% Bernd Berg Nov 10 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 31. Towards the Schwarzschild metric.}
\end{description}

\noindent (a) We have
\begin{eqnarray} \nonumber
  &~& a = a(r) = -e^{B(r)} = a(x^1)\,,~~
      b = b(r) = -r^2 = b(x^1)\,,~~\\ \nonumber
  &~& c = c(r,\theta) = -r^2\,\sin^2\theta = c(x^1,x^2)\,,~~
      d = d(r) = e^{A(r)} = d(x^1)\,.
\end{eqnarray} 
Therefore, $0=a_2=a_3=a_4$, and the remaining part of $R_{11}$ is
\begin{eqnarray} \nonumber
  R_{11}\ =\ &+& \beta\,b_{11} + \gamma\,c_{11} + \delta\,d_{11}
  \\ \nonumber
             &-& \beta^2\,b_1^2 - \gamma^2\,c_1^2 - \delta^2\,d_1^2
  \\ \nonumber - \alpha\,a_1\ (
             &+& \beta\,b_1 + \gamma\,c_1 + \delta\,d_1~~)
\end{eqnarray}

\noindent (b) For the derivatives we find
\begin{eqnarray} \nonumber
   a_1=-B_1\,e^B\,,&~& b_1=-2r\,,~~c_1=-2r\,\sin^2\theta\,,~~
       d_1=A_1\,e^A\,, \\ \nonumber
   &~& b_{11}=-2\,,~~c_{11}=-2\,\sin^2\theta\,,~~
       d_{11}=A_{11}\,e^A+(A_1)^2\,e^A\,,
\end{eqnarray}
and $R_{11}$ becomes
\begin{eqnarray} \nonumber
  R_{11} = R_{rr}\ =\ &+& \frac{1}{r^2} + \frac{1}{r^2} + 
  \frac{A_{11}}{2} + \frac{(A_1)^2}{2} \\ \nonumber
  &-& \frac{1}{r^2} - \frac{1}{r^2}  - \frac{(A_1)^2}{4}\\ \nonumber 
   - \frac{B_1}{2}\,( &+& \frac{1}{r}~+~\frac{1}{r}~+~\frac{A_1}{2}~)
  \\ \nonumber =\ &~& \frac{A_{11}}{2} - \frac{A_1\,B_1}{4} + 
  \frac{(A_1)^2}{4} - \frac{B_1}{r}\,.
\end{eqnarray}


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