% Bernd Berg Oct 13 2017.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 
Solutions} \medskip

\centerline{\bf Set 9} 

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\item {\bf 30. Use the appendix, p.419, 420, of Rindler to show Eq.~(11.4).}
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We choose the identification $(c=1)$
\begin{eqnarray} \nonumber
  d\vec{s}\,^2 &=& a\,(dx^1)^2+b\,(dx^2)^2+(dx^3)^2+(dx^4)^2\\ \nonumber
               &=& a\,(dt)^2+b\,(dr)^2+c\,(d\theta)^2+d\,(d\phi)^2
\end{eqnarray} 
with $a=\exp[+A(r)]$, $b=-\exp[+B(r)]$, $c=-r^2$, $d=-r^2\sin^2\theta$.
The non-zero derivatives are the $'=\partial/\partial x^2=d/dr$ and on
$d$ only $\partial/\partial x^3= \partial/\partial\theta$. Therefore, 
the relevant contribution to the Ricci tensor are
$$ R_{tt} + R_{11} = \beta\,a_{22} - \beta\,a_2\,(\alpha\,a_2+\beta\,b_2
                   -\gamma\,c_2-\delta_2\,d_2)\,, $$
where $\alpha$, $\beta$, $\gamma$, $\delta$ are defined by $\alpha=
1/2a$, $\beta =1/2b$, $\gamma=1/2c$, $\delta=1/2d$. We have
$$ a'=A'\exp(+A)\,,~~b'=-B'\exp(+B)\,~~c'=-2r\,~~d'=-2r\,\sin^2\theta$$
and $a''=A''\exp(A)+(A')^2\exp(A)$. Inserting these gives
\begin{eqnarray} \nonumber
 R_{tt}\ &=& - \exp(A-B)\,\frac{1}{2}\,\left(A''+{A'}^2\right)\\
 \nonumber &~& + \exp(A-B)\,\frac{1}{2}\,A'\,\left(\frac{1}{2}\,A'
           +\frac{1}{2}\,B' -\frac{1}{r}-\frac{1}{r} \right)\\ \nonumber 
        &=& - \exp(A-B)\,\left(\frac{1}{2}\,A''-\frac{1}{4}\,A'\,B'
            +\frac{1}{4}\,{A'}^2+A'/r\right)\,.
\end{eqnarray} 


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