Imagine two Kerr black holes with ring singularities oriented in different axes (e.g. one horizontal and the other one vertical). If they merge, what will happen to these singularities? Will they form a kind of outline of a sphere or get disrupted in some way? And what if hundreds of these black holes merge? Will the ring singularities end up forming a sphere with an inaccessible region inside?

  • $\begingroup$ The end point of the merger will be another Kerr black hole with a different axis. The ring singularity will then be oriented along that new axis. $\endgroup$
    – Prahar
    Commented May 20 at 13:43
  • $\begingroup$ How does that happen, and how is the new axis "decided"? $\endgroup$ Commented May 20 at 13:51
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    $\begingroup$ Kerr black holes are eternal, so they can never merge, or change in any way. They are also the only thing in the universe, so there cannot even be another one to merge with! $\endgroup$
    – m4r35n357
    Commented May 20 at 14:01
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    $\begingroup$ @peterh “I don't think it so” - Are you saying that you are unfamiliar with the mathematical boundary conditions of the popular vacuum solutions, such as Kerr, Schwarzschild, and others? It is not a matter of “thinking”, but rigorous math. It is a common knowledge that the Kerr solution is eternal, never changing, non-evaporating, with nothing ever falling to or merging with it, and nothing else existing anywhere in the infinite, flat, and non-expanding universe. When rotating stars collapse, they may start looking like black holes, change or even radiate, but they are not Kerr black holes. $\endgroup$
    – safesphere
    Commented May 21 at 5:01
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    $\begingroup$ @Flamethrower “what if hundreds of these black holes merge?” - The key in Tim’s correct answer below is that a merger is an outside process. All outside processes complete before the horizon forms in the eternity of the cosmological time. Even if all black holes in the universe hypothetically ended up merging into a single one, still all mergers would complete outside the horizon. The consequent evolution inside is the evolution in time, involving a single event horizon, a single Cauchy horizon, and a single singularity (if you assume it really exists). $\endgroup$
    – safesphere
    Commented May 21 at 5:13

2 Answers 2


The ring singularity in Kerr is something that exists inside the inner horizon. Therefore the ring singularity exists only in the future of anyhting that happens outside of the black holes, including any merger with other black holes. It therefore does not make sense to ask what happens to the ring singularities after the merger, because the ring singularities did not exist at the the time of the merger (for any reasonable definition for "at the time").

More significantly, the inner horizon of the Kerr solution is unstable. Even the slightest perturbation sourced outside the black hole (such as the presence of a second black hole) will become singular at the inner horizon. Consequently, we do not really have a good idea what general relativity predicts inside the inner horizon for realistic situations (quite possibly a singularity at the horizon itself). We should therefore not take any of the predictions from the maximally extended Kerr solution (like the ring singularity) too seriously beyond this point.

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    $\begingroup$ Probably the most reasonable definition is "for the external observer far away". But I think, most likely OP is asking it in the view of some internal guy. $\endgroup$
    – peterh
    Commented May 20 at 23:42
  • $\begingroup$ @peterh Physics doesn’t depend on coordinates. As Tim’s answer correctly explains, everything outside happens before anything at the horizon or beyond (due to the global hyperbolicity of the Kerr spacetime at least outside the horizon). Thus any black hole merger finishes before the horizon comes to existence in any coordinate system. For example, a free falling person, due to his time dilation, would observe the merger dramatically speeding up and completing as he is approaching the horizon. From that point on it is already a single black hole with a single Cauchy horizon ahead in time +1 $\endgroup$
    – safesphere
    Commented May 21 at 4:43
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    $\begingroup$ Also, the black hole decaying into Hawking radiation happens before the singularity forms. $\endgroup$ Commented May 21 at 9:48
  • $\begingroup$ So, before they merge, black holes don't have an event horizon? And is that what happens in practice? $\endgroup$ Commented May 21 at 13:37

Roy Kerr's interest in black holes goes beyond the ringlike singularity which is presented in his mathematical model only as a placeholder for a progenitor physical object like a dead star or collapsed gas cloud.

In Kerr's thinking there is no accessible region within the ring because that's where the progenitor is, rotating in such a way that allows particles to maintain orbits around it. Outside of that force boundary--known as the Cauchy horizon--nothing can prevent particles moving inwards and even light cannot escape (which is akin to saying time stops there from our distant perspective).

  • $\begingroup$ So the material of the object that collapsed into the black hole is in the center of the ring singularity? $\endgroup$ Commented May 21 at 13:39
  • $\begingroup$ @Flamethrower the point is that gravitational collapse is incompatible with ring singularities. $\endgroup$
    – m4r35n357
    Commented May 21 at 14:22
  • $\begingroup$ So there are no real ring singularities? $\endgroup$ Commented May 21 at 14:31
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    $\begingroup$ @Flamethrower Kerr's recent opinion is that there is still spinning matter in there. But that doesn't change the fact that the Kerr BH solution is eternal, non-changing, and solitary. It is a toy model of the simplest possible rotating vacuum spacetime. $\endgroup$
    – m4r35n357
    Commented May 21 at 14:36
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    $\begingroup$ To be more clear: the Kerr metric describes an otherwise empty universe that contains only an eternal black hole and nothing else, while our universe is not eternal in the past and also not empty. Outside of the inner horizon it is assumed to hold good enough though, but inside the inner horizon things are up for debate. $\endgroup$
    – Yukterez
    Commented May 21 at 16:00

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