Physics 7xx: Axially symmetric Green function

Remember that inside of an integration over axially symmetric charge distributions

${1\over |\bm{r}-\bm{r'}|}=\sum_{n=0} {r_<^n\over r_>^{n+1}}P_n(\cos\theta)P_n(\cos\theta')$

and the left-hand side is a Green function for $\mathbb{R}^3$

$\nabla^2 G(x,x')=-4\pi \delta^3(x-x')$

Example 1
A sphere of radius $a$ is permanently and uniformly polarized $\bm{P}=P\bm{k}$ . Find the electric field at it’s center.\\
This is equivalent (in the sense that the fields are the same) to a sphere with the following surface and bulk charge densities

$\rho_{pol}=-\bm{\nabla}\cdot\bm{P}=0\quad\mbox{and}\quad \sigma_{pol}=\bm{P}\cdot\bm{r}=P \cos\theta$

When we have a given charge density, we always use

$V(r)=\int{\rho(r')\over 4\pi\epsilon_0 |\bm{r}-\bm{r}'|} d^3 r, \qquad \int{\sigma(r')\over 4\pi\epsilon_0 |\bm{r}-\bm{r}'|} dA$

to find $V$ at a point, or

$\bm{E}(r)=\int{\rho(r') (\bm{r}-\bm{r'})\over 4\pi\epsilon_0 |\bm{r}-\bm{r}'|^3} d^3 r, \qquad \int{\sigma(r') (\bm{r}-\bm{r'})\over 4\pi\epsilon_0 |\bm{r}-\bm{r}'|^3} dA$

to find the electric field, so in this case

$\bm{E}(0)=-\frac{1}{4\pi\epsilon_0 a^3}\int_0^{2\pi} d\phi \int_0^{\pi} \sin\theta \, d\theta \, a^2 \, \Big(a\bm{k}\cos\theta +a\bm{i}\sin\theta\cos\phi$

$+a\bm{j}\sin\theta \sin\phi\Big)\Big(P \cos\theta\Big)=-\frac{P}{3\epsilon_0}\bm{k}$

Example 2
Find the equivalent magnetization “charge” density $\rho_M$ and equivalent magnetization “surface charge density” $\sigma_m=\bm{M}\cdot\bm{n}$ that has the same magnetic field and scalar potential that a magnetized sphere of radius $a$ , magnetization $\bm{M}=M_0 \bm{k}$ would make.

Because there are no conduction currents and $\bm{\nabla}\times\bm{M}=0$ we can use a scalar potential $\Omega$ , that will satisfy $\nabla^2\Omega=0$ . Uniqueness says

$\Omega_{in}=Ar\cos\theta, \quad \Omega_{out}={B\over r^2}\cos\theta$

Continuity of $H_\theta$ tells us that $Aa={B\over a^2}$ , and of $B_r$ with $\bm{B}=\mu_0(\bm{H}+\bm{M})=\mu_0(-\bm{\nabla}\Omega+\bm{M})$

The field is equivalent to that made by a surface monopole density $\sigma_M=\bm{M}\cdot\bm{n}=M \, \cos\theta=M \, P_1(\cos\theta)$ .

$\Omega={1\over 4\pi}\int_0^{2\pi} a^2 \, d\phi' \int_0^\pi \sin\theta' d\theta' \Big(M \, P_1(\cos\theta')\Big)$

$\cdot \Big( \sum_{n=0}^\infty {a^n\over r^{n+1}} P_n(\cos\theta')P_n(\cos\theta)\Big)$

$={Ma^3\over 3\pi}{P_1(\cos\theta)\over r^2}$

the potential of a magnetic dipole $m={4\pi a^3 M\over 3}$ .

Suppose that you have a region $R$ bounded by a surface $S$ and can construct (by images or variable separation) a Green function for R that vanishes of $S$ . Suppose that you know the potential $V(x')$ on all points $x'\in S$ and want $V(x)$ for $x\in R$ , in which $\nabla^2V=0$ .