# Must collapse to a singularity proceed through a strong curvature singularity?

A strong curvature singularity (s.c.s.) can be defined as one for which a geodesic is incomplete at affine parameter $\lambda=0$, with $\lim_{\lambda\rightarrow0}\lambda^2R_{ab}v^av^b\ne0$, where $v^a$ is the tangent vector. The volume of a cloud of test particles goes to zero as it approaches such a singularity, the interpretation being that infalling matter is crushed, not just spaghettified. A Schwarzschild spacetime's singularity is not an s.c.s., because it's a vacuum spacetime, so the Ricci tensor vanishes.

Is it true that during gravitational collapse to a singularity, there must be an s.c.s.? Does it matter if the singularity is locally naked? My intuition feels wobbly here, since there are some startling claims coming out, e.g., from Joshi and Malafarina, that challenge what I'd imagined about gravitational collapse.

If the answer is yes, then it would be interesting to see a proof. Does it depend on energy conditions? I guess in gravitational collapse to a standard black hole, we would then have have an s.c.s. (possibly naked) that later becomes a non-s.c.s. non-naked singularity.

If the answer is no, then is there a counterexample?

• Well, it is known that exactly in the Lemaitre-Tolman-Bondi models naked singularities show up, they are a canonical example of the violation of cosmic censorship, and it is corroborated by many simulations of dust etc. etc. These singularities do not have the character of gravity inevitably pushing stuff together, they instead seem to form by "accidental degeneracy". This makes people believe that pressure and generally any kind of repulsive interaction will smooth them out. I don't think there is any strict mathematical consensus on that though. – Void Sep 20 '17 at 13:57