Curvature of space in a black hole

Question

This is a very simplistic view from an interested structural design engineer (retired).

Mass curves space. Taking the case of a sphere of uniform density the point at which you have as much mass outside as inside is a spherical shell two thirds the radius of the sphere. Therefore, once you pass through that shell closer to the center should not the curvature of space reduce rather than increase since most of the mass is outside? If so a singularity at the center could not occur. The maximum curvature of space would be a 2D spherical shell not a singularity. Clearly a black hole may not be a uniform density, but even if it increased exponentially, the point at which there would be as much mass outside as inside would still be a 2D shell of some radius. Why is this not the case please.

That is a reasonable thought, but you might look at this to see how curvature works. A new way to visualize General Relativity. ScienceClic make a number of other videos worth watching. What is General Relativity? and Why Time and Space swap in a Black Hole — mmesser314, Apr 12 '23 at 14:38
the point at which you have as much mass outside as inside is a spherical shell at 1/sqrt(2) the radius of the sphere. Try it. This does not invalidate your question, however. — JMLCarter, Apr 12 '23 at 20:26

score 2 · Answer 1 · answered Apr 12 '23 at 14:29

2

A black hole is not like a planet or a star, it does not have any volume filled with mass, all its mass is concentrated in a single point. We can talk, for example, about Schwarzschild radius of a black hole, but it doesn't mean that the black hole is a sphere of Schwarzschild radius filled with matter. Whatever positive radius $r$ you choose, the mass of the matter outside of the 2D sphere with radius $r$ is zero, the whole mass of the black hole is located at a single point at $r=0$. That's what is called a singularity - an infinite density.

answered Apr 12 '23 at 14:29

Alex Sveshnikov

466

In a classical Schwarzschild blackhole, all the non-vacuum curvature is in the singularity, which is not part of spacetime--it's a point defect. The mass resides in the energy of spacetime curvature. The matter that created the blackhole is gone. – JEB Apr 12 '23 at 17:33
$r$ is not radius of 2D sphere but its (Gaussian) curvature radius. – JanG Apr 12 '23 at 19:46
JEB is that right, I thought it is surely mass that is itself required to deform space-time. – JMLCarter Apr 12 '23 at 20:30
The presence of an actual physical singularity is still debated, isn't it, since no observation is possible, and many take exception to physical rather than mathematical infinities. It is just a sufficiently accurate simplification. – JMLCarter Apr 12 '23 at 20:32

John Rennie · Answer 2 · 2023-04-12T15:43:15.847

If a black hole was a sphere of a uniform density then the curvature would indeed decrease as you approached the centre and would indeed be zero at the centre. This geometry is called the Schwarzschild interior metric and in fact exactly this issue is discussed in my answer to What is the general relativity explanation for why objects at the center of the Earth are weightless?

However a black hole is not a uniform sphere or even a sphere with a varying density. The event horizon is not the surface of the black hole. It is just a place in space beyond which light can no longer escape. The black hole is actually all just a vacuum except at the singularity at the centre (where the geometry is undefined) so no matter how close you get to the singularity all the mass of the black hole is still below you. That's why the curvature keeps increasing the farther towards the centre that you go.

Although, the presence of an actual physical singularity is still debated, isn't it, since no observation is possible, and many take exception to physical rather than mathematical infinities, and suspect it is a simplification applied in a region of unknown physics. — JMLCarter, Apr 12 '23 at 20:34

Curvature of space in a black hole

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