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\begin{document}

\centerline{\bf Electrodynamics B (PHY 5347): Spring 2017 
Solutions for the Test on Homework.} \bigskip

\begin{enumerate}

\item {\bf Magnetism in matter.}

\begin{enumerate}
\setcounter{page}{1}
\item For this magnetostatics problem with no free currents the relevant
Maxwell equations are
$$ \nabla\cdot\vec{B}=0~~~{\rm and}~~~\nabla\times\vec{H}=0 $$
with the associated boundary conditions (BCs)
$$ \left.\hat{r}\cdot\left(\vec{B}_2-\vec{B}_1\right)\right|_{r=a,b}
   =0\,,~~~\left.\hat{r}\times\left(\vec{H}_2-\vec{H}_1\right)
   \right|_{r=a,b} = 0\,.$$

\item Since $\nabla\times\vec{H}=0$ we have $\vec{H}=-\nabla\Phi$
and thus $0=\nabla\cdot\vec{B}=\nabla\cdot\left(\mu\vec{H}\right)$.

From the BCs we get
$$ 0=\hat{r}\cdot\left(\vec{B}_2-\vec{B}_1\right)=
   \hat{r}\cdot\left(\mu_2\vec{H}_2-\mu_1\vec{H}_1\right)=
   -\mu_2\frac{\partial\Phi_2}{\partial\,r}
   +\mu_1\frac{\partial\Phi_1}{\partial\,r}               $$
and (because of azimuthal symmetry $\Phi(\vec{x})=\Phi(r,\theta)$ holds)
$$ 0=\hat{r}\times\left(\vec{H}_2-\vec{H}_1\right)
    =\hat{r}\times\left(-\nabla\Phi_2+\nabla\Phi_1\right)
    =\frac{\hat{r}\times\hat{\theta}}{r}\left(
    -\frac{\partial\Phi_2}{\partial\,\theta}
    +\frac{\partial\Phi_1}{\partial\,\theta}\right)\,.     $$
Finally, the behavior of the field at infinity:
$$ \lim_{r\to\infty} \vec{B}(\vec{r}) = \vec{B}_0 = B_0\,\hat{z}\
   \Rightarrow\ \lim_{r\to\infty} \Phi(\vec{r}) = - B_0\,z =
   -B_0\,r\,\cos\theta = -B_0\,r\,P_1(\cos\theta)\,.  $$
\end{enumerate}



\item{\bf Covariant retarded Green function.}
\smallskip

\noindent With $\tau=x^0-y^0$ the zeros of $\delta[(x-y)^2]$ are at
$$ \tau^0=\pm\sqrt{(\vec{x}-\vec{y}\,)^2}=\pm|\vec{x}-\vec{y}\,|\,.$$
Let now $f(\tau)=\tau^2-\left(\vec{x}-\vec{y}\,\right)^2$, so that
$$ \left.\frac{d f}{d \tau}\right|_{\tau=\tau^0}=f'(\tau^0)=2\,\tau^0$$
holds. Therefore,
\begin{eqnarray} \nonumber
 \delta[(x-y)^2]&=& \frac{1}{2\,|f'(\tau^0)|}\left[\delta(\tau-|\vec{x}
     -\vec{y}\,|) + \delta(\tau+|\vec{x}-\vec{y}\,|)\right] \\ \nonumber
 &=& \frac{1}{2\,|\vec{x}-\vec{y}\,|}\left[\delta(x^0-y^0-|\vec{x}
     -\vec{y}\,|)+\delta(x^0-y^0+|\vec{x}-\vec{y}\,|)\right] 
\end{eqnarray}
and the identity of the two expressions for the Green function follows:
$$ \frac{1}{2\pi}\,\theta(x^0-y^0)\,\delta((x-y)^2) = \frac{
   \delta(x^0-y^0-|\vec{x}-\vec{y}\,|)}{4\pi\,|\vec{x}-\vec{y}\,|}\,.$$
\bigskip


\noindent
{\bf Charge conjugation, parity and time reversal in electrodynamics.}
\smallskip

\noindent (1) Using the given values $(-,+,+)$ for $\rho$ as input, 
our $(C,P,T)$ results are obtained as follows:
\begin{eqnarray} \nonumber
  1.~A^0 \sim \rho\ &\Rightarrow&\ (-,+,+)\,,\\ \nonumber
  2.~\vec{J} \sim \vec{v}\,\rho\ &\Rightarrow&\ (-,-,-)\,,\\ \nonumber
  3.~\vec{A} \sim \vec{J}\ &\Rightarrow&\ (-,-,-)\,,\\ \nonumber
  4.~E^i=F^{i0}=\partial^iA^0-\partial^0A^i\ &\Rightarrow&\ (-,-,+)\,,
  \\ \nonumber
  5.~B^k\sim F^{ij}=\partial^iA^j-\partial^jA^i\ &\Rightarrow&\  
     (-,+,-)\,.
\end{eqnarray}
\noindent (2) From the explicit matrix forms we find
$$ F^{\alpha\beta}\,F_{\alpha\beta}=-2\,\vec{E}^{\,2}+2\,\vec{B}^{\,2}
   ~~~{\rm and}~~~^*F^{\alpha\beta}\, F_{\alpha\beta} = 
   -4\,\vec{E}\cdot\vec{B}\,.$$
Using the $(C,P,T)$ results above, we find $(+,+,+)$ for
$F^{\alpha\beta}\,F_{\alpha\beta}$ (proper scalar), and $(+,-,-)$ 
for $^*F^{\alpha\beta}\, F_{\alpha\beta}$ (pseudoscalar).
\bigskip


{\bf TM waves in a rectangular wave guide.}
\smallskip

\begin{enumerate}
\item 
$$ (\nabla_t^2 + \gamma^2)\, E^z = 0\,,\qquad E^z|_S = 0\,, $$
$$ E^z = E^0\, \sin\left(\frac{m\pi x}{a}\right)\,
               \sin\left(\frac{n\pi y}{b}\right)\,
               e^{i\,k\,z}\,e^{-i\,\omega\,t}\,,            $$
$$ \gamma^2_{n\,m} = \pi^2\,\left[\left(\frac{m}{a}\right)^2 +
                            \left(\frac{n}{b}\right)^2 \right]\,,
                   ~~~m=1,2,\dots\,,~~ n=1,2,\dots\,,          $$
\item
$$ \gamma^2_{1\,1} = \pi^2\,\left(a^{-2}+b^{-2}\right)\,,~~~~~
   \omega_{1\,1} = \frac{c\,\pi}{\sqrt{\mu\,\epsilon}}\
   \sqrt{a^{-2}+b^{-2}}                                        $$ 
\end{enumerate}

\end{enumerate}
\end{document}
Behavior of the field at infinity:
