Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

When a small black hole mergers with the larger black hole, Would an observer trapped in the smaller black hole be able to detect any difference? Does the observer still exists in the smaller black holes timeframe? If the larger black hole was built up from millions of smaller black holes do those event horizons merged together? Or do they pile up like BBs in a bowl?

share|improve this question
    
@Manisheath: your first comment sounds spot-on to me, although there are difficulties in the modeling. What happens inside of a black hole system probably does require quantum gravity in these complicated situations, because the spacelike singularity might generically resolve to a Cauchy horizon plus timelike singularity in the perturbed spherically symmetric black hole case. For rotation/charge perturbations, this is naively true, and the arguments for the Cauchy horizon being a singularity are so bad, they might as well be nonexistent, it's basically just "Penrose says so". –  Ron Maimon Mar 22 '12 at 2:42

2 Answers 2

up vote 0 down vote accepted

I will try to answer this question properly--- but the proper answer is difficult because we don't have an exact solution for merging black holes, except for the case where one of the two is infinite in size. We also don't have a full resolution of the Cauchy horizon problem in black holes.

The classical answer, from a physical point of view, is that the black hole horizons merge, and any observer in the interior of a black hole will do what it does normally, without noticing anything from the merger, because their time is pointing in a different direction, into the center. The problem with this answer is that it requires an answer to the question of what is at the center of a black hole--- a spacelike singularity that swallows everything up (as Penrose conjectured), or a wormhole-like Cauchy-horizon pair that leads the observer to turn around and come out of the same black hole (what I personally believe). You would only realistically be able to come out of a spinning or charged black hole, even if Penrose is wrong and I am right. I should point out that absolutely nobody in the world agrees with me regarding this, but they have no real argument. But that doesn't mean anything, it's always like this when you suggest something new.

For a neutral black hole, you will hit the singular center for sure--- there is no Cauchy horizon, or rather, it shrinks to a degenerate point.

The issue of coming out of the black hole complicates the answer, because you could come out inside a bigger black hole, which the original black hole fell into during the intervening time. There is no way to answer this question without knowing how stuff comes out, so I will from now on pretend that this is impossible--- that you can't come out of a black hole. The reason is that I have no idea how long you spend inside a black hole that you come out of, and you would have to sometimes come out antimatter and left handed (if you are righthanded normally).

Anyway, ignoring this, the black holes swallow observers that are killed independently of what happens to the black hole later. Black holes that get close merge, but by each one's interior mushing up on the surface of the new black hole that forms, like soap bubbles hitting together a high-pressure region. Physically, the black hole horizons join together to make a new horizon, but the process of connecting is not classically realizable (it takes an infinite time for each black hole to fall into the other, from the exterior point of view).

Physically all this doesn't matter--- the black holes merge into one like soap bubbles merging. The merging of soap bubbles is also discontinuous from a long-wavelength continuum description.

Preserving discussion regarding the accepted answer

This discussion was interesting, and maybe helpful to see where this idea of rotating BH emissions is coming from, and that it is not mainstream physics (or at least not yet).

RM: You make some not well supported claims that I think are false: 1. An observer will reach a singularity 2. the singularity of another black hole can reach the observer first. Number 1 is only correct for nonrotating unperturbed black holes, and when a black hole is falling into another, I only know how exactly to model the interior solution in the limit that the big black hole is infinite. The observer could just rebound out of the first black hole, going through a Cauchy horizon to the outgoing sheet. As for 2, it is holographically suspect. This smells like an open question.

AB: I might get wrong somewhere of course, but let me disagree with your comment: 1) Even for perturbed and rotating (astrophysical) black holes an observer shall reach the singularity sooner or later, unless the observer gets ejected during the merger. +It has nothing to do with your ability to model the interior solution. 2) Imagine you send and observer to fall into a black hole 1 from a static position, and throw right after him a relativistically moving black hole number 2, so that it is tuned to reach observer when he crosses the horizon of black hole 1. I see no cheating here.

... However, I find very interesting the question you have posed. How much do you know about the possibility of an observer getting outside an event horizon through any process, such as black hole collisions or whatever? Maybe you have any references on that?

RM: There is no evidence that observers can reach a singularity outside of the perfect spherically symmetric Schwartschild solution (which unfortunately is presented as the generic case in books). When a black hole is perturbed, if it is rotating or charged, perhaps also when it is deformed by a strong gravitational field, the central singularity turns into a Cauchy horizon surrounding a timelike singularity. An observer cannot hit a timelike singularity, the observer just bounces past the Cauchy horizon (with tidal forces of course), turns around, and comes back out.

... The problem is that the Cauchy horizon parts are unstable to infalling deformations, and you generically get a lot of crud on the Cauchy horizon which makes a wall of hard-radiation in the limit of an eternal black wall. Some people (meaning Penrose) speculate that this means that you can't cross a Cauchy horizon. If this speculation is correct (I am pretty sure it isn't), then everything will hit a singularity in a real black hole. Whether this speculation is right or not depends on the details of both classical and quantum gravity.

... The classical issue is exactly how singular the Cauchy horizon is. From papers I have seen, and my own seat-of-the-pants intuition, the Cauchy horizon is not terribly singular, it is just like a sudden potential step in time quantum mechanics. It will excite the infalling system, make some anti-particles, but not to infinite energy, and you might be able to survive going through. This is supported by the observation that in an empty universe, the Cauchy horizon is completely regular-- no problem crossing it at all.

Once you cross Cauchy 1, you are in the center region, where you see a timelike singularity. You can't reach this timelike singularity, because it pushes you away, but you can shine light on it. You then cross a second Cauchy horizon, and you end up in the past region of a black hole very similar to the one you fell into, and then you are ejected. This cycle is most pronounced in the extremal case, where you can make things bind to a black hole and make simple harmonic motion by going in and out again and again.

... The reason quantum gravity is needed to make sense of this is that classically, the outgoing region is disconnected from the infalling region--- they are separate universes. In the 1970s, people speculated that the black hole links to another universe for this reason. But we know better today--- the only place you can come out, if this story is correct, is in this universe. But this requires a gluing map which identifies the other universe with this universe, and this gluing map is very difficult to figure out (I tried and never got a sensible answer I trusted).

... You asked about references--- unfortunately I have no references, this is just something I've been thinking about. The closest thing to an argument in the literature is that if you make a stack of D-branes, slide one out, and push it so that it falls into the others, it should oscillate back and forth in a reversible way. Unfortunately, the only literature reference I know is to an article by Gubser, which I think is incorrect and the argument he gives is not sufficiently convincing to me, which says that the brane will not oscillate reversibly, but get trapped in the stack.

PS: There is something wrong with this answer. Consider the phrase "1) The observer is able to feel the collision, provided that he hasn't yet reached the singularity of home black hole." Since, from the point of view of an observer outside the black hole, the infalling observer never actually makes it beyond the event horizon, your "provided" clause is void. – Peter Shor Mar 22 at 11:07

AB(@PS): sorry, but either you are strictly wrong here, or me/you have been unclear. Let us call observer 1 the one who is falling into a black hole and observer 2 the one who is remote. The fact that the observer 2 will never see the observer 1 crossing the horizon does not imply at all that ther observer 1 will not experience crossing the horizon or hitting the singularity. Observer 1 can experience 1) crossing the event horizon, 2) hitting the singularity, 3) feeling the tidal field of another intruding black hole. It is strict, definite and I refer to classical textbooks on GR, say MTW.

AB: I really enjoyed reading your comment and thank you for sharing your ideas. However, I believe the most relevant physics to the question is the one which concerns astrophysical black holes in classical general relativity. Why not quantum? It has not been established yet. Why astrophysical? Because the are the only type of black hole which are known to form in the universe under known physics. I will go through a couple of points where I cannot agree with you in the comments that follow.

... 1) Astrophysical black holes are formed by a gravitational collapse. They don't contain any wormholes and their formation doesn't change the topology of space-time. If an unlucky observer happens to find himself inside the event horizon of such a black hole, he cannot escape it in any way other then just by crossing backwards the horizon, which is impossible for stationary black holes.

...1.1) The question I asked you is if you actually know the following. Given that an observer has crossed an event horizon of any astrophysical black hole (further BH), possible nonstationary, can the observer escape it, under any processes, still in the GR picture.

...2) All perturbed BHs, including rotating ones, are known to settle down by emitting gravitational radiation. This is supported by perturbation theory and numerical relativity. If you throw in an observer, considering him as a perturbation, the system will settle down in the end, and hence the observer shall get static and get absorbed in the black hole solution, hence find himself on a singularity.

AB: Peter, you are definitely right here, probably the time sequencing could be done more careful. However, the whole description can be shrunk to what observer 1 shall see, in a sequence, and what the observer 2 shall see. For the observer 1 the sequence remains valid: he falls into the small black hole, and starts feeling the tidal field of another black hole right after crossing the horizon, and then experiences getting hit by it.

RM: The comments you made are superficially convincing, but are known today to be wrong. Rotating black holes, which are formed astrophysically, settle into a wormhole state without a doubt. The fact that the horizon, at formation, is a pure past horizon is irrelevant. We know today that past horizons and future horizons are dual quantumly, and that the fact that one Penrose diagram says things can only go in doesn't mean that another equally valid Penrose diagram can only have things going out. This is a change in the classical picture due to Susskind complementarity.

AB: Thank you again for your comment, I shall ponder on it. Can you please give me a reference to a research paper, which shows that rotating astrophysical black holes of classical general relativity settle into wormhole states?

RM: I do not use authority to support my positions, but I can explain why. The way to see this is to look at the global structure of the Kerr solution, which is worked out in Hawking and Ellis. It is qualitatively identical to the far simpler Reissner Nordstrom solution, and, unless you declare that the Cauchy horizon is too singular to pass, it makes a wormhole to another disconnected universe. The only wormhole free solution is the Schwartschild, because it is too symmetric. Wormhole to another universe is nonsense, it makes information loss, so you need gluing. – Ron Maimon Mar 23 at 18:42

... I think, from reading over your comments, that you are under the impression that black hole solutions have a singularity which absorbs matter--- this is what people say in popular books, but it is absolutely false. Only unrotating, uncharged Schwartschild black holes have a spacelike singularity which you can hit. There is not a single generic black hole solution with a space-like singularity. The spacelike singularity is just an artifact of spherical symmetry. Generic Penrose singularities are timelike.

AB: Dear Ron, so is it correct to say, that according to Hawking and Ellis (one of the best books on GR, actually) in their book all rotating noncharged black holes, formed by gravitational collapse of ordinary matter, always form also a wormhole singularity?

... I feel great affinity to your position about being critical to scientific sources, such as scientific papers, for example. However, if you make a claim not supported by a sourse you imply that you can provide a scientifically grounded proof of what you say. In other words, you make and original statement with a corresponding grounding -> you are responsible for the correctness of it. If you refer to other persons research, he/she is responsible for the statement and the proof.

RM: Hawking and Ellis say only the following, both of which are correct: 1. the exterior solution is asymptotically Kerr 2. the interior solution of a Kerr solution is a multiple-universe connecting wormhole. They do not say that the wormhole will form during collapse, because the continuation past the Cauchy horizon is suspect, because the Cauchy horizon can become singular. This is well known. It suggests that any astrophysical black hole will link to another universe, and lead to information loss, which Hawking advocated for more than 20 years, because of this property.

... I agree that I am responsible for an argument that I will not cite. But the only original thing in the things I said above is the statement that if you go into a rotating black hole, you go through the Cauchy horizons and come out of the same black hole. This is physically required by unitarity, if the Cauchy horizon is traversible and not singular, but I never found the gluing map. There is no hint for what it should be from classical mechanics, and there are crazy things--- you need to glue time backwards in some places, and you have to make sure you always come out in the future.

AB: well, though you said not a single time, that this and that popular source is incorrect. So, Hawking and Ellis in the end don't ever actually state that the black holes formed by gravitational collapse produce wormholes, right?

... Concerning Hawking and information loss, the whole idea was about no-hair theorem only: you throw information into a black hole, it settles down to a stationary solution, and you find that information has disappeared.

RM: Hawking and Ellis don't take a position on it--- they simply explain that the Kerr solution is a wormhole, and that the Kerr solution forms. You have to check everything for yourself anyway, but they happen to be correct in all their points. Kerr black holes produced by gravitational collapse produce wormholes if their interior Cauchy horizons are not too singular. It is not known if they are or not (I say no). The main difference between what I am saying and what people in the 1970s said is that I am saying the wormhole is to this universe, and you exit the same black hole.

... The no-hair theorem applies at asymptotic times--- it tells you that the black hole will be stable only in a Kerr state. It doesn't imply that the information disappears, only that the maximum entropy end state is Kerr. This is a thermal state (when not extremal) so it is just hiding the information inside like any other thermal body. I am asserting baldly that for near-extremal black holes, the stuff that goes in in the exterior picture spreads out over the horizon, then goes around the hole, and recollects and comes back out, making a harmonic oscillation in and out.

share|improve this answer

The observer falls into the center of a BH in a finite proper time. If you launch a BH and an observer in asymptotically flat background with an ultrarelativistic velocity with respect to some remote observer, the proper time of the person inside the black hole will be significantly delayed with respect to the remote observer. More robustly, you can attach a 'clock' to the black hole by introducing appropriate coordinates near it, set up the correspondance between the proper time (position) of the observer inside the black hole and the time as shown by the black-hole 'clock'. The latter can be put into correspondance with remote clock.

Therefore, many things may happen around, while the observer still falls into the black hole. Observer can also get the information about the events happening around if it is transferred to him before he falls.

Imagine now that relativistically moving black-hole+infalling observer are going to collide with another black hole head-on. During the collision the event horizons shall merge, the observer shall feeling tidal fields from another black hole, and it may even happen that the singularity of another black hole shall reach him first.

Hence: 1) The observer is able to feel the collision, provided that he hasn't yet reached the singularity of home black hole, 2) The observer does exist before reaching a singularity, though the tidal forces may be quite harsh

Concerning the second question, event horizons of many small black holes do merge when the black holes are close enough. You can think of a black hole as a point mass surrounded by spherical-shaped(roughly speaking) horizon. Horizons tend to merge when getting close, and the point masses remain covered by the new horizons, but tend to inspiral onto each other and merge over time. If you imagine millions of small black holes put together, first the horizons shall merge, and then all the masses under the large horizon shall form one bigger mass.

share|improve this answer
    
@Ron: Concerning the gravitational collapse, I shall extend the statement to "All the books on classical GR(cited more than 10 times) either state that the wormholes are not formed or don't state that they are formed in the above context of grav. collapse". Can you prove it wrong? –  Alexey Bobrick Mar 26 '12 at 20:51
    
@AlexeyBobrick: They do not state any such thing--- they usually follow Penrose and say that the Cauchy horizon is singular. There is not a single book that I know which definitively says wormholes are not formed, and even if there was, it would just be wrong. It is annoying to argue over content of sources--- physics is objective, and you can just argue over the matter, not the sources. I haven't read every page of every book, and I don't know what stupid things this or that book says. I only know the interior structure of the black holes themselves, which is enough. –  Ron Maimon Mar 29 '12 at 2:26
    
@all, no problem with preserving the discussion :) –  Alexey Bobrick Mar 29 '12 at 8:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.