% Bernd Berg sep 21 2017.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 
Solutions} \medskip

\centerline{\bf Set 5} 

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

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\item{\bf 14. Relativistic Energy-Momentum Conservation.}
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\begin{figure} \begin{center}
\includegraphics[width=0.70\textwidth]{figures/modphys} 
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\noindent 1. The 4-vector energy-momentum conservation reads
$$ \left(\begin{matrix} m\\ 0\\ 0\\ 0\end{matrix}\right) + 
   \left(\begin{matrix} m\\ 0\\ 0\\ 0\end{matrix}\right) = 
   \left(\begin{matrix}\sqrt{M^2+p^2}\\ -p\\ 0\\ 0\end{matrix}\right)
   \left(\begin{matrix}E_{\gamma}\\ p_{\gamma}\\ 0\\ 0\end{matrix}
\right)$$
Therefore, $p^2=p_{\gamma}^2=E_{\gamma}^2$ implies
$$ 2m - E_{\gamma} = \sqrt{M^2+E_{\gamma}^2},, $$
$$ (2m-E_{\gamma})^2 = 4m^2-4m\,E_{\gamma} + E^2_{\gamma}
    = M^2+E_{\gamma}^2\,, $$
$$ E_{\gamma} = \frac{4m^2-M^2}{4m}\,. $$

\noindent 2. The requested $E_{\gamma}(M)$ values are:
\begin{eqnarray} \nonumber
  E_{\gamma}(0) &=& 1\,m\,,              \\ \nonumber
  E_{\gamma}(m/2) &=& \frac{15}{16}\,m\,,              \\ \nonumber
  E_{\gamma}(1\,m) &=& \frac{3}{4}\,m\,,    \\ \nonumber
  E_{\gamma}(\sqrt{2}\,m) &=& \frac{1}{2}\,m\,,    \\ \nonumber
  E_{\gamma}(\sqrt{3}\,m) &=& \frac{1}{4}\,m\,. 
\end{eqnarray} 

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

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