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\centerline{\bf Electrodynamics A (PHY 5346) Fall 2016 
Solutions} \bigskip

{\bf Set 11:} \bigskip

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% \end{figure} \centerline{Potential as function of $z$.}

\begin{enumerate}

\setcounter{enumi}{33}

\item Exercise E.48: Symmetry relation for spherical harmonics. 
\smallskip

\noindent The proof will be by induction from $m$ to $m+1$. The 
differential operators $L_{\pm}$ are explicitly given by
\begin{eqnarray} \nonumber
  L_+ &=& e^{+i\phi}\,\left( +{\partial\ \over\partial\theta} +
  i\, \cot\theta\, {\partial\ \over\partial\phi} \right)\,,\\ \nonumber
  L_- &=& e^{-i\phi}\,\left( -{\partial\ \over\partial\theta} +
  i\, \cot\theta\, {\partial\ \over\partial\phi} \right)\,.
\end{eqnarray}
Up to a real factor $Y_l^{\ 0}$ is given by
$$(L_+)^l\,Y_l^{\ -l}=(-1)^l\,(L_+)^l\,\sin^l(\theta)\,e^{-il\phi}\,.$$
It follows that $Y_l^{\ 0}$ is real, because the $\partial/\partial
\phi$ differentiation produces $-il\exp(-il\phi)$, so that the 
$i\exp(+i\phi)$ factor in the $\partial/\partial\phi$ part of $L_+$ 
multiplies with $ il\exp(-il\phi)$ to~$-1$. Therefore, the relation
$$ Y_l^{-m}\ =\ (-1)^m\,\overline{Y}_l^{\ m} $$ 
holds for $m=0$, i.e., $Y_l^{\ 0}=\overline{Y}_l^{\ 0}$. By induction 
hypothesis we have (after multiplying both sides by $L_-$)
$$ L_- Y_l^{-m} = (-1)^m L_- \overline{Y}_l^{\ m}\ .$$
The explicit form of the differential operators shows 
$\overline{L}_+=-L_-$. Hence,
$$ L_- Y_l^{-m} = (-1)^{m+1} \overline{L}_+ \overline{Y}_l^{\ m}
   = (-1)^{m+1}\,\sqrt{(l-m)\,(l+m+1)}\ \overline{Y}_l^{\ m+1}  $$
holds together with
$$ L_- Y_l^{-m} = \sqrt{(l-m)\,(l+m+1)}\ Y_l^{-m-1}\ .$$
Comparing the right-hand sides of the last two equations gives
$$ Y_l^{-m-1}\ =\ (-1)^{m+1}\,\overline{Y}_l^{\ m+1} $$ 
and, hence, proves the induction.
\bigskip


\item Exercise E.51:  Dirichlet Green function of a cylinder.
\smallskip

\noindent (a) We want to solve the eigenvalue problem
$$ \frac{\partial^2\Phi}{\partial\rho^2} +
   \frac{1}{\rho} \frac{\partial\Phi}{\partial\rho} +
   \frac{1}{\rho^2} \frac{\partial^2\Phi}{\partial\phi^2} +
   \frac{\partial^2\Phi}{\partial z^2}\,=\,-\lambda^2\,\Phi $$
with $\Phi = 0$ on the surface of the cylinder. We assume the cylinder 
to be of radius $\rho_0$ about the $z$-axis and of length $L$ from
$z=0$ to $z=L$. The separation ansatz $\Phi=R(\rho)Q(\phi)Z(z)$ gives
$$ \frac{d^2 Z}{d z^2} + \alpha^2\,Z \,=\, 0\,,\qquad
   \frac{d^2 Q}{d \phi^2} + m^2\,Q \,=\, 0~~~{\rm and}$$
$$ \frac{d^2 R}{d \rho^2} + \frac{1}{\rho} \frac{d R}{d \rho} 
   + \left(k^2-\frac{m^2}{\rho^2}\right)\,R \,=\, 0 $$
with $m=0,1,\,2,\dots $ and the constraint
$$ \alpha^2 + k^2\ =\ \lambda^2\ . $$
The BC implies then
$$ Z \sim \sin(\alpha_j z),\ \alpha_j = \frac{\pi j}{L},\ 
   j=1,2,\dots\,, \qquad Q \sim e^{\pm im\phi},\ m=0,1,2,\,\dots $$
$${\rm and}~~~R\sim J_m(k_{mn}\rho),\ k_{mn}=\frac{x_{mn}}{\rho_0}\,,$$
where $x_{mn}$ are the roots of the Bessel function of order $m$,
$J_m(x_{mn})=0$, $n=1,2,\dots\,$. The eigenfunctions with 
eigenvalues 
$$ \lambda_{jmn}^2 = \alpha_j^2 + k_{mn}^2 $$ 
are 
$$ \Phi_{jmn} \,=\, A_{jmn}\,\sin(\alpha_j z)\,e^{\pm im\phi}\,
                    J_m(k_{mn}\rho)\,, $$
where the $A_{jmn}$ are normalization constants so that 
$$ \int_0^{\rho_0} \rho\,d\rho \int_0^{2\pi} d\phi \int_0^L dz\,
   \overline{\Phi}_{j'm'n'}(\rho,\phi,z)\,\Phi_{jmn}(\rho,\phi,z)\ =
   \delta_{j'j}\,\delta_{m'm}\,\delta_{n'n} $$
holds. Now for $j'=j$, $m'=m$ and $n'=n$
$$ \int_0^{\rho_0} \rho\,d\rho\,J_m\left(\frac{x_{mn}}{\rho_0}\,\rho
   \right)^2\ =\ \frac{\rho_0^2}{2}\,J_{m+1}\left(x_{mn}\right)^2\,,$$
$$ \int_0^{2\pi}d\phi\,e^{i(m'-m)}\ =\ \int_0^{2\pi}d\phi\ =\ 2\pi\,,$$
$$ \int_0^L dz\,\sin^2\left(\frac{j \pi}{L}\,z\right) = \frac{1}{2}
   \int_0^L dz\,\left[\sin^2\left(\frac{j \pi}{L}\,z\right) +
   \cos^2\left(\frac{j \pi}{L}\,z\right)\right] = \frac{L}{2} $$
and in the convention $\nabla^2 G_D = \delta(\vec{x}^{\,'}-\vec{x})$ 
the properly normalized Green function is
$$ G_D\left(\vec{r}',\vec{r}\right) \,=\, -\frac{2}{\pi\,L\,\rho_0^2}
   \,\sum_{j=1}^{\infty} \sum_{m=-\infty}^{\infty} \sum_{n=1}^{\infty}
   \frac{\lambda^{-2}_{jmn}}{J^2_{m+1}(x_{mn})}\,\times $$
$$ \sin(\alpha_j z')\,
   \sin(\alpha_j z)\, e^{im\,(\phi'-\phi)}\,J_m(k_{mn}\rho')\,
   J_m(k_{mn}\rho)\ .  $$
\bigskip


\item Exercise E.52: Point charge at the center of a box. 
\smallskip

\noindent (a) The separation ansatz 
$$ \Phi_{lmn}(\vec{r})=F_l(x)G_m(y)H_n(z) $$ 
gives the solutions of the eigenvalue problem
$$ (\nabla^2 + k^2_{lmn})\,F_l(x)\,G_m(y)\,F_n(z) = 0 $$
with BCs $\Phi_{lmn}=0$:
$$ F_k(\xi) \,=\, G_k(\xi) \,=\, H_k(\xi) \,=\, 
  \sqrt{\frac{2}{a}}\,\begin{cases} 
  \sin(k\pi\xi/a)~~{\rm for}~~k=2,4,\,\dots\,, \\
  \cos(k\pi\xi/a)~~{\rm for}~~k=1,3,\,\dots\,. \end{cases} $$
In the convention $\nabla^2G_D=\delta(\vec{x}^{\,'}-\vec{x})$ the 
Dirichlet Green function is then given by
$$ G_D\left(\vec{r}',\vec{r}\right)\,=\, -\sum_{l,m,n=1}^{\infty} 
   \frac{F_l(x')\,G_m(y')\,H_n(z')\,F_l(x)\,G_m(y)\,H_n(z)}{k_{lmn}^2}$$
$$ {\rm with}~~~{k_{lmn}^2} = 
  \frac{\pi^2}{a^2}\,\left(l^2+m^2+n^2\right)\ .  $$
\noindent (b) For a given charge distribution is the box and given  
BCs on its surface the potential inside the box is then
$$ \Phi(\vec{r})\,=\, - 4\pi \int_V d^3x'\, \rho(\vec{r}')\,
   G_D\left(\vec{r}',\vec{r}\right) + \int_S d^3a'\, \Phi\, 
   \frac{\partial~}{\partial n'} G_D\ .$$
For our case the BC on the surface is $\Phi=0$ and the second integral 
vanishes. The charge distribution is a point charge at the origin, so
that
\begin{eqnarray} \nonumber
  \Phi(\vec{r}) &=& - 4\pi \int_V d^3x'\, q\,\delta(\vec{r}')\,
  G_D\left(\vec{r}',\vec{r}\right)\\ \nonumber &=& 
  \frac{32\,q}{\pi\,a} \sum_{l,m,n=1,3,\dots} \frac{\cos(l\pi x/a)\,
  \cos(m\pi y/a)\,\cos(n\pi z/a)}{l^2+m^2+n^2} 
\end{eqnarray} 
\noindent (c) The induced charge density on the $z=a/2$ surface is
\begin{eqnarray} \nonumber
   &~& \sigma\left(z=\frac{a}{2}\right)\ =\ \frac{1}{4\pi}\, \left. 
   \frac{\partial\Phi}{\partial z} \right|_{z=a/2}\\ \nonumber &~&=\ 
   \frac{32\,q}{4\pi^2\,a} \sum_{l,m,n=1,3,\dots} \frac{\cos(l\pi x/a)\,
   \cos(m\pi y/a)\,\sin(n\pi/2)\,(-n\pi/a)}{l^2+m^2+n^2}\\ \nonumber
 &~&=\ \frac{8\,q}{\pi\,a^2}\sum_{l,m,n=1,3,\dots} n\,(-1)^{\frac{n+1}{2}}\,
   \frac{\cos(l\pi x/a)\,\cos(m\pi y/a)}{l^2+m^2+n^2}\ .
\end{eqnarray} 
\bigskip

\bigskip \vfil

\end{enumerate}
\end{document}
