Range that the Schwarzschild metric is valid

Question

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.

Answer 1

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.