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Free falling into a black hole that evaporates by Hawking Radiation

  1. Apr 28, 2013 #1
    The solution of Einstein's field equations for a simple black hole show a slowing of time as you get close to the black hole. Time stops at the event horizon. An observer in flat spacetime far from the hole would see an astronaut fall slower and slower as he approaches the event horizon. It would take an infinite amount of time (as seen by the outside observer) for him to reach it. But a black hole will completely evaporate by Hawking radiation in a very long but finite time. So the observer will see that the astronaut never enters the black hole and he is in empty flat space again after the complete evaporation of the black hole trillions of years in the future. The astronaut could then go and shake hands with the outside observer.

    From the astronaut's point of view, he will be in a free fall for only a short time during which he will see time running extremely fast for the outside universe. He will free fall through the event horizon, not noticing anything strange. Once inside the black hole, he will free fall for a short finite time until he is torn apart be tidal forces and his atoms get crushed at the singularity.

    How can these two points of view be compatible? How can the astronaut be killed in the black hole but still survive, see the black hole evaporate, and meet the outside observer in the far future?
    What am I interpreting wrong? Have I made an error in describing either the astronaut's experiences or the outside observer's point of view?

    Jeff
     
    Last edited by a moderator: Apr 29, 2013
  2. jcsd
  3. Apr 29, 2013 #2

    Bill_K

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    Good question! I assume you're familiar with Kruskal coordinates? You've described two scenarios that both take place at t = ∞, but in reality (as depicted by the Kruskal diagram) they occur at different places.

    Each inward-going geodesic crosses the horizon at a different place, labeled by the "advanced" Kruskal coordinate v. One model for Hawking evaporation is that negative energy falls inward, crosses the horizon and gradually reduces the mass. As v increases, more and more of it has fallen in, and more and more of the hole has evaporated. When the astronaut crosses the horizon at v = v0 say, the hole has not yet completely evaporated. So he does get crushed!
     
  4. Apr 29, 2013 #3

    Dale

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    They are not compatible. They describe two different solutions/scenarios, not the same solution/scenario from different points of view.
     
  5. Apr 29, 2013 #4
    There is no crossing of the schwarzschild radius. Impossible for a simple black hole. A black hole is just a star which clocks almost freeze from our frame of reference. Once you go down the gravitational well of the compressed star its clocks become faster again (compared to the falling observers frame of reference). So it becomes all bright and shiny again. What we observe as low wavelength hawking radiation would be cooking hot stuff once you own clock (in relation to our outside frame of reference) goes slower and slower. It would certainly evaporate you long before it evaporates itself.

    ht_cat
     
  6. Apr 30, 2013 #5
    I agree with Bill K in that the Schwarzschild solution for a black hole metric indicates that a free falling astronaut will cross the event horizon at R = 2GM/C^2 in a short proper time and hit the singularity a short proper time later.
    But an outside observer in flat spacetime will see the astronaut almost freeze near the event horizon, taking an infinite amount of time to reach it. He will also see the black hole evaporate in a long but finite time. (Kruskal diagrams do not seem to take Hawking evaporation into account.)
    My question is still "What will the outside observer see happen to the astronaut after the hole completely evaporates?"
    If he sees him still alive, he could ask him how it felt to be crushed at the singularity!
    If he sees that he is dead, then what killed him? (Assume his ship can survive any radiation that is not infinite.) He observed him near but not past the event horizon the whole time!
     
  7. Apr 30, 2013 #6

    pervect

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    Someone posted a link to a couple of papers on this in another thread, but I haven't been able to find the thread or the papers. This question comes up about 3-5 times a day, or it seems like it.

    While I can't find the references, I can tell you what I remember. Maybe the origial poster will repost the references.

    There was a paper by Krauss, I htink, claiming that the black hole evaporated before it formed, and a paper published in rebuttal that said that the black hole DID form, though the calculation was "non-trivial".

    Various FAQ's also claim that the black hole does not evaporate before you fall in, though they don't offer details (and in light of the apparent non-triviality of the claculations, I have to wonder some).

    As far as what you'll see, there are two possibilities. If the spaceship disappears behind the absolute horizon, you'll never see it again, even if you later see the hole does evaporate.

    If the spaceship doesn't disappear behind an absolute horizon, you'll eventually see it again.

    Which of these two possiblities actually happens depends on the details of the calculation.
     
  8. Apr 30, 2013 #7

    Bill_K

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    I'll say it again. :smile: The apparent paradox arises because the Schwarzschild time t is a bad coordinate on the horizon. The outside observer sees the horizon as r = 2M, t = ∞, and thinks that everything that happens at "large times", t = ∞, is simultaneous. In particular he thinks the astronaut's demise and the hole's evaporation are simultaneous. They are not.

    In fact the horizon is a null surface, and although all of it is labeled t = ∞, a more suitable time coordinate to use on it is the advanced Kruskal coordinate v. On the horizon v varies from -∞ to +∞ and describes a time sequence: if for two events on the horizon, v0 < v1, then the first event happens before the second one in every sense, e.g. it can causally influence the second.

    The astronaut reaches the horizon at v = v0 and proceeds onward into the singularity and is demolished. Hawking radiation is a continuous thing: it gradually decreases the mass of the hole, and especially this will go on happening at all v > v0. The hole eventually evaporates, but long after the astronaut is dead. By the time it is gone, the astronaut has long since perished.
     
  9. Feb 13, 2014 #8

    tom.stoer

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    Is it possible to make argument rigorous?

    For the collapse of a sphere of dust we have exact equations. But what about the evaporation phase? How does the spacetime of an evaporating black hole look like approximately? Can one draw Kruskal or Penrose diagrams? Including world lines of different observers?
     
  10. Feb 13, 2014 #9

    pervect

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    Yes.
    I found the paper I mentioned before:

    "Radiation from collapsing shells, semiclassical backreaction and black hole formation" http://arxiv.org/abs/0906.1768

    Compare and contrast to
    T Vachaspati, D Stojkovic and L M Krauss, Phys. Rev. D76, 024005 (2007) [arXiv:gr-qc/0609024];

    So there are published conflicting views in the field at the moment. My personal comments:

    1) Paranjape and Padmanabahn are aware of Krauss' et al work - they even cite it

    2) P&P cite over a dozen papers that they claim support the view (that they describe as a consensus view) that black holes don't evaporate before you fall in.

    3) I haven't seen a published response by Krauss, et al, to P&P's comments. (but it doesn't mean it doesn't exist, I don't have stellar search tools).

    I don't yet feel I understand either paper well enough to comment on the technical parts directly, but I am willing to say that while there are conflicting papers published in the field, I think the consensus view is evaporation of the black hole does not prevent an event horizon from forming.

    I think there may be more papers on this since I wrote my blog entry, I believe there's one by Hawking.
     
  11. Feb 13, 2014 #10

    PeterDonis

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  12. Feb 13, 2014 #11

    PeterDonis

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    The Penrose diagram on this Wikipedia page gives what I understand to be a representation of Hawking's original model (where the BH evaporates, but there is still an event horizon and a black hole region where worldlines end in a singularity):

    http://en.wikipedia.org/wiki/Black_hole_information_paradox

    The big issue with this diagram, of course, is that it indicates information loss--any information carried into the singularity disappears, violating unitarity.

    In previous threads on this topic (including, I think, the one I linked to in my previous post), there's a link to a paper by Padmanabhan (IIRC) which gives Penrose or Penrose-type diagrams for the various possibilities being batted around now for addressing the information loss issue.
     
  13. Feb 13, 2014 #12

    PeterDonis

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    No, he didn't--not if the black hole evaporates. (At least, not if the black hole evaporates but there is still an event horizon and a singularity inside it, as in the Penrose diagram in the Wikipedia page I linked to in my last post.) If the BH evaporates, that changes what the distant observer sees (as compared to the case where the BH is eternal and never evaporates); he now sees the astronaut reach the horizon. The light from the astronaut reaching the horizon gets to the distant observer at the same instant as the light from the final evaporation of the BH; the distant observer basically receives a flash of light containing images of every event that happened on the horizon.

    The distant observer still can't see anything that happened inside the horizon; but he can see that the infalling astronaut fell through it.
     
  14. Feb 17, 2014 #13
    I agree! I previously overlooked the fact that the evaporation of the black hole will indeed change what the distant observer sees. He will see the astronaut reaching the event horizon, reaching the singularity, and the BH final evaporation as all simultaneous.

    Will the falling astronaut observe all 3 events as being simultaneous in his frame of reference too? I believe that he must. His last message an instant before hitting the singularity will be received by the observer, so he could not be inside the BH yet.

    If this is true, then he does not have a few minutes inside the event horizon before hitting the singularity, as he would with an eternal BH. He reaches the event horizon and the singularity simultaneously. In fact, no matter could hit the singularity, except at the last instant of final evaporation. So can an actual singularity really form and exist for a finite time? It seems not! A singularity can only form at the instant of final evaporation (quantum gravity laws would be needed to understand what happens at this instant). Can anything ever enter a BH and exist inside it for a finite time? Again, it seems not!
     
  15. Feb 17, 2014 #14

    PeterDonis

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    Not quite. He won't see the astronaut reaching the singularity. In fact he won't see any event that happened inside the event horizon. No light from any of those events will ever make it out to a distant observer.

    No. To him, he will cross the event horizon, then hit the singularity and be destroyed. He will never see the black hole evaporate, because he falls in and gets destroyed in the singularity before the hole evaporates away.

    No, it won't. See above.

    No, he doesn't. See above.

    Yes. See above.
     
  16. Feb 17, 2014 #15
    Very Interesting discussion. I want to point out a statement in the OP that seems incorrect to me.

    I don't think that is correct. The astronaut does not see the outside universe running extremely fast. The amount of information in his (her) past cone is quite small. The astronaut doesn't see anything particularly exciting as (s)he crosses the horizon.
     
  17. Feb 17, 2014 #16

    PeterDonis

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    Yes, good point. An astronaut free-falling through the horizon will actually see light from the outside universe redshifted, i.e., events whose images come to him via these light signals will seem to be "in slow motion", not speeded up.
     
  18. Feb 17, 2014 #17

    George Jones

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    Actually, if the astronaut started falling from a great distance, the astronaut will see light red-shifted, i.e,, the distant outside universe will appear to run slow for the astronaut.

    Suppose that observer A hovers at a great distance from a black hole, and that observer B hovers very close to the event horizon. The light that B receives from A is tremendously blueshifted. Now suppose that observer C falls freely from a great distance. C whizzes by B with great speed, and, just past B, light sent from B to C is tremendously Doppler redshifted. What about light from A to C? The gravitation blueshift from A to B is less that the Doppler redshift from B to C. As C crosses the event horizon, C sees light from distant stars redshifted, not blueshifted.

    If observer A, who hovers at great distance from the black hole, radially emits light of wavelength [itex]\lambda[/itex], then observer C, who falls from rest freely and radially from A, receives light that has wavelength

    [tex]\lambda' = \lambda \left( 1+\sqrt{\frac{2M}{R}}\right).[/tex]
    The event horizon is at [itex]R = 2M[/itex], and the formula is valid for all [itex]R[/itex], i.e., for [itex]0 < R < \infty[/itex]. In particular, it is valid outside, at, and inside the event horizon.

    See posts 5 and 7 in

    https://www.physicsforums.com/showthread.php?p=861282#post861282

    I have since done the calculations using Painleve-Gullstrand coordinates that are vaild even on the event horizon.
     
  19. Feb 20, 2014 #18

    tom.stoer

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  20. Feb 20, 2014 #19

    PAllen

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    My first reaction is that that paper is likely nonsense because of a number of elemantary errors:

    1) if there is radiation, the exterior is not vaccuum, and Birkhoff does not apply.

    2) Going back to the 1990s at least, it was recognized that the distant exterior of an evaporating BH must be classically similar to a Vaidya metric, not an SC metric. This paper never mentions or shows any awareness of this.

    Those are just the things I noticed in less than a minute.
     
  21. Feb 20, 2014 #20

    PAllen

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    Also, on p. 8, Yi Sun lists a bunch of conditions. One of them, (energy density non-negative everywhers) is simply false for an evaporating BH. This is the same mistake this author makes in another paper of his discussed on these forums.

    I also note that this paper (and none of this author's papers on this topic) are even in arxiv, let alone peer reviewed journals.
     
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