How to prove the absence of naked singularities?

Question

Suppose we have a static, spherically symmetric solutions of Einstein equations coupled to a certain matter source, and we are able to show that the scalars obtained by the stress energy tensor on this solution are all bounded: $$T_{\mu\nu}T^{\mu\nu}\;,\;T_{\mu}^{\mu}\;<\infty \;\;\text{everywhere}$$ This implies that the scalars obtained by contracting the Ricci tensor are bounded. Indeed using the (reverse trace) Einstein equation we find: $$R=-\kappa^2 T$$ and $$R_{\mu\nu}R^{\mu\nu}=\kappa^4\left(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T\right)\left(T^{\mu\nu}-\frac{1}{2}g^{\mu\nu}T\right)=\kappa^4 T_{\mu\nu}T^{\mu\nu}$$

Is this sufficient to argue that the spacetime is free of naked singularities?

I know that we should also check the scalars containing the Riemann tensor: $$R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}\;,\;R_{\mu\nu\rho\sigma}R^{\mu\nu}R^{\rho\sigma}\;,\;R_{\mu\nu\rho\sigma}R^{\mu}_{\tau}R^{\nu}_{\lambda}R^{\rho}_{\alpha}R^{\sigma}_{\beta}(g^{..}g^{..})$$ but I guess that some of these may not be independent from the Ricci-scalars, also due to the symmetries in play (static, spherically symmetric spacetime).

So are the Ricci scalars enough to argue absence of naked singularities in this case or do I need to compute some of the Riemann scalars?

Can at least some of the Riemann scalars be expressed in terms of the Ricci scalars? Or are there inequalities that can be used in order to constrain the Riemann scalars?

Answer 1

There are many possible types of singularities which can occur, even under such benign conditions. Beyond the extremal Nordstrom metric, let's consider :

Minkowski space with the line $\{ (t, 0, 0, 0) | t \in \mathbb{R} \}$ removed. This is a regular boundary point, which is technically a singularity although not terribly interesting.

You can create a quasi-regular singularity with such attributes in $2+1$ dimensions as well, by considering the conical spacetime generated by a point mass, but I'm not 100% sure of how to do such a thing in $3+1$ dimension, since the usual process for this breaks the spherical symmetry.

There are still more types of pathological singularities we can apply there. Take a static, spherically symmetric, for instance :

$$ds^2 = -e^{2\alpha(r)} dt^2 + e^{2\beta(r)} dr^2 + r^2 d\Omega^2$$

There are several ways a singularity can go bad. The general definition of a curvature singularity (a singularity that isn't quasi-regular) is that, given a moving frame along some curve $e^\mu_i$, then the components of the Riemann tensor in that basis are not $C^0$ along that curve. It is entirely possible that all scalar quantities remain well-behaved under such circumstances (so-called non-scalar singularities), although I don't know if this is the case here. So let's just create a rather nasty trick : even if all quantities are bounded, a simple way for quantities to go badly is to have infinite oscillations.

It's not terribly hard to go from there. From Carroll, the Ricci tensor is

\begin{eqnarray} R_{tt} &=& e^{2(\alpha - \beta)} \left[ \alpha'' + (\alpha')^2 - \alpha'\beta' + \frac{2}{r} \alpha' \right]\\ R_{rr} &=& - \left[ \alpha'' + (\alpha')^2 - \alpha' \beta' - \frac{2}{r} \beta' \right]\\ R_{\theta\theta} &=& e^{-2\beta} \left[ r (\beta' - \alpha') - 1 \right] + 1\\ R_{\phi\phi} &=& R_{\theta\theta} \sin^2 \theta \end{eqnarray}

Now the Ricci scalar is simply

\begin{eqnarray} R &=& -e^{2\alpha}e^{2(\alpha - \beta)} \left[ \alpha'' + (\alpha')^2 - \alpha'\beta' + \frac{2}{r} \alpha' \right] \\ &&- e^{2\beta} \left[ \alpha'' + (\alpha')^2 - \alpha' \beta' - \frac{2}{r} \beta' \right]\\ &&+ r^2 (e^{-2\beta} \left[ r (\beta' - \alpha') - 1 \right] + 1)\\ &&+r^2 (e^{-2\beta} \left[ r (\beta' - \alpha') - 1 \right] + 1) \sin^4 \theta \end{eqnarray}

As I am a bit lazy, let's assume that $\alpha = 0$, this simplifies things quite a lot :

\begin{eqnarray} R &=& e^{2\beta} \frac{2}{r} \beta' + r^2 (e^{-2\beta} \left[ r \beta' - 1 \right] + 1) (1 + \sin^4(\theta)) \end{eqnarray}

This way we only have to worry about first derivatives. Let's pick

\begin{equation} \beta = r^3 \sin(\frac{1}{r}) \end{equation}

with

\begin{equation} \beta' = r (3r\sin(\frac{1}{r}) - \cos(1/r)) \end{equation}

Both this function and its derivative are locally bounded, and the Ricci scalar becomes

\begin{eqnarray} R &=& e^{2 r^3 \sin(\frac{1}{r})} 2(3r\sin(\frac{1}{r}) - \cos(1/r)) + r^2 (e^{-2r^3 \sin(\frac{1}{r})} \left[ r^2 (3r\sin(\frac{1}{r}) - \cos(1/r)) - 1 \right] + 1) (1 + \sin^4(\theta)) \end{eqnarray}

itself perfectly locally bounded. You can check at your leisure that all manners of quantities of the stress-energy tensor are bounded as well. However, if you consider the value of $R$ along a fairly simple curve (let's say an infalling curve of the form $(\lambda, -\lambda, 0,0)$), due to the appearance of $\sin(1/x)$, $r = 0$ is a curvature singularity, as the transport of this quantity along a curve is not continuous. I did not check everything for horizons and such, but as far as I can tell the metric components never change signs.

This is a fairly dumb example but you may wish to investigate this type of solution for a more realistic (if not spherically symmetric) version : https://link.springer.com/article/10.1007/BF01651509

Answer 2

For the Einstein vacuum equation, this is true just under the assumption of spherical symmetry. In spherical symmetry, Birkhoff's theorem gives us that the solution is isometric to a subset of the Schwarzschild solution. This is also static and doesn't have a naked singularity as long as $M$ is positive.

For the Einstein-Maxwell system, spherical symmetry gives you the Reissner-Nordstrom spacetime by an analogous result to Birkhoff's theorem. It is also static. Note that for the super-extremal case this black hole has a naked singularity. This may be the unphysical case but one definitely has a counterexample to your claim. You need a rigorous statement in your claim that excludes this case.

To prove that your spacetime is free of naked singularities, you're trying to solve the weak cosmic censorship conjecture. For this, you need to prove that future null infinity is complete. This means that you need to show that the generators of this null hypersurface have an affine parameter that takes all values in $(-\infty,\infty)$.