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\begin{document}

\centerline{\bf Electrodynamics A (PHY 5346) Fall 2016 
Solutions Final December~14.} \bigskip

\begin{enumerate}

\item{Electron-positron annihilation.}  \smallskip 

Let us use natural units with $c=1$ and denote the four-vectors of the
photons by $p$ and $q$. Energy conservation:
$$ p^0+q^0=2E~~{\rm with}~~E=0.511\,[MeV]\,.$$
Momentum conservation and massless photons:
$$ \vec{p}=-\vec{q}~~{\rm and}~~p^2=q^2=0\,.$$
\begin{enumerate}
\item Therefore, we have in the frame where the photon with momentum $p$
moves along the positive $x^1$ direction
$$\left(p^{\alpha}\right) = 
  \left(\matrix{ E\cr E\cr 0\cr 0}\right)~~{\rm and}~~
  \left(q^{\alpha}\right) = 
  \left(\matrix{ E\cr -E\cr 0\cr 0}\right)\,.$$
\item $\beta=3/5\ \rightarrow\ \gamma=1/\sqrt{1-\beta^2}= 5/4$ and the
 energy is the zero component of a four-vector transforms according to
$$ E' = \gamma\,(1-\beta)\,E = \frac{1}{2}\,E = 0.2555\,[MeV]\,.$$
\end{enumerate}  \medskip


\item {\bf Coaxial cable.} \smallskip

\noindent Amp\'ere's law reads
$$ \nabla\times\vec{B}\ =\ \frac{4\pi}{c}\,\vec{J} $$
where in our case $\vec{J}=J\,\hat{z}$ and
$$ J\,=\,J(\rho)\,=\,\cases{
   J_1 = I/(\pi\,\rho_1^2)~~{\rm for}~~\rho\le\rho_1\,,\cr
   0~~{\rm for}~~\rho_1<\rho<\rho_2\,,\cr
   J_2 = - I/(\pi\,\rho_3^2-\pi\,\rho_2^2)
         ~~{\rm for}~~\rho_2\le\rho\le\rho_3\,,\cr
   0~~{\rm for}~~\rho>\rho_3\,.  } $$
Now $\vec{B}=B_{\phi}\,\hat{\phi}$ with $B_{\phi}=B=|\vec{B}|$ and
$$ 2\pi\,\rho\,B_{\phi} = \rho \int_0^{2\pi} d\phi\,B_{\phi} =
   \oint d\vec{s}\cdot\vec{B} = 
   \int_S da\,\hat{a}\cdot(\nabla\times\vec{B}) =
   \frac{4\pi}{c} \int_S da\,\hat{a}\cdot\vec{J} $$
$\rho\le\rho_1$:
$$ 2\pi\,\rho\,B_{\phi} = \frac{4\pi}{c}\,J_1\,\pi\,\rho^2\
   \Rightarrow\ B = \frac{2}{c}\,J_1\,\pi\,\rho = 
   \frac{2\,I\,\rho}{c\,\rho_1^2}\ .  $$
$\rho_1<\rho<\rho_2$:
$$ 2\pi\,\rho\,B_{\phi} = \frac{4\pi}{c}\,J_1\,\pi\,\rho_1^2\ 
   \Rightarrow\ B = \frac{2\,I}{c\,\rho}\ .  $$
$\rho_2\le\rho\le\rho_3$:
$$ B = \frac{2\,I}{c\,\rho} + \frac{2\,J_2}{c\,\rho}\,\pi\,
       (\rho^2-\rho_2^2) = \frac{2\,I}{c\,\rho}\,
       \frac{(\rho_3^2-\rho^2)}{(\rho_3^2-\rho_2^2)}\ .$$
Finally, $\rho>\rho_3$:
$$ B = 0\ .$$
\bigskip


\item{Potential from distinct BCs on half-spheres.} \smallskip

\begin{enumerate}
\item We have to calculate
$$  \Phi(r) = R^2 \int_S d\Omega'\, \Phi_0 \left. 
    {\partial G_D\over \partial r'} \right|_{r'=R}\,,  $$
where the surface is the upper half-sphere and the derivative of the 
Green function is given by
$$ \left. {\partial G_D\over \partial r'} \right|_{r'=R} =
    {R^2-r^2\over 4\pi\,R\,\left(R^2+r^2-2Rr\,\cos\gamma\right)^{3/2}}$$
with $\cos\gamma=\cos (\theta)\,\cos (\theta')+\sin (\theta)\,
\sin(\theta')\,\cos(\phi-\phi')$. For $r=z$ on the $z$-axis we have
$\theta=0$ and, hence, $\cos\gamma=\cos\theta'$. The integral over
the upper half-sphere becomes
$$  \Phi(z) = \Phi_0\,\frac{R}{2}\,\left(R+z\right)\,
    \left(R-z\right)\,\int_0^1 {d\cos\theta'\over 
    \left(R^2+z^2-2R\,z\,\cos\theta'\right)^{3/2}}\,.$$
With $a=R^2+z^2$ and $b=2Rz$ the integral is
\begin{eqnarray} \nonumber 
   \int_0^1 {dx\over\left(a-b\,x\right)^{3/2} } &=& 
   \left. {2\over b\,\left(a-b\,x\right)^{1/2} }\right|_0^1\ 
   =\ {2\over b\,\sqrt{a-b}}-{2\over b\,\sqrt{a}} \\ \nonumber
 &=& {1\over Rz\,\sqrt{(R-z)^2}}-{1\over Rz\,\sqrt{R^2+z^2}}
\end{eqnarray}
and
\begin{eqnarray} \nonumber 
  \Phi(z) = \Phi_0\,\frac{1}{2z}\,(R+z)\,\left(1 -
  \frac{R-z}{\sqrt{R^2+z^2}} \right)\,.
\end{eqnarray}

\item Special values: % EMScript/figures_plt/hsphere/hsphere.f
\begin{eqnarray} \nonumber 
  z\ =\ -R\ &\Rightarrow&\ \Phi(-R)\ =\ 0\,,\\ \nonumber
  z\ =\ - \frac{R}{2}\ &\Rightarrow&\ \Phi\left(-\frac{R}{2}\right)\ 
     =\ \Phi_0\,\left(-\frac{1}{2}\right)\,\left(1-\frac{3}{\sqrt{5}}
      \right)\ =\ 0.17082\,\Phi_0 \\ \nonumber
  z\ =\ \frac{R}{2}\ &\Rightarrow&\ \Phi\left(\frac{R}{2}\right)\ 
     =\ \Phi_0\,\left(\frac{3}{2}\right)\,\left(1-\frac{1}{\sqrt{5}}
      \right)\ =\ 0.82918\,\Phi_0 \\ \nonumber
  z\ =\ R\ &\Rightarrow&\ \Phi(R)\ =\ \Phi_0\,\frac{2R}{2R}
   \ =\ \Phi_0\,,\\ \nonumber
  z\ \to\ 0\ &\Rightarrow&\ \Phi(0)\ =\ \Phi_0\,\frac{R}{2z}\,
     \left(1-\frac{R-z}{R}\right)\ \to\ \frac{1}{2}\,\Phi_0\,.
\end{eqnarray} \bigskip
The plot is shown in the figure. % (next page).
\end{enumerate}  \medskip

\begin{figure}[th] % EMScript/figures_plt/hsphere/hsphere.plt
\centerline{\hbox{ \psfig{figure=hsphere,width=12.5cm} }}
\caption{Potential as function of $z$.} \label{fig_hsphere}
% \vskip -10pt
\end{figure}

\end{enumerate} 
\end{document}
