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Through detection of gravitational waves we observe coalescences of black holes, black holes and neutron stars, or even neutron stars into black hole, although forming of a black hole should take infinitely long time for an outside observer.

Qmechanic
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JanG
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  • You need to be able to distinguish between coordinate time and proper time. Black holes merge in finite proper time. Also, this question has been asked a few times. Try here and here and links therein. – joseph h Nov 16 '22 at 20:10
  • i like the idea of black holes in a state of convalescence. they are just lying around in a hospital, waiting to recover sufficiently to be discharged. ;-) -NN – niels nielsen Nov 17 '22 at 01:08
  • @nielsnielsen good to have you here. I would not notice it, never trust automatic correction. – JanG Nov 17 '22 at 06:56
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    Does this answer your question? Colliding black holes – ProfRob Nov 17 '22 at 07:37
  • Almost, Rennie writes "And the final merged object isn't really a black hole either, for the same reason". Does it means that there are still two event horizons and hence the resulting merger spacetime is not spherically symmetric? – JanG Nov 17 '22 at 12:09
  • @josephh with proper time you mean proper time in frame of infalling object which is finite of course. However, the duration of gravitational impulse we register in our proper time is finite, too. Why does it not suffer the same fate as radiation from event horizon? Only explanation I can see is that the source of black hole merger gravitation waves is not event horizon but its surrounding. – JanG Nov 17 '22 at 12:23
  • Why did you accept a wrong answer? How long the ringdown lasts has nothing to do with the fact that two horizons do merge into one in a finite time by a remote clock. Also timeslices do not represent simultaneity, so indeed a gravitational collapse never ends by the clock of a remote observer based on causality in the globally hyperbolic Schwarzchild spacetime. – safesphere Nov 17 '22 at 19:55
  • Because nobody else proposed an answer and they close my question. I thought to owe "accepted" to the only person who cared to give an answer. – JanG Nov 17 '22 at 20:11

1 Answers1

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Technically, the final phase of a compact binary coalescence, the ringdown, takes infinitely long. The ringdown is an exponentially decaying process, and a well known property of a decay exponential is that it never truly vanishes. More practically, the exponential decay is so quick that within a fraction of second it has become zero for all intents and purposes.

On a more general level I point out that the statement that "forming of a black hole should take infinitely long time for an outside observer" betrays a fundamental misunderstanding of one of the key features of general relativity. In general relativity does not feature a unique notion of global time, and unlike special relativity it isn't even possible to assign a unique global time to the local frame of an individual observer. The more correct physical statement is that the formation of an event horizon does not happen in the causal past of any outside observer. (The astute reader will note that this statement is actually tautology.)

The past light cone of an observer, however, is not what we would usually call "now". E.g. A supernova observed millions of light years away, will typically also be referred to as having happened millions of years ago. The possible choices of a "now" in general relativity are so-called spacelike hypersurfaces. For any observer outside of a black hole there will be many choice of hypersurface that will in fact intersect the event horizon. For these notions of "now" the black hole will have formed in a finite amount of time.

TimRias
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  • Since, in your model, there is no observation that can distinguish between "now" and "never", the distinction is not physical, but merely mathematical. – John Doty Nov 16 '22 at 21:29
  • @JohnDoty "Now" is never observable (unless it is also "here"). We can only ever observe the past. We have no way to observe that the Andromeda galaxy exists now, we can only observe that it existed 150000 years ago. It is however a reasonable inference that it still exists now. – TimRias Nov 16 '22 at 22:29
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    @JohnDoty However, there is an important physical distinction between something never happening in ones causal past versus something never happening. Something never happening also implies that this thing will never happen in ones causal future. One can decide to jump into a black hole and we have a physical prediction of what would happen. This makes statements about this physics and not "mere" mathematics. – TimRias Nov 16 '22 at 22:38
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    If you cannot publish the results of your experiment, it isn't physics. – John Doty Nov 16 '22 at 22:49
  • The difference with the Andromeda galaxy is an ongoing program of observations can verify its continued existence. – John Doty Nov 16 '22 at 22:52
  • @JohnDoty That is a limited view of physics. To illustrate its pointlessness consider this: We have a physical prediction of what would happen in a nuclear apocalypse. (Hopefully), this prediction is enough to ensure that this nuclear apocalypse never happens. If it were however to occur nobody would be around to publish the results of this "experiment". Does this make this prediction less physical? – TimRias Nov 16 '22 at 23:07
  • But your nuclear experiment is publishable, in principle, and perhaps in a few years, in practice. Do the experiment on Earth, publish on Mars. – John Doty Nov 17 '22 at 01:06
  • @JohnDoty The experiment of going into a black hole is “in principle” publishable. Throw the entire earth into a big enough black hole and we all get to live our entire lives ( modulo lack of sunlight) before being torn apart by tidal forces. – TimRias Nov 17 '22 at 07:00
  • @TimRias "The ringdown is an exponentially decaying process" answers my question in the comment (sixth comment after question) I have made to joseph h one. The gravitational wave impulse we observe is from time before the infalling object (second black hole for example) has reached event horizon, or both event horizons have touched themselves. Is that correct? – JanG Nov 17 '22 at 16:52
  • @TimRias "we all get to live our entire lives" - You don't know this. It is your prediction that you want to prove by the experiment. However, if your proper time ends at the horizon, you wouldn't be able to confirm the fact of your own death even to those falling together with you. Your logic here is the same as in life after death. Perhaps when you die you will meet your great grandma again. Or not. Either way it is unfalsifiable and for this reason is not science, but belief. – safesphere Nov 17 '22 at 19:31
  • @safesphere By that logic the statement “the universe will continue to exist tomorrow” is unfalsifiable, and therefore unscientific ( either it is true or it’s not and no one will be able to observe it. ) – TimRias Nov 17 '22 at 19:43
  • @TimRias Science is based on axioms, which in turn are based on undefined concepts. Geometry does not define what "point" or "line" is. Their meaning is based on our practical intuition and trust that it leads to meaningful results. In physics, the concept of "now" has no scientific definition, but is subjective also based on our intuition and our rust that normally time continues, so there will be tomorrow. So yes, tomorrow is unfalsifiable, but on the same level as science itself. Now, a black hole is very different where the coordinate time does stop at the horizon, so not a valid parallel. – safesphere Nov 17 '22 at 20:29