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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?

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    $\begingroup$ The extremal Kerr metric is a vacuum solution and yet has a naked singularity. $\endgroup$ – Slereah Jun 4 '19 at 10:29
  • $\begingroup$ @Slereah Kerr metric is not static though, only stationary. Plus doesn't the singularity comes out of the horizon only for over extremal solutions? $\endgroup$ – AoZora Jun 4 '19 at 11:28
  • $\begingroup$ The point is that an $a > M$ Kerr solution has zero Ricci tensor but a naked singularity, though. But yeah, it's not spherically symmetric. But every static, vacuum, spherically symmetric solution is Schwarzschild, though. $\endgroup$ – Jerry Schirmer Jun 19 '19 at 15:11
  • $\begingroup$ Correct me if I'm wrong, but, astrophysically, the Schwarzschild metric seems to have functioned mainly in a toy model pending its elaboration (which took 48 yrs.) into the Kerr metric. It's possible to imagine a star whose rotation would've been cancelled by gravitational factors cancelling each other, but such an excruciatingly perfect balancing might be rare enough to be "borderline unphysical". $\endgroup$ – Edouard May 19 '20 at 22:38
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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

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  • $\begingroup$ thanks for the effort! So in your example you pick $\beta$ and you compute a stress energy tensor that makes your choice a solution of the equations of motion? And this stress energy tensor has bounded scalar contractions right? $\endgroup$ – AoZora Jun 19 '19 at 15:08
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    $\begingroup$ Quite so, since every component of the Riemann tensor and the metric are locally bounded (I think that by multiplying $\beta$ with a bump function you could even make it bounded). It is very much unphysical and I doubt any reasonable matter model would obey such a distribution (especially as the variations in energy density become infinitely close together near $r = 0$), but it does work as a counterexample. $\endgroup$ – Slereah Jun 19 '19 at 15:12
  • $\begingroup$ I agree, thanks. I am not sure though about whether this singularity corresponds to the formal geodesic incompleteness. Maybe the source you pointed addresses this? $\endgroup$ – AoZora Jun 19 '19 at 20:44
  • $\begingroup$ This is very much the same thing as any other type of singularity : we wish for the various quantities important to GR to be defined everywhere, ideally in a continuous way. If some points do not obey such a constraint, then we remove them from the manifold. It is entirely possible to add back points containing singularities to the manifold, using boundary constructions, but the point is that the manifold ceases to be a spacetime if you do so. Usually this is because we wish for curvature to be defined everywhere. $\endgroup$ – Slereah Jun 19 '19 at 21:19
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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 $(-\infty,\infty)$.

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    $\begingroup$ Is extremal RN solution a naked singularity? Why? We have "a" horizon--the inner horizon and the outer horizon become the same but we still have a horizon. $\endgroup$ – Dvij D.C. Jun 4 '19 at 14:51
  • $\begingroup$ I agree with @FeynmansOutforGrumpyCat, the extremal case should still be acceptable. So maybe as long as we have a horizon, the boundedness of Ricci scalars could still imply that of the Riemann scalars. I am looking for a practical way to prove the absence of naked singularities and I don't think that checking geodesics would be easy, since in many cases it is difficult to have control on the metric when you add matter... If you have tips on how to check geodesics please do tell. Thank you very much for the answer anyways!! $\endgroup$ – AoZora Jun 4 '19 at 15:00
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    $\begingroup$ @FeynmansOutforGrumpyCat Apologies, you're correct you have a horizon. The extremal case is somewhat odd since your initial Cauchy surface `hits' the singularity, and therefore I think this means your initial data is singular. $\endgroup$ – Sam Colley Jun 4 '19 at 15:04
  • $\begingroup$ Thanks for the response. Yes, extremal solutions have interesting properties and I would love to know more. :) $\endgroup$ – Dvij D.C. Jun 4 '19 at 15:05
  • $\begingroup$ Nordstrom black holes have a diverging stress energy tensor towards the singularity, though. $\endgroup$ – Slereah Jun 21 '19 at 14:30

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