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Do naked singularities also appear mathematically if the mass of a black hole is uniformly spread around its Schwarzschild surface?

I recently heard about the concept of a naked singularity and it made me raise an eyebrow. I know near to nothing to about the mathematics behind General relativity. However, I do know that mass can not move faster than the speed of light so, if mass would actually have to rotate around in the Schwarzschild sphere of the black hole, that would put a "nice" mathematical limitation on the maximal rotation of a black hole.

It's probably already thought of, but hey who knows, perhaps it might actually be interesting to look at.

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The mass of a black hole cannot be spread around its event horizon. The solution to Einstein’s equations representing a black hole has zero energy density at the horizon. It also has zero energy density outside the horizon, and zero energy density inside the horizon, except at the singularity where the energy density is infinite. At least in the case of a Schwarzschild black hole, this singularity is a clothed singularity, not a naked one.

So a black hole does not have a mass distribution. Of course, matter and radiation that have not fallen into the hole and are therefore not part of the hole can exist outside the horizon.

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  • $\begingroup$ now I do not want to be a smart ass, but with all due respect, considering blackhole entropy is related to the surface area of a black hole and the fact that for an outside observer sees no time pass at the event horizon, I doubt that that is effectively what is happening in a black hole. Regardless, it was a hypothetical question. What would the mathematics say about how a black hole would spin given this limitation. $\endgroup$
    – Tygo Huurnink
    Commented Jan 13, 2019 at 23:01
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    $\begingroup$ My explanation was for classical black holes, where everyone agrees on the math. There is no consensus on how quantum black holes work. $\endgroup$
    – G. Smith
    Commented Jan 13, 2019 at 23:14
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    $\begingroup$ The movement of particles outside the black hole, and the fact that in a local inertial reference frame they cannot move faster than $c$, has nothing at all to do with the spin of the black hole itself. Classical black holes can spin as fast as they want, but past a certain angular momentum they lose their event horizon, expose a naked singularity, and are no longer proper black holes. In a quantum theory, most physicists assume this either won’t happen or won’t matter because the singularity will no longer be a singularity. $\endgroup$
    – G. Smith
    Commented Jan 13, 2019 at 23:23

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