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The exterior Schwarzschild spacetime

  1. Dec 26, 2015 #1
    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. jcsd
  3. Dec 26, 2015 #2

    Orodruin

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    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.
     
  4. Dec 26, 2015 #3
    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
     
  5. Dec 27, 2015 #4

    bcrowell

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