Chapter 4Boundary Value Problems in Spherical and Cylindrical Coordinates

4.1 Laplace’s Equation in spherical coordinatesLaplace’s equation in spherical coordinates, (r, θ, φ) , has the form

1

r2∂

∂r

·r2

∂

∂rΦ (r, θ, φ)

¸− 1

r2L · LΦ (r, θ, φ) = 0. (4.1)

where

−L · L ≡ 1

sin θ

∂

∂θ

·sin θ

∂

∂θΦ (r, θ, φ)

¸+

1

sin2 θ

∂2

∂φ2(4.2)

Multiplying Eq. 4.1 by r2,dividing by Φ (r, θ, φ) = R (r)Y (θ, φ) and rewriting,

1

R

∂

∂r

·r2

∂

∂rR(r)

¸=1

YL · L Y (θ, φ). (4.3)

This equation must be satisfied for all (r, θ, φ) As r, θ, φ are independent variables, Eq. 4.3 can only be satisfied if bothsides equal a constant which we choose this constant to be ( + 1).Then,

L · LY (θ, φ) = ( + 1)Y (θ, φ) (4.4)

andd

dr

·r2

d

drR(r)

¸− ( + 1)R(r) = 0 (4.5)

Eq. 4.4 can be further separated using Y (θ, φ) = V (θ)W (φ):

1

V (θ)sin θ

∂

∂θ

·sin θ

∂

∂θV (θ)

¸+

1

V (θ)sin2 θ ( + 1)V (θ) = − 1

W

∂2

∂φ2W (4.6)

Again, since θ,φ are independent variables both sides must equal a constant (the second separation constant) which wetake to be −m2

∂2

∂φ2W (φ) = −m2W (φ) or (4.7a)

Wm(φ) = Ameimφ +Bme

−imφ (4.7b)

and

sin θ∂

∂θ

·sin θ

∂

∂θV (θ)

¸+ [sin2 θ ( + 1)−m2]V (θ) = 0 (4.8)

73

Substituting x = cos θ Eq. 4.8 becomes the associated Legendre equation:

d

dx

·(1− x2)

d

dxF (x)

¸+ [ ( + 1)− m2

1− x2]F (x) = 0.; x = cos θ (4.9)

with two linearly independent solutions called the associated Legendre functions of the first, P |m|(x), and second kind,Q|m|(x).

F,|m| (x) = A ,|m|P

|m|(x) +B ,|m|Q

|m|(x) (4.10)

One can show that:the P |m|(cos θ) can be made to converge for all 0 ≤ θ ≤ π if :

0 < m = |m| ≤ = 0, 1, 2. ==> P|m|(cos θ) converge

..The Q|m|(cos θ), however diverge at cos θ = −1 for the above conditions on m and . The Q

|m|(cos θ) are possible

solutions when |cos θ| < 1. In our problem we expect the functions to be well behaved for all |x| = | cos θ| ≤ 1. A specialcase is found when |m| = 0 and these solutions, P 0(x) = P (x)are called Legende functions or Legendre polynomials.The P 0(x) are normalized so that:

P0 (x) = 1 (4.11)P1 (x) = x

P2 (x) =1

2

¡3x2 − 1¢

P3 (x) =x

2

¡5x2 − 3¢

P4 (x) =1

8

¡35x4 − 30x2 + 3¢

The associated Legendre polynomials, P |m|(x) ≡ Pm(x), of degree and order m. are related to theLegendre polynomials by

Pm (x) =¡1− x2

¢m/ 2 dmP (x)

dxm, m ≥ 0; (4.12)

Pm (x) = (−1)m P|m|(x) , m < 0

This particular sign convention is just one possibility. Clearly these are only non-zero when ≥ |m| .

Finally we note that at P 0 (1) = 1 (by definition), P 0 (−1) = (−1) , and, if m 6= 0, Pm (±1) = 0. The generalsymmetry of these functions for x→ −x is Pm (−x) = (−1) +m Pm (x)

Legendre’s functions of the second kind

Q (x) , have singularities at x = ±1. This solution would be one of the set known as Legendre functions of the secondkind. The first two Legendre functions of the second kind are, for |x| ≤ 1,

74

Section 4.2 Spherical harmonics

Q0 (x) =1

2ln

µ1 + x

1− x

¶= tanh−1 (x) (4.13)

Q1 (x) =x

2ln

µ1 + x

1− x

¶− 1

We note the logarithmic divergences for Qn (x) at x = ±1.In general, the associated Legendre functions can be written in terms of the hypergeometric function, F (a, b, c : z) an

infinite series which terminates for a (or b) equal to a negative integer. The termination of the series for a = m− producesthe convergence of the, Pm (x):

Qm (x) = constant·x+ 1

x− 1¸m/2

F (− , + 1, 1−m;1− x

2) (4.14)

Pm (x) = constant·x+ 1

x− 1¸m/2µ

1− x

2

¶mF (m− , +m+ 1,m+ 1;

1− x

2) (4.15)

where (Handbook of Mathematical Functions, M. Abramowitz and I. Stegun, NBS )

F (a, b, c; z) =Γ(1− c)

Γ(a)Γ(b)ΣnΓ(a+ n)Γ(b+ n)

n!Γ(c+ n)zn (4.16)

Γ(a) = (a− 1)!

4.2 Spherical harmonicsLaplace’s and Poisson’s equation in spherical coordinates are encountered in a wide range of problems (E&M,

thermodynamics, mechanics of continuous media, and quantum mechanics). As might be expected the solutions of theseequations have been well studied. The solutions to the angular part of the equation are the spherical harmonics. These arethe appropriately normalized products of the associated Legendre polynomials in cos θ and the exp (imφ)

Y m (θ, φ) = (−1)ms2 + 1

4π

( −m)!

( +m)!Pm (cos (θ)) eimφ (4.17)

The Y m (θ, φ) are solutions to Eq. 4.3:

L · L Y ml = ( + 1)Y m

and

LzYm = mY m; where Lz = z · L

Some useful properties are:

Y −m (θ, φ) = (−1)m Y m (θ, φ)∗ (4.18a)

75

Section 4.3 Poisson’s equation with point charge in (r, θ, φ)

Y m (π − θ, 2π − φ) = (−1) Y −m (θ, φ) (4.18b)

The orthonormality condition: Z 2π

0

Z π

0

Y m00 (θ, φ)

∗Y m (θ, φ) sin θdθdφ = δ 0δmm0 (4.19)

(Y m00 , Y m) = δ 0δmm0

and the completeness condition:

∞X=0

Xm=−

Y m¡θ0, φ0

¢∗Y m (θ, φ) = δ

¡φ− φ0

¢δ¡cos θ − cos θ0¢ (4.20)

Y 00 (θ, φ) =

·1

4π

¸1/2Y 02 (θ, φ) =

·5

4π

¸1/2 ·3

2cos2 θ − 1

2

¸(4.21)

Y 01 (θ, φ) =

·3

4π

¸1/2cos θ Y 1

2 (θ, φ) = −·15

8π

¸1/2sin θ cos θeiφ

Y 11 (θ, φ) = −

·3

4π

¸1/2sin θeiφ Y 2

2 (θ, φ) = −·15

4π

¸1/2sin2 θei2φ

Y 03 (θ, φ) =

·7

4π

¸1/2 ·5

2cos3 θ − 3

2cos θ

¸(4.22)

Y 13 (θ, φ) = −1

4

·21

4π

¸1/2sin θ

£5 cos2 θ − 1¤ eiφ

Y 23 (θ, φ) =

1

4

·105

2π

¸1/2sin2 θ cos θei2φ

Y 33 (θ, φ) = −1

4

·35

4π

¸1/2sin3 θei3φ

4.3 Poisson’s equation with point charge in (r, θ, φ)We will now re-examine the solution to Poisson’s equation with a point charge q. Let the charge lie at coordinates¡

r0, θ0, φ0¢

and let the observation point have coordinates (r, θ, φ) . The solution will be a Green’s function and we considerthe dependence on

¡r0, θ0, φ0

¢and on (r, θ, φ) The charge density is

ρ (r, θ, φ) = qδ(r− r0) Gaussian units (4.23)

=q

r0 2δ (r − r0) δ

¡cos θ − cos θ0¢ δ ¡φ− φ0

¢76

Section 4.3 Poisson’s equation with point charge in (r, θ, φ)

and Poisson’s equation is

∇2Φ (r, θ, φ) = −4πk1ρ (r, θ, φ) (4.24)= −4πδ(r− r0). Gaussian units (4.4)

∇2Φ (r, θ, φ) =−10ρ (r, θ, φ) S.I. units

We look for a solution of the form

Φ (r, θ, φ) =∞X=0

Xm=−

f m

¡r; r0, θ0, φ0

¢Y m (θ, φ) (4.25)

Inserting this into Poisson’s equation we obtain (using Gaussian units)

∇2∞X=0

Xm=−

f m

¡r; r0, θ0, φ0

¢Y m (θ, φ) =

−4πqr0 2

δ (r − r0) δ¡cos θ − cos θ0¢ δ ¡φ− φ0

¢(4.26)

1

r2

∞X=0

Xm=−

·d

dr

µr2

d

drf m

¶− ( + 1) f m

¸Y m (θ, φ) = −4π q

r0 2δ (r − r0) δ

¡cos θ − cos θ0¢ δ ¡φ− φ0

¢Multiply both sides by Y m0∗

0 (θ, φ) and integrate overRR

d(cos θ)dφ Using the orthonormality of the spherical harmonicsone finds: ZZ

1

r2

∞X=0

Xm=−

·d

dr

µr2

d

drf m

¶− ( + 1) f m

¸Y m0∗

0 (θ, φ)Y m (θ, φ) d(cos θ)dφ

=1

r2

∞X=0

Xm=−

·d

dr

µr2

d

drf 0m0

¶− ( + 1) f 0m0

¸δ 0 δm0m =

−4π q

r0 2δ (r − r0)

ZZδ¡φ− φ0

¢δ¡cos θ − cos θ0¢Y m0∗

0 (θ, φ) d(cos θ)dφ

1

r2

·d

dr

µr2

d

drf 0m0

¶− ( + 1) f 0m0

¸= −4π q

r0 2δ (r − r0)Y m0

0¡θ0, φ0

¢∗ (4.27)

Thus the angular dependence of f m

¡r; r0, θ0, φ0

¢is given by Y m

¡θ0, φ0

¢and we define

f m

¡r; r0, θ0, φ0

¢= g (r; r0)Y m

¡θ0, φ0

¢∗. (4.28)

The functions g (r; r0) satisfy

1

r2

·d

dr

µr2

d

drg

¶− ( + 1) g

¸= −4π q

r0 2δ (r − r0) (4.29)

These equations are solved by :(1) obtaining the solutions to the homogeneous equations for r < r0 and r > r0,(2) making g (r; r0) continuous at r = r0, and(3) satisfying the condition placed on g (r; r0) obtained by integrating Eq. 4.29

from r = r0 − ε to r = r0 + ε.

77

Section 4.3 Poisson’s equation with point charge in (r, θ, φ)

The solution to the homogeneous equation in r has the general form

g (r; r0) = α (r0) r + β (r0) r− −1. (4.30)

For r < r0 g (r; r0) = α (r0) r and for r > r0 g (r; r0) = β (r0) r− −1. Continuity at r = r0 requires that

g (r; r0) = ar

r0 +1, r < r0 (4.31)

g (r; r0) = ar0

r +1, r > r0

Integrating Eq. 4.29 from r = r0 − ε to r = r0 + ε we obtainZ r0+ε

r0−ε

1

r2

·d

dr

µr2

d

drg

¶− ( + 1) g

¸dr = −4π q

r0 2

Z r0+ε

r0−εδ (r − r0) dr

Z r0+ε

r0−ε

1

r2

·d

dr

µr2

d

drg

¶− ( + 1) g

¸dr = −4π q

r0 2

The second term on the left is continuous at r = r0 and drops out of the equation.Z r0+ε

r0−ε

1

r2d

dr

µr2

d

drg

¶dr = −4π q

r0 2

Integrating by parts twice,

1

r2

µr2

d

drg

¶|r=r0+εr=r0−ε +

Z r0+ε

r0−ε

2

r

d

drg dr = −4π q

r0 2

1

r2

µr2

d

drg

¶|r=r0+εr=r0−ε +

2

rg |r=r0+εr=r0−ε +

Z r0+ε

r0−ε

2

r2g dr = −4π q

r0 2

Only the first term on the left is non-zero at r = r0so

r0 2"µ

dg (r; r0)dr

¶r=r0+ε

−µdg (r; r0)

dr

¶r=r0−ε

#= −4π q (4.32)

Now use the specific forms for g (r; r0):

r0 2a

"µd

dr

r0

r +1

¶r=r0+ε

−µ

d

dr

r

r0 +1

¶r=r0−ε

#= −4π q

r0 2a[−( + 1) r0

r0 +2− r0 −1

r0 +1] = a[−( + 1)− ] = −a(2 + 1)

Thus

a =4πq

2 + 1(4.33)

78

Section 4.4 The Coulomb potential in spherical coordinates4.4 The Coulomb potential in spherical coordinates

Using Eq. 4.33 we have the expansion of a point charge coulomb potential given by,

Φ (r, θ, φ) = 4πq∞X=0

Xm=−

1

2 + 1

r<r +1>

Y m (θ, φ)Y m¡θ0, φ0

¢∗. (4.33b)

∇2Φ (r, θ, φ) = −4πqδ(r− r0) Gaussian units

where the point charge is at r0 and r< = min (r, r0) and r> = max (r, r0) . Since the position of the point charge is providedin this simple example, one considers r0 = r0 a ”constant vector” and −4πqδ(r− r0) is only a function of .r However, ifwe have a charge density, ρ(r) = −4πq · f(r, θ, ϕ) it is clear that ρ(r) is a function of r, and is specified by f(r, θ, ϕ). Thelatter includes any dependence on known vectors such as r0.

4.5 Green’s Function and Expansion of 1

|r− r0|

Note that the Green’s function for Laplace’s equation (for V= all space) satisfies the following:

∇2 −14π|r− r0| = δ(r− r0)

Thus .

G(r, r0) =−14π

∞X=0

Xm=−

4π

2 + 1

r<r +1>

Y m (θ, φ)Y m¡θ0, φ0

¢∗. (4.34a)

∇2G(r, r0) = δ(r− r0) Gaussian units

Recall that the G(r, r0) can also be written as follows:

G(r, r0) =−14π

1

|r− r0| (4.34b)

The Green’s function differs from the Φ (r) (in Eq 4.33) as Φ (r) describes the electrostatic potential due to a pointcharge, q at a given point, r0 = r0. In contrast, G(r, r0) represents a unit point charge at r0, which is as yet unspecified..

Equations 4.34 yield a convenient expansion for1

|r− r0| in spherical harmonics

1

|r− r0| =∞X=0

Xm=−

4π

2 + 1

r<r +1>

Y m (θ, φ)Y m¡θ0, φ0

¢∗ (4.35)

and provides a convenient way to evaluate integrals of the form:

1

4π

ZZZρ(r0)|r− r0|dV

0 =ZZZ

ρ(r0)G(r, r0)dV

79

Section 4.4 The Coulomb potential in spherical coordinatesExample: Determine the electrostatic potential due to the following distribution of charge (a spherical shell of

charge with a cos θ angular dependence);

ρ(r) =Q

a2cos θδ(r − a)

assuming that the potential→ 0 like1

ras r→∞

Solution: We use the Green’s function for all of space (Eq. 4.34a) as follows in S.I. units:

φ (r) =1

ε

ZZZall space

[−ρ (r0)]G (r, r0) d3r0 +ZZ

r0→∞φ (r0s)∇0G (r, r0s) · dS0 −

ZZr0→∞

G (r, r0s)∇0φ (r0s) · dS0

φ (r) =−1ε

ZZZr0>a

ρ (r0)G (r, r0) d3r0

since the surface integrals vanish like1

r0as r0 →∞. (see page 59).

φ (r) =−1ε

ZZZall space

ρ (r0)−14π

1

|r− r0|d3r0 (note r0 dependence of ρ) SI units

=1

4πε

ZZZall space

Q

a2cos θ0δ(r0 − a)

∞X=0

Xm=−

4π

2 + 1

r<r +1>

Y m (θ, φ)Y m¡θ0, φ0

¢∗d3r0

=1

4πε

Q

a2

∞X=0

Xm=−

4π

2 + 1Y m (θ, φ)

Z ∞0

r<r +1>

r02δ(r0 − a)dr0ZZ

cos θ0Y m¡θ0, φ0

¢∗d(cos θ0)dϕ0

Next, we use the expansion of cos θ0 in spherical harmonics:

cos θ0 =

r4π

3Y 01 (θ

0, φ0)

φ (r) =1

4πε

Q

a2

∞X=0

Xm=−

4π

2 + 1Y m (θ, φ) ·

Z ∞0

r<r +1>

r02δ(r0 − a)dr0 ·r4π

3

ZZY 01 (θ

0, φ0)Y m¡θ0, φ0

¢∗d(cos θ0)dϕ0

=1

4πε

Q

a2

∞X=0

Xm=−

4π

2 + 1Y m (θ, φ) ·

Z ∞0

r<r +1>

r02δ(r0 − a)dr0 ·r4π

3δ 1δm0

=1

ε

Q

a21

2 · 1 + 1

r4π

3Y 01 (θ, φ) ·

Z ∞0

r<r +1>

r02δ(r0 − a)dr0 =−Q3εa2

cos θ · r<r +1>

a2

φ (r) =Q

3εa2r cos θ if r < a

φ (r) =Q

3εa2a3

r2cos θ if r > a

Note that the final angular dependence of φ (r) isr4π

3Y 01 (θ, φ) = cos θ.and the r dependence is linear inside the sphere

of radius a and falls off like1

r2outside the sphere. Continuity at r = a is preserved

80

Section 4.6 Addition Theorem for Spherical Harmonics4.6 Addition Theorem for Spherical Harmonics

Suppose in the expansion of

1

|r− r0| =∞X=0

Xm=−

4π

2 + 1

r<r +1>

Y m (θ, φ)Y m¡θ0, φ0

¢∗ (see 4.33)

we let r0 = r0z. so that θ0 = 0. Then only the m = 0 terms will appear in the sum¡Y m

¡0, φ0

¢= 0 if m 6= 0¢. In this case,

using the expression for Y m (θ, φ) in Eq. 4.17 and P 0(1) = 1

1

|r−r0z| =∞X=0

4π

2 + 1

r<r +1>

Y 0 (θ, φ)Y 0¡0, φ0

¢∗=

∞X=0

4π

2 + 1

r<r +1>

r2 + 1

4πP 0(cos θ)

r2 + 1

4πP 0(1)∗

1

|r−r0z| =∞X=0

r<r +1>

P 0 (cos θ)

1

[r · r− 2r0r cos θ+r02]1/2 =∞X=0

r<r +1>

P 0 (cos θ)

The angular dependence of this expression depends only on the angle, γ, between the vector locating the charge, r0, andthe vector locating the observation point, r. That is, r · r0 = rr0 cos γ. So

1

[r · r− 2r0r cos γ+r02]1/2 =∞X=0

r<r +1>

P 0 (cos γ)

=∞X=0

Xm=−

4π

2 + 1

r<r +1>

Y m (θ, φ)Y m¡θ0, φ0

¢∗Thus, one obtains the addition theorem for spherical harmonics:

P (cos γ) =Xm=−

4π

2 + 1Y m (θ, φ)Y m

¡θ0, φ0

¢∗ (4.36)

r · r0 ≡ cos γ

= sin θ sin θ0[cosφ cosφ0 + sinφ sinφ0] + cos θ cos θ0

= sin θ sin θ0 cos¡φ− φ0

¢+ cos θ cos θ0

4.7 Problems with azimuthal symmetryThe angular dependence of systems in which the boundary conditions and the charge densities have axial symmetry, i.e.,

81

Section 4.6 Addition Theorem for Spherical Harmonicsthey are independent of φ,is given by the Legendre functions Pν (cos θ) . The potential will have the general form

Φ (r, θ) =∞Xn=0

fn (r)Pνn (cos θ) (4.37)

Problems with 0 ≤ θ ≤ πIf the range of θ is from zero to π then the condition that Φ (r, θ) be non-singular requires that νn = n (an integer). In

this case the sum will be over Legendre polynomials. The potential is then given by

Φ (r, θ) =∞Xn=0

£αnr

n + βnr−n−1¤Pn (cos θ) . (4.38)

To evaluate the coefficients we need the orthogonality condition for the Legendre polynomials

2δmn

2n+ 1=

Z π

0

Pm (cos θ)Pn (cos θ) sin θ dθ (4.39)

=

Z π

0

Pm (cos θ)Pn (cos θ) d(cos θ) .

Specifying the potential at r = a and at r = b > a the coefficients satisfy the equations

Φ (a, θ) =∞Xn=0

£αna

n + βna−n−1¤Pn (cos θ) = ∞X

n=0

In (a)Pn (cos θ)

Φ (b, θ) =∞Xn=0

£αnb

n + βnb−n−1¤Pn (cos θ) = ∞X

n=0

In (b)Pn (cos θ)

Mulitply each side byPn0 (cos θ) and integrate over θ:

£αna

n + βna−n−1¤ =

µn+

1

2

¶Z π

0

Pn (cos θ)Φ (a, θ) sin θ dθ = In (a) (4.40)

£αnb

n + βnb−n−1¤ =

µn+

1

2

¶Z π

0

Pn (cos θ)Φ (b, θ) sin θ dθ = In (b)

which have solutions

αn =an+1In (a)− bn+1In (b)

a2n+1 − b2n+1(4.41)

βn = − (ab)n+1 bnIn (a)− anIn (b)

a2n+1 − b2n+1

In the limit that a→ 0 we have that αn → b−nIn (b) and βn → 0, while if b→∞ we have αn → 0 and βn → an+1In (a) .

Problems with 0 ≤ θ ≤ θm < π

If we require that the solution of Legendre’s equation be zero at θ = θm and that it be finite for 0 ≤ θ ≤ θm the mostgeneral solution will be a Legendre function of the first kind, Pν (cos θ) where ν 6= integer (See Abramowitz & Stegun,pp332, 556). Legendre functions of the second kind, which diverges at cos θ = 1,are excluded. The Pν (cos θ) are generallyexpressed in terms of the hypergeometric functions, F (a, b, c;w) (See Eqs. 4.15 and 4.16) where we set a = −ν, b = 1 + ν,

82

Section 4.6 Addition Theorem for Spherical Harmonicsc = 1 and w =

1− cos θ2

= sin2¡θ2

¢. When ν 6= integer ( a is not a negative integer) the F (−ν, 1 + ν; 1, w) is an infinite

series.

Pν (cos θ) = P 0υ (x) = F

µ−ν, 1 + ν; 1, sin2

µθ

2

¶¶=

∞Xn=0

(−ν)n (ν + 1)n[sin2 (θ/2)]n

n!

with the ‘function’ (z)n given by

(z)n = z (z + 1) (z + 2) .... (z + n− 1) . = (z + n− 1)!(z − 1)!

For a given angle, θ = β, the Legendre functions can be plotted as a function of ν to determine the values of ν whichyield Pν (cosβ) = 0. The figure below illustrates the approach with β = π/6, π/3, and π/2. In the case that β = π/2 wenote that the appropriate functions will be the Legendre polynomials of odd order.

0 1 2 3 4 50.5

0.13

0.25

0.63

1

90 deg.30 deg.60 deg

Plot versus order

order

Lege

ndre

func

tion

.Fig. 4.1

Example: the fields near the point of a cone

The results of the last section can be applied to the problem of determining the fields and surface charge density near thetip of a conical conductor. In the region of interest the surface of the cone makes an angle β with the z axis. For β < π/2 theregion of interest is ‘inside’ the cone while for β > π/2 the region is ‘outside’ the conical surface.

If we take the cone to be at zero potential then the leading term in the potential will be

Φ (r, θ, φ) ≈ Φ0 rν1 Pν1 (cos θ) (4.43)

where ν1 is determined by the condition that cosβ is the first zero of Pν1 (cos θ) .

83

Section 4.8 Green’s function in a spherical shellFor example if β = π/6 then ν1 ≈ 4 (see Fig. 4.1). Near the tip of the cone the electric field will have components

Er (r, θ) ≈ −Φ0 ν1rν1−1Pν1 (cos θ)Eθ (r, θ) ≈ − ∂

r∂θΦ0 r

ν1 Pν1 (cos θ)

= −Φ0 rν1−1 sin θ P 0ν1 (cos θ)

and the surface charge density will be given by n ·E = −4πσ (r) where n is normal to the conical surface defined byθ − π/6 = 0. That is, n = θ.

σ (r) ≈ −(4π)−1Φ0rν1−1 sinβ P 0ν1 (cosβ) . (4.45)

We first consider the case with β < π/2. For this system ν1 > 1 and the fields and surface charge vanish as r → 0. Thegeometry of this system would correspond to a conical hole in a conductor. As expected the field does not readily penetrate ahole in a conductor.

The alternate case has β > π/2. From Fig. we deduce that ν1 < 1. Clearly the fields and surface charge diverge asone approaches the point of the cone. This is confirmation of the folk-lore that the fields are strong at pointed regions of acharged conductor.

4.8 Green’s function in a spherical shell

To illustrate the Green’s function techniques we consider systems which lie in a spherical shell a ≤ r ≤ b boundedby conducting (segmented?) spherical shells. In this case the potentials on the conductors are generally specified and theDirichlet green’s function is required. Excluding the value r = r0 gives two homogeneous equations, one for a ≤ r < r0 andthe other for r0 < r ≤ b. The solutions for the r and r0 dependence are of the form in Eq. 4.30. To make the g (r; r0) = 0at the surfaces of the conductor we set

g (r; r0) = α (r0)

"r

r0 +1− a2 +1

(rr0) +1

#, a ≤ r < r0 (4.46)

= 0 at r = a

g (r; r0) = β (r0)

"−(rr

0)b2 +1

+r0

r +1

#, r0 < r ≤ b

= 0 at r = b

Continuity of g (r; r0) at r = r0 requires that

α (r0)

"1

r0− a2 +1

(r0)2 +2

#− β (r0)

"−(r

0)2

b2 +1+1

r0

#= 0 (4.47)

while to satisfy the differential equation, Eq. 4.29, one needs to carry out the procedure in Eq 4.32. (That is one needs tointegrate the differential equation, O2Φ(r, r0) = −4πδ(r− r0) from r = r0 − to r = r0 + ). The procedure (in Gaussianunits) results in the following equation:

α (r0)·r0+( + 1) a2 +1

r0 2 +2

¸+ β (r0)

·r0 2

b2 +1+( + 1)

r0

¸=4π

r0(4.48)

84

Section 4.8 Green’s function in a spherical shellThe solution to these equations is

α (r0) =4π

2 + 1

1− (r0/b)2 +11− (a/b)2 +1

(3-49)

β (r0) =4π

2 + 1

1− (a/r0) 2 +11− (a/b)2 +1

giving the result

Φ (r; r0) =∞X=0

4π£b2 +1 −R2 +1>

¤ £R2 +1< − a2 +1

¤(2 + 1) (b2 +1 − a2 +1) (rr0) +1

+Xm=−

Y m (θ, φ)Y m¡θ0, φ0

¢∗, (4.5)

G (r; r0) = − 1

4π

∞X=0

4π£b2 +1 −R2 +1>

¤ £R2 +1< − a2 +1

¤(2 + 1) (b2 +1 − a2 +1) (rr0) +1

+Xm=−

Y m (θ, φ)Y m¡θ0, φ0

¢∗ (4.6)

R< = min (r, r0) , R> = max (r, r0)

If a = 0 this gives the Green’s function for the system confined to lie inside a spherical shell of radius b while if b→∞this is the Green’s function for the system confined to lie outside a spherical shell of radius a.

Example 1: Uniformly charged annulus in θ = π/2 planeAs an application for the Green’s function of the last section we consider the system consisting of grounded concentric

spheres, radii a < b, separated by a uniformly charged annulus (ring) with total charge Q0. The problem is to determine thetotal charges on the spherical surfaces r = a and r = b. The surface charge density on the ring between the spheres is

σ0 =Q0

π (b2 − a2),

the charge density is

ρ (r) = σ0δ (z) =σ0rδ (cos θ) , a ≤ r ≤ b

and the boundary conditions are Φ (a, θ, φ) = Φ (b, θ, φ) = 0.Solution: With the given boundary conditions and the azimuthal symmetry

Φ (r, θ, φ) =

ZZZa≤r0≤b

G (r; r0) [−4πρ (r0)]d3r0 Gaussian units

= 2πσ0

∞X=0

P (cos θ)P (0)

(b2 +1 − a2 +1)

Z b

a

r0£b2 +1 −R2 +1>

¤ £R2 +1< − a2 +1

¤(rr0) +1

dr0

The charge on the inner sphere is

Qa =

ZZr=a

σ(r)r2dΩ =

ZZr=a

Er

−4πr2dΩ =

a2

4π

ZZ ·∂Φ (r, θ, φ)

∂r

¸r=a

dΩ

= −2πσ0ab− a

Z b

a

(b− r0) dr0 = −πσ0a (b− a)

85

Section 4.9 Laplace’s equation in cylindrical coordinates

and on the inner surface of the outer sphere

Qb = − b2

4π

ZZ ·∂Φ (r, θ, φ)

∂r

¸r=b

dΩ

= −2πσ0bb− a

Z b

a

(r0 − a) dr0 = −πσ0b (b− a)

where we have used the orthogonality of the Legendre polynomialsZ 1

−1P (x) dx = 2δ 0. (4.51)

As a variation on this problem we relax the condition that the two spheres are at the same potential. Instead we requirethat the inner sphere be neutral and determine the potential of the inner sphere minus the potential of the outer sphere. Theresult can be readily obtained using the superposition principle. The inner sphere is made neutral by distributing a chargeπσ0a (b− a) uniformly over its surface. The inner surface of the outer sphere will then have a total charge of π

¡b2 − a2

¢σ0.

Applying the superposition principle yields that

Φ2 (a)− Φ2 (b) = πσ0 (b− a)2

b=

Q0 (b− a)

b (b+ a)

Example 2: Inner sphere potential is Φ0P1 (cos θ) , outer sphere is grounded

To demonstrate another application of the Green’s function we consider a boundary value problem. Again the systemconsists of concentric spheres, radii a < b. The boundary conditions are

Φ (a, θ, φ) = Φ0 cos θ, Φ (b, θ, φ) = 0.

The green’s function solution to this problem is given by

Φ (r, θ, φ) = − a2

4π

ZZΦ0 cos θ

0·− ∂

∂r0G (r; r0)

¸r0=a

dΩ0

Using the orthogonality of the Legendre polynomials this reduces to

Φ (r, θ, φ) =a2

3Φ0 cos θ

∂

∂r0

ãb3 − r3

¤ £r0 3 − a3

¤(b3 − a3) (rr0)2

!r0=a

= a2Φ0 cos θb3 − r3

r2 (b3 − a3)

This result illustrates the general result that the boundary value problems. The boundary conditions are expanded inspherical harmonics and the boundary value problem is solved for each coefficient in the expansion. The solution of theproblem is then given by the superposition principle

4.9 Laplace’s equation in cylindrical coordinates

Problems in which the system is confined to cylindrical volumes are best analyzed in cylindrical coordinates. Incylindrical coordinates Laplace’s equation is

1

ρ

∂

∂ρ

·ρ∂

∂ρΦ (ρ, ϕ, z)

¸+1

ρ2∂2

∂ϕ2Φ (ρ, ϕ, z) +

∂2

∂z2Φ (ρ, ϕ, z) = 0. (4.52)

86

Section 4.9 Laplace’s equation in cylindrical coordinates

As in the case of spherical coordinates, this equation is solved by a series expansion in terms of products of functions of theindividual cylindrical coordinates. That is, we use separation of variables.

Φ (ρ, ϕ, z) =Xl,m,n

αlmnUl (ρ)Vm (ϕ)Wk (z) . (4.53)

Substituting Φlmk (ρ, ϕ, z) = Ul (ρ)Vm (ϕ)Wk (z) .into Eq 4.52 and dividing by Φlmk (ρ, ϕ, z), we can sequentiallyseparate all the variable .dependence:

1

Ulρ

d

dρ

·ρd

dρUl

¸+

1

Vmρ2d2

dϕ2Vm = − 1

Wn

d2

dz2Wk = ±k2 (4.54)

ρ

Ul

d

dρ

·ρd

dρUl

¸∓ ρ2k2 = − 1

Vm

d2

dϕ2Vm = ±m2 (4.55)

These functions satisfy the Eqs. 4.54, 4.55 and 4.56:

d2Wk (z)

dz2= λzWk (z) ; λz = k2 (4.54a)

Wk (z) = Akekz +Bke

−kz (4.54b)

Note that k can be complex, pure imaginary or real.

d2

dϕ2Vm (ϕ) = λϕVm (ϕ) ; λϕ = −m2 (4.55)

Vm (ϕ) = Cmeimϕ +Dme

−imϕ (4.55b)

Note that m could be non-integer and/or complex

ρ2d2

dρ2U (ρ) + ρ

d

dρU (ρ) +

¡k2ρ2 −m2

¢U (ρ) = 0 (4.56)

(kρ)2d2

d(kρ)2Um (kρ) + kρ

d

d(kρ)Um (kρ) +

¡(kρ)2 −m2

¢Um (kρ) = 0

Um (kρ) = amJm (kρ) + bmNm (kρ) (4.56b)

where now the label on Um (kρ) becomes m and the Jm and .Nm are called Bessel functions of the first and second kind,respectively. [Note that one could also use Hankel functions, H±

m(kρ) ≡ Jm(kρ) ± iNm(kρ), rather than Jm(kρ) andNm(kρ).]

The boundary conditions on the system determine the allowed values of k and m. Typically the system fills the regionfrom ϕ = 0 to ϕ = 2π. In this case the continuity of Vm (ϕ) and V 0

m (ϕ) requires that m be an integer. We consider thesesystems first. The values of k are determined by boundary conditions requiring solutions to be periodic in z ( k must be pureimaginary, k = ±i|k|) or exponential in z ( k = ±|k|).

87

Section 4.9 Laplace’s equation in cylindrical coordinates

4.9.1 Potential is zero at ρ = a, ρ = b In the case that the potential on the cylindrical surfaces are zero, Eq. 4.56determines the allowed values for k2. The appropriate solutions, U1(ρ), will be the Bessel functions of the first and secondkind, Jm (kρ) and Nm (kρ) (see Abramowitz & Stegun, Chapts. 9-11) which can be made equal to zero at .ka and kb.

We can obtain the small ρ dependence of the solutions by neglecting the term k2ρ2 in the differential equation. In thiscase, if m2 6= 0,

U1(ρ) = Jm (kρ) ∼ (kρ/2)m

Γ (m+ 1)(4.57)

U2(ρ) = Nm (kρ) ∼ Γ (p) (kρ/2)−m /π.

If m2 = 0,we find that U (ρ) ∼ 1 or U (ρ) ∼ (2/π) ln ρ. For real m the non-singular solutions are Bessel functions of thefirst kind, Jm (kρ) , and the singular solutions are Bessel functions of the second kind, Np (kρ) . For large values of kρ theasymptotic behavior of these functions is

Jm (kρ) ∼r

2

πkρcos³kρ− mπ

2− π

4

´, (4.58)

Nm (kρ) ∼r

2

πkρsin³kρ− mπ

2− π

4

´.

Note: Either of these approximations (4.57 and 4.58) can be used to determine the normalization of the functionsfor the Green’s function. Also, the Nm (kρ) has already been made linearly independent of Jm (kρ) for integer m.

The nth root, in order of magnitude, of the equation Jm (x) = 0 is given by (see Hildebrand, Advance Calculus forEngineers, Prentice-Hall 1949, p. 575-577)

xn = nπ

µ1 +

2m− 14n

¶− c1

k0− 4c23k03

− 32c315k05

− .... (4.59)

where

k0 = 2π (2m+ 4n− 1) , q = 4m2, (4.59b)c1 = q − 1, c2 = (q − 1) (7q − 31) ,c3 = (q − 1) ¡83q2 − 982q + 3779¢ .

Some values for xn are

n m = 0 m = 1 m = 2 m = 3 m = 4 m = 51 2.405 3.832 5.135 6.379 7.586 8.7802 5.520 7.016 8.417 9.76 11.064 12.3393 8.654 10.173 11.620 13.017 14373 15.7004 11.792 13.323 14.796 16.224 17.616 18.9825 14.931 16.470 17.960 19.410 20.827 22.220

It is often useful to have the zeros of J 0m (x) . The nth root, in order of magnitude, of the equation J 0m (x) = 0 is givenby

xn = nπ

µ1 +

2m+ 1

4n

¶− c1

k0− 4c23k03

− 32c315k05

− .... (4.60)

where

k0 = 2π (2m+ 4n+ 1) , q ≡ 4m2, (4.60b)c1 = q + 3, c2 = 7q

2 + 82q − 9,c3 = 83q3 + 2075q2 − 3039q + 3537

88

Section 4.10 Expansion of the Green’s function in cylindrical coordinates

The reference also gives formulae for the solutions of

Jm (x)Nm (ax)− Jm (ax)Nm (x) = 0 (a > 1) (4.61)J 0m (x)N

0m (ax)− J 0m (ax)N

0m (x) = 0 (a > 1)

If the system is contained in a cylinder of radius a and length L, a boundary value problem has the potential given onthe end caps. In this case the solutions are Bessel functions of the first kind which satisfy the condition Jm (ka) = 0. Thisrequirement determines values of k, kn with n = 1, 2, ....These functions satisfy the orthogonality relationshipZ a

0

ρJm (kn0ρ)Jm (knρ) dρ = 0.5 δn0n [aJm+1 (kna)]2 . (4.62)

For a given value of m we can assume that these functions form a complete set of orthogonal functions on the interval0 ≤ ρ ≤ a.

Finally, if a→∞ the orthogonality condition isZ ∞0

ρJm (kρ)Jm (k0ρ) dρ = k−1δ (k − k0) . (4.63)

4.9.2 Potential is zero at z = z1, z = z2 With the potential equal to zero on the ‘end-caps’ of the cylindricalregion Eq. 4.54 provides the eigenvalue problem for λz. The required solutions would be a superposition of cos (|k|z) andsin (|k|z) and λz = (ik)

2 = −|k|2. The ρ dependence is given by the modified Bessel functions, Im(|k|ρ) = i−mJm(i|k|ρ)and Km(|k|ρ) = π

2im+1H+

m(i|k|ρ) The solutions to Bessel’s equation would be the associated Bessel functions Im (|k|z)and Km (|k|z) . For |k|z → 0 these functions behave as

Im (|k|z) ∼ 1

Γ (m+ 1)

µ |k|z2

¶m(4.64)

Km (|k|z) ∼ Γ (m)

2

µ2

|k|z¶m

, m 6= 0

∼ − lnµ |k|z2

¶, m = 0

and for large kz these functions have the asymptotic form

Im (kz) ∼ (2π|k|z)−1/2 exp (|k|z) (4.65)

Km (kz) ∼µ

π

2|k|z¶1/2

exp (−|k|z) .

A general cylindrical problem would expected to involve both types of solutions.

4.10 Expansion of the Green’s function in cylindrical coordinates The following steps illustrate thegeneral procedure for finding a Green’s function expansion.

.1.In cylindrical coordinates the Green’s function satisfies

∇2G (r; r0) = δ (r− r0) = 1

ρδ (ρ− ρ0) δ (z − z0) δ (ϕ− ϕ0) (4.66)

2. Note that there are three representations of delta functions, useful for cylindrical problems:

δ (z − z0) =1

2π

Z ∞−∞

eik(z−z0)dk =

1

π

Z ∞0

cos [k (z − z0)] dk, , (4.67a)

89

Section 4.10 Expansion of the Green’s function in cylindrical coordinates

δ (ϕ− ϕ0) =1

2π

∞Xm=−∞

eim(ϕ−ϕ0) =

1

2π1 + 2 cos [m (ϕ− ϕ0)] (4.67b)

δ (ρ− ρ0) = ρ0Z ∞0

kJp (kρ)Jp (kρ0) dk. (4.67c)

3.Making use of the orthogonal functions appearing in the delta function expansions, write a general form for G (r; r0)leaving one set of variables (in this case z and z0) in terms of an unknown function, gm (z; z0):

G (r; r0) =1

2π

∞Xm=−∞

eim(ϕ−ϕ0)Z ∞0

k gm (z; z0)Jm (kρ)Jm (kρ0) dk. (4.68)

4. Apply∇2 to the general form forG (r; r0) and set the result equal to δ (r− r0) :

∇2G (r; r0) = [1

ρ

∂

∂ρ

·ρ∂

∂ρ

¸+1

ρ2∂2

∂ϕ2+

∂2

∂z2]1

2π

∞Xm=−∞

eim(ϕ−ϕ0)

·Z ∞0

k gm (z; z0) Jm (kρ)Jm (kρ0) dk

=∞X

m=−∞eim(ϕ−ϕ

0)Z ∞0

kJm (kρ0)Jm (kρ)

1

2π[gm (z; z

0)ρJm (kρ)

∂

∂ρ

·ρ∂Jm (kρ)

∂ρ

¸+−m2gm (z; z

0)ρ2

+∂2gm (z; z

0)∂z2

]

=∞X

m=−∞eim(ϕ−ϕ

0)Z ∞0

kJm (kρ0)Jm (kρ)

1

2π[(m2

ρ2− k2 +

−m2

ρ2)gm (z; z

0) +∂2gm (z; z

0)∂z2

]dk

=∞X

m=−∞eim(ϕ−ϕ

0)Z ∞0

kJm (kρ0) Jm (kρ)

1

2π[−k2gm (z; z0) + ∂2gm (z; z

0)∂z2

]dk

= δ (ρ− ρ0) δ (z − z0) δ (ϕ− ϕ0)1

ρ

90

Section 4.10 Expansion of the Green’s function in cylindrical coordinates

5. Multiply both sides byR 2π0

dϕe−im0ϕR∞0

ρJm0 (k0ρ) dρ and integrate over dϕ and dρ :

1

2π

Z 2π

0

∞Xm=−∞

e−i(m0−m)ϕe−imϕ0dϕ

Z ∞0

kJm (kρ0) [Z ∞0

ρJm (kρ)Jm0 (k0ρ) dρ][−k2gm (z; z0)

+∂2gm (z; z

0)∂z2

]dk

=

Z 2π

0

dϕ

Z ∞0

dρδ (ρ− ρ0) δ (z − z0) δ (ϕ− ϕ0) e−im0(ϕ)Jm0 (k0ρ)

= e−im0(ϕ0)Jm0 (k0ρ0) δ (z − z0)

6. On the left hand side, use the orthogonality condition, 12π

R 2π0

eimϕe−im0ϕdϕ = δmm0 to reduce the sum over m to a

single term with m = m0, and the integral,R∞0

ρJm (kρ)Jm (k0ρ) dρ = k−1δ (k − k0) to carry out the integration over k.

e−im0(ϕ0)

Z ∞0

kJm0 (kρ0) k−1δ (k − k0) [−k2gm0 (z; z0) +∂2gm0 (z; z0)

∂z2]dk

= e−im0(ϕ0)Jm0 (k0ρ0) · [−k2gm0 (z; z0) +

∂2gm0 (z; z0)∂z2

] = e−im0(ϕ0)Jm0 (k0ρ0) · δ (z − z0)

7. Finally, extract the differential equation for, gm0 (z; z0), which in this case must satisfy The Dirac delta function on the

right hand side indicates

d2

dz2gm (z; z

0)− k2gm (z; z0) = δ (z − z0) . (4.69)

thatdgmdz

has a discontinuity at z=z’.

8.To determine the overall ”normalization constant” for gm integrate both sides of 4.69 overR z0+z0− dzZ z0+

z0−[d2

dz2gm (z; z

0)− k2gm (z; z0)]dz =

Z z0+

z0−δ (z − z0) .dz

Z z0+

z0−

d2

dz2gm (z; z

0) dz −Z z0+

z0−k2gm (z; z

0) dz =Z z0+

z0−δ (z − z0) .dz = 1

Z z0+

z0−

d

dz[d

dzgm (z; z

0)]dz =d

dzgm (z; z

0) |z0+z0− = 1

9. Assume a general form for gm (z; z0) in terms of the functions given in step 2.and evaluate the limit in step 8.

gm (z; z0) = Cm exp(kz<) exp(−kz>)

d

dzgm (z; z

0) |z0+z0− = Cm[exp(kz0)d

dzexp(−k(z + ))− d

dzexp(kz) exp(−kz0) = 1

Cm[exp(kz0)(−k) exp(−k(z0 + ))− k exp(k(z0 − ) exp(−kz0) = 1

91

Section 4.11 Expansion of 1|r−r0| and the Coulomb potential:

As → 0,

Cm[−2k exp(kz0 − kz0) = 1

Cm =−12k

10. Put the function, gm (z; z0) =−12kexp(kz< − kz0>) into the expansion for the Greens’ function:

G (r; r0) =1

2π

∞Xm=−∞

eim(ϕ−ϕ0)Z ∞0

k−12kexp(kz<) exp(−kz>)Jm (kρ)Jm (kρ0) dk.

G (r; r0) =−14π

∞Xm=−∞

eim(ϕ−ϕ0)Z ∞0

exp(kz<) exp(−kz>)Jm (kρ)Jm (kρ0) dk. (4.70)

=−1

4π |r− r0|z> = max (z, z0) , z< = min (z, z

0)

4.11 Expansion of 1|r−r0| and the Coulomb potential: Thus by comparing G (r; r0)

with its compact form, −14π|r−r0| , the expansion of 1

|r−r0| is obtained

1

|r− r0| =∞X

m=−∞eim(ϕ−ϕ

0)Z ∞0

exp [−k (z> − z<)]Jm (kρ)Jm (kρ0) dk (4.71)

and the ‘Coulomb potential’ (with unit charge) in cylindrical coordinates can be written

q4π |r−r0|=

q4π

P∞m=−∞eim(ϕ−ϕ

0) R∞0exp [−k (z> − z<)]Jm (kρ)Jm (kρ0)dk (4.71b)

If we set ρ0 = 0 and z0 = 0 we can obtain the Laplace transform of J0 (t) from

1pρ2 + z2

=

Z ∞0

e−kzJ0 (kρ) dk or (4.72)

1√1 + s2

=

Z ∞0

e−stJ0 (t) dt

You should refer to Jackson to review the expansion of the Green’s function in terms of the modified Bessel functions.

4.12 Eigenfunction expansions for the Green’s function We consider a system in a volume Ω. A propertyof the system satisfies the differential equation

∇2ψ (r) + [f (r) + λ]ψ (r) = 0 (4.73)

92

Section 4.13 Green’s function in orthogonal functions of spherical coordinates

which with appropriate boundary conditions on ψ (r) becomes an eigenvalue equation for λ. Let λn and ψn (r) be theeigenvalues and corresponding eigenfunctions. The ψn (r) can be assumed to provide a complete set of functions fordescribing the property of the system and, since the ψn satisfy an eigenvalue problem in a finite volume, they form anorthogonal set and Z

Ω

ψn (r)ψm (r) d3r = δmn. (4.74)

Completeness implies that Xn

ψn (r)ψn (r0) = δ (r− r0) (4.75)

for r, r0 in Ω.We now use the functions ψn (r) to expand the Green’s function solution to

∇2G (r; r0) + [f (r) + λ]G (r; r0) = −4πδ (r− r0) (4.76)

= −4πXn

ψn (r)ψn (r0)

The solution is

G (r; r0) = −4πXn

ψn (r)ψn (r0)

λ− λn(4.77a)

In the case that λ = 0 we have

G (r; r0) = 4πXn

ψn (r)ψn (r0)

λn(4.77b)

This is a common form for the Green’s function.

4.13 Green’s function in orthogonal functions of spherical coordinates We consider an infinitesystem, the eigenfunctions are taken to be zero on a sphere of radius R and then R is taken to infinity. The eigenvalueproblem is

∇2ψ (r) + k2ψ (r) = 0 (4.78)

from which we immediately extract the angular dependence obtaining

ψ (r) =∞Xl=0

lXm=−l

Rl (r)Yml (θ, φ) . (4.79)

The radial dependence satisfies

r2d2

dr2Rl (r) + 2r

d

drRl (r)− l (l + 1)Rl (r) + k2r2Rl (r) = 0. (4.80)

For small r the solutions behave as rl and r−l−1. Requiring that the solution is regular at the origin we pick the solutionwhich varies as rl for small r. These solutions are the spherical Bessel functions.

jl (kr) =

rπ

2krJl+ 1

2(kr) (4.81)

93

Section 4.14 Simple application: point charge at originThe first few spherical Bessel functions are

l jl (z)0 z−1 sin z1 z−2 sin z − z−1 cos z2

¡3z−3 − z−1

¢sin z − 3z−2 cos z

The eigenfunctions satisfy the orthogonality condition ( Cohen-Tannoudji, Du, and Laloë, Qunatum Mechanics, Vol. II,p. 925) Z ∞

0

r2jl (kr) jl0 (k0r) dr

ZZY ml (θ, φ)Y m0

l0 (θ, φ)∗dΩ =

π

2kk0δ (k − k0) δll0δmm0 . (4.82)

The (assumed) completeness relationship is given by

∞Xl=0

Z ∞0

k2jl (kr) jl (kr0) dk

lXm=−l

Y ml (θ, φ)∗ Y m

l

¡θ0, φ0

¢=

π

2δ (r− r0) (4.83)

For reference a plane wave can be written as

exp (ik · r) = 4πXlm

iljl (kr)Yml (θr, φr)Y

ml (θk, φk)

∗ (4.84)

Assuming that the Green’s function has the form

G (r, r0) =Xlm

Z ∞0

k2gl (k) jl (kr) jl (kr0)Y m

l (θ, φ)Y ml

¡θ0, φ0

¢∗ (4.85)

we find that gl (k) = 8k−2. It follows that

1

|r− r0| = 8Xlm

Z ∞0

jl (kr) jl (kr0) dk Y m

l (θ, φ)Y ml

¡θ0, φ0

¢∗ (4.86)

This can be compared to the expansion in terms of plane waves

1

|r− r0| =1

2π2

ZZZexp [ik· (r− r0)]

k2d3k (4.87)

Comparing Eq. ?? with Eq. ?? (taking a = 0 and b→∞) we find thatZ ∞0

jl (kr) jl (kr0) dk =

π

2(2l + 1)

rl<rl+1>

(4.88)

4.14 Simple application: point charge at origin To illustrate the application of thisGreen’s function expansion we consider a point charge q0 located on the z axis at z = a. The potential will be evaluated fora→ 0. The charge density for this system is

ρ (r) =q0a2

δ (r0 − a) δ¡cos θ0 − 1¢ δ ¡φ0 − φ0

¢. (4.89)

94

Section 4.15 Multipole expansion of charge distributions

In terms of the Green’s function, expanded in the spherical Bessel functions and spherical harmonics, the potential is

Φ (r) =4q0π

ZZZ µr0

a

¶2dr0 d

¡cos θ0

¢dφ0 δ (r0 − a) δ

¡cos θ0 − 1¢ (4.90)

×Xlm

Z ∞0

jl (kr) jl (kr0) dk Y m

l (θ, φ)Y ml

¡θ0, φ0

¢∗The spatial integration over the delta functions can be performed to obtain

Φ (r) =2q0π

∞Xl=0

(2l + 1)Pl (cos θ)

Z ∞0

jl (kr) jl (ka) dk (4.91)

Since jl (0) = 0 for l 6= 0 and j0 (0) = 1, if a→ 0

Φ (r)→ 2q0π

Z ∞0

j0 (kr) dk =2q0πr

Z ∞0

sin (kr)

kdk =

q0r

(4.92)

It is obvious that the expansion of the Green’s function for Poisson’s equation in terms of spherical Bessel function is notas convenient as the expansion in terms of powers of r. Expansions of Green’s functions in terms of orthogonal functionsis more commonly used when differential equations are converted into integral equations. In that case the solutions areoften expanded in a set of orthogonal functions automatically leading to the expansion of the Green’s function in orthogonalfunctions.

4.15 Multipole expansion of charge distributions In the case that the charge is restricted to a finite volume,say a sphere of radius R, then the potential outside of this volume is obtained using the form of the Green’s function given inEq. ??

Φ (r) = 4π∞Xl=0

lXm=−l

1

2l + 1

Y ml (θ, φ)

rl+1

ZZZρ (r0) r0 lY m

l

¡θ0, φ0

¢∗d3r0 (4.93)

The integral in this expression defines the multipole moments of the charge distribution because of the relationshipY −ml (θ, φ) = (−1)m Y m

l (θ, φ)∗ it follows that ql−m = (−1)m q ∗lm. In terms of the multipole moments of the charge

distribution the potential for r > R is

Φ (r) = 4π∞Xl=0

lXm=−l

qlm2l + 1

Y ml (θ, φ)

rl+1. (4.95)

The set of elements qlm; m = −l, ..., l − 1, l form a spherical tensor of rank l. There are two common representationsof tensors, the familiar rectilinear representation and the spherical representation. For comparison we consider the chargedipole moment. In rectilinear representation

p =3Xi=1

pi ei =

ZZZr ρ (r) d3r

pi =

ZZZr i ρ (r) d

3r

or pi; i = 1, 2, 3 . In the spherical representation we find that

q1±1 =

r3

8π[±p1 − ip2] (4.96)

q10 =

r3

4πp3.

95

Section 4.16 Multipole expansion of the energy of a charge distribution in an external field

4.16 Multipole expansion of the energy of a charge distribution in an external field Consider acharge distribution contained in a volume of radius R. Let there be a potential Φ (r) in this region due to charges external tothe volume. The potential energy of the charge distribution is

W =

ZZZρ (r) Φ (r) d3r (4.97)

For convenience we can chose the origin of the coordinate system to be at the ‘middle’ of the charge distribution. Since thesources of the potential are external to the volume containing the charge, the electric potential can be expanded in a Taylorseries.

Φ (r) = Φ (0) +3X

i=1

ri

·∂Φ (r0)∂r0i

¸r0=0

(4.98)

+1

2

3Xi, j=1

rirj

"∂2Φ (r0)∂r0i∂r

0j

#r0=0

+ ...

The potential energy has the expansion

W = Φ (0)

ZZZρ (r) d3r −E (0) ·

ZZZrρ (r) d3r (4.99)

+1

2

3Xi, j=1

"∂2Φ (r0)∂r0i∂r

0j

#r0=0

ZZZrirjρ (r) d

3r + ...

The first term is a kind of average potential energy, the second is the interaction of the electric dipole distribution with theapplied electric field, and the third term gives the interaction of the quadrupole moment of the charge distribution with thegradient of the electric field.

The quadrupole moment of the charge distribution, which transforms as a second rank tensor under rotations, is

Qij =

ZZZ ¡3rirj − δijr

2¢ρ (r) d3r. (4.100)

The additional diagonal term, which is invariant under rotations, does not contribute to the potential energy.1 Since thequadrupole moment of the charge distribution is a symmetric, traceless second rank tensor. It is therefore specified by fiveparameters. These parameters for this rectilinear tensor are related to the five parameters of the spherical tensor q2m,

The simplest relationship occurs for q20 and Q33This simplicity is obtained because the spherical tensors we are usinghave taken the z or 3 axis as a preferred axis.

4.17 Boundary value problems, conclusions

We have developed the solutions of Laplace’s and Poisson’s equations in rectilinear, cylindrical, and sphericalcoordinates. Most problems which we encounter are most readily solved in one of these coordinate systems.

1

3

i, j=1

δij∂2

∂ri∂rjΦ (r)

r=0

= ∇2Φ (r)r=0

= 0

96

Section 4.16 Multipole expansion of the energy of a charge distribution in an external field

The solutions of Laplace’s equation were obtained by taking a superposition of potentials. Each solution matched thegiven boundary conditions on a different segment of the bounding surface and was zero on the rest of the surfaces.

The solutions to Poisson’s equation, and to Laplace’s equation, were obtained in terms of Green’s functions. We foundthat given the Green’s function for a given boundary geometry the solution to a potential problem with this boundary wasreduced to integration. A number of Green’s functions, in various forms, for rectangular, cylindrical, and spherical volumeswere obtained.

Finally, we have found that the potential generated by a charge distribution is a superposition of potentials due to themultipole moments of the charge distribution.

97