Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What happens when you fall in an evaporating black hole?

  1. Dec 30, 2013 #1
    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 :'(
     
    Last edited: Dec 30, 2013
  2. jcsd
  3. Dec 30, 2013 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    This part is fine.

    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.

    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.)

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

    No, it's moving *outward* at the speed of light, not inward. See above.
     
  4. Dec 31, 2013 #3
    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 thats 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.
     
  5. Dec 31, 2013 #4

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    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.

    Time is relative; time according to whom?

    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.

    Yes.

    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.

    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.

    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).

    Yes.

    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.

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

    ...is not a valid deduction.

    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.

    No, only according to O. O' can't tell whether the hole is evaporating or not. See above.
     
  6. Dec 31, 2013 #5

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook