Exterior Schwarzschild Spacetime Defined

And yes, the end of the manifold is diffeomorphic to ##\mathbb{R}^3 \setminus K##, where ##K## is a compact set (in this case, it is the event horizon).
  • #1
Sasha_Tw
4
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 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!
 
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes bcrowell
  • #3
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
 
  • #4
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.
 

1. What is exterior Schwarzschild spacetime?

Exterior Schwarzschild spacetime is a mathematical model that describes the space surrounding a non-rotating, spherically symmetric object. It is a solution to Einstein's field equations of general relativity and is often used to study the behavior of massive objects like stars and black holes.

2. How is exterior Schwarzschild spacetime defined?

Exterior Schwarzschild spacetime is defined by a set of coordinates known as Schwarzschild coordinates, which describe the curvature of spacetime around a massive object. It is characterized by a central singularity, a boundary known as the event horizon, and a region of spacetime outside the event horizon.

3. What is the significance of the event horizon in exterior Schwarzschild spacetime?

The event horizon marks the point of no return for any object or light rays that enter it. Once an object passes the event horizon, it is impossible to escape the gravitational pull of the massive object at the center. This is a defining feature of black holes, which are often described using exterior Schwarzschild spacetime.

4. How does exterior Schwarzschild spacetime relate to general relativity?

Exterior Schwarzschild spacetime is a solution to Einstein's field equations of general relativity, which describe the gravitational interactions of massive objects. It is one of the first solutions to these equations and has been used extensively to study the behavior of massive objects in the universe.

5. Can exterior Schwarzschild spacetime be observed or measured?

No, exterior Schwarzschild spacetime is a mathematical model and cannot be directly observed or measured. However, its predictions have been tested and confirmed by numerous experiments, making it a useful tool for understanding the behavior of massive objects in the universe.

Similar threads

  • Special and General Relativity
2
Replies
44
Views
1K
  • Special and General Relativity
Replies
9
Views
1K
  • Special and General Relativity
Replies
18
Views
2K
  • Special and General Relativity
2
Replies
50
Views
3K
  • Special and General Relativity
Replies
5
Views
1K
  • Special and General Relativity
Replies
8
Views
1K
  • Special and General Relativity
2
Replies
42
Views
3K
  • Special and General Relativity
Replies
30
Views
3K
  • Special and General Relativity
Replies
5
Views
369
Replies
12
Views
1K
Back
Top