% Bernd Berg Nov 8 2018.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2018 
Solutions} \medskip

\centerline{\bf Set 12} 

\begin{description} 
\item {\bf 29. Electromagnetic current conservation.
Exercise 14.2 of Rindler.}
\end{description}

\noindent We start off with Rindler (14.2), p.298,
\begin{eqnarray} \nonumber
  E_{\mu\nu} = \Phi_{\nu:\mu} - \Phi_{\mu:\nu}\,,
\end{eqnarray} 
where $:$ denotes the covariant derivative. The field equation may
the be written as
\begin{eqnarray} \nonumber
  E_{\mu\nu:}^{~~~\nu} = g^{\nu\hat{\nu}}\,E_{\mu\nu:\hat{\nu}} = 
  g^{\nu\hat{\nu}}\,\Phi_{\nu:\mu\hat{\nu}} - 
  g^{\nu\hat{\nu}}\,\Phi_{\mu:\nu\hat{\nu}} = \frac{4\pi}{c}\,J_{\mu}\,.
\end{eqnarray} 
We want to show that the continuity equation
\begin{eqnarray} \nonumber
  g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}\,E_{\mu\nu:\hat{\nu}\hat{\mu}} &=&
  g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}\,\Phi_{\nu:\mu\hat{\nu}\hat{\mu}} - 
  g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}\,\Phi_{\mu:\nu\hat{\nu}\hat{\mu}}
  \\ \noindent &=& \frac{4\pi}{c}\,g^{\mu\hat{\mu}}\,J_{\mu:\hat{\mu}} =
  \frac{4\pi}{c}\,J_{\mu:}^{~~\mu} = 0\,. \nonumber
\end{eqnarray} 
Here we have used that covariant differentiation commutes with 
contraction. See Rindler (10.34), p.212. Let us interchange the 
$\mu$ and $\nu$ dummies in the second term:
\begin{eqnarray} \nonumber
  g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}\,\Phi_{\mu:\nu\hat{\nu}\hat{\mu}} =
  g^{\mu\hat{\mu}}\,g^{\nu\hat{\nu}}\,\Phi_{\nu:\mu\hat{\mu}\hat{\nu}} 
\end{eqnarray} 
After this we use Rindler (10.57), p.218 and get
\begin{eqnarray} \nonumber
  \frac{4\pi}{c}\,J_{\mu:}^{~~\mu}&=&g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}
  \,\left(\,\Phi_{\nu:\mu\hat{\nu}\hat{\mu}} - 
            \Phi_{\nu:\mu\hat{\mu}\hat{\nu}}\right) = 
  g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}\, \left(
  \Phi_{\alpha:\mu}\,R^{\alpha}_{~\nu\hat{\nu}\hat{\mu}} +
  \Phi_{\nu:\alpha}\,R^{\alpha}_{~\mu\hat{\nu}\hat{\mu}}\right) 
  \\ \nonumber &=& g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}\,
  g^{\alpha\hat{\alpha}} \left(
  \Phi_{\alpha:\mu}\,R_{\hat\alpha\nu\hat{\nu}\hat{\mu}} +
  \Phi_{\nu:\alpha}\,R_{\hat\alpha\mu\hat{\nu}\hat{\mu}} \right)\,.
\end{eqnarray} 
Re-labelling the dummies in the second term by $\alpha\to\beta$,
$\nu\to\alpha$ and $\mu\to\nu$ gives
\begin{eqnarray} \nonumber
  \frac{4\pi}{c}\,J_{\mu:}^{~~\mu} &=&
  g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}\,g^{\alpha\hat{\alpha}}\,
  \Phi_{\alpha:\mu}\,R_{\hat\alpha\nu\hat{\nu}\hat{\mu}} +
  g^{\alpha\hat{\alpha}}\,g^{\nu\hat{\nu}}\,g^{\beta\hat{\beta}}\,
  \Phi_{\alpha:\beta}\,R_{\hat\beta\nu\hat{\alpha}\hat{\nu}} \,.
\end{eqnarray} 
Next, replacing in the second term $\beta\to\mu$ gives
\begin{eqnarray} \nonumber
  \frac{4\pi}{c}\,J_{\mu:}^{~~\mu} &=&
  g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}\,g^{\alpha\hat{\alpha}}\,
  \Phi_{\alpha:\mu}\,\left( R_{\hat\alpha\nu\hat{\nu}\hat{\mu}} +
  R_{\hat\mu\nu\hat{\alpha}\hat{\nu}}
  \right)\,.
\end{eqnarray} 
Using, respectively, identities (10.64), (10.62) of p.218 of Rindler, 
and re-labelling we find
\begin{eqnarray} \nonumber
  g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}\,g^{\alpha\hat{\alpha}}\,
  R_{\hat\mu\nu\hat{\alpha}\hat{\nu}} &=&
  g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}\,g^{\alpha\hat{\alpha}}\,
  R_{\hat{\alpha}\hat{\nu}\hat\mu\nu} = \\ \nonumber -
  g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}\,g^{\alpha\hat{\alpha}}\,
  R_{\hat{\alpha}\hat{\nu}\nu\hat{\mu}} &=& -
  g^{\nu\hat{\nu}}\,g^{\mu\hat{\mu}}\,g^{\alpha\hat{\alpha}}\,
  R_{\hat{\alpha}\nu\hat\nu\hat{\mu}} 
\end{eqnarray} 
implying
$$  \frac{4\pi}{c}\,J_{\mu:}^{~~\mu} = 0\,. $$

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