Schwarzschild Metric Singularity: Why?

[Charles_Xu]

Why does the Schwarzschild metric have a singularity at r=0 if it is only valid outside the spherically symmetric static mass?

 

[PeterDonis]

Why does the Schwarzschild metric have a singularity at r=0 if it is only valid outside the spherically symmetric static mass?

If we are talking about the vacuum region outside a spherically symmetric static mass, that region does not include $r = 0$, and it does not include a singularity.

The singularity at $r = 0$ is only present in a black hole spacetime, where the vacuum Schwarzschild region is not outside a static mass; if a mass (i.e., a region occupied not by vacuum but by matter) is present in the spacetime, it is not static but is a collapsing region (as in the Oppenheimer-Snyder 1939 model of gravitational collapse), and the vacuum region outside it goes all the way down to $r = 0$.

 

[Nugatory]

Why does the Schwarzschild metric have a singularity at r=0 if it is only valid outside the spherically symmetric static mass?

The Schwarzschild solution to the EFE is vacuum everywhere - no mass anywhere, stress-energy tensor is zero everywhere, the $M$ that appears in the metric is a parameter that characterizes the solution not the mass of anything. Thus any point with $r\gt 0$ is an element of the manifold and it makes sense to consider the singularity at $r=0$.

When we apply the Schwarzschild solution to the vacuum outside of a real object of mass $M$ and non-zero radius we’re considering just a subset of the entire manifold, a subset that doesn’t include the singularity.

 

[Dale]

Just to be slightly pedantic: the Schwarzschild manifold does not include r=0.

 

[Nugatory]

Just to be slightly pedantic: the Schwarzschild manifold does not include r=0.

This is true but properly stating it was more work than I wanted to do.