What is the value of Ricci scalar at $r=0$ inside Black Hole? Since $R_{\mu \nu}=0$ is vacuum solution and valid outside event Horizon of black hole where there is no mass energy density. But inside black hole at $r=0$ what is the value of Ricci scalar, where $T_{\mu\nu}\neq 0$. At this point geometry blow up (having infinite curvature) and Kretschmann scalar have value $K=\frac{48 G^2 M^2 }{r^6}$.
[Physics] Ricci scalar for Black Hole
black-holescurvaturedifferential-geometrygeneral-relativitysingularities
Best Answer
If we want to be formal, at $r=0$ there is a manifold singularity. So no tensor/scalar quantities are well defined there (i.e. our mathematical formalism does not work there). On the other hand, as physicists, we can look for different examples that behave in a similar manner. Also the (classical) electromagnetic field generated by a point-like charged particle diverges as $r \rightarrow 0$. There is however a subtle difference: a field can be ill-defined in one point without many head scratches, for gravity what is ill-defined is space-time itself, so the whole Physics becomes hard there (usually Physics is encoded mathematically with functions of position and time).
There have been attempts to formalize mathematically this issue. In Electrodynamics we study distribution (Dirac deltas) of charges why can't we use them in gravity? Starting from the Einstein equations ($G=c=1$, $\Lambda=0$) $$ R_{ij}-\frac{1}{2}g_{ij}R=8\pi \ T_{ij}, $$ and multiplying them by $g^{ij}$ you get the an equation for the Ricci scalar $$R=-8\pi\,T$$ where $T$ is the trace of the Energy-momentum tensor. So, if the matter is concentrated only at $r=0$ you would expect a distributional quantity that describes $T$ and therefore $R$.
In literature, it is possible to find this approach: for example in this paper Distributional Nature of the Energy Momentum Tensor of a Black Hole or What Curves the Schwarzschild Geometry ? by H. Balasin and H. Nachbagauer.
They find that for a Schwarzschild black hole of mass $M$, the Ricci tensor should have the form $$ R = 8 \pi \, M\, \delta^{(3)}(x) $$ where $x$ is the position in space-time and $\delta^{(3)}$ is the Dirac-delta function in space, namely zero everywhere but at $r=0$. If it was in a flat space-time the delta could be defined also as $\bigtriangleup\left(1/r\right)=\bigtriangledown^{2}\left(1/r\right)=\delta^{\left(3\right)}\left(x\right)$, but I am not sure in the general case.
But, we have to be very careful with this distributional approach since at the singularity we don't know what space-time physically is, and the mathematical theory behind it, as far as I know, for pseudo-Riemannian geometries, is not solid, yet.