<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">% Bernd Berg Nov 23 2017.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 
Solutions} \medskip

\centerline{\bf Set 12} 

\begin{description} % Exercise 7.8 of Rindler.
\item {\bf 38. Retarded potential and Lorenz gauge.}
\end{description}

\noindent In the retarded solution 
$$ \Phi^{\mu}(\{x^{\nu}\}) = \frac{1}{c} \int d^3y\,\frac{J^{\mu}
   (x^0-|\vec{x}-\vec{y}|,\vec{y})}{|\vec{x}-\vec{y}|} $$
for the potential we replace $x^{\nu}$ by $x^{\nu}+\epsilon^{\nu}$
\begin{eqnarray} \nonumber
  \Phi^{\mu}(\{x^{\nu}+\epsilon^{\nu}\}) =
  \frac{1}{c} \int d^3y\,\frac{J^{\mu} (x^0+\epsilon^0-|\vec{x} +
  \vec{\epsilon}-\vec{y}|,\vec{y})}{|\vec{x}+\vec{\epsilon}-\vec{y}|}
\end{eqnarray}
and expand to order $\epsilon$ on the left hand side:
\begin{eqnarray} \nonumber
  \Phi^{\mu}(\{x^{\nu}\})+\epsilon^{\rho}\,\frac{\partial~}{\partial 
  x^{\rho}}\Phi^{\mu}(\{x^{\nu}\})=\frac{1}{c} \int d^3y\,\frac{J^{\mu} 
  (x^0+\epsilon^0-|\vec{x} + \vec{\epsilon}-\vec{y}|,\vec{y})}
  {|\vec{x}+\vec{\epsilon}-\vec{y}|}\,.
\end{eqnarray}
On the right hand side we substitute $\vec{y}=\vec{y}\,'+
\vec{\epsilon}$ and get
\begin{eqnarray} \nonumber
  \Phi^{\mu}(\{x^{\nu}\})+\epsilon^{\rho}\,\frac{\partial~}{x^{\rho}}
  \Phi^{\mu}(\{x^{\nu}\}) = \frac{1}{c} \int d^3y'\,\frac{J^{\mu} 
  (x^0+\epsilon^0-|\vec{x}-\vec{y}\,'|,\vec{y}\,'+\vec{\epsilon}\,)}
  {|\vec{x}-\vec{y}\,'|}\,,
\end{eqnarray}
where we used that the Jacobian is one, i.e., $d^3y=d^3y'$. We
now expand to order $\epsilon$ on the right hand side (note
$y'^0+\epsilon^0=x^0+\epsilon^0-|\vec{x}-\vec{y}$) and obtain
\begin{eqnarray} \nonumber
  \epsilon^{\rho}\,\frac{\partial~}{x^{\rho}}\Phi^{\mu}(\{x^{\nu}\}) = 
  \frac{1}{c} \epsilon^{\rho} \int \frac{d^3y'}{|\vec{x}-\vec{y}\,'|}\,
  \frac{\partial~}{\partial y'^{\rho}}\,J^{\mu}(y'^0,\vec{y}\,)
\end{eqnarray}
As the $\epsilon^{\rho}$ are arbitrary, we can drop them, and we
can also eliminate the prime on the integrations variable. So,
\begin{eqnarray} \nonumber
  \frac{\partial~}{x^{\rho}}\Phi^{\mu}(\{x^{\nu}\}) = 
  \frac{1}{c} \int \frac{d^3y}{|\vec{x}-\vec{y}\,|}\,
  \frac{\partial~}{\partial y^{\rho}}\,J^{\mu}(\{y^{\nu}\})
\end{eqnarray}
holds. Contracting $\sigma$ with $\mu$
\begin{eqnarray} \nonumber
  \frac{\partial~}{x^{\mu}}\Phi^{\mu}(\{x^{\nu}\}) = 
  \frac{1}{c} \int \frac{d^3y}{|\vec{x}-\vec{y}\,|}\,
  \frac{\partial~}{\partial y^{\mu}}\,J^{\mu}(\{y^{\nu}\}) = 0
\end{eqnarray}
follows due to current conservation on the right hand side,
and the Lorenz gauge $\partial_{\mu}\Phi^{\mu}=0$ is obtained
on the left hand side.

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