Vector Analysis: A Physicist's Guide to the Mathematics of Fields in Three Dimensions

Vector analysis A physicist's guide to the mathematics of fields in three dimensions

Vector analysis A physicist's guide to the mathematics of fields in three dimensions N. KEMMER Tait Professor of Mathematical Physics University of Edinburgh

CAMBRIDGE UNIVERSITY PRESS CAMBRIDGE

LONDON NEW YORK MELBOURNE

CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, Delhi

Cambridge University Press The Edinburgh Building, Cambridge C132 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521211581

© Cambridge University Press 1977

This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1977 Re-issued in this digitally printed version 2008

A catalogue record for this publication is available from the British Library Library of Congress Catalogue Card Number: 75-36025 ISBN 978-0-521-21158-1 hardback ISBN 978-0-521-29064-7 paperback

Contents

Preface

Introduction

page ix xi

1 Summary of vector algebra

1 Addition and multiplication of vectors 2 The cartesian components of a vector 3 The quotient theorem

7

Exercises A

9

1

5

2 The geometrical background to vector analysis

1 The point 2 The curve 3 The surface 4 The three-dimensional region of space 3 Metric properties of Euclidean space 1 The length of a curve and the total mass on it 2 The area of a surface and the total mass on it 3 The volume of, and mass in a three-dimensional

13 13

17 22

27 28 29

region Exercises B

4 Scalar and vector fields 1 Scalar fields 2 Vector fields Exercises C

5 Spatial integrals of fields 1 The work integral 2 The flux integral 3 The quantity integral v

31

37

40 45

49 54 58

Contents 6 Further spatial integrals

1 Vector integrals 2 More general integrals over fields

60

Exercises D

67

7 Differentiation of fields. Part 1: the gradient 1 The operator V 2 The gradient of a scalar field 3 The field line picture of a gradient field 4 An alternative definition of grad 0 5 The fundamental property of a gradient field 8 Differentiation of fields. Part 2: the curl 1 Definition of the curl of a vector field 2 Evaluation of curl f. Stokes' theorem 3 The fundamental property of a curl field 4 The quantitative picture of a solenoidal field 9 Differentiation of fields. Part 3: the divergence 1 Definition of the divergence of a vector field 2 Evaluation of divf. The divergence theorem 3 A survey of results

65

72 73

74 77 78 81 82 86

89 93

94 96

10 Generalisation of the three principal theorems and some remarks on notation

1 General integral theorems 2 A general notation 3 Generalisation of ranges of integration

98 100 101

Exercises E

107

11 Boundary behaviour of fields

1 Surface discontinuities 2 Singularities at lines and points 3 A special discontinuity: the double layer 4 Discontinuities of scalar fields

115 126 129 133

Exercises F

134

12 Differentiation and integration of products of fields

1 Differentiation 2 Integration by parts

vi

138 142

Contents 13 Second derivatives of vector fields; elements of potential theory

1 The Laplace operator. Poisson's and Laplace's equations 2 The `Newtonian' solution of Poisson's equation -

145

149

a digression Exercises G

155

14 Orthogonal curvilinear coordinates

1 The basic relations 2 Definition of A for scalars and for vectors. An alternative approach 3 Fields with rotational symmetry. The Stokes' stream function 4 Solenoidal fields in two dimensions - a

160 164 165

169

digression Exercises H

171

15 Time-dependent fields

1 The equation of continuity 2 Time-dependent relations involving the velocity

181 184

field v(r) Exercises I

191

Answers and comments Exercises A

195

Exercises B Exercises C Exercises D Exercises E Exercises F Exercises G Exercises H Exercises I

196 202 205 208 218 223 229 247

Index

251

vii

To dearest Margaret

and, with her, to

ARK SD C

Preface

The justification for adding one more to the many available texts on vector analysis cannot be novelty of content. In this book I have included a few topics that are more frequently encountered as part of the discussion of a specific physical topic - in the main in fluid dynamics or electrodynamics - but this in itself does not justify telling the whole story over again, perhaps least of all in the classical way - in bold-type vectors, excluding suffixes, tensors, n-dimensional space etc. However, after giving a course in the standard shape for many years (to Scottish students in their second year - roughly comparable in level to an English first-year University course) I gradually developed a way of presenting the subject which gave the old tale a new look and seemed to me to make a more coherent whole of it. Once my course had taken this shape the students were left without a suitable text to work from. More and more ad hoc handouts became necessary and finally the skeleton of this book emerged.

In preparation for publication I spent more time on devising the sections of exercises than on the main text which was largely in existence from the start. The exercises were indeed devised rather than compiled and I am well aware that they reveal my idiosyncrasies. I see them as a necessary part of learning a language. Their nature changes from the `plume de ma tante' level to excerpts from the Classics and the Moderns - interspersed with lists of idioms. Predominantly they require a hard grind, which I have tried to enliven to my taste. The student who is uncertain about integrating anything beyond x" will not be at a particular disadvantage, but one who fears algebraic manipulation will face a challenge - though not without words of encouragement in appropriate places. He will also be shown by example the advantage of keeping his examples strictly orderly and he should certainly become increasingly aware of how much symmetries help one to simplify calculations. ix

Preface I must remark particularly on exercises A after chapter 1 where I have amused myself by providing - sometimes with considerable

effort - vector relations involving integers only. If I had been seeking physical realism at that stage this would have been utterly inappropriate. In defence I would say that in this age of the pocket calculator any inducement to doing arithmetic by hand should be welcome. As the whole concept of my presentation is linked to the visualisation of parameterised curved surfaces, their edge curves and the spaces enclosed by them, a main condition for the conversion of my

lectures into a book was adequate illustration. I have been extremely fortunate in getting support from an artist who translated the rough sketches used in my course into superb illustrations. I am very happy to express here my deep gratitude to Mr P. Flook for this beautiful artwork. I have special pleasure in expressing my sincerest thanks to my secretary, Mrs Rachel Chester. The preparation of this book would have been delayed by many months, if not altogether frustrated, without her expert and devoted assistance. Through all stages of this venture she has produced the typescript of every page of draft and final manuscript - all the text and all the formulae. My sincere thanks go also to all members of the staff of the Cambridge University Press, who have proved most helpful at all stages of the production of this book. N.K.

May 1976

x

Introduction

If one wishes to express in precise and general terms any statement in physics in which positions, directions and motions in space are involved the most appropriate language to use is the language of vectors. In the mechanics of particles and rigid bodies vectors are used extensively and it is assumed in this monograph that the reader has some prior knowledge of vector algebra, which is the part of vector theory required in mechanics. Nevertheless chapter 1 provides a summary of vector algebra. There the notation to be used is made explicit and a brief survey of the whole field is given with stress laid on a number of particular results that become especially important later. The reader is also given the opportunity to test his understanding of vector algebra and his facility in applying it to detailed problems: a fairly extensive set of examples (exercises A) follows the chapter, with some comments and answers provided at the end of the book. A considerably widened theory of vectors becomes necessary when one turns to such parts of physics as fluid dynamics and electromagnetic theory where one deals not just with things at certain particular points in space but with the physical objects as distributed continuously in space. Quantities that are continuous functions of the coordinates of a general point in space are called fields and some of the fields of greatest interest in physics are vector fields. How such vector fields can be described and interrelated by using the methods of integral and differential calculus is the theme of the book. After the introductory chapter 1 there are two further chapters in which vector fields are not yet mentioned. These deal with geometrical preliminaries. Within mathematics there is an important subject, differential geometry, in which among many other things the topics of these two chapters are fully covered. Here we need only some selected results and these are summarised in precisely the xi

Introduction form wanted for later use. Two non-standard definitions are introduced: the 'quasi-square' and the 'quasi-cube'. A great deal of the later analysis depends on these and it is hoped that readers will not skip this preliminary work. The exercises B are designed to familiarise the student thoroughly with these ideas. The illustrations in the main text should also help to drive home the important ideas. Chapter 4 opens the discussion of vector fields and deals with ideas which are introduced at the very beginning in other books. The reason why a rather slower approach to them has been chosen is because in this way one can develop the whole of vector analysis proper without introducing any further mathematical machinery on the way. The earlier parts of some other presentations may be simpler, but then some later mathematical points must either be taken on trust or be allowed to interrupt the more central part of the development. Since this text is intended primarily for physicists, full mathematical rigour is not aimed at, at any stage; but the assumptions that are necessary to justify the way we proceed later will all have been stated by the end of chapter 3. As far as the mathematics goes the reader will be virtually coasting down hill from then on.

The real core of the book will be found in chapters 5-9; in particular the last three of these chapters form a sub-unit within which the powerful central theorems of vector analysis are established. Then follow the chapters 10 and 11 which are in the nature of addenda, containing material which is not always included as part of the subject. Chapter 12 is then somewhat transitional, leading up to the final three chapters which deal with topics closely related to the central theme but involving other ideas as well. In this part it had to be decided what was worth including and the decisions were strongly influenced by the thought that many lecture courses on physical topics could benefit from saving of time often spent on mathematical preliminaries, which cannot be found so easily in books written for physicists.

The sets of exercises associated with the earlier parts of the book including the central chapters - will be found to be unconnected with physical applications. They are intended to help the student to acquire a clear picture of the properties of any vector fields and not only the special ones that are later of importance. But when it comes to the later groups of exercises, ideas related to particular physical situations are introduced. From these the reader should get a first xii

Introduction impression of how the language of vectors helps to make physics the exact science that it is. At the beginning of each group of exercises there are comments on the contents of the group. The student should bear these in mind and remember that he is not expected to work slavishly through all the exercises. Different people will feel the need for different amounts of explicit work before they feel they understand a technique. The exercises try to cater even for the most dogged characters. But it is hoped that some of the exercises will be attempted by all readers - with or without help from the `Answers and comments' at the end of the book. There is a fair amount of interconnection between the different groups of exercises. In a number of cases the student will encounter the same topic or problem treated with increasing sophistication as the subject is developed and what can only be worked out with hard labour at an earlier stage may be seen as a simple consequence of a general theorem later. Also, the relevance of a particular result to physics may emerge gradually. However, while it is hoped that practising on these examples will benefit the serious student considerably, let him not be under the illusion that solving them will be directly useful to him in answering standard British examination questions on vector analysis. Very few indeed of the exercises are of the traditional format. A few are too brief - very many more are too long and heavy. None have been taken from old examination papers. Understanding of how to solve these exercises will surely be helpful; imitation would be dangerous! Finally a word about related subjects. There are many fields in physics, foremost among them the special and general theories of

relativity, in which the limitations of this book - the restriction to space of not more than three dimensions and the exclusion of tensors (and of spinors) - are unacceptable. In more advanced disciplines one must not only widen the range of mathematical objects utilised but also change drastically the form in which these objects are described: tensor analysis not only is but also looks quite different from vector analysis. But this does not mean that the methods of this book deserve to be forgotten and replaced by more general ones. There are still many parts of physics in which vector methods remain peculiarly appropriate and, what is even more important, no other approach offers the same combinations of analytic precision with geometrical descriptiveness. In electromagnetism the physicist will always need to understand and visualise how electric fields relate to xiii

Introduction charges and magnetic fields to currents - as vector theory tells him even though he knows that from a more advanced point of view he should describe these fields combined as a tensor and the charge and current densities together as a `vector in four-dimensional spacetime'.

xiv

1

Summary of vector algebra

1 Addition and multiplication of vectors The reader of this book is expected to be acquainted with the idea of representing a directed quantity (e.g. a displacement, velocity or force) by a vector a, such that the associated length is denoted by

a = jai

(1)

and the direction by

n (or a) = ala.

(2)

He will also know the meaning of the sum and the difference of two vectors a and b

c = a±b

(3)

and of the product of a vector a with a scalart a,

b = cva.

(4)

(Note that (2) is a special case of (4).) In addition to the product defined by (4), vector algebra recognises two types of product formed from two (and only two) vectors. Firstly, let a and b be two arbitrary vectors and let 0 be the angle between their directions. Then the scalar product a b of the two vectors is defined by

a b = ab cos 6.

(5)

Geometrically a b is the projection of either of the vectors on to the direction of the other multiplied by the length of the latter. One notes that a2

t A scalar may be simply a numerical factor or it may represent a physical quantity (such as mass, density or temperature) which can be defined without reference to orientation in space. 1

Summary of vector algebra

1

Fig. 1. The scalar product a b of two vectors.

and that

a b= 0

if a and b are perpendicular.

The introduction of this product enables one to write down identities like

la ± b l 2 = a2 ± 2a b + b2

(6)

closely resembling theorems of ordinary algebra. The scalar multiplication law is commutative, a distributive law

(8)

is valid. Both (7) and (8) are used in stating (6). As scalar multiplication is definable for two vectors and not more, it cannot be involved alone in any form of associative law. However, one may note that

a(a b) = (aa) b = a (ab),

(9)

where a is a scalar, but that e.g.

The second form of product of two vectors is the vector product.t Its definition is somewhat more complicated: The vector product a x b is the vector c that has the magnitude

c = ab sin B

(10)

and the direction perpendicular to both a and b so that t We are not here concerned with the fact that the possibility of defining a vector in this way is confined entirely to the physically interesting case of three space dimensions and even then requires a conventional choice of the direction into which the product vector shall point. From a more advanced point of view a X b is what is known as an axial vector or pseudo-vector. It is more properly related to the plane of the vectors a and b than to the direction perpendicular to it. 2

1

Addition and multiplication A

C

a Area =jcj=ja x b 1

Fig. 2. The vector product a x b of two vectors.

(11)

Furthermore

a, b, c, in that order, form a right-handed system.

(12)

The most striking and unusual feature of this definition of a product is that it is anticommutative and not commutative,

bxa = -axb.

(13)

This follows from (12) since b, a, - c and not b, a, c form a righthanded system. Either (10) or (13) demonstrates that

axa = 0.

(14)

Of particular importance to us is the fact that the magnitude c of a x b (see (10)) is the area of the parallelogram subtended by the two vectors a and b. We refrain from giving the non-trivial proof that vector multiplication is distributive, i.e. that

ax(b+c) = axb+axc

(15)

but would observe that the particular definition chosen for this vector product is determined by the need to have a law like (15). Without it this product construct would be of little use. On the associative law the same remarks apply as for the scalar product. The identities

a(a x b) _ (aa) x b = a x (ab) are correct. 3

(16)

Summary of vector algebra

1

In addition to the two types of product formed from two vectors, it is useful to single out from among various products that are definable as a consequence of the rules already stated, one product involving three vectors, a, b and c. It is a scalart product and apart from the sign it is the only scalar that can be formed from three vectors. It is

(a,b,c) =

(17)

By virtue of (13) one sees that a non-cyclic interchange among the three vectors reverses the sign of this scalar triple product. If a, b and c form a right-handed triplet (and, in particular are not coplanar) the triple product (16) is positive and equal to the volume of the parallelepiped subtended by the three vectors. In general (using the modulus sign in its non-vectorial algebraic sense) that volume must be stated as j (a, b, c) I.

There are numerous identities that follow from the definitions given above. Without any proofs we provide a list of those that will be of use later in the book (a, b, c, d, e represent arbitrary vectors throughout):

ax (b x c) = (a

x b)

(a c)(b d) - (a d) (b c).

(20)

For ease of reference at a particular point in the text below, it is worth re-writing the right hand side of (20) in a slightly different and rather clumsy looking form:

(a x b) (c x d) = b [(a c)d] - a [(b c)d]

.

(20')

This reformulation involves nothing more than a simple use of relation (9). Two further identities of considerable importance are:

(a,b,c)d =

a)

(21)

b)

and a closely related equation:

(a, b, c)d = [(b x c) d] a + [(c x a) d] b + [(a x Finally, by multiplication of both sides of (21), by e, one finds t In fact a pseudo-scalar; see footnote on p. 2.

4

(22)

The Cartesian components

2

(a, b, c) (d e) = (a d) [(b x c) e] + (b d) [(c x a) e] (23)

As for (20), a trivial re-formulation of (22) will be stated for future reference:

(a, b, c) (d e) _ (b x c) [(a d)e] + (c x a) [(b d)e] + (a x b) [(c d )e] . (23') The fact that one is concerned with space of precisely three dimensions is expressed mathematically by the statement that three, but not more than three, vectors can be linearly independent. Linear independence of three vectors means geometrically that (e.g. when drawn from the same origin) they do not lie in the same plane. Hence

(a, b, c) * 0

(24)

means that a, b, c are linearly independent. If (24) holds, then any further vector d must be expressible as a linear combination of the first three. This is precisely what equation (22) puts in evidence. One is not frequently interested in using (22) in full generality, but a special case of (22) is very well known indeed; this will be the subject of the next section. 2 The Cartesian components of a vector

We introduce for the first time the notion of a set of three orthogonal unit vectors i, j and k, which in particular we choose (in that order) to be a right-handed set. Then we have

lil = Ijl = Ikl = 1,

0,

(i, j, k) _ + 1.

(25)

The theorem that a general vector d may be represented as

d = d1i+d2j+d3k

(26)

is recognised at the very beginning of vector algebra and it is equally well known that

dl =

d2 =

d3 =

(27)

The reader may care to check that (26) follows from the much more general (22) with the use of (25) and (27). Having by (27) represented a general vector by its three `cartesian' 5

Summary of vector algebra

d,i

Fig. 3. Relation between a vector d and a set of cartesian components for it.

components, in a particular `orthonormal reference frame' (25), one can reformulate the whole of the preceding development in terms of cartesian components.t Thus we find that

albs +a2b2 +a3b3 and

(28)

a x b = (a2b3 -a3b2)i + (a3b1 -a1b3)j

+ (a1b2 -a2b1)k.

(29)

Often an alternative way of writing (29) is introduced

axb =

i

j

k

a1

a2

a3

b1

b2

b3

(30)

and in the same determinantal language

(a, b, c) =

a1

a2

a3

bl

b2

b3

C 1

C2

C3 -1

(31)

I

Relations (28)-(31) are of course extremely useful in working out vector relations explicitly, but they do not help much in giving one a physical picture of vectors. However, they have the merit of making it quite obvious that relations like (8) and (9) are true. t As used in physics, all three components of a vector represent quantities of the same dimensions and their numerical values always refer to the same unit.

6

The quotient theorem If one needs to know the cartesian components of a vector it is not always convenient to introduce them in terms of i, j and k by 3

a = al i + a2j + a3k;

(32)

it is often more convenient to summarise the same information in the

formt

(33) a = (a1 , a2, a3 ). We shall adopt this latter form of referring to vector components, so that we shall make statements such as

ax b = (a2b3 -a3b2,a3b1 -a1b3,a,b2 -a2b1). (34) In using this form one has to remember that the triplet of numbers in parentheses, unlike the vector they represent, has meaning only in relation to a particular set of base vectors i, j and k. For this reason a vector component, though a single number, is not a scalar. The beginner in the art of using vectors should particularly watch that he avoids the commonest of all confusions - namely between expressions like (3 3) or (34) for sets of numbers representing vectors and others like (28) which involve sums of numbers and represent a scalar. The notation involved in (29) and (32) is intermediate, involving sums, but of symbols rather than numbers. Equation (30) also falls into this latter category while (31) stands for a sum of six terms forming a scalar. While on the subject of elementary errors, the beginner should also be urgently reminded that there is no such thing as division by a vector, i.e. a single vector symbol can never occur in a denominator! However, an operation which could be loosely termed division by a vector will be the subject of the next section.

3 The quotient theorem There is one further theorem of vector algebra, on a somewhat different level from those discussed in the preceding section, that it is important to explain. We consider an arbitrary vector a. We know that in any particular cartesian frame it can be represented as in (33). Let us consider at the same time another set of three numbers written as t There should be no confusion between the right side of (33), which involves three numbers enclosed in bold type parentheses, and (17), which combines three vectors between ordinary parentheses. 7

Summary of vector algebra

[gl,g2,q31 which we combine with the components of a, to form the single number

algl +a2q2 +a3q3 = K.

1

(35)

(36)

We do not know at the outset that the three numbers (3 5) represent a vector - that is why we have written them enclosed in square brackets, contrary to the convention of equation (33). But we assume that K is a scalar, i.e. that it has a value independent of any reference frame. While the right hand side of (36) remains the same in all frames, the triplet of numbers (a I , a2, a 3) changes if the cartesian frame i, j, k is changed: in a new frame the vector a will be represented by new components

a = (ai,a',a3).

(37)

Clearly then the triplet (3 5) must also be replaced by a new set if

[qi, qi, q3l aigi + a2g2 + a3g3 = K

(38) (39)

is to hold. If one did assume that (35) represents the components of a vector q and that (38) represents the same vector in the new frame, the fact that the left hand sides of (36) and (39) are equal would be obvious: K = a q. The quotient theorem tells us that the converse is true - no three numbers other than the new components of the same vector q will do if (39) is to follow from (36) for allt vectors a. So the numbers (35) or (38) define a vector:

(g1,g2,q3) = (q',, g2,g3) = q.

(40)

Geometrically, this means that a magnitude and a direction can be associated with q, however the three numbers (35) were originally arrived at. This is the precise statement of the theorem but it can be summarised by saying: If a is an arbitrary vector and a q a scalar, then q must be a vector. It may seem that this section makes rather heavy weather of an obvious point. But we shall meet various proofs in the book which start from the knowledge that a q is a scalar when a is any vector. To be sure that then q is a vector requires the above theorem. In it q appears as a quotient in a formal sense - a q undergoes division by a - hence the name of the theorem. t it would be enough to say ' for three non-coplanar vectors a'; to know that (35) is true for only one or two vectors a is not enough to ensure (40).

8

Exercises A

This set of exercises, like the preceding text, is aimed at refreshing the reader's knowledge of vector algebra. With the exception of exercise 9, which will be used later in the text, none of the particular relations are of intrinsic importance, least of all the numerical values in exercises 1 to 8. Exercises 15 to 21 deal with the application of vector algebra to the geometry of the straight line and the plane, a topic not stressed elsewhere in the book. 1.

If a = (6, 3, 2) and b = (-2, -1, 2) calculate (a) (b) (c)

c=a+b. The lengths of the three vectors a, h and c. The cosines of the angles between b and c; c and a; a and b.

2.

Show that the sum of the vectors (6, 6, - 7) and (6, 3, - 2) is perpendicular to the vector (3, 0, 4).

3.

Calculate the cosines of the angles between the vectors (a) a = (9, 6, 2) and b = (2, 6, - 3). (b) a = (12, 4, 3) and b = (1, 2, 2). (c) a = (2, 10, 11) and b = (12, - 12, 1).

4.

Calculate (by one of two possible vector methods) the sine of the angle between the two vectors of each of the following pairs: (a) a = (12, 4, 3), b = (8, 12, 9). (b) a = (36, 8, 3), b = (12, - 1, 12). (c) a = (29, 24, 3), b = (16, 3, - 24).

5.

Evaluate the area of the parallelogram defined by the two vectors a = (48, 19, 12), b = (39, 12, - 4)

and calculate the direction of the unit vector perpendicular to this parallelogram. 6.

Prove that the three vectors a = (52, 12, 81), b = (64, - 15, 60), c = (28, - 24, 3) are linearly dependent and find coefficients a, 0, y such that the relation 9

Exercises A

cta+(3b+7c = 0 is satisfied.

(Hint: Showing that b x c, c x a and a x b are multiples of the same vector enables both parts of the question to be answered together.) 7.

Show that the three vectors

a = (51, 42,-26) b = (18, 19, 66) c = (46, - 54, 3) are mutually orthogonal and form a right-handed system. Construct a set of orthogonal unit vectors (i, j, k) having the same directions. Show that if the numbers given above are taken in columns instead of rows a second right-handed set of orthogonal vectors can be constructed. 8.

The three vectors

a=(12,4,3), 9.

b=(8,-12,-9), c=(33,-4,-24)

define a parallelepiped. Evaluate the lengths of its edges, the areas of its faces and its volume. Prove that if a and b are any two vectors,

b=

-a-2a x (a x b).

What is the meaning of this way of expressing b? 10.

Show that ifaxb=a-bthen a=b.

11.

Show that if a = b x c and b = c x a the three vectors a, b, c are mutually orthogonal and c is a unit vector.

12.

An equilateral triangle lies in the xy-plane. Its vertices are at the points P1 , P2 and P3 whose coordinates are ri = (xi, yi). Given that xl = 0, that the centroid of the triangle is at the origin and that the edges of the triangle are of unit length, find the (two-component) vectors ri. Patterned on the solution of exercise 12 find the four vertices of a regular tetrahedron of unit edge length and centroid at the origin, given that one vertex lies on the positive z-axis and the other three in a plane parallel to the plane z = 0, one of these being in the yz-plane.

13.

14.

How would you verify with least effort the truth of the statement that the end points of the following sets of 12 and 20 vectors respectively define the vertices of a regular icosahedron and a regular dodecahedron? (It may be checked that all the vectors are unit vectors.) (a)

a= (

0,

0,

b=( c=(

0.894, 0.274,

0,

d= (- 0.724,

1

)

0.447) 0.851, 0.447) 0.526, 0.447)

e = (- 0.724, - 0.524, 0.447) f = ( 0.274, - 0.851, 0.447)

plus the six vectors - a, .. 10

,

- f.

(b)

Exercises A a= ( 0.607, b= ( 0.188,

0.795) 0.557, 0.795) 0.357, 0.795) 0,

c=(-0.491, d=(-0.491,-0.357,0.795) e=( 0.188,-0.577,0.795) f = ( 0.982, g=(

0.188) 0.934, 0.188) 0.577, 0.188) 0,

0.304,

b=(-0.795,

i = (- 0.795, - 0.577, 0.188)

j = ( 0.304, - 0.934, 0.188) plus the 10 vectors - a, .. , - j. 15.

If the end points of the two vectors a and c of exercise 1, both assumed drawn from the origin 0, are A and C, how are the angles of the triangle OAC related to the angles calculated in exercise 1(c)? Verify that the sum of these latter angles is n. Indicate in a rough perspective sketch the shape and orientation of the triangle.

16.

Show that if r is the position vector of a point P, n a unit vector and p a scalar, the equation p

is that of a plane. What do n and p signify in relation to that plane? 17.

If r is the position vector of a variable point P, a and b are two constant vectors and A is a variable parameter, what does the equation

r = Aa+b

(-o0 Al or X X2. We then have a curve that goes from or to infinity - or both. Loosely we can talk of infinity as an `endpoint' of a general curve. We also may have self-intersecting curves, namely any time when (4) r(Xa) = r(Xb), X1 GXa < X b -X2 It is a fact that, strictly speaking, none of the theorems that we shall

develop in this book demand that we exclude self-intersecting curves. However, in applications that are natural in a physical context 14

The curve self-intersecting curves are never important and the consequences of applying a theorem to a self-intersecting curve may be difficult to visualise. 2

We are therefore in the happy position of being able both to tell the reader who is interested in the physics that he need not worry about self-intersection of curves and to allow the mathematical reader to include these curves if he wishes, so avoiding the precise mathematical formalism that would be needed to exclude them. Self-intersection will not henceforth be referred to, but the interested reader is welcome to use his imagination to visualise the meaning of our theorems when it is not excluded. We must however refer to the case of closed curves, which may be looked upon as special cases of self-intersection: (5) r(A1) = r(A2 ) Closed curves will be of great importance to us but we shall in fact treat them slightly differently from the way implied by (5) and we shall return to this a little later. To a physicist a natural assumption is that the three functions

x(X), y(X) and z(X) are sufficiently smooth for his purposes. We shall indeed always take them to be continuous and differentiable. In some important proofs the assumption that they can be differentiated twice will be necessary. We shall, therefore, freely form the derivatives

and

dr

dx dy dz

dA

(dX' dA' dA

d2r _

(6)

d2x d2y d2z

dX2 - (dA2'

dA2' dA2

(7)

Functions that are less well behaved than this are only of interest mathematically - with one exception; curves with `kinks' do occur in physical situations and will in fact be prominent later in the book. Mathematically this means that on a curve C there may be discrete points A* for which r* = r(X*) is uniquely defined but dr/dX does not exist. (dr/dX does exist for X < A* and separately for X > )* but these two functions do not tend to the same limit at r*.) One way in which a physicist is entitled to overcome any difficulties that may arise from the behaviour at r* is by noting that to every such curve with a `kink' it is always possible to approximate as closely as is desired by smooth curves. Within the accuracy of any 15

2 Geometrical background to vector analysis physical measurement, one cannot decide whether a curve is `really' smooth or kinked. Therefore any theorem of use to a physicist must be required to hold for smooth and kinked curves alike. Very often it is of advantage at a kink in a curve to consider the two portions on either side of the curve as two separate curves. Instead of putting

dr we say

dA

(8)

Al < A < A2

C : r(A),

undefined at X = A*

C = C1 + C2

Al A1, that integral is positive whichever end of the curve is defined as being the starting point r1 = r(A1). We can then state the important theorem: Let C1 and C2 be two closed curves with a common segment. Let the work integral with respect to a field be taken round each curve 51

5 Spatial integrals of fields separately with such senses of circulation that the common segment is traversed in opposite directions on the two curves. Then the sum of the two work integrals is equal to the work integral over that closed curve C which consists of C1 and C2 less their common segment, the sense of circulation on C remaining the same as in the separate integrals.

C

Fig. 2. The coalescence of two closed curves C1 and C2 into a single closed curve C. (Note the relative senses of circulation.)

The statement of this theorem is involved, but its content is very simple. Its proof can be disposed of by a glance at fig. 2. We can clearly parameterise the common segment in the same way for both curves and it follows from the preceding discussion that its contribution to the work integral is equal and opposite for C1 and C2. This property will prove to be of great importance. It is not shared by the integral (3.3). That it cannot be is best seen from the physical meaning of that integral for r = 1; it gives the length of the curve involved and clearly neither the sum nor the difference of the lengths of C1 and C2 is related to the length of C. The reason that the theorem holds for the work integral is precisely the fact that the work integral is linear in b,\. Brief mention must be made of some other rather trivial properties of the work integral. We have already seen that it is a scalar. It is also linear in the vector field f and hence it follows that if f and g are two vector fields then

fc

fc

fc

(7)

for any curve C. One glance at (4) is sufficient to show that this is true. Even more obvious is a property which we have in fact already used in stating the theorem illustrated by fig. 2 but which might be worth spelling out. It is that if C and C' are (open) curves such that 52

1

The work integral

or:

Fig. 3. Possible quasi-square parametrisations for a curve C as in fig. 2, given such parametrisations for the curves C 1 and C2 .

the end of C coincides with the start of C' (r(X2) = r(X )) then the work integral along the combined curve is the sum of the work integrals along its parts

Jc+cj-dl = fc

(8)

A few words should also be said about the relation of these developments to the quasi-square techniques introduced earlier. The closed curves C1 and C2 of fig. 2 were drawn as if not parametrised 53

5 Spatial integrals of fields as quasi-squares and, even if they had been so drawn, the curve C would in general require some new parametrisation to bring it into quasi-square form. This will not be an obstacle to further discussion but it may be worth illustrating this point by a diagram (fig. 3).

2 The flux integral This is a double integral over an arbitrary surface S (not excluding a closed surface S). Just as in chapter 3, we shall write our equation for a general open surface in terms of a quasi-square parametrisation. In other words we shall ignore the formal complications that arise if the boundary of S is not represented by four curves A = X1, X = AZ ,

µ=µ1P=µ2

Physically the flux of a vector F through a plane area A which is oriented so that its unit normal is n, is given by (9)

An example of the use of this notion may be taken from fluid dynamics. Let F be the product of the uniform flow velocity of a fluid with its density. The total amount of fluid that will cross an area A will be a product of three factors: A, the density of the fluid and the normal component of the flow velocity. Thus the amount of fluid crossing the area in unit time is given exactly by (9). This relation is easy to generalise to non-uniform flow described by a vector field f (r). Let dS be an infinitesimal surface area and n its normal. As shown in chapter 3

dS = n dS = hN x hµ dX dµ and the flux off through this infinitesimal surface is

(10) (11)

Therefore the total flux through a finite surface is f,,2

fs f-dS =

f ux f(r(X,p))-(hx x hµ)dXdp.

(12)

As in (4) for the line integral, it is useful to indicate how the right hand side of (12) must be written explicitly in the evaluation of particular problems:

54

The flux integral

2

Fig. 4. Two representations of the flux of a rapidly varying vector field f through the surface S of fig. 2.8.

f

µ=

S

f- ds = f 'z N, J Al

{fj [x (X, µ), Y (X, µ), Z (X,

)]

aZ ax

+f2[x(X,,U),y(X,µ),z(X,µ)](A

aµ

ax ay +f3[x(X, A),y(X, µ), z(X, µ)] (ax aµ

(ay az as aµ

az

ay)l

ax aµ

ax aZ a

aµ

(13)

ay ax aA aµ)l dXdµ.

We also note that the normal n at any point of S is

n55

hX x hµ

IhxxhMI

(14)

Spatial integrals of fields so that an alternative form of the flux integral is

=f ,,

JMf.nIhxxhpIdXd.

5

(15)

As for the work integral, there are numerous important physical applications of the flux integral. In addition to the example from fluid dynamics referred to already, it is significant in the theory of electromagnetism, particularly in relation to Faraday's law of induction. In that case f is the magnetic induction field. Just like the work integral, the flux integral is essentially simple simpler than the surface integral given by (3.8). This is again due to the linear nature of (13) in contrast to (3.7). In the definition of the normal n contained in (10) the order of factors in the vector product is crucial. The directions of increasing X, increasing µ and of n, in that order, form a right handed system. For an open surface there is no absolute distinction between the two possible directions in which the surface normal may be drawn. To change the sense of the normal one has to replace one (but not the other) parameter X or p by its negative and there cannot be an absolute rule as to which sense of normal one `ought to' use. But things are different for a closed surface S. There is an absolute distinction between the outward and the inward normal. If a closed surface is represented as the surface of a quasi-cube, the sense and sequence of the parameters used in describing each face should be adapted to using an outward normal at every one of the quasi-cube faces. If there is reason to keep to the `wrong' order of parameters the flux integral must be taken with a minus sign. In analogy with the discussion on p. 52, we see that if two closed surfaces coalesce to S then the sum of the flux integrals out of the two component surfaces is equal to the total flux integral out of the resultant outer surface. The contributions from the parts of the original boundaries of S1 and S2 which are internal to S must cancel because they enter the equation with opposite normals. The normal which is outward for S 1 is inward for S2 .

Again this idea is simpler to explain geometrically than in words. Fig. 5 illustrates it. The reader who interprets the operation shown as a `docking manoeuvre' of two space craft is absolutely right. It will be clear that the equivalent of this result does not apply to integrals over closed surfaces of the type discussed in chapter 3 (e.g. (3.8)). 56

2

The flux integral

Fig. 5. The coalescence of two closed surfaces S1 and S2 into a single closed surface S.

Clearly we have not at this point adhered to the quasi-cube picture of a closed surface. To give an illustration equivalent to fig. 3 with a quasi-cube parametrisation would be rather more awkward than for the curves of fig. 3, but in principle the point made there applies here just as well. As additional points concerning the flux integral we must note again that it is a scalar quantity, that its structure (see (13)) guarantees that fs fs (16) JS

and finally that if S and S' are two open surfaces which have a piece of boundary in common and if the sense chosen for the normal is 57

Spatial integrals of fields the same in both, then

5

fs,

J+5,f.ds =,fs

(17)

3 The quantity integral Having discussed line and surface integrals with simple linearity properties we will naturally have a look at the corresponding situation for three-dimensional volume integrals. However, because we are dealing exclusively with the physical space of three dimensions and not with abstract mathematical multi-dimensional spaces, we find that the situation for volume integrals is rather different and a good deal simpler than for line or surface integrals. We begin by noting that while dl and dS were vector quantities

dV = (hx hµ)

dX dµ dv

is a scalar, except thatt there is a sign convention involved in the scalar triple product. To form a scalar integral we must therefore have a scalar field in the integrand and what is natural here is the form of integral that was already shown in (3.14) of chapter 3. There we stressed the desirability of using parameters for which the triple product was everywhere positive. If in the course of developments there were a danger of having to relax this rule - and this does not happen in normal physical applications - one should be aware that two possibilities are open: either one postulates that (ha, hµ, should always be replaced by its modulus or one permits either sign to occur. In the former case one loses the simple linearity properties that were stressed in this chapter and were absent in chapter 3. These matters are only mentioned for completeness. There is no counterpart to closed curves and closed surfaces (i.e. curves without endpoints, surfaces without boundary curves or rims) when one is dealing with three-dimensional volumes and so there is no theorem corresponding to the situations illustrated by fig. 2 and fig. 5. Analogues to the simpler relations (7), (8) and (16), (17) do however exist and are equally simple. We quote them briefly. The typical scalar volume integral has the form

fv OdV = I

t See footnote on p.4. 58

,

(18)

The quantity integral where 0(r) may be any scalar field.t If we have two scalar fields, then clearly 3

Sdv=Jv cdV+fv OdV

(19)

and formally in analogy with (8) and (17), if V and V' are adjacent volumes then

JdV+ J,cv = fV+V, 0dV.

(20)

Once again it is worth remarking that relations like (20) are very obviously true and that the use of quasi-square and quasi-cube techniques is neither needed nor helpful in dealing with them.

t

It may be worth noting that the scalar fields occurring in physical applications divide into two distinct classes: those for which the evaluation of integrals like (18) is very natural and important and those for which it has little meaning. The first class is that of scalar densities. As was discussed in chapter 3, if describes the density of some substance, (18) measures the total amount of the substance present in V. Quantities of this nature are called extensive. On the other hand, there would be little sense in forming an integral like (18) if 0 described, say, the distribution of temperature. The second kind of quantity is called intensive. It is also of general interest to note that on the much more advanced level of general relativity there is actually a mathematical difference between the two kinds of scalar. 59

6

Further spatial integrals

1 Vector integrals The three types of integral introduced in the preceding chapter were all scalars. Recapitulating, they were the work integral, the flux integral and the quantity integral:

I, =Jc

I2 =Js

13 = fV. dVp.

(1)

Th e reader will note that in (1) all three integrals have been written in a slightly unusual form - with the infinitesimal element of line, surface or volume standing before the symbol for the field. Such a notation is unusual in mathematics but often used by physicists (if only to make sure that the infinitesimal is not accidentally omitted!). The advantage of changing to it here will soon become clear and we use it at the slight risk of ambiguity of notation. The conventional notation, by sandwiching the integrand between the integral sign and the infinitesimal, shows explicitly where the integrand ends. In practice, this ambiguity does not seem to cause confusion to physicists - where there is doubt, brackets enclosing this integrand can always be used. It is now easy to form further integrals which have all the simple properties of the integrals in (1) except that now the integrals are not scalars but vectors. In the case of the work and flux integrals we consider a vector

field that has the form f(r) = ao(r) where a is a constant vector and O(r) a scalar field. This f is a very particular vector field for which only the magnitude varies from one point to another while the direction remains constant, so that the field lines are parallel straight lines. In this case we can write

I, = fc

a fc NO,

12 = fs (2)

60

Vector integrals These expressions are scalars and have the form of scalar products

1

of a constant vector with another factor. Since the constant vector may be chosen quite arbitrarily, the `quotient theorem' of chapter 1 holds and the other factor must be a vector. So we have found the two vector integrals

J, = fo

NO

J2 = fS dSO.

and

(3)

Let us make quite explicit what is meant by such integrals. Each expression stands for three quantities

.Ii = (Jc dl,0,fd12,fcdl3 ) and

rIfs

(4a)

dS10, is2 0, JsdS3cb).

=

(4b)

Spelling this out in detail, we get

Ji =

(jx. A 0 [r(X) I A,

f

X2 dz X.

dX cb[r(X)l

J

2

d , [r(X)l dA,

dA)

(Sa)

and for a quasi-square az

J2 =

µ

µaZ ax _

if(

a z aµ

ax aZ

aA aµ

aA

µ , kA ay

aA

)[r(X, µ)l dX dµ,

aµ)

[r (A, µ) l dX dµ,

J2JM2( (5b)

µ)[r(X,µ)]

There is another way in which I, and I2 could be related to vector integrals. We choose f = g x a where again a is an arbitrary constant vector while g is some new vector field. We can then put

I, =

fc

I2 = fs

61

L

dlxg

fs

a fs dSxg

(6)

Further spatial integrals and we may conclude by the quotient theorem that

K1 = fc dl x g

and

K2 = fS dS x g

6

(7)

are vector integrals for any vector field g. We leave to the reader the explicit working out of the component expressions analogous to (4) and (5). Let us now turn to the quantity integral 13. Here the integrand is a scalar p(r) which in particular might be formed from some vector field g(r) and again a constant vector a. Thus we may write

13 = fV

a- fV dVg

(8)

and since a is an arbitrary vector

J3 =

(9)

fV dVg

must be a vector for any vector field g. In components

.I3 = (Jv dVgl, JdVg2,$ dVg3 ) .

(10)

This is in a sense the equation analogous to (4), but the result (10) is in fact very obvious and trivial. Integration, including volume integration, is a generalised form of addition and (10) simply tells us that the sum (i.e. integral) of a lot of vectors, in the usual sense of vector addition, is itself a vector. The equation analogous to (5) is quite trivial also - it simply means replacing dV by the Jacobian form of (3.13) and we omit its statement. The results of these integrations can be summarised rather neatly. If we examine the integrals involved in (1), (3), (7) and (9) and note their vector character, it is natural and correct to say that the symbolic integrals

f dl...

and f s

C

dS...

(11)

are vectors, while the integral

fV dV..

.

(12)

is a scalar.

Standing alone, these expressions are meaningless. To make them 62

1

Vector integrals

Fig. 1. The vector length of a curve C.

significant some field must be inserted in place of the dots with use of one of the multiplication procedures allowed in vector calculus. In this way all our previous results are unified. A particular simple application of the integrals of (11) is to the simplest of all scalar integrands: 0 = 1. We then have J1

(13)

J2 = fS dS.

= fc dl,

These expressions must not be confused with those we have just finished discussing. The present integrals are not symbolic - they have an integrand, namely 1. They also bear a strong resemblance to the integrals (3.1) and (3.6) which determined the length of a curve and the area of a surface. The difference is that the present integrals are vectors. We may call J1 the vector length of C and J2 the vector area of S. Let us evaluate these Az dr

J1 = f7" -dX = r(X2)-r(X1) = r2 -r1.

(14)

The `vector length' of a curve is simply the vector joining its two endpoints. The meaning of J2 is slightly less obvious.

JX2j) L2

=

dr

dr

(15)

dX dµ.

Let us look at one of the components of this vector, say the first: fKAX, "µz

(JZ)1

63

,

J µ,

ay az

az ay

ax aµ

ax aµ

- dX dµ

16

Further spatial integrals

6

Fig. 2. The x-component of the vector area of a surface S.

® + ve FIM zero - ve

Fig. 3. The z-component of vector area of a surface S which `folds over'. The component of vector area must be calculated as an algebraic sum.

64

More general integrals over fields It is easy to see that this represents the surface area of the projection of S on to the yz-plane. Given any surface in space one can define its `shadow' on any coordinate plane, e.g. the yz-plane, (see fig. 2); and the equation of this particular shadow surface is 2

r(A, µ) = (0, y (A, µ), z (A, µ)) and clearly (16) is its area. The same is true for the other two components of J2 and the `vector area' is the vector having these three shadows as its components. But the word `shadow' is used here in a somewhat unusual sense. At two different points of the surface the projections of the normal on a coordinate direction may have opposite signs. Two shadows, if overlapping, may cancel each other (see fig. 3).

A particular result of interest is that the vector length of a closed curve (which has no endpoints) and the vector area of a closed surface (which has no boundary curve) are both zero. 2 More general integrals over fields

We have come to the end of the list of integrals that may be relevant for our purposes. If we are interested in complete generality many other kinds of integral could be named. In a footnote on p. 40 we made brief reference to field functions that behave neither as scalars nor as vectors. Naturally spatial integrals over such quantities do exist. We can ignore them entirely. But there is another direction in which our way of forming integrals can be generalised. This must be mentioned briefly because later on the proof of some important theorems will depend on knowing that integrals outside our list can be excluded. Let us discuss first the case of line integrals along a curve C. All those we have met, including the length integral of chapter 3, depend on the shape of C via r(X) and dr(X)/dX = h;,. Derivatives of h;, do not occur. When mathematicians study curves they have very good reasons for also being interested in higher derivatives. The differential geometry of curves is concerned with quantities like the curvature and the torsion and these are defined in terms of such higher derivatives. But while there are mathematical reasons for introducing such concepts elsewhere, it is a fact that no relationships involving them are needed in vector analysis. Precisely the same is true for surfaces S which, in differential geometry, require derivatives of hx and b,, 65

6 Further spatial integrals for their full characterisation. Again, integrals involving such derivatives are irrelevant for physics. If we stop at integrals which have all the linearity properties discussed in chapter 5, which exclude quantities other than scalars and vectors, and which do not involve derivatives of the hx, our list is complete.

66

Exercises D

The stress in this group of exercises is still on the quasi-square and quasi-cube technique. Though scalar and vector fields are now involved, very few examples relate to `realistic' fields - fields one would encounter in physics. To illustrate the important features of the technique without making the examples too heavy much use is made of symmetries. Acquiring a feel for the utilisation of symmetries should stand the student in good stead in more realistic work. 1.

Find the work integral of the vector field

f = axr taken in the sense of increasing X along the curve defined by

r = Xb+X2c,

0<Xo L

(10)

It is not difficult to see that this definition is completely equivalent to the previous one. We can put

0(r2)-0(ri) _ 0[r(A2)] -O[r(A1)) (N2_d

Here we have merely used the fact of elementary calculus that the integral of the derivative of a function is the function itself. But now d dA

0 [r(A)] =

= and, using (6),

d dA

0 [x(A), y(A), z(A)]

ao do dy + ao dz dx + ax dA ay dA az A

O(r2)-0(ri) =

fz

VO-

dr

dA

(12) (13)

X,

which, by (3.1), transforms to

0(r2) - O(ri) = fC V t dl.

(14)

Inserting this into (10) and noting that (at any non-singular point) VO will tend to a constant value as L tends to zero, we find that (15) 0= 77

Differentiation of fields: the gradient 7 Clearly this result cannot depend on which curve of the set C is selected for the limiting process - because in the limit every curve is replaceable by its tangent. Also, the identity (15) holds whatever the choice of t and so we have proved that the gradient vector, defined in a coordinate-independent way by (10), has the mathematical form (7). This is so if the coordinates used are cartesian. But later on we shall be able to use the same basic definition to define the gradient in other coordinate systems.

5 The fundamental property of a gradient field Let us go back to (13). If we put - grad 0 = f

(16)

and refer back to the definition of the work integral in chapter 5 we see that (17) =

f f-

The minus sign in (16) has been introduced to conform with the most frequently (but not universally) adopted convention in physics that relates 0 to the line integral representing the work done against the field f rather than by the field f. In mechanics if f represents the force acting on a particle when it is at r, 0(r) is the potential energy related to that force. Usually the integral in (17) is studied as a function of its upper limit, so that r2 becomes the variable r, while the lower limit is kept constant. One puts

fo

f dl = - 0(r) + const.

(18)

On the left r is understood to appear as the upper limit of the integral. The lower limit is held fixed, but it can be chosen arbitrarily - hence the arbitrary constant on the right. For a given gradient field the scalar potential defined by (7) is arbitrary to the extent of an additive constant. It is clear that not every vector field f has a scalar potential. Equation (18) implies that the work integral of the field taken from a fixed point to r is a function of r only and otherwise independent of the choice of the path of integration C. That this is not so for all vector fields f can be understood intuitively by glancing back at the 78

5

Fundamental property of a gradient field

Fig. 3. A closed curve C represented as C1 - C2 where these two curves have common endpoints.

lower part of fig. 4.4 or fig. 4.5. In the region of the `whirlpool' depicted there the work integral from a point in the whirlpool to another point half-way round it will actually have opposite sign for a clockwise or an anti-clockwise path. So not every field of force in mechanics need have a potential energy and - saying essentially the same thing in the language of vector analysis - not every vector field is a gradient field. In mechanics a field of force that does have a potential energy is called a `conservative' field, in general vector analytic language a field satisfying (16) is called `irrotational'. The precise justification of the term irrotational has yet to be stated, even though the reader might suspect that remarks such as those just made about a `whirlpool' picture would be enough. We shall see that a little more precision is needed. But there is a useful way of reformulating the same ideas. Instead of comparing work integrals along two open curves between two points, we can reverse the direction of integration along one of them. We know from chapter 5 that this results in a change of sign of the work integral. Equation (17) gives directly

if

f f=

- grad 0

provided that C1 and C2 both connect the same two points. Introducing the closed curve C = C 1 - C2 we can say instead

f, f dl = 0 79

(20)

Differentiation of fields: the gradient

7

if f = - grad 0.

(Alternatively (20) can be verified by composing C out of more than two open curves C1 as in (2.10).) Furthermore we can immediately replace the word `if' in (20) by `if and only if'. To prove this we assume that the integral relation of (20) is satisfied for all closed curves C in some region of space. Picking out two points r1 and r2 on some C we reverse our previous procedure by splitting C into C 1 and - C2 and see that the work integral between r1 and r2 is the same for either curve - and indeed also for any other curve joining these points. For fixed r1 and varying r2 = r this means that the work integral is a function of r only, not of the path. Thus we have returned to our starting point (17). Let us then finally restate this important result: The work integral of a vector field taken round any closed curve that may be drawn in some region of space is zero, i.e.

f, for all C, if and only if

f = - grad in that region. A vector field with this property is called irrotational.

80

(21)

8Differentiation of fields. Part 2: the curl

1 Definition of the curl of a vector field The discussion of the last chapter led to the distinction between fields that did and those that did not satisfy (7.21). That identity relates to general closed curves C of finite length. We now relate the insight gained to a differential property of a vector field. This will involve the idea of contracting a curve C onto a single point. We take a point r in the region in which the vector field is defined. At that point we select arbitrarily a fixed unit vector n and the plane through r normal to n. We then consider the set S of all surfaces S that pass through r and there have n as their normal and therefore the aforementioned plane as their tangent plane. Taking any surface S from S we draw on it a closed curve C that encircles the point r. (We assume that C has a sense of circulation that appears anti-clockwise if seen from the side into which n is pointing.) Part of the original surface S is now the interior of C and as we are not going to be concerned with any part of S except this interior piece, we may use the symbol S to describe the interior alone. Let its surface area (as defined in chapter 3) be A. For every vector field f(r) we can define another vector field, which we shall call curl f, obtained from f by the following process of differentiation (cf. the procedure in section 4 of chapter 7):

(n curl f) = lim A fc f dl.

(1)

Clearly for A - 0 every surface S from S tends to the tangent plane. The process defined by (1) must therefore lead to a unique answer for every fixed unit vector n. By performing the limit process for all possible n at r we can thus define the vector field curl f. If f is irrotational curl f is necessarily zero because the integral in (1) is then always zero. Therefore

curl grad 0 = 0. 81

(2)

Differentiation of fields: the curl

8

Fig. 1.. The definition of curl f.

However, in general neither the integral in (1) nor its limit need necessarily vanish. If at some point r the limit is not zero, this evidently implies some element of `curliness' in the field line picture around r. This was alluded to qualitatively on p. 78. Hence the name curl. But two things must be noted: firstly the `curliness' may be much less in evidence than could be suggested by fig. 4.4 - it may be just a slight trace masked by an overall field line pattern very unlike fig. 4.4 and, secondly, and more importantly, if curl f # 0 at r this requires `curliness' around r and not around some neighbouring `eddy centre' like the one clearly seen in fig. 4.4. That picture tells us nothing about curl f at a point r away from the centre, while (1) shows that curl fat r depends only on f and its first derivatives at that point. In descriptive language the field curl f associated with a field f is often called its vorticity - a term originating in fluid dynamics.

2 Evaluation of curl f. Stokes' theorem To obtain an explicit expression for the vector field curl f, especially in cartesian coordinates, we must transform the integral on the right hand side of (1). As stressed in chapter 2 we may take the curve C in (1) to be a quasi-square as defined in (2.13), without effective loss of generality. The surface S is then given by (2.12) and the line integral has four parts. Using the notation of equation (2.15) we can write

Jc 82

f dl+

f- dl = fc 4

JC

2

Evaluation. Stokes' theorem

f

- J 2f[r(A,µz)] h;, dX +

f[r(Xi, p)] 'h" dµ fX2f[r(A,µi)] hxdX.

(3)

It has not been made explicit, but is understood, that h,, or h", as the case may be, is a function of the same argument as the f that it multiplies. We use the same device that was introduced in evaluating (7.11) and remember that the limits of both integrals are independent constants. Therefore the order in which we perform integration over X and over µ is immaterial. In this way we get

[a(fh)-a a

fcf - dl =

)J dAdµ.

µ

fX1f"12

(4)

Here we have suppressed the independent variables in the integrand. At this stage they are simply X and µ. We calculate further

(5)

But

a (3r)

a

a

all

am

ax all

aX

a

ax

bAl

(6)

so the second and fourth terms cancel. It is essential that such a cancellation should take place: in chapter 6 it was promised that integrals involving second derivatives of r would not bother us! Turning to the first and third integrals we note that f depends on A and µ only through r(X, µ). Therefore

aAf = (ar V)f = (hx V)f and

µf=

am

f

(7)

=

This enables us to write

jf.dl C

JX25M2 Xi

-hx 83

{b" dAdy.

(8)

8 Differentiation of fields: the curl Equation (1.20') was specially designed to deal with the next step.

We find

f f dl = f iX 2f µ12(hx x hµ) (V x f) dX dµ.

(9)

But now the right hand side is precisely a flux integral as defined in (5.12). So we have

fc f-dl = fs

fs (ox

(Ox

(10)

We can now evaluate curl f explicitly: insert the right hand side of (10) into (1), assume that the components off are free of singularities around the point r at which (1) is being evaluated and note that for sufficiently small A the integrand on the right of (10) may be taken outside this integral. This gives us n can be chosen to have any direction,

curl f = V x f.

(11)

Thus the cartesian form of curl f is the cross product of the V symbol with f. Just as for the gradient, we see that here again V behaves like a vector: by (1), n curl f is a scalar, and curl f is then a vector by the quotient theorem. Explicitly we have then, in cartesians,

_ curlf = Vxf

a

(ay f3

a

a

aZf2i aZf1 a

ax fl - ay f2

ax f3,

(12)

At this point it is worth looking back to (2) and noting that in terms of the Cartesian V operator its equivalent V X (V) = 0 (2') t As a mnemonic the alternative form

VXf=

is quite useful.

84

(13)

2

Evaluation. Stokes' theorem

Fig. 2. Stokes' theorem illustrated.

can be very easily verified by direct calculation or by formally applying the vector identity a x a = 0 to the symbol V.

Returning to (10), we can now assert that

f f- dl

= fs curl

(14)

This is the statement of one of the most important theorems of vector analysis, which says in words: The work integral around a closed curve C of a vector field f is equal to the flux integral of curl f - through any surface S of which C is the rim.

(The normal n on S used in the flux integral must be chosen in such a way that the sense of circulation around C appears anticlockwise if seen from the side into which n points.) This theorem is known as Stokes' theorem. Fig. 2 serves to illustrate it.

Let us review what we have achieved so far in this chapter. In the preceding section the physical concept of the curl of a vector field was defined by a limiting process akin to that which serves to find any derivative in calculus (the definition was also strictly parallel to the one given for the gradient in section 4 of chapter 7). In this section we have shown how to evaluate fin cartesian coordinates by forming V x f. We also have the prescription for the corresponding evaluation in other coordinates but will return to that later. In 85

Differentiation of fields: the curl addition, and most important of all, we have been able to formulate Stokes' theorem which enables us to convert the flux integral across a surface of a certain kind of vector field into a work integral round the boundary of that surface. This last statement shifts the emphasis from f to curl f and takes us over to the next section.

3 The fundamental property of a curl field The subject of our study in this section is a vector field that is the curl of another vector field. Changing our notation we call this field

f(r) so that

f(r) = curl a(r),

(15)

where a(r) is another vector field. Equation (15) is similar in structure to (7.16) and we shall proceed with a discussion that is in close parallel to section 5 of chapter 7. In the new notation Stokes' theorem (14) becomes

J5f dS = f.

(16)

(with the same sign convention as stated after (14)). If the field f is the quantity of interest in physical study and it is known that it is of the form (15), the field a is not uniquely determined. In the comparable situation of (7.16) a constant could be added to 0(r) without changingf(r). Here the freedom is much greater. Bearing in mind (7.21) and (2) we see that neither (15) nor (16) is altered if we make the replacement

a -+ a+gradX

(17)

where X(r) is an entirely arbitrary scalar field. Any field a related to f by (38) is called a vector potential for f and (17) shows that the same f can be associated with many different vector potentials. On the other hand, not every field f can be represented in the form (15). Equation (16) follows from (15) and it tells us that a field of this kind has the same flux integral through any two surfaces S that have the same boundary curve C. This is because the right hand side of (16) involves only that boundary curve. Examples of vector fields for which this could not possibly hold are not difficult to find. Take two surfaces S, and S2 with the same boundary or rim C. Fig. 3 illustrates the situation. Note that if our rule, relating the circulation round C to the normal 86

Fundamental property of a curl field S2

Fig. 3. A closed surface S represented as S2 -S1i with the curve C as the common boundary of these two surfaces.

on S 1 and S2 , is observed then in fig. 3 the correct normal on S2 points outwards while the correct normal on S1 points inwards. If we reverse the normal direction on S 1 (for simplicity let us call the surface with reversed normal - S 1 ) then S2 - S 1 is a closed surface S with outward normal. A field satisfying (15) necessarily has zero flux out of S. Physical intuition tells us that any field of flow that has sources inside S will certainly not have zero flux out of S. Such a field would therefore not be expected to satisfy (15) and the name 'source-free field' suggests itself as a name for one that does. In fact the technical name for a field satisfying (15) is `solenoidal' - a name derived from the Greek word for a tube. Why this name is appropriate will need some explaining in due course. The fact that for a solenoidal field the flux out of a closed surface vanishes can also be shown in a slightly different way. We recall our remarks at the beginning of chapter 5 on the work integral over adjoining closed curves. Let us assume that (15) applies separately to two curves C and C2 with a piece of common boundary, and that these curves are the rims of two surfaces S 1 and S2 respectively. Then (15) will also be true for S1 + S2 and the boundary C which consists of C1 and C2 less the piece they have in common. Now a general closed surface S can be thought of as covered by a mosaic of surfaces like S1 and S2 - like an egg with a net of cracks on its shell. In building up S in this way, all pieces of all boundary 1

87

Differentiation of fields: the curl

8

Fig. 4. The coalescence of two closed curves (as in fig. 5.2) and the simple addition of the surfaces Sl and S2 bounded by them.

Fig. 5. Replacement of a single surface S by many `fragment' surfaces; the circulations round all boundary curves are fixed, given the circulation round S.

curves will cancel in the same way as the one central piece did in fig. 4. Also, if the normal is outward on one piece of `eggshell', it will have to be outward on all. Hence

fs

f=

curl a.

Again the word `if' in the equation can be replaced by `if and only if' and the reasoning that proves this is essentially the same as on p. 80 following (7.20). Unfortunately it is now a little more involved. If we know that the flux integral in (18) vanishes we can reverse our previous steps and choose an arbitrary closed curve C on S that divides S into S2 and S, I. Over the second of these we reverse the normal compared to S. Then we can say that the fluxes through S I and S2 must be the same, and they must be the same also for any other surfaces that have C as a boundary. This means that the form 88

4

Quantitative picture of a solenoidal field off anywhere except on C cannot affect the value of the flux integral. This integral therefore must be reducible to an integral round C. Can this integral be of any other form than the right hand side of (16) with some a? Our careful discussions in chapter 2 are sufficient to ensure that it cannot. Whatever else we might allow, the integral over C that we seek must (a) be a scalar, (b) have the properties of linearity expressed by (5.7) and (5.8) and (c) not involve derivatives of the hx . The reason that these derivatives can be excluded is that we know that they do not appear in the flux integral on the left of (16). A process of integration leading from that flux integral to something involving only the single integral over C can certainly not increase the number of differentiations occurring under the remaining integral sign. Now we learnt that the work integral round C was the only type of integral that satisfies (a)-(c) and therefore (16) and hence (15) must be true. This proves the required extension of (18). The flux integral of a vector field over any closed surface that may be drawn in some region of space is zero, i.e.

fs

0

for all S if and only if

f = curl a.

(19)

A vector field with this property is called solenoidal.

4 The quantitative picture of a solenoidal field In the preceding section we have demonstrated the basic analytical property of a curl or solenoidal field. We stressed the parallel between the theorems concerning irrotational fields and those concerning solenoidal fields, but as yet we have far less of an intuitive picture of the solenoidal field than we found for an irrotational field in the level surfaces of its scalar potential. The key to visualising a solenoidal field is (19) applied to a particular type of surface S defined as follows: We start from an arbitrary closed curve C in a region of space in which a field f is defined. We then consider all those field lines off that pass through C. Together all these lines form a tube. (Note that the word `solenoidal' originates with this notion!) We consider a finite length of this tube which is closed at either end by surfaces S 1 and S2 so chosen as to be perpendicular to all the field lines on and encircled 89

Differentiation of fields: the curl

8

Fig. 6. A field line tube.

by C. Having constructed a closed surface S in this way we can forget about the original closed curve used to define S. The closed surface consists of the two `endplates' S1 and S2 and the tube itself which we call S *. As always we associate an outward normal with S* and note that this normal is the same as the one that belongs to S2, according to our conventions and to fig. 6, but opposite to the one we would associate with S 1 . Thus if f is solenoidal

Jf.ds

=

Js

f dS + Js f dS - Js f' dS = 0.

(20)

2

But as S * consists entirely of field lines, f cannot anywhere have a component normal to S*. Therefore the first in the sum of three terms in (20) must be zero so that

Js, f dS = Js

f* dS

(21)

if f is solenoidal. Now let the surface areas of S 1 and S 2 be A 1 and A2 respectively. These are not independent numbers and the smaller of the two must tend to zero if the larger one does. We assume that 90

4

Quantitative picture of a solenoidal field

both are sufficiently small for it to be possible to neglect any change of magnitude or direction off over both S1 and S2. In that case (21) becomes

If1IA1 = If21A2

increased if f is solenoidal. Supposing that we draw a particular number of field lines to pass through S I . In fig. 6 the number drawn through S I is 10. The same ten lines also pass through S2. Because (in the particular case) A 2 happens to be smaller than A 1 the number of lines crossing unit

area has increased in the proportion A I/A2. But by (22) the magnitude If I has increased by the same factor! We therefore have the theorem: In a solenoidal field the magnitude If I at any point is directly proportional to the number of field lines crossing unit perpendicular area there. This assumes that every field line is drawn continuously with no field lines terminating at any increased finite point r. Without the last proviso the theorem would be meaningless. Clearly it would be possible with any vector field to make the average number of lines around any point proportional to f there. But to do this one would in general have to start new lines of force at certain points, lines or surfaces, or - worst of all - `continuously' in space. Normally lines of force are drawn without `loose ends' and then, if f is solenoidal, their number allows one to deduce the strength of the field from their density. If, for example, one knows that a field of flow as illustrated by fig. 7 is solenoidal one can conclude a great deal about the field. There is a strong current flowing from north to south on the left of fig. 7 and there is a vortex flow on the right which has particularly high velocities at its western edge.

Fig. 7. The field line picture of a solenoidal field made quantitative. 91

Differentiation of fields: the curl 8 The reader should note that there are no field lines terminating anywhere in the picture. They all either extend to infinity or form closed lines. We now have two classes of vector field that we can visualise

quantitatively - irrotational fields with the aid of their equipotential surfaces and solenoidal fields with the aid of the last theorem on numbers of lines of force. Unless the field is irrotational there is of course no such thing as an equipotential surface. Are there any other types of field we have to be concerned about? The analysis of the next chapters will help us to answer this question.

92

9

Differentiation of fields. Part 3: the divergence

1 Definition of the divergence of a vector field We know that the flux integral of a solenoidal field out of any closed surface is zero. This is not so for a general vector field and this fact provides a measure of how much a vector field can deviate from being solenoidal. This measure is given by another form of differentiation used in vector analysis. We consider the flux integral of a general vector field out of a closed surface S. Let V denote the volume of the space enclosed by 9. Then at any point r we define a scalar quantity, the divergence of the vector field f as the following limit:

div f = lim

- J.f.dS 1

(1)

where it is assumed that V is made to shrink around a given point r. Because of the scalar nature of divf as defined in (12), this limit process is rather simpler than either of the two earlier ones given by (7.10) and (8.1). When V tends to zero no properties off at points other than r should affect div fat r and the limiting process performed separately at every point r will define a scalar field. We can be sure that

div curl a = 0

(2)

because for a solenoidal field f = curl a, the flux integral on the left of (1) is zero from the start. For a field that is not solenoidal on the other hand, (1) tells us that a net (outward or inward) flux must be present at r if divf does not vanish at that point. Clearly this explains the name divergence (which may be positive or negative). But, just as for the curl, it is important to remember that a non-vanishing divergence is not necessarily clearly visible in the pattern of field lines `radiating' from that point and that the properties off only at r and not at other points determine div fat r. 93

9 Differentiation of fields: the divergence Just as curl f is often referred to - both in fluid dynamics and in other fields - as the vorticity of f, so divf is referred to as the source density of the field, again using hydrodynamic language.

2 Evaluation of div f The divergence theorem Still proceeding in exact parallel to section 2 of chapter 8 we go on to obtain an explicit expression for div f, primarily in cartesian coordinates. In the present situation we take the volume V to be a quasi-cube so that S consists of the six surfaces explicitly introduced in (2.20). The reader should take note again of what was said there concerning directions of normals to the six surfaces. We can write out the flux integral on the left side of (1) as follows:

v2

A,

f (A2 , µ, v) (h,2 x

J

v,

f 'I

dµ dv

v) (hµ x

dµ dv

v2 f'2 f (X, + fv, X, µz , v) (hv x

dv dµ

f µ2 µ,

-f

V2

v

v2

f

f(X, pi, v) (bv x hx) dv dX

X

µi

- f12 fµ2 X1

(3)

2

+ fA2 fµ2 X,

f (X , µ,

µ,

,f(X, µ, v2) (h, x hµ )dX dµ f (X,

µ, vl) (b,, x hµ) dX dµ.

As before each pair of terms can be simply combined and altogether we get a 2 µ2 v2 a =f1, [f (hµ x hr)] + a [f (hU x bx)] Js

a

+ av [ f (hxx h,, )] y dX dµ dv.

(4)

Again the apparent appearance of derivatives of the h vectors is fortunately illusory. For example, we get f (ahµ/aX x as part of 94

2

Evaluation. Divergence theorem

Fig. 1. The divergence theorem illustrated.

the first term and f (hv x aha/aµ) as part of the second term. Clearly the two terms are equal and opposite. Two other pairs cancel each other in the same way and in all we get JX2J2Jv2{(h µi

S

vi

a

+ (hx x bµ) a ldX dp dv.

(5)

Using the result of (8.7) again, we transform this to Jf'dS=

JJ2J{(hfµ x hv) [(hx V)f] (6)

(hx x hµ) [(hv V) f ] } dA dp dv.

Equation (1.23) was stated in anticipation of this development. It allows us to write instead of (6):

Jf.dS

=

fv V.fdV.

(7)

We insert this result into (1), go to the limit and find that

divf =

(8)

in terms of the cartesian operator V, and to be quite explicit

divf = V f =

a

ax f'

a

a

+ ay f2 + az f3 .

Since the limit process (1) clearly defines a scalar we see that again 95

(9)

Differentiation of fields: the divergence 9 the V operator enters an equation just as any vector - to produce a scalar multiplication with another vector. So V behaves like a vector in all three forms of multiplication used in vector calculus and this is enough to say that V is a symbolic vector. At this stage it is also evident that relation (2) has a very simple algebraic equivalent 0.

(10)

Either evaluation in components or comparison with the obvious relation a (a x b) = 0 confirms the correctness of (10) or (2). Finally we combine (7) and (8) to express another central theorem of vector analysis, the divergence theoremt

j...f.dS =f, div f dV.

(11)

In words: the flux integral out of a closed surface of a vector field f is equal to the volume integral of div f over the volume enclosed by

S.t 3 A survey of results At this point we have assembled all the essential ideas of vector analysis. We possess definitions of all the basic differential operations required: for a scalar field the gradient, which is a vector; for a vector field the curl, which is a vector; and the divergence, which is a scalar. On the one hand we have definitions related to the physical picture of these derivative fields, on the other we showed how they can be expressed simply in terms of the cartesian operator V. We also derived the three main theorems governing the integration of these derivative fields. These are the (unnamed) theorem on the work integral of a gradient, Stokes' theorem on the flux integral of a curl, and the divergence theorem on the quantity integral of a divergence. To be precise, we showed that while for general vector fields work and flux integrals can only be evaluated for a given curve or a given surface, as the case may be, the line integral of an irrotational or gradient field and the flux integral of a solenoidal or curl t Nomenclature varies. The same theorem is generally known as Gauss' theorem on the Continent while traditional British usage restricted that name to an electrostatic application of the theorem. Green's theorem is another name sometimes used, but elsewhere that term is associated with certain secondary consequences of (11). $ Note in comparing this statement to that after equation (8.14), the difference between 'any surface' there and 'the volume' here. Compare this with remarks on p. 29. Note also that a theorem analogous to (7.21) and (8.19) cannot be stated - for closely related reasons.

96

3

Survey of results

field can be reduced to a function of the boundary - i.e. of the two endpoints of C in the first case, of the rim curve C of S in the second. In other words they can be formally `integrated', as distinct from just being evaluated. Similarly we have shown that the quantity integral of the divergence of a vector field can be integrated, i.e. expressed in terms of the field itself at the boundary of the volume over which the integral is taken. The basic theorems also helped us to understand and visualise the properties of irrotational vector fields on the one hand, and solenoidal vector fields on the other. A question that has not been answered so far, however, is the nature of fields that fall into neither of these classes. Later discussion will show that we need not worry about more general fields; they can always be represented as sums of irrotational and solenoidal fields. This will be demonstrated at end of chapter 12 and in chapter 13 after a little more mathematical development.

97

10

Generalisation of the three principal theorems and some remarks on notation

1 General integral theorems In chapter 6 it was shown how to define integrals that are similar to the work, flux and quantity integrals but are of vector rather than scalar character. We now show how the three principal theorems on integration can be extended so that they are applicable to certain vector integrals of this kind. This time let us consider the divergence theorem first. We shall confine ourselves for the present to the statement of this theorem in its cartesian form:

JdS.f=J

V

(1)

Notice that we have returned to the convention introduced in chapter 6 of writing the differential to the left of the integrand. We do not need to repeat in detail the reasoning of chapter 6 and leave it to the reader to verify that by looking at f = ao and f = a x g, where a is a constant vector, it is easy to prove

fs dso

fV dV VO

(2)

and

JdSxg=JdVVxg.

(3)

With precisely the same justification as was given in chapter 6 we can therefore generalise the divergence theorem to

fs dS... = JV dVV...

(4)

Essentially the same device enables us to generalise Stokes' theorem. We start from

fc dlf = fS (dSx 98

(5)

General integral theorems The right hand side of (4) is not exactly in the form in which the theorem was stated earlier. It is slightly different not only because the order of factors has been changed but also because we have used the algebraic vector identity 1

c) = (ax It is easy to verify that this relation is true when V is involved by writing the expression in component form. Now again, insertion of act and of a x g leads to

f

NO = fs (dSx V)O,

(6)

fz dlxg = fS (dSx V)xg.

(7)

.

So the generalised Stokes' theorem reads

fc dl ... = fS

(dSx V)

...

(8)

We have left to the last the simplest of the three theorems, simply because this presents us with a small problem of notation. We can begin by stating the work integral identity in the form

c(r2)-q5(rj) =

.lc

(9)

dl-V0

which is essentially what we wrote down before. Chapter 6 has taught us that in the case of a scalar field appearing in vector identities replacement of 0 by a f will lead directly to the corresponding vector identity. This could be stated as

f(r2)-f(rj)

(10)

= fc

Unfortunately passing over from (9) to something analogous to (4) or (8) would force us to write the left hand side in some form such

as (... (r2)) - (... (r1)) which is not very pleasing or readable. But a small notational device gets over this difficulty, which will also be useful later on. We write, e.g. for O(r), 2

0(r2) - O(r,) = 0 The notation I i (or Ib - or particularly 1±) is taken from ordinary 99

10 Generalisation of theorems; notation integral calculus and simply means that one takes whichever function stands at the left of the vertical line at the value of argument characterised by the upper symbol and subtracts from it the same function taken at the value of argument characterised by the lower symbol. In this form the generalised work integral law becomes

...

(12) 12= Jc Here 1 and 2 respectively stand for the values of r at the two endpoints of C. Note that the three generalised integration theorems (12), (8) and (3) required the use of the V operator in a form in which it cannot be replaced automatically by one of the words grad, curl or div. At present therefore we have only shown how to state these generalised theorems in cartesian coordinates. We shall see in due course that this is not a real restriction.

2 A general notation In the last few equations of the preceding section we were detained a little by matters purely of notation. This section will be devoted entirely to notation and may be skipped without danger of losing the thread of the remaining argument. Except, that is, for one remark which really belongs to the preceding section. There we wrote down our various integral relations without full explanation of their meaning simply because the reader could be trusted to have remembered the tiresome bits: that the sense in which the curve C is taken in (5) is related to the direction into which dS points by the right hand rule and that the normal used in the surface integral in (4) is outward. We have also not reiterated the fact that in our notation C stands for an open curve, C for a closed one and the same with S and 9.

The omission of these details should assist rather than hinder the reader of this particular text, but what if these results are used in applications? One device to avoid tedious explanations about paths of integration being closed is the use of the symbol f dl for what we have been writing as f E dl. Some writers have also used fi dS instead of fs dS, but this is less common and perhaps less appealing. But this notation does not solve the problem of succinctly stating the sign conventions. We propose a notation, designed to avoid these things. It is a 100

Generalisation of ranges of integration development of the f dl notation and readers can judge its merits, but it must be admitted that it is a notation to be read in printed texts and not one for quickly scribbled notes. The author did not choose to use it in this text where there is a step by step development of the formulae of vector analysis. But if he could assume a reader's familiarity with it, he would not hesitate to let it be printed to give a succinct statement of, say Faraday's law of electromagnetic induction (which relates the rate of change of a flux integral to a work integral around a closed curve). We shall not spend time on explaining the notation. If that were necessary, it would prove the worthlessness of the notation. It should be entirely self-explanatory. So we simply write down the general integral theorems (1), (2) and (3) in terms of it: 3

=

fdl.v...

(13)

dl... = #dSxV...

(14)

dS... =

(15)

dVV...

3 Generalisation of ranges of integration Though in every important sense the relations just restated in a picturesque form are the most general results on integration of fields that we need consider, there is just one more direction in which some generalisation is possible: do such theorems have to be stated for single curves, surfaces and volumes or can they be extended to sets of curves, surfaces and volumes? The answer to this question has three aspects - one trivial, a second very academic, but a third which is quite important. The trivial side is that if one merely thinks of curves or surfaces or volumes set side by side in space, all three theorems are valid for each piece separately and therefore for the sum of pieces. It is slightly less trivial, but entirely covered by previous discussion, that one can allow the coalescing and cancelling of terminal points in (13), arcs of boundary curves in (14) and portions of boundary surfaces in (15). In all cases the right hand integrals simply add up. 101

Generalisation of theorems; notation

10

Fig. 1. Two interlocked closed curves. How can Stokes' theorem be applied?

Fig. 2. Two interlocked curves and an equivalent single, unknotted curve.

Unfortunately this statement does not cover all possibilities except for theorem (13). In the case of (14) - what about interlocking boundaries? Consider fig. 1. Clearly Stokes' theorem can be applied separately to C1 and a suitable S1 and to C2 and some S.2. If the two rings lie practically flat on a table one might be interested in the flux through the larger dotted curve in fig. 1. This would clearly be practically the same as the flux through S1 + S2 less the flux through the small central surface between A and B, since in S1 + S2 that surface is counted twice over. If we treat the lines C 1 and C2 as wires of finite thickness and do not neglect the dimension perpendicular to the table, the larger dotted curve will have to include two short vertical portions at A and B, both going down, while the 102

3

Generalisation of ranges of integration

H (a)

(b)

Fig. 3. (a) A simple knot: a closed curve. (b) Two simple closed curves as an approximation to the knot.

smaller dotted curve must include exactly the same vertical pieces both going upward. In the sum these vertical pieces cancel out, and we get the same result from Stokes' theorem whether we base it on the two rings or on the other two curves. This gives us a clue on how to apply Stokes' theorem to other interlocking situations. As this is really only a matter of curiosity we shall simply indicate what happens without going into detail. Figure 2 shows two interlocking circles not in a plane and shows that again (by a process of deformation from the previous situation) they can be replaced by two non-interlocking curves - Stokes' theorem can be applied to either pair of curves without altering the total flux integral. Figure 3(a) shows the simplest single closed knotted curve where the situation is exactly the same as for the two circles of fig. 1. Fig. 3(b) indicates how (except for three pairs of equal and opposite vertical joins at A, B and C) the knot, when flat, can be replaced by two unlinked curves. We shall not attempt to give a diagram equivalent to fig. 2 for this case. A final word: about the Mobius strip. This object is famous for being a surface which does not have two distinct sides. Stokes' theorem cannot be directly applicable to such a surface S because no unique sign convention can relate C to a preferred normal direction. But if we examine what this boundary curve C is, the paradoxical aspect of this case disappears. C in fig. 4(b) is not knotted, but appears to have a twist. If C is thought of as made of wire it can soon be untwisted and it is then clear how surfaces S can correctly be associated with C. Imagining an S made of rubber fitted into the untwisted C it can be twisted back 103

Generalisation of theorems; notation

10

(a)

(b)

C.

(c)

Fig. 4. (a) The Mobius strip. (b) The boundary curve of the Mobius strip. (c) The boundary of the cut Mobius strip.

Fig. 5. Two interpenetrating volumes.

into the position of fig. 4(b). Clearly the shape of that S is something quite different from the Mobius strip. Alternatively Stokes' theorem can be applied to the strip itself, if it is cut. Now C' is quite a different curve, while S is the correct Mobius surface. But note that the normal points to opposite directions on the two sides of the cut, and therefore the value of the flux 104

3

Generalisation of ranges of integration

Fig. 6. A volume with an inner boundary surface.

integral through S of any field, will depend critically on where the cut is made. When we pass from surfaces S and Stokes' theorem to volumes V and the divergence theorem we encounter matters of rather greater importance. Now we have two cases: either two intersecting closed bounding surfaces S 1 and S2 or one boundary surface within another. (We eschew Klein bottles!) The former case is not very interesting. Fig. 5 gives an example. If V is the total volume of the object, V1 and V2 the volumes of the separate cylinders and V the volume of their intersection, then clearly

V = V1 +V2-V'

(16)

and using standard notation, with outward normals in every case, we have also

S = S1 +S2 -S'.

(17)

It is clear that the flux of any vector field out of the whole outer surface is correctly described by the sum of the fluxes out of V1 and V2 less the flux out of V. This situation - and its generalisation to more than two intersecting volumes - presents no real points of interest. On the other hand important use will be made of what is in fact a special case of the above, namely when V2 is entirely enclosed by V1. In the notation just used V = V2 and V = V1. The physical situation 105

Generalisation of theorems; notation

10

Fig. 7. The equivalence of a volume as given in fig. 6 with two volumes without internal boundaries.

of greatest interest is when V2 is actually a hole, e.g. when only the volume V = V1 - V2 is the seat of the field. The boundary surface S of V now consists of two parts, S and 9. 2, but the sense or normal that is outward as far as V is concerned is in fact the inward normal 1

on S2 .

Our previous general considerations are sufficient to guarantee that the divergence theorem may be applied to this geometrical situation with the boundary surface separated into several parts but because the theorem is quite important it may be useful to visualise the nature of the proof geometrically. Consider a hard-boiled egg: if the egg is in halves the outward surface of the egg white includes the pair of knife cut surfaces, and the divergence theorem is applicable to each half separately. If the two halves are put together again, the parts of S1 and S2 from the knife cuts cancel. Then the volume of integration is just the egg white and its surface consists of the separate outer and inner parts (fig. 7).

106

Exercises E

The material of chapters 7-9 is so interrelated that it would not have been practical to provide separate exercises for each, or even to arrange the exercises in an order that relates strictly to the topics of the individual chapters. The student should, however, find little difficulty in identifying questions that require the understanding of earlier material only. An attempt at logical progress has been made except that the first three questions clearly relate to chapters 7, 8 and 9 separately. Sections 2 and 3 of chapter 10 do not call for illustration by exercises, but the last few questions in the following relate in part to section 1 of chapter 10. At this stage in the book the idea behind the quasi-square and quasi-cube technique can be taken for granted. The technique is still relevant to some of the exercises, but many are now independent of it - and particular fields that are of importance in physics are now introduced when possible. However, some of the exercises still deal with quite artificial constructs. As before, these serve to illustrate general ideas by non-trivial particular examples. In some case the work is saved from being intolerably cumbersome only by the symmetries built in. Artificial though this may be, training in the exploitation of symmetries is of value in itself. Throughout this group of exercises the symbols a, b, c stand for constant vectors, i, j, k represent an orthonormal triad and n is an integer - not necessarily positive except when this is explicitly stated. In exercises 1- 3 an asterisk attached to a Roman number, e.g. (iii)*, signifies that the field in question is of particular interest in physics. 1.

Evaluate the gradient fields of the following scalar fields-

(ii) Ir-a I2, (iii) r, (iv) Ir-al, (v) * 1/r, (vi)* Mr-al, NO Ir-a In, (viii) * (a - r), (ix) (i) r 2,

(a

(xiii) la x r12, 107

(a r) r-3 , (xii) (a' r)'n(b r)n, (xiv) lax rIn, (xv) log la x rl, (xvi)* log Ik x rI.

Exercises E In which of the cases listed are the scalar fields and/or their gradients infinite or undefined at certain points or on certain lines? 2.

Evaluate the curl fields of the following vector fields: (i) r,

(ii)* a/r,

(iii) ar", (vii) a(b r),

(iv) a x r,

(v) * (a x r)r-3,

(ix) a(b r) (c r), (x) (a X r) (b r), (xi) a lb x rig, (xii) a lb x rl", (xiii) rla x rl", (xiv) a log lb X rl, (xv)* k log lk x rl. In which of the cases listed are the fields given and/or their curls infinite or undefined at certain points or on certain lines? (vi) (a x r)rn,

3.

(viii) r(a r),

Evaluate the divergence (scalar) fields of the following vector fields: (i) r,

(iii)rIr-al",

(ii)*rr°3,

-ar-3,

(vi)*

(viii) a (b r),

(ix) a(a r) (b r),

(xii) alb x r 1 2 ,

(iv)aIr-aI"

(xiii) r log la x r1,

(x) (a x r)r",

(xi) (a x r) (b r),

(xiv) a log la x rl,

(xv)* (k x r)/Ik x r12, (xvi)

_

* kx(kxr) l k x r12

[

r2 - (k r)2

In which of the cases listed are the vector fields and/or their divergences infinite or undefined at certain points or on certain lines? 4.

Show that if F(u) is an arbitrary function the field lines of grad 0, where (a) 0 = F (a r), are parallel straight lines; (b) 4) = F(r), are radial straight lines; and (c) 0 = F(I a x rI ), are straight lines radiating at right angles from a central axis.

5.

If the scalar field 0 is given as 4) = r2 /2z show that the field lines of its gradient field can be given as

r(O) = a(cos all + cos 0), sin all + cos 0), sin 0) where a and a are arbitrary constants and 0 < 0 < 27r. (The answer given can be readily verified, but alternatively it is not difficult to find geometrically first the equipotential surfaces and then the field lines.) 6.

Exercise C.1 gave an instance of a fairly easily visualisable scalar field. What can be deduced without detailed calculation about the nature of the field lines of grad 0? The mathematical treatment of this question is considerably simplified if the three constant vectors a, b and c are chosen to be mutually perpendicular so that in the appropriate cartesian frame we can put

0 = cos ax cos by cos cz. 108

Exercises E Evaluate the gradient field and verify that as functions of a suitable parameter X these lines can be described by Si, (1

r(A) _ \a sin

1

«

(.

b sin-1 0

p

, C

1

\-

/

where a, 0, y are arbitrary constants. What feature of this field pattern is reflected in the fact that the inverse sine function is only defined for arguments between - 1 and + 1 and is multi-valued? 7.

(a) Why in exercise D.2 does the work integral over the two paths differ for f but not for g? (b) Why in exercise D.3 is the work integral independent of the path of integration?

8.

Re-examine exercises D.5, D.6, D.9 and D.10. To what extent does the general theory since covered abbreviate the answers to the questions posed then?

9.

Exercise C.13 requires discussion of the field lines of a particular vector field. Show that that field is in fact irrotational and derive a scalar potential for it.

10.

Verify that the field defined in exercise C.14 is irrotational. What features in the numerical expression for the coefficients in the field components ensure this? Prove that the scalar potential of the field is (apart from an arbitrary additive constant)

0 = A(x2 + 11y2 + 8z2 - 8yz - 2zx + lOxy). Show that if this quadratic form is transformed to principal axes it becomes

0 = A(x'2 + y'2 + 2z'2 ). (If the student is not familiar with the general technique of principal axis transformation he can assume that the relation between old and new cartesian coordinates is given by

x = j(-x' + 2y' + 2z'), y = (2x' -y' + 2z'),

z = '(2x'+2y'-z').) Note that this reduction to principal axis form makes the field of exercise C.14 become a special case of the field of C.13. The expression derived in C.14 for the field lines would therefore be obtained from the general form given in C. 13. The relatively simple results for those field lines depended on the fact that the fields are irrotational. 11.

Let F(u) be an arbitrary function, a a constant vector and L the straight line through the origin in the direction of a. Show that for each of the following types of field f each field line of curl f is a circle with its centre on L and lying in a plane perpendicular to L : 109

Exercises E (i) f = aF(r),

(ii) f = rF(a r),

(iii) f = aF(Ia x rl),

(iv)f=rF(Iaxr1). What can be said about the related forms f = rF(r) and f = aF(a r)? 12.

(a) Why in exercise D.12 does the flux integral across the two surfaces

differ for fl but not for f2? (b) Why in exercise D.13 is the flux integral of the first but not the second field different across the two surfaces? 13.

Apply the general theory now available to the field which, in terms of exercise D.11, can be defined as f2 - 3f1 and show how the question put in that exercise receives an immediate answer. (Consider the flux not only through the hemisphere but also through its plane circular base!)

14.

Stokes' theorem formulates the relation between a vector field f and its curl, and the intuitive picture of that relationship is that the field lines of f `curl around' those of curl f. Calculate curl f for the field

f = (cosy + sin z, cos z + sin x, cos x + sin y). Compare the result with the intuitive picture and comment on it. 15.

(a) Verify that any field of the form

fi = («(x), (3(Y), y(z)) is irrotational, and any field of the form f2 = (Oz (y, z),13(z, x), 'Y (X, y))

is solenoidal.

(b) If the components of a vector field are linear functions of the cartesian coordinates:

f = (ax + by + cz, dx + ey +fz, gx + by + iz) what conditions must the coefficients satisfy for the field to be (i) irrotational, (ii) solenoidal? Find an expression for the scalar potential in case (i) and verify that in case (ii) a possible vector potential is

g = (j (e - i)yz + dzx -gxy, - byz + j(i - a)zx + bxy, cyz - fzx + 4j(a - e)xy). Is the vector potential in general solenoidal? 16.

Which of the four fields

fl = C (2xz, 2yz, 2z2 + x2 + y2 ) f2 = C (- 2xz, - 2yz, 2z2 - x2 - y2 )

f3 = C (- 2xz, - 2yz, 2z2 + x2 + y2 )

110

Exercises E f4 = C (2xz, 2yz, 2z2 - x2 - y2 ) are irrotational and which are solenoidal? Wherever possible find a scalar potential and/or a vector potential for each field. (Verify that g =

C (- yz2 + yx2 , xz2 - xy2,0) is a vector potential for one of the fields: given this fact, construction of a vector potential for another case is not difficult.) (These fields are introduced here to be used in later exercises. This exercise in itself does not introduce anything novel.) 17.

Re-examine exercises D.7 and D.14. The former shows that f3 is not a gradient field, but that the other two may be. Decide whether they are and if so (in either or both cases) construct an appropriate scalar potential. Exercise D.14 shows that f2 is not solenoidal but that fl and f3 may be. Construct a vector potential where possible. (There are many possible vector potentials, but one may be constructed fairly easily which has the form g = (0, gy, ge).) If your construction gives a vector potential in either case which differs from the one given on p. 214, verify that the difference is an irrotational field.

18.

(a) Find grad 0 for the following three scalar fields (a, b and c are constant vectors):

(i)0 _ -a-r, (ii)= (iii) o _ - 5 (a r) (b r) (c r) (b) Find curl g for the following three vector fields:

(i) g = '(a x r),

(ii) g = (a x r)(b r) + (b x r)(a r),

(iii) g = a [(a x r) (b r) (c r) + (b x r) (c r) (a r) + (c x r) (a r) (b

Compare the results of (a) and (b). In what sense do the fields here defined belong to a special class? 19.

Examine whether and for what values of the constants occurring, the following fields are irrotational and/or solenoidal (a, b are constant vectors, n a positive integer). (All the fields below tend to zero at large distances from the origin. They can therefore be used for the description of physical situations in unbounded regions of space. On the other hand, these fields inevitably have singularities at the origin and are not applicable to realistic descriptions of conditions near 0.)

(Of = ar-n, (ii) f = rr', (iii) f = (a x r)r n, (iv) f = a(b r)r-n, (v) f = r(a r)r"n, (vi) f = (a x r)(b r)r-n, (vii) f = r(a r) (b r) r- ". 111

r)]

Exercises E For particular choices of exponents n and constant vectors a, b linear combinations of fields of this and similar structure can be found that are both irrotational and solenoidal. Can you find several of these? (Compare also exercises 2 and 3.) 20.

The preceding exercise drew attention to a class of fields that are both irrotational and solenoidal and which are singular at the origin r = 0 (cf. also the related non-singular fields of exercise 18). Show that this class, of which three members can be found in exercise 19, can be defined in generality as follows: _ (a(1).Q)(a(2).Q)... r and

g(k) = (a(1). V)

(a(2). V)

... (a(k -1). V) (a(k) X V) (a(k+1). p) ...

(a(n), V) 1 . r

Put f(O) = - grad 0 and f (k) = curl g(k) and prove that all k + 1 definitions lead to the same field. (Satisfy yourself that the various vector potential fields g(k) are not in general identical.) 21.

Find the divergence and the curl of the field

fl = (x2yz, y2zx, z2xy). Show that the field

f2 = - j ((y2 + z2 )yz, (z2 + x2 )zx, (x2 + y2 )xy) has the same curl as ft but is solenoidal. Hence find an irrotational field f3 with the same divergence as ft and a scalar potential from which f3 can be derived. Verify that

g = 3 (x3 (y2 - z2 ), y3 (z2 - x2 ), z3 (x2 -y2 )) is a possible vector potential for f2 and hence show that ft can be represented as the sum of an irrotational and a solenoidal field. 22.

Verify that the vector potential field g of exercise 21 can be replaced in that exercise by

g = T'1(x(y4 - z4 ), y(z4 - x4 ), z(x4 - y4 )) and show that both g and g' are solenoidal. The field g" = g - g' is therefore both irrotational and solenoidal everywhere. Construct a scalar potential function and verify that g" " can be represented alternatively as the curl of the solenoidal field

b = 11E(.y3z3 -x2yz(y2 +z2)-ts(y4 +z4)+x4yz, 112

Exercises E

iz3x3 -y2zx(z2

+x2)-1(24 -33(x4 +x4)+y4zx,

Ix3 y3 - z2xy(x2 +.Y2)

23.

+.Y4) + z4x.Y).

Evaluate the curl of the following vector field

f

_

1

3r

yz(z2 - y2)

zx(x2 - z2 )

(x2 +y2)(Z2 +x2)' (r2 +22)(x2 +y2)' xy(y2 -' x2 ) (z2 + x2) (y2 + z2 )

The rather lengthy calculation, if done correctly, should give a very simple result. Reflect on this result in the light of what is said on p. 86. 24.

In the explicit evaluation of work integrals of a field f along a curve C it is obviously a real practical advantage to know if the field is irrotational, in which case the calculation can be reduced to taking the difference between the values of the scalar potential at the endpoints of the curve. When evaluating the flux of a solenoidal field across a surface it is not necessarily of much practical advantage in computation to replace the flux integral by the work integral of a vector potential around the rim of the surface. In the case of the evaluation of the quantity integral of a scalar field p it is very unlikely to be of practical use to replace the direct evaluation by the evaluation of a flux integral of a field f out of the boundary surface of the volume, even though many fields f such that div f = p for a given p are readily found. However, for completeness an exercise is included which the enthusiastic learner may evaluate by this method (with considerable effort!): Evaluate the quantity integral of p = 6xyz over the volume of the quasi-cube defined in exercise B.20:(a) directly; (b) by evaluating the flux of the field fl of exercise 21 through the surface of the quasi-cube.

25.

(a) Show that the volume enclosed by a surface S can be expressed as

(b) Show that

J..(a.r)dV =

`J...(a.r)r.dS.

26.

Show that the vector area (defined on p. 63) of any surface S bounded by the closed curve C can be expressed as I fE r x dl. (Note that the result implicit in the discussion on p. 65, namely that the vector area depends only on the shape of the boundary curve, is here explicitly demonstrated. Note also that for the case of a plane surface S the result is very familiar in connection with Kepler's second law.)

27.

Prove that

(a) f 113

(n + 3)J rndV

Exercises E

= nJ rr' 2dV

I

(c)

f rnrx dS = 0 r

(d) 28.

r"dS

(b)

J

r dF r

F(r)dS = J

ar r dV.

Using the relations (6.1) which proved the vector character of integrals of the form f C 0 dl and fs 0 dS, show that the value of the first integral between any two points is independent of the path of integration only if 0 = const. and that the value of the second is the same for all surfaces bounded by the same rim again only if 0 = const.

114

11

Boundary behaviour of fields

1 Surface discontinuities In the discussion so far we have talked of fields being irrotational or solenoidal without paying much attention to the regions in space where these fields exist and have such properties. All actual fields in physics exist only in confined regions of space and very often one must think of fields as terminating on outer boundary surfaces. Or conditions may change discontinuously within regions of interest an electric field may exist partly in air, partly in glass. In this way, boundary surfaces enter. Also it may happen that the curl or divergence of a vector field - or indeed the gradient of a scalar field - is only different from zero in some confined region. This may be a narrow boundary layer which for practical purposes can be treated as a surface or it may lie effectively on a curve or at a point. Very often the general nature of the field is decisively determined by discontinuities on surfaces or by singularities on lines and points. The study of such behaviour will be the theme of this chapter. We begin by discussing discontinuities of vector fields across surfaces. It is rather natural to find that we need to distinguish between the behaviour of a field component normal to a surface of discontinuity and one tangential to it. We shall see shortly that the behaviour of the former is related to div f and that of the latter to curl f. We shall put these statements into quantitative form. We take a certain surface SD on which we examine possible discontinuities of irrotational and of solenoidal vector fields. Our standard parameter approach allows us, without any restriction of generality, to assume that SD is the surface, v = 0 in a parametrisation of the relevant region of space in the standard form r = r(X, p, v). We talk of a point in space being `just above' SD if for it v = v2 > 0 and v2 is small. Similarly a small value v = vl < 0 characterises a point just below the surface. For any function of v we take F It to mean the limit of F(v2) - F(vl) for vl -* 0, v2 --> 0. 115

Boundary behaviour of fields

11

Fig. 1. The curve C used in defining the surface curl.

Fig. 2. The surface S used in defining the surface div.

If F j± * 0, F possesses a surface discontinuity. We define a particular quasi-square C that cuts through the surface

Sn: r(No, it, vi ),

lai

µ

µ2

r(Xo, 1a2, v),

VI

v

v2

r(Xo,-la',v2),

lag 0, (ii) y > 0, (iii) z > 0 while f = 0 in the other half plane in each case. Show that for (i) f could be described as intrinsically irrotational with a certain surface source density a at z = 0, while for (ii) and (iii) it could be described as intrinsically solenoidal with surface vorticities g at y = 0 or z = 0 respectively. Give expressions for a and for g in the two cases.

3.

Let r1 and r2 be defined as

r, = Ir-di, r2 = Ir+dl, 134

Exercises F i.e. as the distances from the points d = dk and - d. The irrotational field that has the scalar potential 0 = 1/rl, if taken to exist in the whole of space, would describe a single source at d. If, however, the same source is supposed to exist in some confined region of space, this expression for the potential would not satisfy whatever boundary conditions are required by the physical circumstances prevailing. Consider the case that a field with a point source at d exists in the half space r d > 0 (i.e. z > 0 in a suitable coordinate system) and that on the surface r d = 0 (z = 0) there are surface sources such that for r d < 0 the field vanishes, consistently with the boundary conditions for an irrotational field. Show that a field that satisfies these conditions is provided by the scalar potential

rl

r2

and evaluate the source density a on the boundary surface. (Note: In electrostatic terms this a is (proportional to) the charge density `induced' by a point charge on a conducting plane. The technique here described is known as the `method of images'; the potential - 1/r2 is that of an `image charge' at - d. It is important to understand that such image charges must be placed in regions of space in which the expression for 0 or f is not relevant.) 4.

Using the ideas and notation of the preceding question discuss the character of the field that has the potential 1

1

rl

r2

0=0

(z>0) (z < 0).

What are the surface discontinuities on z = 0? Suggest a simple fluid dynamical interpretation for such a field. 5.

Show that the field just described in exercise 4 is also derivable from the vector potential 1

a

= (-, yx+ 0) x2 + y 2

(z - d rl

+

z+d r2

and verify that this choice of vector potential satisfies the condition (6'), p. 125. Show that the field of exercise 3 can also be derived from a vector potential. Is there any advantage in being able to do this? 6.

An irrotational field has no sources except for a uniform distribution of surface sources on a sphere of radius R centred at the origin. Given that the total source strength is q find a suitable scalar potential to describe this field. (The scalar potential for both r > R and r < R should be found; the answer for r < R is a trivial one, but should be justified.)

135

Exercises F 7.

In the interior of the sphere r = R there is a uniform field fi = fk where k is a unit vector and f a constant. Outside the sphere there is a field of the form

fe = Show that with different choices of the constant X either the tangential or the normal components of fi and fe can be made to coincide at r = R. Writing the fields on the sphere as functions of the unit vector n = r/r, find expressions for the surface source density or the surface vorticity as the case may be. Verify that both fi and fe can be represented either as f = - grad 0 or as f = curl a with

¢i =

ai = Jf(kxr)

0e=

ae = AfR3(kXr)/r3

and that the two values of A previously found correspond respectively to continuity of 0 and of a on the sphere.

8.

The pattern brought out in the preceding exercise can be found for other fields of which the following is an example. Show that one and the same field can be represented either as

- grad 0

with

0 = C [-(k r)(1 r) - 3r2(k l )1 /r5

curl a

with

a = C [ (k x r) (1 r) + (I x r) (k r) ] /2r5 .

or as

If the field is present outside the sphere r = R determine the field for r !i: (29) But this is Poisson's equation for a scalar field X. If we have learnt how to solve Poisson's equation - which is what Potential Theory teaches us - we can find X when 0 is given, and hence using (27) and (26) we can satisfy (25). Now we go back to (22), (23) and (24) and find and

_ - A¢ j = curl curl a = grad div a - ha p = - div grad

_ - ha.

(30)

(31)

So the determination of first and a, then f, and f2, and finally f again reduces to solving Poisson's equation: for 0 in terms of p and for each component of a in terms of the corresponding component of j. Only one more step is now needed to show that our theorem is true: the second part of (23) implies that div j = 0. Therefore (31) is clearly compatible with div a = 0. But when the solution of (31) is determined, it must be verified that it satisfies (25). 148

`Newtonian' solution of Poisson's equation What we have given is a programme of how a vector field f can be constructed, given its sources and vortices. In the next section we shall go as far as is possible in this book in actually constructing 0 and a for a typical situation in potential theory. But from the physicist's point of view we can now accept that the theorem is correct in principle. There is another way of summarising our results. We have the identity

2

V2f = VV f-Vx Vx f

= VV-fl _VX VXf2

= VP-Vx j.

(32)

If V were an ordinary vector a one could pass from an equation like

a2f = to

f) f)

f_ a2

(33) (34)

a2

and this would be called splitting the vector f into a component parallel to a and another perpendicular to a. The operation of dividing by a2 must in our case be replaced by the operation of `solving Poisson's equation'. For a vector field we have to solve separately for the `parallel' and the `perpendicular' parts but if p, j are known, each part can be successfully determined by means of potential theory.

2 The `Newtonian'solution of Poisson's equation a digression

We saw in chapter 11 that the field f = r/r3 has vanishing divergence everywhere but has a singularity at r = 0, because of which the flux off out of any surface surrounding the origin is non zero. In fact, since the magnitude of the field is 1/r2, the outward flux out of a sphere of radius r and surface area 47rr2 is clearly 47r. Because the field is solenoidal at every point except the origin, the flux out of any other surface surrounding the origin is also 47r. Thus a unit source at the origin is evidently to be represented by the field r

f or

(35)

47r r 3

_ 0 149

1

1

4;r r ,

f = - grad t.

(36)

13 Second derivatives; potential theory Supposing now we have an irrotational field f whose divergence is

given:

div f = p

0O = - p

or

(37)

and which is assumed to exist through all space. The physical meaning of (37) is that our field now has a continuous distribution of sources where the previous one had a single point source. Since we know that all relations between a field and its sources are linear, a physicist would immediately conclude from (36) and (37) thatt 1

O(r) =

p(r')

fv,

41r

,

r - r'

dV ' ,

f

grad

(38)

where the integral is extended over all space. Moreover, if in addition to the volume source density p, there are surface sources on certain surfaces S' so that (cf. (11.9)) I+

=a

(39)

on these surfaces, we would predict that there will be additional terms in (38) of the form 1

f

Js'

dV/.

(r

y'

(40)

These equations merely say that each point of space contributes to O(r) according to the source strength it carries. Mathematically the assertions (38) and (40) cannot be accepted .without proof and it is worth establishing these results more firmly, particularly since the method of proving them is both very typical of potential theory in general and an excellent illustration of the usefulness of the ideas we have gradually developed. We shall label some of the equations that now follow with the numbers they had when first derived in earlier chapters. This will show how dependent we are in such proofs on what was previously discussed. We start with the divergence theorem

div f dV = fs fV

and apply it to a field f composed of two scalars:

t See footnote on p. 71.

150

(9.11)

2

`Newtonian'solution of Poisson's equation

Fig. 1. The volume V used in the derivation of (38).

f = ¢ grad

- ,1i grad 0.

(41)

Using the general rule

V -(0g) = OV g + g' oo

(12.4)

we find that (42)

We now have

fV (OAO - AOW = J (on grad Ji - On grad O)dS. (43) This result is called Green's formula. Now we specialise (43) in two ways. Firstly we choose for S a surface which encloses all the region of space with which we are concerned and which will soon be made to recede to infinity in all directions, plus a small sphere centred at the origin. Thus we proceed as described on p. 106 and note that the outward normal on the small inner part of S points towards the origin. Clearly the volume V in (43) excludes the small sphere round O. Secondly we specialise i to be 1/r, the reciprocal of the distance from O. By cutting out the little sphere we ensure that A0 = 0 everywhere in V. Therefore using (37), (43) becomes

f 1r p dV = fs`(O nr r+ 1r n- grad O I dS. 3

V

(44)

/)

On the right hand side, though only two terms appear explicitly, there are in fact four because S has two parts. We can now conclude firstly that if the large part of S is made to recede to infinity the surface integral over it will tend to zero. This is because if the sources of the field are localised at finite distances from the origin, sooner or 151

13 Second derivatives; potential theory later, as S expands, they will appear indistinguishable at any point of S from a single point charge at 0. In that region 0 must decrease (at least) as 1/r and grad 0 as 1/r2 . Hence (in both terms) the integrand will decrease at least as 1/r3 , while the surface area S will only increase as r2 . The total surface integral will therefore tend to zero. On the small sphere at the origin things are reversed, as far as ,li = 1/r is concerned. When the sphere shrinks to zero >Ai tends to 00; so does its normal gradient 1/r2 . The surface area S is simply 47rr2 , decreasing as r2 . But the origin is not meant to be a singular point of either 0 or grad 0, so both should remain finite. The second term on the right of (43) will then clearly tend to zero - dS wins by a factor of r over >fi. Not so the first term; n r/r3 is again 1/r2 and, for sufficiently small r, the potential P(r) can be replaced by 0(0). Thus we get

Jv -dV = 47rfi(0). r

(45)

We now observe that the point r = 0 is in no way special for fi and if we shift the origin of our coordinates and call the variable of integration r' instead of r,

Jv

E(r)dV

becomes

fV1

I

p(r') dV'. I

Thus (38) is proved. But how do possible surface sources fit in? If we have to make provisions for these we cannot let S consist just of the two pieces previously discussed. Anywhere where there might be a discontinuity we must apply the divergence theorem separately to the two part volumes Vl and V2 on either side of the discontinuity. Note in fig. 2 that the normals on the dividing surface are opposite in the two integrals. If the assumption is that

-grad 0-nI_ = a (but, as before, fi ± = 0), our expression (44) will retain an additional term

br

fi

dS.

The sign of this term in (44) must be checked carefully. It is immaterial which side of the dividing surface we associate with the 152

2

`Newtonian' solution of Poisson's equation

Fig. 2. The subdivision of the volume of fig. 1 required when surface sources are involved.

positive normal, but (44) tells us that the one we call positive demands no change of sign from that used in (44) while the other does. So we do get - a dS/r on the left, and taking that term over to the other side we get the addition to the potential predicted in (40). We shall not extend the discussion any further, but merely mention that both point and line sources can be dealt with by similar techniques (as long as the latter are confined to lines not extending to infinity). The volume of integration simply has to be chosen to exclude the singular points. We have now established the so-called Newtonian form for the scalar potential of a localised source distribution p. But if this result is correct then the analogue of (3) must clearly be correct for each of the three cartesian components of a vector a related to the `vorticity vector' j by

Da = -j.

(46)

We can assert without further proof that

a(r) =

1

4Tr J

j(r,) ,

Ir - rI

d V'

( 47 )

provided no surface singularities etc. are present. However (as was noted on p. 148) it must be checked that the vector potential (47) is solenoidal. We shall deal with this verification in the exercises at the end of the chapter. We shall also refrain from spelling out how tangential discontinuities of the solenoidal field (surface vortices) can be dealt with in a more general proof. This chapter cannot claim to be more than a brief introduction to the very general techniques of `Green functions' (11j r - r' l is the 153

Second derivatives; potential theory 13 Green function of the problem we have tackled) on which much of potential theory is based. (Even so, we now possess a better mathematical justification for the general theorem on vector fields that we stated on p. 147.)

154

Exercises G

The material of chapter 12, especially of its first half, is of rather a formal nature and the motivationfor including the detailed matters discussed there lies in their usefulness in potential theory. Chapter 13 is concerned with potential theory, but can not be more than a very sketchy introduction to some of the main ideas. The following exercises serve both as illustrations of techniques, and also (far more than the earlier exercises) as an amplification of theoretical thinking. The reader who may have had little patience with some of the lengthy manipulation required in the earlier exercise chapters, might find a careful study of this set more rewarding. 1.

(a)

If a scalar field p and a vector field g are interrelated as follows:

grad p = Xg,

div g = Xp,

where A is a constant, prove that

div (pg) = A(p2 + g2 ). (b)

If two vector fields b and g are related by

curl h = Xj,

curl j = - Ah,

where A is a constant, prove that

div (h x j) = X(b 2 + j2 ). 2.

A scalar field p and three vector fields g, b and j are related as follows (A and p being constants)

grad p = Ag - p j

curl j = - Ah curlh = Aj+pg div g = Xp. Prove that

div(pg+jxh) = A(p2 -j2 +g2 -h2). 155

Exercises G Show that as consequences of the equations just stated

div h = 0

curlg = -th div j

= - µp

and hence that , I-z(p2 + j2 + g2 + h2) _ 3.

div(hxg-pj).

Prove that if g is a vector field and a a constant vector, then

Hence show that the fields of exercise 1(b) satisfy

div [j(a j) - Jaj2 ] = div[b(a b) - Jah2 ] = X(a, j, b). 4.

Using the equations of exercise 2 and the theorem of exercise 3 prove that

+j(a'j)-Ja(g2 -h2 +j2 +p2)] = 0 +h2 +j2 +p2)] =

+h2 -j2 +p2)] = 2ua (gx b) + 2pp(a j). 5.

Two vector fields fl and f2 are defined within a volume V and on its boundary surface S and are free from singularities. Throughout V:

div fl = div f2 (a); and on S: n fl = n f2 (b). The field fl is irrotational. (Thus, in words, both fields have the same sources, but only f2 may, in addition, possess vorticity.) Prove that

fV}idV> fvfidV. (Hint: Apply integration by parts to f2 -f i - (fl -f2 )2 ) 6.

If in exercise 5 the relations (a) and (b) are replaced by curl fl = curl f2 and n x ft = n x f2 respectively and if fl is solenoidal, prove that the same integral identity holds and the same conclusion can be drawn from it.

7.

A scalar field 0 and a vector field f are defined in a volume V and on its boundary S. Neither field has any singularities or discontinuities within V and both are finite on S. Let div f = - p, curl f = j and on S, with the

normal n defined to be pointing outward, let n f = a and n X f = -g. Prove that fv

156

L dV = fs >/u dS + fv > p dV

Exercises G

1vdV = fS /gdS+ fv iij dV. 8.

Consider a generalisation of exercise 7 in which the total volume of interest is Vo, its boundary surface So , but with Vo now subdivided into a finite number of parts V;. Let the boundary surface of Vi be S{. Some of these Si can be closed surfaces entirely in the interior of Vo but some will necessarily be in part coincident with go. We assume that the scalar field is still continuous throughout Vo but that f can have discontinuities across any of the internal surfaces described (but nowhere else). Wherever f is discontinuous we define

g = nxf

a=

(Note that where parts of So are involved there is no field on the + side, if n is taken outward, so that this notation is consistent with exercise 7.) Prove that the identities of exercise 7 still hold in terms of these definitions, provided the volume integration is over the whole of Vo and the surface integration over all the Si. 9.

A solenoidal vector field j exists within the volume V. At the boundary S of the volume j has no normal al component - i.e. there is no flux density of j out of any element of S. Let a and b be two constant vectors; verify

that div [ (a r)

(a r) (b

r)

and prove (a) that 0

J. and hence

Jv

jdV = 0

and (b) that

fV [(a-r)(b-j)+(b-r)(a-j)[ dV = 0 and hence that

fV (a-r)(b-j)dV =

157

j)

Exercises G 10.

If the expressiont O(r)

p(r

1

- 4it v'

Ir-rdV

'

for the scalar potential of a field associated with a distribution p of sources is used in the case that p differs from zero only within a limited volume V' near the origin 0 and if 0(r) is taken at points r much farther from 0 than any point of V', show that approximately q

1

0(r)=4tr

1

p

r

r

where

q=

Jv, p

dV'

and p= fv,r'PdV'.

Show that if q 0 a change from 0 to a neighbouring origin 01 is possible which makes p vanish, but that if q = 0 the `dipole moment' vector p has a value independent of the choice of 0. 11.

If the expression 1

a=

4tr

f j(r) dV'

for the vector potential of a field associated with a distribution of vorticity j is used with the same conditions of `localisation' in a volume V' as in exercise 10, and if in addition j has vanishing normal component at all points of the boundary S' of V' (i.e. if there is no flux density of j into or out of V) show (using the results of exercise 9) that with the same degree of approximation as was assumed in exercise 10

a = -1 Ox

i

4tr

-

where

r

m=

1

2

J

v'

r xj(r)dV'.

Prove that m has a value independent of the choice of origin 0 and check that the expression for the field f = curl a is of exactly the same form as the field f = - grad 0 of exercise 10 when q = 0. 12.

Given that the field of a single dipole of moment p at the origin may be stated as

f = - grad 0

where

0(r) = - 1 p 0 r

define the scalar potential at a general point r of a distribution of dipoles over a volume V' with boundary S' such that the total dipole moment in the volume element d V' is P(r) dW. (Note that the independence of p of the choice of origin - in the absence t See footnote on p. 71. 158

Exercises G of sources q - allows us to define the total moment of a dipole distribution as the vector sum of individual moments.) Use the result of exercise 7 to prove that such a distribution of dipoles is equivalent to a source density p = - div P together with a surface density a = n P on S', n being the outward normal. 13.

Repeat the considerations of the preceding exercise for a distribution of ring currents, the field of each of which is identical with a dipole field (see comments in exercise F. 10) but is now better represented as curl a with a = m x 0(1/r). Show that the ring current ('dipole') density M is equivalent to a volume current density j = curl M and a surface current density g = - n x M on S'.

14.

Prove the result used in exercise F. 10 that I('

15.

dl'

Cr-r

= V X (2r 1

I(

JC'

r x dlJ

approximately.

It follows from exercise F. 6 that the irrotational field outside the sphere r = R on which sources are distributed uniformly with density a (so that the total source strength is 47rR2 a = q) is a radial field of magnitude f = q/47rr2 , while the field inside the sphere is zero. As long as the field is required only at external points any number of concentric shells carrying sources must therefore combine to produce a total field of the form just stated, where q is the aggregate strength of all the sources. Thus the above formula must also be valid for any spherically symmetric volume source distribution. But in the interior of such a volume distribution only the sources at radii r' < r will contribute to the field at r. Find the scalar potential for a field, the only sources of which are uniformly distributed over the sphere r = R with density p (so that q= (a) using the ideas just outlined, and (b) using (13.38) of the text. 3trR3p):

16.

Equation (13.47) defined a vector potential a(r) for a field f with a vorticity given by the solenoidal field j which is different from zero only within a finite volume W. No proof was given in the text of the fact that V a = 0 for the vector potential as given. Give this missing proof with use of the physically obvious condition that the vorticity (or in electrical terms the current density) j has vanishing normal component at all points of the boundary S' of the volume W. (A flux of j out of V' would contradict the assumption of localisation of the vorticity.) The proof should be valid not only for points r outside V' and should therefore not ignore the singularity of the integrand at r' - r. (In the discussion of the singularity the work in the solution of the preceding exercise can be put to good use.)

159

14

Orthogonal curvilinear coordinates

1 The basic relations

In our exposition so far we dealt on the one hand with the three cartesian coordinates x, y, z, on the other hand, when presenting general proofs, with three parameters X, µ and v. It was pointed out on p. 25 that if these parameters are chosen in such a way that the same set can be used throughout all space they can be taken to be coordinates alternative to x, y and z. This is sometimes very useful in practice - but hardly ever with X, µ and v left as general as in our previous work. Certain special systems of `curvilinear' coordinates are very well known - in particular the polar coordinates. Without exception these standard systems have the special property of being orthogonal. In our language this means that

hx'hµ = 0

(1)

bx) =

hx(b,x hxb,,h,,,

(2)

or, introducing

ex _hhx-, e,` =hhµ µ e - hh h,Ax h = hxx h,, = hxhµe,,.

(3)

hx = (4)

Such orthogonal coordinates remain more general than cartesians (though of course cartesians are a special case of them: h,, = by = hZ = 1). The b vectors are not necessarily constant, either in magnitude or in direction. Therefore, in particular, a vector field represented as

f=

(5)

has components referring to different directions at different points. 160

The basic relations This aspect of `curvilinear' coordinates has to be borne in mind constantly when using them. Much confusion can arise from misunderstandings at this point. We first deduce the form of the gradient of a scalar field in such coordinates. We use our general definition of gradient but confine ourselves to defining the three components of grad 0 along the three coordinate directions. Using the notation of (5) we say that f = grad 0 has as one of its components 1

fx = Llo

1

L

[O(r2) - 0(rl

(6)

where r2 and r, are two points on a curve µ = const., v = const. We can put

A = lim 1 JX2dX N, aA

and

L-o L

L =f=

b, dA.

(7)

x1

Taking the limit we get

A

1 a = hx aA

and similarly 1

fµ

-

_

30

1

(8)

a

f° h, av It is not at all surprising that the components of the gradient involve `scaling factors' so that the differentiation in (8) is not with respect to A but in effect with respect to hxX. Since a curvilinear coordinate could be not a length but, e.g. an angle, the presence of the 1/h factors in (8) is in general necessary for dimensional reasons. We now define the curl of a vector field in the same way, again stating only its components in the three coordinate directions. If

hµ aµ '

f = curl g

then, e.g., 1

fx _

A-+o

A 1

Al o A

fc

g-dl

x,r,.

a

Jµ, [ aµ (h"g°)

a

av (hµ8µ) dµ dv

(9)

(cf. (8.4)) and

A = f "f µ'hµh dµ dv. 161

(10)

Orthogonal curvilinear coordinates

14

Therefore hµ by Laµ (h°gU)

a

j x [ LP by fv

(b,&) av

aA

(hvgv )J

(hxgx )J . h hµ I ax (bµgµ) - aµ

There is a very useful mnemonic that one can associate with this formula. It consists of two parts. Firstly, as for the gradient every operator a/ax, a/aµ, a/av is associated with the appropriate factor 1/hx, 1/hµ) 1/hv. If it were not, there would be no guarantee that dimensionally these differentiations represented divisions by lengths. Secondly, factors involving the hx other than the one associated with the differentiation must be inserted in such a way as to ensure that

curl grad 0 = 0, without changing dimensions. That this is true by virtue of (8) and (11) is easy to check. Finally we define as

p = divf p

(12) ff

v

0VJSf-dS

(where V is chosen as a quasi cube with surfaces A =X 1i A = A2 ,

11 =11141 =P2,V=V1,V=V2). Then

p = lim

V -*O

1

2,Iµ= V2 µ,1v1

a (hv hxfµ ) lax (hµ h,ff) + aµ (13)

+ av (hxhµ fv)J dA dµ dv and X J12

V =JX ,

µ,

(14)

f VZhxhµhv dX dµ dv. v,

Hence

p

h

hlµ h v [aA (bµbvfx) +

a (bvbxfµ)+ av (bxbµfvi (15) a

la

Again the mnemonic works: dimensions are correct because the expressions (1/hx) (a/aA) etc. appear correctly and also the factors are such that 162

1

The basic relations

div curl f = 0.

(16)

The definition of grad, curl and div in orthogonal coordinates is thus complete. Readers who compare this text with others will note that the relation (17) d12 = h2 dX2 + hµ dµ2 + by dv2

- `the metric' - did not appear in our definition of curvilinear coordinates, although it was in effect introduced as early as (3.1). Because it makes explicit the presence of scale factors in the definition of distance in curvilinear coordinates, (17) is instructive. But in fact the metric is not very frequently directly useful in elementary vector analysis.

It would be unhelpful to proceed much further with the present development without presenting explicitly the most frequently used non-cartesian expressions for grad, curl and div. Cylindrical polar coordinates are defined by

r = (x, y, z) _ (r cos 0, r sin 0, z )

0-r aefr curl g =

az go' 0' r

a

(30)

11 (rgo)

.a

r sin 0 a0

1

(sin 0 go),

a

r ar

(rgm )' 0

We see therefore that T-fields are necessarily solenoidal (after all, they must have closed field lines!), while P-fields need not be. We .see also that the curl of any field of the one kind is a field of the opposite kind. Of course, if we take a P-field that has been obtained as the curl of a T-field, it must necessarily be solenoidal. The study of such fields is particularly associated with the name of Stokes. Both in fluid dynamics and in electromagnetism, vector fields whose field lines have rotational symmetry about an axis and lie entirely in planes through the axis, i.e. P-fields, are of considerable interest. Our earlier discussion made it evident that the description of general solenoidal fields is on the whole rather more complicated than of irrotational fields - simply because a scalar potential 0 is a simpler concept than a vector potential with its three components and its considerable arbitrariness of definition. Stokes' work shows that at least in the case of these special fields, solenoidal fields 167

.

14 Orthogonal curvilinear coordinates can be dealt with just as simply as irrotational fields. We consider (30) and define

or

'

in cylindrical polars

= - rg ,

(31)

41 = - r sin 0 g4, in spherical polars (note that both equations define the same quantity because the r sin 0 of spherical polars is the r of cylindricals). We find then that a general solenoidal P-field f can be written as

f

I+

f =

(-p2 sin o a9 ' r sin 6 ar ' 0

r

,

r

0,

or as 1

a41

a) 1

in cylindrical polars

a'

(32)

in spherical polars. (33)

The curl of such a field is a T-field:

curl f = j =

0,ora- (1a -r -orJ l+ -1a2oz 2 , 0 r

or

curl f = j

0,

1

r sin o

a2'

1

a

)

in cylindricals 1

dr2 + r3 ao (sin

in sphericals.

a' 0 TOP

(34)

(35)

If the single possible component jp of the T-field j is given, the solution of either (34) or (35) can be tackled on lines similar to those for Poisson's equation in polar coordinates which are well known in potential theory and the subject of exercises H. 12-15. We shall use this fact in some further exercises (H.18-21). The physical situation described by a Stokes' stream function in fluid dynamics is one of a velocity field with a T-field vorticity. An example would be the description of a single circular smoke ring. In magnetism a T-field describes an electric current j - in a single loop, say, or in a solenoid. The field f is then the magnetic field of such a current. If a field is described by a Stokes' stream function it is possible to give an equation for its field lines in a very simple form. The reader may remember from chapter 4 that in general the determination of field lines in three dimensions is fairly complicated. But in the present case the rotational symmetry ensures that all field lines lie on surfaces of rotation about the axis. These surfaces have a very simple equation. Let us evaluate the total flux off through a circular 168

Solenoidal fields in two dimensions 4 disc bounded by the circle C which is centred on the axis of symmetry and has its plane perpendicular to the axis. Using polars we see readily that

fs

fs curl

fa- dl

=

-foiYdo 27r'Y.

(36)

A surface of rotation consisting wholly of field lines has no flux passing through it anywhere. Clearly it is just a surface

' = const.

(37)

This is therefore the equation of the field lines.

4 Solenoidal fields in two dimensions - a digression There is a very strong similarity between the content of the section just completed and the situation in which one can treat a vector field as two-dimensional. This does not mean that one is forsaking the real world for an imaginary two-dimensional one but simply that one assumes that the field has no component pointing into the third dimension (e.g. the z-direction) and that its other two components are the same in all planes z = const. The mathematical form of such a field f is thus

f = (fix (x, y), fy (x, y), 0). (38) There is an analogy between such a field f and a field of the type described by (27). (That is why we have inserted the discussion of this purely cartesian situation in the chapter on curvilinear coordinates!) afy curl f = (39) ay and if f itself is solenoidal a vector potential that will describe it is a = (0, 0, a,). Putting aZ as convention has decreed, we get

(0,0,&_t)

f

alp ao ay ax

0

and the equation

curl curl f = j 169

becomes

a2 0

a2 0

axe

ay

2 = iZ

Orthogonal curvilinear coordinates 14 In this two-dimensional case an essentially solenoidal field may be described exactly in parallel with a two-dimensional irrotational field. The latter possesses equipotential lines, 0 = const., while the former has field lines, >!i = const. In this discussion the fact that ii is really the z-component of a vector potential is quite forgotten! If such a field is both irrotational and solenoidal (which is possible only in limited regions of space, except if f = const.) the two sets of lines intersect orthogonally.t

t The informed reader will recognise that we are now simply stating, in an unorthodox way, things that are well known for analytic functions w(z) = 0 + icy of a complex variable z = x + iy.

170

Exercises H

The preceding chapter was kept brief because it deals with matters of detailed technique rather than with general ideas. However, these techniques are quite essential to the use of vector analysis at the point where it merges into potential theory, and there is no limit to the illustration that might have been chosen. The following examples serve on the one hand to show that the general idea of using curvilinear systems is not in practice only realisable in the two types of polar coordinates defined in the text and, on the other, to demonstrate the power of precisely these polar coordinates. In this latter function what is presented is a selection of applications to potential theory - a selection which reflects the author's individual preference. It is far from a substitute for a systematic text on potential theory. The reader should note that at this point there has occurred an inevitable `clash of symbols'. In vector analysis 0 normally stands for a scalar potential. In polar coordinates it denotes an azimuth angle. In this section only the letter ii rather than 0 has been used to denote all scalar potential functions that occur. 1.

Using the relations (14.20) and (14.21) write down the component forms of the unit base vectors e = b/h in both systems of polar coordinates. Prove that for cylindrical polars aer a¢

= eo,

ae

ar

= - e,.

while all other partial derivatives of this kind are zero and that for spherical polars the only non-vanishing partial derivatives are aer

aee

ae

ae

ae, a0

sin 0 eo,

a

= cos0e,,

a

_ - singer - cosBee.

How can these results be understood geometrically? 171

Exercises H 2.

Using the method of the preceding exercise, check the results given in the text for grad 0, curl f and div f in terms of cylindrical and of spherical polars.

3.

Work out explicit expressions for A>' in cylindrical and in spherical polars.

4.

In terms of the cartesian V evaluate (r x V)2 >G where qi is any scalar field

and prove the identity 02

=

+ 1) (r' V) + (r x 0)2}.

r {[(r'

Hence show that the form that (r x V)2 takes in spherical polars is 1

sine30+

a

a0

sin 0 aB

az0

l

sine B

Show that if , satisfies 01 = 0 and is of the form r'YI(0, 0) then (r x V)2 YI = -1(1 + 1)YI and show also that the new function

x=

r-(I+t) yl

also satisfies AX = 0. 5.

As explained in the text, a correct expression for Of, where f is a vector field, can be obtained either by using the definition

Af = grad div f - curl curl f or by working out V2 (e,\fx

+ e),fµ + efv)

bearing in mind that the e vectors are not constant. Verify by either method that if g = Af then, in cylindrical polars a

g

1

1 a2fr

a

(rfr) + r2 ao2 ar [r ar a

1

a

ar [r ar 1 a Y ar

(r

af, aY

a2fr

2 afo

+ aZ2 - r2

a

+ 1 a2fo+a2fm+ 2 afr r2 a02 1 a2fz r2 a02

az2

r2 30

a2fZ) aZ2

and in spherical polars

r

g = 1a [r

172

ar (r2fr) + r2 in 0 a8

z

(sin B a9)+

r

r2 sin 2 0 a0

Exercises H 2

a

1

a

afe

r2

-a

1

+

ar

r2 ar

2 afo r2 sin 0 ao

(sin, fa)

r2 sin B a0

a

1

r230 sin 9 a0

1

(sin B

a 2 fr

fe J + r2 sin 0 a02 )

2 cos0afo+2afr r2 sin2 0 ao

l a ra r2 ar

+1

aY

+

2

r2 ae a

(sine

r2 ae sin 0 a0

f )+ 1

2

f

sin 2 eaa02 r2

afr + 2 cos e afq

r2 sin a a 6.

1

r2 sin2 0 a

Among possible systems of orthogonal curvilinear coordinates polar coordinates stand out because of their usefulness in so many applications as well as their relative simplicity. The uses of other systems are more limited. In the classical period where there was much stress on the achievement of exact solutions to as many problems as possible, the most general type of orthogonal curvilinear system that was found to be tractable - and relatively simple - was that of general ellipsoidal coordinates. It would be hard to justify inclusion of this system for its usefulness but as an illustration of the power of the method of curvilinear coordinates and to establish the connection between several more useful special cases, we devote this and the next exercise to this topic. General ellipsoidal coordinates may be defined in the usual way by writing down r(X, p, v). But it is more instructive to prepare the way before defining them, as follows. Let a, b and c be constants such that a > b > c > 0. The parameters A, p, v are chosen to lie within the limits c < v < b, b < p < a, a < A < 00. The three equations x2 A2

y

z2

2

-a2+A2 -b2+A2

- x2 _

y

2

a2 - p2 + p2 - b2

_x 2 a2 - v2

y2 b2 - v2

z2

7-:

= 1, c2

z2

v2 - c2 =

1,

then clearly define families of ellipsoids, single sheeted hyperboloids and double sheeted hyperboloids respectively. Choosing fixed values of A, p and v one selects one surface out of each of these families. The intersection of these three particular surfaces determines the cartesian components of r(X, p, v). (Strictly speaking, only Ix 1, ly I and I z I are determined in this way;

there are eight points in which the three surfaces intersect and a one-to-one relation exists between the parameters and the cartesian coordinates of 173

Exercises H only one octant of physical space. This is not a practical drawback to the use of these coordinates.) Given these definitions, show that (a) r can be expressed explicitly as

-a2)(a2 -µ2)(a2

Y

(a2

-v2)l1/2

b2)(a 2 - c2)

(X 2

(X2 -b2)(µ2 -b2)(b2 -v2)11/2 (b2 - c2)(a2 -b 2) (X2 -c2)(µ2 -c2)(v2 -C2) 11/2

/

(a2 - c2)(b2 - c2) bX

(b)

_

(a2 -µ2)(a2 - v2)

(ff

-

11/2

1(X2 - a2)(a 2 - b2) (a2 - c2 ) 1/2

(x,12 - b2)(b2 - v2)

(X2 -b2)(b2 -c2)(a2 -b2) (µ2 - c2) (v2 - C2)

I1/2

)

(X2 - c2) (a2 - c2) (b2 - c2)

(X2 - a2)(a2 - v2)

bµ

11/2

[(a2 - µ2) (a2 - b2) (a2 - c2) (A2 v2)

1/2

-(A2 - c2) (v2 - C2)

1 /2

(µ2

- b2)(b2 2 b2) (b2 - c2)(a - b2 )

I

(µ2 - c2)(a 2 - c2) (62 - c2)

b

1

V

J

11/2

(A2 -a2)(a2 -µ2) (a2 -v2)(a2 -b2)(a2 -c2)I

11/2

(A2 - b2)(µ2 - b2)

(b2 -v2)(b2 -c2)(a2 -b2) 1/2' (A2 -c2)(µ2 -c2) (v2 - c2)(a 2 - c2)(b 2 - c2 )

(c) The three b vectors are mutually orthogonal. 1 /2 (A2 - µ2) (X2 - v2) (d)

bX

_

(A2

f

- a2 )(X2 - b2)(A2 - c2)

- v2) (µ2 - v2) 2- (A2 (62 (v2 174

[(a

I

(µ2 - v2) (A2 - µ2)

v2 )

- v2 )

1/2

1 /2 1

- c2 ) J

.

Exercises H 7.

Using the coordinates whose properties are defined in exercise 6, prove

that AO

(X2

- p2) (A2 - v2)

a) )

aA

+ (p2 - v2) (X2 -/12)V(/') all

ap 1

a

1

+ (A2 - v2) (p2 - v2) w(v) av

av

where u(a)

) (X2

= [(X2 -a2

- b2)(X2 - c2)11/2/X

v(p) = [(a2 - p2) (p2 -b

w(v) = [(a2 -

v2

) (b2 -

2 ) (p2

v2

- C2 )l

) (v2 - C2 )l

112 /p, 1/2

/v.

In the above, as in the whole of exercise 6, the constant c has been retained on equal footing with a and b. This was for the sake of preserving symmetry and making the algebraic work more systematic. However, the parametrisation is not essentially changed if one puts c = 0. Assuming this is done, put «(X) Q(p)

= f J a[(X'2 X dA' -b2)11/2 -a2)(X'2 fµ

dp

b

[(a2 -p,2)(µ2 -b2)11/2 dy'

fo"

[(a2 -v'2)(b2 -v'2)11/2

and prove that the functions Iii = a, 'Li = a and >li = y are all solutions of

AO=0. Show that the first of these enables one in principle to determine an irrotational field which has no sources except surface sources a on an arbitrary ellipsoid, which is zero in the interior of that ellipsoid and which tends to zero as 1/r2 at large distances from the ellipsoid. 8.

To define the ellipsoidal coordinates of exercises 6 and 7 it is essential that a > b > c, (though c may be zero). If a = b orb = c one of the parameters would cease to be definable. But in practice these special cases are more interesting than the general case because they are applicable to the problems with rotational symmetry which are often of interest. We define two sets of coordinates as follows.

r = (p cos 0, p sin 0, z) with (a)

p = a(X2 - 1)1/2(1 -p2)1/2

-1 [Dt Here the convective derivative (9) has been used to combine the first two terms in the last integrand of (11). Instead the ordinary derivative off at a fixed point in space may be introduced. Using a little vector algebra according to the rules of chapter 12, we then llget

ddt

fc(t)

Jc ( :> [a at

f)Jdl. (15)

This last result has not found many applications in physics but one will be mentioned in the next exercises. We now consider a moving surface S(t) which consists of particles of a medium whose motion is described by the velocity field v(r). Using the same notation as above, we examine the flux integral fs(t) f dS and as always take S(t) to be a quasi-square. Then 186

2

Relations involving the velocity field v(r)

Fig. 2. Motion of a surface S in time At.

d

dt JS(t)fds lim

At, r(X, p) + Atv(r(X, µ))l

1

nt_,0 At

[ax

x

_ _

at

[r(X, /a) + Atv(r(X, !1))]

[r(A, µ) + Atv(r(X, µ))] I da dp

fX2 fµ=__

._

._

fa

._

a

.1

._

J2J::(af+

[(v V )fl ' (hx x hµ )

dXdy. µ Then using (12) we find again with the methods of chapter 12

f-

`aa v x hµ + hx x

-

so that

.

Jh,u,

dtJL(t)f-dS =

(16)

a v) µ

JJ

hu)

(17)

J[-f-(f. V)v +

(18)

This identity corresponds to our result (14) for the line integral. Again we can transform to an expression corresponding to (15) which gives.

d JS(t)1 dS = fs0 [ata f-Vx(vx 187

I.dS. (19)

Time-dependent fields 15 This last form is the most widely known and used. It is particularly important in electromagnetism because Faraday's law of induction can only be formulated completely if one considers the flux through a moving surface. However, the alternative expression (18) is in some ways more natural. Both (14) and (18) reduce to integrals over the convective derivative if the velocity is uniform. This fact is obscured in (15) and (19). But then it may be noted that it is the forms (15) and (19) that give us expressions involving only the grad, div and curl operators:

dt

f,(,) f dl = f {-i+ grad (v f) - v x curl f

l J

dl (15')

Jcurl (v x

dt

Finally, is there a corresponding theorem for volume integrals? There is indeed, but it is somewhat simpler than the previous two theorems because, as noted before, a closed surface S(t) has a unique interior volume V(t). We examine an integral of the type JV(t)

O(t, r) dV

which measures the total quantity of some substance in a moving volume V(t). Then d

dt Jv(t)

' nimo at Ja, 1

0 dV I

=

v2

L O(t + At, r + v&t)

(20)

(r + vat), aµ (r + vOt), av (r + vat) I dX dµ dv r'2

JJ'2

,

0(t, r)

a

(a a a r, r, r )dXddP] ax ax

(to abbreviate our notation we have not made it explicit here that r = r(X, µ, v() and v = v(X, u, v))

=

f flitµ,f A,

X,

v

at

+ (b,\, b 188

,

(bv o )v) I dX dµ dv.

Relations involving the velocity field v(r) Using (12) again we can transform this to

2

d

dt Jv(t) 0 dV

D(21)

= f2r2f2

1

[+(vV)0 + oo v] at

hµ, b,) A dµ dv

_ fV(7+)dv or

d

dt SS(t)

0 dV =

fv(t>

I

ao + div (Ov) 1 dV

(22)

(note that in this case both forms may be expressed in terms of the divergence operator). The method of proof was adopted to conform with the other cases - it is not the simplest possible: the final result can at once be transformed to

dV + Js(t) ov dS dt fv(t>O dV = Lo at

(23)

and in this form the content of the theorem is rather obvious. The change of the volume integral is accounted for by two factors - the change of 0 at any fixed point in V and the transport across the surface 3(t). It would be simple to turn this statement into a complete proof. If 0 = p, the fluid dynamical density of matter, the integrand of (22) is just the expression which is zero by the equation of continuity (2). For in fluid dynamics the current density j is equal to pv. This special case therefore provides a re-derivation of that equation. Of more interest is the case that 0 has two terms of which one is the density, the other some other field 4/(r). dt

Jv(t>

P>fi dV = fV,,) I at

+div

dV (24)

fV(t)

The first term vanishes by the equation of continuity so that we find 189

Time-dependent fields dt Jv(t)

p4i dv = JV(t) p dv. Dt

15

(25)

This relation is of central importance in fluid dynamics. It can easily be seen that it does not depend on the nature of ii. Hence the same identity is true if p is replaced by a vector. In fact the most important application of this theorem is to the case where iy is replaced by the velocity vector v. The identity

d

Jv

pv dV

= JV

p Dt dV

enables one to use interchangeably either a differential or an integral form for the equations of motion of fluid dynamics. This is a good point at which to end our account of vector analysis. Further developments are best left for inclusion in courses on particular physical topics.

190

Exercises I

The best exercises for use with chapter 15 can be sought in books on fluid dynamics, electrodynamics, quantum mechanics etc., which use the language of three-dimensional vectors. The few examples provided here originate in these subjects, though in exercises 1- 3 these origins are disguised. (What exercise 3 disguises may be obvious only to particle theorists.) Only exercise 8 may not exist anywhere in a form resembling the one given here. 1.

Show that if two scalar fields 0 and 0 satisfy the differential equations

at

= k, a _ - kAo

the scalar field

P = #(02 + 0) satisfies an equation of continuity. Find the current density j associated with the density p. 2.

Find a density p and a current density j related by an equation of continuity, in terms (a) of a scalar field 0 and a vector field g related by

ag = k grad 0, at

ao at

= k div g;

(b)two vector fields h and f related by

of = k curl h at

ab at

= - k curl f.

(Cf. exercise G.1!) 3.

Three vector fields and a scalar field are connected by the time-dependent equations of

at = 191

- k grad 0 + Kg,

ag = k curl h - Kf

at

Exercises I and by the time-independent relations

k curl f + Kb = 0,

k div g - KO = 0.

Show that provided K * 0 the additional time-dependent relations

ao

at =

- k div f

ab

= - k curl g

at

also hold, and that

div h = 0. Show that an equation of continuity holds with p = (02 + f 2 + g2 + h2 ) and

j = k(gxh+Of).

(Cf. exercise G.2.) 4.

Before the work of James Clerk Maxwell the accepted description of electromagnetism, if translated into modern vector language (and in suitable units), was the following: 1. The electric charge density p and the electric field E are related by

divE=p. 2.

The electric current density j and the magnetic field B are similarly related by

curl B =

-1 j

where c is a constant.

C

3.

The magnetic field B is solenoidal, div B = 0, and consistently with this Faraday's law of induction links the electric and magnetic fields by

curl E+ 4.

aB

at

= 0.

The work done by the electric forces in moving electric charges is

w= Maxwell's modification to this description was by replacement of the term j/c in the second equation by (j + aE/at)lc. Show that the introduction of this change has the result that (a) an equation of continuity for electric charge is ensured; and that (b) with the definition

PC = J(E2 +B2) for the energy density of the electromagnetic field a generalised equation of continuity as defined by (15.7) is valid. Find expressions for the je and Pe in this relation and give a physical interpretation for it. 5.

In theories involving vector fields - such as Maxwell's electromagnetic 192

Exercises I theory - conservation theorems and therefore continuity equations occur that apply to vector quantities such as momentum or angular momentum. In such a case the term in a continuity equation which corresponds to the usual p will be a vector, s say, and j must be replaced by something more complicated than a vector - a tensor in fact, the discussion of which would be outside the scope of our study. But we can avoid the explicit appearance of such complications by looking not at s but at p = a s where a is an arbitrary constant vector. The j corresponding to this density p will be a vector, but one that depends on a. We shall exemplify these general considerations only for the case of Maxwell's equations as defined in the preceding exercise. The identity derived in exercise G. 3 will help us. Prove that if Maxwell's equations hold, then a(E2

at

ll

= It is known that [pE + (j x B)/c] dV is the correct expression for the forces, electric and magnetic, with which an electromagnetic field acts on charges p dV and currents j dV. Given this fact, what physical statement does the generalised continuity equation quantify? 6.

Experimental confirmation of Faraday's law

13B_ at =

cur]E+c

0

(A)

may be obtained by observing the current that flows in a stationary loop of wire while the magnetic flux through the loop is changing. (If Stokes' theorem is applied to equation (A) the change of flux is related to the work integral of E round the loop; the non-vanishing of this integral means that the net effect of the forces eE acting on all the charged particles and of resistive forces is a current.) Quantitatively the same effect of a current flow through a loop can also be produced by moving the loop through a magnetic field B, even if that field is time independent. The current is found to be related to the rate of change of total flux of B in the same way in both cases. Show, using (15.19'), that this result confirms equation (A) provided one remembers that now the force on any charge in the loop has a magnetic component e(v x B)lc. (Cf. exercise 5.) 7.

The classical theory of non-viscous flow when applied to an incompressible homogeneous fluid is governed by the simple equation of motion Dv

p Dt

=

grad p + F,

where p, the density, is a constant, p is the pressure and F stands for any external forces such as gravity. Usually (and always if F represents gravity only) F is irrotational. If this is the case and if w = curl v prove that the 193

Exercises I flux of w through any surface S(t) that consists for all time of the same particles is a constant. (Hint: refer to (12.13).) 8.

In magnetohydrodynamics the motion of fluids of high electric conductivity is studied. The fluid motion is characterised by the velocity field v and the density p which are coupled with the (solenoidal) magnetic field B. Under certain approximations one of the equations linking these fields is

aB

at

_ = curl (v x B) + k.B

where k is a constant which is zero if the fluid is a perfect conductor. (The second dynamical equation which gives av/at in terms of the other fields does not concern us in this exercise.) Using the stated relation and the fact (see p. 189) that for a fluid the equation of continuity relates p and j = pv, discuss the theorem stated in (15.14) or (15.15), applying it to the field f = pB/B2. Show that the integrand of the right hand side of (15.14) can be expressed in the form

BZBx(Vxv)+BpBx (Bx

+Bp[B2LB-2B(B

.B)].

Hence by taking C(t) to be any finite length of a field line of B justify the statement that for a perfectly conducting fluid the magnetic field is `frozen into the fluid' and that k is a measure of the deviation from this condition in a real fluid. If the fluid is incompressible and homogeneous (p = const.) and if k = 0, deduce that if the distance along a field line of two particles lying on it increases in the course of the motion, the strength of the magnetic field must increase.

194

Answers and comments

Exercises A 1.

(a) (4, 2, 4); (b) a = 7, b =

2.

All three vectors also happen to have integral length.

3.

4.

The two methods are by evaluating either [ 1 - (a b/ab)21'2 or (a x blab). (a)4

;(b)S;(c)T673

5.

715, (--h, ij,->3).

6.

The vector products all have the direction Jg (15, 16, - 12);

31a - 44b + 43c = 0. 7.

The length of all three vectors is 71.

8.

Edges: 13, 17, 41; faces: 455, 435, 220; volume: 3696.

9.

Formula is special case of (1.18): splitting of b into components parallel and perpendicular to a.

10.

Multiply vectorially by a.

11.

Proof by geometry simpler than by algebra.

12.

Deduce x3

rl 13.

1

=

(0,

),

rl =

(01

0, 2J2 ),

r3 = 14.

X2, Y2 =y3 =--ly,. 13

`1

1

r2 = (2 r2 = 1

(2,-23,-26)

1

1

2' /3), (0>

\

J6

73

r4 =

1

r3 = ( 2 2.J3

11 \

2,-2 1

11

For a change, some numerical toil is suggested! The simplest method, given that the vectors are of unit length, is to check that the scalar products formed from pairs of vectors pointing to neighbouring vertices are 195

Answers and comments all equal. The exact expressions for the vector components given are not unlike those given above for exercises 11 and 12, but involve -\/5. 15.

The angle at A is the supplement of the first angle in exercise 1(c).

16.

Unit normal to the plane; distance from 0.

17.

This exercise slightly anticipates the topic of chapter 2. Use the result of exercise 9.

n = a/a,

c=-ax(axb)1a2, p =

18.

Direction is parallel to n 1 x n2 ; condition given expresses the fact that the shortest possible vector distance between points on the two lines is zero.

19.

n = n1xn21Injxn2j,

20.

This and the final exercise illustrate quite intricate uses of vector algebra.

21.

Verification of the statements is straightforward. To derive them otherwise (by vector methods) requires care: the direction of n can be found simply; also the relations nl'c = pi,

p = n'c1 =

n2 'C = P2 -

The vector

b1 = n1x(n1xn2) is perpendicular to n and n1; and

b2 = n2x(n2xn1) is perpendicular to n and n2. Putting

c = «b1 +Rb2 leads quickly to the answer given. Exercises B 1.

(a)

A helix of constant pitch lying on the cylinder

x2 +y2 = a2. (b)

A helix on the z > 0 part of the same cylinder, and of varying pitch. The circle

z = 0, (c)

x2 + y2 = a2

is a limiting curve. A helix-like curve on the surface of the double cone

196

x2 + y2

z2

a2

b2

Exercises B 2.

Clearly

x2 + y2 + z2 = a2 . It is a closed loop for (X0 arbitrary).

No < X < Xo + 27r

In such a range z moves between + a and - a once, during three turns of its projection on the plane z = 0. 3.

If b = 0, x2 + y2 = a2.

with b * 0, it takes a range 00

0 was assumed. (Unless the parameters are all restricted to positive values dV is not guaranteed to be positive and will have negative values at some points.) But this means that the cartesian coordinates x, y and z are kept to positive values also. So these parameters do not provide a set of coordinates definable in all space. Exercises C

1.

On all planes

(a r) = 7r(n + 1)

(n integer)

4) vanishes, as also do the two other sets. So all space is divided into cells, the walls of which are the equipotential 0 = 0. The centroid of one cell is the origin, where 0 has its maximum value 00 (provided 00 > 0). Crossing any cell boundary takes one into a cell where 0 < 0 and 0 attains its mini-

mum value - 00 at the centroid. This alternating lattice pattern repeats 202

Exercises C itself. You may verify that the extreme values are attained at the points

r=

IT

(a b

-[1(bxc)+m(cxa)+n(axb)]

where 1, m and n are integers; 0(r) has a maximum if I + m + n is even and a minimum if it is odd. Qualitatively each cell will contain closed level surfaces corresponding to values 0 < I q 5j < 1 which surround the centroid and each other and are surrounded by the cell boundary. 2.

(a) Level surfaces are spheres. If drawn for uniformly spaced values of 0 the spacing decreases with the radius. Values of 0 start from zero at infinity and tend to + - near the singular point. In all the other cases the value of 0 very near one or other of the singular points is dominated by one term and the equipotentials closely surrounding these points are nearly spherical.

(b) Very far from the origin r1 and r2 become almost equal; the most distant level surfaces are approximately spheres round 0. There must be a transition from the 'nearby' shape of level surface - consisting of two separate near-spheres to the distant single surface. The two classes are separated by an hour-glass shaped surface that by symmetry must pass through 0. (c) Here the plane z = 0 is the level surface = 0. Near the two singular points the nearly spherical level surfaces correspond to 0 values of opposite sign. Surfaces for values of 0 near zero will be large closed surfaces, parts of which will in fact pass very close to the plane z = 0. (d) The picture here will be an unsymmetrical distortion of the picture of (b). The level surfaces round the one singular point will crowd closer than round the other. The 'hour glass' will not pass through the origin. (e) A distortion of (c). The role played by the plane z = 0 is now taken over by a non-planar dividing non-closed surface 0 = 0. 3.

4.

0 = (r - a)2 - a2. Minimum of 0 is - a2 . The radius of the sphere

0 = 00 is [q5o + a 2]1/2 .The spheres are concentric centred at r = a.

If0=0o, x2 +y2 + (z -I0o)2 = 1/4(0. The centres of the spheres lie on the positive z-axis if q5o > 0 and on the negative z-axis if Oo < 0. The plane z = 0 is the level surface Oo = 0. These spheres are not concentric. Their pattern resembles that of the not exactly spherical surfaces of exercise 2, case (c).

5.

The only way one can have 0 = 0 is for 0 =fir. The 'equatorial plane' of the polar coordinate system is the level surface 0 = 0. Above that plane (0 < 0 < f ir) the values of 0 are positive, below (47r < 0 < 7r) they are negative. At infinity 0 = 0 in all directions. Therefore the surfaces for which 0o 0 cannot reach out to infinity. So they are closed and 203

Answers and comments although, except for n = 1, they are not spheres, the general pattern is much the same qualitatively as in exercise 4. 6.

The level surface 0 = 0 is no longer the plane 0 it as in exercise 5, but instead the double cone 0 = cos-1(± A). In the interior of the double cone, i.e. surrounding both the positive and the negative z-axis, q5 is positive. Outside these cones 0 is negative. The field 0, since its definition does not involve x and y separately but only x2 + y2 , has rotational symmetry about the z-axis. So the (closed) level surfaces with ¢o > 0 will be `balloons' inside the two cones, while those with 00 < 0 will be 'doughnuts' encircling the axis and keeping outside the double cone. Every doughnut (balloon) encloses and is enclosed by other doughnuts (balloons).

7.

This field line picture tells one that f = no(r), where n = r/r and ¢ may be any scalar field. if I = Arn then must correspond to f = Arrn -1. In the general case the magnitude off may vary in a very general way over points at the same distance from 0. In the special case (and also if f = nO(l r 1)) we have a `spherically symmetric' field.

8.

There is no restriction in taking the constant vector a to lie in the z direction. Then we have

f = a(- y, x, 0). The field lines are easily seen to be circles surrounding the axis. (Formal

proof dy/dx = -x/y or xdx + ydy = 0 so that x2 + y2 = const.) As in exercise 7 it is clear that multiplication by a general scalar field 0(r) does not change the field lines. In case (a) the magnitude of the field is the same in all planes perpendicular to the z-axis but varies with distance from it; in case (b) it does not vary with distance from the axis but does so from one perpendicular plane to another. 9.

The form of f given represents a superposition of the circle pattern of exercise 8 - which is a field without component along the z-direction with a constant field in that direction. Its field lines are evidently helices of constant pitch. In this case again the presence of the scalar factor 0 does not affect the field line picture.

10.

The field is uniform in every plane. z = const. but changes in direction only, not in magnitude, i.e. it `rotates' in passing from one `horizontal' plane to another: quite an unrealistic field which will later be mentioned to bring out a mathematical point.

11.

dr = dr

1

- a(T+a

1

1

r+0'7+'Y

= a(e-x/a e-y/a e-z/a) a e-(x+y+z)la f C

204

Exercises D Particular line: if r = Xa(i, 1, 1)

f = Ce2'(1, 1, 1). 12.

Since the field vector is everywhere perpendicular to the z-axis, the field lines must all lie in planes perpendicular to the z-axis - here z = b. From the first field line equation dx

dy

+y=a

so that

13.

dx

a-y

2ay - 2y2

x2 - y2

dy

x

2xy

2xy

dr = aX-'P

x Ib cX-1

(ax``

_ 14.

``A-'

, by, cz)

=\

fA-

f

fx fy

0C

Inserting the values given for x(X), y(X) and z(A) into f we get 4X4 8X4 2X 4A2 8X4 2X 2X2 4X2 r A f=9A-«+ a a + y ' + y + a - y P

while

dr

1

4X

R

8X3 2

a+i3 + y'a

dX

2A

8X3 2

a +

,

4A

y'a+a

4A3

y

so that dr _ dX

15.

f 9AX

In exercise 12 the field lines were defined in terms of x, y, z directly, in the the other two in terms of a parameter. There is always an additional arbitrary constant when a parameter representation is used. Exercises D

1.

}(a, b, c). If the vectors are coplanar the force will be perpendicular to that plane.

2.

For f the work integrals are 8(a, b, c)2 and a (a, b, c)2 respectively. Forg they are both equal to (a, b, c)3.

3.

In terms of the concepts of mechanics an extremely simple question! As worded:

f = - fr/r 4.

and

f f dl = ffdr.

First integral has the value - *7rAa.

205

5.

Answers and comments (i) p b, (ii) a b + 'b2 , (iii) (p, a, b), (iv) (p-a)(q-b) + (q-a)(p-b) + (p-b)(q-b), (v) (p -b)(q -a) - (p a)(q - b).

6.

Instead of the path r = a + X1, the paths now defined are (a)

r = a+ X(c -a)

and

(b) first

r = a+X(b-a)

and then

r = b + X(c - b).

Inserting these parameters one finds that the result is the same in cases (i), (ii) and (iv) but not in the other two. 7.

For fl: (a) 2C, (b) 2C,

f2: (a) 0,

(b) 0,

.f3: (a) 0,

(b) 2C,

(c) (0 + g + 1g-)C = 2C,

(c) (0 - i + )C = 0, (c) (0 - # + )C =

C.

8.

The flux is Ja(a, b, c) in the direction into which bx x bµ points.

9.

(i) (p, c, d),

(iii) j [p x (2a + c + d)] (c x d). (This last result can be transformed in various ways without notable simplification.) (ii) (a, c, d),

(iv) (v) 10.

Positive flux through Sa as conventionally defined is into the parallelepiped; flux through Sb is outwards, their difference is the net outward flux. One finds for this (i) zero, (ii) 3(b, c, d), (iii) zero, (iv) 2(p - q) (to prove use (1.23)), (v) zero.

11.

The first flux is equal to 47ra2A/5.

12.

Calculation shows that for both surfaces the flux of (0, x2y, 0) is equal to the flux of (0, 0, x2z). This can be found without performing the integration over X or 9 respectively and shows that the flux of the second field is the same, namely zero, through S1 and S2. The flux of f1 is 121127rAa5 through S1 and T921/27rAa5 through S2. (The second value is found by an elementary but rather long integration.)

13.

Flux of f1 through first surface: - 37 (a, b, c). Flux of f1 through second surface: - M9 (a, b, c). Flux of f2 through either surface can be expressed as

(This expression can be re-written in many forms.) If a = i, b =j, c = k the last formula simplifies to

Ad (i -j). Note the relatively simple coefficient, of from the results for f1. 206

that distinguishes this result

Exercises D 14.

For quasi-cube:

Flux out of

F1 +F4 F2 + F5

F3 +F4

Field fl

Field f2

Field f3

-.C

- RC

-C

0

-1C

0

-IC

C

+QC

Total Flux

-4 Sc

For ordinary cube:

Flux out of

Field fl

Field f2

Field f3

F1 +F4

-aC

-a3 C

3 -aC

F2 + F5

+ -12C

F3 +F6

+4C

.2C

-AC

IC AC

Total Flux 0 -1C 0 (The coincidence of the value ( C) for the total flux of field f2 out of the two closed surfaces is, as will be apparent later, accidental.) 15.

(i) b x p,

(ii)bXa,

(iii) b x (p x a) + lb x (p x b),

(iv) (v)

16.

The four integrals around the quasi-square have the values 0, a x c, c x b, 0 in that order. Answer: c x (b'- a).

17.

If the evaluation of the various terms in the integral is undertaken in detail the question becomes tedious, but for both surfaces it is quite readily seen that in the integration over 0 every term one writes down vanishes. This is because the integral of any polynomial in cos 0 and sin 0 taken over the period 0 C 0 G 21r vanishes except for terms of the form cos211 0 sin21 0. If one expresses all the factors cos 20 and sin 20 in terms of cos 0 and sin 0 one can check easily that no non-vanishing integrals occur.

18.

The surface was seen to be closed - a torus. As it has no boundary curve its vector area must be zero.

19.

Since the two surfaces have the same rim, the vector area is the same for

both: -1[(bxc)+(cxa)]. 20.

The vector areas of the pair of faces A = 0 and A = 1 respectively are a (1, - 1, - 1) and 3 (- 1, 1, 1) so that their sum is zero. Corresponding results follow, by symmetry, for the other two pairs.

21.

In (i) the integral involved is f dV = 7ra3. So the answer is I7ra3p. In (ii) (v) the integral is f r d V which is zero because of the symmetry of the volume of integration. To every element r d V there is an element (- r) d V cancelling the former. In polar coordinates

r = (r sin 0 cos 0, r sin 0 sin 0, r cos 0)

207

Answers and comments and (see chapter 14)

(br, ho, h,p) = r2 sin B.

Note further that

cos 0 do = f

sin 0 do = 0.

Joo

f0

We can choose the polar axis in the direction of p, so that We

p r =r pr cos 9. Hence fdV = p f (sin 0 cos 0, sin 0 sin 0, cos 0)r2 cos 0 dV

J

= p(0, 0, 1)

f 21r o

r4 dr -

o

cos2 B sin 6 dO

= - asp(0, 0, 1) = Aasp. 22.

By symmetry all three cartesian components of the vector integral must be equal t

t Cl o fo (

o

+ Xpv)ti2(p2 + v)dX dp dv =

so the integral required is g(1, 1, 1). 23.

If dw is the area element on the unit sphere and dS' = n'dS', then dw =

dS' cos 0

(r - r)'2

where 0 is the angle between n and r - r', i.e. cos B =

Ir -

r,I

Hence the answer, but cos 0 may be negative if dS' is taken to make an obtuse angle with r - r'. If at one point r' the direction of dS' (normal to S') is taken to point towards r, making cos 0 positive, the sign of cos 0 is fixed by continuity for all r' of S'. It will not necessarily be positive for all points r'. Thus 92 has properties similar to those of the vector area - with possible algebraic cancellations of contributions from different parts of S'. The values of SZ for a closed surface S' depend on this fact. To see what they are, go back to the verbal definition of the solid angle: S' has no rim curve and for any r outside it S2 = 0. But if r is inside the surface 92 equals the total surface area of the unit sphere, i.e. E2 = 47r if n' is taken to point inwards.

Exercises E 1.

(i) 2r,

(ii) 2(r - a),

(v)-r/r3, 208

(iii) r/r,

(iv) (r - a)/ r - a 1,

(vi)-(r-a)/Ir-a1, (vii)n(r-a)Ir-a

Exercises E (viii) a, (ix) na(a'r)"

(x) m(a -r)'-'rn + nr(a

(xiii) 2 [rat - a(r a)] = - 2a x (a x r),

(xv)--ax(axr)jaxr1-2,

r)mrn-2

,

(xiv) - n [a x (a x r)] lax r jn -2 ,

-kx(kxr)Ikxrj-2.

(xvi)

(Note that for (xvi) if k lies in the z-direction the original scalar field is log (x2 +y2 )lie and the gradient field is (x, y, 0) (x2 + y2 )-2 (iii) The gradient is indeterminate at r = 0; (iv) at r = a. (v) The scalar field and its gradient are infinite at r = 0; (vi) at r = a. (vii) For negative n, as for (vi); for n = 1, (vii) reduces to (iv). (x) As (vii) at r = 0 (irrespective of value of m). (xi) Special case of (x), scalar and its gradient infinite at r = 0. (xiv) For n = 1, the gradient is indeterminate on the line r x a = 0; for n < 0, potential and field are infinite on the same line. (xv) as (xiv) for n < 0; (xvi) same as (xv) on line k x r = 0,

(x2 +y2)1/2 =0.

2.

(i) 0,

(ii) (a x r)r-3 ,

(v) 3r(a r)r-5 (vii) b x a,

ar-3

,

(viii) a x r,

(iii) - n(a x r)rn

-2

(iv) 2a,

(vi) (2 - n)arn - nr(a r)rn -2

(ix) (b x a) (c r) + (c x a) (b r),

(x) 3a(b r) - r(a b),

=2ax [bx(bxr)], (xii)naxlbx(bxr)]jbxrIn-2

(n * 0; for n = 0 see (iv)),

(xiv) ax[bx(bxr)]lbxrj-1

(xv) - (k x r)Ik x r l-1 (in cartesians (y, - x, 0)(x2 + y2 )-1). (ii) The given field and its curl are infinite at r = 0. (iii) For n < 0, as for (ii); for n = 1, the curl is indeterminate at r = 0. (v) See (vi).

(vi) For n R _

q

1

making

41r r

For r < R rti =

=g1r2 -

2 R2

41r

0(00) = 0.

+ constant.

To make 0 continuous at r = R (no double layers!) we chose the constant so as to get _

(b)

O=

q_

3

1

41rR

2

2 R 2)

P ( 41r

r2

(r < R ).

dV'

v'Ir-rl

This is best evaluated in polar coordinates r', 0', 0' for which (see chapter 14)

d V' = r2 sin 0' dr' d9' dO'.

We also introduce u' = cos 0'. Then P 41r

2 pR rp2 Jr+1 (r2 +r 12du' 2urr)112 dr ' o

1

12 p foRr12 rr1, [(r+r')-Ir-r'I] dr' 227

Answers and comments

= pr for2dr+p (Rr'dr r r2

- P(3 + _

R2

r2

2

2

q (3

1

4trR (2

r2

2 R2

The method (b) is clearly more awkward in this case, but it is much more widely applicable. 16.

If we take a point r outside V' the proof goes through very simply according to methods used widely in the text.

=JV V I;-1 '

Ij(r')dV'

r,

- fV,(v, Ir j(r) fv' Ir - rd

1 1

,I

j(r )dV'

dS,

+Jdiv' j(r)

V'

Ir - rI

dV'.

Both terms vanish because divj = 0 and because j dS' = 0. However this proof cannot be accepted automatically for r < r', because with an integral that has a singularity in its integrand differentiating under the integral sign with respect to a parameter (in this case with respect to r) is not in general legitimate. The technique we shall adopt to overcome this difficulty is one that leads to useful results in many related situations. We proceed to evaluate div a at one specified point r within V'. We must therefore know the value of the integral at points close to that point r. There is no restriction of generality in taking the origin to be at the selected point. We then want to know a(r) for r = 0 and for neighbouring small values of r. We divide V' into Vi and V2 where V i is the sphere r' < e and V2' is the rest of V'. We calculate diva 1 and div a2, the contributions from these two volumes of integration, separately. For V2 the calculation is almost identical with that performed above for the case of r outside V'. The difference is the surface integral over the spherical surface r = e. n'

J

dS' + terms that vanish.

e

Here S' is the surface r' = e. Now if e is sufficiently small we can replace j(r) by j(0). The term j(0) n' then clearly becomes zero when the integration over all directions n' is performed. Thus 0. In calculating we can again use the fact that (throughout V1) j(r) =j(O) with as great an accuracy as we wish. Then 1r

228

r'l

dV

Exercises H But the integral now remaining is exactly one found in the preceding question to be Or

3

1

e

2

2 e2

r2

Hence 41r V -al = - -j(0) r

r < e.

provided

Thus at the centre of the little cut-out sphere V -al = 0, and as the integral over V2 even in the limit e = 0 was seen to be zero, the required result is true. Exercises H

1.

We have, in cylindrical polars,

el, _ (- sin 0, cos 0, 0), eZ = (0, 0, 1)

er = (cos 0, sin ¢, 0), and, in spherical polars,

er = (sin 0 cos 0, sin 0 sin 0, cos 0)

ee = (cos 0 cos 0, cos 0 sin 0, - sin 0)

e,= (-sin0,cos¢,0). The analytic verification of the relations stated is straightforward. Geometrically we see that for cylindrical polars a small displacement of a point in either the z-direction or the r-direction leads to no change in the local orientation of the basic vectors, while a displacement in which 0 increases, i.e. on a circle r = const., z = const., produces a rotation in the ereo-plane, each of these unit vectors acquiring a small increment perpendicular to itself. The relations aer

ao

= e0

ae¢

ao

er

express just this. For spherical polars the base vectors do not change directions if a radial displacement is made. A displacement that changes 0 (a `meridional' displacement) makes er and ee change just as er and eo did in the cylindrical case. A displacement that changes only 0 is one along a line of latitude - a circle of radius r sin 0. In such a displacement the two vectors that undergo rotations are

e' = sin 0 er + cos 0 ee

and

ep

while the vector perpendicular to both these (and in the direction of this displacement), namely e" = sin 0 ee - cos 0 er

229

Answers and comments is unaffected. The formulae given are equivalent to

ae' 30 2.

aeo

= e,

= e,.

36

ae"

-e

=

a + ez

+ eo Or

= 0.

30

in cylindrical polars

a`-

a = er a- + eer as -a - + eo --r sin 9 ao 1

1

in spherical polars.

Or

In either system the verification for grad Li is trivial. In cylindricals

curl f

(er a

a

1

ez

ar + ep r ao

a az

x (erfr + eof(p + ezfz )

a

1a

a

(erxe0)arf0+(erxez)arfz+(eO xer)ra0fr 1

1

1

+(epxeo)rfr+eox(-er)rf(p +(eOxez)r a

a

-fz

a

+ (ez x er) zafr + (ez x ep)z aff er

l a

r a0 +e

f

a

z

az

za fm

1

a

er(1 a

aOfz - az fo)

r

+ ez

la

(

fZ + az fr

ar 1

+r

+e

a

a

fm

/

afr

I

az

1a

a

arfz

1

[rOr (rfo) r a0 fr J

a

er Or + a

a r

r 30 f

(ar

divf =

)

fo+ e0

e rl

a30

al

+ ez az) (erfr +

1a

1

ezfz )

a

aYfr+rfr+ra-fm+aZfz 1

a

r Or

1

a

a

(rfr)+raofo+aZ fz.

These relations clearly agree with the text. The same technique, with a little greater effort, serves also in the case of spherical polars.

230

Exercises H 3.

It is a matter of simple insertion into formulae previously derived to obtain ail 320 a2 a

1

1

r ar

aY

az2

r2

in cylindrical polars and

I' _ 1 a `2 r

a

ar + r2 sin 0 ae

r2 ar

sin B

a0 ae

a2

1

+

2

r2 sin2 0

in spherical polars. Both these expressions are among the most frequently used relations in potential theory. Alternatively, using the results of exercise 1, we can put in cylindrical polars

V

=

(era + eo 1 a2

a2

ep

a2

(aer) a

a

art+r2aO2+az2+r art

+

a

a2

e,9--a

er ar + r 1

a2

ee a ep- a r ae + r sin 9 ao

a

+ r sin 0

ae

a2

a2

1

r2 ae2

are

a

ao

galaoer( ar +

-

a

a2

a

a

e

ar+r

-r ar +-r2 -+-az

and, in spherical polars V2 =

r a + e2 az )

+ ez az (era +

ar

r2 sin2 9 a¢2

ee

aer

a

r

a8

ar

+.aer a + 2.(30a eel a r sin 9

a ar

J ae

r sin 9

(we have omitted various terms that vanish because of the orthogonality of the e vectors). So V2-a2

+1a2

r2 ae2

ar2

+

1

a2 +2a+lea

r2 sin2 a a02

r ar

r2

cot

ao

and this is easily seen to be equivalent to the expression previously obtained. 4.

(r x Q)2 >G = (r x V) (r x VO)

=

(rx vtfi)]

= r2V2,P +(r'Vii)-3(r'VVj)-r'[(r'0)0iP] = r2 The stated identity is thus seen to be true. Now 231

l

a

Answers and comments

r=rer and

+ee l a 1 a +eq r 30 r sin 0 aO or a

So

r2 V2 / - (Y

(r x V)2

r2 V2 41

Y2

ar

+ ) (r

ay )

a2W-2Yo ar

ar2

Comparing the V2, found here with Q2O in spherical polars the stated identification of (r x V)2 is confirmed. Thus if

At = 0

i = r1 Y1(0, 0)

and

we have

1'(0, 0)

r dr

(r2 dr

rl) + r1-2 (r x V)2 Y(0, 0)

= rl-2 [1(1 + 1)Y + (r X V)2 Y] = 0 as stated. Replacing 1 by - (1 + 1) leaves 1(1 + 1) unchanged, hence the final result. This question summarises some of the most important facts about `spherical harmonics' Y. These functions have implicitly appeared in earlier exercises such as E.20. 5.

This exercise involves nothing more than elementary calculus and, although the calculations are quite lengthy, a sketch of how the student should proceed is hardly necessary. It may be useful to record the results in this form, but in fact they are in practice not much used in full generality. Usually the fields one is concerned with are either irrotational or solenoidal; in either case considerable simplifications occur and if, in addition, the field possesses some symmetry as in section 3 of chapter 14, the simplification is even greater. In the case of cylindrical polars the second method of arriving at the given formulae will be seen to be significantly shorter (in this general case) but for spherical polars both methods are lengthy. If the student tries both methods, he is likely to discover some useful identities, such as

1a(2 af1_a2f 23f102 Y2 aY

ar J

ar2 + r Or

r ar2

The difference between this expression and

a [1 a ar

232

r2 ar

(r

(rf).

Exercises H should also be noted (it is 2f/r2) as should the difference between a

1

sin a a0

sin 0

of

a

and

a0

1

ad

a

[sina a0 (sin 0 f)

.

6.

To prove (a) by insertion of the given expression for r into each of the three initial equations involves some algebra but is not too tedious. The differentiation to prove (b) is straightforward. In proving orthogonality (c) and evaluating the magnitudes of the h vectors, it is very helpful to notice that all square roots are eliminated easily and that the lack of symmetry in the formulae, e.g. the occurrence of (a2 - µ2) in one place but (µ2 - c2) in another, can then be rectified. It no longer matters that some of the functions are negative. Once full symmetry is restored cancellations and factorisation lead to the results fairly simply.

7.

It helps in proving the given result to notice that, apart from the denominators written down first in each of the three terms, every term involves one of the parameters separated from the others. It is this separability that makes this type of coordinate system a powerful tool and yields results such as the ones to be proved here: That j = a satisfies z 1, = 0 reduces to the trivial identity d2 a/dal = 0. Thus >G = a is a potential function -- it is constant when X = const., since a is a function of X only. If we define >y = ao, where ao is the value on the boundary ellipsoid, the tangential component of grad >G will be continuous across that boundary, its normal component will be accounted for by the surface source. There remains the proof that >G -+ 0 when r - 00: Using equation (a) of exercise 6 we see that, for X - -, r = Xf(p, v) so r increases linearly with A. But for large X the definition of a shows that

= a ^' const. - 1 ^ const. r

So the field goes as 1/r2. As the sole sources of the field are on the ellipsoid, i.e. near the origin, one expects the field at large distances to be as for a point source. p2

8.

(a)

A = const.; gives

prolate spheroids.

µ = const.;

z2

a µ22

a2 (A2

-

z2

1) + a2a2 X2 = 1:

p2

- a2(1 -p2) = 1:

double sheeted hyperboloids. p2

(b)

A = const.;

z2

+ a2(1 + A2)

a2A2

oblate spheroids.

µ = const.; single sheeted hyperboloids.

233

= 1:

Answers and comments The surfaces 0 = const. are planes containing the z-axis in both cases. The parameters X, p of case (b) can be replaced by

p/ = (1 -p2)1/2

A' = (A2 + 1)1/2

This gives a form to (b) which is just the same as (a) with p and z interchanged.

Thelimit valuesA= 1,p=Oandp=± 1 in(a)A=O,p=0and p = ± 1 in (b) give the following surfaces (this is seen most easily from the original definitions). (a) A = 1: p = 0, z = ap: the segment I z I < a of the axis;

p = 0: p=a(A2 -1)1/2,z=0: the plane z = 0; p = ± 1: p = 0, z = aX: the segments I z I > a of the axis.

(b) A = 0: p = a(1 -p2)1/2: the disc p a of the plane z = 0; p = ± 1: p = 0, z = ± aA: the axis. (Note: in this last case it is also possible and may be convenient to change the ranges of the parameters to - - < A < 00 and 0 < p < 1; thus the whole of the axis corresponds to p = + 1.) (a)

-p2)"2

A(1-p2)1/2

b=

a("(A2 - 1)1/2

bµ =

a(- (1-p2)1/2 cos- (1-pz)1/2 sin&A

-

cos 0 , (A2

p(A2 - 1)1'2

1)1/2 sin 0, p

p(A2 - 1)1/2

bo = a(- (A2 - 1)1/2(1 - p2 )1/2 sin 0, (A2 - 1)1/2(1 _p2)1/2 cos O, 0).

(b) The same expressions with (A2 - 1) replaced by (A2 + 1) everywhere. 0

is easily verified, and for case (a) a(A2 - p2)1/2

a(X2 - /22)1/2

h'

(A2 - 1)1/2

>

hµ = (1 _µ2)1/2

hq = a( X2 - 1)1/2(1 _M2)1/2 .

and for case (b) a( X2

b

-µ2)l/2

(A2 + 1)1/2 '

a(A2 -p2)1/2

bA - (1 -p2)1/2

hq = a(X2 + 1)1/2(1 _p2)1/2 Thus for (a)

1

grad

a

234

-1)1/2 a, 'I (1-p2)1/2 a 'I 2 _ 2 >1/2 ax (A2 - p 2> 1/2 aµ (A µ (o2

'

Exercises H a>y

1

W - 1)1/2(1 _µ2)1/2 34, curl f

1µz)l/2

a 1(A2

am

[(1 -µz1/2ft,] afµ

1

(A2

- 1)1/2(1 _µ2)1/2 aO

af

1

(A2 _ 1)1/2(1 _µ2)1/2 a - (A2 lµ2 )l/2 aA 2 (X2

1/2

2

A2

div f

[(A2

-µ2)

2

)1/2f

aA 1/2

µµ2) aµ [(A2 _µ2)1/2ff] ;

_ a (X2 1µ2)f

3X(

W -12)1/2(A2 - 1)1/2fX)

+ aµ (A_µ2)1/2(1 _p2)1/2(1 _µ2)l/2f a) i

af0

1

1

_µ2 )1/2a +a (A2 - 1)1/2(1

Q_1 a

(Az-1)a

a

1

(A2 - p2 )[a

(

aA

1

+

a

am

(1_p2)

(

am

a2 l)

(A2 -1)(1-p2) am2

For case (b) all these expressions differ from the above solely by (A2 - 1) being replaced by (A2 + 1) wherever it occurs. 9.

Case (a):

[(A2 - 1)

A a(A)

1

= 0,

a=Ccoth-IA=Clog

A

-1

If C > 0, a - - for A -+ 1, a -+ 0 for A -+ oo. Thus this scalar potential iti = a can be associated either with a distribution of sources along the line z I < a, p = 0 or with surface sources on any one of the surrounding 235

Answers and comments prolate spheroids. if one takes the latter case then setting >G = a(A0) _ const. in the interior of the particular spheroid A = No (i.e. assuming a vanishing internal field) is consistent with the assumption that f = - grad has no other sources than these surface sources. Case (b):

aA

[(X2 + 1)

l

a

a(A) I = 0,

a = C cot-' A

(it is slightly more convenient to use cot-'X rather than tan"' A). The potential field >y = a has the value IirC on the disc p = a and becomes zero at infinity, so that its gradient describes the field of a source distribution on this disc or, alternatively, if used only for A > A0 > 0 the irrotational field which has only a distribution of sources on an oblate spheroid with zero field inside the spheroid. The corresponding solutions 0(p) are somewhat less likely to have practical interest as the surfaces on which the boundary behaviour for the field can be simply specified are surfaces that extend to infinity. In both cases the differential equation for (3(p) is the same, d

I(1-p2)dµ/3(p)I = 0.

It has the solution

0(p) = C tanh-' p. In case (a) the potential 0 becomes ± °° on the two line segments z > a and z a of the plane z = 0, where >G = 0 = 0. By symmetry it is clear that the lowest value obtained by the potential must be on the axis of symmetry - and indeed it is - C there. Again, the plane can be replaced as seat of the sources by an `enveloping' hyperboloid - hardly a realistic source distribution. 10.

If ii = a(A, p) we must have ll

AdA [(A2

236

1)aA(Aa+C)I+(Aa+C)dµ(1-p2)a = 0

Exercises H or

d

I

1)

A

( X

(A2

A

(A2

(AZ

- 1) A +

dX

+ a) I - 2 (Not + C) = 0

- 2Aa - 2C = 0 - 1) A + dA I (A2 - 1)a I

0+2(X2 - 1) da+2Aa-2Aa-2C = 0. Similarly

[(l dµ 2(1

l

-p2)µ(pa-C)J+(µa-C)dA(A2-1)dA

= 0

-µ2)aµ-2pj3+ 2µj3-2C = 0.

The entirely analogous solutions for case (b) are

al = p[Aa(A) + C]

where

Qi = A [µj3(µ) - C ]

where

(A2

+ 1) d

= C

and

(1 -M 2)

d

dp

= C.

Such solutions (and other more complicated ones) naturally have more complicated equipotential surfaces and since the coordinate system used is not as easily visualised as a cartesian or polar system, the study of such solutions cannot be easily justified except as a mathematical exercise. It is of course important for a physicist to have the use of all possible solutions of his equations, but general potential theory shows that any solutions of

0 = 0 can be obtained for instance from solutions in ordinary polar coordinates by forming infinite series of such simple solutions. Such expansion in terms of simple solutions is a much more rewarding approach than detailed investigation of functions like the above. z 11. a

p2

= A-4azA

If A = const. this is a paraboloid of revolution. z

p2

a

4a2 p

if p is constant, this is also a paraboloid of revolution. Each of the first family of paraboloids has its vertex on the axis above the plane z = 0 and extends to z = - °°, while the second set are mirror images of this first set, each with its vertex on the axis below z = 0 and extending to z = + co. From the original definition we see that the limiting `surface' A = 0 is the negative axis (p = 0, z < 0) and p = 0 is the positive axis. 237

Answers and comments

)1/2 sin , 11

COs fp,

(µ)1/2 cos fp,

sin (p, 11

ho = 2a0µ)1/2 (- sin 0, cos 0, 0). hx hµ = 0 is simple to check; the other two orthogonality relations are even more obvious.

=

grad

curl

+1/2

=a

h

1

a

f-a

((A+µ 1

(a +

µ

1/2 ao

1

+) as o) (µ 1/2ff) )1/2 aµ 20µ)I/2 a

a>u

aµ'

a

2(xµ)Ii2 x1/2

a0

1

(X +µ)I/2 1 a-

a [( + µ)Irzfµ)

(X

divf =

h= 2a(t)1/2;

µ

Ire a,

A

1

1/2 - (X+M) a

hµ

/If

[x1/2(A+µ)1/2f) +

a(X+µ) aX

A

- (X

1/2

+

µ) p [(X + µ)1r2fXI [µl/2(X+µ)1/2f

1

+ 2a(Aµ)1/2 ao

a (X + µ) [aX (X ax) + aµ (µ aµ)]+ 4aµ aO The field represented by - grad i i would thus have either line sources on the positive or negative half of the axis of symmetry, or on one of the paraboloids, with vanishing field `inside' that paraboloid. 12.

to be a function of r only

If we take

_

dz dr2

-

r dr

1 d2 r dr2

(rtj).

Hence we have d2 dr2

(rO) = rp.

The integral given satisfies this equation as well as the boundary condition 0(0) = 0 if p is sufficiently well behaved at r = 0. (The integral uses a 238

;

Exercises II standard prescription for expressing a function in terms of its second derivative - a form that is easy to verify.) Thus the statement of the question is clearly true - the integral is a particular solution and 11 a complementary function. We have seen earlier that the only solution of A>t, = 0 that depends on r only is C/r, so the general solution of the problem that does not bring in dependence on the angles can differ from the integral given by B + C/r where B and C are constants. If C 0 the volume sources p are not the only ones - there is either a point source at 0 or, possibly, if discontinuities are acceptable, surface sources on spheres surrounding 0. 13.

= (3(r)y(0)

For any >G of the form

L

1 d2 R /3(r) 1 a rsin ay(a)1 0 [r/3(r)] + 11 = Y(0) r dr2 a0 - J r2 sin 0 a0

so that in case (a) 0(r) satisfies

1 d2

r dr

2

r

-Arn.

r

if we put 0 = Crn+2 we get

C[(n+3)(n+2)-2] _ -A or

C(n + 4)(n + 1) _

A.

This gives us the stated solution except for n = - 1 and n = - 4. It also tells us that

= Cr cos 0

and

41 = - cos 0

satisfy

At = 0,

a fact that we have encountered before since these are the expressions for the scalar potentials of a constant field and a dipole field respectively: thus except for A cos 0 A cos 0 p= - and p = r

r4

we have found that

- Arn+2

B + r2 + Cr cos0

(n + 1)(n + 4)

is the general solution of 0> i = - p, except for possible additional terms in i i1 with other angular dependences. The constants B and C are arbitrary - they depend on boundary conditions (see exercise 15). In the exceptional cases

12 dr2 (r2 log r) = r dr (2r log r + r) = r + y log r so that 239

Answers and comments A

_

B

rlogr+ r2 +Cr) core

3

is the required general solution for n = - 1, and

1d2 logr _id

r

logr+ 2 = -r3 + r2

1

rdr

rdr2 ( r

1

r-2

logr

so that

+A

0=

B

3r2 log r +

+ Cr) cos 0

Y2

is the answer for n = - 4. In case (b) the only difference is that

I sin 6 ae (3 cost e - 1)J

sin a a

sin 0 89

(sine 9 cos 9)

6 a (cos 9 - cos3o) sin 0 -30

- 6(3 cos2 6 - 1). So 0 should satisfy

2

1 d2

r dr

6

R

(rN) -

r2

= -Ar n

or, with (3 = Crn+2

Cn(n + 5) = -A.

C[(n + 3)(n + 2) - 6] _

The exceptional values for n are n = 0 and n = - 5. For all others

-A

0_

n(n + 5)

+ B + Cr21(3 cost 6 - 1) r3

I

while for n = 0 we have 2

1 d 2(r3 log r) = 1 d (3r2 log r + r2) = 4 + 6 log r. r dr r dr Hence

= 1-A r2 log r + B + Cr2) (3 cost 6 - 1) r

4

is the one exceptional solution, p = 3 cost e - 1:

and

1 d2

log r_ 1 d

r dr2 (

r2

)

5

rS

240

- 2 log r

r dr

r3

+

6logr r5

1

+ r3

Exercises H

(AlorB22 )

giving

(3 cos 9 - 1)

r+ r3

5r3

for the other exceptional case, 3 cos 2 0

-1

P = 14.

Here 0 1

p.

IfrG

sin e

r ae

r3

-Z

r2 -- = a3

These fields are just a uniform field and a dipole field. For the second case one finds similarly that

41 = r2(3cose0-1) = A(3z2-r2) 243

Answers and comments and

>]i = `3 (3 cost 0 - 1) = -5 (3z2 -r2). 3r

3r

if in the general case we insert either r("" )Z1(B) or r-!Z1(9) into (14.35) and equate to zero we get 1(1 + 1)Z1 + sin 0

- -1 d

dZl

d6

sin B de

0

which is the required equation. The equation for YI of exercise 4 reduces in the case that Y is independent to 1

1(1+1)Y'l+sin8

d d6 \

sinB

dY/l d9

Jf = 0.

Applying sin 0 (d/dO) to this relation and putting sin 0

dY1

= ZI

d6

we obtain the required result. 19.

From (14.35) we have sin B a

a2 W

82 r +

r2

1

a'

a6 sin 0 a6

= r sin B j0.

Putting

' = Crn+3 sin2B

and

Arn sin 0

we get

C[(n + 3)(n + 2) - 2] rn+1 sin 2 0 = Arn+1 sin 2 B

so that C =

A (n + 1)(n + 4)

except for the two exceptional values of n. For n = - 1 put ' = Cr2 log r sin 2 0, then C[(3 + 2 log r) - 2 log r] sin 2 0 = A sin 2 0,

C = A.

For n = - 4 put' = Cr-1 log r sin2 0, then C[(- 3 + 2 log r) - 2 log r] r-3 sin2B = Ar-3 sin2B, If 4r1 is restricted to the same angular dependence

1 = \B + Cr2) r

244

sin 2 6.

C =- AA.

Exercises H 20.

Using the results of the preceding exercise we put

'_(

r4 + Cr2) sin 2 0

4' = Br-1 sin 2 0

rR

and ensure full continuity of the field at r = R by making ' continuous (no surface sources) and d'V/dr continuous (no surface vorticity). We require

1 AR4 + CR2 =

B

10

R

?AR4+2CR2 =

-B R

5

or

B = -7ARs,

C = -*AR2.

The external and internal stream functions are s

r>R

'y = - 1 AR sin *O 15

r

and

q/ = 110Ar4 - 1 AR2r2l sine 0 6

r R they are identical - the same dipole field, with the same numerical coefficient. Internally the parts of the field proportional to C are also qualitatively alike - both are constant fields in the z-direction but with magnitudes in the ratio 2: 1. The other terms in the interior of the sphere are of course different, since the present field is solenoidal and that of exercise 15 is irrotational. These results are closely related to the subject of exercises F.7 and F.B. 21.

If v is a P-field its vector potential can be chosen to be a T-field and

NP = -rsin0ap. Equation (14.35) tells us that curl curl a is another T-field, b say, such that

bo = If

1

a2 q,

1

a

1

a %if

r sin 0 are + r3 a0 sin 0 a0 )

= r" sin2 0,

b4 _ (n - 2)(n + 1)r"

If we now define r sin 0 bp we have

_ -(n-2)(n+1)rn-2sin20 245

sin 0.

Answers and comments and the q5 component of curl curl b can again be calculated from (14.35). The condition is therefore (n-4)(n-1)(n-2)(n+1)rn-s sin20

= 0.

So the fields whose stream functions are r4 sin2 0, r2 sin2 0, r sin2 0 and r-1 sin2 0 all satisfy our condition. The field derived from the first of these expressions increases at infinity as r2 and cannot be used in the physical problem; we put

tY = (-I Vr2 + Ar + Br-1) sin2 0. The first constant is chosen so as to make the magnitude of v at infinity equal to V. If we have

-'VR2 +AR and

+BR-1

=0

-VR+A-BR-2 = 0

both fr and fe will vanish at r = R. Therefore the required stream function is

' = V(- jr2 + jjRr - JR3r-1) sin2 0.

(This field of flow is the basis of Stokes' solution for the problem of a sphere moving steadily in a viscous fluid.) 22.

The equation for f1 may be verified by direct insertion. The rest of the statements are immediately seen to be true if it is shown that g2 is toroidal. Insertion into (14.23) does show immediately that the radial component of g2 is zero. One finds further that, if Ox,

1a a g2 = (0, - -, sinaao W Thus starting from a single scalar x a whole sequence of toroidal and poloidal fields can be generated. Note that it is obvious that a toroidal field is a special kind of solenoidal field; likewise the curl of a general solenoidal field will not have the toroidal form so that a poloidal field is also special - although unlike the toroidal field it has both radial and transverse components. However, see the next exercise. 23.

curl f = (0, - 4z, Sy)

3x3 +2xy2 +xz2, 1(- 9x3 + 16xy2 + 11xz2) - xr2

f=4yz

(r x V)2 X

2x, (r x V )2 yz = - 6yz (r x V)2(- 9x3 + 16xy2 + 11xz2) - 12(- 9x3 + 16xy3 + 11Z2). (Note that the values 2, 6, 12 are 1(l + 1) for l = 1, 2, 3; cf. exercise 4.)

x = 3yz _ k(- 9x3 + 16xy2 + 11xz2) - jxr2

_ - -6 (9x3 + 4xy2 + 5xz2 ). 246

Exercises I Writing down rX and r and working out their curls is lengthy but straightforward. Exercises I 1.

aO+>v

aL+k(wP0O-OAS) = 0

hence

L

ap + k div (iii grad ¢ - grad >fi) = 0

j 2.

(a)

(b) 3.

k (>V grad 0- 0 grad i ).

p =x(02 +g2), j = -kqg. p = (f2 +h2), j = k(fxb).

K at = k div ag = k div (k curl h - Kf) = - kK divf ab K

at

= - k curl (- k grad ¢ + Kg)

kK curl g

K div h = - k div curl f = 0. Hence

at 1' 2 O2) =

-k¢divf

at(Zf2) kg curl b - Kg f

at (2

h) = - kb curl g.

Addition gives the required result. 4.

(a)

LP

(b)

E-

at

=

+ div j aE

at (1 EZ

a div E - div aE = 0. at

at

aB

+-1B 2)

+div[c(ExB)] _ -PE.

If # (E 2 + B2) is interpreted as energy density then clearly c(E x B) signifies the flux density of energy while j E describes the diminution of 247

Answers and comments energy due to `sinks' throughout space. But if the electric field does work to the amount W it is to be expected that there is a corresponding diminution of energy `stored' in the electromagnetic field. 5.

Proof of the identity: 1

E-

aB

C

at

+Ex curlE = 0 - E div E = - pE -B+BX curlB =

C

-c jxB

-BdivB = 0. We add these equations, take the scalar product of both sides with a and use the identity of exercise G.3.

-

I1

a a unit vector and, say, in the x-direction, then the right hand side is the negative of the x-component of force exerted by the electric and magnetic fields on the charges and currents. These must be attached to matter in some form, and the effect of such forces is to impart momentum to the matter. If total momentum is conserved, any momentum lost by the matter must be momentum gained by the electromagnetic field. The right hand side thus represents the gain of (x-component of) field momentum density. Interpreting the relation as an equation of continuity we must take (E x B)/c to represent the field momentum density while, with a in the x-direction, the expression for j represents the flux or transfer of x-component of momentum. The direction of j indicates the direction of a surface across which the x-component of momentum is flowing normally. The same is of course true for y- and z-components. This description circumscribes what is much more naturally stated in tensor language this is as near as we get in this book to tensor ideas. 6.

We have 1

(E+V

fs

aB

v

0.

Using Stokes' theorem we get

fS

(

(curl F. +

1 aB

c at

) dS = 0,

the terms involving v cancel out. This argument for the universal 248

Exercises I applicability of equation (A) for Faraday's law is much strengthened by applying the ideas of the special theory of relativity which lie outside the scope of our discussion. 7.

Dv

curl

=0

Dt

Dv _ av

Dt aw

at

1

grad v2 - v x curl v

at + 2

- curl(vxw) = 0.

Hence by (15.19) and using div w = 0

-d- f S(t)

do

This is the basis of the stability of vortices, e.g. smoke rings. This stability is absolute in the classical picture of frictionless flow. 8.

DP = DB

(vxB)+kLB

Dt =

----

= (B O)v - B V v - kLB Df Dt

=

B Dp B2 Dt

+

p DB

2p

B2 Dt

B4

BB

(using V B = 0) DB

Dt

+LPOB- k

=

p2BX (VXv)+B2 2kp

+ TAB- B4 B(B'OB); this combines to the form given. If dl is parallel to B the terms not containing k vanish and we get d

dt J C(t) B2

-kJ-P-

C(t) B2

provided C(t) is a field line of B. The relation (15.14) or (15.15) concerns curves C that move with the fluid. One cannot give a very precise meaning 249

Answers and comments to the common statement that a field line `moves' in any particular way, since the idea of a field line is an abstract one, describing the momentary direction of the vector field along a line, but a line of fluid particles is a precisely defined concept. If as we have here for k = 0 the work integral of a field f between two endpoints on this line of particles remains constant, there is a good intuitive sense in saying that `the field f moves with the fluid' - or nearly so if k 0 0. By using f we have established a result that holds generally; if we take the case p = const. there is an obvious way of making the result simple quantitatively: ('

dl B

J

IBi

is clearly a measure of the length of the chosen string of particles. If 0 C(t)

B2

an increase in the length must imply an increase in B along C(t). The theorem of this exercise allows further conclusions to be drawn if curves C(t) other than field lines of B are looked at. For instance it shows that for a perfectly conducting fluid the value of (15.14) is generally zero if shear flow is absent. However, adequate discussion of these questions requires the language of tensors.

250

Index

Numbers in parentheses indicate references to exercises or answers Angle between two vectors 1 solid (71)

Approximation of general surface by quasi-square 19 of general volume by quasi-cube 22 Arbitrariness in defining vector potential 86 in defining scalar potential 78 Area element of 28 of a surface 28; vectorial 63 Axial vector 2 Boundary behaviour: of fields 115; of potentials 124 discontinuity, notation 115

Cartesian components of vector 5 Cavity in volume 106 Circulation 51 Closed curve 15 Component functions of vector field 40 Components cartesian 5 orthogonal curvilinear 160 Conservative field 79 Continuity, equation of 181 for electric charge (192) Contour lines 38 Contour map 37 for magnitude of vector field 41 Contour surfaces 39 Convective derivative 184 Coordinates cartesian 13

general ellipsoidal (173) orthogonal curvilinear 160 parabolic (176) polar 162 spheroidal (175)

curl f Cartesian form 84 definition of 81

251

evaluation 82 for P-field 167 for T-field 167 in orthogonal curvilinears 161 in polars 164 Curl

notion of 44 of irrotational field 81 Curliness 82

Current density 182 Curvature 65 Curve

closed 15; boundary of quasi-square 20 closed, general parametric form 16 composed of several curves 16 length of 27;vectorial 63 parametric equation 13 self-intersecting 14 smooth 15 Curvilinear coordinates, orthogonal 164

D/Dt 184 del operator, V 72 V X f84 Vm 73

V' f 146 VZ,p 145

V operation not expressed via grad, curl, div 140 Determination of f from sources and vortices 147 Differential geometry of curves 65 Differentiation in direction of a vector field 140, 141 in direction perpendicular to a vector field 140, 141 of fields 72 of products of fields 138 with respect to parameter 15 Dipole moment 129 distribution (158) scalar potential for 129 Discontinuity of double layer 132

Index of scalar field 133 on surface, notation 115 of vector field, normal component 118, 120 of vector field, tangential component 118, 121 Distance, shortest between straight lines (11) Distortion of quasi-cube shape 22 of quasi-square shape 18 div f cartesian form 95 definition of 93 evaluation: in orthogonal curvilinears 162; in polars 164

Generalised divergence theorems 98 Generalised integral theorems 98 Generalised line integral theorem 100 Generalised Stokes' theorem 99 grad 0

Divergence

cartesian form 73 definition 73; alternative definition 77 grad P in orthogonal curvilinears 161 in polars 164 Gradient, notion of 39 Gradient field line picture 74 Graph of function 37 Green function 153 Green's formula 151 Green's theorem 96

notion of 44 of solenoidal field 93 Division by vector 7 Dodecahedron, regular (10) Double layer potential expressed as solid angle (136) uniform 129

bX defined for curve hA, bµ defined for quasi-square 21 bX, bµ, by defined for quasi-cube 26 Harmonics, spherical (232) related to Stokes' stream function (244)

Electrostatic field, metal boundary 122 Ellipsoidal coordinates general (173)

special (spheroidal) (175) Endpoints of curve 14 Energy density, electromagnetic (192) Faraday's law 188

Field scalar 37 vector 40 Field lines

equation: in terms of 'I' 169; in two dimensions 42; parametric 43 number of 91 of solenoidal field, without ends 91 of vector field 41 tube of 89 Fluid dynamics, equation of motion 142 Fluid flow between solid boundaries 123 Flux integral 54; across quasi-square 54; linearity of 57; out of closed surface 56; properties of 57; time rate of change 187 of field through disc in terms of* 169 of fluid through surface 54 through closed surface of solenoidal field 88 Forces in electromagnetic theory (193) Fundamental property of curl field 86 of gradient field 78 Gauss' theorem 96

252

Icosahedron, regular (10) Images, method of (135) Induced electric charge (135) Integral flux 54 quantity 58 spatial, of fields 49 symbolic 62 theorems, generalised 98 work 49 Integration by parts 142; over infinite space 143 ranges of, generalised 101 Interlocked curves, Stokes' theorem 102 Interpenetrating volumes 105 Intersection of lines, planes (11) Intrinsically irrotational field 117 Intrinsically solenoidal field 118 Irrotational field 79 that is also solenoidal 112 Kinks in curve 15 Knotted curves, Stokes' theorem 102

Laplace operator 145 Length element of: scalar 27; vectorial 27 of a curve 27; vectorial 63 Level surfaces 39 of gradient field 75 Linear independence of vectors 5 Linearity of flux integral 57 of quantity integral 59 of work integral 52

Index Line of flow and of force 41 Line integral v. work integral along interlocking curves 102 theorems, generalised 98 vectorial: of scalar 61; of vector 62 Line singularities 126 on closed curve 132 Lissajous curve (32) Magnetic field

frozen in perfect conductor (194) superconducting boundary 122 Mass, total in volume 30 on curve 28 on surface 29 Maxwell's equations (192) Mesh of parameter lines 18 Metal boundary for electrostatic field 122 Metric in orthogonal curvilinear coordinates 163 Metric properties of space 27 Mobius strip 104 Multi-valued scalar potential 127, 132

Newtonian solution of Poisson equation 149 for vector potential 153 Normal opposite sense on quasi-cube faces 24 to closed surface 56 to surface: definition 54; sense of 56 Notation descriptive for integral theorems 100 for generalised integral theorems 98 Number of field lines 91 Numerical integration of field line equations 43

Operator 72 Orthogonal curvilinear coordinates 160 Orthogonality of gradient field to level surfaces 74 Orthonormal reference frame 5 Parabolic coordinates (176) Parameter change of 13 limits of: for curve 14; for quasi-cube 22; for quasi-square 17 Parametrisation, standard, for boundary of quasi-square 20 Parts, integration by 142 P-fields 166 Picture quantitative of gradient field 76 quantitative of solenoidal field 89 of vector field not quantitative 44 Plane, equation of (11) Point, position vector of 13

253

Point singularities 127 Point source, scalar potential for 128 Poisson's equation for vector field 147 for scalar field 146 Newtonian solution 149 Polar coordinates 162 Poloidal vector fields 166, (179) Potential scalar: boundary condition for 124; of gradient field 78 scalar multi-valued 127, 132 vector: boundary condition for 125; of curl field 86 Product of fields, differentiation 138 of scalar and vector 1 scalar triple 4; in cartesians 6

of two vectors: scalar 1; vector 2; in cartesians 6 Proof of V a = 0 for Newtonian expression for vector potential (159) Pseudo-vector 2

Quantitative picture of curl field 89 of gradient field 76. Quantity integral linearity of 59 over volume 58 sign of 58 time rate of change 188 Quasi-cube

boundary of 24 parametrisation of volume 20 Quasi-square

boundary of 20 parametrisation of surface 17 Quotient theorem 7 Rate of change with time of flux integral 187 of quantity integral 188 of work integral 186 Reference frame Cartesian 5

change of 8 Reversal of direction in work integral 51 Rotational symmetry, fields with 165

Saddle point of surface (33) Scalar 1 Scalar field 37

discontinuities of 133 time-dependent 181 Scalar potential 78 boundary conditions 124 multi-valued 127, 132 Scalar product 1

Index in cartesians 6

Scale factors bx, hµ, by for orthogonal coordinates 160 Second derivatives of fields 145 Self-intersecting curves 14 Separation of irrotational and solenoidal parts of f 148 of poloidal and toroidal parts of solenoidal f (179) Shadow 65 Singularities line 126 point 127 Smooth curve, replaced by kinked one 15 Solenoidal field definition 87 quantitative picture 89 sum of poloidal and toroidal field (179) that is also irrotational 112 Solid angle potential for double layer (136) subtended by surface (71) Source density 94 surface 120 Sources and vortices determine f 147 Spatial integrals of fields 49 Spherical harmonics (232)

related to ' (244) Spheroidal coordinates (175) Splitting of general surface into quasi-squares 19 of general volume into quasi-cubes 24 Stokes' theorem 85 generalised 99 Stream function, Stokes' 165 Substantive derivative 184 Superconducting boundary for magnetic field 122 Surface area of 28 curl of vector field 121 described as quasi-square 17 divergence of a vector field 110 element, vectorial and scalar 28 source density 120 sources in Newtonian solution of Poisson's equation 150, 153 vorticity 121 Surface integral v. flux integral 54 vectorial: of scalar 61; of vector 62 Symbolic integrals 62 Tangent vector, definition 51 Tetrahedron, regular (10) T-fields 166 Time-dependent fields 181 Time-dependent solenoidal flow Toroidal vector fields 166, (179)

254

Torsion 65 Triangle, equilateral (10) Triple product 4 in cartesians 6 Tube of field lines 89 Two-dimensional vector fields 169

Unit vector n I Vector algebra 1 area of surface 63

bracket notation for 7 cartesian components of 5 direction of 1 field 40; in two dimensions 169; quantitative picture 76, 89; time-dependent 181 integrals 60 length of 1

length of curve 63 linear independence of 5 multiplied by scalar 1 potential: boundary condition for 125; of curl field 86; chosen solenoidal 148 scalar product of two 1; in cartesians 6 sum of two 1 unit 1 vector product of two 2; in cartesians 6 Vectorial line integral: of a scalar 61; of a vector 62 Vectorial surface integral: of a scalar 61; of a vector 62 Velocity field solid boundary 124 time-dependent relations 184 Viscous flow round sphere (178), (246) Volume element 30, 58; sign of 58 of region of space 30 Volume integral v. quantity integral 58 Vortex strength constant in classical fluids (193) Vortices and sources determine f 147 Vorticity 82 surface 121

Work by force on moving body 49 Work integral along: curve 50; closed curve 51 linearity of 52 properties of 52 round closed curve, for irrotational field 80 time rate of change 186 Y1(6, 0) spherical harmonics (232)