I Schwarzschild Metric Singularity: Why?

Charles_Xu
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Why does the Schwarzschild metric have a singularity at r=0 if it is only valid outside the spherically symmetric static mass?
 
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Charles_Xu said:
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##.
 
Charles_Xu said:
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.
 
Just to be slightly pedantic: the Schwarzschild manifold does not include r=0.
 
Dale said:
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.
 
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