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

C: r(Xo, t , 0), fs

-Js

dS (4)

S: r(A,

0)

IX1<X

X2

I Al <-A-< µ2

The minus sign in (3) comes from the fact that the direction of C is defined to agree with the direction of the `bottom' (v < 0) part of C. Now if f is an irrotational field the left hand side of (3) is zero before the limit is taken, and this must be so for any curve C that can be described by (1). Therefore the integrand on the right must vanish whatever the direction of t. We can therefore conclude that: A vector field which by its physical definition must be irrotational everywhere cannot have any discontinuity of its tangential component on any possible surface of discontinuity. In mathematical terms: If a field is intrinsicallyt irrotational then both t The reader will note that both in this statement and the corresponding one for solenoidal fields immediately below the word intrinsic is essential. It is perfectly possible to describe a particular field as having 0 X f = 0 everywhere but not n X f1* = 0. Such a field would not be `intrinsically' irrotational. 117

Boundary behaviour of fields

Vxf = 0

and

nxf

11

1+

If, on the other hand, f is solenoidal, then the left hand side of (4) is zero even before the limit is taken. In the limit the S of (2) is an arbitrary part of So . We can conclude that: A vector field which by its physical definition must be solenoidal everywhere cannot have any discontinuity of its normal component on any possible surface of discontinuity. Mathematically: If a field is intrinsically solenoidal then both

V -f = 0

and

On the other hand, discontinuities may be possible in some situations in normal components of irrotational fields and in tangential components of solenoidal fields. We can easily define these discontinuities analytically, but the way they come about is quite easy to see qualitatively simply from rough line pictures. We saw that a field line pattern, such as in fig. 3(a), is in a general way associated with the field having a curl somewhere in the diagram. If this pattern is

(a)

(b)

(c)

Fig. 3. The surface curl as a limit of the ordinary curl. 118

1

Surface discontinuities

modified to fig. 3(b), the presence of the curl is most pronounced in a thin flat volume and it is a small step to go to the limiting picture (fig. 3(c)) where one may say that the curl of the field is in the boundary surface. But this is just the picture of a discontinuous tangential component of f. In the same way a field line picture typically associated with a divergence would be as in fig. 4(a) which might flatten out to fig. 4(b), and in the limit become fig. 4(c). This clearly represents a

(a)

//1111 (b)

1111111 t1littt (c)

Fig. 4. The surface div as a limit of the ordinary div. 119

11 Boundary behaviour of fields discontinuity of the normal field component.t Let us now seek a quantitative measure for the discontinuities depicted in fig. 3 and fig. 4. We shall look first at (4) for a field that is not solenoidal so that div f = p, say. There the left hand side can be written as

fs

ff

f 2 p(hx x hµ) h dA dµ dv

=fs (s:2Phndv) dS

(7)

where S is defined as in (4). If we go to the limit vl -+ 0, v2 -+ 0 the last expression will tend to zero unless we postulate that div f = p increases to some infinite value as the volume enclosed by S collapses onto S. This we shall do and we put lim

Jv2ph.ndv = a.

(8)

We call a the surface divergence or surface source density of f. It does not need to be a constant since it can vary from point to point on S. With this definition, which can be made however far we extend S, we see, going back to (4), that

surf. div f = n f I = a.

(9)

This is clearly the limiting form of the ordinary divergence equation

In other words we say that a discontinuity of the normal component of a (non-solenoidal) field can be described as a surface divergence (9).

Equation (3) can now be dealt with quite similarly. We consider a field that is not irrotational and for which curl f = j.

At this point the remarks on p. 82 should be remembered: the appearance of fields with a curl or fields with a divergence is normally not so striking because the patterns may be masked by other parts of the field that do not contribute to the curl or divergence.

120

1

Surface discontinuities

fc

fs curl M2

i'2

= fCYPj

.

h x j dvl

(10)

dl

lJ

where C is defined by (3). We can then conclude, as in the previous case, that _ (11) lim fllv

Pt

where t may be any direction perpendicular to n. So we may write

nxf =g

(12)

2=

where

g = lim f-P, n x (j x

dv

lim fv2

(13)

dv.

Now the second term in the integrand can be dropped. If it were present the vector field j = curl f would vanish outside a thin layer yet would have a normal component on that layer. Not only does this not correspond to any known physical situation, but further mathematical development would show such an assumption to be untenable. (It would mean div j -* oo, but we shall not go into this proof.) Thus we may define

g = lim fv2

dV

0)

(14)

in full analogy to (8). g is again in general a function of position on S. Again, we visualise a limiting process in which the curl or vortex density off increases to some infinite value as the volume enclosed by S collapses onto S. We call g the surface curl or surface vorticity, and have the result

surf. curl f = n x f

l

=g

as a limiting form of the curl equation 121

(15)

Boundary behaviour of fields

11

curl f=Vxf=j. In words - the discontinuity of the tangential component of a (nonirrotational) field can be described as a surface curl. To a physicist there is no absolute difference between a field with either divergence or curl in a thin three-dimensional layer and one with surface divergence or surface curl. It is satisfactory to know that the mathematical descriptions are also interchangeable, as this section has shown.t There is however one place where it would be very inconvenient indeed to avoid using a `discontinuous' description at a surface namely when the surface is the boundary of the region of space to which the field is confined. Not surprisingly, the language we have just developed is particularly well suited to deal with this case. It will be enough to give a few examples. If an electric field does not vary in time, theory tells us that the field must be intrinsically irrotational. Hence its surface curl should also always be zero. If the field is confined to a volume V outside which we are sure that no electrostatic field can exist, then (5) tells us that the field must also have zero tengential component just inside V. We know that no electrostatic fields can exist inside metals. Therefore the field lines at the surface of a metal must always start off at right angles. This then shows that any metal surface must be a level (equipotential) surface. A typical electrostatic field picture of field lines is shown in fig. 5(a). It represents a charged metal cylinder between two infinite metal plates. Clearly there is a discontinuity of normal components of the field on the boundaries, if there is to be any field at all. But by (9) this means that there exists a surface divergence at the boundaries - in electrical language the conductors carry surface charges. The field line picture for a magnetic field is quite different because one learns that that field must be free from all divergences - volume or surface (intrinsically solenoidal). There also happen to be materials in physics inside which no magnetic fields can exist: superconductors at very low temperatures. We can use the same boundary outline as before to give an example of this case and get quite a different field line picture (fig. 5(b)). To get any field at all while satisfying continuity of the normal field component (vanishing of surface t There exist vector fields in physics for which curl for div f, as the case may be, are not zero but are related to other fields in such a way that one can he sure that a surface curl or surface divergence off must be absent. In such cases the boundary conditions are the same as for intrinsically irrotational or intrinsically solenoidal fields.

122

1

Surface discontinuities

(a)

(b)

(c)

Fig. 5. Vector fields in the space between a cylinder and two infinite planes with: (a) surface sources; (h) surface vorticity: field vanishes at large distance from cylinder; (c) surface vorticity: field constant at large distance from cylinder.

div) at the boundaries, we must allow a surface curl to be present. In physics the curl of a time independent magnetic field must be due to electric currents. Therefore fig. 5(b) illustrates a superconducting cylinder and a pair of planes carrying surface currents. Indeed, superconductors are known to conduct electricity in this way. The ideas illustrated so far by electromagnetic theory must also have applications to other fields, e.g. to fluid flow. Naturally the details of a field line picture likely to exist in fluid dynamics will be rather different from the previous case. With the same boundary 123

11 Boundary behaviour of fields outline one such picture could be fig. 5(c). Clearly it represents a flow between two walls with a cylindrical obstacle in the way. Let us analyse it in the language we have developed: the normal component of that field is zero on all boundaries. As there is no flow outside these limits, the surface divergence is zero. This is quite unlike fig. 5(a) which could only exist in a fluid flow picture with the rather unrealistic assumption that the cylinder was exuding liquid and the two plane boundaries sucking it in. The pattern of fig. 5(b) is more closely related to that of fig. 5(c). In both there is a discontinuity in the tangential component of field on the surfaces, and indeed fig. 5(b) could represent a different pattern of flow in the same geometry - one involving circulation (see next section) round the cylinder. It is rather easier to imagine how the flow of fig. 5(c) could be pro-

duced, but that of fig. 5(b) - or a superposition of the two - would not be forbidden by the laws of classical fluid dynamics. Stated in our language this kind of flow involves a surface curl at solid boundaries. This seems rather unphysical but in fact it is related to a very important feature of real fluids, its viscosity. In a viscous fluid a flow picture such as fig. 5(c) could at best exist as an approximation in the interior or a region, excluding a narrow layer at every boundary of the flow. In that boundary layer friction produces vorticity. It is therefore not surprising that in the simplified classical picture one has to put up with the presence of a surface curl. Returning to the mathematical treatment of surface discontinuities, it is useful to discuss rather briefly the possibility of a reformulation of the second relation of (5) and, up to a point, also the second relation of (6). We take (5) first. If we have an intrinsically irrotational field f with a surface of discontinuity SD , then in the separate regions of space on either side of SD the field f can be expressed as the (negative) gradient of a scalar potential 0. But we must expect some kind of a break in the character of 0 as we pass from one side of SD to the other. This break is almost determined by our previous discussion. Equation (5) tells us that the component of grad 0 tangential to SD cannot be discontinuous (unlike its normal component which is determined by (9)). Now we know that mathematically a scalar potential is only defined apart from an additive constant. Therefore there is no mathematically necessary unique way of relating the values of 0 on either side of SD. However, it is clearly always possible to relate the arbitrary constants in the expressions for 0 on either side of the surface in such a way that 0 1± = 0 at one point of SD. But then the fact that the second 124

Surface discontinuities statement of (5) is true ensures that

1

(5) at all points of SD. This is because the change of 0 in passing from the one point to any other on SD can be found by integrating grad 0 along a path lying wholly on SD - i.e. by use of n x f only. So instead of postulating the continuity of this component off it is simpler to postulate (5'). The characteristic of an intrinsically irrotational field can be stated as the continuity of 0 everywhere. But strictly this is only a sufficient condition for the second part of (5) to hold. Use of a 0 with a different constant on either side of SD is quite permissible. But it is a rather unphysical thing to do as it introduces discontinuities in the form of work integrals taken across surfaces SD . (See a discussion of related matter in section 3 of this chapter.) A physicist can therefore treat (5') as a both sufficient and necessary condition. Its use decidedly simplifies many discussions. The last paragraph may seem to be labouring a rather formal point, but it may help to make clear the corresponding situation for intrinsically solenoidal fields. Here the discontinuity behaviour described by (6) and (15) is involved. The steps corresponding to those just taken for irrotational fields involve introducing a vector potential a on either side of a surface SD and replacing (6) by an equation similar to (5') involving a. We shall only sketch the derivation of such an equation. For an intrinsically solenoidal field the normal component of f, i.e. of curl a, is determined by tangential components of a itself differentiated along tangential directions. So the condition corresponding to (5') would have to be

nxa However, for two reasons the significance of (6') is rather limited. Firstly, (6) is far from being a necessary condition; it is only a sufficient one. This is because the arbitrariness in the definition of a vector potential amounts not just to an additive constant but to an additive function grad X. Secondly, while (5') can claim to be a simpler statement than the second part of (5), the new relation (6') is in general more complicated than the corresponding statement in (6). There is therefore in general little virtue in introducing (6'). However, there are some special cases, particularly those discussed in 125

Boundary behaviour of fields sections 3 and 4 of chapter 14, where in effect, if not explicitly, relation (6') is relevant and useful.

11

2 Singularities at lines and points We have examined the behaviour of vector fields at surfaces in great detail particularly because we can never really avoid boundary surfaces. We do not need to deal equally systematically with exceptional behaviour of vector fields along lines or at points. It is quite enough at this stage to look at a few examples to illustrate the situation. Some of the results will however be of very great general significance. Taking cartesian coordinates, consider the line x = y = 0, i.e. the z-axis, and the field

y x f = (x2y2' (x2 +y2)' 0

(16)

It is very easy to see that V x f = 0 and hence to conclude that the field is irrotational. This seems to be entirely belied by the intuitive picture of the field. Figure 6 sketches some lines of force of the field which in fact provide a correct description of the magnetic field of a currentcarrying wire.

Fig. 6. Field lines of the irrotational vector field f = -grad 0 where 0 = tan-1(y/x). 126

Singularities at lines and points The `curliness' of the pattern seems obvious. And yet the field has no curl! The answer to this paradox is simple. The field is not defined everywhere within any volume through which the z-axis passes. But where the field seems to curl is precisely the z-axis. Going round a closed curve C which does not encircle the z-axis one does

2

find that fjf dl = 0. The `curliness' is a property only of contours encircling the z-axis.t In the language of vortices one says that there is no vorticity in space but a `line vortex on the z-axis'. The field can still be called irrotational but irrotational with circulation around the axis.$ If however we were to examine the field

g=

(x2

-y

x

+ y2 )n '

(x2 + y2 )n , 0

(17)

the lines of force would have exactly the same shape as illustrated in fig. 6, but as long as n 1 such a field is not irrotational. Its vorticity curl g is nowhere zero. The reader should at this stage refer back to the remarks on this very point made on p. 82. In the same way the field

f

(') Xy

z

r3

(18)

has a field line picture, shown in fig. 7 (in two dimensions), which clearly looks as if it possesses a divergence. But the calculation proves that it has not. If however one took

f=

x y

z

rn ' rn rn

(19)

then divf would be 0 0 for n =/= 3. In the former case the field may be termed solenoidal but only in a region of space excluding the origin. One says that (18) is a field that is solenoidal, except for a `point source'. The point source potential occurs very frequently in quite varied contexts. We have introduced it in terms of a particular

t

t

It is very significant, and will be particularly important to the discussion in the next section of this chapter, that the work integral in a case like this can be zero for one class of curves, but have a non-zero value for another class, this value being the same for all curves of that class. The difference depends on whether the curve does not or does encircle some line - here the line x = y = 0. Incidentally it is easy to see that along any curve that encircles the singularity line n times (in the same sense) the work integral has n times its value for one encirclement. tan-' (y/x). This is a If one looks for the scalar potential one finds 0 multi-valued function and the potential does not return to its starting value if traced round a closed path. 127

Boundary behaviour of fields

11

Fig. 7. Field lines of the vector field f = r/r3 for which div f = 0.

set of Cartesian coordinates. Using vectorial notation, taking the singularity to be at the general point a instead of the origin and, for good measure, introducing an arbitrary multiplicative constant q we can express the general point source potential as

_

q

1

47r fr-al

t

(18 )

These examples are by no means exhaustive. Singularities on curves other than straight lines and other types of singularities at individual points are of common occurrence in applications of vector calculus. Of these, we mention only one more, one that is less fundamental than the point source (18) or (18') but which nevertheless occurs very frequently in physical applications. Starting from (18') we examine the field of two neighbouring sources of equal and opposite strength. If their vectorial distance is d and the negative one is at the origin, the scalar potential of the combination is _ q 1 1 _ r (20) I rd l 4ir

and if we treat d as a small quantity, then by a MacLaurin expansion about 0 we find

q (d V) 1 + higher terms.

(21)

r A physicist would say that the expression (21) for the potential of 41r

t The factor 1/4n in (18') is not necessary, merely conventional in the unit systems now in greatest favour. For its justification see p. 149. 128

The double layer such a dipole can be treated as exact as long as one does not study it at small distances, i.e. as long as one stays many multiples of d away from the sources. A mathematician can make (21) exact by defining the limit (22) qd-p for d-+ 0. 3

Clearly this limit implies that one lets the pair of sources move together while simultaneously increasing their strength as required by (22). The result of this limit is termed a dipole; the vector p is the measure of its strength and orientation and is called its moment. For a physicist it is a meaningless question to ask whether dipoles exist with exactly the potential (21), which can also be written as 47r

(p 0)

1

r

1

+ 4ir

r3

(23)

All he can note is that the use of (23) provides a very good description in a number of physical situations. As for the point source (see (18')) it is simple to write down an expression for the potential of a dipole at points other than the origin and one can also contemplate without difficulty the potential and field of a distribution of dipoles. There is a special case of this which we shall discuss in the next section.

3 A special discontinuity - the double layer In section 1 of this chapter we looked at surface discontinuities of importance in physics - the surface curl and the surface divergence. A mathematician could envisage many other discontinuities across a surface - e.g. some vector field component becoming infinite there. We need not discuss general situations that are of no relevance to physical applications. However there is one type of discontinuity not yet introduced that it is useful to understand, although it enters into physical theory in a rather formal way: this is the double layer. The picture of a double layer is closely related to the picture of the dipole introduced at the end of the preceding section. It may be defined as a surface carrying a uniform distribution of dipoles with moments normal to the surface. However, for our purposes it is more convenient to define the double layer slightly differently. (The relation between the two definitions is quite obvious and need not be elaborated.) First let us consider the entirely unrealistic situation of two parallel infinite planes, say the planes x = x 1 and x = x2 , of which the 129

Boundary behaviour of fields

11

xo

Fig. 8. (a) The potential ¢ outside and between two infinite planes carrying equal and opposite uniform surface sources. (b) The potential 0 on both sides of an infinite plane uniform double layer.

first carries a uniform negative source density (surface divergence) and the second an equal positive one. Fig. 8(a) shows the scalar potential 0 of the vector field with these sources. The field, - grad 0, is constant and points in the + x direction in the whole of the space x < x 1i and also has exactly the same value and direction in the space x > X2. Between the two planes (for x 1 < x < x2) the field is constant again, but has the opposite direction. If the two planes are close together its magnitude in the middle region will be much greater than outside. Starting from this situation we go to the limit x1 -> x0, x2 - xo (fig. 8(b)) (a limit that is clearly related to the limit taken in defining a single dipole!). Clearly in this limit the `intermediate' region ceases to be represented in the diagram. The remarkable feature of fig. 8(b) is that although the potential 0 shows a surface discontinuity the field, its (negative) gradient, is perfectly continuous - as if the surface were not there. When introducing the scalar potential of an irrotational field we laid stress upon the fact that only the field, not its potential, is a uniquely definable quantity and that an arbitrary constant can be added to any potential 0 without changing the physical situation. Why then is it worth discussing a discontinuity of potential that leaves the field continuous? The answer can be given by looking at the slightly more realistic situation of a double layer over a finite disc. This new situation 130

3

The double layer 41

0

xa

Fig. 9. (a) Schematic representation of the potential 0 around two symmetrically placed circular discs carrying equal and opposite sources. (b) Schematic representation of the potential 0 of a circular disc carrying a uniform double layer.

might, in principle, be realisable in nature. We need not go into matters of detailed mathematical form. It is clear that if one starts with two flat discs of any shape, provided they carry exactly matching opposite charges, the potential pictures corresponding to figs. 8(a) and (b) will have the shapes given in figs. 9(a) and (b), of which the main feature is still that the gradient of 0 is equal at x = x 1 and x = x2 in fig. 9(a) and on both sides of x = xo in fig. 9(b). But now the arbitrary constant in the potential is clearly seen to be something quite unrelated to the discontinuity 0 of the potential at x = xo : the former constant can be fixed with reference to the behaviour of 0 at x = ± oo (and will normally be chosen to be 0). So given any double layer one has to put up with a `real' discontinuity of potential! Physically, the discontinuity represents the work integral through the double layer. But now let us look at the same situation from another point of view. The only directly observable quantity is the field and this field has no discontinuity, either off the disc or across it. It was expressed as - grad 0 for some 0, so it is irrotational. If originally its only sources were the surface divergences on the two discs no other sources can have arisen in the process of going to the limit of fig. 9(b). But in this limit it is even continuous across its only possible surface of discontinuity. What sort of a field can it be? 131

Boundary behaviour of fields

11

The answer is implicitly contained in the discussion conducted on p. 127 in the preceding section. There we met a field g, (17), that was intrinsically solenoidal but also irrotational except for its curllike singularity on the z-axis. If a scalar potential was used to describe it, it had to be multi-valued - and the work integral of the field taken round any curve encircling the axis was non-zero. --------------

r

Fig. 10. Vanishing (C1) and non-vanishing (C2) work integrals of a vector field that has a `line curl' on C.

The existence of such fields is the clue we need. In saying that the field we are discussing now is not discontinuous across the double layer we are ignoring what happens on the boundary curve C of the disc that carries the double layer. Excluding the unrealistic case of fig. 8, some boundary must exist and it is evident without any further mathematics that this boundary stands in exactly the same relation to the field f as the z-axis for the field g (17). For clearly, remembering the irrotational nature of the fields, a work integral around any curve C1 not linked with C (see fig. 10) will be zero while a work integral round a curve C2 would cross the double layer bounded by C and can therefore only vanish if - AO, the work of going through the double layer, is included. Thinking of the same field but without the double layer the work integral round C2 will thus be equal to 0O while the same integral round, e.g. C3, a curve that encircles C twice, will equal 2AO and so on. Thus the physical picture of f need not refer to the double layer at all - and the original double layer can be replaced by any other bounded by the same curve C and having the same `dipole density'. The important thing for the field is what happens to it near C and we see that in the 132

4

Discontinuities of scalar fields

terms of section 2, it has a circulation about C. Physically such an f describes the magnetic field of a steady current flowing around C just as (17) describes the magnetic field of a current flowing along the z-axis. One can call such a field `irrotational but with circulation', but it is physically better regarded as a special solenoidal field having no volume or surface curl but instead `line' curls - singularities - on curves such as our C. Another way of expressing the result of this section is to say that if we have an intrinsically solenoidal field, but one whose vortices are all line vortices, then it is almost possible to get away with a description of the field as an irrotational one - one with a scalar potential. We can write down such a potential - only it is not a single valued function. A particular geometric way of looking at the double layer potential is the subject of exercises D. 23 and F. 9. 4 Discontinuities of scalar fields

In this chapter so far all the discussion has been on vector fields, the only scalar field appearing being the scalar potential, an auxiliary in the description of an irrotational vector field. In the case of physical scalar fields the role of singularities is incomparably simpler, in fact almost trivial. The situation in which a scalar is discontinuous across a surface S can of course be described as the presence of a surface gradient normal to the surface, but unless there is also some ordinary gradient present the field must just be a constant on either side of S. So a scalar field posessing only a `surface gradient' is rather a useless concept. There is no analogue to the irrotational vector field with only surface sources or the solenoidal vector field with only surface vortices. The section immediately preceding this one dealt with the only really important example of a surface discontinuity in a scalar field that enters general theory, but point and line singularities occur frequently. There would be little point in discussing them in systematic detail.

133

Exercises F

The subject matter of section 1 of the preceding chapter is illustrated by a selection of exercises in the following, but since section 2 consists of an enumeration of special cases there is no scope for further exercises. However, exercise 10 relates to the end of that section. It is also related to exercise 9 which amplified section 3. 1.

A vector field f exists on either side of an infinite plane surface S. On each side the field is constant. On the two sides the magnitudes off are fl and f2 respectively and its directions n 1 and n2. The normal on S is k and its direction is into the half space in which f = f2 n2. The three vectors n1, n2 and k are coplanar. The angles al and a2 are defined by cos 01 = n1 k and cos 02 = n2 k. The discontinuity off on S is due to (i) a surface source density a, and (ii) a surface vorticity g. Prove that in case (i)

fl

and in case (ii)

f1

sin (O 1 - 02) sin 02

sin (02 - 01) cos 02

= f2 sin (01 -02) = a sin B1

_ = f2

sin (02 - 01) cos B1

=

IgI.

What is the role of the coplanarity condition in the two cases (i) and (ii)? 2.

The field fl of exercise D. 7 was tacitly assumed to be present throughout the whole of space. It was seen (exercise 17) to be irrotational and solenoidal but its magnitude increases at large distances from 0. In any application of an expression such as fl to a real field in physics the form of field fl could only be valid up to certain boundaries at or beyond which the physical field would necessarily have sources and/or vortices. Consider (separately) the three cases that f = fl for (i) x > 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
source density on r = R ; f is intrinsically solenoidal and has no vorticity other than a surface vorticity on r = R.

Show that the field in the interior of the sphere differs in the two cases only by an overall factor and find the surface source density or vorticity, as the case may be, in terms of r = r/r. 9.

Show that the expression S2(r) for the solid angle subtended by a surface S' at a point r given in exercise D. 23 provides the mathematical form for the potential of a uniform double layer of strength r, on the surface S', as described in the text: fi(r) =

0(r).

4a

Prove that an alternative way of describing the field f = - grad 0 is as f = curl a where r a(r)

dl

41rJC'Ir-rl

where C' is the rim curve of S'.

136

Exercises F 10.

Verify that the dipole field f = - grad 0 with 0 given by (11.23) can also be represented as f = curl a where

JC'a

41T

vx ( r )

(see exercise E. 20).

If this way of expressing f is used the interpretation off as the field of a doublet of equal and opposite sources is much less immediate than in terms of the original scalar potential. What alternative physical interpretation is more appropriate? The key to the answer lies in exercise 9 and in the approximate identity dl'

JC'

I('

1r-r( 2r1 JC' =VX

l r x C11)

(for large r) which will be proved in exercise G. 14.

137

12

Differentiation and integration of products of fields

In the preceding chapters the basic concepts of vector analysis have been explained, but we have yet to discuss many details which together constitute the art of using vector methods. This chapter as well as some parts of later ones is devoted to such details. Inevitably the discussion now becomes almost entirely mathematical, without reference to applications.

1 Differentiation One of the basic rules one learns in ordinary differential calculus is the product rule. The differential operations grad, curl and div of vector calculus frequently have to be applied to products of fields and the product rule has to be taken over and adapted. Several types of products of two fields occur; the fields may be scalars or vectors f, g, . . . , and the products may be scalar fields

0, fg or vector fields

of, fxg. We can form the gradient of either member of the first pair and either the curl or the divergence of either member of the second pair. There are thus six product rules to examine. We shall see that four of the six can be disposed of very simply, while the remaining two will lead us to some interesting new considerations. We know that the operators grad, curl and div are most simply defined in cartesian coordinates and begin our discussion in terms of these. Then, every one of the products involves taking a scalar or some vector component which may be the sum of several terms all

of the same structure: one component of o - either a/ax, a/ay or a/az - acting on the product of two field components. To give but one example, the first component of V x (f x g) is explicitly a sum of 138

Differentiation four terms:

1

a

(f192) -

a a

(f291)

-

a

a

(f391) +

(flg3 ). (1) aZ y Applying the product rule in vector calculus therefore simply amounts to using the ordinary rule several times in separate terms in various sums and then reassembling the terms in some order. In principle every one of the six expressions listed above could be evalay

aZ

uated in this way, but in at least five of the cases a little thought simplifies the problem greatly. If, for instance, we take the first term in (1) the rule gives

ay (,f1g2) = fl (ay g2) + g2(ay J 1)

(2)

Note that we have written the answer with the differentiated factor last in each term. What we have done in (2) can be described in words as follows: The derivative of a product is found by forming a sum of two terms. The first term is what the whole product would be if the first factor were a constant, the second similarly would be the whole product if the second factor were constant. Interpreting the product rule in this way and remembering to keep the factors in their correct order, we can write down most of the required expression very readily. Naturally the simplest case is (3) V(0i) = OVO + OVO. This is simply the product rule applied, in separate equations, with a/ax, a/ay and a/az. We next take

V.(of) = OV.f+f.V

(4)

.

This is again a very straightforward case; the only thing to be noted here is that the V operator forms a divergence in the first term on the right and a gradient in the second. We go on: (5) V x (of) = OV x f-f X VO. Here the minus sign in the second term must be noted. If f were constant that term would indeed be equal to the left side because of the anticommuting property of the vector product.

Last among the simple cases:

V (fxg) = 139

f)

(6)

12 Products of fields This result uses the properties of the scalar triple product (chapter 1). The three factors in the first term on the right are in non-cyclically permuted order, hence the minus sign. The four cases dealt with so far all allow transcription into our alternative coordinate independent notation:

grad (04i) = 0 grad i/i + ti grad 0 div (of)

= cp div f + f grad 0

curl (of)

_ 0 curl f - f x grad o

div(fxg) =

(7)

f.

The fact that the proof was conducted in cartesian coordinates does not prevent statement of the result without reference to any special coordinates. But this is precisely where a difficulty arises in the next case, which is also somewhat more complicated. We examine V x (f x g). One way of evaluating it is by using the identity (1.18). There will be four terms in all in the answer, two in which f is treated as a constant, two in which g is treated that way:

Vx (fxg) = fV

(8)

One could say that this is a complete evaluation, but could the result be added to the equations listed in (7)? The answer is no, because in the two terms - (f V)g and (g V) f, we are using the V operator in a form in which it produces neither the gradient of a scalar nor the curl or divergence of a vector. Let us see whether we can escape this dilemma by using not (1.18), but instead (1.19):

Vx (fxg) = fx (Vxg)-(fx V)xg

+(gx0)xf-gx(Vxf).

(9)

This gives us an expression alternative to (8), but we are no better off than before because the terms - (f x V) x g and (g x V) x f again involve the V operator in an unusual form. The unusual terms in (8) involve derivatives of one of the vectors in the direction of the other and those in (9) derivatives of one in a direction perpendicular to the other. These notions, especially the first, are by no means unimportant in developments of the vector language. But they are not so easy to relate at present to our differential operators grad, curl and div. For the time being we must therefore leave both (8) and (9) in the cartesian form. We have left to the last the second of the six listed product differentiations: V(f g). It is easy enough to write down the 140

Differentiation components of this gradient and it is also easy to see that application of the product rule to it does not simplify it. Nevertheless it is interesting to see whether it can be re-expressed somehow, even if not

1

simplified.

This can in fact be done in two different ways, each of which needs a little manipulation. Let us evaluate the sum f x (V x g) + g x (V x f) by (1.18). If we were to try to apply that relation separately to either of the above terms we would get into a difficulty of notation. How should one write the first term in (1.18) if b must stand behind a but before c? But if we take the sum of both terms our difficulty disappears and we can write or

fx(Vxg)+gx(Vxf) = fx (Vx g) +

(Vx f).

(10)

This is a form very similar to (8) with two of the four terms on the right not expressible in terms of grad, curl or div. In fact the same new terms as in (8) reappear in (10). The second expression for V(f - g) is found similarly but starting from the `new' construct (f x V) x g + (g x V) x f and again using (1.18)

or

(fx V)xg+(gx V)x f =

(11)

V)xg+(gx 0)x

V(f.g) =

Clearly neither (10) nor (11) is in any practical sense a simplification of V(f g) whereas at least (8), if not (9), may be called simpler than V x (f x g). Nevertheless there is some value in deriving all four identities. Combining (8) and (10) we get

(f V)g = J [grad (f g) - curl (f x g) + f div g - f x curl g--gdivf - g x curl f]

-fx curlg+gdivf -gx curl f]

(12)

(fxV)x g =

-fx curlg -gdivf+gx curlfl (gx V) x f = I[grad

(fxg)-fdivg

+fx curlg-gdivf-gx curl f]

.

The reader will notice that on the right hand sides of these equations it has been possible to express everything in terms of grad, curl and div. This shows that the new constructs - derivatives of one 141

Products of fields 12 vector field in a direction parallel or perpendicular to another - can with some effort be expressed in a form not restricted to cartesian coordinates. For the general case the point made is rather academic, but there is at least one special case contained in (12) which is very frequently used. In the first (or second) equation of (12) put f = g = v. This gives

(v V) v =

grad v2 - v x curl v.

(13)

We have used the symbol v for the vector because this identity is most familiar from its application to the velocity field v in fluid dynamics. In the basic equations of motion there the transformation (13) is standard. When in chapter 14 we turn to examine certain non-cartesian systems of coordinates we shall look again at the results just obtained. partly from a different aspect. We shall show that quantities like (f V)g and (f x V) x g can also be evaluated in non-cartesian coordinates by a different method and that that method can be much handier in practice. But it is worth stressing that these differential operations are a new element in vector analysis and that, at least in principle, (12) can be used to give them a coordinate-independent meaning.

2 Integration by parts The work of the preceding subsection was based almost entirely on the use of the ordinary rule for differentiating a product. That rule is very closely related to the familiar technique of integration by parts in integral calculus. We can write the ordinary product rule of differential calculus as dg

f dx

_

d

dx (fg)

_

df

(14)

g dx

and can immediately deduce that x2

J:: (f d) dx = (fg)

- JX2(g d

df)dx.

(15)

X,

(We are not concerned so much with the use of integration by parts for evaluating indefinite integrals.) Equation (15) becomes particularly simple if one can choose values x1 and x2 at which either f org (or both) vanish. In that case the rule amounts to `throwing the differentiation on to the other factor' - and this is accompanied by a 142

2

Integration by parts

change of sign: fX2fdx.dx = -

J gdf dx

(16)

X

(if fg=0forx=x1 andx=x2). A particular case of (16) deserves special attention: if f and g and their derivatives are known to tend to zero rapidly enough at ± 00 (we need not investigate the precise mathematical conditions) the limits of integration in (16) may be taken as - 00 and + 0. This case in particular - but also the general equation (15) - can be easily generalised to vector calculus. Mathematically it would be quite easy to develop generalisations connected with work integrals along lines or flux integrals over surfaces, but the important use in physical theory of integration by parts involves quantity integrals and the divergence theorem. So we use the second and fourth identities of (7) and can immediately generalise (15) to

fV 0divfdV = fs

ff.curlgdV

=

_j'

fv

(17) (18)

We note that the `integrated' term of (15) has a surface integral as its counterpart here - one of the three integrations in the volume integral has been performed in getting it. Equations (17) and (18) are often useful as they stand. Separating a surface integral may be of direct physical use. But the relations are of particular interest in establishing certain general theorems on integration over the whole of space. No real fields in physics extend over all space. Either one has to assume that they are bounded by certain surfaces of discontinuity of the kind studied in chapter 11 or, if such discontinuities are not present, the fields must die away rapidly in whatever direction one moves to infinity. Systems for which the latter is true can be called `isolated systems'. For these we can go to the limit of assuming that in (17) or (18) the volume of integration increases and is made to extend over all space. The surface S can then be pictured as being blown up like a balloon receding to infinity in all directions. It is then often possible to prove that the surface integral in (17) or (18) tends to zero. In this way one gets the identities 143

Products of fields

12

fV 0divfdV

=-

Jf.curlgdV

=

0dV

(19)

J, g- curl f dV.

(20)

fV

and

Here V- denotes that the integrals are to be taken over all space. As with (16) one can talk of `throwing the derivative over to the other factor'. It is noteworthy that there is a change of sign involved in (19) but none in (20). A particularly interesting special case exists both for (19) and for (20): Assume in (19) that

f = - grad and in (20) that

g = curl f. Then JV_

f2dV =

fV_ g 2 d V =

-J

V_

0divgrad 0dV

fv f curl curl f dV

(21) .

(22)

On the right hand side we see almost for the first time in this book some second derivatives of fields. These will be the theme of the next chapter. Identities of this type are of particular interest in electromagnetic theory. Both (21) and (22) are used to describe the field energy there - (21) for electricity, (22) for magnetism.

144

13

Second derivatives of vector fields; elements

of potential theory

1 The Laplace operator, Poisson's and Laplace's equations In the course of developments we have mentioned the identities and

curl grad

=0

(1)

div curl f

= 0.

(2)

At the end of the last chapter expressions occurred involving and

div grad

(3)

curl curl f.

(4)

If we search for other possible second derivative expressions involving only the operators grad, curl and div and either a scalar or a vector field, only one more can be added to the list: (5) grad divf. The most interesting thing about the quantity (3) is that it gives us, for the first time, a differential operation which takes us from one scalar field 0 to another, div grad 0. We can therefore conclude that div grad is a scalar operator. In its cartesian realisation, its scalar

character is very obvious:

(V' V)0 = 020

div grad 0 =

V2 ax2 = -+-+-. ay2 az2 a2

a2

a2

(6) (7)

Here we meet for the first time the Laplace operator V2, one of the most powerful and important tools of theoretical physics. It has extremely wide application and is both the starting point and the centre of the subject called potential theory. In (7) V2 is defined in terms of a set of cartesian coordinates, but it can always be translated to mean div grad, in which form its definition is coordinate independent. It is sometimes convenient to have a shorter but still coordinate-independent name for the Laplace operator and we put 145

Second derivatives; potential theory div grad

13

00.

(8)

or, in Cartesian coordinates,

o0 = V20.

(9)

(The distinction between 0 and its cartesian realisation V2 is by no means universally made.) The first and most fundamental problem of potential theory is how to find 0 if 0q5 is given. The equation

0q = -P

(10) is known as Poisson's equation and the problem is to find 0 if p is given, i.e. to invert or integrate Poisson's equation. Very often the physical situation is such that p vanishes except in small localised regions or even only on boundary surfaces. There-

fore the solution of

00 = 0

(11)

with appropriate boundary conditions is another central topic of potential theory. In this book we can only dwell on those aspects of potential theory that are most closely related to our main subject of vector analysis. The next section will be devoted to this and in the following section some special topics will be touched on. These remarks referred to scalar fields. What is the position for vector fields? First we may note that the operators (4) and (5) both convert one vector field into another one. Thus they too are scalar operators.t Let us examine their cartesian forms

graddivf = V(V.f) curl curl f = Vx(Vx f) _

(12) V)f.

(13)

Hence we find that

grad div f - curl curl f = V2 f

(14)

where again V2 is given by (7). If we want to give a meaning to V2 f, which is independent of a Cartesian representation, the definition must rely on (14). So for a vector field (8) should be replaced by

grad div f - curl curl f = A f

(15)

t One might suggest that the first-order operator curl deserves to be called a scalar operator because the equation g = curl f links two vectors. But that relation is of the form b = c X a rather than b = aa.

146

The Laplace operator or, in Cartesian coordinates, 1

Af = V2f

(16)

The fact that a separate definition of A appears necessary for scalar and for vector fields needs to be discussed with care. Why should an operation applied to a scalar and to a vector component appear to be the same in cartesian coordinates but not in other coordinates? We shall have to defer giving an answer to this question to the next chapter. Here we shall do no more than draw the reader's attention to this distinction. It is clear, though, from our results so far that any methods of solution of (10) or (11) which are performed in cartesian coordinates should be generally applicable also to the solution of or

Af = -j

(17)

Af =

(18)

0.

We can look at these equations as (10) or (11) three times over, once for each vector component. We see then that the problem of solving Poisson's equation must be important in the theory of vector fields as well as for scalar fields. In fact, the techniques of potential theory (which will be the subject of the next section) enable us to answer a central question that has been in the foreground throughout this book: `what features of a vector field define it completely?' It will be shown in the remainder of the present section that once one accepts that (8) and (17) can be solvedt it can be proved that: A vector field is completely determined if all its sources and all its vortices are given. We proceed to demonstrate this. Given a vector field f we can

take its sources to be determined by

divf=p

(19)

and its vortices by

curl f = j.

(20)

Instead of attempting to define f directly in terms of p and j we put t Mathematically (8) or (17) can only be solved uniquely after specification of the behaviour of the solution at boundary surfaces and at any singularities that may be present. To a physicist the differential equation itself deals with those sources and vortices that happen to be treated as continuous, while the other conditions fix the surface, line and point sources and vortices. In order not to make this exposition too formal, we assume in the main text that only continuous sources and vortices are present. 147

Second derivatives; potential theory

f = fl +f2 so that

curl fl = 0, div f, = p,

13

div f2 = 0

(21) (22)

curl f2 = j.

(23)

Thus f, is an irrotational field and f2 a solenoidal field and we can find a scalar potential for the former and a vector potential for the latter: (24) f, = - V0, f2 = curl a.

We know that 0 can only be defined up to arbitrary constant and a only up to an arbitrary gradient field grad X. We show now that this second arbitrariness allows us to select a vector potential which is itself solenoidal

div a = 0.

(25)

This may be proved as follows: Suppose the a' found originally is not solenoidal, then

div a' = i/i 0 0

(26)

and given a', 0 can be calculated. If we put then

a = a'+gradx

(27)

div a = Vi + div grad X

(28)

and to satisfy (25) we must have

OX = - >!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
0<0<27r,

-oo
(18)

and spherical polars are defined by

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

0
0<0
0<0<27r.

(19)

We treat both systems in parallel and obtain h

=

ar

=

ar

(cos 0, sin 0, 0)

(sin 0 cos 0, sin 0 sin 0, cos 0)

ho =

ar

a

(-rsin0,rcos0,0)

(20)

(r cos 0 cos 0,rcos0 sink-r sin 0)

ar

bz =

az =

(- r sin 0 sin 0, r sin 0 cos 0, 0)

(0, 0, 1) 163

14 Orthogonal curvilinear coordinates We leave to the reader to verify the orthogonality properties (1)-(4), merely noting that

hr=1, ho=r, hZ=1

_l

be = r,

h7 = 1,

(21)

hp = r sin 0.

It follows that

grad >L =

ao 1 _aO' a 30 _r

(ar r

az )

ar '

(23) A

1

r sin 0 a0 1

ar fZ,

1

(sin 0 fo) -

a

1a

r ar (rfe)

r 30

r sin 0 ao fe

rar (rf 1a

-ra

efr

(24)

div f =

1a

rar

(rfr)

1a

+r ao

(r2fr)+rsn0a0(sin0 fo)

r ar

fm

1

a

+ a

az

a

a

1

rsin0aipfr

la la-(rfm)---fr

rar

r sin 0 a0

curl f =

r ao

az

Sip

1

a0 '

'

(fzfb fr

(22)

a>!ila

a

-f0

+ r sin 0 30

fZ

.

2 Definition of A for scalars and for vectors. An alternative approach

If we now look back at (13.8) and (13.15) we see that we can define Aqi for an arbitrary scalar field iii and also A f for any vector field f, and the two definitions are different. We remarked on the circumstance at the time and proposed to explain it later. The time has now come to do so. First however let us note that this difference referred to is real. Whereas a combination of (8) and (15) of this chapter immediately gives us the very important relation O

= div

grad

[Tx 1

a(-hµh aP

hµh ax

hx

a (b bx a>! ax + aµ hµ aµl

+ a (bxbjA mill

av h av

164

(25) '

Fields with rotational symmetry the corresponding result for (13.15) would be tedious to put down; it gives a different result. How does one explain this? The answer is to be found in (5). What we have given in (12.8) and (12.10) are not the results of applying the curl and div operators to (5) but of applying them to the components fx, J, and f,,. But, in general, the factors that multiply these components are not constants. Geometrically the components refer to different directions at different points in space. The statement that can be taken over directly from cartesian to curvilinear coordinates is that 3

0=

+kaza iax+ja ay

= e,,

--+e hx aA 1

a

1

a

hµ aµ

1

a

+e, h av

(26)

whether applied to a scalar >!i or to a vector

f = exff +e,,fµ +e f

(5)

and that from this point of view 0 = V V in all circumstances. But now the left hand V acts partly on the e factors of the right hand V. Using these ideas, a different approach from ours can be developed quite consistently. We have chosen not to use it for two reasons. Firstly, in the proof of general identities the methods used in this book seem to us more direct and, secondly, the use of the unit vector symbols e is anyhow usually abandoned in tensor calculus, which is a generalisation of vector calculus. We have preferred throughout this book to keep the unit vector symbols in the background. It must be admitted however that for working out particular sums, in one of the standard polar coordinate systems, differentiation of the unit vectors is very simple and, combined with (5) and (26), becomes easy to handle. We shall illustrate these techniques in some of the exercises.

3 Fields with rotational symmetry. The Stokes' stream function A rather special application of curvilinear coordinates that nevertheless deserves attention relates to vector fields, particularly solenoidal fields, of a certain symmetry. They are fields with rotational symmetry about an axis. Either cylindrical or spherical polars are very suitable to describe such fields. In either system of polar coordinates these fields have components that do not depend on the azimuth 165

Orthogonal curvilinear coordinates

14

'

Fig. 1. A P-field.

Ci Fig. 2. A T-field.

angle 0. The fields will be of two kinds; we shall call them P-fields and T-fields.t t These names are introduced here for convenience of expression and analogy with the names'poloidal' and 'toroidal' which refer to a related, more general situation. We shall introduce these more general concepts in exercises H. 22 and H. 23 after this chapter. Our T-field could be defined as a '-independent toroidal field, our P-field, if solenoidal, as a 4-independent poloidal field. 166

Fields with rotational symmetry

3

P-field:

Cylindrical polar

Spherical polar

f = (fr(r, z), 0, f, (r, z))

f = (fr(r, 0), fe (r, 0), 0) (27)

g = (0,go(r, z), 0)

g = (0, 0,gm(r, 0))

T-field:

(28)

Using (23) and (24) we find

div f =

r ar

a

(rfr)

az

fz

(29)

r ar

(r2fr) + r sin 0 ae

(sin 0 fe )

divg=0 curl f =

and

(OfrfzO)

C°'0'Y a (rfe>-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
z = AM 1<X<°°, 0<0<21r.

Exercises H = a(A2 + 1)1/2(1 _µ2)1/2

(b)

--1<µ<1, 0<X<-, 0<0<21T. (Verify that, by a simple redefinition of the parameters, case (b) can be reduced to case (a) with p and z interchanged.) Establish the nature of the surfaces X = const., µ = const., and = const. Evaluate the b vectors and their magnitudes, show that they are orthogonal and express grad >Ji, curl f, div f and A> in both systems. 9.

10.

Show that in terms of each of the two curvilinear coordinate systems of the preceding exercise there exist solutions of AO = 0 of the forms 0 = a(A) and , = /i(µ). Suggest source distributions that would correspond to these fields, in particular to a(X), in both cases. Show that if, in terms of the coordinates (a) of exercise 8, ti = a(A) is a solution of AO = 0, i.e. if (A2-1)

da = C dX

where C is a constant, then another solution of A Ii is

= a1(A, µ) = µ[Aa(A) + C] and that if i = (3(µ) satisfies AO, i.e. if (1

then

-µ2)a= C,

0=01(A,t)=X443-C)

also satisfies AO = 0. Find also the corresponding solutions for case (b). 11.

A final example of an orthogonal curvilinear system of coordinates with rotational symmetry is defined as follows. p = 2a(Xp)"2,

z = a(A - µ)

(0 < A < O°,

0 < µ <

Prove that the coordinates p, 0, z, where 0 is the usual azimuth angle, are orthogonal and justify the name `parabolic' coordinates for this system. Evaluate hx, hµ, hcb and derive expressions for grad 0, curl f, div f and A0. Show that the functions i i = log A and 0 = log µ both satisfy AV/ = 0 and find what sources are associated with the fields - grad for these two cases. (Note the precise definition of parabolic coordinates is not always as above. For example, the parameters A' = A2 , µ = p2 may be used.) 12.

Show that if A4 = p, where p is a function of the radius vector r only, the solution can be written as 176

Exercises H >

=

1

r

+'

(r--r)rp(r)dr +

o

where 1 is any solution of O4i 1 = 0. If it is assumed that V1 depends on r only, what is the most general form possible for 1 1 ? 13.

(a) If O> i = - p, where p = Arn cos 0 and A is a constant, show that the general solution is lp = Crn+2 cos 0 +

except for n = - 1 when

,1i = Crlogrcos0+1G1 and for n = - 4 when tG = Cr-2 log r cos 0 +

1

where in each case C is a constant and any solution of 1 = 0. If 01 is restricted to the form >!i, = a(r) cos 0 what is the most general

form of 0? (b) Answer the corresponding questions given that

p = Arn(3 cost 0 - 1). What are the exceptional values of n in this case? 14.

Use the result of exercise 12 to re-derive the results of exercises F.6 and G.15.

15.

Using the general form of i found in part (a) of exercise 13 solve the following problem. Within a sphere r < R the irrotational field f has the source density p = Az, where A is a constant. Outside the sphere the field has no volume sources and there is no discontinuity (no surface sources or vortices) on r = R. Show that the scalar potential 0 of the field outside the sphere (if chosen so that ,(oo) = 0) is

=-115 A Rsr2 cos 8. Find also the potential in the interior of the sphere. 16.

The coordinate systems defined in exercises 8 and 11 are as well adapted as either kind of polar coordinate system to the definition of T- and Pfields and of Stokes' stream function. The translation of the relations summarised on p. 167 of the text to these coordinates need not be repeated since it is merely a specialisation of the general formulae to be obtained in exercises 8 and 11 with gX = gµ = 0 for T-fields and fo = 0 for P-fields. To complete the translation define the Stokes' stream function ' in terms of a T-field vector potential for a solenoidal P-field f and express j = curl fin terms of it for all three types of coordinates defined in exercises 8 and 11. 177

Exercises H 17.

What is the field line picture for the field for which the Stokes' stream function expressed in spherical polar coordinates is 41 = 1 - cos 0? Verify your conclusion by calculating the field from (14.3 3), and prove by use of (14.35) that this field is irrotational.

18.

Investigate the nature of the following (necessarily solenoidal) fields defined in terms of spherical polar coordinates by the following Stokes' stream functions

= r2 sin2 0,

r-1 sin2 0,

r3 sin2 0 cos 0,

r-2 sin2 0 cos 0.

Verify that these fields are also irrotational and determine scalar potentials from which they can consequently be derived. Compare your results with those of exercise 4 for the case that the YI are functions of 0 only. Deduce in analogy to the results found there that the above fields belong to a general category for which IF = rI+1 ZI(0)

or

'P = r-IZI(0)

(1 > 0).

Find, using (14.35), the differential equation satisfied by the Z1. Show that the ZI are proportional to sin 0 (dYI/d0) where the YI are those of the functions defined in exercise 4, which are independent of 0. 19.

Proceeding in analogy with exercise 13 show that if ' is the Stokes' stream function of a solenoidal field with vorticity

j = (0, O, j,),

jo = Ar" sin 0

then, except if n = - 1 or n = - 4, ' is of the form

' = ACrn+3 sin2 0 + '1 where 'I' is the stream function of an irrotational field. Determine C in terms of A. Show also that for the two exceptional values of n the first term in the above expression for must be replaced by Cr2 log r and Cr-1 log r respectively. 20.

In analogy with exercise 15 find the Stokes' stream function for the solenoidal field that has the vortex density

j = (0, 0, Ar sin 0) within the sphere r < R and no vortices elsewhere. Show that the stream function outside the sphere is 1

15

RS

A - sin2 0 r

and find the stream function within the sphere. 21.

In the theory of the motion of incompressible (div v = 0) viscous fluids the equations of motion reduce in a certain approximation to Av = grad X, where v is the velocity field vector and the right hand side summarises the 178

Exercises H action of the forces on the fluid. Show that if v is expressed as the curl of a vector potential a, then curl curl curl curl a = 0. For the case that v has the rotational symmetry of a P-field, translate this statement into an equation for the Stokes' stream function (use (14.35)). Find the values of n for which a stream function of the form Ar" sin2 0 satisfies the relation derived and show that by a superposition of these solutions a flow can be found which is uniform at large distances from 0 and which is zero at all points of a sphere r = R. 22.

Given a vector field of the special form fo = rX(r), where X is an arbitrary scalar field, we define gl = curlfo.

This is obviously a solenoidal field. In spherical polars it has the form

_ ax

ax

1

g' _ (°' sin 0 ao'

a0

A field of this form is known as toroidal. (Note that in the special case that X does not depend on the azimuth angle 0 we have rotational symmetry and g is what in chapter 14 was termed a T-field.) Defining f, = curl g1, we call f, a poloidal field. (In the language of chapter 14 it is clearly a generalisation of a solenoidal P-field.) Show that

fl =

1

a

r sin 0 a0 sin 0 30

Ir

L (sino a0)

a2 X

a

1

a

r sin2 0 aq52 ' 30 [r ar (rX)

ar (rXI

We further define fields g and f as follows

gn+1 = curlf, ,

fl =

Prove that all the g are toroidal and all the f (n > 1) poloidal. 23.

It can be proved by general methods of potential theory that (subject to certain mathematical conditions) a general solenoidal vector field can be expressed as the sum of a toroidal and a poloidal field, so that a general vector field (see section 1 of chapter 13) is a sum of an irrotational field - grad 0, a toroidal field and a poloidal field and thus determined by three scalars 0, X and . The proof of this fact is essentially a construction similar to that of section 1 of chapter 13, but relies on the use of spherical polar coordinates and the properties of the spherical harmonics defined in exercise 4. Rather than give the general proof we give the construction in a particular example: Let f be the field

f = (- 3x2 - 2y2 - z2 , 4xy, 2xz). This is essentially the field first introduced as f3 in exercise D.7 - a solenoidal field. 179

Exercises H Evaluate curl f, r f = and r curl f = Show that if it is possible to represent fin the form f = curl (rX) + curl curl (r x V)2,q = X.

(r Evaluate (see exercise 4) (r x V)Z

z y 1 Ly ) 1 Y = sin 0 a0 sine e + sin 0 a¢2

for the functions x, yz and - 9x3 + 16xy2 + 11xz2 . Prove that for suitable values of a and 0

t1 = a(- 9x3 + 16xy2 + 11xz2) + (3xr2 . Solve equations (A) proving that

x = Jyz

and

-i(9x3 +4xy2 +5xz2).

Verify that

f = curl (rX) + curl curl

180

(A)

15

Time-dependent fields

Throughout this book so far we have talked about fields as functions of position only. In most physical applications of this theory the fields are also functions of time. To physicists fields are dynamical things that both experience and effect interactions with other systems, so the study of their development in time is of great importance. As the time dependence of any physical quantity is usually governed by some basic dynamical law and the laws are different in different branches of physics, it might be thought that the ideas of vector analysis would be too general and all embracing to have anything to say about the time dependence of fields. To some extent this is true and we cannot here go far into dynamical problems. But there are two aspects of time dependence that can be very profitably discussed as aspects of vector analysis. It is fitting to conclude any study with a chapter on these topics, particularly because they provide an excellent illustration of how our general ideas can actually be used in a real physical context. In this chapter we shall talk of scalar and vector fields 0(t, r) and f(t, r) and, in addition to the differentiation symbolised by o, we shall frequently form derivatives of the fields with respect to time: ap/at and of/at. We assume that differentiation with respect to time does not alter the vector character of a field: like 0 itself, a /at is a scalar and, like f, of/at is a vector. .

1 The equation of continuity A physicist is often concerned with relating the amount of substance contained in a given volume at any instant to the amounts supplied to or from the space outside that volume. Let us describe an amount of substance by the density p(t, r). Then volume V contains the total amount 181

Time-dependent fields

15

fV p(t, r) dV. When one speaks of `substance' one normally has the picture of something that is permanently conserved - it cannot be created or annihilated but it may enter or depart through the boundaries of a volume. Sometimes substance might be taken to emerge from just one `point' (e.g. a small nozzle of a hose-pipe emitting liquid). Such a point would be called a point source, and mathematically the relevant fields would be singular there. But if the volume V contains no singularities, any diminution of total amount of substance in V must be accounted for by a flow out of V. Such a flow will be described by a vector function j, the current density, which is such that if S is the boundary of V,

fs j - dS is the total rate of flow out of V per second. Clearly the density p and the current density j must be related by

f pdV = Jj.dS. V

(1)

dt In words: if the amount of substance in V decreases there must be a net outward flux out of S. This implies that we take S and hence V to be fixed in space, so that the limits of integration on both sides are constant in time. In that case the time differentiation may be performed under the integral sign. Further, the right hand side of (1) is a clear candidate for application of the divergence theorem, so that

-fv a! dV = fv divjdV

(2)

and since V can be any source-free volume we must have

ap+divj = 0

(3)

in any source-free region. This is the equation of continuity. If

ap0 at we have a special case of the equation of continuity that brings us back formally to the situation of previous chapters. If (4) is true 182

(4)

1

then also

The equation of continuity

div j = 0.

(5)

We mention this apparently trivial case here because it may not be as trivial as it looks. Situations may occur in physics where not only (4) holds (representing stationary flow), but actually p = 0 everywhere and nevertheless j 0. The density (of charge) is for instance always zero in metallic conduction of electricity. A current may nevertheless exist - but by (5) it must be solenoidal. It is interesting that (5) is entirely consistent with j varying with time! If the substance whose flow is described by (3) is produced at certain sources and annihilated in certain sinks which may be points, lines or surfaces, (3) will be true in the space in between - which is V. But there are comparatively rare cases when continuous sources exist. These cannot be excluded from the volume V and must be dealt with differently. A physical example may be taken from radioactivity. Radioactive elements decay into other elements, among which radon gas is an important one. That gas in its turn decays into other materials. If p is defined to be the amount of radon in a volume then (3) is certainly not true. Instead we must say that the change in the amount of radon in V is partly accounted for by a possible current j of radon leaving (or entering) V but partly also by the appearance and disappearance of radon by radioactive decay, which may take place at any point in V. This can be expressed in the form

dt JvpdV =

Jj.dS-JqdV

(6)

where q is the scalar field of `source density' which must be determined by the special laws that govern the appearance and disappearance of the substance - in the case above the laws of radioactivity. If q(r) is positive at a point r there is a net production of material at that point. In this case (3) generalises to

ap + V-j = q.

(7)

We cannot here discuss the various forms that the source density q may take. It may be a known scalar field, it may be proportional to p or it may be quite a complicated quantity to define. In any case, the value of ap/at + V j may be looked upon as a measure of the non-conservation of whatever substance is under discussion.

183

Time-dependent fields 15 2 Time-dependent relations involving the velocity field v(r)

A vector field of considerable special importance in theoretical physics is the velocity field v(r). Not surprisingly it is one of the principal objects of study in fluid dynamics, but there are other phenomena that can best be understood in terms of such a field. Among them, the electromagnetic behaviour of moving bodies is specially important. Such physical laws as Faraday's law of induction are related to observation by studying material bodies in motion and it would be restrictive to consider only the motion of rigid bodies (for which v(r) would have a rather special form). A statement of the laws in full generality needs the use of the general velocity field, as in fluid dynamics. In describing the motion of a single particle one uses the timedependent position vector r(t). The velocity v(t) of that particle is dr/dt. But when we study flow in a continuous medium, v(t, r) is defined at a general point in space, which in the course of time is occupied by different particles. Therefore, although in fluid dynamics the connection between paths of individual particles and the velocity field can be established, it is not trivial. We shall not require it in our work and the velocity field will be considered as a thing in its own right. Even so, in questions such as conservation of momentum we shall be interested in comparing the value of some field quantity 0(t, r) or f(t, r) at two successive times at the points occupied at those times by the same fluid particles. If these two times are a small interval At apart, the two values of the field, 0 say, are approximately 0(t + At, r + v(r)It) and 0(t, r). Clearly after a short time the position of the particle originally at r will be r + v(r)At. It is natural to define the so-called substantive or convective derivative of 0 as DO = lim (8) Dt to of [¢(t + At, r + v(r)&t) - 0(t, 01,

1

or, taking the limit, as Do

e0 + (v(r) - V)O(t, r).

(9)

The second term, by which the convective derivative differs from the ordinary time derivative, is of a form discussed on p. 140. In structure and physical significance it is closely related to a quantity that one can define in terms of the path r(t) of a single particle through a field. The field at the position of the particle is O(t, r(t)) and its rate 184

Relations involving the velocity field v(r) of change is

2

a

do + Idr = dt at \\dt

114

(10)

V J

being taken at t, r(t) in both terms). As explained a little earlier v cannot in general be replaced by dr/dt in a continuum theory. The quantity Dq/Dt is a scalar field while do/dt of (10) is a function of t only. But the distinction of notation between DO/Dt and do/dt is not universally observed. The definition of the convective derivative can be applied immediately to vector fields also. The considerations leading to (9) are not at all changed if 0(t, r) is replaced by f(t, r); we can therefore consider D

Dt

=

a

+v(t,r)-V

at

to be the convective derivative operator, irrespective of what field function, scalar or vector, it is applied to. Of course this operator can only be defined if a velocity field exists.t Supposing that in a medium where motion is characterised by v(t, r) a curve C(t) consists of the same particles at all times, and supposing we are interested in the line integral

along this curve. What is its rate of change? If at time t the curve C(t) has the form r(X), then at time t + At it will be described by r(X) + itv[r(A)] . Therefore

d JC(t)f dl JX2f[t

giro At

+ At, r(k) + Otv(r(A))]

a [r(A) + Otv(r(A))] dA

f [t' r(A)] aA

r(A)dX

t In conduction of electricity through metals, to give but one example, a unique velocity field cannot be defined. There negative electrons move with one velocity, positive ions with another, usually zero. 185

Time-dependent fields

15

Fig. 1. Motion of a curve C in time At.

=

-r+f --v]dx.

(""z

But

v=

(rV)v

(11)

= (bv)v

(12)

and, using the techniques of chapter 12, a

so that

(13)

OA

f f dl fc(t)

D

1l

dt c(r> [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<+47r co

(two turns round the circle) to close the curve. The curve forms a single non-self-intersecting loop folded over on itself, rather like the rim of the Mobius strip of fig. 10.4. 4.

The remark in brackets indicates that this curve describes a possible motion of a particle subject to a force of the form

F = C(nix, n2 2y, n23Z)

- an anisotropic oscillator. For integer n, let N be the greatest common factor of the n1. Over any range 00 < 0 < 00 + 21r/N

the curve closes. The curve is enclosed in the box

-a2
-a1 <x
-a3
and if the integers are large (and N small) the curve wanders around in a complicated way inside the box. 5.

bx = (- sin 3A, - 61 /2 sin 2),, - 31/2 sin A), hX = 2 sin A(3 - 2 sin 2 A) = sin 3X+ 3 sin A,

L = 10/3. It needs very specially chosen numbers to make the calculation so simple! 6. 7.

Lf m,

[f2 +

2

(a)]

1 /2

dc.

Special case of preceding exercise.

L = 21/2 a eO-, hence answer. 197

hm = 21/2 a e0

h_ = const., r

Answers and comments 8.

Put p = [x2 + y2 11/2 ; (p - a)2 + z2 = b2 - the equation of a torus. The curve coils round this, making n turns before it closes: f2n

[(a + b cos n¢)2 + n2 b2 ] 1/2 dO.

0

9.

A closed curve consisting of six edges of a cube,

-a <x
and (- a, - a, - a). 10.

The typical edge can be defined in the form or as

r = A(rk-ri)+ri

(0
r = Ark + (1 - A)ri = lki.

The same edge traversed in the opposite direction is

r = Ari+(1-A)rk = lik. In this notation the closed curves asked for are: C1: 132, 143, 124, C3: 114,121,142,

C2: 141,134,113, C4: 123,112,131

The statement claimed is clearly true and this is mainly an exercise in tidy grouping. 11.

The point 01 = it is clearly the same as 02 = 0 etc. So the curve is closed. Each part of it is a semicircle perpendicular to a coordinate plane.

12.

Both bx and bµ are constant - hence normal to S is in constant direction n

13.

_ bxc

Ibxci' p

_ (a,b,c) ibxcl

If M=po, r = (a + poc) + X(b + Mod) - clearly a straight line. Similarly for X = Ao;

b,N = b+pd,

bµ = a+Ad

and the normal to the surface will in general be a linear function of X and p so that S is not a plane. 14.

The line X = constant passing through the point A = M = 0 (which is the origin) is r = (A, 0, JA2) and the line M = const. is r = (0, M, - Ip2 ), two 198

Exercises B intersecting parabolas. The first has a minimum at 0, the second a maximum, hence a saddle point.

=1f +1

A

15.

+1

[1 +X2 +µ2)1/2dAdµ.

1

Restriction ensures real coordinate z. The surface is part of the sphere's surface - its (upper) intersection with the square column - b < x G b,

-b
(b b

dAdµ

A = 4aJo fo [a2 -x2 16.

A cone of height b and base radius a. Flattened out, its area becomes a section of a circular disc of radius (a2 + b2)1/2.

A = 7r(a2 + b2)Or:

(a2 +ab21/2 ) =

lra(a2 +b2)1/2

dS = (- Aal cos 0, - Aal sin 0, Xa2 ) whence 2n

A = a(a2 + b2 )1/2 17.

o

1

dO fo X dA = 1ra(a2 +

b2)1/2.

See exercise 8. Proof that the surface is a torus is identical with calculation there (with no replaced by i).

bo = (a + b cos 0)(- sin 0, cos 0, 0)

b,y = M- sin 0 cos

- sin , cos 0, cos 0)

dS = Ibmx by I = b(a + b cos >!i), 18.

A = 47r2ab.

The boundary is

C: r(¢) = a(cos 0, sin 0, cos 20/2 1/2). What is said about the circle in the context of (2.14) applies here also. The curves r = aX(cos 01, sin 01, cos 201 /21 /2) for any constant 01 are the straight lines in S 1.

dS, =

21 /2 a2 X(- sin 20 sin 0 - I cos 20 cos 0,

sin 20 cos 0 - I cos 20 sin ¢, 1/2 1/2) dA dO

dS2 = a2 (- 4 cos 20 sin 0 sin 2¢ sin 0 - sin 20 cos 0 cos 20 cos 0, 4 cos 20 sin 0 sin 20 cos 0 - sin 20 cos 0 cos 20 sin ¢, sin 40) d0 dO

199

Answers and comments 19.

The four pieces of the two quasi-squares can be given as

0<x <1 0

C4: r = - bµ',

- 1 < µ' < 0.

The surface elements are

S1: dS = (bxc)(µ-2 -2Aµ+A2µ-3i2µ2) +(cxa)(A-X2 -2Aµ+Aµ2 -3A2µ2) + (a x b) (1 - A2 - µ2 - 3A2µ2 )

S2: dS = -2(bxc)(xµ2 +A2µ3)-2(cxa)(x2µ+A3µ2)

+(axb)(1-A-µ+Op -3A2µ-3Aµ2 + 3A2µ2 -2 X3M2 -2 A2µ3 ).

20.

This and the next exercises are included for extensive use in later contexts.

(1, 5, 5), (5, 0, 5), (i, 1, 5),

(a)

(0, 0, 0), (1, 1, 1), (0,

(b)

(k, k, 0), (1,1,1). In order given (read along horizontal rows) r = 1 (0, x, x2 ) ,

(c)

21.

(a)

r = 4 (2 + v, v2 , µ), 4(v,v2 +x,x2), 5(µ2,x,x2 +µ),

5(1 +µ2, 2, 1 +µ), 5(1 +v, 1+v2,2) 5(v, I +v2, 1) I(1+µ2, 1,µ), 5(1+v,v2,1). O(µ2,1,1+µ) 5 (2 + v, v2 + 1, µ + 1)

5(v+ 1,v2 +A, x2 + i)

5(2 + 1,x+ 1,x2 +µ).

In the sense implied in the definition of the edges under exercise 20

dl = (0, #, x) dA, (b)

5 (v, v2 , 0)

5 (M2, 0, p),

1(2,1 +A, 1 +A2), 5(1,x, 1 +A2), 511,1+A,A2),

(µ, 0, 1) dµ,

(#, v, 0) dv.

The same expressions hold for the other three triplets. The result of (a) shows that the pair F1 and F4 has identical surface elements dS as do the pairs F2 and F5 and F3 and F6. In that order:

dS = (- } v, J, µv) dµ dv,

(vx, - J X, *) dv dx,

(4, Xu, - Ill) dX dµ. (c)

dV = (* + Aµv)dx dµ dv.

(d)

L = fo ['a+x2] 112dx

1

[=AV5+Jlog (2+\/5))

The length is the same for all 12 edges. 200

Exercises B (e)

A = fo,

f' [A

+ 4v2 + µ2v2 ]' 2dµ dv

0

(the same value for all faces). t

i

(f)

V = 0

22.

23.

(} +Aµv)dAdpdv =

'

.

0

0

dV = (sin A sin µ sin v + cos A cos µ cos v) d'n dA dµ dv.

This is obviously a variant of exercise 21 (but much less tedious).

dV = (Ab+c,µc+a,va+b)dAdµ = (a, b, c) (Aµv + 1) dA dµ. 24.

By the standard method

dS = a2 (sin A, sin µ, 1) rail (3 - coszA - coszµ)liz sin 2X sin 2p dX dµ. M = a2 J 0 0

J

This happens to be an integral that can be evaluated simply to give M = 'j+$ U0 a2 (35/2 -2 7/2 + 1). Checking this would be useful practice. 25.

For a change, the volume is an ordinary cube, so that dV = dx dy dz

Q = AJ

J

'f + (2z2 - x2 _Y2 ) dx dy dz.

The simplest way to see the truth of the statement is to note that by symmetry the integral over x2 must be the same if calculated over y2 or z2 so that

Q = 21-1-1=0. The same idea serves to simplify the next part: noting that

r4 = x4 + y4 + z4 +2 (Y2 Z2 + 22x2 + x2 y2 ),

Q = 3(1 - a) f x4 dV - 6a f y2 z2 dV J

=

8[j-(jJ+J)a]

which becomes zero if a = 26.

.

Cylindrical:

r

= (r cos 0, r sin

hr = (cos 0, sin 0, 0) 201

z)

Answers and comments bo = (- r sin 0, r cos 0, 0) hz = (0, 0, 1). Surface element on a cylinder r = a:

h0xb2 = (r sin 0,r sin 0, 0) dS = r dr do. Volume element of cylinder:

(hr, h0, hz) = r dV = rdr d¢ dz. Spherical:

hr = (sin 0 cos 0, sin 0 sin 0, cos 0)

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

b0_ (-rsin0 sin0,rsin0 cos0,0). Surface element on sphere: be x h0 = r2 (sin2 0 cos 4), sin2 0 sin 0, sin 0 cos 0)

dS = r2 sin 0 dO do

(hr, he, hp) = r2 sin 0 dV = r2 sin 0 dr dO d¢ All these are the standard results that can be derived much more directly (chapter 14). See p. 163 of text. 27.

Assuming, as we always do, that the parametrisation makes d V positive in the range in which we use it, meant in exercise 23 that (a, b, c) > 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 <- 1, as (ii); for n = - 1, the field is indeterminate, its curl infinite at r = 0. (xii) For n = 1, the field is zero on the line b x r = 0 but its curl is indeterminate on the same line; for n < 0 the field and its curl are infinite on the line b x r = 0. (xiii) For n <- 1, as (xii); for n = - 1 the field is indeterminate, its curl infinite on the line a x r = 0. (xiv) The field and its curl are infinite on line b x r = 0. (xv) A special case of (xiv). 3.

(i) 3,

(ii) 0,

(iii) 3Ir-aIn + n(r2

(iv) n(a r - a2 )I r -aIn -2 (Note that the difference of (iii) and (iv) is (3 + n)Ir - aIn and that this confirms (ii)), (vi) 0, (vii) n(n + (v) (4 + (ix) alb r + (a b)(a r), (x) 0, (xi) (r, a, b), (viii) a - b, (xii) (a, b, r x b) = (a r)b2 - (a b) (b r), (xiii) 3 log ja x rI + 1, (xiv) 0 (Note that if a = k one would be calculating the divergence of (0, 0, log (x2 + y2 )1'2) which is obviously zero), 209

Answers and comments (xv) 0 (in the special coordinate frame the given field is

(- y, x, 0)/(x2 + y2 )), (xvi) 0 (in the special coordinate frame the given field is -- (x, y, 0)/(x 2 +y 2)). (iii) For n < 0, the field and its divergence are infinite at r = a. (iv) As (iii).

(v) For n = - 2, the field is indeterminate and its divergence is infinite

at r = 0. For n <- 2, both are infinite. (vi) The field is infinite at r = 0. (vii) n = - 3 is the previous case. For other n < 0, the field and its divergence are infinite.

(x) For n = - 1, the field is indeterminate at r = 0. For n <- 1, it is infinite at r = 0. (xiii) The field and its divergence are infinite on the line a x r = 0. (xiv) The field is infinite on the line a x r = 0. (xv) The field is infinite on the line k x r = 0. (xvi) As (xv). 4.

The gradient is (a) aF', (b) rF'/r, (c) - a x (a x r)F'/I r x a I. If in (c) a lies in the z-direction the gradient is proportional to (x, y, 0).

5.

Equipotentials satisfy

x2 + y2 + (z - c)2 = c2 .

x2 + y2 + z2 = 2cz,

They are spheres centred on the z-axis and passing through 0. The picture is one of rotational symmetry around the z-axis. Hence we need only examine the field lines of grad 0 in the plane y = 0. Such a line is given as being

x = all + cos 0),

z = a sin 0.

Hence

(x-a)2 +z2 = a2. It is a well-known property that for any a such a circle intersects all circles

x2 + (z - c)2 = C2 (i.e. all equipotential surfaces) orthogonally. Hence it is a field line of grad (A. To prove the same analytically: grad ¢ =

xy1 z

z

1x2+y2 2

2

2

z

Hence a field line in they = 0 plane satisfies dz

dx -

fz/fx

-

z2 - x2 2xz

With the r(O) given dx

_

d0

210

- a sin 0,

dz

d0

= a cos 0

Exercises E z2 - x2 _ sin 2 0 - (1 + cost 0) _ 2 sin 0 (1 + cos 0)

2xz 6.

- cot o =

dz

dx

dO

d0

_

dz

dx

As the equipotentials are sets of closed surfaces, one set per cell each set identical in shape with every other and with the cell boundaries themselves as the largest surfaces, the qualitative picture of the field lines can be easily visualised in general terms as roughly circular closed curves, in appearance very similar to the intersections of the equipotential surfaces with some plane parallel to a cell boundary. In the special case we have

fx :fy :fz = a tan ax : b tan by : c tan cz

_dxdydz dA dA dA

(A)

(A a' -i

0

We are given e.g.l 21

= sin ax(X).

-I 1

-

Hence

a

or

- sin ax = a cos ax -

/II

a cos ax

1

dx

A

dX

dx

dA

and the relation (A) is verified. On each field line the values of X run over a limited range up to A = « or A = 0 or A = y, whichever of the three constants is numerically smallest; the field line crosses into another cell, and because of the multi-valuedness of the inverse sine function the pattern of closed field lines repeats itself over the lattice. The qualitative description given above is confirmed. 7.

(a) f2 is clearly = grad ¢ for 4 _ (a r) (b r)(c r). No such 0 can be found for fl . That this is so follows from

curlfl = Oxft = bxa+cxb+axc

0

while V x f2 = (b x a) (c r) - (c x a) (b r) + cyclic permutations which taken together cancel. (b) The field defined in exercise D.3 is

f = - frlr so that

f = f grad r. Again a potential 0 exists and curl f = 0. 8.

Scalar potentials clearly exist for, in exercise D.5, (i) p r, (ii) 2r2 and (iv) (p r) (q r) and, since for the other cases curl f 0 0, the result of exercise D.6 was predictable. The divergences of (i), (iii) and (v) are easily seen to vanish. This makes the result of exercise D.10 predictable. 211

Answers and comments (If for any of the other cases equality of integrals along the particular paths or over the particular surfaces given had been found, this would not have been a contradiction, but merely an accident.) 9.

The potential is, very simply,

_ - I (axe + bye + cz2)

(f = - grad 0).

(The orthogonality of the field lines to the surfaces 0 = constant is easily verified!) 10.

Verification of the field's irrotational property means evaluating its curl:

curlf = (-8+8,-2+2,10-10) = 0. The coefficient of z in fy is the same as the coefficient of y on f2, etc. In such a linear field this symmetry is what determines the irrotational character - and makes it easy to spot the given form of 0. Note that the coefficients of x2 , y2 and z2 in 0 and the related coefficients in f, may have any values! The rest of the `solution' formed part of the question. 11.

f = (i) -a x rF'/r, (ii) a x rF', (iii) (a x r)a2F'lIa x r1, (iv) (a X r)(a r)F'/ja x r1. Thus the direction of each of the fields at any point r is ± a x r - perpendicular to r and to a, which means tangential to some circle centred at, and in a plane perpendicular to, the axis r = Xa. Hence the answer. (Note that the magnitude of the field has quite a different dependence on the position coordinates in the three cases.)

12.

(a)

div fl = 2x2,

divf2 = 0.

Therefore the fluxes through the two surfaces must be the same for f2 but need not be so for fl . (b)

div f1 = 3,

div f2 = 0.

Same conclusion as for (a). 13.

divfl = A(y2 +z2 +x2) = A r2 divf2 = 3Ar2. Hence the flux of f2 - 3f1 must be zero out of any closed surface (noting that neither of the fields has any singularities for finite r). The fluxes in exercise D.11 are through an open surface, S, the hemisphere. But it can be made into a closed surface S by adding the flat disc

x2 + y2 < a2 ,

z = 0.

On that surface the outward normal component is the negative third component, (fl )3 and (f2)3 respectively, which at z = 0 are both zero. Thus neither field has any flux through this base and the assertion of the exercise is true. 14.

Evaluation shows that curl f = f, which seems to belie the intuitive picture; but, although a detailed visualisation of this strange field is not easy, one 212

Exercises E can with effort see that for any field line the neighbouring field lines do have the property of `curling' round it in the sense discussed in the text. 15.

(a) Verification of curl f1 = 0 and div f1 = 0 is very straightforward.

(b)(i)h=f,c=g,d=b; (ii)a+c+i=0.

Part (i) is a straightforward generalisation of exercise 10:

0 = ax2 + Iey2 + Iiz2 + hyz + czx + dxy. For (ii), curl g = (cz + (a - e)x + by - A (i - a)x, dx + A (e - i)y + fz - J (a - e)y, hx + " (i - a)z + gy - J (e - i)z)

A(a - e) - A(i - a) = 31(2a - e - i) = a,

etc.

Hence curl g = f. Further

divg = dz -gy - bz+hx+cy -fx * 0. This g is not solenoidal. It will be shown in chapter 13 how one finds a new vector potential g = g + grad x which is solenoidal.

curl f

div f

fi

0

8zC

f2

0

0 0

16.

f3

A

C(4y, - 4x, 0) C(- 4y, 4x, 0)

8zC

Thus fl and f2 are irrotational fields, f2 and f3 are solenoidal, while f4 has neither of these properties. Note that f4 is a field with the same divergence as fl and the same curl as - f3, but f4 is not identical with

fl -f3 = C(4xz, 4yz, 0). At various points in the text the idea has been expressed that a field is determined by its sources and vortices. This exercise shows that such a statement cannot be accepted without qualifications (which relate to boundary conditions and will be discussed in chapters 11 and 13). Note that in the present example

f4 -(1 -f3) = f2

-

The fact that fields such as f2 which have no source density and no vorticity (and no singularities at any finite points) exist is just another aspect of the point made above. The scalar potentials (fi = - grad 01) can be easily found as follows:

= C[-JZ3 -z(x2 +y2)] 02 = C[-2Z3 +Z(x2 +y2)]. 01

The vector potential given is 92 ; in the other case

g3 = C (- yz2 - yx2 , xz2 + xy2 , 0).

213

Answers and comments curl f

17.

div f

f

0

0

f2 f3

0

- 4xC

C(0, - 4z, 8y)

0

Thus the work integrals of fl and f2 (exercise D.7) are necessarily independent of path and the flux integrals of fl and f3 (exercise D.14) are likewise necessarily the same across all surfaces with common rim. As in the preceding question one can construct 01

= C(-X3 + 2xy2 + xz2 )

02

= C(-X3 + 2xy2 -xz2)

81

=

93

= C(0, x2z + 2y2z, - 2x2 y -y2z).

C(0,x2z-2y2z,-2x2y+y2z)

This example lacks the symmetries of the previous one. It is worth noting that vector potentials with one vanishing component can still be constructed. 18.

(a) (i) - a, (ii) - 3a(b r) - 3b(a r) + 2r(a b) (iii) - 5 [a(b r) (c r) + b(c r) (a r) + c(a r) (b r)] + 2r [(a r) (b c)

+r2 (b) The fields are the negatives of the fields calculated under (a). The special class referred to is that already exemplified by f2 in exercise 16 and fl in exercise 17 - fields which are everywhere irrotational and solenoidal.

divf

curl f

19.

(i) n(a x (ii) 0

r)r-(n+2)

(iii) (2 (iv) - (a x (v) (a x r)r-n

n)ar_n

b)r_n

- n(a Or-(n+2) (3 - n)r-(n+2)

+ nr(a r)r-(n+2) + n(a x r)(b

(vi) (3 - n)a(b r)r-" -r (a

0 r)r'(n+2)

(a b)r_n

+ nr(a r) (b r)r-(n+2 ) (vii) (a x r) (b r)r-n + (b x r) (a r)r-n

r)r-n

(r, a, b)r-n

(5 - n) (a r) (b r)r-"

The following linear combinations of fields of this type are both irrotational and solenoidal. (a) The field given by (ii) with n = 3;

fl = r/r3. (b) 3 X (v) with n = 5 minus (i) with n = 3;

f 2 = (3r(a r) - r2a] /rs . (c) 5 x (vii) with n = 7 minus the three terms: ((iv) for n = 5 + the same with a and b interchanged + (a b) x (i)); 214

Exercises E

f3 = The exercise immediately following indicates how such fields can be found systematically. 20.

As all the a-vectors are constant the order in which the various differentiations (a V) and (a x V) are performed is immaterial. So f(O) can be evaluated as V r

(a(1).V) while

r

...

r

3

f(k) = r

But

VX (a(k)X

Cr)1 = - Vra(k)X r r

r

11

- (v4) =

Y a(k)

(aX r - -

-

r(a).r)

r

=

Vx(a(k)Xr)

(a(k). V) r

.

So that

fQt)=f(o) Since in general 1

(ax

r

(bx

r

the various potentials are indeed in general different although they all define the same field. The definition in terms of 0 is the simplest. The first members of this class of fields have the scalar potentials 1

-, r

1

r

sr

3

and

- 3{5(a(1).r)(a(2).r)(a(3).r)-r2 (2). 0 (a (3). a('))

(3).

[(a(1).r)(a(2).a(3))

(1). a (2)

1

r

These expressions should be compared with the (everywhere non-singular) potentials of exercise 18 as well as with the working of exercise 19.

215

Answers and comments 21.

curl ft = ((z2 -y2 )x, (x2 - z2 )y, (y2 - x2 )z) = curl f2 div fi = 6xyz, div f2 = 0 (see exercise 15) f3 = ft -f2 = ((3x2 + y2 + z2 )yz, (3y2 + z2 + x2 )zx, (3z2 +x2 +y2)xy)

curlg=(-23y-y3Z,-x3z -z3x,-y3x-x3y) =f2

f, = f2 + f3 = curl g - grad ¢ where 0 = J(x3yz + y3zx + z3xy). 22.

curl g' =j(.y3Z-z3y,-23x-x3Z,-x3y-y3x)=f2 div g = I [x2 (y2 - z2) + y2 (z2 - x2) + z2 (x2 -y2)] = 0 div g' = Th (y4 - z4 + z4 - x4 + x4 - y4) = 0 g' = -6 ((2x2 - y2 - z2) (y2 - z2 )x, (2y2 - z2 - x2) (z2 - x2 )y,

(2z2 -x2 -y2)(x2 -y2)z) = I (x4y2 +y4z2 +24x2 -x422 -y4x2 -Z4y2).

Verifying that curl h = g" is a somewhat lengthy calculation. 23.

Mercifully we need only evaluate one component. Symmetry determines the others. X

(curl f)t = 3

X 3

y(y2 -x2) z(x2 -z2) ayr(z2 +x2)(y2 +z2) az r(y2 +22)(x2 +y2) a

a y _ ay r(x2 + z2)

+ az

x r 3r

\r(x2

+ y2)

_

1

+ z2

r(y2 + z2)

+

1

x2 + z2 y2 + z2 1

)1

r(y2y

+

2y2

(y2 + z2 )2

1

x2 + y2

2z2

y2 + Z2 + (y2 + Z2 )2 'x I y2 3r3 x2 + z2

X

3r3

I

1

(x2 + z2) (x2

- y2 y2+ z2

Z2

x2 + y2

- y2 + z2 Z2

-y2

+Y 2

+ z2

)

+(x2 +y2)(x2 +y2 +z2 -z2)+1 x r3

This resulting field is a very familiar one and can be much more simply defined as - grad (1/r) (it is both solenoidal and irrotational). If one chooses to define it as the curl of a vector potential there are many simpler forms for the vector potential than the one given here. This 216

Exercises E particular vector potential is the one with greatest possible symmetry. It has `cubic' symmetry (none of the six directions ± x, ± y, ± z is preferred). But even though curl f is the simple radial field with full spherical symmetry, no vector potential with spherical symmetry exists! 24.

(a)

I

J(p2

v pdV = a f

o Jo

+ v) (v2 + X) (X2 + µ) (Xµv + $) A dµ dv

313 - IM.

(b) Outward flux through the surface X = 1 is

fS f dS =bJ

Jo

(µ2 + v)(v2 + 1)(1 + µ)

X 1µ2v - v2 + a + pv) dp dv =

966-

The outward fluxes through the surfaces p = 1 and v = 1 are the same, by symmetry. The inward flux through the surface X = 0 is

fo f0' (p2 +v)v2µ(1µ2v-av2)dµdv

JS

=

TAZ .

Again the inward fluxes through µ = 0 and v = 0 are the same. Thus the total outward flux is 79

3x 960

_

1

_ 79x4-3 _

313

1280

1280

1280

in agreement with (a). 25.

Use divergence theorem, noting that

(a) div r = 3 and (b) div [r(a r)] = 4(a r). 26.

frxdi=- fS (dS X V) X r ja f, rxdl = - fS

dSV- r

a JS

a is arbitrary the statement is proved.

27.

(a) follows from the divergence theorem proper, the other three results from its vectorial generalisation. (a) div Or") = (n + 3)rn. . (b) Orn = nrrn-2

217

Answers and comments (c) V x (rrn) = 0. dF r (d) VF(r) = dr r 28.

Consider fc f dI and fs f dS for f = aO. They will be independent of path or surface of integration respectively if and only if curl f and div f respectively vanish. But

curl f = curl W) = - a x grad div f = div (acb) = a grad 0. Unless ¢ = const., grad 0 cannot be zero everywhere on the path (surface) of integration. Exercises F 1.

Putting

f1 = ali+f31k, f2 = a2i+(32k which is possible because of the coplanarity condition, we have (a) from (1.1.5) and (11.19), with n = k 132 - 01 = a,

a1 = a2

or

f2 cos e2 - f1 cos e 1 = a,

f2 sin 92 = f, sin 01

and the given result follows. In case (b) we have from (11.6) and (11.12) 132-13

= 0,

(«2-«1)j = g.

The second condition implies that g must be in the j direction and then

a2 -al = 191These relations are equivalent to

f2 Cos 02 - f1 cos 01 = 0,

f2 sin 02 -fl sin 01 = Ig1,

and the second given result follows. Whereas the continuity of the tangential components of field ensures that k, n1 and n2 are coplanar so that the coplanarity condition is redundant, this is not so for (b) for which in general the relations between f1 and f2 are more complicated. 2.

On x = 0: f1 = C(2y2 + Z2 , 0, 0).

Ony=0: f1 =C(-3x2 +z2,0,2xz). On z = 0: f1 =C(- 3x2 + 2y2, 4xy, 0). In the first case f1 is normal to the surface and since the field is supposed to vanish for x < 0, a = C(2y2 + z2 ), 218

Exercises F while condition (11.5) is clearly satisfied. In the other two cases the field is tangential to the surface (y = 0 and z = 0 respectively), condition (11.6) holds and g is in each case equal to the expression stated above for

f, on the boundary. 3.

If 1

1

r1

r2

condition (11.5') on p. 125 is clearly satisfied at points for which d - r = 0 so that r1 = r2. Then

f = - grad l =

r-d r+d 3

3 r2

r1

and

fn = f d/d =

(r2

2d +d 2 )3/2'

In the cartesian system indicated

- 2d

Q(x,.Y) = (x2 + y2 + d2)312

_

f=

r-d r+d 3

rl

+

3

r2

and on the boundary d f = 0 (because r1 = r2 there). Thus f is parallel to the boundary so that, in hydrodynamic terms, no fluid crosses the boundary. The physical picture is thus one of a source facing an impenetrable wall. All the fluid emerging from the source must be pictured as flowing to infinity but remaining in the one half space. (A more practical situation would be described, for example, by a source and a sink in the half space each with its own image source or sink.) Note that this field is neither intrinsically solenoidal (it has a point source) nor intrinsically irrotational because its discontinuity at the wall formally constitutes a surface vorticity. The surface vorticity vector g(x, y) is easily determined from f at z = 0. 1

f(x+.Y, 0 = (x2 +y2 + d2 )3"2 (x, y, 0) k x f = g(x, .Y) = (x2 + y2 + d2 )3i2 (- y, x, 0). 5.

Note first that unlike the two preceding exercises this one uses the special cartesian frame from the start. This case can also be dealt with in terms of general vector symbols but not to any advantage 219

Answers and comments ay a2

a z-d+z+d

x

a az ay

a

fx

x2 + y2az (

-x

r2

rl

-(z-d)2

Iri

ri

x2 + y2

+

r2 -(z+d)2 3

and similarly

fy = 3+ 2 fz =

a

ax

a

ay

ay "x

z-d+z+d

=(

rl

x ax x2 + y2 a

r2

a

1

a

z-d(x2

1

x2 + y2

r1

y

ay x2 + y2

z-d z+d

x2 + y2 (x ax + y ay) ( r,

= 0+

a

+y2)

+

r2

+z+d(x2 r2

+2) y

z-d z+d fz =

3

r1

+

3

r2

This shows that the field is the same as in exercise 4. In terms of a the absence of a normal discontinuity off on z = 0 can be inferred without this differentiation by using (11.6'), i.e. the fact that a = 0 on z = 0. Thus for the field of this exercise one can maintain that there is some value in using the otherwise awkward vector potential representation. Formally it allows one to call the field f `intrinsically solenoidal' but this implies ignoring the simple source picture for the singularity at a. (With a change of sign in the second term one can immediately write down the vector potential for the field of exercise 3. But in that case this procedure cannot claim to have any usefulness.) 6.

As the only sources of this field are distributed with spherical symmetry we must have f = rF(r). Volume sources are absent, so

div f = 3F + rF' = 0. Only two solutions are possible F = 0 or F = Cr-3 . The solution f = Cr/r3 cannot apply inside the sphere as it implies a point source at the origin. So f = 0 for r < R. To have anything we must have f = Cr/r-3 outside with a = C/R2 on the sphere and a total source 220

Exercises F strength of 47rR2 Cr = 47rC. To an observer outside the sphere the spherical source is indistinguishable from a point source! Note that we have used

the fact that f is irrotational only in so far as we did not try to introduce a surface vorticity. In fact every field of the form rF(r) must be irrotational. A potential 0 is related to F by

F 7.

l dO = ----. r dr

On the sphere

n-fi = fn-k, nXfi= fnxk,

nxfe = -Xfnxk.

Thus X = - 1 ensures continuity of the tangential component of f (absence of surface vorticity) and X = # ensures continuity of the normal component (absence of a surface source density). The verification that the scalar and vector potentials given are correct repeats parts of exercises E.1 and E.2. It is immediately evident that q5 is continuous for X = - 1 and a for X = 1. The interest in this exercise lies in the fact that the field fe is just the dipole field discussed at the end of section 2 of chapter 11. It is here seen as being relateable to two entirely different situations near the origin. In electromagnetic theory the first situation relates to electric dipole fields, the second to magnetic dipole fields. These results are also closely connected with those of the following exercises 9 and 10.

fe = -grad

8.

= curl a

= Following the pattern of exercise 7 we examine

¢ = XC[(k-r)(1-r)-3'r2(k-1)J and

- grads _ a= curl a =

So with p = - X/3 these two fields are identical. Retaining the constant X, we call this field fi

nfe = C[3(kn)(1n)-(k1)] nfi = Ca[-2(kn)(1n)+ (k1)] nxfe = C[-(kxn)(1n)+(Ixn)(kn)J nxfi = CX[(kxn)(1n)+(lxn)(kn)]. Thus continuity of the tangential components requires X = 1 and continuity of the normal component requires A = - I. It is immediately clear that these are just the conditions for continuity of 0 and a respectively. 221

Answers and comments 9.

Exercise D.23 defines the solid angle as

S2(r) = f dS'

(r - r ) Ir-rI3

S

For a single dipole at r', rather than at 0 as in the text,

p (r-r')

fi(r) =

I r - r I3

47r

This makes the stated interpretation of r1 /47r obvious. Noting that the above integral is over a fixed surface and that r enters the integrand only as a parameter, we can calculate in terms of an arbitrary constant vector c: 41r JS

l r

47r

f S,dS

dS'

Ir-r13J c

,

c

4ir S'

Ir-r'I3J

Ir-r'I5 `

L

11.

I r- r' I f J

(Use exercise E.1 (v) and (xi) and exercise E.2 (ii) and (v).) Because the integrand is of the form Vx g(I r - /1) we can replace it by

- V'xg(Ir-r'I) so that _ r

f

4ir Js'

=-r

I\\ I

4rr JC'

dl'-

(

Oxc _ r- r' I

px

r 41rJ s'

IV L r f c x ' 41r C' Ir - r) c

c

(dS'x \\\

Ir - r'I /

dl'

Ir - rI

=

This is true for all c, hence result. 10.

The derivation of

f

_

p1

1

47r

r5

r3 J

from a vector potential can be found from exercise E. 20. The double layer field of exercise 9 and the present field should be indistinguishable if the double layer becomes very small in extension compared with the distance of r from any point on it. (It should also be the same as the external field in exercise 7.) But for such points an approximate expression for a is given. This is just of the required form Vx p/r, provided

p=Ifrxdl'.

In this way the field is associated with a singularity which, before going to the limit, is on a closed curve C'. In the same kind of approximation that lets two neighbouring opposite sources tend to a doublet, this 222

Exercises G singularity on j' contracts into the origin and the field becomes the doublet field, though now more naturally represented by a vector potential. This view of the doublet field (and of the field of a finite double layer) underlies the accepted picture of current magnetism. The finite double layer (or `magnetic shell') does as a substitute for a finite current loop and the small current loop may be treated as a dipole. Exercises G 1.

(a)

X(p2 +g2) =

(b)

X(j2 + g2) = (j curl b - b curl j) = div (b x j).

div(pg).

X(p2 -j2 +g2 -h2) =

2.

div(pg+jxh); - X div h = div curl j = 0 hence

div h = 0; X curl g = curl grad p - u curl j = 0 - jXh hence

curl g = - ph; X div j = div curl h - p div g = 0 - /Ap hence

divj = -Xp; µ(p2 +j2 +g2 + h2) _ -pdivj -j grad g

= div (h x g - pj). 3.

-a[gx(Vxg)1 = -g[(aV)g] +(gV)(ag) _ -(aV)jg2 +V[(ag)gl -(ag)V g and the stated result follows. For the fields of exercise 1, divj = div b = 0. The relation given follows immediately.

4.

agdivg-a(gxcurlg) = X(ag)p+pa(gxh) (ah)divh-a(hxcurl h) = -Xa(hxj)-pa(hxg) (aj)divj-a(jxcurlg) = Xa(jxh)-p(aj)p pagradp = Xp(ag) --µp(aj) 223

Answers and comments Using the identity of exercise 3 and

pa grad p = gdiv (pea) all three relations follow immediately. 5.

P2 -f2 - (fl -f2 )2 = 2f1 (f2 -fi ) 2 grad 01 (f2 -fl) 2 div[01(f2 -fl)]

+201(divf2 - divfl). The last term is zero and, applying the divergence theorem, the first term reduces to a surface integral over S, whose integrand vanishes. 6.

P -f1 -(f1 -f2)2 = 2 curl a,- (f2 -fl) = 2 div[aIx(f2 - fl)] + 2a1-(curl f2 - curl f1 ). This time the tangential components of fl and f2 enter the surface integral, as needed in this case to make that integral vanish.

>G = div(lf)-zP divf

7.

fxgrad 41 = -curl (if)+lcurlf. The divergence theorem and, for the second case its vectorial generalisation (10.4), immediately provide the required answer. 8.

The technique of exercise 7 can be applied without change to each of the Vi and the volume integrals can be added together. The surface integrals coming from the separate Si in fact include integration over each interior piece of surface twice, one for each of the two Vi which that piece of surface separates. Initially these pairs of integrals appear with opposite normals. One must be defined to be the positive one. It is immaterial which is so chosen (see the related discussion on p. 152): redefining the normal and exchanging the meanings of I' and I_ leaves everything unchanged. The required result then follows immediately.

9.

The differential identities are, it is hoped, obvious if divj = 0 is used. Then J'"dV

= fs

and it is presupposed that j dS = 0. Since a is any constant vector

fV jdV=0. (Intuitively, since in this case the field lines of j must be closed curves in V, it is not surprising that the vector `sum' of j over V vanishes.) 224

Exercises G Similarly

J[(a.r)(b.j)+(b.r)(a.j)]

fV

dV

+I f The first integral on the right vanishes because j dS = 0, the second integral transforms by simple algebra to the required form. 10.

By Taylor expansion

Ir - rI

r

approximately. Successive terms in the expansion go as 1

1

r

r

1

each multiplied by some function of the angles. (These functions of the angles are the `spherical harmonics' which are very prominent in potential theory.) Thus the expansion is certainly good if r' <
r, = r1 -d

and

p, = p-qd.

Ifq* 0,choose d=p/q.Ifq=0,p1 =p. Note that formally this result does not depend on d being small; but for large d the approximation is a bad one! 11.

Using the same Taylor expansion applied to c a where c is a constant vector,

--b n r

r

where

n=

V

jdV'

b=f

and

.1

Using the results of exercise 9, n = 0 and

r

(V!xc)m r

=

m r

Hence the result. Since passage from 0 to a new origin 01 involves subtracting a term proportional to n, the `magnetic dipole moment' m is independent of the position of the origin 0. 225

Answers and comments 12.

The application of the result of exercise 7 to the case gives the answer almost immediately, but there is an important matter of sign. In that exercise the scalar was ti(r) and the volume integration related to the variable r. Now iti -+ 1/jr - r 'l and the integration relates to r'. But 1 1 ir - r'I = -G it - ri

Hence

dV'P(r) V

47rgXr) = IV,

1 ,

Ir -rj

-

f dV'V' P(r)

r - rI

Jv'

('

_+

JV'

dV'P(r) V'Ir-rI 1-

div'P(r')

JV' Ir-r'i

P(r) dS, div P(r) - J V' Ir - ri JS' Ir - r'I n'

dV'

dV,

which proves the statement given. 13.

Since for a single `dipole' m

1

r

r

41ra = Vx - = -m x Vwe can put here

47ra(r) = - J V'dV'M(r)xVit 1rI= fV, dV'M(r)xVit

= -J ,dV'V'x v

_

n ' X M(r)

- fS,

Ir-ri

M(r)

it - r'i ds,

+ fV,

r1

+ (' dV, V'xM(r') V' it - r'I V' X M(r)

Ir-rI

,

dV W.

Hence the statement is true. 14.

Having solved the earlier exercises in the group the student should need little explanation for this exercise although it deals with a line integral and not a volume integral. Using the Taylor expansion

Se'

Ir-r'I fe,

f

r

r

dl' = 0

(see end of section 1 of chapter 6), and 0

226

Exercises G for any vectors a, b independent of r'. This is analogous to exercise 9; the integrand is grad' [(a r') (b r')] . Thus

-V r

2

VY fC,

]

1 = `lax -- rC rxdl'

2r

J'

if

= a Vx 1 J_rxdl' 2r C' for any a. 15.

(a) For a point at r < R the total source strength at smaller radius is q(r3/R 3 ), hence r3

1

41r

q

_

1

R3 r2

q

r

R3

41r

q 41rr2

Scalar potentials are obtained by integration off = - grad 0. For r > 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 and is zero on z = 0. Thus if one imagines equal and opposite sources on these two segments an irrotational field with these and no other sources will be correctly described by this 0. (Again one can alternatively think of the sources as existing on two opposite hyperboloidal sheets with zero field `inside' the hyperboloids.) In case (b) a useful solution can be stated only after the slight redefinition of variables referred to the solution to the preceding exercise. It is also convenient to alter the sign of the potential, putting

(3(p) = - C tanh-' p. Then this scalar potential describes a distribution of sources over the part p > 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.

Ifr
_P f ll

o

r2

=-P

2

lr'dr +0,

r

-3/ r2

and ifr>R

Pfor(l-rJrdr+>fi1 r 3r R3/

2 = -.\RZ

We can absorb the (same) constant into >y, to get 2

-P R

=

2

2

P

6

+

or

3

__

or

3r

2l

q

3

4aR(2 q

2r

1

47rr

+01

Since we may certainly assume here that 0 1 has no angular dependence and a term C/r in it would signify an additional point charge, we get exactly the answer here as previously, if we postulate as always that 15.

This is just the case (a) of exercise 13 specialised to n = 1, so that our solution will be of the general form

(-A r3 + B

1

+ Cr l cos 6.

fl

10

r2

f)

The term with B cannot be present for r < R because there cannot be a singularity at r = 0. On the other hand, for r > R the constant A must be replaced by zero and C must also vanish if the field is to be finite at infinity. So we have

_

-Ar 10

241

3

+Cr IcosO

(r
Answers and comments B

= 2cos0

(r>R).

r

If there are to be no surface sources or vortices, a1p/ar and >G must be continuous at r = R. So

-AR3

B

10

R2

- + CR

-3AR2+C = -2B 10

R3

Hence

B = lARS

and

15

+

1

C = -AR2 6

1AR5 15

r

_ -AI to r3

COS 0

for

r>R

- 6 R2rl cos 0

for

r
The field outside the source distribution is thus a dipole field. Clearly the same technique could be applied to a source function with any power of r and also with angular dependences of p other than cos 0, e.g. with the angular dependence of case (b) of.exercise 14. One of the most widely used results of potential theory is that in this way, and by forming sums or infinite series of solutions of this type, the potential of a general source distribution p can, at least in principle, be constructed. 16.

Prolate spheroidal coordinates:

41 = -a(X2 - 1)1/2(1 _µ2)1/ego

f = curl g

a2I (A2 -112)1/202 - 1)12aµ

a 0)

1

(A2 -p2)1/ 2

curl curl g =

0, 0,

1

a3

(1 1

(Az

z)1/2 aA

(A2 - 1)12 µ2 > (1

- µ2 )1/2 aAi

µz)12 a2'i,1

+

(A2 - 1)1/2 am

Oblate spheroidal coordinates:

'If = - a(X2 + 1)1/2(1 -µz)1/ego 1

f = curlg = a2 242

1

1

a

0z -,2)1/2(X2 + 1)112 µa '

Exercises H

to curl curl g

1

1

a

1

1

(X2 - µ2 )1/2 (1 1

0,

a

0 b2)'

'

2

(X2 + 1)1/2 a2

1

3 -2-2 -I2) (1-µ 2)1/2

ax e

+ (1 -42)1/2 a2 W + 1)1/2 aµ2 Parabolic coordinates:

\11 = - 2a(Xu)1/2g0 f

axp

1

(X+ µ)1/2µ1/2 ax

x1 /2 aµ'

= Za2

curl curl g = (0, 0,

1

17.

1

= curl g

l1/2 a2,p

1

+

2a3

1

1-I

axe + A

0

4f1/2

a2

The equation 0 = const. defines a cone and

27r(1 - cos 0) = 47r sin2 I0 defines the corresponding solid angle. The field therefore has purely radial field lines and their total flux through any cone 0 = const. is proportional to its aperture. This precisely characterises the familiar point source field. We can easily verify that (14.3 3) gives f,. (1 /r2) and (14.3 5) gives

curl f=0. 18.

In the first two cases we get

f

f=

(- 2 cos 0, 2 sin 0, 0),

and curl f = 0 is readily verified by use of (14.35). Hence a scalar potential

, must exist such that

- a = - 2 cos 0,

r

aB

2 sin 0

giving VG = r cos 0 = z ; alp

2 cos 0

ar

r3

- cos 0 giving Il =

'

1 a>G

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
If the field of exercise 15 is compared with the present one the result, which we shall not write out, is that for 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)