The Schwarzschild spacetime is defined by the following line element(adsbygoogle = window.adsbygoogle || []).push({});

\begin{equation*}

ds^2 = - \left( 1 - \frac{2m}{r} \right)dt^2 + \frac{1}{1-\frac{2m}{r}}dr^2 + r^2 d\theta^2 + r^2\sin \theta^2 d\phi^2.

\end{equation*}

We can use the isotropic coordinates, obtained from the Schwarzschild coordinates by the following transformation

\begin{equation*}

r = \overline{r}\left( 1 + \frac{m}{2\overline{r}}\right)^2

\end{equation*}

to obtain a new form for the Schwarzschild metric given by

\begin{equation*}

ds^2 = -\left( \frac{\overline{r} - \frac{m}{2}}{\overline{r}+\frac{m}{2}} \right)^2dt^2 +

\left( \frac{\overline{r} + \frac{m}{2}}{\overline{r}}\right)^4(d\overline{r}^2 + \overline{r}^2d\Omega^2),

\end{equation*}

where $\Omega = d\theta^2 + \sin \theta^2d\phi^2$.

I read in various books that the exterior Schwarschild spacetime is defined as follows

\begin{equation*}

\left( \mathbb{R} \times (\mathbb{R}^3 \setminus B_{m/2}(0), (1 + \frac{m}{2|x|})^4 (dx^2 + dy^2 + dz^2) - \left( \frac{1-\frac{m}{2|x|}}{1 + \frac{m}{2|x|}} \right)^2 dt^2 \right)

\end{equation*}

with $|x| = \sqrt{x^2 + y^2 + z^2}$.

I don't understand how we obtained the last form ? Why are we considering the manifold to be specifically $\mathbb{R} \times (\mathbb{R}^3 \setminus B_{m/2}(0)$ and why did the $d\Omega^2$ term disappear ?

thanks!

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# The exterior Schwarzschild spacetime

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