What happens when you fall in an evaporating black hole?

[Quantum Immortal]

Here's the best of my understanding of the official explanation of what happens when O' fall in a black hole.

eternal black hole

For an outside observer O, O' slow down more and more as you approach the event horizon, at the event horizon time seems to completely stop. In the reference frame O' of the one that falls in the black hole, nothing special seams to happen, you just cross the event horizon and reach the singularity at a finite time. It's just the usual whacked up relativity, like the twin paradox or the barn experiment or simultaneity. Simply here, the thing doesn't even know if its inside or out side the event horizon

thats fine for me.

evaporating black hole

Because the black hole evaporates, the reference frame O' is different from the one in the case of an eternal black hole.
O' falls in, at the time of evaporation of the black hole, according to O ?
O' sees the event horizon receding as it falls towards the black hole, and eventually reach it when the black hole evaporates?
At the limit of reaching the event horizon, the event horizon isn't receding at the speed of light? thus never reaching it? O' doesn't just see the black hole evaporate before it ever crosses the event horizon?
Saying that you fall in at the time of evaporation, isn't a weird way of saying, that you don't fall in?

I don't get it :'(

 

[PeterDonis]

eternal black hole

This part is fine.

O' falls in, at the time of evaporation of the black hole, according to O ?

Yes. But note that this "time" according to O does not label a single event; it labels an infinite number of events, all of which take place on the hole's horizon before it evaporates, and all of which are distinct. A very large number of people could all fall into the hole separately, and all of them would fall in at the time of the hole's evaporation according to O.

O' sees the event horizon receding as it falls towards the black hole

No. Why do you think this? According to O', the EH is moving outward at the speed of light; it is not receding. (That's true for both the eternal case and the evaporating case. It's true that in the evaporating case, the area of the horizon is slowly decreasing; but that doesn't make any significant difference to what O' sees as he falls through, because a hole of any significant size evaporates *much* more slowly than O' falls in.)

and eventually reach it when the black hole evaporates?

No; according to O', he falls through the horizon before the hole evaporates. (Very long before if the hole is of any significant size.)

At the limit of reaching the event horizon, the event horizon isn't receding at the speed of light?

No, it's moving *outward* at the speed of light, not inward. See above.

 

[Quantum Immortal]

Wait, you are basically saying, that for O', things are basicaly the same in both cases?

Why, does O' crosses the EH at the time of disintegration? What changes relative to all the previous time O' was falling? I don't get it, but i assume its true bellow.
Time is supposed to be frozen at the EH, according to O, i think that's the point that i disrupts my understanding.

According to O.
O' is going slower and slower, seen the black hole evaporating faster and faster?
At some point O' is so slow that it sees (according to O) the event horizon receding faster and faster as it approaches it. Eventually, O sees O' reaching the EH at the time of disintegration.
Why O sees O' reaching the EH? The EH still has extream time dilation.

put it differently
according to O, the BH evaporates at some rate.
according to O', the BH also evaporates, but at some different rate.
according to O, O' goes slower and slower, so according to O, O' should see the rate of evaporation of the BH becoming faster and faster. So fast, that for O, O' reaches the EH at the time of disintegration of the black hole.

put it an other way still.
How you define simultaneity here? According to O and O' the black hole evaporates, so in theory they could deduce what's the time for the other by seen the evaporation rate of the BH. In theory, O' could deduce at what time for O, it crossed the EH, by looking how fast the BH evaporates at that instant.

 

[PeterDonis]

Wait, you are basically saying, that for O', things are basicaly the same in both cases?

Yes. When O' falls through the horizon, he has no way of telling whether the hole he's falling into is going to evaporate or not, just from his local observations.

Why, does O' crosses the EH at the time of disintegration?

Time is relative; time according to whom?

Time is supposed to be frozen at the EH, according to O

Yes, but time is relative; time according to O is different from time according to O'. For one thing, as I pointed out before, the "time of evaporation" according to O labels an infinite number of events, including the event where O' falls through the horizon; but according to O', the time at which he falls through the horizon is much earlier than the time the hole evaporates (in fact, as I noted above, O' can't even tell when he falls through the horizon that the hole is going to evaporate, or if so when, just from his local observations). See further comments below.

According to O.
O' is going slower and slower

Yes.

seen the black hole evaporating faster and faster?

O sees the BH evaporating faster and faster, and the light he sees from this is intermingled with the light he sees from O' falling towards the horizon. But O' does not see this.

At some point O' is so slow that it sees (according to O) the event horizon receding faster and faster as it approaches it.

No. O sees light rays from the evaporating hole and O' falling in all mingled together, but he cannot conclude from that that O' sees the horizon receding. He can only conclude that the coordinates he is using are highly distorted near the horizon.

Eventually, O sees O' reaching the EH at the time of disintegration.
Why O sees O' reaching the EH? The EH still has extream time dilation.

Not once the hole has evaporated; that's the point. Once the hole has evaporated, there is no EH, and all the light that was trapped at the EH can now escape, including the light that contains the image of O' crossing the horizon (while there still was a horizon).

according to O, the BH evaporates at some rate.

Yes.

according to O', the BH also evaporates, but at some different rate.

Not according to O's local observations; he can't tell whether he is falling into an evaporating hole or an eternal hole just from his local observations.

according to O, O' goes slower and slower

Yes, but this is an effect of the distortion of O's coordinates near the horizon, so this...

so according to O, O' should see the rate of evaporation of the BH becoming faster and faster.

...is not a valid deduction.

How you define simultaneity here?

It depends on the coordinates you want to adopt; simultaneity is a coordinate convention. The "natural" sense of simultaneity for O to adopt is highly distorted close to the horizon, to the point of infinite distortion at the horizon, as I said before: that's why, if many different people all fall through the horizon separately, O assigns the same "time" (i.e., the same surface of simultaneity) to *all* of them falling through the horizon.

But O's "natural" sense of simultaneity is different: according to his sense of simultaneity, if many different people all fall through the horizon, they will all fall in at different times--some could fall in earlier than him, some later. As I noted above, O' can't tell from his local observations whether the hole is evaporating or not, but if he does know, by some other means, that the hole is evaporating, the time he assigns to the hole's final disintegration, according to his "natural" sense of simultaneity, will be later than the time that anyone falls through the horizon.

This is just an extreme case of the general observation that simultaneity is relative. That's why simultaneity is, in general, not a good thing to focus on in relativity problems.

According to O and O' the black hole evaporates

No, only according to O. O' can't tell whether the hole is evaporating or not. See above.

 

[PeterDonis]

I'm having a similar discussion in another thread, and I posted a link there to a spacetime diagram that may help to illustrate what I've been saying. The post is here (with some explanation of the diagram after the link to it):

https://www.physicsforums.com/showpost.php?p=4617010&postcount=31