% Bernd Berg Nov 23 2017.
 \documentclass{ws-book9x6}  
\begin{document}

\setcounter{page}{1}

\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 
Solutions} \medskip

\centerline{\bf Set 12} 

\begin{description} 
\item {\bf 39. Lorenz gauge with covariant derivatives.
Exercise 14.1 of Rindler.}
\end{description}

\noindent According to (10.55) of Rindler we have 
$$ \Phi^{\mu}_{~;\rho\sigma}-\Phi^{\mu}_{~;\sigma\rho} 
   = - \Phi^{\alpha}\,R^{\mu}_{~\alpha\rho\sigma}\,. $$
Therefore, the definition (10.68) of the Ricci tensor yields
$$ \Phi^{\mu}_{~;\rho\mu}-\Phi^{\mu}_{~;\mu\rho} = - 
   \Phi^{\alpha}\,R^{\mu}_{~\alpha\rho\mu} = -
   \Phi^{\alpha}\,R_{\alpha\rho}\,.  $$
In the Lorenz gauge $\Phi^{\mu}_{~;\mu\rho}=0$ holds and
we are left with
$$ \Phi^{\mu}_{~;\rho\mu}=-\Phi^{\alpha}\,R_{\alpha\rho}~~{\rm or}
 ~~\Phi^{\mu~~\rho}_{~~;~~\mu}=-\Phi^{\alpha}\,R_{\alpha}^{~\rho}\,.$$
Renaming the dummy indices $\nu$ and the free index $\rho\to\mu$ we
obtain the desired result
$$ \Phi^{\nu~~\mu}_{~~;~~\nu}=-\Phi^{\nu}\,R_{\nu}^{~\mu}\,.$$
As the Ricci tensor is symmetric, we can as well write $R_{\nu}^{\mu}$.

\end{document}
