Assume the weak energy condition (technically the Casmir effect violates it slightly, but it may not be enough to save us) and we make a black hole. The Penrose-Hawking singularity theorems apply to this case so we have singularit(ies) (i.e. GR breaks down somewhere). We don't know what type of singularity (space-like, time-like, orbifold, jump discontinuity, etc) will be created.

Now add the assumption that the limit of pressure/density for any matter or energy is positive as density grows to infinity. This is almost guaranteed given the uncertainty principle: to confine a particle to a smaller and smaller volume will take more and more energy, and in the limit of zero-volume the pressure/density ratio is that of light.

This matter can only reach infinite density with infinite space-time curvature or a perfect implosion. An isolated ball of radius r will explode with an acceleration of ~ c^2/r, so for any finite curvature there is some r below which the ball can't compress. An implosion (i.e. cavitation bubble collapse) requires perfect symmetry to make infinite densities (I believe), so can be discounted in any real scenario. Without infinite densities there are no singularities.

It seems the only way choice for our singularity is to have a space-like one, dooming any in-falling cyborg.

Most studies of charged and rotating blackhole interiors start with the hole and perturb it, rather than creating a hole through the collapse of matter starting with a regular spacetime. But this is unrealistic: the charged hole has a repulsive singularity that is a "Dirac-delta" violation of the weak energy condition and generates the divergence. The rotating hole has closed-timelike-curves. Without either of these, I believe no inner horizon can form.

Is my fatalistic reasoning correct?

This question's answer has a discussion of the singularity theorems but no mention of why the singularities would be space-like rather than time-like, etc.


1 Answer 1


The Schwarzschild singularity is spacelike. With Hawking radiation it is coincident with the end of the black hole as seen by exterior observer. Vafa has proposed the singularity is a condensate of tachyons which is where the vacuum is completely unstable. The observer passing into the black hole interior has this singularity in their future, and it represents the complete quantum instability of the black hole vacuum in the interior.

The Kerr, Kerr-Newman and Reissnor-Nordstrom metrics have time-like singularities. I include a large image of this below. This is the "ideal" situation for the eternal black hole. It also assumes the black holes in the opposing exterior regions I are completely entangled with each other. Susskind and Maldacena have illustrated how this corresponds to two black holes composed of EPR pairs. In this case the singularity at $r~=~0$ is a repelling sort of field.

The physical black hole has some departures from this ideal. The first problematic issue is with the inner horizon $r_-~=~m~-~\sqrt{m^2~-~a^2}$. It is coincident with $r~=~\infty$ or ${\scr I}^+$ which means an observer crossing this horizon encounters all possible radiation than entered the black hole. This is a Cauchy horizon, named for its structure similar to Cauchy sequences, and represent a sort of radiation wall. This appears to have singular properties. In addition, the Susskind-Maldacena result may be correct, but a black hole here could have been made form EPR pairs in a vast number of other black holes. This means that crossing the inner horizon might be a bit like being crushed in a "quantum garlic press"; every particle in your body gets sent into a different nontraversable bridge for each black hole. That would be rough. This means the inner horizon, which is a split null horizon, could be indistinguishable from a spacelike horizon in the Schwarzschild metric.

As a rule it is a good idea not to jump into black holes. Fortunately there do not appear to be any nearby, and the closest known is as I recall about $700$ light years away. The Milky Way black hole is large enough so you could enter it without tidal forces tearing you up, but you still would not last long in the interior --- less than $30$ seconds. Besides, you have to travel $27,000$ light years to get there.

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  • $\begingroup$ This answer is a mass-inflation argument: it starts with a black hole and then perturbs it, showing instability. I am looking for an answer that starts with regular spacetime and has matter collapse, forming a hole, and that shows that a spacelike singularity forms without needing any symmetry. $\endgroup$ Mar 4, 2017 at 5:00
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    $\begingroup$ This is similar to mass inflation. The inner horizon might only be passable if one enters a black hole composed of states entangled with every state that composes another black hole. The inner singularity as a Cauchy horizon could be problematic. It also as a boundary to the inner timelike region could be an Einstein-Rosen bridge leading to a huge number of possible inner regions you get fragmented into. Rough! $\endgroup$ Mar 5, 2017 at 12:50

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