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If two black holes collide and then evaporate, do they leave behind two naked sigularities ore? If there are two, can we know how they interact?

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They will not leave anything. –  Anixx Feb 2 '11 at 23:15
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Since classical gravity is much stronger than quantum gravity, the collision will take place on much faster time scales than the eventual Hawking radiation. So, the two black hole will experience a fairly violent process, and eventually settle down to a static situation which is most likely a new black hole. That black hole will evaporate on much longer time scales, eventually either evaporating completely, or leaving behind some long-lived "remnant". Nobody really knows what the black hole leaves behind after evaporating, but I don't think it will be a naked singularity.

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The answer is neither two nor one. Black holes that collide and subsequently evaporate leave just a bunch of radiation. There is no remaining 'naked singularities'.

In the past there has been a lot of confusion about black hole evaporation. But questions like "what happens with the information describing the micro state of the black hole" have been answered. Key idea is the holographic principle and black hole complementarity. If you want to learn more about this, Lenny Susskind's book 'The Black Hole War' is a good starting point.

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they leave a single black hole plus some radiation, not just radiation. –  Jerry Schirmer Jan 16 '11 at 18:11
    
@Jerry Schirmer - References? –  Johannes Jan 16 '11 at 18:23
    
arxiv.org/abs/0710.1338 , for one. an arXiv search for "binary black hole coalescence" will give a lot of hits. –  Jerry Schirmer Jan 16 '11 at 18:51
    
You are wrong. Nowhere in this paper the authors claim that the black hole remaining after coalescence is immune to Hawking radiation. –  Johannes Jan 16 '11 at 19:40
    
The answer is good with the caveat that any question like this that relates to quantum gravity is not settled. However, the most widely held opinion given what we think we know, is that black holes would eventually evaporate in an explosion of elementary particles leaving no other remnant. The holographic principle and its realisation in string theory has provided new understanding of how this can work without violating the laws of quantum mechanics and thermodynamics. –  Philip Gibbs Jan 18 '11 at 13:02
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you can't produce any naked singularity in any process in 3+1 dimensions. This statement was originally coined by Roger Penrose under the slogan "Cosmic Censorship Conjecture" (CCC). A naked singularity would be a singularity that could affect the world around it because it would have no horizon; I will explain these words later.

Because dynamics seems problematic near a singularity - it's hard to calculate it by classical general relativity or even the descriptions of quantum gravity that we know - it looked like the Universe would be unpredictable if there could be "naked singularities", i.e. singularities that are unprotected by event horizons. That was viewed as a problem by Penrose (and others), a potential inconsistency that the Universe should avoid.

This thinking due to Penrose has been showed partly incorrect, at least in higher dimensions. A proper theory of quantum gravity simply can (and has to) produce predictions - at least in principle - what happens to an observer who can observe any singularity. And indeed, there have been counterexamples - thought experiments in higher dimensions - in which one can produce a naked singularity at the end.

As far as I know, it is still true that the CCC holds in 3+1 dimensions. One can't create naked singularities by any process - whether it involves 0 black holes, 1 black hole, or 2 black holes. Black holes usually have a singularity inside them but they are not naked exactly because they're black holes. A defining feature of a black hole is that they're holes that are black - which means that light can't escape from them. It can't escape from them because it is confined by the event horizon. The event horizon are the "clothes" in which the singularity is "dressed" - so it is not naked.

So despite some popular misconceptions, the defining feature of the black holes is the event horizon, and not a singularity. A singularity is just a typical addition that comes with the black hole "package". In Schwarzschild black holes, the singularity is spacelike. Nevertheless, once the black hole evaporates, nothing is left. At the very final stages of a black hole decay, the tiny black hole looks pretty much indistinguishable from a heavy particle species in ordinary particle physics (or string theory), and it decays to a few particles that we know from the colliders.

Two stabilized black holes, when they collide, emit some gravitational waves, and when they're sufficiently close (so that they don't just continue as two black holes), the rest of their mass=energy that hasn't been radiated away collapses into a single black hole that quickly gets stabilized as well. This black hole will proceed with the Hawking radiation and after a very long time, it disappears as well. Nothing is left. The research in the last 20 years or so has made it pretty much clear that there can be no remnants left etc.

All the best Lubos

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As I read the question, the implicit assumption is that black hole evaporation produces the naked singularity, not the collision of the black holes...I like the part about the singularity/horizon distinction, as you know this causes infinite amount of confusion. –  user566 Jan 16 '11 at 18:30
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An extreme fine-tuning of initial conditions can produce naked singularities, as shown in the critical collapse simulations of the '90s. The modern rephrasing of the Cosmic Censor Conjecture states that evolve to horizon-ed solutions form a dense set in the phase space of physicially reasonable initial conditions, so it would take an implausibly precise fine-tuning to produce a naked singularity. –  Jerry Schirmer Jan 16 '11 at 18:54
    
I believe the suitably refined but still general statement is still as of today a conjecture. You can prove it in certain scenarios with extra symmetry. But the black ring solutions in higher D definitely leaves the door open in 3+1D for exotic possibilities –  Columbia May 25 '11 at 20:44
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The rules of general relativity give us a very specific set of rules for the interaction of the two black holes. We can integrate thee equations to know what would happen if, for example, we had two black holes orbiting each other, or on a head-on collision course.

The result that we get is that the two colliding black holes will not evaporate or form a naked singularity. Their horizons will coalesce into a single distorted horizon, which will then smooth itself out by emitting gravitational radiation. At late times, you will have a single black hole with a mass less than the combined masses of the two initial black holes (the difference in mass will be the energy that is radiated away as gravitational radiation). According to numerical simulations of collisions, no naked singularity should be formed for a generic collisions.

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I assume that the "evaporation" in the question is Hawking radiation. The end stage of Hawking radiation is not well-understood, but nobody expects any naked singularities to be involved. –  Ted Bunn Jan 16 '11 at 17:59
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@Ted Bunn: Yes. I didn't quite understand what colliding black holes had to do with Hawking radiation, so I answered the question from the classical GR side. The only way that I could see a naked singularity involved is from the fact that the critical solution in critical collapse phenomena is a zero mass naked singularity. But in any regard, we'd need a full theory of quantum gravity to be able to talk about the late stages of Hawking radiation. –  Jerry Schirmer Jan 16 '11 at 18:10
    
it is ok to say that we can't have absolute certainty about the late stages of the Hawking radiation process without a full QG theory, but to deny Hawking radiation (see your comment on my answer), is obviously incorrect. –  Johannes Jan 17 '11 at 0:23
    
@Johannes: sure, but we can't be absolutely sure what the final state is. It's completely plausible that the radiation will eventually reach equilibrium with the black hole, for one example. If the horizon evaporates quickly enough, a naked singularity is certainly possible for some length of time. I also hesitate to say how sure we are about an effect that has never been observed, however much it is supported by simple generalizations of existing theory. Perhaps Hawking radiation ends up suppressed, like how QM ultimately prevents electrons from spiraling into protons. –  Jerry Schirmer Jan 17 '11 at 0:35
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The answer is zero. When black holes evaporate due to Hawking radiation they will not leave any singularity. The singularities in black holes are inside the event horizon.

However, when two black holes collide their singularities do merge to form one singularity inside the event-horizon of the merged black hole. Perhaps that is what you really wanted to know.

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