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I have an intuitive scenario. Consider we have a spaceship just below the event horizon of a BH, which is merging with another black hole.

Finally, the singularities merge and we have a single black hole again.

But, in the transient stage, it is unclear to me if a timelike world-line would exist to leave the system.

I suspect, the metric is probably far too complex for an analytical solution, but in the worst case, it could be maybe solved numerically.

As far I know, black hole merges are examined mainly in an inspiral scenario. I suspect, maybe the escape is possible only if they have a hyperbolic-like orbit (i.e. there is no inspiral, but they simply collide).

Is it possible?

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  • $\begingroup$ You have a worse scenario, from outside the black holes an observer will never see them to merge, or will he? $\endgroup$
    – user65081
    Jul 25, 2016 at 23:26
  • $\begingroup$ @Wolphramjonny I think, it is visible (recently we have "seen" one by the LIGO, maybe optically it had been also visible). $\endgroup$
    – peterh
    Jul 25, 2016 at 23:31
  • $\begingroup$ yes it is, due to what you mention is your question, a very complex warping of spacetime that differs a lot from the approximation that as you fall into a black hole, everything freezes $\endgroup$
    – user65081
    Jul 25, 2016 at 23:38
  • $\begingroup$ But I do not know the answer to your specific question $\endgroup$
    – user65081
    Jul 25, 2016 at 23:40
  • $\begingroup$ @Wolphramjonny It would be a theoretical possibility to get out information (measurement data) below the EH. $\endgroup$
    – peterh
    Jul 25, 2016 at 23:41

2 Answers 2

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No. When they merge their horizons will change shape, and eventually become the static or stationary shape of a BH horizon. Nothing inside either horizon while this is happening can escape. At all times the timelike curves stay inside, and the deformed horizons are where the lightlike curves end up. In each, and after they merge.

The area of each horizon right before they merge can not be smaller than before, as area is proportional to entropy which must increase or stay the same. All deformations will increase it (or be the same, but probably increase). At no times can lightlike Curves escape because of some deformation, and much less timelike curves.

I assume you meant you were right inside before. If you meant right outside anything can happen, now you'd have to take the ergospehereergosphere into account, and if inside it also probably no but I am a not sure.

There was similar question posted maybe 3 or so months ago, not in my saved list so I can't give you the reference. There were some answers.

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  • $\begingroup$ Sorry, "just below" instead "right inside" was a translation error :-) Thanks! $\endgroup$
    – peterh
    Jul 26, 2016 at 6:31
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    $\begingroup$ No problem, I guessed. $\endgroup$
    – Bob Bee
    Jul 27, 2016 at 0:40
  • $\begingroup$ Is there any proof backing this? The escape velocity at the event horizon must change if another black hole approaches it. That would intuitively mean the horizon would "recede" closer to the central point. Which contradicts the entropy requirement you mentioned. $\endgroup$
    – Tomáš Zato
    Oct 2, 2018 at 12:56
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Classically speaking, since total BH evaporation takes finite time, yet being at the event horizon stops all time (in your IRF), the black hole will evaporate before you can get to the event horizon.

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  • $\begingroup$ Thanks, but what is the case in the time of the ship? $\endgroup$
    – peterh
    Jul 26, 2016 at 0:54
  • $\begingroup$ For the ship, the universe's time increases proportionally. So the the black hole evporation increases, too. $\endgroup$
    – Digiproc
    Jul 26, 2016 at 2:36
  • $\begingroup$ Classically speaking, black holes do not evaporate. Hawking radiation is a quantum mechanical process. $\endgroup$
    – PM 2Ring
    Mar 28, 2020 at 5:52

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