Exterior Schwarzschild Spacetime Defined

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Discussion Overview

The discussion centers on the definition and properties of the exterior Schwarzschild spacetime, including its line element, transformations to isotropic coordinates, and the implications of these definitions in the context of general relativity. Participants explore concepts such as asymptotic flatness and the nature of the manifold involved.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents the Schwarzschild spacetime line element and discusses the transformation to isotropic coordinates, questioning the derivation of a specific form of the metric.
  • Another participant asserts that the manifold is defined as ##\mathbb{R} \times (\mathbb{R}^3 \setminus B_{m/2}(0)## because it represents the exterior region outside the event horizon.
  • There is a claim that the ##d\Omega^2## term did not disappear but rather is transformed in the transition from spherical to linear coordinates, with an equation provided to support this.
  • A later reply raises the question of whether the exterior region is asymptotically flat, referencing a definition involving diffeomorphism to ##\mathbb{R}^3 \setminus K##.
  • Another participant confirms that the Schwarzschild spacetime is indeed an example of an asymptotically flat spacetime.

Areas of Agreement / Disagreement

Participants express some agreement on the definition of the exterior region and its asymptotic flatness. However, there are unresolved questions regarding the transformation of the metric and the implications of the definitions presented.

Contextual Notes

Participants discuss the transformation of coordinates and the implications of the definitions without reaching a consensus on the specifics of the metric transformation process.

Sasha_Tw
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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 Schwarzschild 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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Sasha_Tw said:
Why are we considering the manifold to be specifically ##\mathbb{R} \times (\mathbb{R}^3 \setminus B_{m/2}(0)##

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

Sasha_Tw said:
and why did the ##d\Omega^2## term disappear ?

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.
 
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Orodruin said:
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.
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
 
Sasha_Tw said:
Thanks for the response! Concerning the exterior region, is it asymptotically flat ?

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

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