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I hope this question isn't too naive, but would it theoretically be possible to have a black hole with 2 singularities (or 2 black holes at the same location). If this is possible, would there be any significant differences on the effects of space time?

What I am thinking of is if you were to have a spinning black hole that has a ring as your first singularity and in the center of that ring was yet another singularity (with no angular momentum). Would it be theoretically possible for this structure to exist or would the inner black hole just begin to spin the same direction gaining some the momentum?

Similarly, what if you have two spinning black holes with perpendicular ring singularities at the same location? (And if you wanted to be really crazy, add another singularity into the middle of that)

If there is something I am missing or misinterpreting about black holes or singularities, please feel free to let me know as I am no expert.

John Rennie
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Mark N
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    Forces impossible to balance - collision would happen immediately. – Mithoron Apr 17 '15 at 15:30
  • You can have additional black hole topologies in 5 or more dimensions. Specifically, Black Saturns have two singularities. 4 dimensional black holes are topologically unique (the black hole uniqueness theorem). – botsina Oct 03 '20 at 02:07

3 Answers3

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There are only four known stable black hole geometries: Schwarzschild, Reissner-Nordstrom, Kerr and Kerr-Newman. We expect that any random assemblage of matter dense enough to form a black hole will relax into one of these four geometries by emission of gravitational waves. None of these geometries has two distinct singularities, so (as far as we know) it is impossible to have a black hole with two distinct singularities.

You suggest possible arrangements of matter in your question, and as Jerry says in his answer these could occur as transient states in a black hole collision. However in time (actually very quickly :-) they will relax into one of the four known geometries with just a single singularity.

John Rennie
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  • I guess the follow up question to this is why would these formations be considered unstable if they were ideally symmetric (which is often associated with stability)? – Mark N Apr 17 '15 at 15:44
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    @MarkN: Google for black hole uniqueness theorem or see this paper. However be warned that it is exceedingly hard going. I don't understand a word of it :-) – John Rennie Apr 17 '15 at 15:55
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    @JohnRennie, I'm not sure this is entirely correct. While Kerr-Newman is the unique stationary solution (at least in D=4), the question of stability is always framed regarding the exterior of the black hole and the event horizon. I don't think there is any reason to consider the interior metric to be stable. As a matter of fact the ring singularity in Kerr is inside a Cauchy horizon, which is notably untable by linear perturbation, so I don't see how to say something meaningful about the singularities – cesaruliana Apr 17 '15 at 17:32
  • And there are other, vague restrictions you have to make for physicality -- the hole could have a nonzero Taub charge or something, but this answer is right enough for the puroposes of the question. – Zo the Relativist Apr 17 '15 at 19:55
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I think the closest model to what you're talking about would be two colliding black holes, during the intermediate period where their horizons had merged, but the central objects had not yet collided. These systems are very different from isolated black holes, as they give off significant gravitational radiation, and have horizons that rapidly change in shape as they give off this radiation.

Zo the Relativist
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    Keeping in mind, of course, that in this context phrases like "had not yet" are more than a little imprecise, particularly when applied to something happening inside the event horizon. :-) – Harry Johnston Apr 18 '15 at 00:26
  • They can, of course, be made precise, given an appropriate foliation of the spacetime. – Zo the Relativist Apr 20 '15 at 02:22
  • Wouldn't such a foliation have to be arbitrary? That is, I would have thought you would usually be able to choose a foliation that gave either answer ("have already collided" or "have not already collided") even if you required the foliations to be equivalent to a given foliation everywhere outside the horizon. – Harry Johnston Apr 20 '15 at 03:48
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    @HarryJohnston: there's a topology change in the spacetime-- you go from two black holes to one. Any spacelike foliation will have this state. Similarly, the horizon is a coordinate-independent thing. How long the period lasts and the like may be coordinate-dependent, but a middle period where the horizons have merged, but the singularities have not IS something that can be defined in a foliation-independent way. – Zo the Relativist Apr 20 '15 at 12:13
  • I think I was confused by the fact that the merger of the singularities can't be in the past cone of any spacetime point outside the horizon, and that's the usual definition of "already happened". But what you're saying is that there are some spacetime points outside the horizon whose future cones include both the individual singularities and the merged singularity, and some whose future cones only include the merged singularity? – Harry Johnston Apr 20 '15 at 21:02
  • @HarryJohnston: OOOOHH. Yes. That's true. I'm kind of suggesting a bit of a cheat -- break up the spacetime into a foliation of spacelike surfaces, and then call each of these surfaces "an instant". It's an unphysical construction, but it works mathematically. – Zo the Relativist Apr 20 '15 at 21:18
  • To be honest, I still don't understand why black hole evaporation doesn't mean that the singularity never forms. But that's a whole other question. :-) – Harry Johnston Apr 20 '15 at 21:35
  • @HarryJohnston: the short answer is because the time of an internal observer and an external observer is different. The long answer would look at HOW the horizon evaporates and would involve drawing a Kruskal diagram of this spacetime, and singularity formation would depend on this answer. – Zo the Relativist Apr 21 '15 at 19:45
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Quite apart from whether a point singularity can exist inside a ring singularity there's the little detail that a solid ring (and a ring singularity will behave as one) in orbit about a central body is unstable. If one point is infinitesimally closer to the center (and it will--quantum uncertainty) there will be more attraction there and the difference will be magnified. Given the forces involved the ring and the point will very quickly collide and you'll be left with one singularity.

Your other configurations fare no better in the stability department.

  • How do you know that you won't be left with two (or three, four, five ... ) singularities rotating around each other? We don't know the analytical solutions for the inside of a stable Kerr black hole. – Peter Shor Jul 07 '18 at 01:54