book
The Mathematical Theory of Electricity and Magnetism (5th ed, 1927) — part 15 of 39
1 January 1927
is clearly a rational integral function of 6 of degree 3, the coefficient of 6s being — 1. It vanishes when 6 is equal to A., /x or v, these being the curvi- linear coordinates of the point x, y, z. Hence the expression must be equal,
identically, to
- (0 - ) (0 - fi) (6 - v).
Putting 6 = — a2 in the identity obtained in this way, we get the relation
x2 (b2 - a2) (c2 - a2) = (a2 + A) (a2 + fi) (a2 + v),
279-282] Confocal Coordinates 245
so that x, y, z are given as functions of A, /x, v by the relations
. (a2 + A)(a2 + /z)(a2 + ^) "- (b2-a2)(o2-a2) GtC <201>-
- To examine changes as we move along the normal to the surface A = cons., we must keep /* and v constant. Thus we have, on logarithmic differentiation of equation (201),
„ dx _ dX
*j — ==
x a2 + A, '
and there are of course similar equations giving dy and dz. Thus for the length ds of an element of the normal to X = constant, we have
(ds)2 = (dx)2 + (dy)* + (dz)*
tV \tc(a2 + VK&2-a2)(c2-a2)
= i /V7\ Y> (A-^)(A-y)
tK > (a2 + A)(62 + A)(c2 + A)'
The quantity ds is, however, identical with the quantity called -=— in
§ 277, so that we have
4(q2 + A)(62+A)(C2 + A)
- ~ (X-f,)(X-v) {^2)>
and clearly h2 and A3 can be obtained by cyclic interchange of the letters A, fi and v.
-
If for brevity we write
AA = V(a2 + A) (b* + A) (c2 + A), we find that
Ms 2AMA,
so that by substitution in equation (198), Laplace's equation in the present coordinates is seen to be
^r-^A^^}+a7r-x)A^^r3-4(^~^A^^r°
(203).
On multiplying throughout by AAAMA„, this equation becomes
(204).
246 Methods for the Solution of Special Problems [ch. viii
Let us now introduce new variables a, /3, 7, given by
"KdX
A*'
*-/'£'
f" dv
f") A.'
then we have — = AA — ;
and equation (204) becomes
d2V ?2V ?PV (/1_1/)|^ + (l/x)|Z + (x/i)|I = 0 (205).
Distribution of Electricity on a freely-charged Ellipsoid.
-
Before discussing the general solution of Laplace's equation, it will
be advantageous to examine a few special problems.
In the first place, it is clear that a particular solution of equation (205) is
V=A+Ba (206),
where A, B are arbitrary constants. The equipotentials are the surfaces a = constant, and are therefore confocal ellipsoids. Thus we can, from this solution, obtain the field when an ellipsoidal conductor is freely electrified.
For instance, if the ellipsoid
x2 1/2 z2 ,
h — -I — = 1
a2 b2 c2
is raised to unit potential, the potential at any external point will be given by equation (206) provided we choose A and B so as to have V =1 when A, = 0, and V=0 when X = 00 . In this way we obtain
["dX V = ^-^ (207).
Jo a;
The surface density at any point on the ellipsoid is given by
- J—- -?Z?±--h,— dn d\ dn dX
/,
dX
0 aI
dX
abc -r- 0 AA
.(208).
282-285] Confocal Coordinates 247
Thus the surface density at different points of the ellipsoid is proportional to hy.
-
The quantity h^ admits of a simple geometrical interpretation.
Let I, m, n be the direction-cosines of the tangent plane to the ellipsoid at
Fig. 79.
any point X, fi, v, and let p be the perpendicular from the origin on to this tangent plane. Then from the geometry of the ellipsoid we have
p2 = (a2 + X)l2 + (b2 + X)m2 + (c2+X)n2 (209).
Moving along the normal, we shall come to the point X + dX, fi, v. The tangent plane at this point has the same direction-cosines I, m, n as before,
but the perpendicular from the origin will be p+dp, where dp = -r-. To
obtain dp we differentiate equation (209), allowing X alone to vary, and so
have
2pdp = dX (I2 + m2 + n2) = dX.
Comparing this with dp = -j- , we see that hx = 2p.
Thus the surface density at any point is proportional to the perpendicular from the centre on to the tangent plane at the point.
In fig. 79, the thickness of the shading at any point is proportional to the perpendicular from the centre on to the tangent plane, so that the shading represents the distribution of electricity on a freely electrified ellipsoid.
It will be easily verified that the outer boundary of this shading must be an ellipsoid, similar to and concentric with the original ellipsoid.
- Replacing h^ by 2p in equation (208), we find for the total charge E on the ellipsoid,
ZTTUOC -r-
Jo AA Since I IpdS is three times the volume of the ellipsoid, and therefore
equal to 4nra.bc, this reduces to
e= 2
f
Jo
248 Methods for the Solution of Special Problems [ch. vm
Since the ellipsoid is supposed to be raised to unit potential, this quantity E gives the capacity of an ellipsoidal conductor electrified in free space.
The capacity can however be obtained more readily by examining the form of the potential at infinity. At points which are at a distance r from the centre of the ellipsoid so great that a, b, c may be neglected in
comparison with r, X, becomes equal to r2, so that AK = r2 , and
dX_2
Thus at infinity the limiting form assumed by equation (207) is
2
/;
V
f^dx' Jo Al
E and since the value of V at infinity must be — the value of E follows at
T
once.
A freely -charged spheroid.
-
The integral I -r- is integrable if any two of the semi-axes
Jo A\
become equal to one another.
If b = c, the ellipsoid is a prolate spheroid, and its capacity is found to be
2 2ae
E
r ^ — , kg (i±^y
where e is the eccentricity.
If a = b, the ellipsoid is an oblate spheroid, and its capacity is found to be
2 ae
E =
r-
Jo (a?
dX sin_1e
(a2 + ) (c2 + X,)i
Elliptic Disc.
-
In the preceding analysis, let a become vanishingly small, then
the conductor becomes an elliptic disc of semi-axes b and c.
The perpendicular from the origin on to the tangent-plane is given, as in the ellipsoid, by
p"2 =
x2 y2 z*
h— -I —
a4 b* c4
285-289]
Confocal Coordinates
249
and when a is made very small in the limit, this becomes
1_ a-
a*
V
2 —
-yi-*L
2 '
so that the surface density at any point x, y in the disc is proportional to
b* *, (210).
(l-y--Z-)
Circular Disc.
- On further simplifying by putting b = c, we arrive at the case of a circular disc. The density of electrification is seen at once from expression (210) to be proportional to
1-
c-
-*
and therefore varies inversely as the shortest chord which can be drawn through the point.
Moreover, when a = 0 and b = c, we have Ax = (c2 + X) Va,, so that
r^tan-^andf0^ J a AA c WxJ Jo Ax
IT C
Thus the capacity of a circular disc is — , and when the disc is raised to potential unity, the potential at any external point is
2
- tan-1 ,
t VvV
where \ is the positive root of
tf £ + £»
X c2 + X
-
Lord Kelvin* quotes some interesting experiments by Coulomb on the density
at different points on a circular plate of radius 5 inches. The results are given in the
following table :
Distances from the plate's edge
Observed Densities
l Calculated Densities
5 ins.
1
1
4
1-001
1-020
3
1-005
1-090
2
1-17
1-250
1
1-52
1-667
05
2-07
2-294
0
2-90
oo
Papers on Elect, and Mag. p. 179.
250 Methods for the Solution of Special Problems [ch. vni
Much more remarkable is Cavendish's experimental determination of the capacity of a circular disc. Cavendish found this to be -=-^= times that of a sphere of equal radius,
while theory shews the true value of the denominator to be tj or 1-5708!
- By inverting the distribution of electricity on a circular disc, taking the origin of inversion to be a point in the plane of the disc, Kelvin* has obtained the distribution of electricity on a disc influenced by a point charge in its plane, a problem previously solved by another method by Green. The general Green's function for a circular disc has been obtained by Hobsonf.
Spherical Bowl.
- Lord Kelvin has also, by inversion, obtained the solution for a spherical bowl of any angle freely electrified. Let the bowl be a piece of a sphere of diameter /. Let the distance from the middle point of the bowl to any point of the bowl be r, and let the greatest value of r, i.e. the dis- tance from a point on the edge to the middle point of the bowl, be a. Then Kelvin finds for the elec- tric densities inside and outside the bowl :
Pi
V
2tt2/
P
a'
a?
r2
— tan-
r
a
a-
)•
Po= Pi +
V 2*/-
Some numerical results calculated from these formulae are of interest. The six values in the following tables refer to the middle point and the five points dividing the arc from the middle point to the edge into six equal parts.
Plane disc
Curved disc arc 10°
Curved disc arc 20°
1-00 1-01 1-06 1-15 1-34 1-81
Po
Mean
Pi
Po
Mean
Pi
P0
1-00
1-0000
•91
1-06
1-0000
•86
1-14
1-01
1-0142
•95
1-08
1-0141
•88
1-15
1-06
1-0607
•99
113
1-0605
•92
1-20
1-15
1-1547
1-09
1-22
1-1542
1-02
1-29
1-34
1-3416
1-27
1-41
1-3407
1-29
1-56
1-81
1-8091
1-74
1-88
1-8071
1-67
1-94
Mean
1-0000 1-0010 1-0369 1-1106 1-2606 1-6474
- Papers on Elect, and Mag. p. 183. t Trans. Camb. Phil. Soc. xvni. p. 277.
289-292]
Ellipsoidal Harmonics
251
Bowl arc 270°
Bowl arc 340°
Pi
•013
•014 •018 •025 •045 •120
Po
Mean
Pi
PO
1-986
1-0000
•0001
1-9999
1-987
1-0009
•0002
1-9999
1-991
1-0041
•0002
2-0000
1-998
1-0118
•0004
2-0001
2-018
1-0316
•0009
2-0006
2-093
1-1060
•0042
2 0040
Mean
1-0000 1-0000 1-0001 1-0002 1-0007 1-0041
Discussing these results, Lord Kelvin says : " It is remarkable how slight an amount of curvature produces a very sensible excess of density on the convex side in the first two cases (10° and 20°), yet how nearly the mean of the densities on the convex and concave sides at any point agrees with that at the corresponding point on a plane disc shewn in the first column. The results for bowls of 270° and 340° illustrate the tendency of the whole charge to the convex surface, as the case of a thin spherical conducting surface with an infinitely small aperture is approached."
.(211),
Ellipsoidal Harmonics. 292. We now return to the general equations (205), namely
(^-v)^Hv-Vw+fr-f*)w=o
and examine the nature of the general solutions of this equation. Let us assume a tentative solution
V=LMN,
in which L is a function of X only, M a function of /a only, and N a function of v only. Substituting this solution the equation reduces to
1 7)2T
Since a is a function of A, only, y ^ is a function of \ only, and the equa- tion may be written in the form
(ft - v)f() + (v-) F(/x) + (-ft)& (v) = 0, where /, F and <I> are functions whose form we have to determine.
This functional equation must hold for all values of A,, fi, v. Putting /x = v we find that F(v) = <&(v), and since this is true for all values of v, F and <t>
252 Methods for the Solution of Special Problems [ch. viii
must be the same function. By a similar procedure, it follows that f must also be the same function, so that the equation can be written
(fi - v)f(X) + (v- X)/(/i) + (X - fi)f(v) = 0. To find the form of the function / we put X = 0 and obtain
/00-/(o)=/fr)-/«>\
fl v
Thus a function of /x, is equal to the same function of v, so that each must be a constant. Calling this B, and writing A for/(0), we find that
f(X) = A + BX. 293. Restoring its value to /(A) we see that we must have
~ = (A+B)L (212),
and similar equations, with the same constants A and B, must be satisfied by M and N.
Equation (212), on substituting for a in terms of X, becomes
a differential equation of the second order in X, while M and N satisfy equations which are identical except that fju and v are the variables.
,7(A,£j = (A+BX)L (213),
The solution of equation (213) is known as a Lamp's function, or ellip- soidal harmonic. The function is commonly written as E^(X), where p, n are new arbitrary constants, connected with the constants A and B by the
relations
n(n+l) = B, and (b2 + c2)p = - A.
Thus El (A.) is a solution of
d^={n(n + l)X-p(b* + c>)}L,
and a solution of equation (211) is
V=*X2E>i{)E><ji)E>i(v) (214).
p n
- Equation (213) being of the second order, must have two inde- pendent solutions. Denoting one by L, let the other be supposed to be Lu. Then we must have
dot
82 (Lu)
da?
^ = (A + BX)L,
= (A +Bx)Lu;
293-295] Ellipsoidal Harmonics 253
so that on multiplying the former equation by u, and subtracting from the latter,
r d-u _ dL du _
da? dot da
mi [do. f d\
Thus u =
L2 J Z2AX ' and the complete solution is seen to be
OL + DLffa,
where C and D are arbitrary constants.
Accordingly, the complete solution of equation (211) can be written as
V^(GnpE^) + DnpE^)J{^^^
(cnp"E*(v) + DnP"Ei{v){E^)Y^ .
This corresponds exactly to the general solution in rational integral spherical harmonics, namely
V=XZ(Gnpr"+Dnpr-^)
p n
(Onp'e?p* + Dnp'erW) (Cnp"P»(cos0)+Dnp"P*(cos6)).
Ellipsoid in uniform field of force.
- As an illustration of the use of confocal coordinates, let us examine the field produced by placing an uninsulated ellipsoid in a uniform field of force.
The potential of the undisturbed field of force may be taken to be V=Fx, or in confocal coordinates (cf. equation (201))
V (b*~ - a2) (c2 - a2) This is of the form V= GLMN,
where G is the constant F (62 — a2) ~~ - (c2 — a2) ~ * , and L, M, N are functions of A. only, fx only and v only, respectively, namely L = va2 + , etc.
Since V= LMN is a solution of Laplace's equation, there must, as in § 294, be a second solution V— Lu . MN, where
dX f dX u
D\K J (a2 + X) Ax
254 Methods for the Solution of Special Problems [ch. viii
The upper limit of integration is arbitrary : if we take it to be infinite, both u and Lu will vanish at infinity, while M and JV are in any case finite at infinity. Thus Lu . MN is a potential which vanishes at infinity and is proportional (since u is a function of X only) at every point of any one of the surfaces X = cons., to the potential of the original field. Thus the solution
V=CLMN+DLu.MN .(215)
can be made to give zero potential over any one of the surfaces X = cons., by a suitable choice of the constant D.
For instance if the conductor is X = 0, we have, on the conductor,
dX
u
H (a2
! + X)A> Thus on the conductor we have
V= LMN (g+dT , , d\ A ) . V Jo (a2 + X)AA/
The condition for this to vanish gives the value of D, and on substituting this value of D, equation (215) becomes
V=CLMN fl-
K I
d\
o (a2 + X)AX/ dX
JK (a3 + X)A, = Jtx\ 1 —
dX
o (a2 + X)AA/ dX
= ^(a2 + X)A. (gl6)
Jo (a2 + X) AA
This gives the field when the original field is parallel to the major axis of the ellipsoid. If the original field is in any other direction we can resolve it into three fields parallel to the three axes of the ellipsoid, and the final field is then found by the superposition of three fields of the type of that given by equation (216).
Spheroidal Harmonics.
-
When any two semi-axes of the standard ellipsoid become equal
the method of confocal coordinates breaks down. For the equation
+^+^ = i (2m
a2 + 6 &2 + e cn~ + e
295-297] Ellipsoidal Harmonics 255
reduces to a quadratic, and has therefore only two roots, say , ft. The surfaces \ = cons, and /j, = cons, are now confocal ellipsoids and hyperboloids of revolution, but obviously a third family of surfaces is required before the position of a point can be fixed. Such a family of surfaces, orthogonal to the two present families, is supplied by the system of diametral planes through the axis of revolution of the standard ellipsoid.
The two cases in which the standard ellipsoid is a prolate spheroid and an oblate spheroid require separate examination.
Prolate Spheroids. 297. Let the standard surface be the prolate spheroid
a2_t" 62 ~ '
in which a >b. If we write
y = •or cos (j), z = -ST sin <f>,
then the curvilinear coordinates may be taken to be , /u, <£>, where X, fj, are the roots of
x*
- 7^^=1 (218).
a2 + e fr + 0
In this equation, put tf—fr^c* and a2+0 = c262, then the equation becomes
x2 . CT2
If £2, rf are the roots of this equation in 0'2, we readily find that ,x2= £2t?2c2, so that we may take
x = c%v (219),
ct = cv/(1-P)(7?2-1) (220)
in which r\ is taken to be the greater of the two roots.
The surfaces £ = cons., 77 = cons, are identical with the surfaces # = cons., and are accordingly confocal ellipsoids and hyperboloids. The coordinates £, tj, <f> may now be taken to be orthogonal curvilinear coordinates.
It is easily found that h A /TEE j, -I /ZZI h 1
from which Laplace's equation is obtained in the form
8 in w3Fl d \n *,dr) 1 "'~?' 3°F n
256 Methods for the Solution of Special Problems [ch. vin
-
Let us search for solutions of the form
F=EH3>,
where 3, H, <£ are solutions solely of f , rj and <f> respectively. On substituting this tentative solution and simplifying, we obtain
a-r)(T-i)
iJU(W')i}-4i>-^
+i?5=o.
772-p L3S£l 9£) Ha77{w ' 877JJ 4> 302
As in the theory of spherical harmonics, the only possible solution results from taking
where — m2 is a constant, and m must be an integer if the solution is to be single valued. The solution is
<I> = G cos mfy + D sin m<fi (221).
We must now have
1 1 in _ « dM + 1 1 k. - 1) !5l = m'("'~^) 3 3f r «;8fJ+H3,lW 1;S,f (l-f>)(,= _l)
m- m*
"1-P ^-1' and this can only be satisfied by taking
together with
Jj^-^S-S^-0 (223)-
Equations (222) and (223) are identical with the equation already dis- cussed in §§ 273, 274. The solutions are known to be
B = AP^) + BQ^), n = A'P%(v)+B'Q%(v),
where s = n (n + 1) and P™, Q% are the associated Legendrian functions already investigated. Combining the values just obtained for 3, H with the value for <£> given by equation (221), we obtain the general solution
F=2S3H<S>
mn
= XS {AP™{%) + BQ£(®} {A'P2(V) + B'Q:(V)} {C cos mtf> + D sin m(/>}.
mn
At infinity it is easily found that
77 = 00 , f = .- = COS 0,
vV + ot2
while at the origin <q = 1, f = 0.
Thus in the space outside any spheroid, the solution P™(£) Q™(>/) is finite everywhere, while, in the space inside, the finite solution is Pjj '(£) P," l(rj).
298-301] Problems in two Dimensions 257
Oblate Spheroids.
- For an oblate spheroid, a2- b2 is negative, so that in equation (218) we replace b2- a2 by «2, so that k = ic, and obtain, in place of equations (219) and (220),
x = K^irj,
■& = K V(l - f2) (1 - rf).
Replacing iv by £ we may take £, f and <f> as real orthogonal curvilinear coordinates, connected with Cartesian coordinates by the relations
x = «f£
vr = * V(l-£2)(l + £2).
We proceed to search for solutions of the type
F=EZ<D,
and find that H, 3> must satisfy the same equations as before, while Z must satisfy
-||(1 + ^|}-rfi2Z+7l(ri+1)Z=a
The solution of this is
Z = A'P™(iO + B'Q%(ia and the most general solution may now be written down as before.
Problems in two Dimensions.
- Often when a solution of a three-dimensional problem cannot be obtained, it is found possible to solve a similar but simpler two-dimensional problem, and to infer the main physical features of the three-dimensional problem from those of the two-dimensional problem. We are accordingly led to examine methods for the solution of electrostatic problems in two dimensions.
At the outset we notice that the unit is no longer the point-charge, but the uniform line-charge, a line-charge of line-density cr having a potential (cf. § 75)
(7—2cr log r.
Method of Images.
- The method of images is available in two dimensions, but presents no special features. An example of its use has already been given in § 220.
j. 17
258 Methods for the Solution of Special Problems [ch. vin
Method of Inversion.
- In two dimensions the inversion is of course about a line. Let this be represented by the point 0 in fig. 81.
Let PP', QQ' be two pairs of inverse points. Let a line-charge e at Q produce potential Vp at P, and let a line-charge e' at Q produce potential Vp at P', so that
VP = C-2e\ogPQ;
Vjy = C'-2e'\ogP'Q'.
If we take e = e', we obtain FlG> 81>
Vj,-VP, = C"-2e\og^
= C"-2e\og^ (224).
Let P be a point on an equipotential when there are charges ex at Ql} e2 at Q2, etc., and let V denote the potential of this equipotential. Let V denote the potential at P' under the influence of charges e1} e2, ••• a^ the inverse points of Q1} Q2, .... Then, by summation of equations such as (224),
V- V= - S (2e log OP') + 2 (2e log OQ) + constants, or V= constants- 2 (Xe) log OP' (225).
The potential at P' of charges e1} e2, ... at the inverse points of Qlt Q2, ... plus a charge — 2e at 0 is
V+C+2($e)\og0P',
and this by equation (225) is a constant. This result gives the method of inversion in two dimensions :
If a surface S is an equipotential under the influence of line-charges elf e2, ... at Q1} Q2> ..., then the surface which is the inverse of S about a line 0 will be an equipotential under the influence of line-charges e1} e2, ... on the lines inverse to Q1} Q2, ... together with a charge — Xe at the line 0.
Tw o - dimensiona I Harmonics.
- A solution of Laplace's equation can be obtained which is the analogue in two dimensions of the three-dimensional solution in spherical harmonics.
In two dimensions we have two coordinates, r, 6, these becoming identical with ordinary two-dimensional polar coordinates. Laplace's equa- tion becomes
ld_foV\ d*V
302-304] Problems in two Dimensions 259
and on assuming the form
in which R is a function of r only, and © a function of 6 only, we obtain the solution in the form
V = "5°° (Arn + —J (C cos n<J>+D sin n<£).
M=0 V ? /
Thus the " harmonic-functions " in two dimensions are the familiar sine and cosine functions. The functions which correspond to rational integral harmonics are the functions
rn sin nd, rn cos n6.
In x, y coordinates these are obviously rational integral functions of x and y of degree n.
Corresponding to the theorem of § 240, that any function of position on the surface of a sphere can (subject to certain restrictions) be expanded in a series of rational integral harmonics, we have the famous theorem of Fourier, that any function of position on the circumference of a circle can (subject to certain restrictions) be expanded in a series of sines and cosines. In the proof which follows (as also in the proof of § 240), no attempt is made at absolute mathematical rigour : as before, the form of proof given is that which seems best suited to the needs of the student of electrical theory.
Fourier s Theorem.
- The value of any function F of position on the circumference of a circle can be expressed, at every point of the circumference at which the function is continuous, as a series of sines and cosines, provided the function is single-valued, and has only a finite number of discontinuities and of maxima and minima on the circumference of the circle.
Let P (/, a.) be any point outside the circle, then if R is the distance from P to the element ds of the circle r ^^p^/> a)
(a, 6) we have
/
2iraRi
a ds = 1.
This result can easily be obtained by inte- gration, or can be seen at once from physical considerations, for the integrand is the charge induced on a conducting cylinder by unit line- charge at P,
Fig. 82.
17—2
260 Methods for the Solution of Special Problems [ch. viii
Let us now introduce a function u defined by
u =
p-o? [F
h
ds
.(226).
2ira J R2
Then, subject to the conditions stated for F we find, as in § 240, that on the circumference of the circle, the function u becomes identical with F. Also we have
1_ 1
B? ~p + a? - 2a/ cos (0 - a)
1
(/- ael <ea>) (/- ae-^e-a))
f2-ai\f-aei {ea) a - f& {e~a) J
=7^2ii+2!(7rcosw('-a)}-
Hence u = = — / F \ 1 + 2 2 ( 4 ) cos w (0 - a)[ c?s
2ttJ 0=o 77" 1 /
n r8=2ir
e=o
Fcosn(0-a)dd,
and on passing to the limit and putting a =f, this becomes
^=^-| ^d<? + -$ Fcosn(0-ct)d0 (227),
expressing F as a series of sines and cosines of multiples of cl We can put this result in the form
00
F = F + X (an cos not + bn sin not),
where
1 f2lT an = - I F cos nddO, ttj o
hn = -** Famnddd,
•2tt
-
1 /"2,r
and F = ^-\ Fd9,
so that F is the mean value of F.
If F has a discontinuity at any point 0 = ft of the circle, and if F1} i£ are the values of F at the discontinuity, then obviously at the point 0 = fi on the circle, equation (226) becomes
u = ^(F1 + F2), so that the value of the series (227) at a discontinuity is the arithmetic mean of the two values of F at the discontinuity (cf. § 256).
304-307] Conjugate Functions 261
- We could go on to develop the theory of ellipsoidal harmonics etc. in two dimensions, but all such theories are simply particular cases of a very general theory which will now be explained.
Conjugate Functions. General Theory.
- In two-dimensional problems, the equation to be satisfied by the
potential is
fty ^y
w+W=0 (228);
and this has a general solution in finite terms, namely
V=f(x + iy) + F(x-iy) (229),
where / and F are arbitrary functions, in which the coefficients may of course involve the imaginary i.
For V to be wholly real, F must be the function obtained from f on changing i into — *. Let f (x + iy) be equal to u + iv where u and v are real, then F(x + iy) must be equal to u — iv, so that we must have V=2u. If we introduce a second function U equal to — 2v, we have
U+iV=-2v + 2iu
= 2i (u + iv)
= 2if(x + iy)
= j>(x + iy) (230),
where <f>(x + iy) is a completely general function of the single variable x + iy.
Thus the most general form of the potential which is wholly real, can be derived from the most general arbitrary function of the single variable x + iy, on taking the potential to be the imaginary part of this function.
- If (f) (x -f iy) is a function of x + iy, then i<j) (x + iy) will also be a function, and the imaginary part of this function will also give a possible potential. We have, however, from equation (230),
i<j> (x + iy) = i(U+iV) = -V+iU, shewing that U is a possible potential.
Thus when we have a relation of the type expressed by equation (230), either U or V will be a possible potential.
262 Methods for the Solution of Special Problems [ch. vm
-
Taking V to be the potential, we have by differentiation of
equation (230),
dU , .dV .,,. . ,
and hence
.fd_U .d_V \dx dx
_dU ,dV " dy dy'
Equating real and imaginary parts in the above equation, we obtain
dU=d_V
dx dy '
dU= _d_V
dy dx '
so that
djjdv dUd_y
dx dx dy dy
.(231).
This however is the condition that the families of curves U = cons., V = cons., should cut orthogonally at every point. Thus the curves JJ = cons, are the orthogonal trajectories of the equipotentials — i.e. are the lines of force.
Representation of complex quantities.
If we write
z = x + iy
so that z is a complex quantity, we can suppose the position of the point P indicated by the value of the single complex variable z. If z is expressed in Demoivre's form
z = reie = r (cos 6 + i sin 6), then we find that r = */x2 + y2 and 0 = tan-1 y~. The
x
Fig. 83.
quantity r is known as the modulus of z and is denoted by \z, while 6 is known as the argument of z and is denoted by arg z. The representation of a complex quantity in a plane in this way is known as an Argand diagram.
- Addition of complex quantities. Let P be z = x + iy, and let P' be z = x' + iy'. The value of z + z is (x + x') + i(y + y'), so that if Q represents the value z + z it is clear that OPQP' will be a parallelogram. Thus to add together the complex quantities z and z we complete the parallelogram OPP', and the fourth point of this parallelogram will represent z + z' .
308-311]
Conjugate Functions
263
The matter may be put more simply by supposing the complex quantity z = x + iy represented by the direction and length of a line, such that its projections on two rectangular axes are x, y. For instance in fig. 83, the value of z will be represented equally by either OP or P'Q. We now have the following rule for the addition of complex quantities.
To find z + z, describe a path from the origin representing z in magnitude and direction, and from the extremity of this describe a path representing z. The line joining the origin to the extremity of this second path will repre- sent z + z'
-
Multiplication of complex quantities. If
z = x + iy = r (cos 6 + i sin 6 ), and z' = x' + iy' = r (cos & + i sin 0'),
then, by multiplication
zz' = rr {cos (0 + 0') + ism (0 + 0')}, so that | zz' | = rr' = \ z | \z',
arg {zz') = 6 + 6' = arg z + arg z', and clearly we can extend this result to any number of factors. Thus we have the important rules :
The modulus of a product is the product of the moduli of the factors.
The argument of a product is the sum of the arguments of the factors.
There is a geometrical interpretation of multiplication.
In fig. 84, let OA = 1, OP - *, OP' = *' and OQ = zz\
Then the angles QOA, P'OA being equal to 6 + 0' and 9' respectively, the angle QOP' must be equal to 6, and therefore to POA.
Moreover
OQ OP OP' ~ OA '
each ratio being equal to r, so that the triangles QOP' and POA are similar. Thus to multiply the vector OP' by the vector OP, we simply construct on OP' a triangle similar to AOP.
The same result can be more shortly ex- pressed by saying that to multiply / (= OP') by z (= OP), we multiply the length OP' by | z \ and turn it through an angle arg z.
So also to divide by z, we divide the length of the line representing the dividend by | z \ and turn through an angle — arg z. In either case an angle is positive when the turning is in the direction which brings us from the axis x to that of y after an angle tt/2.
264 Methods for the Solution of Special Problems [ch. vni
Gonformal Representation.
-
We can now consider more fully the meaning of the relation
JJ + iV = <f> (as + iy).
Let us write z = x + iy, and W = U + iV, z and W being complex imaginaries, which we must now suppose in accordance with equation (230) to be connected by the relation
W=<f>(z) (232).
We can represent values of z in one Argand diagram, and values of W in another. The plane in which values of z are represented will be called the 2-plane, the other will be called the W -plane. Any point P in the .z-plane corresponds to a definite value of z and this, by equation (232), may give one or more values of W, according as <p is or is not a single-valued function. If Q is a point in the W -plane which represents one of these values of W, the points P and Q are said to correspond.
As P describes any curve S in the 2-plane, the point Q in the TT-plane
which corresponds to P will describe some curve T in the W-plane, and the
curve T is said to correspond to the curve S. In particular, corresponding
to any infinitesimal linear path PP' in the s-plane, there will correspond
a small linear element QQ' in the Tf-plane. If OP, OP' represent the values
z, z + dz respectively, then the element PP' will represent dz. Similarly the
dW element QQ' will represent d W or —,— dz.
Hence we can get the element QQ' from the element PP' on multiplying
it by -T- , i.e. by ^- <f> (z), or by <f>' (x + iy). This multiplier depends solely
&Z oz
on the position of the point P in the 2-plane, and not on the length or
dW direction of the element dz. If we express -5— or <£' (x + iy) in the form
dW
-j- =$' \x + iy) = p (cos % + isinx),
we find that the element dW can be obtained from the corresponding
dW element dz by multiplying its length by p or
dz
dW
, and turning it through
an angle %, or arg f ^- ) . It follows that any element of area in the 2-plane
is represented in the W -plane by an element of area of which the shape is exactly similar to that of the original element, the linear dimensions are p times as great, and the orientation is obtained by turning the original element through an angle %.
312-315] Conjugate Functions 265
From the circumstance that the shapes of two corresponding elements in the two planes are the same, the process of passing from one plane to the other is known as conformed representation.
-
Let us examine the value of the quantity p which, as we have
seen, measures the linear magnification produced in a small area on passing
from the ^-plane to the IP-plane.
dW We have p (cos % + i sin %) = ~y— = <f>' (x + iy)
= du .d_v
dx dx
dv .dV
dy dx
dV .dV
dy dx
ox J \dy J
dW
is called the "modulus of transformation.
so that p =
The quantity p, or We now see that if V is the potential, this modulus measures the electric
/TdVy /dVy
intensity R, or a / f -^— J -M j— J . Since R = 4nra, this circumstance pro- vides a simple means of finding <r, the surface-density of electricity at any point of a conducting surface.
-
If jr- denote differentiation along the surface of a conductor, on
which the potential V is constant, we have
dW dz
ds '
so that <r = -j— .ft = -j— -~- .
47T 47T OS
The total charge on a strip of unit width between any two points P, Q of the conductor is accordingly
hs=llQMds=l^-u^ <233>-
-
If, on equating real and imaginary parts of any transformation of
the form
U+iV=cf>(x + iy) (234),
Provenance
- Shelf
- Reference library
- Author
- James Hopwood Jeans
- Rights
- Published in 1927, before 1929, and therefore in the public domain in the United States.
- Collected By
- StanBot reference library