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The Schwarzschild metric is the metric calculated from the field equation outside of the black hole. This condition of region (outside of the matter) was the reason why we could use $T_{\mu\nu}=0$.

But we can tell some properties of the singularity of the black hole, which is at $r=0$, from the schwarzschild metric. For example, calculating the curvature tensor from the metric yields infinite curvature and tidal force which means that everything will be destroied at the singularity. But the singularity is obviously inside the range of matter. Because if it is not, it would not be the singularity. Then, why can we apply schwarzschild metric to analyze the geomety at the singularity?

What is the range where we can use schwarzschild solution? And how can we analyze the geometry where we can't apply it.

Mauricio
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Zjjorsia
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    Related: https://physics.stackexchange.com/a/300265/123208 https://physics.stackexchange.com/a/144458/123208 – PM 2Ring Mar 09 '24 at 07:21
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    You can use it only at r>0, so the singularity is not part of the metric. – Yukterez Mar 09 '24 at 09:29
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    But the singularity is obviously inside the range of matter. Because if it is not, it would not be the singularity. That is a non sequitur. – Ghoster Mar 09 '24 at 21:54
  • @Yukterez “the singularity is not part of the metric” - Why not? The metric is perfectly regular at $r=0$ as $d\tau^2=-dt^2$ defining a spacelike Euclidean line: https://math.stackexchange.com/questions/2929400 - I know the singularity typically is excluded from the manifold, but in this case I don’t see any physical difference this would make. In other words, the Schwarzschild singularity is a moment of time with a zero duration when time ends everywhere along the (hyper)cylindrical space that shrinks to a line. Considering this line a part of spacetime or not makes no physical difference. – safesphere Mar 10 '24 at 06:42
  • @PM2Ring Both answers of Ben Crowell linked above are incorrect. – safesphere Mar 10 '24 at 06:50
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    @safesphere wrote: "The metric is perfectly regular at r=0" - it is not, since you get multiple divisions by zero there, especially for the tidal forces and relative velocities. – Yukterez Mar 10 '24 at 13:00
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    @Yukterez Tidal forces and velocities are not components of the metric. Take the equation of the metric, set $r=0$ and you’ll see that the metric is regular. On the tidal forces, if you take two particles following their geodesics, as shown in my link above, then the tidal forces between them are zero. So “the infinite tidal forces” are not physical. The same about “infinite velocities”. The gravitational acceleration at the horizon is infinite too, why are you not complaining about it? Or about massive objects crossing the horizon exactly at the speed of light? These are much bigger issues. – safesphere Mar 11 '24 at 03:46

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The Schwarzschild metric is the metric calculated from the field equation outside of the black hole. This condition of region (outside of the matter) was the reason why we could use $T_{μν}=0$.

You mean Schwarzschild exterior or outer solution. I translate from Schwarzschild Metrik (German version is different to English one):

“The full Schwarzschild model consists of the outer Schwarzschild solution for the space outside the mass distribution and the inner Schwarzschild solution, which solves the field equations inside the mass distribution with the additional assumption that the mass is a homogeneous fluid.”

But we can tell some properties of the singularity of the black hole, which is at r=0, from the Schwarzschild metric?

No, we cannot. However, it has become established to assume that after gravitational collapse all matter disappears in so called “singularity” (which is not a part of the spacetime manifold!) leaving empty ($T_{\mu\nu=0}$) spherically symmetric universe.

But the singularity is obviously inside the range of matter.

You are right. To study what is singularity one should better use the full Schwarzschild solution for Buchdahl limit ($r_{S}/R=8/9$). One can easily calculate that at the center the energy density is finite, but the pressure diverges as $4/\kappa~r^{-2} (\kappa \equiv 8\pi G/c^4$). It looks like it is the pressure and not the geometry that behaves singularly there. By the way, $r=0$ is not a point but a two sphere in the limit of zero surface area.

JanG
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