# The exterior Schwarzschild spacetime

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1. Dec 26, 2015

### Sasha_Tw

The Schwarzschild spacetime is defined by the following line element
\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!

2. Dec 26, 2015

### Orodruin

Staff Emeritus
Because this is the exterior by definition, i.e., the space-time outside of the event horizon.

It did not, it is just a transformation from spherical to linear coordinates, $dx^2 + dy^2 + dz^2 = dr^2 + r^2d\Omega^2$.

Edit: Oh, and you cannot use \$ for LaTeX in text, please use $or the itex tags instead. 3. Dec 26, 2015 ### Sasha_Tw Thanks for the response! Concerning the exterior region, is it asymptotically flat ? I am not too familiar with those notions in general but I was reading that for an asymptotically flat manifold, we would have the end of the manifold diffeomorphic to$\mathbb{R}^3 \setminus K$where$K## is a compact and this is exactly the form we have

4. Dec 27, 2015

### bcrowell

Staff Emeritus
Yes. The Schwarzschild spacetime is the prototypical example of an asymptotically flat spacetime.