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I have read this question:

What I have not seen is a purely classical argument for the non-separation of a black hole merger. One can obviously take the time reversed spacetime manifold of a merger and get a valid manifold that splits - but this is a black hole disturbed by a complex ingoing set of strong gravitational waves that splits, a bit like how a loud crash sound converging on some shards can in principle make them reassemble into a vase jumping back up on a shelf. Possible, but thermodynamically impossible.

If you mean overlap, and if you are asking if it's possible that merging can be avoided even if the event horizons of both black holes were to overlap, then the answer is still no. They will merge (regardless of how fast the black holes are moving). The escape velocity at an event horizon is equal to the speed of light, and nothing (with mass2) can have this speed, so even if two black holes approach with a great speed, they will merge if their event horizons overlap.

Black holes: Is merger inevitable when horizons touch?

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Imagine a setup where two black holes of similar size and energy moving both at relativistic speeds, but in opposite directions so, that when they pass by, their event horizons slightly overlap. Now as far as I understand, gravitational fields here are extremely strong, but not infinitely strong. In my understanding, the kinetic energy of the black holes could overcome the gravitational attraction, and even if the event horizons (which are not physical objects, but just boundaries) slightly overlap, the black holes themselves could continue their way in the opposite directions, and separate the event horizons again (maybe by exchanging part of their energies, so that they change mass).

The first answer says so (it is possible), and the second answer only applies the speed of light limit to say no. The second one is arguing about the escape velocity, but that is not what I am asking about. I am not denying that the horizons will inevitably overlap. I think it is a given that the effects of gravity travel at the speed of light (faster then the black holes), and the horizons will overlap. But just because two boundaries overlap where the escape velocity is exceeding the speed of light, why would that affect the whole objects' (black holes) interior completely? Just because the two boundaries overlap (meaning a partial unification), that does not necessarily mean that the interiors of the objects will completely unify. The rest of the objects could have kinetic energies that overcome gravity. I believe that it is not correct to think that black holes are some kind of rigid solid objects (where if the boundaries touch it will grag the whole object inevitably), whereas in reality they are objects like stars but with a boundaries that signal the extreme gravitational field. Extreme does not mean infinitely strong, that is very important. What I am asking about is the possible domination of the kinetic energies over the attraction of gravity.

So there are two thought coming to mind:

  1. black holes are not solid rigid objects (we are not talking about two billiard balls touching). Why would the overlap of boundaries drag the whole object inevitably? Are we realistically saying that two black holes would slow down in a matter of seconds from 0.9c to close to 0 (or possibly go into a spiral)?

  2. gravity is extreme, but extreme does not mean infinite. Kinetic energies could dominate gravity.

Just to clarify, I am asking whether the kinetic energies of the black holes could be enough to overcome the attraction of the gravitational fields, and yes, especially in the case when the horizons slightly overlap. The gravitational fields of the black holes is not infinitely strong. Theoretically, the kinetic energies of the black holes could overcome this. So no question, the horizons will overlap, and maybe the two black holes will exchange energies, and change mass, but could depart and continue on their opposite ways.

Question:

  1. Can two relativistic black holes' event horizons overlap and separate again?
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    $\begingroup$ By definition, the answer is no, because the definition of the event horizon is that once something is inside it, it can never get out again. What you really may want to ask physically is how the event horizons of two black holes that pass close to one-another are deformed during the passage. $\endgroup$
    – Buzz
    Dec 23, 2021 at 0:41
  • $\begingroup$ @Buzz thank you, but I am asking, because they only partially overlap, most of the interior of the black holes is not inside the other one, why can't those parts depart? 1. Now by your definition, only part of the event horizons overlap, and those should not be able to escape any more, but what about the rest of the interior of the black holes? They are not rigid objects, what would keep them intact? The parts that are never inside the other one should be able to escape. 2. The definition does not talk about kinetic energy overcoming gravity (which is not infinite). $\endgroup$ Dec 23, 2021 at 2:23
  • $\begingroup$ related physics.stackexchange.com/questions/45448/… $\endgroup$
    – MBN
    Dec 23, 2021 at 10:10
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    $\begingroup$ One can not simply superimpose or add solutions like that; General Relativity is wildly nonlinear. $\endgroup$
    – m4r35n357
    Dec 23, 2021 at 10:24
  • $\begingroup$ @safesphere thank you so much! "There is no “volume” in the radial direction inside a BH. This direction is not a dimension of space, but a direction in time measuring the process of shrinking of the thin sphere starting from the horizon and becoming smaller until it vanishes.", this is very interesting, I might ask a another question about this. $\endgroup$ Dec 24, 2021 at 16:33

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Short Answer

No, they can't. One two black holes touch, they won't split again.

I'll sketch a little formal argument which can provide some intuition, but should not be taken too seriously. Nevertheless, I believe it provides an intuitive contradiction in the situation you presented. I invite other people to comment on possible issues and I will raise a few of them myself. At the end of this answer, I'll provide a more solid answer by referring to a rigorous theorem on black hole physics.

Intuitive Approach

Let us suppose one can somehow realize the situation you mentioned, but let me add a minor thing: suppose there is also some point particle of negligible mass on the spacetime. Suppose further this particle is in the intersection of the two black holes when they are superposed. Now, once the black holes separate (assuming the situation you presented is possible), in which of them should the particle stay? It should not be able to leave either black hole, but since they are separating and the particle is in the intersection it will need to somehow choose which black hole it will stay in.

Let me provide an alternative, but very similar, argument. for a Schwarzschild black hole, once a particle is inside the black hole, it must necessarily fall into the singularity within finite proper time. More specifically, one can show (see Problem 6 of Chap. 6 of Wald's General Relativity) that the maximum lifetime of of any observer within the event horizon is $\tau = \pi M$ ($c = G = 1$), where $M$ is the black hole's mass. However, this holds for both black holes (if we assume them to be Schwarzschild), so the particle should crash into singularity $r_i = 0$ within proper time $\tau_i = \pi M_i$. Since this must happen for both black holes and they are (by assuming the situation you proposed) getting further apart, we've reached a contradiction. If the black holes do not merge, then a particle would need to be able to escape the black holes (to be fair, I think the ).

Now for some issues with these lines of thought.

  1. As mentioned in the comments, General Relativity is wildly non-linear. One can't simply superpose solutions. In particular, the metric of the system is far different than just "summing" two black holes and there are gonna be effects of "gravitational energy" making the gravitational field even stronger (I'm writing "gravitational energy" to refer to the non-linear effects of the Einstein Equations, even though one can't properly define gravitational energy in a sensible and local manner). This also means that we can't treat both black holes as independent and just think of a collision with each other: as they get closer, the gravitational field changes in a non-trivial way due to their interaction;
  2. The result I mentioned concerning the maximum lifetime within a black hole uses the Schwarzschild metric and I'm being sloppy when applying it here, since this spacetime is not even stationary. If I recall correctly, more general black holes, such as the Reissner–Nordstrom and Kerr black holes, do not necessarily have these sorts of results, as one can tell from their conformal diagrams;
  3. Once the black holes are together, the pair is the black hole region and, at least in principle, the rule is nothing escapes the black hole region. Since the event horizon is a global construction, we also can't really define where one of them starts and the other one ends, and hence it doesn't really make that much sense to say the particle is within both black holes at the same time (we can't even properly define which black hole is which). This also gives a different hint on why the black holes can't split. I'd say this point is a double-edged sword for the argument haha.

Rigorous Approach

I can't really provide much more intuition, so in this section I'll just provide a statement and a reference for the detailed discussions. Proposition 9.2.5 of Hawing & Ellis' The Large Scale Structure of Space-time reads (up to notation)

Let $\mathcal{B}_1(\tau_1)$ be a black hole on $\mathcal{S}(\tau_1)$. Let $\mathcal{B}_2(\tau_2)$ and $\mathcal{B}_3(\tau_2)$ be black holes on a later surface $\mathcal{S}(\tau_2)$. If $\mathcal{B}_2(\tau_2)$ and $\mathcal{B}_3(\tau_2)$ both intersect $J^+(\mathcal{B}_1(\tau_1))$, then $\mathcal{B}_2(\tau_2) = \mathcal{B}_3(\tau_2)$.

In the above, $\mathcal{S}(\tau)$ is a partial Cauchy surface for the space-time at time $\tau$, properly defined on Proposition 9.2.3. Intuitively, it is a "photograph" of spacetime at time $\tau$. A black hole at time $\tau$ is defined at p. 317 and is the usual definition one would expect: a connected component of the region from $\mathcal{S}(\tau)$ from which nothing can escape to the null infinity. $J^+$ is the causal future.

Notice that, from these definitions, once the black holes touch, they are then one black hole only, and from the proposition I mentioned it follows that it can no longer split.

Hawking & Ellis is, of course, not the only reference discussing this result. Wald also proves it as Theorem 12.2.1 on the book I mentioned above. It is also shown on p. 73 of the Lecture Notes on Black Holes by H. S. Reall.

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  • $\begingroup$ Thank you so much! I understand your argument about a particle being in the overlapping area, and it has to move towards the singularity (which is more a point in the future then in space), but which (there exist two) singularity is it moving towards? $\endgroup$ Dec 27, 2021 at 4:58
  • $\begingroup$ @ÁrpádSzendrei That's exactly the issue: it must move to both, and that can only happen if the singularities end up merging into a single one. This means that if the black holes could continue their separate ways, we'd reach a contradiction, so we conclude they have to merge $\endgroup$ Dec 27, 2021 at 5:07
  • $\begingroup$ (of course, this argument is a bit cartoonish due to the points I raised in the answer, a more adequate argument is the theorem stating that black hole bifurcation is forbidden) $\endgroup$ Dec 27, 2021 at 5:07
  • $\begingroup$ @NíckolasAlves “the singularities end up merging into a single one” - This is incorrect. A singularity simply is the future of the horizon. Both are the same thing at different moments of time, not two things existing separately. If the horizons merge, it happens first, then the future of the merged horizon is one singularity. Two singularities never merge, because they don’t exist before the horizons merge. $\endgroup$
    – safesphere
    Mar 11 at 15:38
  • $\begingroup$ @safesphere Thanks for pointing it out! As I mentioned, that is just a cartoonish interpretation that shouldn't be taken too seriously, but might provide some intuition. The rigorously correct answer is the non-bifurcation theorem $\endgroup$ Mar 14 at 12:40

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