2
$\begingroup$

In the RI talk https://www.youtube.com/watch?v=_8bhtEgB8Mo at 1:00:30 a member of the audience asks about the race between the big crunch and the rate at which someone falls into a black hole. The argument is that since it take Alice an infinite amount of time from Bob's point of view to fall beyond the event horizon, then if the big crunch comes at a finite time, the crunch will happen first, and Alice will never experience falling into the black hole. Sean Carroll's answer appeared to me to miss the point entirely - as I heard it, the core of his answer was "Alice's point of view is valid too". That does not seem to provide justification for anything. The question is - what exactly is Alice's point of view.

I am looking for a more clear and justified answer to the question as asked on this video.

Note: I wonder whether the real answer requires a solution to the GR field equations that has both a black hole in it and a big crunch, in that the structure of the hole might change as the crunch arrives. Or is there a simple answer based on a bit of logic and the two separate solutions?


I have been effectively asked "what is the question". So ...

What is the experience of Bob and Alice in this scenario of a black hole and the big crunch?

Does Alice never reach the singularity in the black hole because something else gets her?

Or does it mean that Alice actually reaches the black hole singularity in a finite amount of Bob's time.

Or perhaps the singularity disappears?


This question has developed into a long interaction between myself and safesphere in the comments. The core of the response is that in the big crunch Alice will experience the big crunch coming to get her before she crosses the event horizon. Alice ends at the big crunch, as does Bob. For me that is the answer to my question. If safesphere does not put this into an answer then I suspect that I eventually will.

But, the discussion with safesphere has shifted into a tug of war over the Eddington Finkelstein (Penrose) coordinates. I want to see where this goes before I make a definite conclusion. I feel this should be opened as another question, but I want to think about it for a bit before doing that.

My day job sees me using distributions, non convergent series, and asymptotically convergent operators in practical engineering contexts, and I use non local operators in my studies of possible field theories. As such, I see nothing fundamentally non physical about partial coordinate transforms with boundary singularities. This is most likely the core of why I currently accept the point, but do not see it as denial of the interior solution. Alice knows what happens. But, I would like to know what more people think.

$\endgroup$
1
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Commented May 30, 2020 at 6:36

1 Answer 1

1
$\begingroup$

The far future evolution of the universe is irrelevant to what Alice experiences. If there's a clock orbiting outside the hole, Alice will still be able to see it from inside. The time she sees on it just before hitting the singularity will be just slightly later than the time she saw on it when crossing the horizon. For small black holes, the difference will be a few microseconds; for a huge hole like the M87 hole, it might be a day or so. It definitely won't be the billions of years it would take for the recollapse of the universe to become relevant.

Bob doesn't see Alice as such, but light that travels from Alice to his retina. Light from Alice when she's inside the hole never reaches Bob's retina, and light from Alice when she's outside but close to the horizon takes a long time to reach him. That's the only reason Alice appears to freeze at the horizon. It has nothing to do with what happens to Alice, just the light coming from her.

If the hole Alice fell into doesn't evaporate before the Big Crunch, then it will coalesce with other black holes, and their shared singularity is the Big Crunch singularity. In this scenario, since everything everywhere eventually hits the singularity, there is technically no "outside" and no event horizon. Bob may see a bit of Alice's history beyond what he incorrectly thought was the event horizon, very shortly before he hits the singularity. But this doesn't change Alice's experience, or Bob's experience long before the Big Crunch.

The discussion between you and safesphere that you mentioned in the question was moved to chat, and it appears that most comments from safesphere are missing. As far as I can tell from the one remaining comment, and some comments elsewhere, safesphere believes that Schwarzschild coordinates are the correct coordinates for Schwarzschild black holes, and other coordinate charts like Eddington-Finkelstein and Kruskal-Szekeres are somehow illegitimate. That's just wrong.

Schwarzschild coordinates are confusing because $r=2m$ does not map to any physically relevant part of the black hole geometry, even though the $0<r<2m$ and $r>2m$ regions both do. It is not the boundary between those two regions, even though there is a boundary between them (the event horizon). The reason worldlines that cross the horizon seem to diverge to infinity in these coordinates is that they have to go "around" the unphysical $r=2m$ line to get from the exterior to the interior. Eddington-Finkelstein coordinates don't have that problem: they cover the same interior and exterior regions as Schwarzschild coordinates, but in addition $r=2m$ is the boundary between those regions.

$\endgroup$
2
  • $\begingroup$ Thanks for responding to a stale question. I had given up on this and on stack exchange as a result of my earlier experience. You are definitely tackling the question and made some interesting suggestions. But, regarding your first line - you seem to say that for a big crunch Alice and Bob end up in the same singularity. But, if the universe goes on forever, it seems this would not be the case. Can you clarify your meaning? $\endgroup$ Commented Mar 21, 2021 at 10:55
  • $\begingroup$ On consideration, I found several severe problems with this answer. $\endgroup$ Commented Mar 23, 2021 at 1:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.