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Why wouldn't black hole singularity evaporate before it can form?

  1. Apr 7, 2013 #1
    I assume people much more knowledgeable then me must have already thought this through, but the following line of thought has me very curious. I'm making a lot of intuitive leaps here though, so I am sure there is a lot of places I could have gone drastically wrong.

    Hawking radiation is predicted to be observed from out side the event horizon of a black hole, however, I would expect somebody just inside the horizon to also observe something similar to hawking radiation. There will still be another event horizon between the observer and the center of the BH from from which out side of, information will be able to be sent back to the observer (although, obviously it can't move away from the center of the BH, it can just move slower towards the center then the observer does so that the observer catches up to it)

    I would expect this would work similarly to what happens optically if you cross a BH horizon, you would still optically see the BH in front of you, although it would appear to wrap around you as you fall towards the center. So, I am expecting you would also continue to be able to observe some hawking like radiation as you fall all the way to the center.

    If this is so, then it may get more interesting towards the center. I think the amount of hawking radiation is generally calculated via the mass of a black hole, but for an observed hawking like radiation inside at a radius inside the horizon, I would expect that it would have to be increase exponentially as the radius decreases since the curvature of space would increase.

    Maybe I am confused, but it seems to me that the rate of increase of this radiation would tend towards infinity faster then an observer would be accelerated towards the center.

    By this logic, I start to think that an observer should see everything in the BH evaporate before they could reach the singularity.

    Following this logic further no particle should ever be able to reach a singularity, so is it possible for one to form?

    It seems a little counter intuitive that a singularity couldn't form given that things would have to be moving out from the center at faster then the speed of light trough space tiem to stop a singularity from forming, but the mechanism of hawking radiation allows for FTL transfer of mass from the BH at the horizon, so could it not allow a trickle up effect from all the way from the center, especially considering the much higher stress on space time towards the center effectively causing space time too pull its self apart that much stronger.

    Like I said, I assume many much more knowledgeable people have spent much more time then me pondering such things, but it does seem like such an elegant solution to many problems to my naive mind that I probably need somebody else to tell me why it isn't so.
  2. jcsd
  3. Apr 7, 2013 #2

    Simon Bridge

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    What leads you to expect this?

    The math does not predict one.

    What leads you to expect this?
    One of the ways to gain understanding is to investigate your expectations - to do that, you need to make the basis for them explicit.

    What does this mean? "Wrap around you"?
    Do you know what the event horizon is?

    What leads you to expect this?
    Do you know how Hawking radiation is expected to happen?

    Things should get interesting towards the center anyway.


    It gets a little tricky - you are talking about time as experienced by an observer falling towards a black-hole? The link above discusses this with math. It will help you sort out these ideas.

    Of course it is.
    The black hole forms as the mass concentration passes a critical value.
    Then is is expected to radiate via the Hawking mechanism.

    You are thinking, perhaps, that the matter forming the Schwarzschild black hole "evaporates" (by the Hawking mechanism) before it can totally collapse - even by that logic, the black hole can still form without all the matter being concentrated in the center.
  4. Apr 7, 2013 #3
    I tried to explain below but obviously could have done better.

    Do you then think the radiation observed coming from a black hole just disappears when you cross the horizon? Isn't a free falling observer meant to see nothing special as they cross the horizon?

    If that is so, then the rest of my post is irrelevant but perhaps I should should try to explain my logic better.

    If you are free falling into a Black Hole, and there is an object in front of you, if it is out side a certain radius of the black hole, you will be able to see light from it, however, any object with in that radius you will not. I am calling that an effective event horizon. (The light from that object of course would have been emitted from a position further out from the Black Hole center then where you are currently located.

    Should you not also not also experience some apparent hawking like radiation in a similar way?

    As you move through the horizon and towards the center, the everything outside the black hole will optically appear to be compressed into a single point, which could look like the black hole was wrapping around you.
  5. Apr 7, 2013 #4


    Staff: Mentor

    Don't forget, the event horizon is a null surface, meaning that it is moving at the speed of light wrt any free-falling local frame. So the fact that you cannot see light from an object which has crossed the horizon until you cross the horizon is no different from the fact that you cannot see light from any event until a time d/c has passed for the light to reach you.

    If Jack and Jill were crossing a supermassive black hole's event horizon and they were 1 light-second apart (ladies first, of course). Then when they were both outside the event horizon Jack would have to wait 1 s to receive a signal from Jill, when Jill was inside and Jack was out Jack would have to wait 1 s to receive a signal from Jill (by which time Jack would be in since the horizon moves at c), and when they are both inside Jack would have to wait 1 s to receive a signal from Jill. And vice versa.
  6. Apr 7, 2013 #5
    What radiation do you think is observed?
    Yes, to the second part.

    Even Einstein did not believe the mathematical results thinking black holes did not REALLY exist. Math can give us insights 'logic' cannot.

    A distant stationary observer will never receive a signal from a source at or within the event horizon even after an infinite amount of proper time; An inertially falling observer will receive a signal from at or within the event horizon after a finite amount of proper time.

    I'd call that a 'logical contradiction'.... at least from classical reasoning.

    [This should be consistent with Dalespam's post just above.]

    edit: for a related view, try reading about the Unruh effect.

  7. Apr 7, 2013 #6
    The simple answer:
    because once mass collapses to within the Schwarszchild radius, r = 2M,
    nothing can get out....gravity is just too strong.
  8. Apr 7, 2013 #7

    Time d/c? Sorry, I can't figure out that term, although, I assume I still understand your meaning.

    I'm not quite sure if you are trying to explain something I am missing here, however, I am curious, what is keeping them 1 light second apart? Would 2 objects in free fall following the exact same motion, starting 1 light second stay 1 light second apart? Talking about a 1 second delay for 1 way communication is very easy to get confused with. Would it be better to talk about a 2 second delay for Jack to have to wait for an instantly returned message from Jill?

    Something that would appear exactly identical to hawking radiation with respect to the type of particles being detected, except, I expect it would need a different name.

    You are getting into a semantic discussion. When I use the word logic, the logic is trying to follow the rules of GR (and QM with regard to QM), so if any of my logic is contradicted by the mathematics, please feel free to let me know.

    I am aware of the unruh affect, although, I am not sure weather that would be considered the same effect I am talking about or not, because from what I have previously read, it occures due to an object constantly accelerating in SR, where as I am talking about objects free falling in GR. Still, I am talking about something similar. It is interesting to consider that if you sit just outside the EH accelerating to stay near the Speed of light to hover, that you would see a lot of Unruh radiation, but by another logic, because of time dialation, you would expect to see the BH evaporate a lot faster via hawking radiation. Does that mean that in this case the Unrah and Hawking radiation are the same thing?

    Are you saying that mass can not be transferred out via hawking radiation?
    Last edited: Apr 7, 2013
  9. Apr 7, 2013 #8


    Staff: Mentor

    Sorry, I should have defined my variables. d is distance, distance divided by speed is time.


    Yes. Remember that the black hole is supermassive so tidal effects are negligible.

    It doesn't matter which you are talking about. The point is that it all behaves exactly the same as normal in a regular local inertial frame without an event horizon.
  10. Apr 7, 2013 #9
    Sorry, I didn't pay proper attention to the word super massive, but thanks for clearing up that fact that you where ignoring small effects due to curvature of space in this case. This is something I already understand well, and I don't think contradicts my like of reasoning.

    Anyway, I've been informed I've crossed the line of being too speculative for the forum rules, so I will try to constrain my discussion to weather any hawking-like radiation may exist inside the EH of a BH.

    If an observer outside the EH of a BH detects Hawking radiation, and they cross the EH,
    1) does the observer simply see the hawking abruptly disappear after crossing the horizon and replaced by nothing, or
    2) would the observer continue to detect some "hawking-like radiation" during and after crossing, that would, from the observers point of view, be very difficult to distinguished from hawking radiation, so much so that no device would be able to measure a sudden change in radiation levels to detect the crossing of the EH.
  11. Apr 7, 2013 #10


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    This strikes me as a situation where Penrose diagrams come in very handy: http://en.wikipedia.org/wiki/Penrose_diagram . Personally I can't reason at all about the cause and effect relationships discussed in #1 without a Penrose diagram in front of me. For example:

    AFAICT from a Penrose diagram, the second observer can receive the information if and only if s/he dives in after the first observer within a short enough time. The WP article is not that great. Anyone who wants to go to Penrose-diagram bootcamp should read the treatment in Penrose's popularization Cycles of Time.

    An event horizon is usually defined as a surface from which light can't escape to an infinitely distant observer. The definition seems to make it clear that horizons have no special properties except in relation to some observer. Therefore I doubt that it's correct to expect Hawking radiation to be a special phenomenon that only exists at an event horizon. This is probably a case where those of us (including me) whose understanding of quantum gravity is at the pop-sci level should not expect popularizations to get us beyond a certain point. Baez has a writeup about his doubts re this particular way of explaining Hawking radiation in popularizations: http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html
    Last edited: Apr 7, 2013
  12. Apr 7, 2013 #11
    Excuse me, I've used that term incorrectly before creating all types of confusion. Perhaps apparent horizon is better term but I am not sure that is right either, I should have just said horizon.
    Last edited: Apr 7, 2013
  13. Apr 8, 2013 #12

    Simon Bridge

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    That would be correct.
    ... that was the gist, but it was possible I misunderstood.
    ... no, that would just have lead to people asking you to be specific.
    The "event horizon" is the distance from the center of mass where the escape velocity is equal to the speed of light[*].

    I think the main thing to take away from bcrowells comment there is the bit about how far pop-science can take your understanding. Did you have a look at the links I gave you at all?

    [*] A more accurate description could be that within this horizon, all lightlike paths are warped so as to fall farther into the hole.
  14. Apr 8, 2013 #13
    I like to visualize curved space as a flowing river with special relativity occurring over the top. With this analogy, inside a BH EH, space flows faster then the speed of light. I can stay pretty close to intuitive classical thinking this way, which makes things much easier to visualize for me, except when time dilation confuses everything, and I think I understand SR reasonably well.

    Which, from my point of view, would be a much better out come. I find it is usually much better to be asked for clarification then completely miss understood.

    I was actually aware of this kind of definition for an EH of a BH. It makes sense to me that there would always be some kind of observer dependent apparent horizon between an observer and the singularity, however, I have previously been told that an "apparent horizon" has some different definition. I think I need to describe better exactly how the horizon in my mind is defined, however, I might leave that for another occasion.

    I've also taken a similar line of reasoning which brings us close back to the point.

    Perhaps we could avoid this horizon talk and get back to the question of hawking-like radiation inside the event horizon?

    To Quote my self:
  15. Apr 8, 2013 #14


    Staff: Mentor

    One of the nice things about discussing supermassive black holes is that the tidal gravity is negligible at the EH, so you can analyze the situation using accelerating reference frames (per the equivalence principle).

    If Hawking radiation emanates from the black hole's horizon then there must be some equivalent from the Rindler horizon of the accelerating reference frame. This is known as Unruh radiation. Unruh radiation is not seen by an inertial observer. So, by the equivalence principle, I would expect that an observer free-falling across the EH of a supermassive BH would also not detect any Hawking radiation.

    What I am less certain of is if an observer were not free falling across the EH, but passed the EH while accelerating (but not enough to "hover", or even accelerating towards the singularity).
  16. Apr 8, 2013 #15
    I just read something on a similar note elsewhere. It also made me wonder if a free falling observer may not observe Hawlking radiation, however I'm not sure that makes sense either.

    One problem with using Supermassive Black Holes is that you have to be careful not to end up ignoring the effect you are looking for. The in falling observer may not see Hawking radiation because it is too little.

    I'm just starting to find some better search times to find stuff on google on this topic. The below link looks interesting, but I suspect that even if I was subscribed, I would need to learn a lot before having much chance of even getting an intuitive understanding of what it is talking about.
    Hawking radiation as perceived by different observers.
  17. Apr 8, 2013 #16


    Staff: Mentor

    That is a good point. The temperature of a black hole is lower the more massive it is, so a supermassive BH would be very cold, perhaps undetectably cold.

    However, IMO, the equivalence principle trumps Hawking radiation, and you can always make the EH of a Schwarzschild BH equivalent to a Rindler horizon for a sufficiently small region of spacetime. However, this is a very handwavy argument that I recognize as being somewhat weak, so I won't do more than merely present it.

    I understand the equivalence principle much more than I understand Hawking radiation, so I rely on it more and tend to be skeptical of Hawking radiation where it seems to conflict. I.e. I would need to see some rigorous proof before believing that a free-falling observer sees Hawking radiation at all, let alone the Hawking-like radiation you are discussing.
  18. Apr 8, 2013 #17
    I'd like to know if a free falling observer is expected to see Hawking radiation or not. Reading much else where, I have seen a lot of conflicting information about this.

    I noticed a similar thread here on physicsforums. however, there seems to be a lot of contradictory answers.

    Consider a small black hole, radiating a large amount of hawking radiation, and two observers, at a distance where gravity is small, one observer hovering, one free falling, both with the same instantaneous velocity. The hovering observer, would not need to accelerate much to stay hovering due to the low gravity, so would not see much Unruh radiation. The free falling observer sees Unruh radiation. But, because it is a small BH it should be radiating a lot, so one might expect the hovering observer to see a large amount of hawking radiation, not a small amount, and have trouble understanding why the free falling observer sees none, leading to think that curvature of space is very important to take into consideration with hawking radiation.

    I can think of some holes in this line of thinking, such as, perhaps a freefaller observes hawlking radiation out side the horizon, but not at the horizon, or maybe there the hawking radiation smoothly changes direction at the horizon, but those ideas bring up a lot of paradoxes such as, shouldn't the hawking radiation cool down the further it gets from a BH, not increase?

    hmm. the way Wikipedia tries to describe it is interesting.
    Can a distant observer use the equivalence principle and claim that they are very quickly accelerating away from the EV and make calculations based off that about what they would observe at the EV, even though, they are actually accelerating very slowly, or even free falling, and there is a large amount of curved space between them and the EV? They would have to then compensate for what ever effect to pass back though curved space to them to be observed though.
  19. Apr 9, 2013 #18
    I've been doing some extra reading.

    These all link directly to pdf files. The first has a good summery about observations of halwking radiation from different observes in the introduction. The others confirm the some of same by different methods. I don't pretend to have tried to understand the maths, so I have to trust the conclusions.

    (1) Taking the Temperature of a Black Hole

    (2) Hawking radiation as seen by an in-falling observer

    (3) Hawking radiation as perceived by di different

    From these, the temperature observed of hawking radiation seems to depend on the observers location, velocity, and acceleration. Whether or not a free falling observer sees hawking radiation or not, depends on there.

    According to [2], the only a free falling observer with zero instantaneous radial velocity at the horizon will see no hawking radiation. I'm a little confused what zero radial velocity means in this context. I guess it is a I guess it means that if we imagine space moving past the EV at the speed of light, the observer would be not moving relative that space.

    If read quite a few claims in other places that a free falling observer sees no hawking radiation, but it sounds like the standard lore may have been over generalized.

    Anyway, I think I will keep reading more of these papers.

    I'm not quite sure what a zero radial velocity observer sees from just out side the EV, and just inside yet.

    I'm wondering if all observers see hawking radiation out side the BH EV, but only some inside do.
  20. Apr 10, 2013 #19
    I saved this post from another discussion:

    According to the book Quantum Fields in Curved Space by Birrell and Davies, pages 268-269,

  21. Apr 10, 2013 #20
    no way things will ever be THAT straightforward!!....Observations are generally 'relative'.

    That's my understanding for a radially free falling observer....Apparently for those observers on an arcing trajectory, some radiation might be observed. But the above quote [prior post] suggests not. A poster in another discussion said the energy of observers also plays a role.

    I have been looking for the following statement for sometime....a prior post of mine....

    This confirms that Hawking radiation as described by 'particle separation' at the horizon is not mathematically based...just one way to think about what is happening. [Penrose is a collaborator of Hawking on many BH/GR issues.]

    Kip Thorne, BLACK HOLES AND TIME WARPS, PGS 435-440

    As soon as one is separated from its virtual partner, it becomes "real" (observable)......but only to an accelerating (stationary) observer outside the horizon.

    Since black holes are still considerably colder than our universe, they radiate nothing but do absorb energy (radiation) from the universe and slowly grow. So nobody has ever seen Hawking radiation.

    I don't personally any longer believe any of Thorne's statements are completely correct....but who am I to contradict him.

    As an example, in a discussion in these forums I recall we concluded that a blackbody radiates as long as it is above absolute zero....and a BH is a black body.....so it does radiate, but not evaporate in our life times, because it absorbs way more than it emits.....

    A key to all understanding all this are posts above discussing global and local constructs. As is always the case in relativity, different mathematical perspectives [different models] offer different [complementary] effects.
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