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I've read a report about LIGO saying that the ring-down oscillation showed that it took time for the pair of BHs to fully merge, implying there was a momentary bulge as one circled around and into the other.

I imagine two black holes going past each other, but coming close enough that the event horizons intersect slightly. Would they have sufficient momentum to NOT merge? Effectively to pull in on each other at something less than acceleration at C?

At the intersection zone I picture gravity pulling in both directions, so momentarily the intersection would have zero local gravity. So perhaps this zone would no longer be behind the event horizon. And moving fast enough the black holes would seemingly clip each other before they merged. Perhaps some ring-down noise would happen as each event horizon got perturbed in the close pass.

Nick
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  • If the event horizons touch then they will never separate. then it's only a matter of time before the event horizon arranges itself in a more stable state. – shai horowitz Mar 09 '22 at 23:23
  • You can probably argue that they have to merge using the laws of black hole thermodynamics; the surface area of a black hole's event horizon cannot decrease. If the total area changes during the fly-by, then it has to have decreased and cannot return to its original value after the black holes have passed each other. This would imply that they have to end up merging. But you would need to argue that the area has to decrease in all 'fly-by scenarios' (as opposed to staying constant). Not sure how to argue that generically. – user34722 Mar 10 '22 at 00:00
  • I wonder if this could be well modeled by comparing it to two water drops falling together and merging, but where the intermolecular cohesion forces are set much higher than surface tension. So once they get close, there is a "tubelike" bridge structure between them very briefly – RC_23 Mar 10 '22 at 03:41
  • this calculation is https://ec.europa.eu/research/participants/documents/downloadPublic/c2V4NHVSdjNHU3kyRVA0dHhsUzVPZjhtNTFjRnlXdnBxRW9wV2RGRSs2ZHhhSXRWZ2xPU2tnPT0=/attachment/VFEyQTQ4M3ptUWZDcXNmbjYrQUN0TjFBUGdEWnJFNzI=about grazing black holes – anna v Mar 10 '22 at 05:19

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Once something crosses a black hole event horizon, it can never escape. This means that if the event horizons of two black holes touch or overlap, they will merge.

joseph h
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  • If this is a valid argument, I don't fully understand the logic. If a star flies past a black hole, and a small portion of the star enters the event horizon, that portion will have to enter the black hole, but not necessarily the entire star. One could argue that the mass of the black hole is 'located at the singularity' and therefore hasn't crossed the event horizon. – user34722 Mar 10 '22 at 00:00
  • that portion will have to enter the black hole, but not necessarily the entire star Can you provide a citation to that? Thanks. – joseph h Mar 10 '22 at 01:19
  • I'm not sure it's ever a good idea to say "escape velocity equals speed of light," since the former concept is entirely Newtonian. And if anything this would correspond to the photon sphere not the event horizon, no? You have more physics education than I do by your profile, so I'm open to being wrong in this opinion. – RC_23 Mar 10 '22 at 03:44
  • In what sense would you consider the event horizon of a black hole to be "something"? – TimRias Mar 10 '22 at 07:52
  • I wouldn't be able to find a paper easily, but here is I think a pretty solid argument: the escape velocity outside the horizon is <c, so it's certainly possible for most of the stars mass to always be travelling at greater than the local escape velocity. Also recall that the gravity from a black hole at a distance R is the same as that of the original star at that same distance, so most of the 'fly-by' star isn't experiencing particularly strong gravity actually. – user34722 Mar 11 '22 at 05:07