% Bernd Berg Oct 14 2017.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 
Solutions} \medskip

\centerline{\bf Set 10} 

\begin{description} 
\item {\bf 33. Isotropic from of the Scharzschild metric.}
\end{description}

Let us transform the crucial factor $(1-2m/r)$ of the Schwarzschild 
metric by substituting $r=(1+m/2\overline{r})^2\,\overline{r}$:
\begin{eqnarray} \nonumber
  &~& 1-\frac{2m}{r} = 1-2m\,\left(1+\frac{m}{2\overline{r}}\right)^{-2}
  \frac{1}{\overline{r}} = \frac{(1+m/2\overline{r})^2\,\overline{r}-2m}
  {(1+m/2\overline{r})^2\,\overline{r}}\\ \nonumber
  &=& \frac{(1+m/\overline{r}+m^2/4\overline{r}^2)\,\overline{r}-2m}
  {(1-m/2\overline{r})^2\,\overline{r}} = \frac{(1-m/2\overline{r})^2\,
  \overline{r}}{(1+m/2\overline{r})^2\,\overline{r}}
  = \frac{(1-m/2\overline{r})^2}{(1+m/2\overline{r})^2}
\end{eqnarray} 
Therefore, we have the new coefficient of $dt^2$. To transform $dr^2$
we calculate
$$ \frac{dr}{d\overline{r}} = 2\,\left(1+\frac{m}{2\overline{r}}\right)\,
   \left(-\frac{m}{2\overline{r}^2}\right)\,\overline{r}+\left(1+\frac{m}
   {2\overline{r}}\right)^2 = \left(1+\frac{m}{2\overline{r}}\right)\, 
   \left(1-\frac{m}{2\overline{r}}\right) $$
and find
$$ \left(1-\frac{2m}{r}\right)^{-1}\,dr^2 = 
   \left(1+\frac{m}{2\overline{r}}\right)^4\,d\overline{r}^2\, $$
From substituint $r^2$ the factor $(1+m/2\overline{r})^4\,\overline{r}^2$
applies to the solid angle, so that our result is
$$ d\vec{s}\,^2 = \frac{(1-m/2\overline{r})^2}{(1+m/2\overline{r})^2}\,dt^2
   - \left(1+\frac{m}{2\overline{r}}\right)^4\,\left[d\overline{r}^2+
   \overline{r}^2\,(d\theta^2+\sin^2\theta\,d\phi^2)\right]\,.  $$
Cartesian coordinates are then defined in the usual way,
$$\overline{z}=\overline{r}\,\cos\\theta\,,~~
  \overline{x}=\overline{r}\,\sin\theta\,\cos\phi\,,~~
  \overline{y}=\overline{r}\,\sin\theta\,\sin\phi\,, $$
and our final result is
$$ d\vec{s}\,^2 = \frac{(1+m/2\overline{r})^2}{(1+m/2\overline{r})^2}
   \,dt^2 - \left(1+\frac{m}{2\overline{r}}\right)^4\,\left(
   d\overline{x}^2+d\overline{y}^2+d\overline{z}^2\right)\,.  $$

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