What happens at r ≤ rS in the Schwarzschild metric?

[Tio Barnabe]

If the Schwarzschild metric is, by construction, valid for $r > r_S$, where $r_S$ is the Schwarzschild radius, so it does not make sense to talk about what happens at $r \leq r_S$, because there will be no vacuum anymore. What am I getting wrong?

 

[martinbn]

The first part is wrong. It is valid for $0 < r < r_S$ as well.

 

[Orodruin]

The Schwarzschild solution is a vacuum solution also for $r < r_S$ (for $r = r_S$ the Schwarzschild coordinates are singular and not well suited to describe the Schwarzschild space-time). However, for $r < r_S$, the $r$ coordinate is time-like and not space-like.

 

[Tio Barnabe]

How can it be? It is a vacuum metric. If a star has radius $r_0$, then the solution is valid only for $r > r_0$, and the range of $r$ would be restricted to $r_0 < r < \infty$.

 

[martinbn]

And if you don't have a star but a black hole, then there is no problem.

 

[Orodruin]

We are talking about a Schwarzschild black hole, not a star.

If you have an actual star, the exterior Schwarzschild solution is only valid outside the star's radius. Inside the star's radius you would have a different behaviour of the metric.

 

[Tio Barnabe]

Oh, I got it.