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Why doesn't hawking radiation prevent a space-like singularity

  1. Aug 31, 2012 #1
    As I understand it, as a particle is observed approaching an event horizon, will would never be observed to cross the horizon, and it's red shift would tend towards infinity.

    I've read conflicting things about this, A, that that red shift WOULD NOT reach infinity in a finite time, and B, that red shift WOULD read infinity in a finite time. I'm assuming the WOULD NOT case here, but could somebody set this straight?

    Enter Hawking Radiation and black hole evaporation. When considering that it isn't possible to observe a particle crossing an event horizon, I would imagine that the process which creates Hawking radiation to prevent any particle from crossing the event horizon, and that the black hole would eventually be observed evaporating, before any particle could be observed crossing the horizon. This leads me to imagine if you where to fall into a black hole, at some point, your matter would be destroyed by virtual anti particles.

    Now, I can't see a way around this, however, considering that I haven't heard this put forward many times, but people often talk about space-like or time-like singularities, and even talk about being able to cross the location of where an out side observer sees the event horizon, with out noticing anything special, I am wondering where I am going wrong here?

    How can a space like singularity form inside a black hole if an outside observer sees no particle crossing the horizon?
     
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  3. Aug 31, 2012 #2

    phinds

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    My understanding is that is would not, but I'm not an expert on this.

    That would seem to be confusing far-away observations with local reality. The EH is irrelevant to what ACTUALLY happens to particles falling into the BH regardless of what it looks like to a far-away observer.
     
  4. Aug 31, 2012 #3
    An observer far away from the hole sees the redshift increasing rapidly, but it only approaches infinity after infinite (coordinate) time. However the proper time required to pass through the horizon (and even to hit the center) is finite -- an astronaut may jump into the black hole at 8 in the morning and he can hit the singularity before his watch shows 8:10.

    What makes you think the singularity HAS formed in the frame of the observer? Your common sense thinking of simultaneity betrays you here. The clocks tick completely differently (hell, they tick in a different direction!) inside and outside.
     
  5. Aug 31, 2012 #4
    I am aware that as you approach an even horizon that you will observe it receding.

    But if you can observe the path of a particle falling into a BH, until the point that the BH evaporates, then you would observe that he particle never crossed the horizon. Are you claiming that 2 observers can disagree on the path that the particle took?

    I just realized a way the particle might be observed moving close to the singularity. As the black hole evaporates, the radius of the event horizon might shrink down to close to infinitely small, during which time, you could observe a particle moving close to the singularity. I don't know if is actually how BH evaporation works though.
     
    Last edited: Aug 31, 2012
  6. Aug 31, 2012 #5

    phinds

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    OF COURSE. They not only can but DO disagree. Again, you are confusing far-away observation with local reality. Do you understand the concept of frames of reference?
     
  7. Aug 31, 2012 #6

    George Jones

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    According to the book Quantum Fields in Curved Space by Birrell and Davies, pages 268-269,
    This finite amount of radiation is negligible for observers freely falling into an astrophysical black hole.
     
  8. Aug 31, 2012 #7

    PeterDonis

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    If by "never be observed" you mean "never be observed by an observer who remains at a constant, very large distance from the hole", then yes, this is true for the case of an "eternal" black hole (one that never evaporates). As other posters have noted, the fact that the distant observer never sees the free-falling particle cross the horizon, does not mean it doesn't actually cross the horizon. It does, in a finite amount of proper time by the particle's own clock, and in a further finite amount of proper time, the particle hits the singularity at r = 0 and is destroyed; but the light rays from events on the particle's worldline at or beneath the horizon can't get back out to the distant observer.

    However, it's important to note that this is a *different* case from that of an evaporating BH. See below.

    If we're talking about an evaporating BH, that's no longer true. The key thing to remember is that, as I said above, a spacetime with an evaporating BH is a *different* spacetime from that of an "eternal" BH. The simplest way to put it is that an "eternal" BH spacetime is a purely classical spacetime, where the stress-energy tensor is literally zero everywhere--it's a "classical vacuum" spacetime. An evaporating BH spacetime is a "quantum vacuum" spacetime; the stress-energy tensor is *not* zero. Rather, the SET is that of the vacuum state of whatever quantum field is present. That changes the global properties of the spacetime.

    In an evaporating BH spacetime, the distant observer will see light rays emitted by objects as they fall past the horizon, at the same time as he sees the BH evaporate. So light emitted by objects falling past the horizon no longer takes a time approaching infinity to get out; the evaporation of the BH lets the light escape faster. Roughly speaking, that's because the evaporation decreases the mass of the BH, so it decreases the strength of its gravity, which is what "holds back" the light and slows it down.

    I should note that the above picture is our best understanding of how an evaporating BH works, but AFAIK nobody actually has an exact solution that describes this; the picture is put together from various approximations.

    This question is actually still valid even for an evaporating BH spacetime, because there is still a span of time where objects can fall past the event horizon, and any objects that do so will be destroyed in the singularity, even though the light they emit as they fall past the horizon will eventually get back out. (The question is valid for an "eternal" BH spacetime too, of course.) The short answer is that the infalling objects themselves "see" the region of spacetime inside the horizon, and that's sufficient. There are plenty of threads already in this forum that give longer answers. :wink:
     
  9. Aug 31, 2012 #8

    PeterDonis

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    No, you won't. You will observe it moving towards you at the speed of light.

    No; when you see the final evaporation of the BH, you also see light emitted by objects as they crossed the horizon. But you do not see light from *inside* the horizon; that light gets swallowed up in the singularity, just as the infalling objects do. See further comments below.

    Only if the distant observer insists on claiming that, because he can't *see* objects beneath the horizon, they can't exist there. The two observers will agree on every portion of the path that both of them can observe; and the distant observer will also be able to calculate that the proper time experienced by the infalling observer, for the portion of his path that the distant observer can see, is finite.

    It is, sort of. As the hole evaporates, the radius of the EH does shrink. However, that doesn't help in observing particles that fell through the horizon earlier. Take any radius that is inside the horizon at the point where some particular particle passes it; then that particle, and any light it emits at that radius, is guaranteed to hit the singularity before the BH's evaporation shrinks the horizon to that radius. Only if a given radius is exactly *at* the horizon when a particle passes it, will light emitted outward by that particle be seen by the distant observer when the BH evaporates.
     
  10. Aug 31, 2012 #9
    Just re-quoting my self, from a late edit to my last post, so it isn't missed.
    I think this explanation makes more sense to my naive thinking, but can anybody tell me if this actually fits with GR and evaporation BHs?

    I think it should be quite clear from what I have written that I at least know something about different frames of reference, however, can somebody else confirm this claim that 2 observers can disagree on the path a particle takes? It is very confusing in GR, but I think it more likely that everybody would agree on the path, just not the speed or time, or order.

    Thanks, I have been considering this paradox too, however, the resolution is a lot of jargon to me, but, at least it is sounds impressive;)
     
  11. Aug 31, 2012 #10

    bcrowell

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    A commonly posed paradox runs something like this. As infalling matter accretes onto a black hole such as Sagittarius A*, gravitational time dilation affects it more and more extremely as it approaches the event horizon. The time dilation approaches infinity as the matter approaches the horizon, so that it never actually passes through. We therefore conclude that the black hole could never have formed in the first place by gravitational collapse.

    The main mistake involved in this paradox is that it treats time dilation as something that happens to the infalling matter. Gravitational time dilation, like special-relativistic time dilation, is not a physical process but a difference between observers. When we say that there is infinite time dilation at the event horizon we don't mean that something dramatic happens there. Instead we mean that something dramatic *appears* to happen according to an observer infinitely far away. An observer in a spacesuit who falls through the event horizon doesn't experience anything special there, sees her own wristwatch continue to run normally, and does not take infinite time on her own clock to get to the horizon and pass on through. Once she passes through the horizon, she only takes a finite amount of clock time to reach the singularity and be annihilated. (In fact, this ending of observers' world-lines after a finite amount of their own clock time, called geodesic incompleteness, is a common way of *defining* the concept of a singularity.)

    We should also be careful about ascribing the characteristics of the Schwarzschild metric to an astrophysical black hole. The Schwarzschild metric describes a permanent black hole: one that has always existed and always will. By definition, it did not form by gravitational collapse. While an astrophysical black hole is forming, its structure is more complicated. For example, it has an event horizon (a surface from which light can't escape to infinity) which forms before the "apparent horizon" (a surface from which an outward-directed wavefront of light has decreasing volume).

    [EDIT] Deleted claim that Vaidya metric shows collapse in finite time. This is actually more subtle than I thought.
     
    Last edited: Aug 31, 2012
  12. Aug 31, 2012 #11

    bcrowell

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    It sounds like you're getting into a discussion of black hole complementarity, which AFAIK is still controversial, so although Birrell and Davies may believe that this is a satisfactory resolution, there may not be a consensus to that effect.

    In any case, I don't think the quantum features of the OP's question are essential. Quantum mechanics would be relevant if there were enough time during collapse for Hawking radiation to be significant -- the OP has in mind the idea that the time for collapse is infinite. We have exact solutions to the classical Einstein field equations for which formation takes finite time according to a distant observer. (By "formation" I mean the formation of an event horizon and a trapped null surface, which are the only events that can be timed by a distant observer.) This finite time is many orders of magnitude less than the time scale for black hole evaporation, so the quantum effects are completely irrelevant.
     
  13. Aug 31, 2012 #12
    Umm..... I'm going to assume you had some brain freeze there or miss understood completely. From your own reference frame, if you move towards an event horizon are you saying you will observe it moving towards you at the speed of light?

    Edit: perhaps there is some confusion here on my part about the difference between an Event Horizon and an Apparent Horizon, and by approaching, I meant moving towards, still not 100% sure where this speed of light thing came form though. I'm quite confused about the different between an event horizon, an apparent horizon, and the cosmic event horizon now. Its sounds like in BH talk, the event horizon usually refers to the absolute horizon, although, I guess this is one of the reasons so many people get confused... you would think that the default terms used in general relativity would be relative ones, not absolute ones.

    So, if you observe a particle sitting just before the Event Horizon until the BH evaporates, will your observation of the particle become discontinuous?

    Again, does this mean this is a discontinuity in the outside observers observation of the particle at the point of evaporation of the black hole?

    I'm not convinced here yet. If I take any point inside the horizon, I can just place another observer closer who observers the the particle outside the horizon.

    Can we please try to keep things in the frame of the distant observer, I am trying to understand what the distant observer *sees*. It is easy to understand what is happening from the point of view of the particle which is falling in, because there is no horizon crossing in this case.
     
    Last edited: Aug 31, 2012
  14. Aug 31, 2012 #13
    Thanks you everybody,

    What I would like to understand now is what to to different observers *see* as a particle falls into a black hole, then the black hole evaporates.

    The first observer follows the particle.

    The second observer stays at a fixed distance from the Black Hole.
     
  15. Aug 31, 2012 #14

    Bill_K

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    The event horizon is a null surface. It moves radially outward at the speed of light, even though its radial coordinate remains at R = 2M. This is seen to be true by near observers, distant observers, observers who are standing still and observers who are falling in.
     
  16. Aug 31, 2012 #15

    bcrowell

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    The first observer, A, sees the particle free-falling just ahead of the faceplate of her spacesuit, and finally sees it impact the singularity after a finite amount of time on her wristwatch. (She dies a nanosecond later, assuming she has somehow avoided already dying from tidal forces.) She can't observe the evaporation of the black hole, because she's dead -- her dead, compressed body is part of what eventually will evaporate back out.

    If the second observer, B, is following a radio signal being emitted by the particle, he observes that the signal becomes shifted to longer and longer wavelengths and become weaker and weaker in wattage. It becomes undetectable within a very short time. Gazillions of years later, he observes that the black hole starts to emit Hawking radiation at a rate that is detectable and greater than the rate at which it absorbs infalling matter and radiation.
     
  17. Aug 31, 2012 #16

    Bill_K

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    It's getting into the question of information loss in a black hole, to which Susskind's "complementarity" is a proposed solution. B & D take the position that information is lost.
     
  18. Aug 31, 2012 #17
    I knew somebody would give this answer, but it is the easy way out. I should have asked if we please assume no technical limitations on these observations and stick to the mathematics.

    It sounds like that for Observer A, they hit a point is space time that is discontinuous, so what happens after that may not be well defined.

    But, what happens from observer Bs point of view?
     
  19. Aug 31, 2012 #18

    George Jones

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    I agree that black hole complementarity is controversial, but I don't think that at least part of the passage that I quoted form Birrell and Davies is particularly controversial. I think that some folks in the complementary camp also take this view. For example, Susskind and Lindesay, in their semi-technical book "An Introduction to Black Holes, Information and the String Theory Revolution: The Holographic Universe," write "As the proton is observed over shorter and shorter time intrevals, the region it is localized must grow."
    To which solutions do you refer?
    For these solutions: 1) does Hawking radiation stop once the event horizon has formed; 2) does an external observer see a free falling particle fall through the event horizon in a finite time? Could what you have in mind be, at least partly, an intuitive version of the more technical passage by Birrell and Davies?

    In my mind, the passage from Birrell and Davies directly addressed
     
  20. Aug 31, 2012 #19
    Thanks, I was bad terminology, I was thinking of a apparent horizon, not an absolute horizon.

    Edit: Actually, given the context of what I was saying, I assume that the person originally correcting me was knowledgeable enough for it to be obvious that I had my understanding of the terminology could be better. If so, pointing out what my mistake actually was would have been much more appreciated.
     
    Last edited: Aug 31, 2012
  21. Aug 31, 2012 #20

    PeterDonis

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    In "your own reference frame", you are at rest--hence, you are not moving.

    In order to "observe" the horizon at all, you have to fall past it; someone who stays forever outside the horizon will never "observe" it, because light emitted outward at the horizon stays at the horizon--that's part of what the horizon *is*. If you fall past the horizon, then in your own reference frame, you will see the horizon moving outward past you at the speed of light.

    Yes, strictly speaking, that's true, and strictly speaking, the absolute horizon is not the same as the apparent horizon. The distinction is not really important for the cases we're discussing, which is why I've just been saying "the horizon". However, to be clear about what the terms mean:

    The apparent horizon is the surface at which outgoing light stays at the same radius. This can be observed locally (at least in principle), so it's a locally defined concept.

    The absolute horizon is the boundary between the region of spacetime that can send signals out to infinity, and the region that can't. This concept can only be defined globally; you have to know the entire structure of the spacetime to know for sure where the absolute horizon is.

    Strictly speaking, the description of the horizon moving outward past you at the speed of light as you fall only applies to the absolute horizon, since that's always a null surface (meaning it moves at the speed of light relative to local observers), whereas the apparent horizon is not necessarily always null. However, in the two cases we're discussing, as I said, the distinction isn't really important: for an "eternal" black hole (one that never evaporates and never absorbs any matter, so it stays the same size forever), the two horizons coincide; and for an evaporating BH, the evaporation is slow enough (at least until the very late stages, when the BH is almost gone) that the difference between the two horizons is not significant.

    Actually, the default terms used in "relativity" are those describing invariants, i.e., absolute ones. The real confusion comes from the name "relativity"; I forget which physicist it was who said that really the theory should be called the "theory of invariants", since the whole point is to focus on the things that *aren't* relative.

    No. You see the particle continually fall towards the horizon, so that it appears to you to arrive at the horizon at the same instant that the BH finally evaporates. The particle will appear to slow down as it approaches the horizon, but not infinitely, as it would if the BH were eternal.

    No; see above. Also, the distant observer still doesn't see any of the particle's worldline after it crosses the horizon, even after the BH evaporates; all of the light the particle emitted once it got inside the horizon was swallowed up by the singularity before the BH evaporated.

    Huh? I don't understand what you're describing here.

    See above.

    Sure there is. The infalling particle does cross the horizon; how are you concluding that it doesn't?

    If you mean the infalling particle doesn't *know* that he's crossed the horizon, that's not true either. He can directly test whether or not he has crossed the apparent horizon, as noted above: he just checks to see whether light that is emitted directly outward in his vicinity is able to reach a larger radius that that at which it was emitted. It's true that he can't directly test, locally, whether he has crossed the absolute horizon, because that requires global knowledge of the entire spacetime, as I said above. However, since the difference between the two is small for the case we're discussing, he can conclude that he will have crossed the absolute horizon within a very short time after he has crossed the apparent horizon.
     
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