=0
(6)
SEC. 61
THE ALGEBRA OF VECTORS
5
We have proved our first important theorem. A necessary and sufficient condition that the end points of any three vectors with common origin be on a straight line is that real constants it m, n exist such that
la+mb+nc=0
l+m+n=O
(7)
with l2+m2+n2p,, 0. We shall, however, find (5) more useful for solving problems. Example 1. Let us prove that the medians of a triangle meet at a point P which divides each median in the ratio 1:2. C
0 FIG. 9.
Let ABC be the given triangle and let A', B', C' be the midChoose 0 anywhere in space and construct the vectors from 0 to A, B, C, A', B', C', calling them a, b, c, a', b', c' (see Fig. 9). From (5) we have points.
a'=4b+ic b' = Ja + 4c
(8)
Now P (the intersection of two of the medians) lies on the line joining A and A' and on the line joining B and B'. We shall thus find it expedient to find a relationship between the four vectors a, b, a', b' associated with A, B, A', B'. From (8) we eliminate the vector c and obtain 2a' + a = 2b' + b or
*a' + '}a = *b' + b
(9)
6
VECTOR AND TENSOR ANALYSIS
[SFC. 6
But from (5), -sa' + -ia represents a vector whose origin is at 0 and whose end point lies on the line joining A to A. Similarly, ,'jb' + *b represents a vector whose origin is at 0 and whose end point lies on the line joining B to B'. There can only be one
y
vector having both these properties, and this is the vector p = OP. Hence p = tea' + is = -b' -fib. Note that P divides AA' and
BB' in the ratios 2: 1. Had we considered the median CC' in connection with AA', we would have obtained that p + J e, and this completes the proof of the theorem. Example 2. To prove that the diagonals of a parallelogram bisect each other. Let ABCD be the parallelogram and 0 any C
FIG. 10.
point in space (see Fig.
10).
The equation d - a = c - b
implies that ABCD is a parallelogram.
Hence
Ja+4c=lb+Id =p so that P bisects AC and BD. Problems
1. Interpret I a a 2. Give a geometric proof of (3). 3. a, b, c are consecutive vectors forming a triangle. What is the vector sum a + b + c? Generalize this result for any closed polygon.
4. Vectors are drawn from the center of a regular polygon to its vertices. From symmetry considerations show that the vector sum is zero.
SEc. 61
THE ALGEBRA OF VECTORS
7
5. a and bare consecutive vectors of a parallelogram. Express the diagonal vectors in terms of a and b. 6. a, b, c, d are consecutive vector sides of a quadrilateral. Show that a necessary and sufficient condition that the figure be
a parallelogram is that a + c = 0 and show that this implies
b + d = 0. 7. Show graphically that lal + lbl >_ la + bl. From this show that la - bl >_ lal - lbi. 8. a, b, c, d are vectors from 0 to A, B, C, D.
If
b-a=2(d-c) show that the intersection point of the two lines joining A and D and B and C trisects these lines. 9. a, b, c, d are four vectors with a common origin. Find a necessary and sufficient condition that their end points lie in a plane.
10. What is the vector condition that the end points of the vectors of Prob. 9 form the vertices of a parallelogram? 11. Show that the mid-points of the lines which join the midpoints of the opposite sides of a quadrilateral coincide. The four sides of the quadrilateral are not necessarily coplanar. 12. Show that the line which joins one vertex of a parallelogram to the mid-point of an opposite side trisects the diagonal. 13. A line from a vertex of a triangle trisects the opposite side. It intersects a similar line issuing from another vertex. In what ratio do these lines intersect one another? 14. A line from a vertex of a triangle bisects the opposite side. It is trisected by a similar line issuing from another vertex. How does this latter line intersect the opposite side? 15. Show that the bisectors of a triangle meet in a point. 16. Show that if two triangles in space are so situated that the three points of intersection of corresponding sides lie on a line, then the lines joining the corresponding vertices pass through a common point, and conversely. This is Desargues's theorem.
17. b = (sin t)a is a variable vector which always remains parallel to the fixed vector a. What is rically the meaning of
A.
fib?
Explain geomet-
8
VECTOR AND TENSOR ANALYSIS
[SEC. 7
18. Let v, be the velocity of A relative to B and let v2 be the velocity of B relative to C. What is the velocity of A relative to C? Of C relative to A? Are these results obvious? 19. Let a, b be constant vectors and let c be defined by the equation c = (cos t)a + (sin t)b When is c parallel to a? Parallel to b? Can c ever be parallel do d2c If a and b to a + b? Perpendicular to a + b? Find 9
dt dt2
are unit orthogonal vectors with common origin, describe the positions of c and show that c is perpendicular to
dc
20. If a and b are not parallel, show that ma + nb = ka + 3b implies m = k, n = j. 21. Theorem of Ceva. A necessary and sufficient condition that the lines which join three points, one on each side of a triangle,
to the opposite vertices be concurrent is that the product of the algebraic ratios in which the three points divide the sides be -1. 22. Theorem of Menelaus. Three points, one on each side of a
triangle ABC, are collinear if and only if the product of the algebraic ratios in which they divide the sides BC, CA, AB is unity. 7. Coordinate Systems. For a considerable portion of the text we shall deal with the Euclidean space of three dimensions. This is the ordinary space encountered by students of analytic geometry and the calculus. We choose a right-handed coordinate system. If we rotate the x axis into the y axis, a right-hand screw will advance along the positive z axis. We let i, j, k be the three unit vectors along the positive x, y, and z axes, respectively. The vectors i, j, k form a very simple and elegant basis for our three-dimensional Euclidean space. From Fig. 11 we observe that
r=xi+yj+zk
(10)
The numbers x, y, z are called the components of the vector r. Note that they represent the projections of the vector r on the x, y, and z axes. r is called the position vector of the point P
SEC. 71
THE ALGEBRA OF VECTORS
9
and will be used quite frequently in what follows. The most general space-time vector that we shall encounter will be of the form
u =u(x,y,z,t) =a(x,y,z,t)1+#(x,y,z,t)j + y(x, y, z, t)k
(11)
It is of the utmost importance that the student understand the meaning of (11). To be more specific, let us consider a fluid in motion. At any time t the particle which happens to be at the z
Fm. 11.
point P(x, y, z) will have a velocity which depends on the coordinates x, y, z and on the time t. As time goes on, various particles arrive at P(x, y, z) and have the velocities u(x, y, z, t) with components along the x, y, z axes given by a(x, y, z, t), Xx, y) z, t), y(x) y, z, t).
Whenever we have a vector of the type (11), we say that we An elementary example would be the vector u = yi - xj. This vector field is time-independent and so is have a vector field.
called a steady field.
At the point P(1, -2, 3) it has the value
- 2i - j. Another example would be u = 3xzeq - xyztj + 5xk. We shall have more to say about this type of vector in later
VECTOR AND TENSOR ANALYSIS
10
[SEC. 8
chapters and will, for the present, be interested only in constant vectors (uniform fields).
A moment's reflection shows that if a = ali + a2j + ask, b = b1i + ba + b3k, then a + b = (ai + b1)i + (a2 + b2)j + (as + b3)k (12) xa + yb = (xai + yb1)i + (xaz + ybz)j + (xaa + yb3)k 8. Scalar, or Dot, Product. We define the scalar or dot product of two vectors by the identity a b = JaJJbJ cos 0
(13)
where 0 is the angle between the two vectors when drawn from a common origin. It makes no difference whether we choose 0 or - 0 since cos 0 = cos (- 0). This definition of the scalar product
arose in physics and will play a dominant role in the development of the text.
Fm. 12.
From (13) we at once verify that
a a
ab
then
the
b
b=
a a
a
b
b
ab
= (proj a)b (bI = (proj b)a (al
(17)
SEC. 8J
THE ALGEBRA OF VECTORS
11
With this in mind we proceed to prove the distributive law, which
states that (18)
From Fig. 13 it is apparent that
[proj (b+c)]aIaF f
I
= (proj b)a Iai + (proj C)a Ial
Fm. 13.
since the projection of the sum is the sum of the projections. Let the reader now prove that
Example 3. To prove that the median to the base of an isosceles triangle is perpendicular to the base (see Fig. 14). From (5) we see that
m=1
1
so that
0 FIG. 14.
which proves that OM is perpendicular to AB. Example 4. To prove that an angle inscribed in a semicircle is a right angle (see Fig. 15).
_j.k:g=o
1'1
_
C2
co$B
2ab
-b2+a2_.
6
Example
that
$o
.]a
PIG.
cb
_a)
(b
a -a Ab
b2
_
C
C.
.
T'`iq°n°metr
of
Lau,
Cone
&
xample
E
18.
I°
j'i
c,a
c
z a
0
angle
aright
4$CAis
that
so
0
_
AC
BC
-,
$
+c
s
[SEc.
BC AND
c a
'
yAC
YECr?,oR
12
THE ALGEBRA OF VECTORS
SEC. 8]
13
Hence if a = a1i + a2j + ask, b = b1i + b2j + b3k, then
a b = albs + a2b2 + a,b3 Formula (19) is of the utmost importance. Example 7. Cauchy's Inequality
(19)
Notice that 0 a = 0.
(a - b)(a - b) _ 1a121b12 cost 0 5 1a12Ib12
so that from (19) (albs + aab2 + aab3)2 S (a12 + a22 + as2)(b12 + b22 + b 32)
In general n
n
n
I aaba s (Z aa2)1(I ba2)1
a-1
a-1
a-1
(20)
Let i' be a unit vector making angles a, fl, 7 with the x, y, z axes. The projections of i' on the x, y, z axes are cos a, cos ft, cos -y, so that Example 8.
i' = cosai+cos0 j+cos7k
=pli+qlj+rlk
(21)
pi, q1, ri are called the direction cosines of the vector Y. Similarly, let j' and k' be unit vectors with direction cosines p2, qa, r2 and pa, qa, r,. Thus Notice that p12 + q12 + r12 = 1.
j' = p2i + q2j + r2k
k' = pai+qsj+rak
()
We also impose the condition that i', j', k' be mutually orthogonal, so that the x', y', z' axes form a coordinate system similar to the x-y-z coordinate system with common origin 0 (see Fig. 17).
We have r = r' so that xi + yj + zk = x'i' + y'j' + z'k', where x, y, z are the coordinates of a point P as measured in the x-y-z coordinate system and x', y', z' are the coordinates of the
same point P as measured in the x'-y'-z' coordinate system. Making use of (21) and (22) and equating components, we find
that
x = p1x' + ply' + pgz' y = q1x' + qty' + qaz' z = r1x' + r2y' + raz'
(23)
14
VECTOR AND TENSOR ANALYSIS
1Src. 8
We now find it more convenient to rename the x-y-z coordinate system. Let x = x1, y = x2, z = x3, where the superscripts do not designate powers but are just labels which enable us to differentiate between the various axes. Similarly, let x' = x1, y' = x2'
y
FIG. 17.
Z' = 28.
Now let a,a represent the cosine of the angle between the xa and V axes. We can write (23) as 3
xa = I a0a.V,
a = 1, 2, 3
(24)
a-1
By making use of the fact that i' j' = j' k' = k' i' = 0, we can prove that 3
xa = I Asax#, 0-1
a = 1, 2, 3
(25)
where Asa = a). We leave this as an exercise for the reader. Let us notice that differentiating (24) yields axa
- = a,a,
of, or = 1, 2, 3
(26)
SEC. 91
THE ALGEBRA OF VECTORS
15
Example 9. The vector a = a'i + a2j + a'k may be represented by the number triple (a', a2, a'). Hence, without appealing to geometry we could develop an algebraic theory of vectors.
If b = (b', b2, b3), then a + b is defined by the number triple (a' + b', a2 + b2, a3 + b3), and xa = x(al, a2, a') is defined by the number triple (xa', xa2, xa3). From this the reader can prove
that (a', a2, a3) = a'(1, 0, 0) + a2(0, 1, 0) + a3(0, 0, 1)
The triples (1, 0, 0), (0, 1, 0), (0, 0, 1) form a basis for our linear vector space, that is, the space of number triples. We note that the determinant formed from these triples, namely, 1
0
0
1
00
0
0 =1 1
does not vanish. Any three triples whose determinant does not vanish can be used to form a basis. Let the reader prove this result. We can define the scalar product (inner product) of two triples by the law (a - b) = a'b' + a2b2 + a3b3. 9. Applications of the Scalar Product to Space Geometry (a) We define a plane as the locus of lines passing through a
fixed point perpendicular to a fixed direction. Let the fixed point be Po(xo, yo, zo) and let the fixed direction be given by the vector N = Ai + Bj + Ck. Let r be the position vector to any -4
point P(x, y, z) on the plane (Fig. 18). Now POP = r - ro is perpendicular to N so that or
[(x - xo)i + (y - yo)j + (z - zo)kl - (Ai + Bj + Ck) = 0 and
A(x - xo) + B(y - yo) + C(z - zo) = 0
(27)
This is the equation of the plane. The point Po(xo, yo, zo) obviously lies in the plane since its coordinates satisfy (27). Equation (27) is linear in x, y, Z. (b) Consider the surface Ax + By + Cz + D = 0. Let P(xo, yo, zo) be any point on the surface. Of necessity,
Axo+Byo+Czo+D=0.
16
VECTOR AND TENSOR ANALYSIS
[SEC.9
Subtracting we have
A(x - xo) + B(y - yo) + C(z - zo) = 0
(28)
Now consider the two vectors Ai + Bj + Ck and
(x - xo)i + (y - yo)j + (z - zo)k
Fia. 18.
Equation (28) shows that these two vectors are perpendicular. Hence the constant vector Ai + Bj + Ck is normal to the surface at every point so that the surface is a plane.
Fia. 19.
(c) Distance from a point to a plane. Let the equation of the plane be Ax + By + Cz + D = 0, and let P(E, % r) be any point in space. We wish to determine the shortest distance from P
THE ALGEBRA OF VECTORS
SEC. 91
17
to the plane. Choose any point Po lying in the plane. It is apparent that the shortest distance will be the projection of
PoP on N, where N is a unit vector normal to the plane (see Fig. 19).
Now
d = JPoP NJ =
IAZ+B, +Cr+DI
(29)
(A2 + B2 + C2)f
where use has been made of the
fact that
Axo+Byo+Czo+D=O.
FIG. 20.
(d) Equation of a straight line through the point Po(xo, yo, zo)
parallel to the vector T = li + mj + A. From Fig. 20 it is z
Fia. 21.
VECTOR AND TENSOR ANALYSIS
18
[SEC. 9
apparent that r - ro is parallel to T so that r - ro = AT,
--
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).
THE ALGEBRA OF VECTORS
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-
VECTOR AND TENSOR ANALYSIS
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.
VECTOR AND TENSOR ANALYSIS
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).
THE ALGEBRA OF VECTORS
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
VECTOR AND TENSOR ANALYSIS
[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
THE ALGEBRA OF VECTORS
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.
VECTOR AND TENSOR ANALYSIS
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.
THE ALGEBRA OF VECTORS
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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
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).
VECTOR AND TENSOR ANALYSIS
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
DIFFERENTIAL VECTOR CALCULUS
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
VECTOR AND TENSOR ANALYSIS
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-
DIFFERENTIAL VECTOR CALCULUS
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
VECTOR AND TENSOR ANALYSIS
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)
DIFFERENTIAL VECTOR CALCULUS
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
VECTOR AND TENSOR ANALYSIS
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
DIFFERENTIAL VECTOR CALCULUS
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
VECTOR AND TENSOR ANALYSIS
[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
DIFFERENTIAL VECTOR CALCULUS
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
VECTOR AND TENSOR ANALYSIS
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)
DIFFERENTIAL VECTOR CALCULUS
43
The loss of mass per unit time is thus seen to be equal to a(pv)
ay
dx dy dz
If we also take into consideration the other two faces, we find that the total loss of mass per unit time is
[a(Pu)+o(Pv)+a(Pw)]dddz z
A
dy
E
ry
Fia. 35.
so tha t a(Pu) ax
+
a(pv)
ay
+
(66)
a(Pw) az
represents the loss of mass per unit time per unit volume. This quantity is called the divergence of the vector pf. We see at once that VV
. (pf )
= di v (pf) =
a(Pu) ax
a(Pv)
+
ay
+
a(Pw) az
= 1 dM V dt
(67 )
VECTOR AND TENSOR ANALYSIS
44
[SEc. 20
since i, j, k are constant vectors. M and V are the mass and volume of the fluid.
The divergence of any vector f is defined as V f. We now calculate the divergence of p(x, y, z)f. V
a(q.u)
+ a(te) +
ax
ay
_
(ax
au
av
x
+ay+
aw az
a(cyw) az
arp
a'p
sip
+ uax +v ay +u'aa
(68)
We remember this result easily enough if we consider V as a vector differential operator. Thus, when operating on spf, we first keep p fixed and let V operate on f, and then we keep f fixed and let V operate on V(V V is nonsense), and since f and Vp are
vectors we complete their multiplication by taking their dot product. Example 25.
Compute V f if f = r/r' (inverse-square force).
V (r-'r) = r-3V r+r Vr-'
V (r- 3r) = 0
(69)
This is an important result. The divergence of an inverse-square force is zero. We note that y
Example 26.
What is the divergence of a gradient?
k/
axe+ayz+az,
DIFFERENTIAL VECTOR CALCULUS
SEc.21]
45
This important quantity is called the Laplacian of 0.
= v2 = axe
v
(70)
aye
az2
21. The Curl of a Vector. We postpone the physical meaning of the curl and define
curl f=vxf=
_ -J
Caw
vxf = i
ay
+
j
a
a
ax
ay
az
u
v
w
_ awl
1
az
i
a
+k
ax
az
_ au
1
ax
(71)
ay,
Example 27
=0
Vxr= Example 28 i
j
k
a
a
a
ax
ay
az
cpu
vv cv
1
v x (wf) =
[a(te) _ a()1 ay
az
JJ
[a(sou) az
a(caw)
ax
+ k [a(;)
a(,Pu)]
ay
ax v x (cOf) = cO
aw
[i (ay
av
az
+j
au
Ow
az
ax
U
Vx('vf)=(Pvxf+Via xf
v
w
(72)
46
VECTOR AND TENSOR ANALYSIS
[SEC. 21
This result is easily obtained by considering V as a vector differential operator. Example 29. To show that the curl of a gradient is zero. i a V X (VP) = ax
j
k
a ay
a az
i
av av av ax
ay
- az ay ay az a2
a2IP
az 02,p
a2,p
+ i
\ az ax
ax az/ + k \ax ay a2,
C1 2,p
ay ax/
Hence
Vxvio =0
(73)
provided (p has continuous second derivatives. Example 30. To show that the divergence of a curl is zero.
a au_aw'
a aw_av
V (v x f) = ax ay
az
a2u
a2ul
ay az
az ay
+ ay az a2V
ax /
a(av_au) + az \ax a2w
a2v
+ az ax
ax az
+
ay/ a2w
ax ay - ay ax
Thus (74)
Example 31. V.
What does (u V)v mean? We first dot u with
This yields the scalar differential operator a
a
a
ux ax + ua ay + U. az Then we operate on v obtaining uxax+Uyay+u$az
Thus
df=axdx+af dy+azdz y
=dxa+dya+dzaz y
DIFFERENTIAL VECTOR CALCULUS
SFC. 211
47
and
df = (dr V)f
(75)
since dr = dx i + dy j + dz k. If f = f(x, y, z, t),
df = (dr V)f + a dt
(76)
Example 32
+vsa = vJ + vyj + vZk (v - V)r = v
(77)
where r is the position vector xi + yj + zk. Example 33. Let us expand V(u v). Now u x (V x v) = Vti(u
v) - (u V)v
Here we have applied the rule of the middle factor, noting also that V operates only on v. Vn(u v) means that we keep the components of u fixed and differentiate only the components of v. v) - (v V)u. Adding, we Similarly, v x (V x u) = obtain
Vu(u v) + Vn(u v) = u x (V x v) + v
x u)
+ (u V)v + (v V)u and
u x(V xv) + v x(V xu) +
(78)
Example 34
V x (u x v) = (v V)u - v(V u) + u(V v) - (u
V)v
(79)
48
VECTOR AND TENSOR ANALYSIS
[SEc. 22
Example 35
V. (u xv) = Vu (u xv) +Vv (u xv) (V
(80)
(V
Example 36
V X (V x v) = V(V v) - V2v Let A = V x ((pi) where V2,p = 0. pute A V x A. Since A = V(p x i, we obtain Example 37.
V x A = V x V xi =
V
iV2
(81)
We now com-
a cix
from (7), Sec. 22. Thus
AVxA=
i
j
k
app
app
app
a2
ax
ay
az
ax2+j ay ax+kazax
1
0
0
app a2 = az ay ax
a2y
()2
-P
app a2g
ft azax
If also sp = X(x)Y(y)Z(z), we can immediately conclude that 22. Recapitulation. We relist the above results: 1. V (UV) = u Vv + v Vu 2. V
V
xv+V(pxv
4. V x (V(p) = 0
5. V. (Vxv)=0 6. V. (u x v) _ (V x u) - v - (V x v) - u
7. V x (u x v) _ (v V)u - v(V u) + u(V v) - (u V)v 8.
V
Vxr
=0 13. df = (dr V)f + a dt
Sxc. 22]
DIFFERENTIAL VECTOR CALCULUS
14. dco = dr VV +
49
LP dt
15. V- (r 3r) =0 Problems
1. Show that V2(1/r) = 0 where r = (x2 + y2 + z2)}. 2. Compute V2 r, V2r2, V2(1/r2) where r = (x2 + y2 + z2)3. 3. Expand V(uvw).
4. Find the divergence and curl of (xi - yj)/(x + y); of
xcoszi+ylogxj -z2k.
5. If a = axi + fiyj + yzk, show that V(a r) = 2a. 6. Show that V x V(r)r] = 0 when r = (x2 + y2 + Z2) i and
r = xi + yj + zk. 7. Let u = u(x, y, z), v = v(x, y, z). Suppose u and v satisfy an equation of the form f(u, v) = 0. Show that Vu x Vv = 0. 8. Assume Vu x Vv = 0 and assume that we move on the surface u (x, y, z) = constant. Show that v remains constant and hence v = f(u) or F(u, v) = 0. 9. Prove that a necessary and sufficient condition that u, v, w satisfy an equation f(u, v, w) = 0 is that Vu Vv x Vw = 0, or au
au
au
ax
ay
az
av
ft
av
ax
ay
az
0
aw aw aw ax
ay
az
This determinant is called the Jacobian of (u, v, w) with respect to (x, y, z), written J[(u, v, w)/(x, y, z)].
10. If w is a constant vector, prove that V x (w x r) = 2w,
where r=xi+yJ+zk.
11. If pf = Vp, prove that f V x f = 0. 12. Prove that (v V)v = 4 Vv2 - v x (V x v). 13. If A is a constant unit vector, show that V x (v xA)] = 14. If fl, f2, f3 are the components of the vector f in one set of
rectangular axes and 11, /, / are the components of f after a
50
VECTOR AND TENSOR ANALYSIS
[SEC. 23
rotation of axes (see Example 8), show that a fa
a=1
xa
aa_
=
a=1 x
so that V f is a scalar invariant under a rotation of axes.
Also
see Prob. 21, Sec. 9. 15. Prove (79), (80), (81). 16. Let f = f1i + f2j + f3k and consider nine quantities afi axi
9is
of, axis
i, j = 1, 2, 3
Show that gi; = -gi and that three of the nine quantities yield the three components of V x f. Use this result to show that
V x((pf) =('V xf+Vcpxf. 23. Curvilinear Coordinates. Often the mathematician, physicist, or engineer finds it convenient to use a coordinate system other than the familiar rectangular cartesian coordinate system. If he is dealing with spheres, he will probably find it expedient to describe the position of a point in space by the spherical coordinates r, 8, sp (see Fig. 31). Let us note the following: The sphere x2 + y2 + z2 = r2, the cone z/(x2 + y2 + z2)'1 = cos 8, and the plane y/x = tan sp pass through the point P(r, 0, Sp). We may consider the transformations
r = (x2+y2+Z2)} a = cos- ' Sp =
z
(x2 + y2 + z2)
tan-1 y
X
as a change of coordinates from the x-y-z coordinate system to the r-B-#p coordinate system. The surfaces r = (x2 + y2 + z2)} = cl 8 = cos-' [z/(x2 + y2 + z2)}1 = c2, O = tan-' y/x = c3 are respect tively, a sphere, cone, and plane. Through any point P in space, except the origin, there will pass exactly one surface of each type, the coordinates of the point P determining the constants c1, c2, c3. The intersection of the sphere and the cone is a circle, the circle of latitude, having e, as its unit tangent vector at P. This circle is called the (p-curve since r and 0 remain constant on this curve
SEC. 23]
DIFFERENTIAL VECTOR CALCULUS
51
so that only the coordinate p changes as we move along this curve. The intersection of the sphere and the plane yields the 0-curve, the circle of longitude, while the intersection of the cone and plane yields the straight line from the origin through. P, the r curve. ee and er are the unit tangent vectors to the 0- and r curves, respectively. The three unit vectors at P, er, ee, e, are mutually perpendicular to each other and can be considered as forming a basis for a coordinate system in the neighborhood of P. Unlike i, j, k, they are not fixed, for as we move from point to point their directions change. Thus we may expect to find more
complicated formulas for the gradient, divergence, curl, and Laplacian when dealing with spherical coordinates. Since Vv is perpendicular to the plane cp = constant, we must have V(p parallel to e, Hence e9, = h3 Vsp, where hs is the scalar factor of proportionality between e,, and Vp. If drs is a vector tangent to the p-curve, of length dss = jdr3j, we have from (58)
dip = drs VP = dr3
e,p
hs
=
ds3
hs
so that ds3 = hs dip.
Hence h3 is
that quantity which must be multiplied into the differential change of coordinate gyp, namely, d-p, to yield arc length along the p-curve.
Thus e,, = r sin 0 V(p, while similarly er = Vr and ee = r VB. We note that er = ee x e,, = r2 sin 0 VO X VV,
eB=e,xe,.=rsin 0V(pxVr
e,=erxee=rVrxVB
Any vector at P may be represented as f = flex + flee + f3ec. The scalars f1, f2, fs can be functions of r, 0, p. We may also
represent f as f = f1 Vr + f2r VO + far sin 0 VV and also by f = f1r2 sin O VO x V(p + f2r sin BV(p x Vr + far Vr x V8.
We also
note that the triple scalar product Vr V9 x Vtp is equal to (r2 sin 0)-1 and that dV = dsl ds2 ds3 = r2 sin 0 dr d9 dp. Spherical coordinates are special cases of orthogonal curvilinear coordinate systems so that we will proceed to discuss these more general coordinate systems in order to obtain expressions for the gradient, divergence, curl, and Laplacian. Let us make a change of coordinates from the x-y-z system to a u1-u2-u3 system as given by the equations Ul = u1(x, y, z) U2 = u2(x, y, z) ug = u3(x, y, z)
(82)
52
VECTOR AND TENSOR ANALYSIS
[SEC. 23
We assume that the Jacobian J[(ul, u2, u3)/(x, y, z)] ;P,-, 0 so that
the transformation (82) is one to one in the neighborhood of a point. A point in space is determined when x, y, z are known and hence when ui, u2, u3 are known. By considering ui(x, y, z) = ci
u2(x, y, z) = C2, u3(x, y, z) = es, we obtain a family of surfaces. Through a point P(xo, yo, zo) will pass the three surfaces ui(x, y, z) = ui(xo, yo, zo) z
y
FIG. 36.
u2(x, y) z) = u2(xo, yo, Zo), and u3(x, y, z) = u8(xo, yo, zo). Let us
assume that the three surfaces intersect one another orthogonally. The surfaces will intersect in pairs, yielding three curves which intersect orthogonally at the point P(xo, yo, zo). The curve of intersection of the surfaces uI = ci and u2 = C2 we shall call the us curve, since along this curve only the variable us is allowed to change. Let u1, u2, us be three unit vectors issuing from P tangent to the ui, us, us curves, respectively (see Fig. 36).
DIFFERENTIAL VECTOR CALCULUS
SEc. 231
53
Now Vu3 is perpendicular to the surface u3(x, y, z) = u3(xo, yo, zo)
so that Vu3 is parallel to the unit vector u3. Hence u3 = h3 Vu3 where h3 is the scalar factor of proportionality between us and Vu3.
Now let dr3 be a tangent vector along the us curve,
dr3' = dss.
Obviously dr3 us = dss, and
so that from (58) dss = h3 du3
(83)
We see that hs is that quantity which must be multiplied into the differential coordinate dus so that are length will result. For example, in polar coordinates ds = r do if we move on the 0-curve,
so that r = h2. Similarly, ul = h1 Vu1j u2 = h2 Du2, so that u1 = U2 X u3 = h2h3 Vu2 X Vu3 u2 = us x u1 = h3h1 Vu3 X Vu1 113 = u1 x u2 = h1h2 Vu1 x out
(84)
and
Du1 Du2 X Vu3 =
u1 U2
h1 h g
x
Us s
= (h1h2ha)-1
(85)
Note that the differential of volume is dV = ds1 ds2 ds3 = h1h2h3 du1 due du3
and making use of (85) as well as Prob. 9, Sec. 22,
dV = j (_x, y, z ) du1 due du3 u1, U2, u3
Example 38.
In cylindrical coordinates ds2 = dr2 + r2 d02 + dz2
so that h1 = 1, h2 = r, h3 = 1.
(86)
VECTOR AND TENSOR ANALYSIS
54
[SEC. 23
If f = f (U1, u2, u3), then from Example 21,
Example 39.
of
yf =
vu3 Vul + au2 ` f vu2 + of au3
au1
of
of
of
Vf = h1 au1 u1+----u2+-''U3 h3 au3 h2 49u2 1
1
1
(87)
In cylindrical coordinates
of
ar
R+p+k r a9
az
Our next attempt is to obtain an expression for the divergence
of a vector when its components are known in an orthogonal curvilinear coordinate system. Now f = flul + f2u2 + f3u3
= f lh2ha Vu2 x vu3 + f 2h3h1 Vu3 x Vul + fahlh2 Vu1 x vu2
from (84).
Consequently
V f = V (f 1h2h3) vu2 x Vu3 + f 1h2h3 V (Vu2 X Vu3) + V(f 2h3h1) Dua x Vu1 + f 2h3hly (Vu3 X Vu1) + V(f3h1h2) Vu1 X Vu2 + f3h1h2V (Vul x vu2)
Now V(f1h2h3) VU2 X Vu3 =
(88)
a(flh2ha ) Vu1 Vu2 x Vu3, and out
V (vu2 x vua) = 0 ao that (88) reduces to [o(h2h3fi)
1
h1h2h3
49U,
+
a(h3h1f2) au2
h,h2f3)
+
(89
49u3
)
If we apply (89) to the vector VV as given by (87), we obtain
VV2
1
a
h1h2h3 au1
h2h3 a
{
h1 49U,
a (hAi a +au2
h2 au2
a (h,h2 a
+ aus
h3
au] (90)
55
DIFFERENTIAL VECTOR CALCULUS
SEC. 23]
This is the Laplacian in any orthogonal curvilinear coordinate system. Example 40. V2V
In cylindrical coordinates
l r
a
ar
r
a
ar
-I-
a
1a_
aB
r aB
+az- r aza
a
(91)
Example 41. Solve V2V = 0 assuming V = V(r),
r = (x2 + y2)i From (91)
1 d (LV r J =0 dr r dr
r
or
ddV = c1
and
V =cllogr+C2 Finally we obtain the curl of f. f = f1u1 + f2u2 + faun = f 1h1 Vu1 + f2h2 Due + fshs Vua and
V x f = V(f1hi) x Vul + V(f2h2) X Vu2 + V(faha) X Vus
since V x (Vul) = V X (Vu2) = V x (Vua) = 0. Now V(f1h1) X VU, =
a(f1h1)
49U,
Vul X Vul +
a(f1h1)
Vu2 X Vu1
49U2
+ a(flhl aus
Vus X Vul
us
Replacing Vu2 x Vul by -- hlh2, etc., we obtain
Vxf=
ul f a(hsf3) h2h3 L
au2
u2 ra(hlfl) ^ a(hsfa) )1 - a(h2f2 aul aua J + hsh1 L aua ua ` a(h1f1) (92) La(h2f2)
+
1h2
49U1
49U2
56
VECTOR AND TENSOR ANALYSIS
[Sec. 23
Problems 1. For spherical coordinates, ds2 = dr2 + r2 d92 + r2 sine 9 d(p2 where B is the colatitude and (p the azimuthal angle. Show that V Z V=
a I r2 sin B a r2 sin 9 ar \\\ ar 1
I
a
}-
(sin a9 `
6
aV aB/J
+ a (sin 9 2. Solve V2V = 0 in spherical coordinates if V = V(r). 3. Express V - f and V x f in cylindrical coordinates. 4. Express V f and V x f in spherical coordinates by letting a, b, c be unit vectors in the r, 9, (p directions, respectively. 5. Write Eq. (92) in terms of a determinant. 6. Show that V x [(r V9)/sin 0) = V(p where r, 0, (p are spherical coordinates. 7. If a, b, c are the vectors of Prob. 4, show that as
as
ar=0'
a9= b,
ab
=0
ab
as =sin9c
= -a
ar
a9
ac
ac
ar=0'
a9=0'
ab
= cos 9 c
a(P
_
ac
= - sin0a - cos0b
8. If x = r sin 9 cos (p, y = r sin 9 sin (p, z = r cos 9, then the form ds2 = dx2 + dy2 + dz2 becomes ds2 = dr2 + r2 d92 + r2 sin2 9 d(p2 3
Prove this. If, in general, ds2 = I (dxa) 2, and if a-1
xa = xa(y', y2, y3)
a = 1, 2, 3, show that axa axa
ds2 = a,
..ydyOdy
_ I go, dys X-f
Y
dyo dyr dy''
SEC. 231
DIFFERENTIAL VECTOR CALCULUS
57
where 3
90Y = aI
ax" axa
a" ay"
Check this result for the transformation to cylindrical coordinates:
x = r cos 0 y = r sin 0 z=z and obtain ds2 = dr2 + r2 d02 + dz2.
9. By making use of V2V = V(V V) - V x (V x V), find V2V for V = v(r)e,, V being purely radial (spherical coordinates). Find V2V for V = f(r)e, + (p(z)e, in cylindrical coordinates. 10. Find V IV if V = w(r)k x r. 11. Consider the equations a2S
(X -f 'U)V(V
X, µ, p constants.
s) + u V2s = p ate
Assume s = eiP'sl, p constant, and show that
(X + p)V(V s1) + (A + pp2)sl = 0
Next show that [V2 + (µ + pp2)/(X + µ)](V s1) = 0, X+µ
0.
12. If A = V x (¢r), V2¢ = 0, show that 1
a a2Y'
sin 0 appaOar
a¢ a2Y' a0acpar
so that A V x A = 0 if, moreover, ¢ = 13. Show that Cpi = Ae9 + Bey + Ce° satisfies V2sp1 = ci, and show that if 402 satisfies V°02 = 0, then rp = Cpl + 4p2 also satisfies V2ip = gyp. Find a solution of V2Sp = -(p.
CHAPTER 3 DIFFERENTIAL GEOMETRY
24. Frenet-Serret Formulas. A three-dimensional curve in a Euclidean space can be represented by the locus of the end point of the position vector given by
r(t) = x(t)i + y(t)j + z(t)k
(93)
where t is a parameter ranging over a set of values to < t < ti. We assume that x(t), y(t), z(t) have continuous derivatives of all orders and that they can be expanded in a Taylor series in the neighborhood of any point of the curve.
We have seen in Chap. 2, Sec. 16, that ds is the unit tangent vector to the curve. Let t =
ds-
Now t is a unit vector so that
its derivative is perpendicular to t. Moreover, this derivative, dse
tells us how fast the unit tangent vector is changing direction
as we move along the curve. The principal normal to the curve is consequently defined by the equation dt
= Kn
(94)
ds
where K is the magnitude of ds and is called the curvature.
The
reciprocal of the curvature, p =I 1K, is called the radius of curvature. It is important to note that (94) defines both K and n,
K being the length of ds while n is the unit vector parallel to dt At any point P of our curve we now have two vectors t, n at ds right angles to each other (see Fig. 37). This enables us to set up 58
DIFFERENTIAL GEOMETRY
SEC. 24]
59
a local coordinate system at P by defining a third vector at right angles to t and n. We define as the binormal the vector
b = t xn All vectors associated with the curve at the point P can be written as a linear combination of the three fundamental vectors t, n, b, which form a trihedral at P. z
0
Y
x Fia. 37.
Let us now evaluate - -- and
dn
Since b is a unit vector, its
derivative is perpendicular to b and so lies in the plane of t and n. Moreover, b t = 0 so that on differentiating we obtain
t=
0.
Hence `
must be parallel to n.
is also perpendicular to t so that
dd_b
Consequently, ds = rn, where r by defini-
tion is the magnitude of -. r is called the torsion of the curve.
VECTOR AND TENSOR ANALYSIS
60
[SEC. 24
Finally, to obtain dn, we note that n = b x t so that b
ds
b x ds +
x t = b x Kn + rn x t =- Kt- rb
The famous Frenet-Serret formulas are dt
Kn ds
do
- (Kt + Tb)
ds
95)
db rn ds
Successive derivatives are functions of t, n, b and the derivatives of K and r. Example 42.
The circular helix is given by
r = a cos ti+asintj+btk t =ds = (-a sin ti+acostj + bk) st and
(dl (a2sin2t+a2cos2t+b2) 2
1=
(.)2 (a2 + b2)
Hence
t = (-a sin ti+acostj+bk)(a2+b2) 4
Now
Kn = d = (-a cos t i - a sin t j)(a2 + b2)-t so that K = a(a2 +
b2)-i
Also
i
j
b = t x n = -a sin t a cost -cos t -sin t
k b
(a2 + b2)-1
0
= (b sin t i - b cos t j + ak) (a2 +
b2)-+
DIFFERENTIAL GEOMETRY
SEC. 241
61
and
db ds
= rn = (b cos t i + b sin t j)(a2 + b2)-1
so that T = b(a2 + b2)-1
Problems
1. Show that the radius of curvature of the twisted curve x = log cos 0, y = log sin 0, z = V2 0 is p = csc 20. 2. Show that r = 0 is a necessary and sufficient condition that a curve be a plane curve.
(r'r"r"').
3. Prove that T = K
4. For the curve xz = a(3t - te), y = 3at2, z = a(3t + t3), show that K = T = 1/3a(1 + t2)2.
5. Prove that
AA _= ds . ds
Kr,
do db d8 - ds
=
0,
dt dn _ d8 . ds
= 0.
6. Prove that r"' = -K2t + K'n - rKb, where the primes mean differentiation with respect to are length.
7. Prove that the shortest distance between the principal normals at consecutive points at a distance ds apart (s measured along the arc) is ds p(p2 + 8. Find the curvature and torsion of the curve r`2)_;
y = all - cos u),
x = a(u - sin u),
z = bu 9. For a plane curve given by r = x(t)i + y(t)j, show that
x,y - y,x
[(x')2 + (y1)2]1
10. Prove that (t't"t"') = K5
ds (K)
11. Show that the line element ds2 = dx2 + dy2 + dz2 - c2 dt2 remains invariant in form under the Lorentz transformation
x=
- yt [1
- (V2/c2)1I
y=I
z=2 t=
- (V/c2)x [1 - (V2/c2)J*
62
VECTOR AND TENSOR ANALYSIS
[SEC. 25
V, c are constants. The transformation ct = iT, i leads to the four-dimensional Euclidean line element
ds2=dx2+dy2+dz2+dr2 12. If xa = xa(s), a = 1, 2, . . . , n, represents a curve in an n-dimensional Euclidean space for which d82
= (d_-1)2 + (dx2) 2 +
.
. + (dxn) 2
define the unit tangent vector to this curve, this definition being a generalization of the definition of the tangent vector for the case
n = 3. Show that the vector
d2xa
ds2
) a = 1, 2,
.
.
.
, n, is normal
to the tangent vector, and define the unit principal normal n, and curvature K, by the equations d2xa
dta
d82 - ds =
Klnla,
a = 1, 2,
..,n
a
Show that to
a-1
ds
a = 1, 2,
dnla = - K1. !Is
. . .
Define the second curvature K2 and unit
normal n2 by th e equati ons d
..
, n, is normal to nl and that
d
la
= -Klt a + K2n2 a, a =
1, 2,
i
. , n, and show that n2a is normal to to and nla if K2 ;P'- 0. Continue in this manner and obtain the generalization of the
Frenet-Serret formulas. 25. Fundamental Planes. The plane containing the tangent and principal normal is called the osculating plane. Let s be a variable vector to any point in this plane and let r be the vector to the point P on the curve. s - r lies in the plane and is conse-
quently perpendicular to the binormal. The equation of the osculating plane is
(s - r) b = 0
(96)
The normal plane to the curve at P is defined as the plane through P perpendicular to the tangent vector. Its equation is
DIFFERENTIAL GEOMETRY
SEC. 26]
63
easily seen to be
(s - r) t = 0
(97)
The third fundamental plane is the rectifying plane through P perpendicular to the normal n. Its equation is
(s - r) n = 0
(98)
Problems
1. Find the equations of the three fundamental planes for the curve
x = at,
y=bt2,
z=cta
2. Show that the limiting position of the line of intersection of two adjacent normal planes is given by (s - r) n = p where s is the vector to any point on the line. 26. Intrinsic Equations of a Curve. The curvature and torsion of a curve depend on the point P of the curve and consequently on the are parameter s. Let is = f(s), r = F(s). These two equations are called the intrinsic equations of the curve. They owe their name to the fact that two curves with the same intrinsic equations are identical except possibly for orientation in space. Assume two curves with the same intrinsic equations. Let the trihedrals at a corresponding point P coincide; this can be done by a rigid motion. Now ds
(t1. t2) = t1
,cn2 + xni t2
T (nl n2) = n1 (-Kt - rb2) + n2 (--Kt, -- rbi)
d Adding, we obtain
s
0
(99)
64
VECTOR AND TENSOR ANALYSIS
(SEC. 27
so that
constant = 3 since at P
tl=t2,
n1=n2,
(100)
b1=b2
Since (100) always maintains its maximum value, we must have dr, _ dr2 tl = t2, n1 n2, bi = b2 so that or r1 = r2 locally. ds
ds
Hence the two curves are identical in a small neighborhood of P. Since we have assumed analyticity of the curves, they are identical everywhere.
Problems
1. Show that the intrinsic equations of x = a(9 - sin 8), y = a(l - cos 8), z = 0 are p2 + s2 = 16a2, 7- = 0, where s is measured from the top of the are of the cycloid. 2. Show that the intrinsic equation for the catenary y=a'(ex/a+e-(sla)) 2
is ap = s2 + a2, where 8 is measured from the vertex of the catenary.
Fla. 38.
27. Involutes. Let us consider the space curve r. We construct the tangents to every point of r and define an involute
DIFFERENTIAL GEOMETRY
SEc. 271
65
as any curve which is normal to every tangent of r (see Fig. 38). From Fig. 39, it is evident that
r,=r+ut
(101)
is the equation of the involute, u unknown. Differentiating (101), we obtain
dr,
ds,^lt
A
Cdr
d_u \ ds
ds+ u ds+dst
ds,
where s is are length along r and s, is arc length along r'. (95), (102) becomes r
ti =
(t+uua+dut) d ds ds,
(102)
Using
(103)
Now t t, = 0 from the definition of the involute so that du 1 +=0
and
u=c - s Fta. 39.
(104)
Therefore r, = r + (c - s)t, and there exists an infinite family of involutes, one involute for each constant c. The distance between corresponding involutes remains a constant. An invo-
r
Fia. 40.
lute can be generated by unrolling a taut string of length c which has been wrapped along the curve. The end point of the string generates the involute (see Fig. 40). What are some properties of the involute?
VECTOR AND TENSOR ANALYSIS
66
[SEc. 28
r1=r+(c-s)t ti = dr,
dr / dt Ids + lC S) ds ds, -
(C - s)
t]
ds dsl
ds K - n ds1
Hence the tangent to the involute is parallel to the corresponding normal of the curve. Since ti and n are unit vectors, we must have (c - s) K d = 1. The curvature of the involute is l
do ds
dt1
obtained from - = Kin, ds= ds1 --= dsl
(-Kt -7-b)
Hence
(c - s)K
+ r2
r K12
K2
K2(C - 8)2
(105)
28. Evolutes. The curve t' whose tangents are perpendicular to a given curve is called the evolute of the curve. The tan-
gent to r' must lie in the plane of b and n of r since it is perpendicular to t. Consequently
Fla. 41.
rl=r+un+vb is the equation of the evolute.
tl
=
dr1
= dr
dsl -
{ds
Differentiating, we obtain
d_n db d_u dv A + u ds + v ds + ds n + d8 bl ds,
= It + u(-Kt -rb) +vTn+ds n +d8b]dsl Now t t1 = 0, which implies I - uK = 0 or u = dv) tl=L(-ru+as , b+( +ds)n
1
= p. Thus
K
d8
Also t1 is parallel to r1 - r = un + vb (see Fig. 41). Therefore (dv/ds) - UT
(du/ds) + Pr
u
V
DIFFERENTIAL GEOMETRY
SEC. 28]
67
or
T=
uv' - vu' u2+v2
= dsd- tan-' -vu)
Therefore
=
ra
o
Tds=tan--'U -C -V
and v = p tan (,p - c) since u = p. Therefore r1 = r + pn + p tan (,p - c)b
(106)
and again we have a one-parameter family of evolutes to the curve T.
Problems
1. Show that the unit binormal to the involute is Kb - Tt
b1 = (C - S)KK1
2. Show that the torsion of an involute has the value = T1
T-K
dS
] [K(K2 + r2)(C - 8)I-1
3. Show that the principal `normal to the evolute is parallel to the tangent of the curve 1'. 4. Show that the ratio of the torsion of the evolute to its curva-
ture is tan (,p - c). 5. Show that if the principal normals of a curve are binormals (equal vectors not necessarily coincident) of another curve, then c(K2 +,r2) = K where c is a constant. 6. On the binormal of a curve of constant torsion T, a point Q is taken at a constant distance c from the curve. Show that the binormal to the locus of Q is inclined to the binormal of the given curve at an angle
tan-'
CT2
K(C2r2 + 1)}
7. Consider two curves which have the same principal normals (equal vectors not necessarily coincident). Show that the tangents to the two curves are inclined at a constant angle.
VECTOR AND TENSOR ANALYSIS
68
[SEC. 29
29. Spherical Indicatrices
(a) When dealing with a family of unit vectors, it is often convenient to give them a common origin and then to consider
the locus of their end points. This locus obviously lies on a unit sphere. Let us now consider the spherical indicatrix of the tan-
gent vectors to a curve r = r(s).
The unit tangent vectors are t(s) = ds- Let r, = t. Then =
t'
dr,
= dt ds
ds1
ds dal -
ds
ds,
Thus the tangent to the spherical indicatrix P is parallel to the normal of the curve at the corresponding point. Moreover, Fla. 42.
1 = K d1' t, = n. Let us now find the curvature K, of the indicatrix.
We obtain do ds =K,n,='--=-(-Kt-rn) ds ds, as,
dt,
1
K
and K2 + r2 K12
2
K
(b) The spherical indicatrix of the binormal, r1 = b. entiating, dr, ds A ds = = re
t,=--- ds ds1
ds,
Therefore
rds=1
ds1
t,=n
and
ds,
Differentiating,
dt, dS1
= x,n1 =
do ds A dS;
_
1
r
and K12 =
K2 + r2 T2
`-Kt - rn)
Differ-
SEC. 30J
DIFFERENTIAL. GEOMETRY
69
Problems
1. Show that the torsion of the tangent indicatrix is T(dK/ds) - K(dr/ds) T2)
Ti
K(K2 +
2. Show that the torsion of the binormal indicatrix is Ti
T(dK/ds) - K(dT/ds) T(K2 r2)
+
3. Find the curvature of the spherical indicatrix of the principal normal of a given curve. 30. Envelopes. Consider the one-parameter family of surfaces F(x, y, z, c) = 0. Two neighboring surfaces are
F(x, y, z, c) = 0 and F(x, y, z, c + Ac) = 0. These two surfaces will, in general, intersect in a curve. But these equations are equivalent to the equations F(x, y, z, c) = 0 and F(x, y, z, c + Ac) - F(x, y, z, c) Ac
=0
where Ac # 0. As Ac --> 0, the curve of intersection approaches a limiting position, called the characteristic curve, given by
F(x, y, z, c) = 0 aF(x, y, z, c) ac
-0
(107)
Each c determines a characteristic curve. The locus of all these curves [obtained by eliminating c from (107)] gives us a surface called the envelope of the one-parameter family. Now consider two neighboring characteristics
F(x,y,z,c) = 0
aF(x,,z,c) ac
=0
and
(108)
F(x, y, z, c + Ac) = 0
aF(x, y, z, c + Ac) = 0
which, in general, intersect at a point.
ac
The locus of these points
is the envelope of the characteristics and is called the edge of
70
VECTOR AND TENSOR ANALYSIS
[SEc.31
The edge of regression is given by the three simultaneous equations F(x, y, z, c) = 0 regression.
aF(x, y, z, c) = 0 ac
(109)
a2F(x,y,z,c) = 0 ace
Example 43. Let us consider the osculating plane at a point P.
From (96) we have [s - r(s)] b(s) = 0. If we let P vary, we obtain the one-parameter family of osculating planes given by
f(x, y, z, s) = [s - r(s)] b(s) = 0 where s is the parameter and s = xi + yj + A. Now
of = as
dr ds
A = (s - r) b + (s -r).ds
of = 0, we obtain (s - r) n = 0. as plane.
This locus is the rectifying
The intersection of f = 0 and as = 0 obviously yields
the tangent lines which are the characteristics. a If
as2
in, and setting
Now
= -t n + (s - r) (-Kt - A) + (s - r) n da
It is easy Y to verify Y that s = r satisfies f =
a
as
=
az 4982
-
= 0, so that
the edge of regression is the original curve r = r(s). A developable surface, by definition, is the envelope of a oneparameter family of planes. The characteristics are straight lines, called generators. We have seen that the envelope of the
osculating planes is the locus of the tangent line to the space curve P.
In general, a developable surface is the tangent surface
of a twisted curve. A contradiction to this is the case of a cyiinder or cone.
31. Surfaces and Curvilinear Coordinates. the equations x = x(u, v) y = y(u, v) z = z(u, v)
Let us consider (110)
DIFFERENTIAL GEOMETRY
SEc. 32]
71
where u and v are parameters ranging over a certain set of values. If we keep v fixed, the locus of (110) is a space curve. For each v, one such space curve exists, and if we let v vary, we shall obtain
a locus of space curves which collectively form a surface. We shall consider those surfaces (110) for which x, y, z have continuous second-order derivatives. Equation (110) may be written
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k
(111)
where the end point of r generates the surface. The curves obtained by setting v = constant are called the u curves, and similarly the v curves are obtained by setting u = constant. The parameters u and v are called curvilinear coordinates, and the two curves are called the parametric curves. 32. Length of Arc on a Surface. If we move from the point r to the point r + dr on the surface, the distance ds is given by
(arudu+-dvN 2
ds2 =
or
2
C_ J
due + 2
9r
o9r
au . av
du dv + (av)2 dv2
or
ds2 = E du2 + 2F du dv + G dv2
(112)
where
E
\12
au/ '
F
ao9r .
u av'
2
G
(Ov)
Equation (112) is called the first fundamental form for the surface
r = r(u, v).
In particular, along the u curve, dv = 0, so that
(ds) = 1'E du and similarly
(113)
(ds), = VG_ dv Now
Or
and av are tangent vectors to the u and v curves, so
that the parametric curves form an orthogonal system if and only
Or
Or
au av
72
VECTOR AND TENSOR ANALYSIS
Example 44.
1Szc. 33
Consider the surface given by
r = r sin 8 cos v i+ r sin 8 sin V j+ r cos 8 k,
r = constant
Differentiating,
ar = r cos 0 cos sP i + r cos 0 sin V j - r sin 8 k d0 ar
a st
= -r sin6sinpi+rsin8coscpj
and is
E
a8/ =
c1r
r2,
F=-e
c1r
2
= 0,
G=
r2 Sin2 0
so that ds2 = r2 d82 + r2 sin2 0 dcp2 and the 0-curves are orthogonal to the 9-curves. Of course the surface is a sphere.
33. Surface Curves. By letting u and v be functions of a single variable t, we obtain
r = r[u(t), v(t)]
(114)
which represents a curve on the surface (111). Along this curve, (arau ar dvl dr = ` du + Wt dt. dr is completely determined when du dt av 1 and dv are specified, so that we will use the notation (du, dv) to specify a given direction on the surface. Now consider another
curve such that ar =
su au + av av, where su and av are the differential changes of u(t) and v(t) for this new curve. Now
dr or = E du au + F(du av + dv au) + G dv av
(115)
so that two curves are orthogonal if and only if
Edu su +F(duav +dv &u) +Gdv av = 0 or
dv E+FCa6Vu+au/+Gd -=0
(116)
If we have a system of curves on the surface given by the differential equation P(u, v) su + Q(u, v) av = 0, the differential equa-
DIFFERENTIAL GEOMETRY
SEC. 341
73
tion for the orthogonal trajectories is given by
E+F,(_P+dv1 - GPdv Q
``
by
P
bu
Q
since -
dull
Q du
=
0
(117)
Problems 1. Find the envelope and edge of regression of the one-param-
eter family of planes x sin c - y cos c + z tan 8 = c, where c is the parameter and 0 is a constant. 2. Show that any two v curves on the surface r = u cos v i + u sin v j + (v + log cos u)k cut equal segments from all the u curves. 3. Find the envelope and edge of regression of the family of x2
ellipsoids c2 (a2
+
y
z2
+ c2 = 1 where c is the parameter.
b2
4. If 8 is the angle between the two directions given by P due + Q du dv + R dv2 = 0
show that tan 0 = H(Q2 - 4PR)}/(ER - FQ + GP), where H
ar
8 avl.
8u
tvhl
5. Prove at the differential equations of the curves which bisect the angles between the parametric curves are
VEdu-VGdv=0 and 1/E du + 1i dv - 0. 6. Given the curves uv = constant on the surface r = ui + vj, find the orthogonal trajectories. 7. Show that the area of a surface is given by f f (EG
- F2)1 du dv
34. Normal to a Surface. The vectors
or
and av are tangent
to the surface r(u, v) along the u and v curves, respectively. Consequently,
ar
8r
au x 49V
,
is a vector normal to the surface. Note
VECTOR AND TENSOR ANALYSIS
74
[SEC. 35
that au need not be a unit tangent vector to the u curve since the
parameter u may not represent are length. Since
(ds),, = 1/E du a necessary and sufficient condition for u to be arc length is that E = 1. We define the unit normal to the surface as n _
(Or/au) x (ar/av) (118)
(ar/au) x (ar/av) 35. The Second Fundamental Form.
Consider all the planes
through a point P of the surface r = r(u, v) which contain the normal n. These planes intersect the surface in a family of curves, the normals to the curves being parallel to n. We now compute the curvature of any one of these curves in the direction (du, dv). Let ds be length of are along this curve. Now
dr Or du Or dv t=ds=auds+avds Therefore
d2r
dt
ds2
ds
_
492r
du 2
au2 (ds)
K"n
2
a2r d_u dv
a2r dv 2
+ au av A ds +
av2 (ds)
Or d2u
Or d2v
+
au ds2
+
av ds2
119)
and
(du)2
n)
since n
-
Or
au
(n au2)
n
- = 0. av
Kn =
ds+ 2 (n au2av) ds ds )z
Therefore
edue+2fdudv+gdv2 ds2
Kn
edue+2fdudv+gdv2 Edue+2Fdudv+Gdv2
(120)
DIFFERENTIAL GEOMETRY
SEC. 361
75
where we define a2r
e=n-au2'
a2r
a2r _n. auav' f
(121)
The quantity e due + 2f du dv + g dv2 is called the second fundamental form. Now consider any curve t on the surface and let its normal be n, at a point P, the direction of r being (du, dv) at P. Let r' be
Fia. 43.
the normal curve in the same direction (du, dv) with normal n at P (Fig. 43). We have
r" K
=n r"K
since n rl" = r" n for two curves with the same (du, dv) [see (119)].
Therefore COS0=
-
Kn K
so that K = K,, sec 0
(122)
This is Meusnier's theorem.
36. Geometrical Significance of the Second Fundamental Form. We construct a tangent plane to the surface at the point r(uo, vo). What is the distance D of a neighboring point r(uo + Du, vo + Av)
VECTOR AND TENSOR ANALYSIS
76
on the surface, to the plane? It is D = Ar n. r(uo + Au, vo + AV) = r(uo, vo) + 1
+
2!
from the calculus.
Au2 + 2
Now
ar
+ av
a2r
AV
a2r
+ av2 Av2 + au av Au AV
Consequently
1/
D=
-
(a2r
ar Au au
[SEC. 36
z
z
Av2
except for infinitesimals of higher order.
Thus
2D = e du2 + 2f du dv + g dv2
(123)
Problems 1. For the paraboloid of revolution
r=ucosvi+usinvj+u2k show that E = 1 + 4u2, F = 0, G = u2, e = 2(1 + 4u)`}, f = 0, g = 2u2(1 + 4u2)-}, and find the normals to the surface and the normal curvature for the direction (du, dv).
2. What are the normal curvatures for directions along the parametric curves? 3. Find the second fundamental form for the sphere
r=rsin0coscpi+rsin6sincpj+rcosOk r = constant.
4. Show that the curvature K at any point P of the curve of intersection of two surfaces is given by K2 Sln2 0 = K12 + K22 - 2K1K2 COS 0
where 9j, K2 are the normal curvatures of the surfaces in the direc-
tion of the curve at P, and 0 is the angle between their normals. 5. Let us make a change of variable u = u(u, v), v = v(u, 1). Show that E, F, G transform according to the law
E
lz Caul
+ 2F
au au
+G(av \au
2
DIFFERENTIAL GEOMETRY
SEC. 37]
F
au au
au av
77
av au
av av
=Eaudo +F (aft do +audo )+Gag do UP-
(avl 2
au av
au\2
E
Ii + 2F avav
+ G `avl
and that E due + 2F du dv + G dv2 = E due + 2F du do + G W. Also show that
e=±
(
e
au + (au)2
au au au av g
-±
[e
(au)2
av
au av
(OV\2 1
au au + g au J au av av au au av au av au av
+ 2f avav+ g
37. Principal Directions.
(K .E - e) due +
Zf
av av
au CIO]
(av12 av
J
From (120) we have
f) du dv +
g) dv2 = 0
(124)
or
A due + 2B du dv + C dv2 = 0
This quadratic equation has two directions (du, dv), (Su, Sv), which give the same value for x,,. These two directions will coincide if the quadratic equation (124) has a double root. is true if and only if
B2-AC= (K.F'-f)2-
This
0
or ,
2(F2 - EG) +
Moreover, we have
so that
du
gE - 2fF) + (f2 - eg) = 0
(125)
- B and d = - C if B2 - AC = 0, A
(K. E - e) du + (x F - f) dv = 0 f) du + (xnG - g) dv = 0
6126)
The solutions of (124) give the two directions for a given x,,. When x is eliminated between (124) and (125), the two directions coincide and satisfy
(Ef - Fe) due + (Eg - Ge) du dv + (Fg - Gf) dv2 = 0 (127)
VECTOR AND TENSOR ANALYSIS
78
[SEc. 37
The two directions, solutions of (127), are called principal directions and are the only ones with a unique normal curvature, that is, no other direction can have the same curvature. The normal curvatures in these two directions are called the principal curvatures at the point. The average of the two principal curvatures is
HeG+gE-2fF
(128)
2(EG - F2)
which is obtained by taking one-half of the sum of the roots of (125). The Gaussian curvature K is defined as the product of the curvatures, that is,
f2eg K= F2EG
(129)
A line of curvature is a curve whose tangent at any point has a
direction coinciding with a principal direction at that point. The lines of curvature are obtained by solving the differential equation (126). The curvature of a line of curvature is not a principal curvature since the line of curvature need not be a normal curve. Example 45. Let us consider the right helicoid
r= ucospi+usine'j+cook We have Or =cosSPi+sinrpj,
au 2
zr
a2G2-0'
8ua
= -u sins
= -sin+Cos(pj, a2r
Q= -ucosipi - usinrpj
Hence
(ar)2
au-
1,
F=
(Or\2 au - am
0'
G=
am
= u2 + c2
DIFFERENTIAL GEOMETRY
SEC. 371
79
Also
n = (ar/au) x (,9r/app) = (c sin p i - c cos v j + uk) (C2 +
u2)-}
(ar/au) x (ar/ap)j 82r
a2r
f=n
au ago
= -c(c2 +
u2)-#
2
Equation (125) yields - (u2 + C2)K,, + C2(C2 + u2)-1 = 0
whence C
u2+C2 The average curvature is H = 2
the Gaussian curvature is K =
(u2
2
C2)2'
+ C2) =
0, and
The differential
equation (126) for the lines of curvature becomes u2)-}
-c(c2 +
dug + c(c2 + U2)Id tp2 = 0
so that d(p = ±
du
= ± log (u + \/u2 + c2) + a
and
(C2 + u2) b
and the lines of curvature are given by
r = u cos [± log (u +
u2 + c2) + aji + u sin -p j + cspk.
Referring to (126) for the two principal directions, we have dv
du
av
Eg - Ge
+
Fg - Gf _ Ef - Fe du au Fg - Gf au dv av
(130)
Substituting (130) into (116) we obtain
E - F(Fg-Gf/ +G \F'g -Gf/ so that the principal directions are orthogonal.
0
VECTOR AND TENSOR ANALYSIS
80
[SEC. 38
Now let us choose the principal curves as the parametric lines.
Thus u = constant, v constant are to represent the principal curves. These two curves satisfy the equation du dv = 0, so that from (127) we must have
Ef - Fe = 0
Fg-Gf=0
Eg - Ge
0
From these equations we conclude that
f(Eg - Ge) = gfE - feG=Feg - eFg =0 and F(Eg - Ge) = 0, so that f = F = 0. We have shown that a necessary and sufficient condition that the lines of curvature be parametric curves is that
f=F=0
(131)
Problems 1. Find the lines of curvature on the surface
x=a(u+v),
y=b(u-v),
z=uv
2. Show that the principal radii of curvature of the right conoid x = u cos v, y = u sin v, z = {f (v) are given by the roots of f'=K2
- of"(u2 + fF )iK - (u2 + ft )2 = 0
3. The surface generated by the binormals of the curve r = r(s)
is given by R = r + ub. Show that the Gauss curvature is K = -.r2/(1 + r2u2)2. Also show that the differential equation of the lines of curvature is
-T2 du2 -
(K + Kr2u2 + d3 u)
du ds + (1 + r2u2)T M = 0
38. Conjugate Directions. Let P and Q be neighboring points on a surface. The tangent planes at P and Q will intersect in a straight line 1. Now let Q approach P along some fixed direction. The line 1 will approach a limiting position 1'. The directions PQ and 1' are called conjugate directions. We now compute the analytical expression for two directions to be conjugate. Let n be the normal at P and n + do the
DIFFERENTIAL GEOMETRY
SFc. 38]
normal at Q, where dr = PQ = tion of 1' be given by Sr =
Or
au
Su
Or
an
du + ar
Sv.
+ av
Or
-
81
dv.
Let the direc-
Since Sr lies in both
planes, we must have Sr n = 0 and Sr (n + dn) = 0. These two equations imply Sr do = 0, or (au
Su
+
av
Sv)
Can du +
a dv)
=0
(132)
Expanding, we obtain far
(_au
-) du bu + I (-av -) Sv du + (auau Or
anav
Or an aU) au dvJ
anau
ar an
+
av
av
Sv dv = 0
(133)
Now n au = 0, so that by differenuiating we see that ar
an
a2r
which implies
an ar
au au
a2r -n- au, - = -e
an
Or
an Or
av au
au av
Similarly
-f
an Or 9
so that (133) becomes e du Su + f (du Sv + dv Au) + g Sv dv = 0
(134)
If the direction (du, dv) is given, there is only one corresponding conjugate direction (Su, Sv), obtained by solving (134). Now consider the lines of curvature taken as parametric curves. Their directions are (du, 0), (0, Sv). Equation (134) is satisfied by these directions since f = 0 for lines of curvature, so that the lines of curvature are conjugate directions.
82
VECTOR AND TENSOR ANALYSIS
[SFc. 39
39. Asymptotic Lines. The directions which are self-conjugate are called asymptotic directions. Those curves whose tangents are asymptotic directions are called asymptotic lines. If a direcdv
tion is self-conjugate,
=
du
Sv,
Su
so that (134) becomes
e due + 2f du dv + g dv2 = 0
(135)
We see that the asymptotic directions are those for which the second fundamental form vanishes. ture rc, vanishes for this direction.
If e = g = 0, f
Moreover, the normal curva-
0, the solution of (135) is u = constant,
v = constant, so that the parametric curves are asymptotic lines if and only if e = g = 0,f3PK 0. Example 46. Let us find the lines of curvature and asymptotic lines of the surface of revolution z = x2 + y2. Let x = u cos v, y = u sin v, z = u2, and
r=ucosvi+usinvj+u2k We obtain Or
=cosvi-{-sinvj+2uk,
ar= -u sinvi+ucosvj av
au azr
8u2
a2r
=2k' a2r av2
au av
= -sinvi+cosvj
- -ucosvi - usinvj
n= (ar/au) x (ar/av)
(ar/au) x (ar/av)I = (-2u2 cos v i - 2u2 sin v j + uk)u-1(1 + 4u2)'I
Therefore z
=2(1+4u)-},
au2 z
g=n
f=n aua2rav =0
= 2u2(1 +4 U2)-f 2
DIFFERENTIAL GEOMETRY
SEC. 40] ar
83
ar
Also F = -- - = 0, so that f = F = 0, and from (131) the au av parametric curves are the lines of curvature. The asymptotic lines are given by due + u2 dv2 = 0. These are imaginary, so that the surface possesses no asymptotic lines. Problems 1. Show that the asymptotic lines of the hyperboloid
r=acos0seeipi+bsin0sec4, j+ctan 4k are given by 0 ± ¢ = constant. 2. The parametric equations of the helicoid are
x=ucosv,
y=usinv,
z=cv
Show that the asymptotic lines are the parametric curves, and
that the lines of curvature are u + V 'u2+ c2 c2 = Ae}°. Show that the principal radii of curvature are ± (u2 + c2)c-1. 3. Prove that, at any point of a surface, the sum of the normal curvature in conjugate directions is constant. 4. Find the asymptotic lines on the surface z = y sin x. 40. Geodesics. The distance between two points on a surface (we are allowed to move only on the surface) is given by
8 = I.1 [E (dt/2 + 2F dt dt + G (i)]1 dt
(136)
Among the many curves on the surface that join the two fixed points will be those that make (136) an extremal. Such curves are called geodesics. We wish now to determine the geodesics. To do this, we require the use of the calculus of variations, and so we say a few words about this important method. Let us first consider the integral fQ(xt,yi) (1 (x,, VS)
+ y'dx
(137)
We might ask what must be the function y = y(x) joining the two points P and Q which will make (137) a minimum. The reader might be tempted to say, y' = 0 or y = constant, since the integrand is then a minimum. But we find that y = constant will
84
VECTOR AND TENSOR ANALYSIS
[SEC. 40
not, in general, pass through the two fixed points. Hence the solution to this problem is not trivial. We now formulate a more general problem: to find
y
y = y(x) such that
j'f(x, y, y') dx is an extremal. ry
Ar
}' M
The function
f(x, y,)y ' is given. It is y(x) and so also y' (x) that are unknown . Let y = y (x) be that function which makes
I.-, B
,
(138)
It
(138) an extremal Now 1 t Y(x, a) = y(x) + arp(x), where
a is arbitrary and independ-
P.x
b
a
ent of x and jp(x) is any function with continuous first
Fzo. 44.
derivative having the property that jp(a) = p(b) = 0 (see Fig. 44). Under our assumption,
J(a) = j,'f(x, Y, ?') dx is an extremal for a = 0. dJ
=
da a-0
(139)
dJ
da0 = 0 or
Consequently
Ib\ay,+a-
(140)
since
of as
of 81' of 81" aY as + aY' as
of
if
+
of C7F
'p,
and for a=0, of
a> r
of
of
of ay'
Tay ,
e
We now integrate the right-hand term of (140) by parts and obtain P] b
fb ay
dx
+
Lay
-
Jab
dx ay'/ `p
dx = 0
(141)
DIFFERENTIAL GEOMETRY
SEc. 40]
85
Now (p(a) = p(b) = 0 by construction of v(x), so that (142)
Now let us assume that
a (-'
ay
ay
dx
(ay')
is continuous.
If
is not identically zero on the interval (a, b), it
will be positive or negative at some point. If it is positive at x = c, it will be positive in a neighborhood of x = c from continuity (see Sees. 42 and 43). We can construct (p to be positive on this interval and zero elsewhere. Then
Jb[afd(of)]
ay`p dx > 0
dx
y
so that we have a contradiction to (142). Consequently, the function of y(x) must satisfy the Euler-Lagrange differential equation of
d of dx Cay'
ay
0
(143)
If f = f(y, y'), we can immediately arrive at an integral of (143).
Let us consider
f dx (
y yl
ay
1J, +
a y
y [Of
_
lay
f-y
aof y,
yy
of ay
d dx ay'
(t\1 = 0
= constant
d (af 1 y/ dx TO from (143)
(144)
is an integral of (143) if f = f(y, y'). Example 47. To extremalize (137), we have f = (1 + y'2)1, which is independent of x. From (144), f - y
of
1
= constant =
a
VECTOR AND TENSOR ANALYSIS
86
[SEC. 40
so that
(1 +y') - Y'(1
J
a
y'2),
and
y'= and finally
y= ±
av-1x+ 0
(145)
The constants a and fi are determined by noting that the straight line (145) passes through the two given fixed points. Example 48.
If f = f Cx',x 2, ... , x",
dt2) ...
dtl,
dtn?
t);
fhfdt is an extremal when the xa(t) satisfy
then
d (ftaf dt a
for a = 1, 2,
. .. , n with is =
of axa = o X.
(146)
The superscripts are not
powers but labels that enable us to distinguish between the various variables. The formulas (146) are a consequence of the fact
that
fo f dt must be an extremal when x`(1) is allowed to vary
while we keep all other x1 fixed, j = 1, 2, ... , i - 1, i + 1, ,n. Let us now try to find the differential equations that u(t) and v(t) must satisfy to make (136) an extremal. We write
s= f"
(E,42 + 2Fuv + Gv2)} dt
and apply (146), where x' = u and x2 = v. We thus obtain
dt \a4/
au - 0
off`
of
dt avl
av
d
o
(147)
(148)
where
f = (E,42+ 2Fuv + Gv2)i = dt'
E = E(u, v), etc.
---
DIFFERENTIAL GEOMETRY
SEc. 401
Now
af au
Eic + Fv
ice
of au
f
CIE
au
+ 2uv
OF
au 2f
+ v2
87
aG
au
so that (147) becomes
d Eu + Fvl _ 2t2 + 2"(W/au) + v2(aG/au) dt
ds/dt /
2 ds/dt
(149)
and if we choose for the parameter t the are length s, then t = a
and dt = 1, so that (149) reduces to d ds
(Eic + Fv) = 2( u2 aE + 2uv
OF
+ 62
-n/A opu
while similarly (148) yields
d (Fu+Gv)
\
(150)
_2(u2av +24vav +v2av/
In Chap. 8 we shall derive by tensor methods a slightly different system of differential equations. Example 49. Consider the sphere given by
r = a sin 0cos(pi+a sin 0 sin cpj+acosOk where ds2 = a2 do2 + a2 sin2 0 dp2 so that E = a, F = 0, G = a2 Sin2 0, and OE
80
_
aEE
ac
_
OF
_
OF
a9
av
_
aG
_
ap
8G
0'
aB
_
- a sin 29 2
Hence (150) reduces to 2
ds a ds/
2 ds
sin 2B
d (a sin2 0 ds) = 0 Integrating (151) we have sine 0 LIP = constant.
d
(151)
We can choose
our coordinate system so that the coordinates of the fixed points
VECTOR AND TENSOR ANALYSIS
88
are a, 0, 0 and a, 0, 0. Hence sin' 8
ds
= 0, and
[SEC. 40
ds
= 0, so that
0. Hence the geodesic is the are of the great circle joining the two fixed points. Example 50. Let us find y(x) which extremalizes
f y(1 + y")} dx
Since f = y(1 + y")# = f(y, y'), we can apply (144) to obtain a first integral. We obtain y(1 + y'')t -- y''y(1 + y'')-l = a-1, and simplifying this expression yields y' = ± (a2y2 - 1)I. A further integration yields ay = cosh (0 ± ax). These are the curves (catenoids) which have minimum surfaces of revolution.
Problems 1. Find the geodesics on the ellipsoid of revolution 2 x2+ z2 b2 = 1 + a2
Hint: Let x = u cos v, z = u sin v. 2. Show that the differential equation of the geodesics for the right helicoid x = u cos v, y = u sin v, z = cv is du TV
+
1
[(u2 + C2)(u2 + C2
h2)1],
h = constant
-h
3. Prove that the geodesics on a right circular cylinder are helices.
4. Show that the perpendicular from the vertex of a right circular cone to the tangents of a given geodesic is of constant length. 5. Find y(x) which extremalizes
f'[(l + y')/y)]} dx.
CHAPTER 4 INTEGRATION
41. Point-set Theory. In geometry and analysis the student has frequently made use of the concept of a point and of the notion of a set or collection of elements (objects, points, numbers, etc.). We shall not define these concepts, but shall take their notion as intuitive. We may be interested in the points subject
to the condition x2 + y2 < r2. These will be the points interior to the circle of radius r with center at the origin. We might also be interested in the rational points of the one-dimensional continuum. For convenience, we shall consider only points of the real-number axis in what follows. Any set of real numbers will be called a linear set. The integers form such a set, as do the
rationals and irrationals. All the definitions and theorems proved for linear sets can easily be extended to any finitedimensional space. Closed Interval. The set of points { x j satisfying a < x S b will be called a closed interval. If we omit the end points, that
is, consider those x that satisfy a < x < b, we say that the interval is open (open at both ends). For example, 0 S x 5 1 is a closed interval, while 0 < x < 1 is an open interval. Bounded Set. A linear set of points will be said to be bounded if there exists an open interval containing the set. It must be emphasized that the ends of the interval are to be finite numbers, which thus excludes - oo, + oo. An alternative definition would be the following: A set of numbers S is bounded if there exists a finite number N such that
-N <x
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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
[SEC. 41
this by postulating Every nonempty set of points has a supremum
(152)
Hence the rationals whose squares are less than 3 have a supreIt is obvious that we should define this supremum as the square root of 3. The supremum of a set may be + oo as in the case of the set of all integers, or it may be - oo as in the case of the null set. The infemum of a set S is the number s such that 1. x in S implies s < x mum.
2. t > s implies an x in S such that x < t Example 53. Let a > 0 and consider the sequence a, 2a, 3a, ... , na, ... . If this set is bounded, there exists a finite supremum s. Hence an integer r exists such that ra > s - (a/2) so that (r + 1)a > s + (a/2) > s, a contradiction, since na 5 s for all n. Hence the sequence Ina} is unbounded. This is the Archimedean ordering postulate. Example 54. To prove that a rational exists between any two numbers a, b. Assumed: a > b > 0, so that a - b > 0. From
example 53, an integer q exists such that q(a - b) > 1, or qa > qb + 1. Also an integer p exists such that p 1 = p > qb. Choose the smallest p.
Thus p > qb ? p - 1. Hence
qa>gb+1zp>qb, and a > p/q > b. Q.E.D. With the aid of (152) we are in a position to prove the wellknown Weierstrass-Bolzano theorem. "Every infinite bounded set of points S has a limit point."
The proof proceeds as follows: We construct a new set T. Into T we place all points which are less than an infinite number of S. T is
not empty since S is bounded below. From (152), T has a supremum; call it s. We now show that s is a limit point of S. Consider any neighborhood N of s. The points in N which are less than s are less than an infinite number of points of S, whereas
those points in N which are greater than s are less than a finite number of points of S. Hence N contains an infinite number of
S, so that the theorem is proved. We have at the same time proved that s is the greatest limit point of S. A limit point may or may not belong to the set.
SEC. 41)
INTEGRATION
93
Problems 1, 2, 3, . . . is 1. The set 1/2, 1/3,. . . , 1/n, . . unbounded. Does it have any limit points? Does this violate the Weierstrass-Bolzano theorem? 2. Prove that every bounded monotonic (either decreasing or increasing) sequence has a unique limit. 3. Prove that lim r" = 0 if Irl < 1. Hint: The sequence r, r2, . . . , r", . . . is bounded and monotonic decreasing for r > 0, and r"+1 = rr" 4. Show that if P is a limit point of a set S, we can pick out a subsequence of 8 which converges to P. 5. Show that (152) implies that every set has an infemum. 6. Show that removing a finite number of elements from a set cannot affect the limit points. 7. Prove the Weierstrass-Bolzano theorem for a bounded set of points lying in a two-dimensional plane. 8. Let the sequence of numbers sl, 82, .. . , sn, . . . satisfy the following criterion: for any e > 0 there exists an integer N such that Is,,+F - snI < e for n ? N, p > 0. Prove that a unique limit point exists for the sequence.
.,
9. A set of numbers is said to be countable if they can be written as a sequence, that is, if the set can be put into one-to-one correspondence with the positive integers. Show that a countable collection of countable sets is countable. Prove that the rationals are countable.
10. Show that the set S consisting of x satisfying 0 5 x 5 1 is uncountable by assuming that the set S is countable, the numbers x being written in decimal form. Theorem of Nested Sets. Consider an infinite sequence of nonempty closed and bounded sets S1, S2, . . . , Sn, . such There that Sn contains that is, S1 M S2 D Sa M .
exists a point P which belongs to every Si, i = 1, 2, 3, . . . . The proof is easy. Let P1 be any point of S1, P2 any point of P,, S2j etc. Now consider the sequence of points P1, P2, . . . . This infinite set belongs to S1 and has a limit point P which belongs to 8, since Sl is closed. But P is also a limit of P, Pn+1, ... , so that P belongs to Sn. Hence P is in all Sn, n = 1, 2, . . . .
... ,
Diameter of a Set. The diameter of a set S is the supremum of all distances between points of the set. For example, if S is the
VECTOR AND TENSOR ANALYSIS
94
[SEc. 41
set of numbers x which satisfy I < x < 1, 3 < x < 7, the diameter of S is 7 - = 61. There are pairs of points in S whose distance apart can be made as close to 6,91 as we please.
Problems 1.
If a set is closed and bounded, the diameter is actually
attained by the set. 2.
Prove this.
If, in the theorem of nested sets, the diameters of the S
approach zero, then P is unique. Prove this. The Heine-Borel Theorem. Let S be any closed and bounded
set, and let T be any collection of open intervals having the property that if x is any element of S, then there exists an open interval 7' of the collection T such that x is contained in T.
The theorem states that there exists a subcollection T' of T which is finite in number and such that every element x of S is contained in one of the finite collection of open intervals that comprise T'. Before proceeding to the proof we point out that (1) both the set S and the collection T are given beforehand, since it is no great feat to pick out a single open interval which completely covers a bounded set S; (2) S must be closed, for consider the set S(1, 1/2, 1/3, . . . , 1/n, . . .) and let T consist of the following set of open intervals: TIIx such that Ix
- 11 < 22
T21x such that Ix
- _f < 32
1
T,,I x such that Ix - 1 n < (n + 1)2
It is very easy to see that we cannot reduce the covering of S by eliminating any of the given T,,, for there is no overlapping of these open intervals. Each T,, is required to cover the point 1/n of S contained in it.
Src. 421
INTEGRATION
95
Proof of the theorem: Let S be contained in the interval
-N < x < N.
This is possible since S is bounded. Now divide this closed interval into two equal intervals (1) -N < x _:5- 0 and
(2) 0 < x < N. Any element x of S will belong to either (1) or (2).
Now if the theorem is false, it will not be possible to cover
the points of S in both (1) and (2) by a finite number of the given collection T, so that the points of S in either (1) or (2) require an infinite covering. Assume that the elements of S in (1) still require an infinite covering. We subdivide this interval
into two equal parts and repeat the above argument. In this way we construct a sequence of sets S,
S2
S3
,
such
that each Si is closed and bounded and such that the diameters of the Si -* 0. From the theorem of nested sets there exists a unique point P which is contained in each Si. Since P is in S, one of the open intervals of T, say T, will cover P. rt'his Tp has a finite nonzero diameter so that eventually one of the Si will be contained in TD, since the diameters of the Si --> 0. But
by assumption all the elements of this Si require an infinite number of the { T } to cover them.
This is a direct contradiction
to the fact that a single Tp covers them. Hence our original assumption is wrong, and the theorem is proved. 42. Uniform Continuity. A real, single-valued function f (x)
is said to be continuous at a point x = c if, given any positive number e > 0, there exists a positive number a > 0 such that If(x) - f(c)I < e whenever Ix - cI < S. The a may well depend on the a and the point x = c. The function f(x) is said to be continuous over a set of points S if it is continuous at every point of S.
We now prove that if f (x) is continuous on a closed and bounded interval, it is uniformly continuous. We define uniform continuity as follows: If, for any e > 0 there exists a S > 0 such that If(XI) - f(X2)1 < e whenever Ix1 - x21 < S, then f(x) is said to be uniformly continuous. It is important to notice
that 8 is independent of any x on the interval. We make use of the Heine-Borel theorem to prove uniform continuity.
Choose
an e/2 > 0. Then at every point c of our closed and bounded set there exists a S(c, a/2) such that If(s) - f (c)I < e/2 for c - a < x < c + S. Hence every point of S is covered by a 26-neighborhood, and so also by a a-neighborhood. By the Heine-Borel theorem we can pick out a finite number of these
96
VECTOR AND TENSOR ANALYSIS
[SEc. 43
Let S, be the diameter of the smallest of this finite collection of 6-neighborhoods. Now consider any two points x, and x2 of S whose distance apart is less neighborhoods which will cover S.
than bl. x1.
Let xo be the center of the 6-neighborhood which covers
From continuity lf(xi) - f(xo)I < e/2. But also x2 differs
from xo by less than 26, so that I f (x2) - f (xo) I < e/2. Consequently If(xl) - f(x2)1 < e. Q.E.D. 43. Some Properties of Continuous Functions (a) Assume f(x) continuous on the closed and bounded interval a < x < b. As a consequence of uniform continuity, we can prove that f (x) is bounded. Choose any e > 0 and consider the corresponding S > 0 such that If(XI) - f(X2)1 < e whenever
-
x21 < S. Now subdivide the interval (a, b) into a finite number of S-intervals, say n of them. It is easily seen that Ix1
max If(x) I < If (a) I + ne. (b) If f (a) < 0 and f (b) > 0, there exists a c such that f (c) = 0,
a < c < b. Let {x} be the set of all points on (a, b) for which f (x) < 0. The set is bounded and nonvacuous since a belongs to {x}, for f(a) < 0. The set {x} will have a supremum; call it c. Assume f (c) > 0. Choose e = f (c) /2. From continuity, a 8 > 0 exists such that If(x) - f (c) I < f(c)12 if Ix cI < 6. Hence f(x) ? Jf(c) for all x in some neighborhood of x = c. Hence c is not the supremum of f x 1. Similarly f (c) < 0 is impossible, so that f (c) = 0. (c) We prove that f (x) attains its maximum. In (a) we showed that f (x) was bounded. Let s be the supremum of f (x),
-
a 5 x S b. As a consequence, f (x) <= 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
VECTOR AND TENSOR ANALYSIS
[SEC. 45
I xm - x,,l < e/2 for all m, n > N. Hence IL - xml < e for all m > N. Q.E.D. Problems
1. Show how the convergence of a series can be transformed into a problem involving the convergence of a sequence.
2. Show that the Cauchy criterion implies that the nth term of a convergent series must approach zero as n -- oo. 3. Show that the sequence 1, 1/2, 1/3, . . . , 1/n, . . . converges by applying the Cauchy test.
45. Regular Arcs in the Plane. Consider the set of points in the two-dimensional plane such that the set can be repreP2 sented
in
some coordinate
system by x = f (t), y = sp(t), a S t < 0, where f(t) and ap(t) are continuous and have continuous first derivatives.
Such
curves are called regular arcs.
A regular curve is a set of points consisting of a finite number of regular arcs joined
one after the other (see Fig. P4
Fia. 45.
45).
PoPI, P1P2, P2t 3, P3P4 are
the regular arcs joined at P1, P2, P3. Notice that there are at most a finite number of discontinuities of the first derivatives. In Fig. 45 the derivatives are discontinuous at P1j P2, P3.
46. Jordan Curves. The locus i x = f (t) a < t 5 fl, will be y = P(t), called a Jordan curve provided that f (t) and (p(t) are continuous
l curve correspond to two and that two distinct points on the distinct values of the parameter t (no multiple points). A closed Jordan curve is a continuous curve having f(a) ,o(a) = 9(fl) but otherwise no multiple points.
From this we see that a Jordan curve is always "oriented," that is, it is always clear which part of the curve lies between two points on the arc, and which points precede a given point. We
shall be interested in those curves which are rectifiable, or, in
IN TkGRA TION
Sr:c. 47]
99
other words, we shall attempt to assign a definite length to a given Jordan are or curve. 47. Functions of Bounded Variation. Let f(x) be defined on
the interval a < x < b. Subdivide this interval into a finite number of parts, say a = xo, x,, . . . x;, . . . , x,-1) x = b. Now consider the sum
- f(xo)I + I f(x2) - f(xi)1 + .
I f(xi)
. + l f(xn) - f(xn - 1)I n
_ i_1 ± I f(xi) - f (xt-1) If the sums of this type for all possible finite subdivisions are bounded, that is, if n
s=x
I f(xi) - f(xc-1)I < A < -o
(153)
we say that f(x) is of bounded variation on (a, b).
A finite, monotonic nondecreasing function is always of bounded variation since n
n
i=1
If(x=) - f(x=-,)1 = I If(x=) - f(xi-1)) = f(b) - f(a) = A t=1
An example of a continuous function that is not of bounded variation is the following: f(x) = x sin 2x,
0< x 5 1
f (O) = 0
Let us subdivide 0 5 x <_ 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.
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
[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)
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
112
ff
[SEC. 54
x a) dd
s
ff s
a
f
f f= a
dd x
a
we have S
ff [a(V s
g) - (a V)gl dd
=a- f f (V.g)dd-a f f dd (summed).
V
There-
fore
xdr = a- f f (ddxv) xg and dr x g = f f (dd x V) x g S
(174)
We notice that in all cases
0 dr * f=
Jf(doxv)*f
(175)
S
The star (*) can denote dot or cross or ordinary multiplication. In the latter case, f becomes a scalar f.
INTEGRATION
Sec. 541
113
Problems
1. Prove that fdr = 0 from (175). 2. Show that L, 'dr x ri taken around a curve in thex-y plane is twice the area enclosed by the curve.
3. If f = cos y i + x(1 + sin y)j, find the value of ff - dr around a circle of radius r in the x-y plane.
4. Prove that fr dr = 0. 5. Prove that ff dd x r =
fir 2 dr.
s
6. Prove that f u Vv dr = - ,f v V u dr. 7. Prove that u Vv - dr = JJvu x Vv dd. s
8. If a vector is normal to a surface at each point, show that its curl either is zero or is tangent to the surface at each point. 9. If a vector is zero at each point of a surface, show that its curl either is zero or is tangent to the surface.
10. Show that fi a x r dr = 2a. Jf dd, if a is constant. S
11. If fi E dr = that V x E =
B dd for all closed curves, show -cat Jf s
---
C at 12. By Stokes's theorem prove that V x (V(p) = 0.
13. Show that fi dr/r = f f (r/r3) x dd where r = Id. 8
14. Find the vector f such that xy = f(O,(X'o, Y,) f dr. o)
15. If f = r/r3, show that V x f = 0 and find the potential yo such that f = Vsp.
16. Show that the vector f = (- yi + xj)/(x2 + y2) is irrotational and that (f r f dr = 2w, where r is a circle containing the origin.
Does this contradict Stokes's theorem? Explain.
17. Show that ff v x f dd = 0, where S is a closed surface. s
18. Let C1 and C2 be two closed curves bounding the surfaces Show that
S1 and S2.
114
VECTOR AND TENSOR ANALYSIS
k. ki r122 dr1
[SEC. 55
dr2 = -4 fJ dal f f dd2 St
s2
f, fc, r122 dr1 x dr2 = - 2 f f dot x ff dd2 Si
s2
where r12 is the distance between points on the two curves.
55. The Divergence Theorem (Gauss). Let us consider a region V over which f and V f are continuous. We shall assume that V is bounded by a finite number of surfaces such that at each surface there is a well-defined continuous normal. We shall also assume that f can be integrated over the total surface bounding V. Now, no matter what physical significance f has, if any,
we can always imagine f to be the product of the density and velocity of some fluid. We have seen previously (Sec. 20) that the net loss of fluid per unit volume per unit time is given by V f. Consequently, the total loss per unit time is given by
JfJ(v.f)dr
(176)
V
Now since f and V f are continuous, there cannot be any point in the region V at which fluid is being manufactured or destroyed;
that is, no sources or sinks appear. Consequently, the total loss of fluid must be due to the flow of fluid through the boundary
S of the region V. We might station a great many observers on the boundary S, let each observer measure the outward flow, and then sum up each observer's recorded data. At a point on the surface with normal vector area dd, the component of the velocity perpendicular to the surface is V N, where N is the unit outward normal vector. It is at once apparent that pV dd = f dd represents the outward flow of mass per unit time. Hence the total loss of mass per unit time is given by
Jff.dd
(177)
Equating (176) and (177), we have the divergence theorem:
JJf.do=fJf(v.f)d.r 8
V
(178)
INTEGRATION
SEC. 55]
115
For a more detailed and rigorous proof, see Kellogg, "Foundations of Potential Theory." We now derive Gauss's theorem by a different method. Let f be a differentiable vector inside a connected region R with rectifiable surface S. Surround any point P of R by a small element of volume dr having a surface area AS. Form the surface integral JJf. dd and consider the limit, As
lira
f
As
Ar-.0
DT
If this limit exists independent of the approach of Ar to zero, we define
Jff.dd div f = lim
As
AT-40
(179)
AT
We can write
tardivf= JJ
(180)
where e -40 as Ar -* 0. If we now subdivide our region into many elementary volumes, we obtain formulas of the type (180) for all of these regions. Summing up (180) for all volumes and then passing to the limit, we have
fffdivfdr__
(181)
Jffdd
Inthe derivation of (181) use has been made of the fact that for
each internal dd there is a -dd, so that all interior surface integrals cancel in pairs, leaving only the boundary surface S as a contributing factor. The sum of the ei Ar; vanishes in the limit, for I Zet Aril 5 ZIel... OT S IEIo,.=V, and if div f is continu-
ous, IeI,. - 0 as 2r --i 0. By choosing rectangular parallelepipeds and using the method of Sec. 20, we can show that
fff.do div f = slim
AS
AT
clu
c3v
aw
ax
+ ay +
49Z
(182)
116
VECTOR AND TENSOR ANALYSIS
[SFC. 55
for f = ui + vj + wk, which corresponds to our original definition of the divergence. Example 65. Consider a sphere with center at 0 and radius a.
Take f=r=xi+yj+zk, dd =
(r) dS = a
(ai)
(
xi+yj+zk)dS
Now V f =3, f dd = (1/a)(x2 + y2 + z2) dS = a dS on the sphere.
Applying (178), f _f f 3 dr = f f a dS, or 3V = aS, where
S is the surface area of the sphere. If V is known to be g 47a8, then S = 4aa2.
Example 6 6. Let f = qr/rs and let V be a region surround-
ing the origin and let S be its We cannot apply the divergence theorem to this resurface.
gion since f is discontinuous at
r = 0. We overcome this difficulty by surrounding the origin by a small sphere E of radius e (see Fig. 50). The divergence theorem can be applied to the connected region V'. The region V' has two boundaries, S and Fzo. 50.
E.
Applying (178),
ffJ V. () dr
f
8
r ad+J1 E
r"d
(183)
In Example 25 we saw that V (r/r3) = 0. This implies that (183) reduces to
ff.!.do=
(184)
Now for the sphere 1, dd = -r dS/e, since the outward normal to the region V' is directed toward the origin and is parallel to the radius vector. Hence
JJ.dd= fff.dd= ffds=4fq
INTEGRATION
SFe. 551
117
The integral f f f dd is called the flux of
since f f dS =
s
the vector field f over the surface S. square force f = qr/r3,
fJ f S
We have, for an inverse-
dd = 4irq
(185)
Example 67. A vector field f whose flux over every closed surface vanishes is called a solenoidal vector field.
From (178)
it is easy to verify that V f = 0 for such fields. Hence a solenoidal vector is characterized by either /f dd = 0 or 0.
Now assume f = V x g, which implies V f = V (V x g) = 0. Is the converse true? If V f = 0, can we write f as the curl of some vector g? The answer is "Yes"! We call g the vector potential of f. This theorem is of importance in electricity theory, as we shall see later. Notice that g is not uniquely determined since V x (g + Vp) = V x g. We now show the existence of g. Let f = Xi + Yj + Zk and assume
g=ai+ij+yk We wish to find a, $, y such that f = V x g, and hence
x
=---
Y= Z
ay
a#
ay
az
as
ay
az
ax
as
as
ax
ay
(186)
=---
Now assume a = 0. Then X =
ay ay
- 8-, Y = - aye Z = as. 8z
ax
Consequently if there is a solution with a = 0, of necessity $6=
fXzdx+o.(y,z)
y=Now V f = 0 or
ax
o xos
Ydx+r(y,z)
aY y
+ az) by assumption.
ax
VECTOR AND TENSOR ANALYSIS
118
f
Hence
ay
aY
JZo
az
dx
,"
+
+ + ar
- as
ay
az
ay
fx ax ax
aZ az /
ar ay
= X (x, y, z) - X (xo, y, z) +
_
[SEC. 55
aa
az
or
ao
ay -
az
so that (186) is satisfied by a = 0, S = I o Z dx + or (y, z), jX
y
Y dx by choosing r = 0, o(y, z) = - JZX(xo,y,z) dz.
Hence f =Vxgwhere
g=[jz
Z dx + c(y, z)] j - faVdxk.
In general
g=[ faZdx+v(y,z)]j- foYdxk+Vp
(187)
For example, if f = V(1/r), then V f = 0. Now
X=--
Y=-y,
s,
Z=-raz
where r2 = x2 + y2 + z2. Applying (187)
g=
x -z dx y dx (x2+y2+z2), j+1 (x2+y2+z2)$k+V
,
v arbitrary
and
g = (x2 + y2 + z2)#(y2 + Example 68. Green's theorem. and f2 = v Vu and obtain
z2)
We apply (178) to fl = u Vv
f f f V. (u Vv)dr= f f f R
(-zj + yk) + VSP
ff
R
S
(188)
f f f V V. (v Vu) dr = f f f (v V2u + VV. VU) d7 = Jf v Vu dd R
S
R
Subtracting, we obtain
f f f (u V2v - v V2u) dr = f f (u Vv - v Vu) dd (189) R
S
INTEGRATION
SEC. 55]
119
Example 69. Uniqueness theorem. Assume two functions which satisfy Laplace's equation everywhere inside a region and
which take on the same values over the boundary surface S. The functions are identical. Let V2(P = 1724, = 0 inside R and (p ='p on S.
Now from (188) we have
JJJov2odr+JfJvo.vodr=JJove.do R
R
Define 0 - (p - ¢ so that 1720 = 0 over R and 0=0 on S. Hence f f f (V0)2 dT = 0, which implies V0 = 0 inside R. R
Hence 0 = constant = p -'p, and since sc
on S, we must
have cp ='G. We have assumed the existence of Vp, V¢ on S. Example 70. Another uniqueness theorem. Let f be a vector
whose curl and divergence are known in a simply connected region R, and whose normal components are given on the surface S which bounds R. We now prove that f is unique. Let f1 be
another vector such that V f = V f1, V x f = V x f1, and f dd = f, dd on S. We now construct the vector g = f - f1.
We immediately have that V g = V x g = g dd = 0. In Sec. 52 we saw that if V x g = 0, then g is the gradient of a scalar Consequently, V g = V2ip = 0. Applying (188) cp, g = V
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
VECTOR AND TENSOR ANALYSIS
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.
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
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;
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
[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]
STATIC AND DYNAMIC ELECTRICITY
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.
VECTOR AND TENSOR ANALYSIS
132
[Sec. 59
4. Let ql, q2, . . . , q be a set of collinear electric charges residing on the line L. Let C be a circle whose plane is normal
to L and whose center lies on L. Show that the electric flux n
through this circle is N = 121rga(1 - cos Sa), where Na is the
a-I
angle between L and any line from qa to the circumference of C. 5. Let the line L of Prob. 4 be the x axis, and rotate a line of
force r in the x-y plane about the x axis (see Fig. 58). If no Y
xa q3
0
x2
xl
qa
qi
x
I
II
`
I
I
1
1
/
Fia. 58.
charges exist between the planes x = A, x = B, show that the equation of a line of force is n
a-1
qa(x - xa)[(x - xa)z + y2]i = constant
6. Point charges +q, -q are placed at the points A, B. The line of force that leaves A making an angle a with AB meets the plane that bisects AB at right angles in P. Show that
sin 2 = 59. Poisson's Formula.
E_
sin (+
PA B)
In Sec. 57, we saw that
-v
a-1 r,
STATIC AND DYNAMIC ELECTRICITY
SEC. 591
133
For a continuous distribution of charge density p, we postulate that the potential is =
fff
pdr
(207)
where the integration exists over all of space. At any point P where no charges exist, r > 0, and we need not worry about the convergence of the integral. Now let us consider what happens at a point P where charges exist, that is, r = 0. Let us surround the point P by a small sphere R of radius e. The integral
f f f (p dr/r) exists if p is continuous. We define 9 at P Y-R
as lim f f f (p dr/r). This limit exists, for using spherical y-R
coordinates,
If f f "d*T
=I
fo2r fo=
0 pr sin a dr d9 d
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
VECTOR AND TENSOR ANALYSIS
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)
STATIC AND DYNAMIC ELECTRICITY
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
VECTOR AND TENSOR ANALYSIS
[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
STATIC AND DYNAMIC ELECTRICITY
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
VECTOR AND TENSOR ANALYSIS
[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]
STATIC AND DYNAMIC ELECTRICITY
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
VECTOR AND TENSOR ANALYSIS
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]
STATIC AND DYNAMIC ELECTRICITY
141
is grounded, and a point charge q is placed at a distance r from
its center, r > b > a. Show that the charge induced on the sphere is Q = -Kabq{r[b + (K - 1)a])-1. 2. A pair of concentric conductors of radii a and b are connected by a wire. A point charge q is detached from the inner one and moved radially with uniform speed V to the outer one. Show that the rate of transfer of the induced charge (due to q) from the inner to the outer sphere is dQ
dt
= -gab(b - a)-'V(a + Vt)-2
3. A spherical condenser with inner radius a and outer radius b is filled with two spherical layers of dielectrics Kl and K2, the boundary between being given by r = J(a + b). If, when both
shells are earthed, a point charge on the dielectric boundary induces equal charges on the inner and outer shells, show that K1/K2 = b/a.
4. A conductor has a charge e, and V1, V2 are the potentials
of two equipotential surfaces which completely surround it (V1 > V2). The space between these two surfaces is now filled with a dielectric of inductive capacity K. Show that the change in the energy of the system is - e(V l - V2) (K - 1 )K 1. 5. The inner sphere of a spherical condenser (radii a, b) has a constant charge E, and the outer conductor is at zero potential.
Under the internal forces, the outer conductor contracts from radius b to radius b1. Prove that the work done by the electric forces is E2(b - b1)b-'b1-1.
64. Method of Images. We consider a charge q placed at a point P(b, 0, 0) and ask if it is possible to find a point Q(z, 0, 0) such that a certain charge q' at Q will cause the potential over the sphere x2 + y2 + z2 = a2, a < b, to vanish. The answer is "Yes"! We proceed as follows : From Fig. 63 we have 82 = z2 + a2 - 2az cos B
t== b2+a2-2abcos6 We choose z so that zb = a2, and call Q(a2/b, 0, 0) the image point of P(b, 0, 0) with respect to the sphere. Thus a2
s2 = a (a2 + b2 - 2ab cos 0) =
a2 b2
0
VECTOR AND TENSOR ANALYSIS
142
[SEC. 64
and
The potential at S due to charges q and q' at P and Q is
P=
s,-I- q
t
=-t Cbq'+qJ
and p = 0 if we choose q' _ - (alb)q.
FIG. 63.
The potential at any point R with spherical coordinates r, 0, (p is
_
(a/b)q
q
(r2 + b2 - 2rb cos 0)#
[r2
+ (a4/b2) - (2a2/b)r cos 0]1 (223)
with = 0 on S and V24 = 0 where no charges exist. Now let us consider the sphere of Example 83. The function of (223) satisfies Laplace's equation and is zero on the sphere. From the uniqueness theorem of Example 69, F of (223) is the potential function for the problem of Example 83. The radial field is given by Er
8
q(r-bcos9)
or
(r2 + b2 - 2rb cos B)' (a/b)q[r - (a2/b) cos 0] [r2 + (a4/b2) - (2a2/b)r cos B];
SEC. 651
STATIC AND DYNAMIC ELECTRICITY
143
and the surface distribution is given by
_ (Er)rs _
b2 - a2
q
4ir a(a2 + b2 - 2ab cos 9)'
47r
Problems 1. A charge q is placed at a distance a from an infinite grounded plane. Find the image point, the field, and the induced surface density. 2. Two semiinfinite grounded planes intersect at right angles. A charge q is placed on the bisector of the planes. What distribution of charges is equivalent to this system? Find the field and the surface distribution induced on the planes.
3. An infinite plate with a hemispherical boss of radius a is at zero potential under the influence of a point charge q on the axis of the boss at a distance f from the plate. Find the surface density at any point of the plate, and show that the charge is attracted toward the plate with a force q2 4f2
4g2a3 fa
+
(f' - a')2
65. Conjugate Harmonic Functions. If we are dealing with a two-dimensional problem in electrostatics, we look for a solution of Laplace's equation V2V = 0. The curves V (x, y) = con-
stant represent the equipotential lines. We know that these curves are orthogonal to the lines of force, so that the conjugate function U(x, y) (see Sec. 56) will represent the lines of force. We know that V2V = V2U = 0. We now give an example of the use of conjugate harmonic functions. In Example 73 we saw that
U(x, y) = A tan-' y x
(224)
V (X, y) = 2 log (x2 + y2)
are conjugate functions satisfying Laplace's equation. If we take V (x, y) as the potential function, then the equipotentials are the circles (A/2) log (x2 + y2) = C, or x2 + y2 = e2cie.
144
VECTOR AND TENSOR ANALYSIS
[SEC. 65
Hence the potential due to an infinite charged conducting cylinder is
log r2 = A log r, r2 = x2 + y2
V (X, y) = 2 log (x2 + y2) =
since A log r satisfies Laplace's2equation and satisfies the bound-
ary condition that V = constant for r = a, the radius of the charged cylinder. If q is the charge per unit length, then
_ aV q
(Er)r..a
ar
24ra
41r
41r
r-a
A 4ara
so that A = -2q and V = -2q log r.
1e=Jr)
0
U=U1
U=U0 (0=0)
Frs. 64.
If we choose U(x, y) = A tan-1 (y/x) as our potential function, then the equipotentials U = constant are the straight lines A tan-1 (y/x) = C, or y = x tan (C/A). As a special case we may take the straight lines 0 = 0, 0 = v as conducting planes raised to different potentials (see Fig. 64). The lines of force are the circles (A/2) log (x2 + y2) = V. The theory of conjugate functions belongs properly to the theory of functions of a complex variable. With the aid of the Schwarz transformation it is possible to find the conjugate functions associated with more difficult problems involving the twodimensional Laplace equation.
Problems
1. By considering Example 72, find the potential function and lines of force for two semiinfinite planes intersecting at right angles.
STATIC AND DYNAMIC ELECTRICITY
SEC. 66J
145
2. What physical problems can be solved by the transformation
x = a cosh U cos V, y = a sinh U sin V? Show that V2U = V2V = O
66. Integration of Laplace's Equation. Let S be the surface of a region R for which V24p = 0. Let P be any point of R, and let r be the distance from P to any point of the surface S. We make use of Green's formula
fJf (,p 02' - ¢ V2V) dr = f f (9 Vi'
DAP)
dd
R
We choose ¢ = 1/r, and this produces a discontinuity inside R, namely, at P, where r = 0. In order to overcome this difficulty, we proceed as in Example 66. Surround P by a sphere Z of radius e. Using the fact that V2,P = V24, = 0 inside R' (R minus the Z sphere), we obtain
0=
f8 f (9 V r
r
V-r) - dd +
f f (p V 1r - r Vp) - dd
(225)
//
B
//
Now V(1/r) = -r/r', and on the sphere X,
r
r8
r
e$
and (1/r) VV dd is of the order eIV(pf, so that by letting e --j 0, (225) reduces to
v(p) = 4
ffir
VV - sa V
r)
dd
(226)
This remarkable formula states that the value of Sp at any point P is determined by the value of (p and V(p on the surface S.
Problems
f
1. If P is any point outside the closed surface S, show that f V(1/r)] dd = 0, where V2cp = 0 inside S and
r is the distance from P to any point of S.
VECTOR AND TENSOR ANALYSIS
146
a2
2. Let rp satisfy V2cp =
aX2 +
[SEC. 67
2
2 = 0. Let r be the closed
1P
y°
boundary of a simply connected region in the x-y plane. If P is an interior point of t, show that
a[log (1/r)]
1) app
1
,P(P) - 2x
`0 an
1(`log r an
ds
where use is made of the fact that
ff
(u an -van) da `
(u V 2V - v V 2U) dA
A
n being the normal to the curve. 3. Let (p be harmonic outside the closed surface S and assume that ip --> 0 and rI V pl -> 0 as r -' oo. If P is a point outside 8, show that
P(P) = 4x Ifs (1r V p - ip v
T
dd
where the normal dd is inward on S. 4. Let so be harmonic and regular inside sphere 7. Show that the value of p at the center of is the average of its values over the surface of the sphere. Use (226). 67. Solution of Laplace's Equation in Spherical Coordinates. From Sec. 23, Prob. 1,
+ aB (sin
+
49(p
o
(sin
0
(227)
B
To solve (227), we assume a solution of the form (228)
V(r, 0, gyp) = R(r)6(0),P(,p)
Substituting (228) into (227) and dividing by V, we obtain sin 0
d8) d( r2 dR) - J +-A (sin O -+
R dr
1
dr
0 d9
d8
1
d2 -- =
4) sin 0 d(p2
0
STATIC AND DYNAMIC ELECTRICITY
SEc. 67j
147
Consequently
_-
_1 d dR\ r R dr dr 2
d
1
0 sin 0 dB
sin
0
d9 d9
1
d4
- sin 2 0 4, dp2
(229)
The left-hand side of (229) depends only on r, while the righthand side of (229) depends on 0 and gyp. This is possible only if both quantities are constant, for on differentiating (229) with
respect to r, we obtain
d
d (r2 LR
i
J
= 0. We choose as
the constant of integration c = -n(n + 1), so that R dr (r2 dR)
n(n + 1)
or
r2
4+24+n(n+1)R=0
(230)
It is easy to integrate (230), and we leave it to the reader to show
that R = Ar" + Br-n-1 is the most general solution of (230). Returning to (229), we have
1d
'e
= n(n -{- 1) sing 0 --
sin
(sin 0 de)
(231)
d0
41)
Since we have again separated the variables, both sides of (231) are constant. We choose the constant to be negative, -m2, m being an integer. This choice guarantees that the solution of d24 d(p2
-I-mq =0
is single-valued when p is increased by tar.
(232)
The solution of
(232) is 4) = A cos mtp + B sin mcp.
Finally, we obtain that 0(0) satisfies sin 0 d0 (sino)+[n(n+1)sin2o_m2Je=o (233) We make a change of variable by letting µ = cos 0,
dµ= -sin0d0
VECTOR AND TENSOR ANALYSIS
148
[SEC. 67
so that (233) becomes
[(1 - µ2)
(1 - µ2)
+ [(1
µ2)n(n + 1) - m2]0
dµ
= 0
(234)
If we assume that V is independent of p (symmetry about the z axis), we have m = 0, so that (234) becomes (235)
d
dIA
This is Legendre's differential equation. By the method of series solution, it can be shown that 0 = P"(14)
1
d"(µ2 - 1)n
2 -n!
dµn
satisfies (235); the P (µ) are called Legendre polynomials.
Two important properties of Legendre polynomials are
the
following: f 11 P,n(;&)Pn(p) dµ
=0
if m : n
J 11 P.'(µ) dµ = 2n + 1
(236) (237)
We give a proof of (236). P. and P. satisfy
[(1 dµ
- p!) dµn1 + n(n + 1)Pn = 0
(238) (239)
Multiplying (238) by P. and (239) by P. and subtracting, we obtain
P. d- [(1 - 2) a ,, - Pn d [(1
_ u2) d µ"'
+[n(n+1) -m(m+l)PPm=0
SFC. 671
STATIC AND DYNAMIC ELECTRICITY
149
or
-M2)(Pmddn-PndPm)]
µ[(1 + [n(n + 1) - m(m + 1)]P,aPm = 0 (240; Integrating between the limits -1 and + 1, we obtain
[n(n + 1) - m(m + 1)J f i P,nPn dµ = 0 and ifm -d n, 1
PmPn dµ = 0
A particular solution of (227) which is independent of gyp, that
is, aV = 0, is given by V(r, 0) = (Anrn + Bnr-' 1)Pn(cos 0). aip
Now it is easy to show that any sum of solutions of (227) is also a solution, since (227) is linear in V. Consequently a more general solution is
V=
n-0
(Anrn + Bn7rn-1)Pn(cos 0)
(241)
provided that the series converges. If we wish to solve a problem involving V2V = 0 with spherical
boundaries, we try (241) as our solution. If we can find the constants An, B. so that the boundary conditions are fulfilled, then (241) will represent the only solution, from our previous uniqueness theorems involving Laplace's equation. We list a few Legendre polynomials:
Po(µ) = 1 Pi(µ) = A P2(p) = . (3µs - 1) P,(µ) = 4(5Aa - 3µ)
P,(µ) = 9(35."4 - 30u$ + 3) P.(0) = 0, n odd P-(0) = (_1)n/x 1-3-5 ... (n - 1)
2.4.6 ... n
Pn(1) = 1 Pn(-µ) _ (-1)"Pn(µ)
(242)
uneven
VECTOR AND TENSOR ANALYSIS
150
[SEC. 68
Problems
1. Prove (237). 2. Solve V2V = 0 for rectangular coordinates by the method of Sec. 67, assuming V = X(x)Y(y)Z(z). 3. Investigate the solution of VI V = 0 in cylindrical coordinates. 68. Applications Example 84. A dielectric sphere of radius a is placed in a uni-
form field Eo = Eok. We calculate the field inside the sphere. The potential due to the uniform field is p = -Eoz = -Eor cos 0. There will be an additional potential due to the presence of the dielectric sphere. Assume it to be of the form ArPI = Ar cos 0 inside the sphere and Br-2PI ° Br-2 cos 0 outside the sphere. We cannot have a term of the type Cr-2 cos 0 inside the sphere, for at the origin we would have an infinite field caused by the presence of the dielectric. Similarly, if a term of the type Dr cos a occurred outside the sphere, we would have an infinite field at infinity due to the presence of the sphere. If we let VI be the potential inside and V11 the potential outside the sphere, we have
VI = --Eor cos 0 + Ar cos 0 (243)
Vu = --Eor cos B + B cos 0 r2
Notice that VI and Vn are special cases of (241). We have two unknown constants, A, B, and two boundary conditions,
VI= Vu at r=a or
DN, = D.Y.
(see Sec. 62).
a I
K
=
II
a
at r = a
(244)
From (243) and (244) we obtain
A=K-'Eo,
x+2
B_aaK-1Eo K+ 2
so that
VI = I
-'V+2 Eor cos 0 = -
2 Eoz K
+
(245)
STATIC AND DYNAMIC ELECTRICITY
SEC. 681
151
We see that the field inside the dielectric sphere is
E_ -vVI=K+2Eo and E is uniform of intensity less than E0 since K > 1. the sphere
VII = -Eor cos B +
K - 1 _ Eo cos 8 K + 2 r2
Outside (246)
The radial field outside the sphere is given by
aVII=Eocos0+2K +
E,
ocos9
2aa
For a given r the maximum E, is found at 0 = 0.
Example 85. A conducting sphere of radius a and charge Q is surrounded by a spherical dielectric layer up
to r = b (Fig. 65). Let us calculate the potential distribution.
From
spherical
symmetry V = V (r), so that we try
FlG. 65.
The boundary conditions are
VI= VIIatr=b
(i)
aVI
_
aVII
at r = b Or = Or J2rJT (a D dd = K
(11)
(iii)
Q= 41r
ff s
47
T a2 sin 0 d9 dp
J
From (i) Alb = (B/b) + C; from (ii) -A/b2 = -KB/b2; from (iii) Q = (K/4x) (B/a2) f f dS = KB.
Hence
VECTOR AND TENSOR ANALYSIS
152
VI = Q'
VII =
Q +90C
1
(SEC. 68
(247)
K
Example 86. A conducting sphere of radius a and charge Q is placed in a uniform field. We calculate the potential and the distribution of charge on the sphere. We assume a solution
V = -Eor cos 0 +
B
r
The boundary condition is
Q = 4v
Jf - avlr-a dS
so that Q
=
1
4
,
f2v
f-
(E cos 8 + a a2 sin 8 d8 dsp = B
and
V= -Eorcos8+Qr For the charge distribution aV
a-
cIr
4"
4I(E0cos8+Q
Example 87. Consider a charge q placed at A (b, 0, 0). Let us compute the potential at any point P(r, 8, gyp) (see Fig. 66). The potential at P is 4
V=
= q(r2 + b$ - 2rb cos
8)_;
There are two cases to consider:
(a) r < b. Let µ = cos 8, x = r/b, so that (1 - 2µx
V= b
+2)-}
Now (1 - 2µx + x=)-} can be expanded in a Maclaurin series in powers of x, yielding
SEC. 68]
STATIC AND DYNAMIC ELECTRICITY _
m
V=b
n= b
n0 I
153
m ()fl
n0
P.(--)
( 248)
The proof is omitted here that the Pn(p) are actually the Legendre polynomials. However, we might expect this, since V satisfies Laplace's equation and P,,(µ)rn is a solution of VI V = 0. z
Fia. 66.
(b) r > b. In this case
V = q I Pn6u) rn=0
/b n
(249)
r
Notice that each term is of the form Pn(µ)r-n-1, which satisfies Laplace's equation.
Example 88. A point charge +q is placed at a distance b from the center of two concentric, earthed, conducting spheres of radii a and c, a < b < c. We find the potential at a point P
for a
For r > b, V = (q/r) 0
and for r < b, V = (q/b) 2 (r/b)'Pn(cos 0). Moreover, we have 0
VECTOR AND TENSOR ANALYSIS
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
STATIC AND DYNAMIC ELECTRICITY
SEC. 691
155
n(n + 1) 2n+1 ,410n+K(n+ 1)(a/
(K - 1)g2'' c2
3. A point charge q is placed a distance c from the center of an
earthed conducting sphere of radius a, on which a dielectric layer of outer radius b and constant K exists. Show that the potential of this layer is m
V
=
(2n + 1 )b2n+1(rn - a2n+lr-n-1) q -c n c cn { [(K +1)n + 1]b 2n+1 + (n +1)(K -1)a 2n+1 } Pn(cos e) 0
4. Show that the potential inside a dielectric shell of internal and external radii a and b, placed in a uniform field of strength E, is
V=
9KE
9K - 2(1 - K2)[(b/a)3 - 1]
r cos 0
5. The walls of an earthed rectangular conducting tube of infinite length are given by x = 0, x = a, y = 0, y = b. A point charge is placed at x = xo, y = yo, z = zo inside the tube. Show that the potential is given by V = 8q
I (m2a2 + nil m-1
n2b2)-e-a-'b-'(mta'+n$t)i*(s-ss)
to
sin
n1rx0
a sin
nrx a
sin
mTryo
b
sin
miry b
69. Integration of Poisson's Equation. Instead of assuming that p is harmonic, let us consider that 1p satisfies V2co = -47rp. By applying Green's formula as in Sec. 66, we immediately obtain
,P(P) = Jff rdr+Jf (rVp
- jP 0
(254)
If we make the further assumption that rp is of the order of 1/r for large r and that jVjpj - 1/r2, we see that by pushing S out to infinity the surface integral will tend to zero. Our assumption is valid, for if we assume the charge distribution to be bounded by some sphere, then at large distances the potential will be of the order of I /r, since we may consider all the charges as essentially
156
VECTOR AND TENSOR ANALYSIS
[SEc. 70
concentrated at a point. Thus
V(P) = JJf e dr
(255)
70. Decomposition of a Vector into the Sum of Solenoidal and Irrotational Vectors. In Example 70, we saw that if Ifl tends to
zero like 1/r2 as r --> oo, then f as uniquely determined by its curl and divergence. We now proceed to write f as the sum of irrotational and solenoidal vectors. Let f dr
W(P) = JfJ
(256)
where r is the distance from P to the element of integration dr. If we write f = fli + f j + f k, W = WA + W,,j + Wek, then
f
ffrldr
W2 = f f f
f2
dr
(257)
M
W3=fffr$dr
00
We assume that the components of f are such that the integrals of (257) converge and that I Wl - 1/r, I V W,,l '' 1/r2, n = 1, 2, 3. From Sec. 69, V2W. = -4xfn, so that
V2W = -4,rf
(258)
From (256)
(259)
VxW= JJJf xVdr 40
Now
V x(V xW) =
--V2W
SEc. 711
STATIC AND DYNAMIC ELECTRICITY
157
so that f
=4-V
-4
X
X
and hence
f = V x A + Vcp
260)
where
A=4-VxW, = Problems 1. Show that (256) is a special case of (254). 2. Find an expression for (p(P) if V2(p = -41rp inside S and if P is on the surface S.
3. f = yzi + xzj + (xy - xz)k. Express f as the sum of an irrotational and a solenoidal vector. 71. Dipoles. Let us consider two neighboring charges -q and +q situated at P(x, y, z) and Q(x + dx, y, z). The potential at
the origin 0(0, 0, 0) due to -q is -q/r, and that due to +q is q/(r + dr), where r = (x2 + y2 + Z2)I and
r+dr= [(x + dx)2 + y2 + Z2]1 The potential at 0(0, 0, 0) due to both charges is q
r + dr
q
r
g2dr r2
Now dr = x dx/r, so that rp - qx dx/r3. If we now let q -p co and dx -+ 0 in such a way that q dx remains finite, we have formed what is known as a dipole. Let r be the position vector from the origin to the dipole, and let M = q dr, where dr is the vector from
the negative charge to the positive charge; !drl = dx. We have (M r)/r8 = (qr dr)/ra = (qx dx)/r3, so that
M is called the strength or moment of the dipole.
than one dipole, the potential at a point P is given by
For more
VECTOR AND TENSOR ANALYSIS
158
i= i
[SEC. 72
(Mi.ri) ri3
where r; is the vector from P to the dipole having strength M. Example 89. The field strength due to a dipole is E _ --Vv so that r (262) E = V rs ) rs ra
-
-
Example 90.
Potential energy of a dipole in a field of potential V.
Let (p, be the potential at the charge q and c2 the potential at the charge -q. The energy of the dipole is
L
app
ds = M W = spiq + IP2(-q) = q(spl - IP2) = g as ao where ds is the distance between the charges. Now
d,p = ds V p
so that W = M as VV = M V(p. 72. Electric Polarization. Let us consider a volume filled with dipoles. The potential due to any single dipole is given by (261). If we let P be the dipole moment per unit volume, that is, P = lim (AM/or), then the total potential due to the dipoles is A"o
= I f f r ra- dr
(263)
Now V V. (P/r) = (1/r)V P - [(P r)/r3]. The reason that we have taken V(1/r) t= r/r3 instead of -r/r3 is that
r = [( - x)2 + (n - y)2 + (J
- z)2]'
and V performs the differentiations with respect to x, y, z, the coordinates of the point P at which V is being evaluated. The coordinates t, ot, t belong to the region R and are the variables of
integration, and r = (E - x)i + (n - y)j + (r - z)k. (263) becomes `P=fffv.(!)dr R
fff R
Hence
SEC. 72]
STATIC AND DYNAMIC ELECTRICITY
159
and
(p -
ff!.dd_ fff_!d
r
(264)
by applying the divergence theorem. Example 91. Let us find the electric intensity at the center of a uniformly polarized sphere. Here P = Pok, so that V P = 0 inside R. Hence (264) becomes sp(x, y, z) =
Pok dd
ff
x)2 + (71 - y)2 + ( - z) 2]i
S
and
E=
fj
Po(k
dd)[(E - x)i + (n
- y)j + (1' - z)k]
[(E-x)2+(7] -y)2+(J -z)2]$
S
E(0,0,0)
f
f
ni + 'k)Po(k dd)
(265)
Now for points on the sphere, 52 +' 72+-2 = a2,
and letting t = a sin B cos -o, n = a sin 0 sin tp, is easily seen that (265) reduces to
E(0, 0, 0) = -frPok
a cos 0, it (266)
E is independent of the radius of the sphere. By superimposing (concentrically) a sphere with an equal but negative polarization, we see that the field at the center of a uniformly polarized shell is zero. Problems 1. Prove (262). 2. Prove (266). 3. If M1 and M2 are the vector moments of two dipoles at A and B, and if r is the vector from A to B, show that the energy - 3(M1 r)(M2 r)r-6. of the system is W = M1 4. The dipole-moment density is given by P = r over a sphere of radius a. Calculate the field at the center of the sphere. Mgr-a
160
VECTOR AND TENSOR ANALYSIS
[SEc. 73
73. Magnetostatics. The same laws that have held for electrostatics are true for magnetostatics with the exception that OZcp.. = 0 always, since we cannot isolate a magnetic charge. We make the following correspondences, since all the laws of electrostatics were derived on the assumption of the inversesquare force law, which applies equally well for stationary magnets.
EH
Magnetostatics
Electrostatics q
qm
D ---) B (magnetic induction) K <---- 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
STATIC AND DYNAMIC ELECTRICITY
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
VECTOR AND TENSOR ANALYSIS
[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
STATIC AND DYNAMIC ELECTRICITY
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
VECTOR AND TENSOR ANALYSIS
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]
STATIC AND DYNAMIC ELECTRICITY
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
VECTOR AND TENSOR ANALYSIS
[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
STATIC AND DYNAMIC ELECTRICITY
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.
VECTOR AND TENSOR ANALYSIS
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
STATIC AND DYNAMIC ELECTRICITY
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
VECTOR AND TENSOR ANALYSIS
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).
STATIC AND DYNAMIC ELECTRICITY
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
VECTOR AND TENSOR ANALYSIS
[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
STATIC AND DYNAMIC ELECTRICITY
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)
VECTOR AND TENSOR ANALYSIS
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'`i`
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)
STATIC AND DYNAMIC ELECTRICITY
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
VECTOR AND TENSOR ANALYSIS
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
STATIC AND DYNAMIC ELECTRICITY
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
VECTOR AND TENSOR ANALYSIS
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)
STATIC AND DYNAMIC ELECTRICITY
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)
VECTOR AND TENSOR ANALYSIS
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
STATIC AND DYNAMIC ELECTRICITY
SBc. 831
1
r z=o (F)
V
IL f(x)
dx =
181
1
V
(r
t_-r/v
Hence the left-hand side of (3' 09) reduces to 4V
(311)
()t_-r/vdr
Now considering the right-hand side of (309), we see that
lim e-.0
ff
E
rf(E + Vt) Dtp L
E
(pf'(E + Vt)
+
R] dd = 0
E
since dd is of the order E2, and f, f , tp, Dtp are bounded for a fixed a.
We also have that
lim -
t: jI
dt f f
since f f dS =
'pE, f(E + Vt) . dS = -
1
v 47rcp(P)
(312)
and for small 6,
E
Finally,
fs f i:'
dtlf(r+ Vt)D
(rf'
Dr] dd
2
r2
L
ii fl'
= s
dt{f(r+Vt)Dp-Vr].dd
r
r
f
- f f j,t dt r f'(Dr . dd) = ff f .t dt
[f(r + Vt) r
+ -f Dr] . dd + f f j i s rV f at dt (Dr dd)
(313)
s
on integrating by parts and noticing that f} = 0. Finally the right-hand side of (313) becomes equivalent to
V sf f(rD t - -r/V -,p t--r/V
1
r
1
t3(o
rV at t- -r/V (314)
VECTOR AND TENSOR ANALYSIS
182
[SEC. 83
Combining (311), (312), and (314), we obtain V(P)
` 1 JR J
F=
dr
-r/ v 1
- i fJ
(v , ~
S
V r + rV at
V
r)
dd
(315)
t= -r/'V
Now let S recede to infinity and assume that (p,
when
evaluated at t = -r/V on the surface S, have the value zero until a definite time T. For large r, t = -r/V is negative and so is always less than T. Hence the surface element vanishes, and V
(P)=
JJJ
t- -r/V
di
-
(316)
The solutions to (305) are thus seen to be A (P,
t) =
f fJ W
4
pv
Ie
r t-(r/V)
,v(P,t)'JLI x t - (r/V) whe re V = Finally,
dr 317)
dr
B=VxA l aA E=-'cat-V
(318)
The physical interpretation of these results is simple. The values of the magnetic and electric intensities at any particular point P at any instant t are, in general, determined not by the state of the rest of the field (p, v) at that particular instant, but by its previous history. The effects at P, due to elements at a distance r from P, depend on the state of the element at a previous time t -- (r/V). This is just the difference in time required for the waves to travel from the element to P with the velocity V = c/ V AK, hence the name retarded potential. Had we considered the function f(r - Vt), we should have obtained a solution depending on the advanced potentials. Physically this is impossible, since future events cannot affect past eventsl
SEC. 831
STATIC AND DYNAMIC ELECTRICITY
183
Problem
1. A short length of wire carries an alternating current, j = pv = Io (sin wt)k, -1/2 5 z 5 1/2. (a) At distances far removed from the wire, show that
A=
j-o1
cr
sin w
(t
- rc//J k
and that in spherical coordinates A,.
=jot sin wit - r1 cos 0
\ c/l / Isin0 Ae= - Iotsinwltc// cr cr
A, =O (b) Show that H, = He = 0, and that Hw =
cr sin 0
cos w Ct - c) + r sin w IC
t - -) 1
(c) Find p from the equation of gauge invariance, and then
E,, E8, E from E + c A - V o.
CHAPTER 6 MECHANICS
84. Kinematics of a Particle. We shall describe the motion of a particle relative to a cartesian coordinate system. The motion of any particle is known when r = x(t)i + y(t)j + z(t)k is known, where t is the time. We have seen that the velocity and acceleration, relative to this frame of reference, will be given by
v' dt i +dt a=
d2y
d2x dtz 1
+
dt
j
dtz
k
d zz
+
dt2
k
The velocity may also be given by v = vt, where v is the speed
and t is the unit tangent vector to the curve r = r(t). Differentiating, we obtain
a=dt
dtt+vdsdt
by making use of (95).
=dtt+Kv2n
(319)
Analyzing (319), we see that the accelera-
tion of the particle can be resolved into two components: a tangential acceleration of magnitude dt, and a normal accelera-
This latter acceleration is called centripetal acceleration and is due to the fact that the velocity vector is changing direction, and so we expect the curvature to play a role here. For a particle moving in a plane, we have seen in Sec. 17, Example 18, that the acceleration may be given by tion of magnitude v2rc = v2/p.
2r
2l
- r ` de]R -}- r d (2 de P = [ dt r
a
184
MECHANICS
SEC. 84J
Example 101.
185
Let us assume that a particle moves in a plane
and that its a cceleration is only radial. In this case we must have r d [ r2 de1 = 0, and integrating, +r2 dte = h = constant.
From the calculus we know that the sectoral area is given by dA = }r2 dB (see Fig. 72). Thus dA = constant, so that
equal areas are swept out in equal intervals of time. Example 102. For a particle
Fie. 72.
moving around a circle r = b with constant angular speed coo = d
we have dt = 0 and
(r2wo) = 0, so that a = -bwo2R.
Example 103. To find the tangential and normal components of the acceleration if the velocity and acceleration are known.
v=vt, and
a v = vat Also
so that
as =
v
axv=va,nxt= -vanb
and
an = I$)V,i°I
Problems 1. A particle moves in a plane with no radial acceleration and constant angular speed wo. Show that r = Ae"o' + Be at. 2. A particle moves according to the law
r = cos t i + sin t j + t2k Find the tangential and normal components of the acceleration.
VECTOR AND TENSOR ANALYST °
186
[SEc. 85
3. A particle describes the circle r = a cos 0 with constant Show that the acceleration is constant in magnitude and directed toward the center of the circle. 4. A particle P moves in a plane with constant angular speed w about 0. If the rate of increase of its acceleration is parallel speed.
d2r
to OP, prove that ate = 4rw2 5. If the tangential and normal components of the acceleration
of a particle moving in a plane are constant, show that the particle describes a spiral. 85. Motion about a Fixed Axis. In Sec. 10, Example 12, we saw that the velocity is given by v = w x r. Differentiating, we obtain dw dr xr a=wxa+dt
a=wxv+axr
(320)
Since v = w x r, we where a is the angular acceleration --' at have also
wl
a=wx(wxr)+axr
_ (w - r)w - w2r + a x r
If we take the origin on the line of w in the plane of the motion, then w is perpendicular
tororor = 0, so that
a= -w2r+axr a x r is the tangential acceleraFm. 73.
tion, and w x (w x r) is the
centripetal acceleration. If we assume that a particle P is rotating about two intersecting lines simultaneously, with angular velocities cal, w2 (Fig. 73), we can choose our origin at the point of intersection so that
vi = wl x r, and the total velocity is
v2 = 632 x r
V = V1 + V2 = (wi + (02) x r
MECHANICS
SFC. 86]
187
A particle on a spinning top that is also precessing experiences such motion. 86. Relative Motion. Let A and B be two particles traversing curves r, and r2 (Fig. 74). r, and r2 are the vectors from a point 0 to A and B, respectively.
r2=r+rl
(321)
Definition: dt is the relative velocity of B with respect to A, written V4(B).
Fia. 74.
Differentiating (321), we have dr2
dr
dt - dt
dri
+
dt
or
Vo(B) = VA(B) + Vo(A)
(322)
More generally, we have
Vo(A) = V4,(A) + VA,(A1) + VA.(A,) + ..
.
+ Vo(A.)
It is important to note that V4(B) _ -VB(A). Example 104. A man walks eastward at 3 miles per hour, and the wind appears to come from the north. He then decreases his speed to 1 mile per hour and notices that the wind comes from the northwest (Fig. 75). What is the velocity of the wind? We have V0(W) = VM(W) + V0(M) G(ground)
VECTOR AND TENSOR ANALYSIS
188
ISEc. 86
In the first case
VM(W) = -kj,
V0(M) = 3i
so that
VG(JV) = -kj + 3i In the second case,
VM(W) = h(i - j),
VG(M) = i
so that
V0(W) = h(i - j) + i = (h + 1)i - hj, and
3=h+1,
-k= -h,
VG(W) =3i-2j
miles per hour, and its direction The speed of the wind is makes an angle of tan-' I with the south line.
N
x.
E
S Fia. 75.
Fzo. 76.
Example 105. To find the relative motion of two particles moving with the same speed v, one of which describes a circle of radius a while the other moves along the diameter (Fig. 76). We have
P=acos0i-}-asin0j,
adze=v
Q = (a - vt)i This assumes that both particles started together.
T -dQ=(-a sin0dO+v)i+acos0doi VQ(P) = v(1 - sin 0)i + v cos 6 j
MECHANICS
Sec. 87]
189
The relative speed is IVQ(P)I = [v2(1
- sin
0)2 + v2 cos2 B]} = 2}v(1 - sin B)}
Maximum IVQ(P)I occurs at 0 = 3x/2, minimum at 0 = x/2. Problems 1. A man traveling east at 8 miles per hour finds that the wind seems to blow from the north. On doubling his speed, he finds that it appears to come from the northeast. Find the velocity of the wind. 2. A, B, C are on a straight line, B midway between A and C. It then takes A 4 minutes to catch C, and B catches C in 6 minutes. How long does it take A to catch up to B?
3. An airplane has a true course west and an air speed of 200 miles per hour. The wind speed is 50 miles per hour from 1300. Find the heading and ground speed of the plane. 87. Dynamics of a Particle. Up to the present, nothing has
been said of the forces that produce or cause the motion of a particle. Experiment shows that for a particle to acquire an acceleration relative to certain types of reference frames, there must be a force acting on the particle. The types of forces encountered most frequently are (1) mechanical (push, pull), (2) gravitational, (3) electrical, (4) magnetic, (5) electromagnetic. We shall be chiefly concerned with forces of the types (1) and (2). For the present we shall assume Newton's laws of motion hold for motion relative to the earth. Afterward we shall modify this. Newton's laws are: (a) A particle free from the action of forces will remain fixed or will continue to move in a straight line with constant speed. (b) Force is proportional to time rate of change of momentum,
that is, f = dt (mv). In general, m = constant, so that
The factor m is found by experiment to be an invariant for a given particle and is called the mass of the particle. In the theory of
relativity, m is not a constant. my is called the momentum. (c) If A exerts a force on B, then B exerts an equal and opposite force on A. This is the law of action and reaction: fAB = -fha
VECTOR AND TENSOR ANALYSIS
190
[SEC. 88
By a particle we mean a finite mass occupying a point in our Euclidean space. This is a purely mathematical concept, and physically we mean a mass occupying negligible volume as compared to the distance between masses. For example, the earth
and sun may be thought of as particles in comparison to their distance apart, to a first approximation. 88. Equations of Motion for a Particle.
Newton's second law
may be written f = m dt = ma. We postulate that the forces acting on a particle behave as vectors. This is an experimental fact. Hence if fl, f2j . . . , f act on m, its acceleration is given by a
1
(fl+f2+....+fn)_
m
1I
mti_1
We may also write f = m d
dt2
,
f,_ mff
where r is the position vector from
the origin of our coordinate system to the particle. If the particle d
is at rest or is moving with constant velocity, then
dt2
= 0, and
so f = 0, and conversely. Hence a necessary and sufficient condition that a particle be in static equilibrium is that the vector sum of the forces acting on it be zero.
A standard body is taken as the unit mass (pound mass). A poundal is the force required to accelerate a one-pound mass one foot per second per second. The mass of any other body can be compared with the unit mass by comparing the weights (force of f12/ m2 gravit y at mean sea leve l ) o f th e two objects. This assumes the equivalence of gravitational mass and inertial mass. Example 106. Newton's law of gravi-
tation for two particles is that every pair of particles in the universe exerts Fzo. 77.
a mutual attraction with a force directed along the line joining the particles, the magnitude of the force being inversely
proportional to the square of the distance between them and directly proportional to the product of their masses. f12 = (Gmim2/r2)R (see Fig. 77). G is a universal constant.
Let
191
MECHANICS
SEC. 88]
the mass of the sun be M and that of the earth be m. We shall assume that the sun is fixed at the origin of a given coordinate system (Fig. 78). The force act-
ing on the earth due to the sun is f = - (Gm.M/r3)r From the second law GmM rs r
dv dt
d2r m dt2
so that dv
GM
dt
r3
M
(323)
Fia. 78.
Now d v) dt (r x
dv
= r x dt
and hence d
dt(rxv)=rx(-GMr) =0
This implies
r x v = h = constant vector or
rxa = dr
(324)
h
Since Ir x drl = twice sectoral area, we have 2 dA = Ihl, or equal
areas are swept out in equal intervals of time. This is Kepler's first law of planetary motion. Moreover, r I r x dt ] = r h = 0,
so that r remains perpendicular to the fixed vector h, and the motion is planar. Now
-GMrxh= -GMrx(rxv)
a xh from (324), and
d
d
(v x h) = d x h, so that (v x h)
GM
r x (r x v)
(325)
VECTOR AND TENSOR ANALYSIS
192
Now r = rR, where R is a unit vector.
SEC. 88
Hence R
v
so that (325) becomes
\ d(vxh)=-GMrx (rxrR) / =
-GMRx(RxdR) -GM K R .
ddR
t/
R - R2 dRJ
= GM ddR
(326)
since R is a unit vector. Integrating (326), we obtain
vxh=GMR+k and
h' = GMr + rk cos (R, k)
(327)
Thus h2/GM
r
(328)
1 + (k/GM) cos (r, k)
We choose the direction of the constant vector k as the polar axis, so that
r
h'/GM 1 + (k/GM) cos 8
(329)
This is the polar equation of a conic section. For the planets these conic sections are closed curves, so that we obtain Kepler's second law, which states that the orbits of the planets are ellipses with the sun at one of the foci. Let us now write the ellipse in the form S
r
1+ e cos 9
where e=
GM' p
k
MECHANICS
SEC. 88]
193
The curve ci osses the polar axis at 0 = 0, 0 = it so that the length of the major axis is
ep + ep 2, - 1+e 1-e
2p
2h2
1-e2
GM(1-e2
For an ellipse, b2 = a2 - c2 = a2 - e2a2, orb = a(l - ,2)1. area of the ellipse is A = Tab = ,ra2(1 - e2)1, and since
The
dA = Jh, dt
the period for one complete revolution is
T
2A
2ara2(1 - e2)i Zral - e2)1 = G1Mi
- h - a1GiM1(1
Thus
_
T2
47r2
0
GM =
constant, for all planets
(330)
This is Kepler's third law, which states that the squares of the periods of revolution of the planets are proportional to the cubes of the mean distances from the sun. Problems
1. A particle of mass m is attracted toward the origin with the force f = - (k2m/r6)r. If it starts from the point (a, 0) with the speed vo = k/21a2 perpendicular to the x axis, show that the path is given by r = a cos 0. 2. A bead of mass m slides along a smooth rod which is rotating
with constant angular speed w, the rod always lying in a horizontal plane. Find the reaction between bead and rod. 3. A particle of mass m is attracted toward the origin with a force - (mk2/r3)R._ If it starts from the point (a, 0) with velocity vo > k/a perpendicular to the x axis, show that the equation of the path is (OW k2)i r = a sec C 0
-
avo
I
4. In a uniform gravitational field (earth), a 16-pound shot leaves the putter's fingers 7 feet from the ground. At what angle should the shot leave to attain a maximum horizontal distance?
194
VECTOR AND TENSOR ANALYSIS
[SEC. 89
5. Assume a comet starts from infinity at rest and is attracted toward the sun. Let ro be its least distance to the sun. Show that the motion of the comet is given by r = 2ro/(1 + cos 0). 89. System of Particles. Let us consider a system consisting of a finite number of particles moving under the action of various forces. A given particle will be under the influence of two types of forces: (1) internal forces, that is, forces due to the interaction of the particle with the other particles of the system, and (2) all
other forces acting on the particle, said forces being called external forces.
If r; is the position vector to the particle of mass m,, then we shall designate f,(e) as the sum of the external forces acting on the jth particle, and f,(i) as the sum of the internal forces acting on this particle. Newton's second law becomes for this particle
f1( + f1( = m; d2r'
(331)
Unfortunately, we do not know, in general, f;('), so that we shall not try to find the motion of each particle but shall look rather for the motion of the system as a whole. Since Eq. (331) is true for each j, we can sum up j for all the particles. This yields n
J=1
f'c6> +
d 2r,
f'(i) ;=1
1-1
m' dt2
From Newton's third law we know that for every internal force n
there is an equal and opposite reaction, so that This leaves
A
f ®= 0.
n
n
l m,
f,
dt2
(332)
We now define a new vector, called the center-of-mass vector, by the equation n
Im,r,
n
ra =1' =
(333) n
;11
7-1
MECHANICS
SEC. 89)
195
The end point of r. is called the center of mass of the system. It is a geometric property and depends only on the position of the particles. Differentiating (333) twice with respect to time, we obtain Md2r` _ dt2
n
md?r, ' dt2
;=1
so that (332) becomes n
f
f.(e)=M
2_1°
(334)
z
)ml
Equation (334) states that the center of mass of the system accelerates as if the total mass were concentrated there and all the external forces acted at that point.
Fia. 79.
Example 107. If our system is composed of two particles in free space and if they are originally at rest, then the center of
mass will always remain at rest, since f = 0 so that d2° = 0, and r. = constant satisfies the equation of motion and the initial condition
0.
For the earth and sun we may choose the center
of mass as the origin of our coordinate system (Fig. 79). The equations of motion for earth and sun are m
d2r1 dt2
r2)2f = - (rlGmMR
M d2r2 dt2
_ (rlGmMR - 12)2
Since r. = 0, we have mr, + Mr2 = 0, and d2r1 d12
_
-GM
rl
[1 + (m/M))2 rig
196
VECTOR AND TENSOR ANALYSIS
[SEc. 91
This shows that m is attracted toward the center of mass by an inverse-square force.
The results of Example 106 hold by replac-
ing M by M[1 + (m/M)1-2. Problems
1. Show that the center of mass is independent of the origin of our coordinate system. 2. Particles of masses 1, 2, 3, 4, 5, 6, 7, 8 are placed at the corners of a unit cube. Find the center of mass. 3. Find the center of mass of a uniform hemisphere. 4. Find the force of attraction of a hemisphere on another hemisphere, the two hemispheres forming a full sphere. 90. Momentum and Angular Momentum. The momentum of a particle of mass m and velocity v is defined as M = mv. The total momentum of a system of particles is given by M = j m;v1. j-1
We have at once that dM dt
n
I mj j-1
dv1
dt =
n
I f; (e) = f
(335)
j-1
We emphasize again that the mass of each particle is assumed constant throughout the motion. The vector quantity r x my is defined as the angular momentum, or moment of momentum, of the particle about the origin 0. f The total angular momentum is given by n
H = E r; x mjvj
(336)
j-1 91. Torque, or Force Moment. Let Fla. 80.
f be a force acting in a given direction and let r be any vector from the origin
whose end point lies on the line of action of the force (see Fig. 80). The vector quantity r x f is defined as the force moment, or torque, of f about 0. For a system of forces, n
L = I r, x f, j-1
(337)
MECHANICS
SEC. 921
197
We immediately ask if the torque is different if we use a differ-
ent vector ri to the line of action of f.
The answer is in the
negative, for
(r, - r) x f = 0 since ri - r is parallel to f. Hence rl x f = r x f. What of the torque due to two
equal and opposite forces both acting along the same line? It is zero, for
r1xf+r1x(-f)
=r1x(f-f)=0
Two equal and opposite forces
with different lines of action constitute a couple (see Fig. 81).
Let ri be a vector to f and r2 a vector to -f. The torque due
Fio. 81.
to this couple is
L = r1xf+r2x(-f) = (r1 - r2) x f
The couple depends only on f and on any vector from the line of action of -f to the line of action of f. Problems 1. Show that if the resultant of a system of forces is zero, the total torque about one point is the same as that about any other point. 2. Show that the torques about two different points are equal, provided that the resultant of the forces is parallel to the vector joining the two origins.
3. Show that any set of forces acting on a body can be replaced by a single force, acting at an arbitrary point, plus a Prove this first for a single force. 4. Prove that the torque due to internal forces vanishes.
suitable couple.
92. A Theorem Relating Angular Momentum with Torque. We are now in a position to prove that the time rate of change of angular momentum is equal to the sum of the external torques for a system of particles.
198
VECTOR AND TENSOR ANALYSIS
[SEC. 93
Since n
H = Ir3xmjv
r, x
j=1
j=1
dri
m'dt
we have on differentiating
dH _ dt
n
Z r, x m,
dt2
n
I
j-1 r' x
aii = dt
n
d2r,
dri
dri
+ j=1 dt x m' dt
(f1" + f, (') ) n
I r; x
j=1
f;(e)
=L
(338)
93. Moment of Momentum (Continued). It is occasionally more useful to choose a moving point Q as the origin of our
FIG. 82.
coordinate system. space. We define
Let 0 be a fixed point and Q any point in n
dri
HQa = I (r, -- rQ) x m; di j-1
(339)
The superscript a stands for absolute momentum, that is, the velocity of m; is taken relative to 0, whereas the subscript Q stands for the fact that the lever arm is measured from Q to the particle m; (Fig. 82).
Differentiating (339), we obtain
MECHANICS
SEc. 941
dHQa
_
dt
drQl
(dri _ drQ
dl
dr,
dt /
+ L (r. -
dt
dr1
n
j-1
m;r,, so that M dt =
m1
dr,
-, and
l1
j=1
j-1
dt2
(r1 - rQ) x (f.(e) + f .(i))
xLm'dt+I
n
Now Mr =
rQ) x "ni
j=1
n
j=1
d2r;
nn`
X 7n,
(dt
j=1
199
n f,(i)=0,
j=1
r1 x f;(') = 0 from Sec. 91, so that dHQa
M
dt =
drQ
I
dF
n
(r, - rQ) dt x dt + j-1
x fj(e)
or dtQa
LQ (e)
-M
Qx dt
dtc
(340)
We can simplify (340) under three conditions:
1. Q at rest, so that
drQ
dt
=0
2. Center of mass at rest, dt` = 0
3. Velocity of Q is parallel to velocity of center of mass, drQ
dre
dt
dt
0
In all three cases
dHe
= LQ(e)
dt
(341)
In particular, if LQ(e) = 0, then He - constant, and this is the law of conservation of angular momentum. 94. Moment of Relative Momentum about Q. In Sec. 93 we assumed that the absolute velocity of each particle was known.
It is often more convenient to calculate the velocity of each particle relative to Q.
This is
dr, dt
- drQdt
We now define rela-
VECTOR AND TENSOR ANALYSIS
200
[SEC. 94
tive moment of momentum about Q as n
HQr =
j-1
(rj - rQ) x mj
(rj - rQ)
dl
(342)
Differentiating, dHQ*
dt
j-1 =
(d2rj
(r' - rQ) x m'
(r7 - rQ) x
dt2
dl2
d2rQ
`
(fj(ei
/
d2rQ
f3(i))
n
x
dt2
I m, (rj - rQ) j-1
We see that dtiQr
a,
LQ(' +
'rQ d12
n
x I mj (rj - rQ)
(343)
js1
Under what conditions does dd Q' = Lo d2rQ
dt2
?
We need
n
x I mj(rj - rQ) = 0
or
M
d2rQ
dt2
x (r, - rQ) = 0
(344)
Now (344) holds if 1. rQ = rQ, or Q is at the center of mass.
2. Q moves with constant velocity,
dQ l = 0.
2
3. r. - rQ is parallel to
2 dt'
Problems n
1. Show that
jet
a rQ
m4(r1 - rQ) x dt2 = M(r. - rQ) x
d2rQ dt2
2. A system of particles lies in a plane, and each particle remains at a fixed distance from a point 0 in this plane, each
MECHANICS
SEC. 95]
201
particle rotating about 0 with angular velocity w. n
Show that
Ho = Iw, where I = I m;r;2, and show that Lo = I at j=1
3. A hoop rolls down an inclined plane. What point can be
taken as Q so that the equation of motion (343) would be simplified?
95. Kinetic Energy. We define the kinetic energy of a particle of mass m and velocity v as T = jmv v. C M For a system of particles, r.-r n
T
n
1
2 mv,2 f=1
fdrs\ 2
1
2 m'
.1-1
(345)
dt 1
P;
`r
r
Now let r, be the vector to the center of 0 mass C (Fig. 83). It is obvious that
Fia. 83.
r;=rr+(r,-r-) so that dr; dt dr; dr; _ (dro)2 di dt dt
_ dr, dt
+
d dt
(r; - r-)
(r, - rc) -{+2 drd dt dt
I d. L dt
(r; - rc) ]
2
Hence 1 (dr12 dr" T=2M dtJ +dt
d
n
2
+ jn
Now Mr. =
1
2 m'
n
(346)
n
r,, so that
m;r, j-1
r`)] [dt (r' -
j-1
m1 alt (r; - r-) = 0, ;-1
and (346) reduces to
T=IM
()2
-}-
-1 2
m; [-d
dt
(r; - r.)]2 a)(347) 1
This proves that the kinetic energy of a system of particles is equal to the kinetic energy of a particle having the total mass
202
VECTOR AND TENSOR ANALYSIS
[SEC. 96
of the system and moving with the center of mass, plus the kinetic energy of the particles in their motion relative to the center of mass. 96. Work. If a particle moves along a curve r with velocity v
under the action of a force f, we define the work done by this force as
W=
fr f t
fr f
f acts at right angles to the path, no work is done. If the field is conservative, f = -V(p, the work done in taking
the particle from a point A to a point B is independent of the path (see Sec. 52). Now dvi = fj(a) + fi(.) dt dvi = fi(e) . V1. + mivi . dt m,
Vi
and integrating and summing over all particles, n
J`o
mivi
dt' dt =
La
i:1
fi(a)
. vi dt +
fee:' fi('i
. v; dt
or n
i=1
lmi[vi2(ti) - vi2(to)l = W(e)
W°
(349)
This is the principle of work and energy. The change in the kinetic energy of a system of particles is equal to the total work done by both the external and internal forces.
If the particles always remain at a constant distance apart, (ri - rk)2 = constant, the internal forces do no work. Let r, and r2 be the position vectors of two particles whose distance apart remains constant, and let f and -f be the internal forces of one particle on the other and conversely. Now
(r, - r2) (r, - r2) = constant
MECHANICS
Sec. 97]
203
so that
(r, - r2)
dr, di
- dr2dt/
0
(350)
Also
W(o =
f ff
f is parallel to r, - r2, Ave have f = a(r, - r2) and
f (v, - v2) = a(r, - r2) (v, - v2) = 0
from (350). Thus W(° = 0. Problems 1. A system of particles has an angular velocity w. n
T=
i-i
Show that
Jm;lw x r,J2.
2. If to of Prob. 1 has a constant direction, show that T = }Iw2, n
where I =
md;2, d; being the shortest distance from m; to
line of w.
3. Show that dT = w L, by using the fact that T =
4-mv;2
and that v, = to x r;. 4. Show that the kinetic energy of a system of rotating particles is constant if the system is subjected to no torques. What if L is perpendicular to w?
5. A particle falls from infinity to the earth. Show that it strikes the earth with a speed of approximately 7.0 miles per Use the principle of work and energy. 97. Rigid Bodies. By a rigid body we mean a system of particles such that the relative distances between pairs of points remain constant during the discussion of our problem. Actually no such systems exist, but for practical purposes there do exist such rigid bodies, at least to a first approximation. Moreover, the rigid body may not consist of a finite number of particles, but rather will have a continuous distribution, at least to the unaided eye. We postulate that we can subdivide the body into a great many small parts so that we can apply our laws of motion for particles to this system, this postulate implying that we can use second.
VECTOR AND TENSOR ANALYSIS
204
the integral calculus. the following form:
[SEC. 98
Our laws of motion as derived above take
T = f f f -pv2 dr,
p = density
R
ff pr dr f f f p dr R
dr,
( 351)
2
J
fR f f(e)
dt2
H = f f f prxvdr R
-it ff r x f(e) dr where f(.) is the external force per unit volume. 98. Kinematics of a Rigid Body. Let 0 be a point of a rigid body for which 0 happens to be fixed.
It is easy to prove that the velocity f V (P1 = yr
P
of any other point P of the body must be perpendicular to the line joining 0
to P, for if r is the position vector from 0 to P, we have r - r = constant throughout the motion so that 0.
Q.E.D.
r dt = We next prove that if two points of a rigid body are fixed, then all other particles of the body are rotating around the line joining these two points.
Let A and B be the fixed
points and P any other point of the body.
From above we have
so that P is always moving perpendicular to the plane ABP. Moreover, since the body is rigid, the shortest distance from P to the line A B remains constant, so that P moves in a circle
Sec. 981
MECHANICS
around AB (Fig. 84). could be written
205
We saw in Sec. 10 that the velocity of P
VP =wxrp Is w the same for all particles? Yes! Assume Q is rotating about AB with angular velocity w,, so that vQ = to, x rQ. Now (rP - rQ)2 = constant, so that (rP - rQ) (vp - vQ) = 0 or
(rP - rQ) ((a x rp - w, x rQ) = 0 Thus
x rP -
xrQ = 0
and rQ x rp
(w, - w) = 0
We leave it to the reader to conclude that w, = w. VA
B
1
FIG 85.'
If one point of a rigid body is fixed, we cannot, in general, hope to find a fixed line about which the body is rotating. However, there does exist a moving line passing through the fixed point so that at any instant the body is actually rotating around this line. The proof proceeds as follows: Let 0 be the fixed point of our rigid body and let r` be the position vector to a point A. From above we know that the velocity of A, VA, is perpendicular to rA. Construct the plane through 0 and A perpendicular to VA (Fig. 85). Now choose a point B not in the plane. We also have that vB rB = 0, so that we can construct the plane through 0 and B perpendicular to vB. Both planes pass through 0, so
206
VECTOR AND TENSOR ANALYSIS
[SEC. 98
that their line of intersection, 1, passes through 0. Now consider any point C on this line. We have vc rc = 0. Moreover, (rc - rA) - (rc - rA) ° constant, so that
(re-rA).(vc-VA) =0 and (rc - rA) VC = 0, since vA is perpendicular to (rc - rA). Similarly (re - rB) vc = 0. Hence the projections of vc in three directions which are nonplanar are zero.
This means that
FIG. 86.
Vc - 0, so that we have two fixed points at this particular instant. Hence from the previous paragraph the motion is that of a rotation about the line 1. If w is the angular-velocity vector, then v; = w x r;, where r; is the vector from 0 to the jth particle. Now let us consider the most general type of motion of a rigid -r represent a fixed coordinate system in space, body. Let and let 0-x-y-z represent a coordinate system fixed in the rigid body (see Fig. 86). Let ,o; and r; represent the vectors from 0'
and 0 to the jth particle, and let a be the vector from 0' to 0. We have ei = a + r,, and differentiating, dLD;
dt
- da dt
+
dr, dt
MECHANICS
SEC. 98]
207
Now l represents the velocity of Pi relative to 0.
This means
0 is fixed as far as Pi is concerned, and from above we know that dri
dt - w
Thus
x ri.
_dei =
da
dt
dt
v'
(352)
+wxr;
that is, the most general type of motion of a rigid body is that of
a translation
A dt
plus a rotation w x r;.
We next ask the following question: If we change our origin
from 0 to, say, 0" does w change? (Fig. 87.) The answer is
"No"! Let b be the vector
0
from 0' to 0". Then v;
rf
=t=' + w,xr;
But
and
db
da
dt
dt
ri = (a
+w x (b - a) -- b) + ri
Fio. 87.
Thus
vi = d +w x(b - a)+wi x (a -b)+w, xri
(353)
Subtracting (352) from (353), we obtain
(w - wi) x (b-a)+(w,-w) xr;=0 or
(w - wl) x (b-a-r,)=
xr;"=0
0 and not parallel to the vector w - wl, at any particular instant. Hence w, ° w. We can certainly choose an r;"
Problems
1. Show that if r, and r2 are two position vectors from the origin of the moving system of coordinates to two points in the rigid body, then r, dt2 + r2 - dtl = 0.
VECTOR AND TENSOR ANALYSIS
208
[SEC. 99
2. A plane body is moving in its own plane. Find the point in the body which is instantaneously at rest.
3. Show that the most general motion of a rigid body is
a translation plus a rotation about a line parallel to the translation.
99. Relative Time Rate of Change of Vectors. Let S be an y vector measured in
the
moving system of coordinates (Fig. 88).
S = S i + Sj + S,k
Fin. 88.
(354)
To find out how S changes with time as measured by an observer at 0', we differentiate (354), dS
dS=.
dt = dt 1+
dS
.
dt'+
dk dS, di _ dj dt k+S= dt+Sydt+S'dt
(355)
We do not keep i, j, k fixed since i, j, k suffer motions relative to
0'. But we do know that dt is the velocity of a point one unit along the x axis, relative to 0. Hence dk = w x k.
at dS
d
= w x is dt = w x it
Hence (355) becomes
dSs
dt - dt
i -}
dS dt
dS,
j + d k + w x (3j +
S1t)
and dS
dt
DS
S dt + w x
(356)
where DS represents the time rate of change of S relative to the moving frame, for S= is measured in the moving frame and so d _ t
is the time rate of change of S. as measured by an observer in the moving frame.
MECHANICS
SEC. 1001
209
Intuitively, we expected the result of (356), for not only does S change relative to 0, but to this change we must add the change in S because of the rotating frame. The reader might well ask, What of the motion of 0 itself? Will not this motion have to be considered? The answer is "No," for a translation of 0 only
pulls S along, that is, S does not change length or direction if 0 is translated. It is the motion of S relative to the frame O-x-y-z and the rotation about 0 that produce changes in S. Problems
1. Show that d dl 2. For a pure translation show that
dS
DS
dt =
dt
3. From (356) show that di = w x i.
Let P be any point in space and let 9 and r be the position vectors to P 100. Velocity.
from 0' and 0, respectively (see Fig. 89). Obviously a + r, so that _ dp _ A dr v
dt
dt
+
dt
Now r is a vector measured in
the O-x-y-z system, so that (356) applies to r. This yields dr _ Dr + to x r and
7t
dt
v
dt
This result is expected.
t
-dt
FIG. 89. xr+Dr
(357)
A is the drag velocity of P, co x r is
the velocity due to the rotation of the 0-x-y-z frame, and
Dr
is
the velocity of P relative to the 0-x-y-z frame. The vector sum is the velocity of P relative to the frame 0'--i-r.
VECTOR AND TENSOR ANALYSIS
210
(SEC. J 01
101. Acceleration. In Sec. 100 we saw that
v=ac
A dt+wxr+ Dr dt
To find the acceleration, we differentiate (357) and obtain
dv_d2p_d2a dt
dt2
dt2
d
+ dt (wxr) +
d(Drl dt
dt
(358)
We apply (356) to w x r and obtain
d
(wxr) = w x (wxr) + D (w x r)
Similarly d
dldtl d2p
dt 2
_
d2a dt2
+ w x (wxr) +
D xdt+d`dtl
Dr D2r do x r + 2w x dt dt + dt2
(359)
If P were fixed relative Dr = D2r to the moving frame, we would have 0 and consedt2 = dt quently P would still suffer the acceleration Let us analyze each term of (359).
d2a dt2
&
+wx(wxr)+dt xr
This vector sum is appropriately called the drag acceleration of the particle. Now let us analyze each term of the drag acceleration. If the moving frame were not rotating, we would have 2
0, and the drag acceleration reduces to the single term dta This is the translational acceleration of 0 relative to 0'. Now in Sec. 84 we saw that w x (w x r) represented the centripetal accel-
eration due to rotation and d x r represented the tangential component of acceleration due to the angular-acceleration vector
MECHANICS
SEC. 1011
d We easily explain the term relative to the O-x-y-z frame.
D2r
211
as the acceleration of P
What, then, of the term 2w x dr?
This term is called the Coriolis acceleration, named after its discoverer. We do not try to give a geometrical or physical reason for its existence. Suffice to say, it occurs in Eq. (359) and must be considered when we discuss the motion of bodies moving over the earth's surface. Notice that the term disappears for par-
ticles at rest relative to the moving frame, for then dt = 0.
It
also does not exist for nonrotating frames.
Now Newton's second law states that force is proportional to the acceleration when the mass of the particle remains constant. It is found that the frame of reference for which this law holds best is that of the so-called "fixed stars." We call such a frame of reference an inertial frame. Any other coordinate system moving relative to an inertial frame with constant velocity D d2
is also an inertial frame, since from (359) we have d e = because w = Of
66
1
0,
dt2r
da d1 constant, dta = 0. dt =
Let us now consider the motion of a particle relative to the earth. If f is the vector sum of the external forces (real forces, 2
that is, gravitation, push, pull, etc.), then d p = m, and (359) becomes D2r
d2a
Dr
d4,3
f
dt2 - w x (w x r) - d x r - 2w x dt + m (360)
dt2
This is the differential equation of motion for a particle of mass m with external force f applied to it. Example 108. Let us consider the earth as our rotating frame. The quantity w x ((a x r) is small, since jwl 27r/86,164 rad/sec,
and for a particle near the earth's surface, lrl , (4,000)(5,280) feet.
Also
d2
dt
- 0 over a short time; da ,= 0 over a short time;
212
VECTOR AND TENSOR ANALYSIS
[SEC. 101
so that (360) becomes D2r
-2w x dt +
dt2
m
(361)
Now consider a freely falling body starting from a point P at rest relative to the earth. Let the z axis be taken as the line joining the center of the earth
to P, and let the x axis be taken perpendicular to the z axis in the eastward direction.
We shall denote the latitude of the place by A, assuming A > 0. The equation of motion in the eastward direction is given by
yd1x
_
t2
S
(fm/, = 0.
tion it is
(see Fig. 90).
dj dt
, + Cfm
/
has no component eastward, so
We do not know
-gtk +
x
Now f (force of attraction)
Fio. 90.
that
-2 (w
at
but to a first approxima-
Moreover, w = w sin A k + w cos A j
i.
Hence (w x
at)
d2x
dt' =
0
= -wgt cos A, and
2wgt cos A
(362)
If the particle remains in the vicinity of latitude A, we can keep A constant, so that on integrating (362), we obtain dx dt
X=
=wgt2cosA 3ts cos A
(363)
(363) is to a first approximation the eastward deflection of a shot if it is dropped in the Northern Hemisphere. If h is the
MECHANICS
SEc. 1011
213
distance the shot falls, then h = Jgt2 approximately, so that 2
x = 3 wh
cos X
2gh1}
( J
Problems
1. Show that the winds in the Northern Hemisphere have a horizontal deflecting Coriolis acceleration 2wv sin X at right angles to v. 2. A body is thrown vertically upward. Show that it strikes the ground 'jwh cos X (2h/g)1 to the west. 3. Choose the x axis east, y axis south, z axis along the plumb line, and show that the equations of motion for a freely falling body are d2x
W
+2wsinUdt z- 2wcos0dty=0 dt +2w cos0dy =0
dtz-g-2w sin0dx =0 2
where 0 is the colatitude. CO
Fio. 91.
4. Using the coordinate system of Prob. 3, let us consider the motion of the Foucault pendulum (see Fig. 91). Let ii, i2, i$ be the unit tangent vectors to the spherical curves r, 8, p. We leave it to the reader to show that the acceleration
VECTOR AND TENSOR ANALYSIS
214
[SEC. 101
along the is vector is 2 cos e .06 + sin 0 0 when the string is of unit length.
The two external forces are mgk along the z axis and
the tension in the string, T = - Tr = - Tit. We wish to find the component of these forces along the i3 direction. T has no component in the is direction. Now k is = 0, so that mgk has no component along the is direction. Finally, we must compute Dr The velocity vector is the is component of -2w x Dr dt
_ Bit + sin 6 rpis
Also w = w(- cos A j - sin X k), so that we must find the relationship between i i2, is and i, j, k. Now
r = i, = sin0cosci+sin Bsin cpj+cos0k ail i2
cos 0 cos v i + cos 0 sin
= aB =
_
ail
1
sin B ai
1s
sin 0 k
- sin sp i + coo sp j
Thus
i = (i il)i, + (i i2)i2 + (i
i3)is
= sin 0 cos Tp it + cos 0 cos jP is -- sin rp is
j = sin 0 sin Sp i, + cos B sin 'P i2 + C0803 k = cos 0 it - sin B i2 11
is
i2
-2wx-=2w cos A sin 0 sin P + sin A cos 0 0
-sinAsin0 6
sine c
and
(-
r
2w x -D dt
= #(sin A sin 0 sin ip + sin A cos 0)
Equation (361) yields 2 cos 006+ sin 8;p = 2w(B sin A sin B sin rp + # sin A cos 0) (364)
MECHANICS
SEC. 102]
For small oscillations, sin 0
215
0, and (364) reduces to
, = w sin X
(365)
Hence the pendulum rotates about the vertical in the clockwise sense when viewed from the point of suspension with an angular speed w sin X. At latitude 30° the time for one complete oscillation is 48 hours.
5. Find the equation of motion by considering the i2 components of (361) for the Foucault pendulum. 102. Motion of a Rigid Body with One Point Fixed. The motion of a rigid body with one point fixed will depend on the forces acting on the body. Let O-x-y-z be a coordinate system fixed in the moving body, and let O-- be the coordinate system fixed in space. 0 is the fixed point of the body. In Sec. 94 we
saw that
-Or
= Lo.
Now Ho* = f if r x p dt dr. We can
replace dt by to x r (w unknown). Thus
H0? = f f f pr x (w x r) dr R
= fJf p[r2w - (r
w)r] dr
(366)
R
Let
w=w=i+wyj+w
r=xi+yj+zk
so that
r2w - (r w)r = (x2 + y2 + z2) (w i + wyj + W k) + (xwx + ywy + zws) (xi + yj + zk) = [(y2 + z2)wz - xywy - xzws]i
+ [-xyw. + (22 + x2)wy - yzw=lj + [-xzwz - yzwy -- (x2 + y2)w]k We thus obtain H0? = i[w= f f f p(y2 + z2) dr - w f f f pxy dr - wz f f f pxz dr]
+ j[ - w=f f f pxy dr + wyf f f p(z2 + x2) dr - w: f f f pyz dr] + k[ -w=f f f pxz dr - wyf f f pyz dr + w=f f f p(x2 + y2) dr] (367)
216
VECTOR AND TENSOR ANALYSIS
[SEC. 102
The quantities A = f f f p(y2 + Z2) dr
B=fffp(z2+x2)dr C = fffp(x2 + y2) dr D = f f fpyzdr E = f f fpzxdr
(368)
F = f f f pxy dr
are independent of the motion and are constants of the body. That they are independent of the motion is seen from the fact that for a particle with coordinates x, y, z, the scalars x, y, z remain invariant because the O-x-y-z frame is fixed in the body.
The quantities A, B, C are the moments of inertia about the x, y, z axes, and D, E, F are called the products of inertia. We assume the student has studied these integrals in the integral calculus.
Now from Sec. 99 we have
Lo =
dHor dt
_
DHor + w x Hor so that dt
DRO, dt + to x Hor
Hence
L=i+Lvj+LA=i(A
+
ds-F
+j(-F
w
+ k \-E
dt
dtE
dts/ +Bdty-Ddt;/
- D dt + C dta/
i
k
wy
wt
Aws - Fwv - Ewz,
WV
-Fws+Bwv - Dw=,
-Ews - Dwv+Ccos (369)
In the special case when the axes are so chosen that the products of inertia vanish (see Sec. 107), we have Euler's celebrated equations of motion :
MECHANICS
SEc. 1031
217.
L. = A d[ + (C - B)wYwz
L = B
dwY
(370)
+ (A - C)w,wZ dt
L. = C
dw, dt
+ (B
A)wzwy
103. Applications. If no torques are applied to the body of Sec. 102, Euler's equations reduce to A
(i)
dwz
dt-
+ (C
B)wYw, = 0
B dt +(A-C)w,wx=0
(371)
C d,+(B-A}wZwY=0
Multiplying (i), (ii), (iii) by
w,, respectively, and adding,
we obtain Aws
dws
dt
+ Bw dwY + Cw, dwa = 0 dt dt
Integrating yields Aws2 + Bw,,2 + Cw,2 = constant
(372)
This is one of the integrals of the motion. We obtain another integral by multiplying (i), (ii), (iii) by Aw,z, Bw,,, Cw,, and adding. This yields AZ w,
dws
dt
+
BZwy'dw'
dt
+
C2w,
dw, dt
=
0
so that A 2w12 + B2wV2 + C2w,2 =constant
(373)
If originally the motion was that of a rotation of angular velocity w about a principal axis (x axis), then initially
VECTOR AND TENSOR ANALYSIS
218
[SEc. 103
wz(0) = coo
0
(374)
ws(0) = 0
and we notice that (371) and the boundary condition (374) are satisfied by wt(t) = -coo
0
wZ(t) = 0
so that the motion continues to be one of constant angular velocity about the x axis. Here we have used a theorem on the uniqueness of solutions for a system of differential equations.
Now suppose the body to be rotating this way and then slightly disturbed, so that now the body has acquired the very We can neglect ww: as compared Euler's equations now become
small angular velocities w,,, wt.
to
and wswo.
B dty + (A - C)wswo = 0
Cdt
+(B0
(375)
wo = constant
Differentiating the first equation of (375) with respect to time and eliminating d s' we obtain B
z
dtzy +
(- C)((A - B) w02w = 0
(376)
If A is greater than B and C or smaller than B and C, then (A - C) (A - B) a2 =
C
> 0, and the solution to (376) is
wy = L cos (at + a)
Also ws =
aBL sin (at + a) by replacing w,, in (375). wo(A - C)
SEC. 1041
MECHANICS
219
Problems 1. Solve the free body with A = B for wzj co,, co,.
2. A disk (B = C) rotates about its x axis (perpendicular to the plane of the disk) with constant angular speed wo. A constant torque Lo is applied constantly in the y direction. Find w and W..
3. Show that a necessary and sufficient condition that a rigid
body be in static equilibrium is that the sum of the external forces and external torques vanish. 4. A sphere rotates about its fixed center. If the only forces acting on the sphere are applied at the center, show that the initial motion continues.
5. In Prob. 2 a constant torque Lo is also applied in the z direction.
Find w,, and ws.
104. Euler's Angular Coordinates. More complicated problems can be solved by use of Euler's angular coordinates. Let O-x'-y'-z' be a cartesian coordinate system fixed in space, and let O-x-y-z be fixed in the moving body (Fig. 92). The x-y plane will intersect the x'-y' plane in a line, called the nodal line N. Let 0 be the angle between the z and z' axes, L' the angle between the x' and N axes, and (p the angle between the nodal line and the x axis. The positive directions of these angles are indicated in the figure. The three angles &, 0, rp completely specify the configuration
of the body. Now d represents the rotation of the O-z'-N-T' frame relative to the O-x'-y'-z' frame; de represents the rotation of the O-z-N-T frame relative to the O-z'-N-T' frame, and finally, d(P
represents the rotation of the O-x-y-z frame relative to the
O-z-N-T frame. Therefore
+ dO + dt gives us the angular
velocity of the O-x-y-z framed* relative to the fixed O-x'-y'-z' frame, and
d1 + dO + d
(377)
220
VECTOR AND TENSOR ANALYSIS
[SEC. 104
The three angular velocities are not mutually perpendicular. We now define i, j, k, i', j', k', N, T', T as unit vectors along the x, y, z, x', y', z', N, T', T axes, respectively. Thus w =
d
k
wzi + w,j + wLk wz'i' + wy j' + wZ k'
Fia. 92.
Now it is easy to verify that i = cos rp N + sin cp T
j = - sinpN+cosjpT i' = cos 4, N - sin ' T'
(378)
MECHANICS
SEC. 1051
221
sin 4, N+cos#T' sin psin B cosspsin 0
sin
so that
ki
ws at
at
at
sin p sin 0 d + cos s dB day
do
at dp
at
w = cos (p sin 8 --- - sin (p day
W: = cos B d +
dt
Rewriting this, we have
dt
wy =
d4,
sin 0 sin (P + sin B cos
wI = d cos 8 +Also
wi=wss+w2+(O2=
cos
- 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
VECTOR AND TENSOR ANALYSIS
222
(i)
A
(ii)
A d' + (A - C)cvxwZ = 0
[SEC. 106
(382)
Cdts=O
(iii)
Integrating (iii), we obtain wz = w, = constant. and add to (i). We obtain by i =
Multiply
A dt (w= + iwv) + (C - A)wo(wy - iwz) = 0 or
A
d (co., + iwy) = iwo(C - A)(wx + i
)
Integrating, wz + 2Wy =
ae<'(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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
ap + V (Pv) = #(x, y, z, t)
[SEC. 109
(407)
This is the equation of continuity. For no source and sink, (407) reduces to 0 at + V - (Pv) =
(408)
If furthermore the liquid is incompressible, p = constant, aP
at
= 0, and (408) becomes
Vv=0
(409)
If the motion is irrotational, that is, if f v dr = 0, then V=
so that the equation of continuity for an incompressible
fluid possessing no sources and sinks and having irrotational motion is given by V2V = 0
(410)
We call p the velocity potential. We solve Laplace's equation for rp, then compute the velocity from v = Vrp. Problems 1. If the velocity of a fluid is radial, u = u(r, t), show that the equation of continuity is ap
at
LP
+ u ar +
P a
r2 ar
(r2u)
Solve this equation for an incompressible fluid, if '(r, t) = 1/r2. z _2xyz x z - 1<)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
HYDRODYNAMICS AND ELASTICITY
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
VECTOR AND TENSOR ANALYSIS
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]
HYDRODYNAMICS AND ELASTICITY
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
VECTOR AND TENSOR ANALYSIS
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
HYDRODYNAMICS AND ELASTICITY
-- 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
VECTOR AND TENSOR ANALYSIS
(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
HYDRODYNAMICS AND ELASTICITY
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
VECTOR AND TENSOR ANALYSIS
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]
HYDRODYNAMICS AND ELASTICITY
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
VECTOR AND TENSOR ANALYSIS
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
HYDRODYNAMICS AND ELASTICITY
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
VECTOR AND TENSOR ANALYSIS
[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 `ay +
(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
HYDRODYNAMICS AND ELASTICITY
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
VECTOR AND TENSOR ANALYSIS
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
HYDRODYNAMICS AND ELASTICITY
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
VECTOR AND TENSOR ANALYSIS
[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
HYDRODYNAMICS AND ELASTICITY
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)
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
[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
HYDRODYNAMICS AND ELASTICITY
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
VECTOR AND TENSOR ANALYSIS
[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]
HYDRODYNAMICS AND ELASTICITY
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
VECTOR AND TENSOR ANALYSIS
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 a`i - axi a+i axi _ 'q a au; au) 2 a(V v) +1 axi 3 = axi axi + ax'
-
-
ap a; - axi
257
HYDRODYNAMICS AND ELASTICITY
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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
. + -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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
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"-'.
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
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'
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
[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 ` $,k - Bay axp avk
=0
Since A' is arbitrary, we must have f k = Bay result. Example 122. tensor, for
a2 ax
the desired
The Kronecker delta, S, is a mixed absolute =
axi axa
' aga axi' Example 123.
axa axk
axi axa = 80 a
axa axs
= asa
If Ai and Bi are the components of a contra-
variant and a covariant vector, then C = AiB, are the components of a mixed tensor, for s
Ai = Aa ax , axa
axa B' _ B a ali
-
277
TENSOR ANALYSIS
SEC. 1261
CO that AaB#
C
axa axe =
Cs
axa Ox'
Let gij be the components of a covariant tensor
Example 124.
ax-,9x#
Taking determinants and applying
so that gi, = gap - =
axt ox' Example 115 twice, we obtain lax
ax
I#I} = IgIl
or
Igl =1g1 l ax
a2
Now if Ai are the components of an absolute contravariant vector, then A{ = Aa ax Y so that ax
B, = IgI}A' =
B.
ax
aax
IgI'Aa
Of
axa
Ba2i
ax-
are the components of a vector density. This method affords a means of changing absolute tensors into relative tensors. Example 125. Assume gap dxa dxO an invariant, that is,
so that Bi °
IgI#Ai
9aP d7a d2O = gp dxa dxfi a
Now d2a = axu dxµ, so that gap
axa aaxu ax,
dxu dx = g,,, dxu dx'
or (9aO
a0
°1° ^ 9u axu ax
dxu dx' = 0
(496)
If we assume gap = gaa, then since (496) is identically zero for arbitrary dxi, we must have (see Example 114)
278
VECTOR AND TENSOR ANALYSIS
ISEC. 126
axa axs 9µY = 9a$
(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)
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
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).
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
(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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
(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)
`oar"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
VECTOR AND TENSOR ANALYSIS
(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
VECTOR AND TENSOR ANALYSIS
{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
VECTOR AND TENSOR ANALYSIS
[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)
VECTOR AND TENSOR ANALYSIS
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 + B`i - (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.
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
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)
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
[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
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
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)
SEC. 139] FURTHER APPLICATIONS OF TENSOR ANALYSIS 321
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
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
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)
SEC. 140] FURTHER APPLICATIONS OF TENSOR ANALYSIS 325
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
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
[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)
VECTOR AND TENSOR ANALYSIS
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
SEC. 141] FURTHER APPLICATIONS OF TENSOR ANALYSIS 331
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
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
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
VECTOR AND TENSOR ANALYSIS
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
