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There have been some lengthy discussions, here and here (among others), about whether, to the distant observer, black holes form in a finite time or an object falls into a black hole in a finite time. To the extent that readers understood and addressed the OPs' questions, it would seem that things can collapse to a point that it's very much like a black hole, a "collapsar", and can be treated like one and astronomers talk about it as if it were a black hole. But still stuck outside the event horizon.

But nobody seems to have explored some of the implications of that. The free-falling observer falls through the event horizon and hits the singularity in a finite amount of proper time. The distant observer never actually "sees" (or could infer by measurement) that the free-faller reached the event horizon, but eventually will see the stars run out of fuel and wink out as the universe continues to expand, and the black holes evaporate by Hawking radiation and disappear, and all of that before, in the external flat-space reference frame, the free-faller crosses the event horizon.

At first glance, maybe we can't just assume that, from the free-faller's perspective, the event horizon is going to statically exist long enough to reach it. But GR is a classical theory that exists without that. So if, in distant coordinates, it takes an infinite amount of time for the free-faller to go through the event horizon, how long does it take for the free-faller to reach the singularity? More than infinite?

It seems like a lot has been said about what's going on inside the event horizon that might just be null and void. Would quantum gravity explain what happens at the singularity? Irrelevant, because there is no singularity. Is information lost when it crosses the event horizon? Irrelevant, because it doesn't cross the event horizon--we might not be able to get it back, but it's still stuck outside. What does it actually mean for time to become spacelike and the spatial dimensions to become timelike? Maybe it literally is nonsense, like the advanced-wave solutions in EM which are mathematically valid but physically meaningless. Maybe it's a clue.

Understand that I'm not questioning the validity of GR, just the popular understanding of it vis-a-vis black holes and event horizons.

To make this a specific question rather than a call for discussion, is it just physically meaningless to talk about what happens "inside" of a black hole?

I suspect this will be an unpopular perspective.

Qmechanic
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Greg
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    Is this a question or just a misunderstood rant? "Stopping on horizon" is nonsense. It's not a matter of perspective, but facts. Time flows slower for infalling observer, but it only makes him fall faster from his pov. – Mithoron Feb 28 '20 at 23:59
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    What does it actually mean for time to become spacelike and the spatial dimensions to become timelike? My impression is that you do not understand the meaning of spacelike and timelike intervals and you may need to look more closely at the fundamentals of SR to get a firm hold on these concepts before looking at GR problems. See this Q&A. – StephenG - Help Ukraine Feb 29 '20 at 00:00
  • @Mithoron If it wasn't clear from the question, let's talk about the pov of the external observer. Of course it takes a finite amount of time for the free-faller, I said that. But is that a contradiction? – Greg Feb 29 '20 at 00:08
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    @Greg that is true for some metrics, but not all: https://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates – m4r35n357 Feb 29 '20 at 10:05
  • @m4r35n357 I know there is a coordinate singularity that is overcome by a change in coordinates for an in-falling observer. I haven't studied those. But considering coordinate time in terms of the new coordinates, it doesn't look easy for the external observer. – Greg Feb 29 '20 at 16:36
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    @Greg see my comment on the answer below. Any infinities in the maths are due to the particular form of the metric used, and I maintain that a stationary metric says nothing about accumulation of mass or its effects over time. It is simply too mathematically over-simplified to adequately describe the physical situation. Same for the metric I pointed you to ;) – m4r35n357 Mar 01 '20 at 10:38

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It seems that nobody can answer your question. Here are some elements which might help for the answer. Many open questions are left, but you may decide yourself if some of the following elements may be interesting for your reflection:

  1. The infalling observer reaches the event horizon of the Schwarzschild metric not earlier than at the "end of time" of the universe. That means, if from his point of view he crosses the event horizon within finite time, all outside observers will agree that he will cross the event horizon "never" (= at the end of our time).

  2. Intuitively, "the end of time" seems to be no well specified time indication, but it might be interesting to consider the end of time as a moment which is quite precise (at least from the point of view of the reference frame of the infalling observer). If so, we could say that all infalling observers are crossing the event horizon exactly at the same moment, and that the existence beyond the event horizons (from the point of view of all infalling observers) begins after our time.

  3. A consequence of the fact that the infalling observer is reaching the event horizon only at the end of time might be that all information of the infalling observer is remaining outside of the event horizon, and that there is no infalling information (the infalling observer is "stripping off" all information before entering beyond the event horizon).

  4. Another question which is important is the one if the entropy of infalling particles is zero at the event horizon. This question arises also in models of the Big Crunch.

Moonraker
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  • The Swartzschild metric simply does not account for infalling matter. That is the real reason behind all of these duplicate questions. The question cannot be answered with a stationary metric. Simple as that. – m4r35n357 Feb 29 '20 at 10:08
  • How would one "strip off all information before entering beyond the event horizon", do you really think that at coordinate time infinity+1 you have some esoteric bodyless information hovering above the event horizon? By the same logic I would have to strip off all my information when I am behind someone else's cosmic event horizon, since it also takes an infinite amount of coordinate time to cross it when using standard De Sitter coordinates. 1&2 are correct though, but the gish galopp at 3 messes it all up – Yukterez Feb 29 '20 at 11:35
  • Thank you, Moonraker. I wasn't sure if there was a coordinate transformation that fixes it, but those seem to specifically concern the free-faller, not trying to relate to the external observer. An answer that begins like "Yeah, but the way they look at it is..." would have been acceptable. It just seems funny that we can tell a physically reasonable story about the free-faller that, well, seems to take more than infinite external time. I might have found a better way to express myself in the original post. – Greg Feb 29 '20 at 16:58
  • @m4r35n357 “The Swartzschild metric simply does not account for infalling matter.” - A typical objection is perturbation. A tiny spec of dust falling to a supermassive Schwarzschild black hole cannot measurably change the global spacetime. So the logic is that the solution in this case must be only very slightly changed (perturbed), but still essentially Schwarzschild. Thus your statement is insufficient to rule out this solution. A stronger statement is needed, such as that the Schwarzschild solution inside the horizon is unphysical. – safesphere Mar 01 '20 at 18:50
  • @safesphere My viewpoint (such as it is) is that the "distance travelled" to the horizon would be shortened as the horizon must expand (in global coordinates) before that infinite time has elapsed. Please correct me if this is wrong. I am not familiar with the perturbation approach, so I should leave that with you! – m4r35n357 Mar 01 '20 at 22:32
  • @m4r35n357 When an object $m$ approaches a black hole $M$, its horizon expands toward the object until they meet at $r_s=2(M+m)$ (not at $r_s=2M$).The coordinate time of the falling object stops at the horizon $r_s=2(M+m)$. So, despite the expansion of the horizon, we "see" the falling object frozen there forever. – safesphere Mar 02 '20 at 04:18
  • In which case I stand corrected. – m4r35n357 Mar 02 '20 at 10:02