Essential Graduate Physics
EM: Classical Electrodynamics
and similarly for v. This means that the sum of second-order partial derivatives of each of the real functions u( x, y) and v( x, y) is zero, i.e. that both functions obey the 2D Laplace equation. This mathematical fact opens a nice way of solving problems of electrostatics for (relatively simple) 2D
geometries. Imagine that for a particular boundary problem we have found a function w(z) for that either u( x, y) or v( x, y) is constant on all electrode surfaces. Then all lines of constant u (or v) represent equipotential surfaces, i.e. the problem of the potential distribution has been essentially solved.
As a simple example, let us consider a problem important for practice: the quadrupole electrostatic lens – a system of four cylindrical electrodes with hyperbolic cross-sections, whose boundaries are described by the following relations:
2
a ,
,
electrodes
right
and
left
for the
2
x 2
y
(2.74)
2
a ,
,
electrodes
bottom
and
top
for the
voltage-biased as shown in Fig. 9a.
(a)
(b)
y
plane z
plane w
y
v
V / 2
a
V / 2
a
2
a
0
a 0 a
x
0 a
2
x
a
u
V / 2
a
V / 2
Fig. 2.9. (a) The quadrupole electrostatic lens’ cross-section and (b) its conformal mapping.
Comparing these relations with Eqs. (72), we see that each electrode surface corresponds to a constant value of the real part u( x, y) of the function given by Eq. (71): u = a 2. Moreover, the potentials of both surfaces with u = + a 2 are equal to + V/2, while those with u = – a 2 are equal to – V/2. Hence we may conjecture that the electrostatic potential at each point is a function of u alone; moreover, a simple linear function,
2
2
c u c c ( x y ) c ,
(2.75)
1
2
1
2
is a valid (and hence the unique) solution of our boundary problem. Indeed, it does satisfy the Laplace equation, while the constants c 1,2 may be readily selected in a way to satisfy all the boundary conditions shown in Fig. 9a:
2
2
V x y
.
(2.76)
2
2
a
so that the boundary problem has been solved.
According to Eq. (76), all equipotential surfaces are hyperbolic cylinders, similar to those of the electrode surfaces. What remains is to find the electric field at an arbitrary point inside the system:
x
y
E
V
, E
V
.
(2.77)
x
2
y
2
x
a
y
a
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These formulas show, in particular, that if charged particles (e.g., electrons in an electron-optics system) are launched to fly ballistically through such a lens, along the z-axis, they experience a force pushing them toward the symmetry axis and proportional to the particle’s deviation from the axis (and thus equivalent in action to an optical lens with a positive refraction power) in one direction, and a force pushing them out (negative refractive power) in the perpendicular direction. One can show that letting the particles fly through several such lenses, with alternating voltage polarities, in series, enables beam focusing.30
Hence, we have reduced the 2D Laplace boundary problem to that of finding the proper analytic function w(z). This task may be also understood as that of finding a conformal map, i.e. a correspondence between components of any point pair, { x, y} and { u, v}, residing, respectively, on the initial Cartesian plane z and the plane w of the new variables. For example, Eq. (71) maps the real electrode configuration onto a plane capacitor of an infinite area (Fig. 9b), and the simplicity of Eq. (75) is due to the fact that for the latter system the equipotential surfaces are just parallel planes u = const.
For more complex geometries, the suitable analytic function w(z) may be hard to find. However, for conductors with piece-linear cross-section boundaries, substantial help may be obtained from the following Schwarz-Christoffel integral
dz
w(z ) const
.
(2.78)
1
k
k 2
kN
(z x ) (z x ) ...(z
1
x
)
1
2
N 1
that provides a conformal mapping of the interior of an arbitrary N-sided polygon on the plane w = u +
iv, onto the upper-half ( y > 0) of the plane z = x + iy. In Eq. (78), xj ( j = 1, 2, N – 1) are the points of the y = 0 axis (i.e., of the boundary of the mapped region on plane z) to which the corresponding polygon vertices are mapped, while kj are the exterior angles at the polygon vertices, measured in the units of , with –1 kj +1 – see Fig. 10.31 Of the points xj, two may be selected arbitrarily (because their effects may be compensated by the multiplicative constant in Eq. (78), and the additive constant of integration), while all the others have to be adjusted to provide the correct mapping.
y
plane w
plane z
k
3
w
3
k
x
2
x
x
0
x
1
2
3
w2
w k
Fig. 2.10. The Schwartz-Christoffel mapping of
1
1
a polygon’s interior onto the upper half-plane.
30 See, e.g., textbook by P. Grivet, Electron Optics, 2nd ed., Pergamon, 1972.
31 The integral (78) includes only ( N – 1) rather than N poles because a polygon’s shape is fully determined by ( N
– 1) positions w j of its vertices and ( N – 1) angles kj. In particular, since the algebraic sum of all external angles of a polygon equals 2, the last angle parameter kj = kN is uniquely determined by the set of the previous ones.
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In the general case, the complex integral (78) may be hard to tackle. However, in some important cases, in particular those with right angles ( kj = ½) and/or with some points w j at infinity, the integrals may be readily worked out, giving explicit analytical expressions for the mapping functions w(z). For example, let us consider a semi-infinite strip defined by restrictions –1 u +1 and 0 v, on the w-
plane – see the left panel of Fig. 11.
plane w
plane z
v
y
w i
3
Fig. 2.11. A semi-
infinite strip mapped
w
w
1
2
onto the upper half-
1
0
1
u
x 1
0
x 1
x
plane.
1
2
The strip may be considered as a triangle, with one vertex at the infinitely distant vertical point w3 = 0 + i . Let us map the polygon on the upper half of plane z, shown on the right panel of Fig. 11, with the vertex w1 = –1 + i 0 mapped onto the point z1 = –1 + i 0, and the vertex w2 = +1 + i 0 mapped onto the point z2 = +1 + i 0. Since the external angles at both these vertices are equal to +/2, and hence k 1 = k 2 = +½, Eq. (78) is reduced to
z
d
z
d
z
d
w(z) const
const
const i
.
(2.79)
1/ 2
1/ 2
2
1/ 2
(z )
1
(z )
1
(z )
1
1
( 2 1/ 2
z )
This complex integral may be worked out, just as for real z, with the substitution z = sin , giving
-1
sin z
(
w z) const '
d
(2.80)
c sin 1- c .
z
1
2
Determining the constants c 1,2 from the required mapping, i.e. from the conditions w(-1 + i 0) = –1 + i 0
and w(+1+ i 0)= +1 + i 0 (see the arrows in Fig. 11), we finally get32
2
πw
w(z)
sin 1 z
i.e.
,
z sin
.
(2.81a)
2
Using the well-known expression for the sine of a complex argument,33 we may rewrite this elegant result in either of the following two forms for the real and imaginary components of z and w: 2
1
2 x
2
1
- ( x
)
1 2
2
y 1/2 ( x
)
1 2
2
y 1/2
u
sin
,
v
( x )
1 2
2
y 1/2 ( x )
1 2
2
y ,
cosh
1/ 2
2
32 Note that this function differs only by a linear transformation of variables from the function z = c cosh w, which is the canonical form of the definition of the so-called elliptic (not ellipsoidal!) orthogonal coordinates.
33 See, e.g., MA Eq. (3.5).
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EM: Classical Electrodynamics
u
v
u
v
x sin
cosh
,
y cos
sinh
.
(2.81b)
2
2
2
2
It is amazing how perfectly does the last formula manage to keep y 0 at the different borders of our w-
region (Fig. 11): at its side borders ( u = 1, 0 v < ), this is performed by the first multiplier, while at the bottom border (–1 u +1, v = 0), the equality is enforced by the second multiplier.
This mapping may be used to solve several electrostatics problems with the geometry shown in Fig. 11a; probably the most surprising of them is the following one. A straight gap of width 2 t is cut in a very thin conducting plane, and voltage V is applied between the resulting half-planes – see the bold straight lines in Fig. 12.
y
V / 2
V / 2
t
t
x
Fig. 2.12. The equipotential surfaces of
the electric field between two thin
conducting semi-planes (or rather their
cross-sections by the plane z = const).
Selecting a Cartesian coordinate system with the z-axis directed along the cut, the y-axis normal to the plane, and the origin in the middle of the cut (Fig. 12), we can write the boundary conditions of this Laplace problem as
V / ,
2
for x t, y
,
0
(2.82)
V / ,
2
for x t, y 0.
(Due to the problem’s symmetry, we may expect that in the middle of the gap, i.e. at – t < x < + t and y =
0, the electric field is parallel to the plane and hence / y = 0.) The comparison of Figs. 11 and 12
shows that if we normalize our coordinates { x, y} to t, Eqs. (81) provide the conformal mapping of our system on the plane z to a plane capacitor on the plane w, with the voltage V between two conducting planes located at u = 1. Since we already know that in that case = ( V/2) u, we may immediately use the first of Eqs. (81b) to write the final solution of the problem:34
V
V
2 x
1
u sin
.
(2.83)
2
2
2
( x t) y 1/2
2
2
( x t) y 1/2
The thin lines in Fig. 12 show the corresponding equipotential surfaces;35 it is evident that the electric field concentrates at the gap edges, just as it did at the edge of the thin disk (Fig. 8). Let me 34 This result may be also obtained by the Green’s function method, to be discussed in Sec. 10 below.
35 Another graphical representation of the electric field distribution, by field lines, is less convenient. (It is more useful for the magnetic field, which may be represented by a scalar potential only in particular cases, so there is no surprise that the field lines were introduced only by Michael Faraday in the 1830s.) As a reminder, the field Chapter 2
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leave the remaining calculation of the surface charge distribution and the mutual capacitance between the half-planes (per unit length of the system in the z-direction) for the reader’s exercise.
2.5. Variable separation – Cartesian coordinates
The general approach of the methods discussed in the last two sections was to satisfy the Laplace equation by a function of a single variable that also satisfies the boundary conditions. Unfortunately, in many cases this cannot be done – at least, using reasonably simple functions. In this case, a very powerful method called the variable separation,36 may work, typically producing “semi-analytical”
results in the form of series (infinite sums) of either elementary or well-studied special functions. Its main idea is to look for the solution of the boundary problem (35) as the sum of partial solutions,
c ,
(2.84)
k k
k
where each function k satisfies the Laplace equation, and then select the set of coefficients ck to satisfy the boundary conditions. More specifically, in the variable separation method, the partial solutions k are looked for in the form of a product of functions, each depending on just one spatial coordinate.
Let us discuss this approach on the classical example of a rectangular box with conducting walls (Fig. 13), with the same potential (that I will take for zero) at all its sidewalls and the lower lid, but a different potential V at the top lid ( z = c). Moreover, to demonstrate the power of the variable separation method, let us carry out all the calculations for a more general case when the top lid’s potential is an arbitrary 2D function V( x, y).37
z
c
V ( x, y)
Fig. 2.13. The standard playground for the
0
b
variable separation discussion: a rectangular box
y
0
with five conducting, grounded walls and a fixed
potential distribution V( x, y) on the top lid.
a
x
For this geometry, it is natural to use the Cartesian coordinates { x, y, z}, representing each of the partial solutions in Eq. (84) as the following product
line is the curve to which the field vectors are tangential at each point. Hence the electric field lines are always normal to the equipotential surfaces, so that it is always straightforward to sketch them, if desirable, from the equipotential surface pattern – like the one shown in Fig. 12.
36 This method was already discussed in CM Sec. 6.5 and then used also in Secs. 6.6 and 8.4 of that course.
However, it is so important that I need to repeat its discussion in this part of my series, for the benefit of the readers who have skipped the Classical Mechanics course for whatever reason.
37 Such voltage distributions may be implemented in practice, for example, using the so-called mosaic electrodes consisting of many electrically-insulated and individually-biased panels.
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X ( x) Y ( y) Z ( z) .
(2.85)
k
Plugging it into the Laplace equation expressed in the Cartesian coordinates,
2
2
2
k
k
k 0 ,
(2.86)
2
2
2
x
y
z
and dividing the result by XYZ, we get
1
2
d X
1 2
d Y
1 2
d Z
0 .
(2.87)
2
2
2
X dx
Y dy
Z dz
Here comes the punch line of the variable separation method: since the first term of this sum may depend only on x, the second one only of y, etc., Eq. (87) may be satisfied everywhere in the volume only if each of these terms equals a constant. In a minute we will see that for our current problem (Fig.
13), these constant x- and y-terms have to be negative; hence let us denote these variable separation constants as (-2) and (-2), respectively. Now Eq. (87) shows that the constant z-term has to be positive; denoting it as 2 we get the following relation:
2
2
2
.
(2.88)
Now the variables are separated in the sense that for the functions X( x), Y( y), and Z( z) we have got separate ordinary differential equations,
2
2
2
d X
d Y
d Z
2
X ,
0
2
Y ,
0
2
Z ,
0
(2.89)
2
2
2
dx
dy
dz
which are related only by Eq. (88) for their constant parameters.
Let us start from the equation for X( x). Its general solution is the sum of functions sin x and cos x, multiplied by arbitrary coefficients. Let us select these coefficients to satisfy our boundary conditions. First, since X should vanish at the back vertical wall of the box (i.e., with the coordinate origin choice shown in Fig. 13, at x = 0 for any y and z), the coefficient at cos x should be zero. The remaining coefficient (at sin x) may be included in the general factor ck in Eq. (84), so that we may take X in the form
X sin x .
(2.90)
This solution satisfies the boundary condition at the opposite wall ( x = a) only if the product a is a multiple of , i.e. if is equal to any of the following numbers (commonly called eigenvalues):38
,
n
with n ,
1 ,...
2
(2.91)
n
a
(Terms with negative values of n would not be linearly-independent from those with positive n, and may be dropped from the sum (84). The value n = 0 is formally possible, but would give X = 0, i.e. k = 0, at 38 Note that according to Eqs. (91)-(92), as the spatial dimensions a and b of the system are increased, the distances between the adjacent eigenvalues tend to zero. This fact implies that for spatially-infinite systems, the eigenvalue spectra are continuous, so that the sums of the type (84) become integrals; however, the general approach remains the same. A few problems of this type are provided in Sec. 9 for the reader’s exercise.
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any x, i.e. no contribution to sum (84), so it may be dropped as well.) Now we see that we indeed had to take real, i.e. 2 positive – otherwise, instead of the oscillating function (90) we would have a sum of two exponential functions, which cannot equal zero at two independent points of the x-axis.
Since the equation (89) for function Y( y) is similar to that for X( x), and the boundary conditions on the walls perpendicular to axis y ( y = 0 and y = b) are similar to those for x-walls, the absolutely similar reasoning gives
Y sin y,
,
m
with m ,
1 ,...
2
,
(2.92)
m
b
where the integer m may be selected independently of n. Now we see that according to Eq. (88), the separation constant depends on two integers n and m, so that the relationship may be rewritten as 1/ 2
2
2
n
m
.
(2.93)
nm
2 2
n
m 1/ 2
a
b
The corresponding solution of the differential equation for Z may be represented as a linear combination of two exponents exp{ nmz}, or alternatively of two hyperbolic functions, sinh nmz and cosh nmz, with arbitrary coefficients. At our choice of coordinate origin, the latter option is preferable because cosh nmz cannot satisfy the zero boundary condition at the bottom lid of the box ( z = 0). Hence we may take Z in the form
Z sinh z ,
(2.94)
nm
which automatically satisfies that condition.
Now it is the right time to merge Eqs. (84)-(85) and (90)-(94), replacing the temporary index k with the full set of possible eigenvalues, in our current case of two integer indices n and m: Variable
separation
nx
my
in Cartesian
( x, y, z) c sin
sin
sinh z ,
(2.95)
nm
coordinates
n,
1
a
b
nm
m
(example)
where nm is given by Eq. (93). This solution satisfies not only the Laplace equation but also the boundary conditions on all walls of the box, besides the top lid, for arbitrary coefficients cnm. The only job left is to choose these coefficients from the top-lid requirement:
nx
my
( x, y, c)
c sin
sin
sinh c V ( x, y) .
(2.96)
nm
nm
n, m 1
a
b
It may look bad to have just one equation for the infinite set of coefficients cnm. However, the decisive help comes from the fact that the functions of x and y that participate in Eq. (96), form full, orthogonal sets of 1D functions. The last term means that the integrals of the products of the functions with different integer indices over the region of interest equal zero. Indeed, a direct integration gives a
nx
n'x
a
sin
sin
dx ,
(2.97)
a
a
2 nn'
0
where nn’ is the Kronecker symbol, and similarly for y (with the evident replacements a b, and n
m). Hence, a fruitful way to proceed is to multiply both sides of Eq. (96) by the product of the basis functions, with arbitrary indices n’ and m’, and integrate the result over x and y: Chapter 2
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a
nx
n'x
b
my
m'y
a
b
n'x
m'y
c sinh c sin
sin
dx sin
sin
dy dx dyV ( x, y)sin
sin
. (2.98)
nm
nm
n, m 1
a
a
b
b
a
b
0
0
0
0
Due to Eq. (97), all terms on the left-hand side of the last equation, besides those with n = n’ and m =
m’, vanish, and (replacing n’ with n, and m’ with m, for notation brevity) we finally get a
b
4
nx
my
c
dx dyV ( x, y)sin
sin
.
(2.99)
nm
ab sinh c
a
b
nm
0
0
The relations (93), (95), and (99) give the complete solution of the posed boundary problem; we can see both good and bad news here. The first bit of bad news is that in the general case we still need to work out the integrals (99) – formally, the infinite number of them. In some cases, it is possible to do this analytically, in one shot. For example, if the top lid in our problem is a single conductor, i.e. has a constant potential V 0, we may take V( x,y) = V 0 = const, and both 1D integrations are elementary; for example
a
nx
a n
a
2 , for
odd,
n
sin
dx
sin d
(2.100)
a
n
n
0
0
for
,
0
even,
n
and similarly for the integral over y, so that
16 V
,
1
if
both
n
and
m
odd,
are
c
0
(2.101)
nm
2
nm sinh c
nm
otherwise.
0,
The second bad news is that even on such a happy occasion, we still have to sum up the series (95), so that our result may only be called analytical with some reservations because in most cases we need to perform numerical summation to get the final numbers or plots.
Now the first good news. Computers are very efficient for both operations (95) and (99), i.e. for the summation and integration. (As was discussed in Sec. 1.2, random errors are averaged out at these operations.) As an example, Fig. 14 shows the plots of the electrostatic potential in a cubic box ( a = b =
c), with an equipotential top lid ( V = V 0 = const), obtained by a numerical summation of the series (95), using the analytical expression (101). The remarkable feature of this calculation is a very fast convergence of the series; for the middle cross-section of the cubic box ( z/ c = 0.5), already the first term (with n = m = 1) gives an accuracy of about 6%, while the sum of four leading terms (with n, m = 1, 3) reduces the error to just 0.2%. (For a longer box, c > a, b, the convergence is even faster – see the discussion below.) Only very close to the corners between the top lid and the sidewalls, where the potential changes rapidly, several more terms are necessary to get a reasonable accuracy.
The related piece of good news is that our “semi-analytical” result allows its ultimate limits to be explored analytically. For example, Eq. (93) shows that for a very flat box (with c << a, b), n,mz n,mc
<< 1 at least for the lowest terms of series (95), with n, m << c/ a, c/ b. In this case, the sinh functions in Eqs. (96) and (99) may be well approximated with their arguments, and their ratio by z/c. So if we limit the summation to these terms, Eq. (95) gives a very simple result
z
( x, y) V ( x, y) ,
(2.102)
c
which means that each elementary segment of the flat box behaves just as a plane capacitor. Only near the sidewalls, the higher terms in the series (95) are important, producing some deviations from Eq.
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(102). (For the general problem with an arbitrary function V( x,y), this is also true in all regions where this function changes sharply.)
y b / 2
y b / 2
z / c 95
.
0
0.8
0.8
0.6
0.6
( ,
x y, z)
75
.
0
V 0
0.4
0.4
x
5
.
0
a
2
.
0
1
.
0
5
.
0
0.2
0.2
05
.
0
25
.
0
0
0
0
0.2
0.4
0.6
0.8
0
0.2
0.4
0.6
0.8
x / a
z / c
Fig. 2.14. The electrostatic potential’s distribution inside a cubic box ( a = b = c) with a constant voltage V 0
on the top lid (Fig. 13), calculated numerically from Eqs. (93), (95), and (101). The dashed line on the left panel shows the contribution of the main term of the series (with n = m = 1) to the full result, for z/ c = 0.5.
In the opposite limit ( a, b << c), Eq. (93) shows that on the contrary, n,mc >> 1 for all n and m.
Moreover, the ratio sinh n,mz/sinh n,mc drops sharply if either n or m is increased, provided that z is not too close to c. Hence in this case a very good approximation may be obtained by keeping just the leading term, with n = m = 1, in Eq. (95), so that the challenge of summation disappears. (As was discussed above, this approximation works reasonably well even for a cubic box.) In particular, for the constant potential of the upper lid, we can use Eq. (101) and the exponential asymptotic for both sinh functions, to get a very simple formula:
16
x
y
2 a 2 b1/2
sin
sin
exp
c z .
(2.103)
2
( )
a
b
ab
These results may be readily generalized to some other problems. For example, if all walls of the box shown in Fig. 13 have an arbitrary potential distribution, we may use the linear superposition principle to represent the electrostatic potential distribution as the sum of six partial solutions of the type of Eq. (95), each with one wall biased by the corresponding voltage, and all other grounded ( = 0).
To summarize, the results given by the variable separation method in the Cartesian coordinates are closer to what we could call a genuinely analytical solution than to a purely numerical solution.
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Now, let us explore the issues that arise when this method is applied in other orthogonal coordinate systems.
2.6. Variable separation – polar coordinates
If a system of conductors is cylindrical, the potential distribution is independent of the z-
coordinate along the cylinder axis: / z =0, and the Laplace equation becomes two-dimensional. If the conductor’s cross-section is rectangular, the variable separation method works best in Cartesian coordinates { x, y}, and is just a particular case of the 3D solution discussed above. However, if the cross-section is circular, much more compact results may be obtained by using the polar coordinates {,
}. As we already know from Sec. 3(ii), these 2D coordinates are orthogonal, so that the two-dimensional Laplace operator is a sum of two separable terms.39 Requiring, just as we have done above, each component of the sum (84) to satisfy the Laplace equation, we get
1
k
1
2
k 0
.
(2.104)
2
2
In a full analogy with Eq. (85), let us represent each particular solution k as a product R() F().
Plugging this expression into Eq. (104) and then dividing all its parts by RF /2, we get
d dR 1 2
d
F 0
.
(2.105)
2
R
d
d F
d
Following the same reasoning as for the Cartesian coordinates, we get two separated ordinary differential equations
d d
R
2
R ,
(2.106)
d
d
2
d
F
2
F ,
0
(2.107)
2
d
where 2 is the variable separation constant.
Let us start their analysis from Eq. (106), plugging into it a probe solution R = c where c and
are some constants. The elementary differentiation shows that if 0, the equation is indeed satisfied for any c, with just one requirement imposed on the constant , namely 2 = 2. This means that the following linear superposition
R
a
b ,
for ,
0
(2.108)
with any constant coefficients a and b, is also a solution of Eq. (106). Moreover, the general theory of linear ordinary differential equations tells us that the solution of a second-order equation like Eq. (106) may only depend on just two constant factors that scale two linearly-independent functions. Hence, for all values 2 0, Eq. (108) presents the general solution of that equation. The case when = 0, in which the functions + and – are just constants and hence are not linearly independent, is special, but in this case, the integration of Eq. (106) is straightforward,40 giving
39 See, e.g., MA Eq. (10.3) with / z = 0.
40 Actually, we have already performed it in Sec. 3 – see Eq. (43).
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R a b ln ,
for 0.
(2.109)
0
0
In order to specify the separation constant, let us explore Eq. (107), whose general solution is
c cos s sin, for 0,
F
(2.110)
c s ,
for
.
0
0
0
There are two possible cases here. In many boundary problems solvable in cylindrical coordinates, the free-space region, in which the Laplace equation is valid, extends continuously around the origin point
= 0. In this region, the potential has to be continuous and uniquely defined, so that F has to be a 2-
periodic function of . For that, one needs the product ( +2) to equal + 2 n, with n being an integer, immediately giving us a discrete spectrum of possible values of the variable separation constant:
n ,
0 ,
1 ,...
2
(2.111)
In this case, both functions R and F may be labeled with the integer index n. Taking into account that the terms with negative values of n may be summed up with those with positive n, and that s 0 has to equal zero (otherwise the 2-periodicity of function F would be violated), we see that the general solution of the 2D Laplace equation for such geometries may be represented as
Variable
separation
n
b
in polar
(,) a b ln
a
n
cos
sin
.
(2.112)
0
0
n
c
n s
n
n
n
n
coordinates
n 1
Let us see how all this machinery works on the famous problem of a round cylindrical conductor placed into an electric field that is uniform and perpendicular to the cylinder’s axis at large distances (see Fig. 15a), as if it is created by a large plane capacitor. First of all, let us explore the effect of the system’s symmetries on the coefficients in Eq. (112). Selecting the coordinate system as shown in Fig.
15a, and taking the cylinder’s potential for zero, we immediately get a 0 = 0.
(a)
(b)
E
0
y
R
0
x
Fig. 2.15. A conducting cylinder inserted into an initially uniform electric field perpendicular to its axis: (a) the problem’s geometry, and (b) the equipotential surfaces given by Eq. (117).
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Moreover, due to the mirror symmetry about the plane [ x, z], the solution has to be an even function of the angle , and hence all coefficients sn should also equal zero. Also, at large distances (
>> R) from the cylinder, its effect on the electric field should vanish, and the potential should approach that of the uniform external field E = E 0n x:
E x E cos,
for .
(2.113)
0
0
This is only possible if in Eq. (112), b 0 = 0, and also all coefficients an with n 1 vanish, while the product a 1 c 1 should be equal to (– E 0). Thus the solution is reduced to the following form B
(,) E cos
n
cos n ,
(2.114)
0
n
n 1
in which the coefficients Bn bncn should be found from the boundary condition at = R:
( R,) 0 .
(2.115)
This requirement yields the following equation,
B
B
1
E R
cos n cos
n ,
0
(2.116)
0
R
n
n2 R
which should be satisfied for all . This equality, read backward, may be considered as an expansion of a function identically equal to zero into a series over mutually orthogonal functions cos n. It is evidently valid if all coefficients of the expansion, including (– E 0 R + B 1/ R), and all Bn for n 2 are equal to zero. Moreover, mathematics tells us that such expansions are unique, so this is the only possible solution of Eq. (116). So, B 1 = E 0 R 2, and our final answer (valid only outside of the cylinder, i.e. for R), is
R 2
R 2
(,) E
cos E 1
x .
(2.117)
0
0
x 2 y 2
This result, which may be graphically represented with the equipotential surfaces shown in Fig.
15b, shows a smooth transition between the uniform field (113) far from the cylinder, to the equipotential surface of the cylinder (with = 0). Such smoothening is very typical for Laplace equation solutions. Indeed, as we know from Chapter 1, these solutions correspond to the lowest integral of the potential gradient’s square, i.e. to the lowest potential energy (1.65) possible at the given boundary conditions.
To complete the problem, let us use Eq. (3) to calculate the distribution of the surface charge density over the cylinder’s cross-section:
2
R
E
E cos
2
cos.
(2.118)
0
n surface
0
E
0
0
0
0
R
R
This very simple formula shows that with the field direction shown in Fig. 15a ( E 0 > 0), the surface charge is positive on the right-hand side of the cylinder and negative on its left-hand side, thus creating a field directed from the right to the left, which exactly compensates the external field inside the conductor, where the net field is zero. (Please take one more look at the schematic Fig. 1a.) Note also that the net electric charge of the cylinder is zero, in correspondence with the problem symmetry.
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Another useful by-product of the calculation (118) is that the surface electric field equals 2 E 0cos, and hence its largest magnitude is twice the field far from the cylinder. Such electric field concentration is very typical for all convex conducting surfaces.
The last observation gets additional confirmation from the second possible topology, when Eq.
(110) is used to describe problems with no angular periodicity. A typical example of this situation is a cylindrical conductor with a cross-section that features a corner limited by two straight-lines segments (Fig. 16). Indeed, we may argue that at < R (where R is the radial extension of the planar sides of the corner, see Fig. 16), the Laplace equation may be satisfied by a sum of partial solutions R() F(), if the angular components of the products satisfy the boundary conditions on the corner sides. Taking (just for the simplicity of notation) the conductor’s potential to be zero, and one of the corner’s sides as the x-
axis ( = 0), these boundary conditions are
F ( )
0 F ( ) 0 ,
(2.119)
where the angle may be anywhere between 0 and 2 – see Fig. 16.
(a)
(b)
R
R
0
Fig. 2.16. The cross-sections
0
of cylindrical conductors with
(a) a corner and (b) a wedge.
Comparing this condition with Eq. (110), we see that it requires s 0 and all c to vanish, and to take one of the values of the following discrete spectrum:
m
,
with m ,
1 ,...
2
.
(2.120)
m
Hence the full solution of the Laplace equation for this geometry takes the form
m/
m
a
sin
,
for R, 0 ,
(2.121)
m
m1
where the constants s have been incorporated into am. The set of coefficients am cannot be universally determined, because it depends on the exact shape of the conductor outside the corner, and the externally applied electric field. However, whatever the set is, in the limit 0, the solution (121) is almost41 always dominated by the term with the lowest m = 1:
/
a
sin ,
(2.122)
1
because the higher terms go to zero faster. This potential distribution corresponds to the surface charge density
41 Exceptions are possible only for highly symmetric configurations when the external field is specially crafted to make a 1 = 0. In this case, the solution at 0 is dominated by the first nonzero term of the series (121).
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a 1 /
1
E
.
(2.123)
0
n surface
0
const, 0
0
()
(It is similar, with the opposite sign, on the opposite face of the angle.)
The result (123) shows that if we are dealing with a concave corner ( < , see Fig. 16a), the charge density (and the surface electric field) tends to zero. On the other case, at a “convex corner” with
> (actually, a wedge – see Fig. 16b), both the charge and the field’s strength concentrate, formally diverging at 0. (So, do not sit on a roof’s ridge during a thunderstorm; rather hide in a ditch!) We have already seen qualitatively similar effects for the thin round disk and the split plane.
2. 7. Variable separation – cylindrical coordinates
Now, let us discuss how to generalize the approach discussed in the previous section to problems whose geometry is still axially-symmetric, but where the electrostatic potential depends not only on the radial and angular coordinates but also on the axial coordinate: / z 0. The classical example of such a problem is shown in Fig. 17. Here the sidewall and the bottom lid of a hollow round cylinder are kept at a fixed potential (say, = 0), but the potential V fixed at the top lid is different. Evidently, this problem is qualitatively similar to the rectangular box problem solved above (Fig. 13), and we will also try to solve it first for the case of arbitrary voltage distribution over the top lid: V = V( , ).
z
V (,)
l
Fig. 2.17. A cylindrical volume
0
R
y
with conducting walls.
x
0
Following the main idea of the variable separation method, let us require that each partial function k in Eq. (84) satisfies the Laplace equation, now in the full cylindrical coordinates { , , z}:42
1
k
1
2
2
k
k 0
.
(2.124)
2
2
2
z
Plugging k in the form of the product R() F()Z( z) into Eq. (124) and then dividing all resulting terms by this product, we get
1
d dR
1
2
d F
1 2
d Z
0
.
(2.125)
2
2
2
R d d F d
Z dz
Since the first two terms of Eq. (125) can only depend on the polar variables and , while the third term, only on z, at least that term should equal a constant. Denoting it (just like we did in the rectangular box problem) by 2, we get the following set of two equations:
42 See, e.g., MA Eq. (10.3).
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2
d Z
2
Z ,
(2.126)
2
dz
1
d dR
d
2
1
2
F 0
.
(2.127)
2
2
R
d
d
F
d
Now, multiplying all the terms of Eq. (127) by 2, we see that the last term of the result, ( d 2 F/ d2)/ F, may depend only on , and thus should equal a constant. Calling that constant 2 (just as in Sec. 6
above), we separate Eq. (127) into an angular equation,
2
d
F
2
F 0,
(2.128)
2
d
and a radial equation:
2
d R
1
2
d
R ( 2
) R 0 .
(2.129)
2
2
d
d
We see that the ordinary differential equations for the functions Z( z) and F() (and hence their solutions) are identical to those discussed earlier in this chapter. However, Eq. (129) for the radial function R() (called the Bessel equation) is more complex than in the 2D case and depends on two independent constant parameters, and . The latter challenge may be readily overcome if we notice that any change of may be reduced to the corresponding re-scaling of the radial coordinate . Indeed, introducing a dimensionless variable ,43 Eq. (129) may be reduced to an equation with just one parameter, :
2
d
Bessel
R
1
2
dR
1
R 0 .
(2.130)
equation
2
2
d
d
Moreover, we already know that for angle-periodic problems, the spectrum of eigenvalues of Eq. (128) is discrete: = n, with integer n.
Unfortunately, even in this case, Eq. (130), which is the canonical form of the Bessel equation, cannot be satisfied by a single “elementary” function. Its solutions that we need for our current problem are called the Bessel function of the first kind of order , commonly denoted as J(). Let me review in brief those properties of these functions that are most relevant for our problem – and many other problems discussed in this series.44
First of all, the Bessel function of a negative integer order is very simply related to that with the positive order:
J ( ) ( )
1 n J ( ) ,
(2.131)
n
n
enabling us to limit our discussion to the functions with n 0. Figure 18 shows four of these functions with the lowest positive n.
43 Note that this normalization is specific for each value of the variable separation parameter . Also, please notice that the normalization is meaningless for = 0, i.e. for the case Z( z) = const. However, if we need the partial solutions for this particular value of , we can always use Eqs. (108)-(109).
44 For a more complete discussion of these functions, see the literature listed in MA Sec. 16, for example, Chapter 6 (written by F. Olver) in the famous collection compiled and edited by Abramowitz and Stegun, available online.
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1
0.5
J ()
n
0
n
1
0
3
2
0.5
Fig. 2.18. Several Bessel functions
Jn() of integer order. The dashed
lines show the envelope of the
1
asymptotes (135).
0
5
10
15
20
As its argument is increased, each function is initially close to a power law: J 0() 1, J 1()
/2, J 2() 2/8, etc. This behavior follows from the Taylor series
n
2 k
k
( )
1
J ( )
,
(2.132)
n
2
k 0 k!( n
k)! 2
which is formally valid for any , and may even serve as an alternative definition of the functions Jn().
However, the series is converging fast only at small arguments, << n, where its leading term is 1
n
J ( )
(2.133)
.
n
0
!
n 2
At n + 1.86 n 1/3, the Bessel function reaches its maximum45
675
.
0
max J ( )
,
(2.134)
n
1/ 3
n
and then starts to oscillate with a period gradually approaching 2, a phase shift that increases by /2
with each unit increment of n, and an amplitude that decreases as –1/2. All these features are described by the following asymptotic formula:
2 1/2
n
J ( )
cos
(2.135)
n
,
4
2
which starts to give a reasonable approximation soon after the function peaks – see Fig. 18.46
45 These two approximations for the Bessel function peak are strictly valid for n >> 1, but may be used for reasonable estimates starting already from n = 1. For example, max [ J 1()] is close to 0.58 and is reached at
2.4, just about 30% away from the values given by the asymptotic formulas.
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Now we are ready for our case study (Fig. 17). Since the functions the Z( z) have to satisfy not only Eq. (126) but also the bottom-lid boundary condition Z(0) = 0, they are proportional to sinh z – cf.
Eq. (94). Then Eq. (84) becomes
J c
n s
n
z .
(2.136)
n
cos sin
n
n
sinh
n0
Next, we need to satisfy the zero boundary condition at the cylinder’s side wall ( = R). This may be ensured by taking
J ( R
) 0.
(2.137)
n
Since each function Jn( x) has an infinite number of positive zeros (see Fig. 18 again), which may be numbered by an integer index m = 1, 2, …, Eq. (137) may be satisfied with an infinite number of discrete values of the parameter :
nm
,
(2.138)
nm
R
where nm is the m-th zero of the function Jn( x) – see the top numbers in the cells of Table 1. (Very soon we will see what we need the bottom numbers for.)
Table 2.1. Approximate values of a few first zeros, nm, of a few lowest-order Bessel functions Jn() (the top number in each cell), and the values of dJn() /d at these points (the bottom number).
m = 1
2
3
4
5
6
n = 0
2.40482
5.52008
8.65372
11.79215
14.93091
18.07106
-0.51914
+0.34026
-0.27145
+0.23245
-0.20654
+0.18773
7.01559
10.17347
13.32369
16.47063
19.61586
1
3.83171
-0.40276
+0.30012
-0.24970
+0.21836
-0.19647
+0.18006
2
5.13562
8.41724
11.61984
14.79595
17.95982
21.11700
-0.33967
+0.27138
-0.23244
+0.20654
-0.18773
+0.17326
3
6.38016
9.76102
13.01520
16.22347
19.40942
22.58273
-0.29827
+0.24942
-0.21828
+0.19644
-0.18005
+0.16718
11.06471
14.37254
17.61597
20.82693
24.01902
4
7.58834
-0.26836
+0.23188
-0.20636
+0.18766
-0.17323
+0.16168
5
8.77148
12.33860
15.70017
18.98013
22.21780
25.43034
-0.24543
+0.21743
-0.19615
+0.17993
-0.16712
+0.15669
Hence, Eq. (136) may be represented in a more explicit form:
Variable
separation in
z
cylindrical
(,, z)
J
cos
sin sinh
.
(2.139)
n nm
c
n s
n
nm
nm
nm
coordinates
n0 m1
R
R
(example)
46 Eq. (135) and Fig. 18 clearly show the close analogy between the Bessel functions and the usual trigonometric functions, sine and cosine. To emphasize this similarity, and help the reader to develop more gut feeling of the Bessel functions, let me mention one result of the elasticity theory: while the sinusoidal functions describe, in particular, transverse standing waves on a guitar string, the functions Jn() describe, in particular, transverse standing waves on an elastic round membrane (say, a round drum), with J 0() describing their lowest (fundamental) mode – the only mode with a nonzero amplitude of the membrane center’s oscillations.
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Here the coefficients cnm and snm have to be selected to satisfy the only remaining boundary condition –
that on the top lid:
l
(,, l)
J
.
(2.140)
n
nm
c cos
n
s sin
n
nm
nm
sinh
V (,)
nm
n0 m 1
R
R
To use it, let us multiply both sides of Eq. (140) by the product Jn( nm’/ R) cos n’, integrate the result over the lid area, and use the following property of the Bessel functions:
1
1
J s J s sds
J
.
(2.141)
n nm n
(
)
nm'
n nm 2
1
mm '
2
0
As a small but important detour, the last relation expresses a very specific (“2D”) orthogonality of the Bessel functions with different indices m – do not confuse them with the function orders n, please!47 Since it relates two Bessel functions of the same order n, it is natural to ask why its right-hand side contains the function with a different order ( n + 1). Some gut feeling of that may come from one more very important property of the Bessel functions, the so-called recurrence relations:48
2 nJ ( )
J
( ) J
( )
n
,
(2.142a)
n 1
n 1
dJ ( )
J
( ) J
( ) 2
n
,
n 1
(2.142b)
n 1
d
which in particular yield the following formula (convenient for working out some Bessel function integrals):
d
nJ ()
J
.
(2.143)
n
n ()
n 1
d
Let us apply the recurrence relations at the special points nm. At these points, Jn vanishes, and the system of two equations (142) may be readily solved to get, in particular,
dJ
J
,
(2.144)
n1
n
nm
nm
d
so that the square bracket on the right-hand side of Eq. (141) is just ( dJn/ d)2 at = nm. Thus the values of the Bessel function derivatives at the zero points of the function, given by the lower numbers in the cells of Table 1, are as important for boundary problem solutions as the zeros themselves.
Now returning to our problem: since the angular functions cos n are also orthogonal – both to each other,
47 The Bessel functions of the same argument but different orders are also orthogonal, but differently:
d
1
J
( ) J
( )
.
n
n'
nn'
n n'
0
48 These relations provide, in particular, a convenient way for numerical computation of all Jn() – after J 0() has been computed. (The latter task is usually performed using Eq. (132) for smaller and an extension of Eq. (135) for larger .) Note that most mathematical software packages, including all those listed in MA Sec. 16(iv), include ready subroutines for calculation of the functions Jn() and other special functions used in this lecture series. In this sense, the conditional line separating these “special functions” from “elementary functions” is rather fine.
Chapter 2
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2
cos( n) cos( n' ) d
,
(2.145)
nn'
0
and to all functions sin n, the integration over the lid area kills all terms of both series in Eq. (140), besides just one term proportional to cn’m’, and hence gives an explicit expression for that coefficient.
The counterpart coefficients sn’m’ may be found by repeating the same procedure with the replacement of cos n’ by sin n’. This evaluation (left for the reader’s exercise) completes the solution of our problem for an arbitrary lid potential V(,).
Still, before leaving the Bessel functions behind (for a while only :-), let me address two important issues. First, we have seen that in our cylinder problem (Fig. 17), the set of functions Jn( nm/ R) with different indices m (which characterize the degree of Bessel function’s stretch along axis ) play a role similar to that of functions sin( nx/ a) in the rectangular box problem shown in Fig.
13. In this context, what is the analog of functions cos( nx/ a) – which may be important for some boundary problems? In a more formal language, are there any functions of the same argument
nm/ R, that would be linearly independent of the Bessel functions of the first kind, while satisfying the same Bessel equation (130)?
The answer is yes. For the definition of such functions, we first need to generalize our prior formulas for Jn(), and in particular Eq. (132), to the case of arbitrary, not necessarily real order .
Mathematics says that the generalization may be performed in the following way:
2 k
k
( )
1
J ( )
,
(2.146)
2
k 0 k! (
k
)
1 2
where ( s) is the so-called gamma function that may be defined as49
( s) s1
e d .
(2.147)
0
The simplest, and the most important property of the gamma function is that for integer values of its argument, it gives the factorial of the number smaller by one:
( n )
1 n! 1 2 ... n ,
(2.148)
so it is essentially a generalization of the notion of the factorial to all real numbers.
The Bessel functions defined by Eq. (146) satisfy, after the replacements n and n! ( n +
1), virtually all the relations discussed above, including the Bessel equation (130), the asymptotic formula (135), the orthogonality condition (141), and the recurrence relations (142). Moreover, it may be shown that n, functions J() and J-() are linearly independent of each other, and hence their linear combination may be used to represent the general solution of the Bessel equation. Unfortunately, as Eq. (131) shows, for = n this is not true, and a solution linearly independent of Jn() has to be formed differently. The most common way to do that is first to define, for all n, the following functions:
J
( )
cos
( )
