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 $K=\frac{48 G^2 M^2 }{r^6}$.
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1$\begingroup$ Note that a similar question was already posed. $\endgroup$– HaroldCommented Jul 23, 2016 at 14:09
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3 Answers
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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.
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$\begingroup$ PS Not to mention that the "resolution of the singularity" is one of the targets of many theories of quantum gravity. In fact, for the quantum electromagnetic field, its classical singularity is not an issue. $\endgroup$– DDdCommented Jul 21, 2016 at 15:30
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$\begingroup$ "the resolution of the singularity" issues arise very unambiguously in the components of $R_{abc}{}^{d}$ that don't contribute to $R_{ab}$, all of which contain factors of $\frac{1}{r}$, and whose singularities are clearly physical, because they are coordinate invariant. $\endgroup$ Commented Jul 21, 2016 at 15:40
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$\begingroup$ @JerrySchirmer Yes, I completely agree. You can find that most of the components of the curvatures blow up, even the Weyl curvature. It is not only a problem of the Ricci scalar, but of the manifold itself at that point. $\endgroup$– DDdCommented Jul 21, 2016 at 21:56
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This depends on how you do the analysis -- If you do it in the Kruskal spacetime, which is the more physical way, It's hard to talk about what "$r = 0$" is, and the question becomes somewhat ill-posed.
If you naïvely do it in Schwarzschild coordinates, you'll find that $R$ has terms involving constants multiplied $\nabla^{2}\frac{1}{r}$, which an analysis in, for example, Jackson's Electrodynamics book, will tell you are proportional to $\delta^{3}(\vec{r})$. So, I guess it depends on what you're talking about. But certainly, it's not crazy to think of the mass being compressed into an area of infinite density, which has to happen as you paste the matter distribution in something like an Oppenheimer-Snyder "collapsing dust star" solution onto the Schwarzschild solution after the collapse is complete.
answered Jul 21, 2016 at 15:27
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When you solve the Einstein equations to find the black hole the energy-momentum tensor is vanishing everywhere, even at $r = 0$. At no moment you are imposing that it is a point source $T_{\mu\nu} \sim \delta(r)$, this would change the solution. The interpretation of the solution as a black hole with a central singularity comes by inspecting the properties of the spacetime that you get after solving the equations. Hence there is really nothing in this spacetime. Recall that you did not put any matter in your spacetime so there is nothing that can gives mass/energy except the gravitational field itself (whose energy-momentum "tensor" is hidden in $G_{\mu\nu}$).
To conclude one has $R = 0$ everywhere and in particular at $r = 0$: this is why one needs to turn one's attention to the more complicated Kretschmann invariant in order to find the singularity.
