%% Bernd Berg sep 14 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 4:} \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}{11}

\item Exercise: End of Spacetrip.  \smallskip

We use energy-momentum conservation
$$ 0 = dp = \left(
      \begin{matrix}dp^0_r\cr dp^1_r\cr 0\cr 0\end{matrix}\right) +
\left(\begin{matrix}dp^0_e\cr dp^1_e\cr 0\cr 0\end{matrix}\right)\,,$$
where the subscript $r$ stands for rocket and $e$ for exhaust. In 
the temporary rest frame of the rocket we have 
$$  dp^1_e=-v\,dm\,,\qquad dp^1_r=m\,du=m\,g\,d\tau\,,$$
where $v$ is the velocity of the exhaust, $dm$ the infinitesimal 
change of rest mass of the rocket  and $du$ the infinitesimal velocity 
of the rocket. Note that the exhaust may have no rest mass and our 
equation for $dp^1_e$ (no $\gamma$) is not obvious. It follows from
$$dp^0_e=-dp^0_r=c\,dm~~{\rm and}~~-\beta=\frac{dp^1_e}{dp^0_e}\,.$$
With this definition $\beta=v/c$ is positive, because $dp^1_e$ is
negative when we choose $dp^1_r$ positive.
Now, separation of variables gives
$$ \frac{dm}{m} = - \frac{g}{v}\,d\tau\ \Rightarrow\ 
   \int_{m_0}^{m(\tau)} \frac{dm}{m} = 
   - \frac{g}{v} \int_0^{\tau}d\tau'\ \Rightarrow
   \ \ln\left(\frac{m(\tau)}{m_0}\right) = -\frac{g\,\tau}{v}\,. $$
(1) The mass of the spaceship decreases with its proper time according to
$$ m(\tau) = m_0\,\exp\left(-\frac{g\tau}{v}\right)\,.$$
(2) With $v=0.66\,c$ and after a trip of twenty years the remaining 
fraction of the mass is $m(20\,{\rm years})/m_0=2.68\times 10^{-14}$.

\noindent (3) With $v=c$ it becomes $m(20\,{\rm years})/m_0=1.1\times 
10^{-9}$.
\bigskip


\item Exercise: Redshift. \smallskip

\noindent The equation for the redshift is
\begin{eqnarray} \nonumber
  \lambda'\ =\ \lambda\,\sqrt{\frac{1+\beta}{1-\beta}}\ &\Rightarrow&\
  \left(\frac{\lambda'}{\lambda}\right)^2\ =\ {\frac{1+\beta}{1-\beta}}
  \\ \nonumber
  \left(\frac{\lambda'}{\lambda}\right)^2 -\beta\,
  \left(\frac{\lambda'}{\lambda}\right)^2 &=& 1+\beta
  \\ \nonumber
  \left(\frac{\lambda'}{\lambda}\right)^2 - 1\ =\ \beta\,\left[ 1
  + \left(\frac{\lambda'}{\lambda}\right)^2 \right]\ &\Rightarrow&\
  \beta\ =\ \frac{(\lambda'/\lambda)^2-1}{(\lambda'/\lambda)^2+1}\,.
\end{eqnarray}
With $\lambda'=(729.2\,[nm])\,m^2/(m^2-4)$ and $\lambda=(364.56\,[nm])
\,m^2/(m^2-4)$ we find
$$ \beta\ =\ \frac{v}{c}\ =\ 0.6\,.$$
\bigskip


\item Exercise: Relativistic energy-momentum conservation.

\begin{figure} \begin{center} % figures/modphys.f, modphys.eps.
\includegraphics[width=0.90\textwidth]{figures/modphys} 
\caption{Comparison of $f(M)=2m-M$ with photon energy $E_{\gamma}(M)$.} 
\label{fig_emog} 
\end{center} \end{figure} 

\noindent 1. The 4-vector energy-momentum conservation reads
$$ \left(\begin{matrix} m\cr 0\cr 0\cr 0 \end{matrix}\right) + 
   \left(\begin{matrix} m\cr 0\cr 0\cr 0 \end{matrix}\right) = 
   \left(\begin{matrix}\sqrt{M^2+p^2}\cr -p\cr0\cr 0\end{matrix}\right) 
   + \left(\begin{matrix}E_{\gamma}\cr p_{\gamma}\cr 0\cr 0\end{matrix} 
   \right)  $$
Therefore (eliminating $p$ via $p=p_{\gamma}=E_{\gamma}$),
$$ 2m - E_{\gamma} = \sqrt{M^2+E_{\gamma}^2}, $$
$$ (2m-E_{\gamma})^2 = 4m^2-4m\,E_{\gamma} + E^2_{\gamma}
    = M^2+E_{\gamma}^2\,, $$
$$ E_{\gamma} = \frac{4m^2-M^2}{4m}\,. $$

\noindent 2. The requested $E_{\gamma}(M)$ values are: 
$E_{\gamma}(0)=m$,
\begin{eqnarray} \nonumber
  E_{\gamma}\left(\frac{m}{2}\right)=\frac{15\,m}{16}\,,~~
  E_{\gamma}(m)=\frac{3\,m}{4}\,,~~
  E_{\gamma}(\sqrt{2}\,m)=\frac{m}{2}\,,~~
  E_{\gamma}(\sqrt{3}\,m) &=& \frac{m}{4}\,. 
\end{eqnarray} 

\noindent 3. The sketch is given in the figure.
\bigskip \bigskip


\item{\bf Electromagnetic field tensor in $\vec{E}$ and $\vec{B}$ fields.}

\noindent A. From the $J^0$ component we get
$$ \partial_0F^{00}+\partial_1F^{10}+\partial_2F^{20}+\partial_3F^{30}=
   \frac{4\pi}{c}\,J^0 = 4\pi\,\rho = \nabla\cdot\vec{E} =
   \partial_1E^1+\partial_2E^2+\partial_3E^3\,. $$
Therefore, $F^{00}=0$ by anti-symmetry, $F^{10}=E^1$, $F^{20}=E^2$ and
$F^{30}=E^3$.

From the $J^1$ component we get
$$ \partial_0F^{01}+\partial_1F^{11}+\partial_2F^{21}+\partial_3F^{31}=
   \frac{4\pi}{c}\,J^1 = \partial_2B^3-\partial_3B^2 - \partial_0E^1\,.$$
Therefore, $F^{01}=-E^1$ consistent with $F^{10}=E^1$, $F^{11}=0$ by 
anti-symmetry, $F^{21}=B^3$ and $F^{31}=-B^2$.

From the $J^2$ component we get
$$ \partial_0F^{02}+\partial_1F^{12}+\partial_2F^{22}+\partial_3F^{32}=
   \frac{4\pi}{c}\,J^2 = \partial_3B^1-\partial_1B^3 - \partial_0E^2\,.$$
Therefore, $F^{02}=-E^2$, $F^{12}=-B^3$ both consistent with anti-symmetry,
$F^{22}=0$ by anti-symmetry and (new) $F^{32}=B^1$.

Using anti-symmetry all components are now determined and the $F^{\alpha\beta}$
field tensor reads
$$ \left(F^{\alpha\beta}\right)=\left(\begin{matrix} 
    0  & -E^1& -E^2& -E^3\cr
    E^1&  0  & -B^3&  B^2\cr
    E^2&  B^3&  0  & -B^1\cr
    E^3& -B^2&  B^1&  0  \cr\end{matrix}\right)\,.$$
The remaining equations from $\beta=3$ can be used for consistency checks.
\bigskip

\noindent B. For $\beta=0$: 
$$\partial_{\alpha}\,^*F^{\alpha 0}=\partial_1\,B^1
+\partial_2\,B^2+\partial_2\,B^2=\nabla\cdot\vec{B}=0\,.$$
For $\beta=1$: 
$$\partial_{\alpha}\,^*F^{\alpha 1}=-\partial_0\,B^1-\partial_2\,E^3
+\partial_3\,E^2=-\partial_0\,B^1-(\nabla\times\vec{E}\,)^1=0\,.  $$
For $\beta=2$: 
$$\partial_{\alpha}\,^*F^{\alpha 2}=-\partial_0\,B^2+\partial_1\,E^3
-\partial_3\,E^1=-\partial_0\,B^2-(\nabla\times\vec{E}\,)^2=0\,.  $$
For $\beta=3$: 
$$\partial_{\alpha}\,^*F^{\alpha 3}=-\partial_0\,B^3-\partial_1\,E^2
+\partial_2\,E^1=-\partial_0\,B^3-(\nabla\times\vec{E}\,)^3=0\,.  $$
So,
$$ \nabla\cdot\vec{B}=0~~{\rm and}~~-\partial_0\vec{B}-\nabla\times
   \vec{E} = 0 \,.$$
\vfil

\end{enumerate}
\end{document}

\item{\bf Homogeneous Maxwell equations.}
\smallskip \noindent

We choose $\beta =0$ in 
$\partial_{\alpha}\,^*F^{\alpha\beta}=0$ and obtain:
$\partial_{\alpha}\,^*F^{\alpha 0}=0$. Using the definition of the 
dual tensor, this reads
$$ \frac{1}{2}\,\left(\partial_{\alpha}\,
   \epsilon^{\alpha 0\gamma\delta}\,F_{\gamma\delta}\right)=0\,$$
or
$$ \partial_{\alpha}\,F_{\gamma\delta} +
  \partial_{\gamma}\,F_{\delta\alpha} +
  \partial_{\delta}\,F_{\alpha\gamma}\ =\ 0 $$
for all three indices different and not 0. Repeating this for 
$\beta=1,\,2$ and 3, we find that the equation 
hold for all combinations of distinct indices $\alpha,\,\beta$
and $\gamma$. Due to the antisymmetry of electromagnetic 
tensor, the equation still holds when some of the indices
agree. Finally, all indices can of course be raised. This
proves the equivalence of the two versions of the homogeneous
Maxwell equations. More by brute force, one could just write
out all 64 equations.

More elegantly: Use
$$ \frac{1}{2}\,\partial_{\alpha}\,
   \epsilon_{\hat\alpha\beta\hat\gamma\hat\delta}\,
   \epsilon^{\alpha\beta\gamma\delta}\,F_{\gamma\delta} = 0 $$
and expand the contraction in $\beta$ of the two Levi-Civita symbols 
into Kronecker deltas.
