% Bernd Berg Nov 24 2016.
% \documentclass{ws-jnmp}  
% \documentclass{sfp-ws}  
 \documentclass{ws-book9x6}  

\begin{document}

% \setcounter{page}{1}

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

{\bf Set 12:} \bigskip

% \begin{figure}[tb] % \vspace{5pc}
% \centerline{\hbox{ \psfig{figure=hsphere,width=12.5cm} }}
% \caption{for assignment~14.} \label{fig_lader} \vskip -10pt
% \end{figure} \centerline{Potential as function of $z$.}

\begin{enumerate}

\setcounter{enumi}{36}

\item{\bf Expansion up to Cartesian quadrupoles.}
\medskip

\begin{enumerate}
\item For $|\vec{x}|>|\vec{x}\,'|$ we write
\begin{eqnarray} \nonumber
  \frac{1}{|\vec{x}-\vec{x}\,'|}\ =\ (r^2-2\,\vec{x}\cdot\vec{x}\,'
  +r'^2)^{-1/2} &=& \\ \nonumber 
  r^{-1}\,(1-2\,\vec{x}\cdot\vec{x}\,'/r^2 +r'^2/r^2)^{-1/2}
  &=& r^{-1}\,(1+\epsilon)^{-1/2}
\end{eqnarray}
with $\epsilon=-2\,\vec{x}\cdot\vec{x}\,'/r^2 +r'^2/r^2$ and 
$|\epsilon|<1$. The Taylor expansion of $f(1+\epsilon)$ about
$f(1)$ reads
$$ f(1+\epsilon)\ =\ f(1) + \epsilon\,f'(1) + \frac{\epsilon}{2}\,
   f''(1) + {\cal O}(\epsilon^3)    $$
and in out situation
\begin{eqnarray} \nonumber
  f(1+\epsilon)\ =\ (1+\epsilon)^{-1/2}\,,~~
   f'(1+\epsilon)&=&-\frac{1}{2}\,(1+\epsilon)^{-3/2}\\ \nonumber
   {\rm and}~~f''(1+\epsilon)&=&\frac{3}{4}\,(1+\epsilon)^{-5/2}\,.
\end{eqnarray}
Therefore, $f(1)=1$, $f'(1)=-1/2$ and $f''(1)=3/4$ and up to order 
$\epsilon$ we have
$$ \frac{1}{|\vec{x}-\vec{x}\,'|}\ =\ \frac{f}{r}\ =\ 
   \frac{1}{r}+\frac{\vec{x}\cdot\vec{x}\,'}{r^3}
   -\frac{1}{2}\,\frac{r'^2}{r^3}  $$ 
and integration over the first two terms gives
$$ \int d^3x'\,\frac{\rho(\vec{x}\,')}{|\vec{x}-\vec{x}\,'|}\ =\ 
   \frac{q}{r} + \frac{\vec{p}\cdot\vec{x}}{r^3}\,. $$
From the order $\epsilon^2$ we keep only the term $(2\,\vec{x}\cdot
\vec{x}\,')^2/r^4$. So we are left with the task to integrate
$$ -\frac{1}{2}\,\frac{r'^2}{r^3} + \frac{3}{4}\,\frac{(2\,\vec{x}
   \cdot\vec{x}\,')^2}{r^4}\,,  $$ 
which results in order of the terms in
$$ -\frac{1}{2}\sum_{i=1}^3\sum_{j=1}^3Q^{ij}_D\,\frac{x^ix^j}{r^5}
   +\frac{1}{2}\sum_{i=1}^3\sum_{j=1}^3Q^{ij}_S\,\frac{x^ix^j}{r^5}\,.$$
Note:
\begin{eqnarray} \nonumber
  \sum_{i=1}^3\sum_{j=1}^3 x^i\,Q_D^{ij}\,x^j &=& r^2\,\int d^3x'\,r'^2\
  \rho(\vec{x}\,')\,,\\ \nonumber 
  \int d^3x'\,(\vec{x}\cdot\vec{x}\,')^2\,\rho(\vec{x}\,') &=&
  \vec{x}\cdot \int d^3x'\,\vec{x}\,'\,\rho(\vec{x}\,')\,\vec{x}\,'
  \cdot\vec{x}\ =\ \sum_{i=1}^3\sum_{j=1}^3x^i\,Q_S^{ij}\,x^j\,.
\end{eqnarray}
\medskip
\end{enumerate}
\medskip

\item We write the quadrupole moments as
\begin{eqnarray} \nonumber
  Q^{ij} &=& Q^{ij}_S - Q^{ij}_D~~{\rm with}~~
  Q^{ij}_S = 3\,\int d^3x\,x^i\,x^j\,\rho(\vec{x})\,,~~
  Q^{ij}_D = \int d^3x\,r^2\,\delta^{ij}\,\rho(\vec{x})\,.
\end{eqnarray}
Results:
\begin{eqnarray} \nonumber
  q_{22} &=& \frac{1}{4}\,\sqrt{\frac{15}{2\pi}} \int d^3x\,
  (x-iy)^2\rho(\vec{x})\ =\ \frac{1}{4}\,\sqrt{\frac{15}{2\pi}} 
  \int d^3x\, (x^2-ixy-y^2)^2\rho(\vec{x}) \\ \nonumber &=&
  \frac{1}{12}\,\sqrt{\frac{15}{2\pi}}\,\left(Q^{11}_S-2i\,Q^{12}
  -Q^{22}_S\right) = \frac{1}{12}\,\sqrt{\frac{15}{2\pi}}\,
  \left(Q^{11}-2i\,Q^{12}-Q^{22}\right)\,, \\ \nonumber
  q_{21} &=& -\sqrt{\frac{15}{8\pi}} \int d^3x\,z\,(x-iy)\,
  \rho(\vec{x})\ =\ -\frac{1}{3}\,\sqrt{\frac{15}{8\pi}}\,
  \left(Q^{13}-i\,Q^{23}\right)\,,\\ \nonumber
  q_{20} &=& \frac{1}{2}\,\sqrt{\frac{5}{4\pi}} \int d^3x\,(3z^2-r^2)
  \,\rho(\vec{x})\ =\ \frac{1}{2}\,\sqrt{\frac{5}{4\pi}} \left(Q^{33}_S
  - Q^{33}_D\right)\ =\ \frac{1}{2}\,\sqrt{\frac{5}{4\pi}}\,Q^{33}\,.
\end{eqnarray}
\bigskip

\item Exercise E.55: Number of Cartesian multipoles.
\smallskip

There are $l+1$ ways to distribute $l$ entries on two histograms. 
So, for three histograms we have
$$ 1 + 2 + \dots + l + (l+1) = (l+1)\,(l+2)/2 $$
possibilities.
\bigskip


\item{\bf Cartesian quadrupole moment of an ellipsoid.} 
\medskip

\noindent (a) We have
\begin{eqnarray} \nonumber
  \left(\frac{x^1}{c^1}\right)^2 + \left(\frac{x^2}{c^2}\right)^2 +
  \left(\frac{x^3}{c^3}\right)^2 &=& 1~~~{\rm with}~~~
  c^1=c^2=a\,,~~c_3=b \\ \nonumber
  {\rm and\ charge\ density}~~ \rho(\vec{x}) &=& 
  \begin{cases} \rho = {\rm const}~~{\rm within\ the\
            ellipsoid,}\\ 0~~~{\rm otherwise.} \end{cases} 
\end{eqnarray}
The total charge is 
\begin{eqnarray} \nonumber
  q&=&\int_{\sum_{i=1}^3\left(\frac{x^i}{c^i}\right)^2\le 1}d^3x\,\rho
  \,=\, c^1\,c^2\,c^3\,\rho \int_{\sum_{i=1}^3 (x'^i)^2 \le 1} d^3x'\\
  \nonumber &=& \frac{4\pi}{3}\, c^1\,c^2\,c^3\,\rho 
  \,=\, \frac{4\pi}{3}\, a^2\,b\,\rho 
\end{eqnarray}
We write the quadrupole moments of the ellipsoid as
\begin{eqnarray} \nonumber
  Q^{ij} &=& Q^{ij}_S - Q^{ij}_D~~{\rm with}~~
  Q^{ij}_S = 3\,\rho\int_{\sum_{i=1}^3\left(\frac{x^i}{c^i}\right)^2\le 1}
          d^3x\,x^i\,x^j\,~~
  Q^{ij}_D = \rho\int_{\sum_{i=1}^3\left(\frac{x^i}{c^i}\right)^2\le 1}
          d^3x\,r^2\,\delta^{ij}\,.
\end{eqnarray}
These moments transform into
\begin{eqnarray} \nonumber
  Q^{ij}_S &=& 3\,c^1\,c^2\,c^3\,\rho \int_{\sum_{i=1}^3(x'^i)^2\le 1} 
  d^3x'\,c^i\,c^j\,x'^i\,x'^j \\ \nonumber
  Q^{ij}_D &=& 
   c^1\,c^2\,c^3\,\rho \int_{\sum_{i=1}^3 (x'^i)^2 \le 1} d^3x' 
   \delta^{ij}\sum_{k=1}^3 (c^k\,x'^k)^2\ .
\end{eqnarray}
We have $Q^{ij}_D=0$ for $i\ne j$ and the relevant integrals for
$Q^{ij}_S$ are of the form
$$ \int_{\sum_{i=1}^3 (x^i)^2 \le 1} d^3x\,x^i\,x^j = \delta^{ij} 
   \int_{\sum_{i=1}^3 (x^i)^2 \le 1} d^3x\,(x^i)^2\ 
   \Rightarrow\  Q^{ij} = 0~~{\rm for}~~i\ne j $$
and
$$ \int_{\sum_{i=1}^3 (x^i)^2 \le 1} d^3x\,(x^i)^2 = 
   \int_{r=|\vec{x}|\le 1} d^3x\,(x^3)^2 = 2\pi\int_0^1 r^4\,dr\,
   \int_{-1}^{+1} (\cos\theta)^2\,d\cos\theta = \frac{2\pi}{5}\,
   \frac{2}{3} = \frac{4\pi}{15} $$
$$ \Rightarrow~~ Q^{ii}_S = \frac{q}{5}\,3\,(c^i)^2\,,~~
   Q^{ii}_D = \frac{q}{5}\,\sum_{k=1}^3 (c^k)^2\,.          $$
With $c^1=c^2=a$ and $c_3=b$:
$$ Q^{11}_S=Q^{22}_S=\frac{3\,q}{5}\,a^2\,,~~ Q^{33}_S=
   \frac{3\,q}{5}\,b^2\,,~~Q^{11}_D=Q^{22}_D=Q^{33}_D=
   \frac{q}{5}\,(2\,b^2+a^2)\,                              $$
resulting into the traceless tensor given by
$$ Q^{11}=Q^{22}=\frac{q}{5}\,\left(a^2-b^2\right),\qquad
   Q^{33}=\frac{2\,q}{5}\,\left(b^2-a^2\right)\,.  $$
(b) The energy is
$$ W = q\,\Phi(0) - \vec{p}\cdot\vec{E}(0) - \frac{1}{6}\sum_{i=1}^3
   \sum_{j=1}^3 Q^{ij}_S\,\frac{\partial E^j}{\partial x^i}\ . $$
With the normalization $\Phi(0)=0$ and
$$ \vec{p} = \rho \int_{\sum_{i=1}^3\left(\frac{x^i}{c^i}\right)^2\le 1}
   d^3x\,\vec{x} = 0 $$
the result for an external electric field $\vec{E}=E(z)\,\hat z$ becomes
$$ W = -\frac{1}{6}\,Q^{33}_S\,\frac{d\,E}{d\,z} = -
    \frac{q\,b^2}{10}\,\frac{d\,E}{d\,z}\ .  $$
\bigskip 


\item Exercise E.58: Dielectric sphere.  \smallskip

\noindent (a) We expand the potential into spherical harmonics. Due 
to the axial symmetry we have only $m=0$ contributions, which are
Legendre polynomials.
\begin{eqnarray} \nonumber
  {\rm Inside:}~~~~\Phi_{\rm in} &=& \sum_{l=0}^{\infty} A_l\,r^l\,
  {P}_l(\cos\theta)\,, \\ \nonumber
  {\rm Outside:}~~~~\Phi_{\rm out} &=& \sum_{l=0}^{\infty} \left[
  \,B_l\,r^l + C_l\,r^{-(l+1)}\,\right]\,{P}_l(\cos\theta)\,.
\end{eqnarray}
For $r\to\infty$: $\Phi\to -E_0\,z = -E_0\,r\,\cos\theta$. This
implies that the only non-vanishing coefficient $B_l$ is
$$ B_1\ =\ -E_0\ .$$
BCs at $r=R$:
\begin{eqnarray} \nonumber
  {\rm Tangential:}~~~~\frac{1}{R}\,\left. \frac{\partial\Phi_{\rm in}}
  {\partial\theta}\right|_{r=R} &=& \frac{1}{R}\,\left. 
  \frac{\partial\Phi_{\rm out}}{\partial\theta}\right|_{r=R}\,, 
  \\ \nonumber
  {\rm Normal:}~~~~\epsilon_1\,\left. \frac{\partial\Phi_{\rm in}}
  {\partial r}\right|_{r=R} &=& \epsilon_2\,\left. 
  \frac{\partial\Phi_{\rm out}}{\partial r}\right|_{r=R}\,. 
\end{eqnarray}
Matching for the tangential BC $\partial{P}_l/\partial\theta$ 
term by term (they are independent functions of $\theta$), we find:
\begin{eqnarray} \nonumber
  A_1\,R\,=\,B_1\,R + \frac{C_1}{R^2}\ \Rightarrow\
  A_1 &=& -E_0 + \frac{C_1}{R^3}\\ \nonumber
{\rm and\ for}~~l\ge 2:~~ A_l &=& \frac{C_l}{R^{(2l+1)}}\ .
\end{eqnarray}
% 
\begin{figure}[tb]
\vspace{5pc}
 \centerline{\hbox{ \psfig{figure=figures/sol01a.pdf,width=9cm} }}
% \caption{ }  \label{fig_fort}
\end{figure}
%
\begin{figure}[tb]
\vspace{5pc}
 \centerline{\hbox{ \psfig{figure=figures/sol01b.pdf,width=9cm} }}
% \caption{ }  \label{fig_fort}
\end{figure}
%
Similarly we match for the normal BC $P_l$ term by term and find:
\begin{eqnarray} \nonumber
  \epsilon_1\,A_1 &=& -\epsilon_2\,E_0-2\,\epsilon_2\,\frac{C_1}{R^3}
  \\ \nonumber {\rm and\ for}~~l\ge 2:~~ \epsilon_1\,l\,A_l &=& 
                  - (l+1)\,\epsilon_2\,\frac{C_l}{R^{(2l+1)}}\ .
\end{eqnarray}
Putting the $l\ge 2$ equations together we get
$$ \frac{C_l}{R^{(2l+1)}} = - \frac{(l+1)\,\epsilon_2}{l\,\epsilon_1}\,
   \frac{C_l}{R^{(2l+1)}}\ \Rightarrow\ C_l=0,~~l\ge 2\
   \Rightarrow\ A_l=0,~~l\ge 2\ .  $$
Let us define
\begin{eqnarray} \nonumber
  \epsilon\ =\ \frac{\epsilon_1}{\epsilon_2}\ .
\end{eqnarray}
With this notation we have
$$  \epsilon\,A_1\,=\,-E_0-2\,\frac{C_1}{R^3}\ .$$
Combining this with our other equation for $A_1$ gives
$$ A_1\,=\, -\left(\frac{3}{\epsilon+2}\right)\,E_0\,,\qquad
   C_1\,=\, \left(\frac{\epsilon-1}{\epsilon+2}\right)\,R^3\,E_0\,.$$
Therefore,
\begin{eqnarray} \nonumber
  \Phi_{\rm in} &=& -\left(\frac{3}{\epsilon+2}\right)\,E_0\,r\,
  \cos(\theta) \,=\, -\left(\frac{3}{\epsilon+2}\right)\,E_0\,z\,,
  \\ \nonumber \Phi_{\rm out} &=& -E_0\,z + 
  \left(\frac{\epsilon-1}{\epsilon+2}\right)\,E_0\,\frac{R^3}{r^2}\,
  \cos(\theta)\,.
\end{eqnarray}
Note that the last term is the potential of a dipole. The electric
fields are then
\begin{eqnarray} \nonumber
  \vec{E}_{\rm in} &=& -\nabla\,\Phi_{\rm in} \,=\,
  \left(\frac{3}{\epsilon+2}\right)\,\vec{E}_0\,, \\ \nonumber
  \vec{E}_{\rm out} &=& -\nabla\,\Phi_{\rm out} \,=\, \vec{E}_0 - 
  \left(\frac{\epsilon-1}{\epsilon+2}\right)\,R^3\,\left(
  \frac{r^2\,\vec{E}_0-3\vec{r}\,(\vec{r}\cdot\vec{E}_0)}{r^5}
  \right)\,.
\end{eqnarray}
(b) Examples are given in the two figures.

\noindent (c) The surface charge density is
\begin{eqnarray} \nonumber
  \sigma_{\rm pol} &=& \frac{1}{4\pi}
  \left( E^r_{\rm out}-E^r_{\rm in} \right) \,=\, \frac{1}{4\pi}\,
  \hat{r}\cdot\left(\vec{E}_{\rm out}-\vec{E}_{\rm in}\right) 
  \\ \nonumber &=& \frac{1}{4\pi}\,E_0\,\cos(\theta)\,
  \left[\left(1-\frac{3}{\epsilon+2}\right) - \left(
  \frac{\epsilon-1}{\epsilon+2}\right)\,\left(1-3\right)\right]
  \\ \nonumber &=& \frac{3}{4\pi}\,\left(
  \frac{\epsilon-1}{\epsilon+2}\right)\,E_0\,\cos(\theta)\ .
\end{eqnarray}
\medskip

\end{enumerate}
\end{document}
