2

There are many questions on this forum about objects falling into black holes. Most of them describe that to distant observers, the objects never quite reach the event horizons and appear to be frozen just outside. For example: How can anything ever fall into a black hole as seen from an outside observer?

Answers often refer to the Schwarzchild metric, even though that describes a static situation which isn't strictly applicable to in-falling objects. We know from studies of black-hole mergers (Revisiting Event Horizon Finders ) that in more realistic scenarios, event horizons can start growing before objects actually cross them.

This seems to suggest (perhaps naively) that outside observers could possibly see objects disappear behind event horizons in a finite time, since the event horizon may be larger than previously assumed.

That view might be countered by noting that the gravitational time dilation would also be changing along with the event horizon, with the result that the disappearance would still take an infinite time.

I don't expect that such a complicated situation could have a definitive answer without resorting to numerical relativity calculations, but would like to know what conclusions have been reached about the effects of event horizon growth.

It has been suggested in comments, that this question is a possible duplicate of Does an expanding event horizon “swallow” nearby objects? However, that question involves the merger of two black holes, and therefore has two event horizons, while this question is for the somewhat simpler case of aa object falling into a single black hole. I'm not sure if the answers would be different, but I believe this is still a separate question.

D. Halsey
  • 2,143
  • 4
    Possible duplicate of Does an expanding event horizon “swallow” nearby objects? – John Rennie Apr 16 '19 at 19:06
  • Actually you might have a point: I think the definition of horizon (null closed hypersurface) does not imply that the time-time component of the metric would be zero in a non static case. I will think about it. First I will check the possible duplicate of the question posted above. – AoZora Apr 16 '19 at 19:47
  • "event horizons can start growing before objects actually cross them" - Correct. "This seems to suggest (perhaps naively) that outside observers could possibly see objects disappear behind event horizons in a finite time" - Incorrect. Everything "frozen" at the horizon moves out with it when the horizon expands. This is self evident, if you consider a collision of a static rock with a fast flying black hole. The rock will be pushed by the moving horizon and will move with the horizon. – safesphere Apr 17 '19 at 03:58
  • @John Rennie: I had forgotten about that question & might not have posted this one if I had remembered it. Reading it again however, I think I would call it a related question, not an exact duplicate. It's actually a comical situation. According to a comment from the OP, his question involved merging black holes, but you wrote your answer assuming it was about an object falling into a black hole (like this question). Then, even though he didn't seem to agree with your answer, he accepted it anyway. – D. Halsey Apr 17 '19 at 14:54
  • @D.Halsey yes it was a bit of a mix up :-) – John Rennie Apr 17 '19 at 14:56
  • @John Rennie: Given some of the comments in that case, would you still answer the question the same way now? – D. Halsey Apr 17 '19 at 15:01
  • Possible duplicate of Does an expanding event horizon "swallow" nearby objects? – Jon Custer Apr 17 '19 at 15:08
  • @D.Halsey yes I would. While the objections made have a point, if you simply asking whether anything crosses the horizon in the external observer's coordinates the answer is still no. – John Rennie Apr 17 '19 at 15:19

1 Answers1

1

In relativity one needs to make a distinction between “an outside event happening at a time $t$ by the clock of an observer $\beta$” and “an observer $\beta$ sees an event happening at time $t$ by her clock”. The first situation means that event happens within the slice of a constant time $t$ in the observer's frame of reference, while the second implies that there is a null geodesics connecting the event with the observer at time $t$.

With that in mind,

  • By taking the backreaction into consideration (in this situation it means that the horizon would be growing) an object would cross the event horizon in finite time by the clock of an outside observer (with any reasonable scheme of defining slices of constant outside time in the vicinity of black hole, such as Fermi–Walker frames)

  • Signals propagating with the speed of light from the event of “object crosses the event horizon” would by definition never reach an outside observer. Signals from events immediately preceding would reach the outside observer arbitrarily late and extremely redshifted, so in practice an observer would only see an object outside the event horizon for some time until the signals from it get redshifted beyond any possibility of detection.

A.V.S.
  • 15,687
  • 2
  • 17
  • 42
  • I was under the impression that it was only in SR that observations of events were corrected for the travel-time of light between the event & the observer. Are you saying that is also appropriate in GR? – D. Halsey Apr 17 '19 at 14:21
  • @D.Halsey: Special relativity is a special case of general relativity (flat spacetime plus restricted choice of coordinates). So relativity of simultaneity is not only present in GR but much enhanced since causal structure could be much more complex. – A.V.S. Apr 17 '19 at 15:39
  • @ A.V.S. :OK, that makes sense. But in this case, where the time dilation may be infinite, can you still make sense of any notions of simultaneity? – D. Halsey Apr 17 '19 at 15:44
  • @D.Halsey: That's where the growth of the event horizon comes in. If BH is growing, then EH crossing does not happen at the surface of infinite time dilation but strictly outside it. So simultaneity there could still have some ambiguity but there are no infinities. In more technical terms, apparent horizon is the surface of infinite dilation of “outside time”, while for a black hole that is gaining mass, event horizon would be strictly outside it. – A.V.S. Apr 17 '19 at 16:00
  • @ A.V.S. : +1 That's the kind of thing I was hoping to learn. – D. Halsey Apr 17 '19 at 16:10
  • Are you sure that "an object would cross the event horizon in finite time by the clock of an outside observer"? This makes no intuitive sense. An expanding horizon is locally equivalent to a moving horizon. Consider a flying black hole colliding with a static rock. The moving horizon will push the rock, but never swallow it in the view of an outside observer. Similarly, an expanding horizon will simply push an object out, so the object would not cross the horizon in a finite outside time. What is the flaw in this reasoning? – safesphere Apr 18 '19 at 08:19
  • @safesphere: I am not at all sure what insight boosting the black hole would bring. Boosted black hole has a frame-dragging effect and an ergosphere (even when BH is non-spinning). So the interpretation of physics in terms of constant time slices is different. And it is the metric that determines the motion of an object not the horizon, same for the “time of outside observers”. The difference is that position of an object $x^\mu(t)$ is a function of the past only, while the EH position is dependent on both past and future of the spacetime. – A.V.S. Apr 18 '19 at 19:23
  • @A.V.S. Understood, but I'm not really boosting a BH. The BH may be static in, say, the galactic frame like Sagittarius A*, but I am moving relative to the galaxy with a constant velocity matching the initial velocity of the rock. For a remote static observer the rock never crosses the horizon. I can easily sync my clocks with this observer, so for me the rock never crosses the horizon either, but in my frame the BH looks boosted. So boosting a BH does not change the fact that the horizon is impenetrable. A moving horizon pushes nearby objects away simply due to the relativity principle. No? – safesphere Apr 18 '19 at 20:34
  • @A.V.S. This question has been closed as "not about physics..." for some reason & I don't know if that prevents me from sending comments, but I'm going to try following up on your answer a bit more. – D. Halsey Apr 24 '19 at 16:27
  • @A.V.S. OK, the site seems to be allowing more comments, so here's my question: In SR, the Terrell effect (or Penrose-Terrell rotation) is an especially interesting example of the difference between what is observed (after accounting for the finite speed of light) and what is actually seen (in the eyes of the observer). Even though what is actually seen seems to contradict the reality of the Lorentz contraction, nobody interprets it that way. It is just dismissed as an optical effect... – D. Halsey Apr 24 '19 at 16:44
  • @A.V.S. (continued) In the case of the object falling into a black hole: Can we make an analogy to the Terrell effect and consider the perception that the object is almost frozen at the event horizon to also be just an optical effect? – D. Halsey Apr 24 '19 at 16:47
  • @D.Halsey: Can we … consider the perception … to also be just an optical effect? Not just the optical effect. The “freezing” of falling object relative to the time of outside observer is quite an objective phenomenon (gravitational time dilation), its just the return trip of a light signal also introduces time delay and it is this optical delay that gives infinite time from the moment of horizon crossing, while time dilation for that event remains large but finite. – A.V.S. Apr 24 '19 at 19:48
  • @A.V.S. You never replied to my comment above, but your statement that an object can cross a growing horizon in a finite outside time seems incorrect. Even with the backreaction, the Schwarzschild spacetime is globally hyperbolic with well defined causality. Therefore, if your statement were correct, then an outside observer would stop receiving signals (ignoring the redshift) from the falling object in a finite time. This is easily disproven by the diverging time dilation at the expanding horizon for the falling object. The horizon is impenetrable, because it stops the coordinate time. ... – safesphere Nov 01 '22 at 06:28
  • @A.V.S. (continued) It doesn't matter if the horizon moves, grows, or merges. And if your statement is true in some scheme (like Fermi–Walker), then such a scheme would be unreasonable, as it would contradict causality. – safesphere Nov 01 '22 at 06:28
  • @A.V.S. Please correct me if I am wrong, but your argument seems to apply to a static horizon as well - an outside observer can be on a spacelike slice that crosses the horizon. If so, then the backreaction condition in the first bullet of your answer is irrelevant and misleading. To say that an object can cross the Schwarzschild horizon along a spacelike slice is the same as to say that crossing the horizon requires moving faster than light, which is unphysical. Here is a physically meaningful protocol for a Schwarzschild black hole. Consider a hovering mirror. We send a signal at noon and … – safesphere Nov 01 '22 at 14:32
  • … receive a reflection at 2 pm. Due to symmetry, we can say that the reflection at the mirror happened at 1 pm on our clock. Similarly, if a falling particle hits a heavy hovering object and bounces back, it can return with no time limit. If a particle physically crossed the horizon on a clock of an external observer, then there would be a time limit, after which no signal from the particle can be received or the bounced particle can no longer return. This is not the case, so your spacelike time slicing does not physically represent the clock of an external observer. – safesphere Nov 01 '22 at 14:33
  • @safesphere: To say that an object can cross the Schwarzschild horizon along a spacelike slice There is no material “object” that moves along a spatial slice, this is more like a constraint equation, completely nondynamical and $t=t_0$ simply defines a spacelike hypersurface solving it for a specific boundary condition at a distance. But static event horizon would not work for this specific choice of time synchronization protocol, because a set of observers moving along the timelike Killing vector field stops at static horizon. – A.V.S. Nov 01 '22 at 17:50
  • Here is a physically meaningful protocol … Due to symmetry, we can say .But that is the problem: we can only penetrate the event horizon that is growing, which means that the situation is not symmetrical under time reversal. – A.V.S. Nov 01 '22 at 17:53
  • @A.V.S. I understand, but respectfully disagree that anything can physically cross a growing horizon on any external clock simply because of the diverging time dilation - a falling object can bounce off a hovering mass and return with no external time limit. So in the external physical sense the falling object is outside forever. Your symmetry objection doesn’t hold, because a growing horizon is locally equivalent to a boosted static horizon, which is impenetrable (see my earlier comment). The object is repelled by linear frame dragging. Thank you for your insight! I will do more research :) – safesphere Nov 01 '22 at 20:29
  • @A.V.S. You also recently claimed that the speed of a free falling object doesn’t have to approach the speed of light at the horizon. To make my argument clear, I have done the math. Please see my “EDIT” here: https://physics.stackexchange.com/questions/450032#524855 - it is always asymptotically the speed of light for hovering observers and consequently for all external observers. – safesphere Nov 01 '22 at 20:49