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

lukesfn
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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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?

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

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.

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.

clamtrox
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?

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.

How can a space like singularity form inside a black hole if an outside observer sees no particle crossing the horizon?

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.

lukesfn
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.
I am aware that as you approach an even horizon that you will observe it receding.

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.

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.

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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?

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?

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According to the book Quantum Fields in Curved Space by Birrell and Davies, pages 268-269,
These consideration resolve an apparent paradox concerning the Hawking effect. The proper time for a freely-falling observer to reach the event horizon is finite, yet the free-fall time as measured at infinity is infinite. Ignoring back-reaction, the black hole will emit an infinite amount of radiation during the time that the falling observer is seen, from a distance to reach the event horizon. Hence it would appear that, in the falling frame, the observer should encounter an infinite amount of radiation in a finite time, and so be destroyed. On the other hand, the event horizon is a global construct, and has no local significance, so it is absurd too conclude that it acts as physical barrier to the falling observer.

The paradox is resolved when a careful distinction is made between particle number and energy density. When the observer approaches the horizon, the notion of a well-defined particle number loses its meaning at the wavelengths of interest in the Hawking radiation; the observer is 'inside' the particles. We need not, therefore, worry about the observer encountering an infinite number of particles. On the other hand, energy does have a local significance. In this case, however, although the Hawking flux does diverge as the horizon is approached, so does the static vacuum polarization, and the latter is negative. The falling observer cannot distinguish operationally between the energy flux due to oncoming Hawking radiation and that due to the fact that he is sweeping through the cloud of vacuum polarization. The net result is to cancel the divergence on the event horizon, and yield a finite result, ...

This finite amount of radiation is negligible for observers freely falling into an astrophysical black hole.

Mentor
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.

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.

Enter Hawking Radiation and black hole evaporation. When considering that it isn't possible to observe a particle crossing an event horizon

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.

How can a space like singularity form inside a black hole if an outside observer sees no particle crossing the horizon?

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.

Mentor
I am aware that as you approach an even horizon that you will observe it receding.

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

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.

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.

Are you claiming that 2 observers can disagree on the path that the particle took?

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.

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.

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.

lukesfn
Just re-quoting my self, from a late edit to my last post, so it isn't missed.
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.
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?

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?
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.

These consideration resolve an apparent paradox concerning the Hawking effect. The proper time for a freely-falling observer to reach the event horizon is finite, yet the free-fall time as measured at infinity is infinite. Ignoring back-reaction, the black hole will emit an infinite amount of radiation during the time that the falling observer is seen, from a distance to reach the event horizon. Hence it would appear that, in the falling frame, the observer should encounter an infinite amount of radiation in a finite time, and so be destroyed. On the other hand, the event horizon is a global construct, and has no local significance, so it is absurd too conclude that it acts as physical barrier to the falling observer.

The paradox is resolved when a careful distinction is made between particle number and energy density. When the observer approaches the horizon, the notion of a well-defined particle number loses its meaning at the wavelengths of interest in the Hawking radiation; the observer is 'inside' the particles. We need not, therefore, worry about the observer encountering an infinite number of particles. On the other hand, energy does have a local significance. In this case, however, although the Hawking flux does diverge as the horizon is approached, so does the static vacuum polarization, and the latter is negative. The falling observer cannot distinguish operationally between the energy flux due to oncoming Hawking radiation and that due to the fact that he is sweeping through the cloud of vacuum polarization. The net result is to cancel the divergence on the event horizon, and yield a finite result, ...
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;)

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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.

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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.

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.

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

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

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.
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?

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.
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.

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lukesfn
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.

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?
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.

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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.

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.

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.
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.

lukesfn
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.
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?

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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.

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."
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.)

To which solutions do you refer?
This finite time is many orders of magnitude less than the time scale for black hole evaporation, so the quantum effects are completely irrelevant.

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
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.

lukesfn
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.
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.

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Mentor
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?

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.

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

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.

you would think that the default terms used in general relativity would be relative ones, not absolute ones.

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.

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

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.

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?

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.

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.

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

Can we please try to keep things in the frame of the distant observer, I am trying to understand what the distant observer *sees*.

See above.

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.

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.

lukesfn
Sure there is. The infalling particle does cross the horizon; how are you concluding that it doesn't?
Now, obviously I was talking about the apparent horizon. Now, maybe I am wrong again, but isn't the apparent horizon relative to the observer? So an observer free falling into the black hole will always see the apparent horizon in front of them? Or am I still using the wrong term?

Mentor
Now, obviously I was talking about the apparent horizon. Now, maybe I am wrong again, but isn't the apparent horizon relative to the observer?

Not really. The apparent horizon is not a completely invariant global feature of the spacetime, like the absolute horizon is; but it's also not completely relative either. In the case we're discussing, each infalling observer will pass the apparent horizon at some particular instant by their clock, and that instant will be very close to the instant at which they pass the absolute horizon.

So an observer free falling into the black hole will always see the apparent horizon in front of them?

If you mean "see" as in "receive light from", then yes, an infalling observer will continue to receive light from the horizon, coming from in front of him, even after he falls past it. This is true of both kinds of horizons (apparent and absolute), and it happens because light emitted *inward* from the horizon is bent by the curvature of spacetime inside the hole so that it appears to be coming from in front of the observer as he falls. None of which changes the fact that the observer actually falls past each horizon at some definite instant, by his clock.

lukesfn
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.
Does the horizon shrink down to nothing? or does it evaporate at a certain radius?

So, the partial reaches the horizon at the same instant that the BH evaporates? Then? if there is no discontinuity as asserted? What is observed happening to the particle?

I am having trouble seeing how from the particles frame, there is a discontinuity as it hits the singularity, but from the distant observers frame, there is no discontinuity, but it is never observed falling into the hole? Surly either, there is a discontinuity some where, from the distant observers point of view, or the particle doesn't hit the singularity. Am I wrong?

lukesfn
Not really. The apparent horizon is not a completely invariant global feature of the spacetime, like the absolute horizon is; but it's also not completely relative either. In the case we're discussing, each infalling observer will pass the apparent horizon at some particular instant by their clock, and that instant will be very close to the instant at which they pass the absolute horizon.

The below is from http://en.wikipedia.org/wiki/Apparent_horizon, not that I necessarily trust Wikipedia 100%
Apparent horizons depend on the "slicing" of a spacetime. That is, the location and even existence of an apparent horizon depends on the way spacetime is divided into space and time. For example, it is possible to slice the Schwarzschild geometry in such a way that there is no apparent horizon, ever, despite the fact that there is certainly an event horizon.

Anyway, as I have been tought, if you fall into a bh, you will see a bh in front of you, all the way to the singularity. There will always be a horizon in front of you from which light cannot reach you. Or is that wrong? If not then you could see the singularity couldn't you?

audioloop
Mentor
Does the horizon shrink down to nothing? or does it evaporate at a certain radius?

As the BH evaporates, the horizon shrinks; eventually, it shrinks to zero radius, and at that instant, the BH has completely evaporated.

However, all this happens long *after* the infalling particle has crossed the horizon, reached the singularity, and been destroyed.

So, the partial reaches the horizon at the same instant that the BH evaporates?

No. The distant observer *sees* the particle reaching the horizon at the same time that he *sees* the BH finally evaporate. But that's because the light that was emitted outward by the particle as it crossed the horizon ends up on the same outgoing trajectory as the light from the final evaporation of the BH. It is *not* because the BH's final evaporation happens at the same event as the particle crossing the horizon.

It is *really* important to keep clear the distinction between what the distant observer sees, as in what light signals he receives, and what happens down at the horizon, because the horizon is an outgoing null surface, so light that is emitted outward at the horizon stays at the horizon, for as long as the horizon exists. That means that, when the BH finally evaporates, all the light that was emitted at the horizon, for the entire time of the horizon's existence, is now mingled with the light from the BH's final evaporation, and all of that light goes outward together.

What is observed happening to the particle?

The distant observer never sees anything of the particle once it has crossed the horizon, as I said before. The actual light images he receives will show the BH evaporating away along with the particle falling through the horizon; but as noted above, that does *not* mean that is what actually happened. The light from events that happened in distinct portions of spacetime gets mingled together because of the way the spacetime is curved.

I am having trouble seeing how from the particles frame, there is a discontinuity as it hits the singularity

Because the singularity has infinite curvature. But as noted, the distant observer can never see that, or indeed anything inside the horizon. See below.

but from the distant observers frame, there is no discontinuity, but it is never observed falling into the hole?

Because the distant observer can never see anything inside the horizon. That's what the horizon *is*--the boundary of a region of spacetime that can't send signals out to distant observers.

Surly either, there is a discontinuity some where, from the distant observers point of view, or the particle doesn't hit the singularity. Am I wrong?

Yes. You've left out a third possibility: that the region of spacetime inside the horizon, even though it exists, can't send light signals out to the distant observer. The reason it can't is that the curvature of spacetime in that region is great enough to force even outgoing light to fall inward towards the singularity.

The below is from http://en.wikipedia.org/wiki/Apparent_horizon, not that I necessarily trust Wikipedia 100%

A wise policy. In this case, the Wiki page is correct in what it says, but what it says does not contradict anything I've said.

I have not been able to read the actual papers that are referenced in that article, since they're behind paywalls, but AFAIK, whatever spacetime slicings would not show an apparent horizon, the ones that are relevant to our discussion--the spacetime slicing of the distant observer, and the spacetime slicing of the infalling observer--both *do* show apparent horizons. (Others who have commented in this thread may have read the actual papers and be able to correct me if I've misstated something, or add more clarification.)

Also, as I've noted before, the boundary of the region that can't send light signals to the distant observer is the *absolute* horizon, not the apparent horizon; and strictly speaking, what I said above about light emitted outward at the horizon mingling with light from the BH's final evaporation applies to the absolute horizon. (As I've noted before, for the case we're discussing the difference between the two kinds of horizons, for the obsevers we're talking about, is not significant.)

Finally, just to clear up a possible misunderstanding of the term "apparent horizon": the word "apparent" in this term does not refer directly to whether, or for how long, the observer actually sees light from the horizon. It is just a way of distinguishing it from the absolute horizon. What has been said about the horizon "appearing" to be in front of the infalling observer, even after he has actually fallen past it, applies to both kinds of horizons, apparent and absolute.

Anyway, as I have been tought, if you fall into a bh, you will see a bh in front of you, all the way to the singularity.

You will *see* the horizon in front of you, yes, because of the bending of light emitted *inward* from the horizon by the curvature of spacetime inside the hole. That does not mean you are actually outside the horizon.

There will always be a horizon in front of you from which light cannot reach you. Or is that wrong?

Yes. Light from the horizon appears to be coming from in front of you, but the curvature of spacetime inside the horizon is extreme enough that you can't trust your intuitions about what that light is telling you.

If not then you could see the singularity couldn't you?

No. You can never see the singularity, because it's in your future; you can't see it for the same reason you can't see tomorrow.

Naty1
George:

"As the proton is observed over shorter and shorter time intrevals, the region it is localized must grow."

I agree that black hole complementarity is controversial,

George, shame on you for lacking the 'Susskind faith'.....

I can't vouch for his mathematics, but his physical inetrpretations are facinating...

That's an insight that Susskind uses frequently....I immediately recognized your description from Susskind's THE BLACK HOLE WAR where he describes 'Alice's last
visible propeller'...imagining rotating propellers falling into a black hole...just as a local propeller slows down you can see more and more of it ,redshift resulting from falling into a black hole slows stuff down bringing larger portions of the object into view....

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To which solutions do you refer?
I had in mind the Vaidya metric, not because it's necessarily particularly important but because it's simple and I happen to have heard of it.

For these solutions: 1) does Hawking radiation stop once the event horizon has formed
It's a solution to the classical field equations, so it doesn't describe Hawking radiation. But isn't it pretty clear that in real-life cases, the time-scales for formation and evaporation are wildly mismatched?

2) does an external observer see a free falling particle fall through the event horizon in a finite time?
Interesting question, and I don't know the answer. I could cop out and say that this question is ill-defined, because we only have a well-defined notion of how long a process takes to a distant observer if the spacetime is static, but the Vaidya metric isn't static. We could also worry about whether there are both an event horizon and an apparent horizon (boundary of the union of all null timelike surfaces) in the case of the Vaidya metric.

If we want to make this discussion precise, probably the difficult issue is how to define the notion that according to a distant observer, a certain black hole has "now" formed an event horizon, in a case where the spacetime is nowhere static. By definition, a distant observer can't receive a signal from an event on the event horizon. But of course by any reasonable definition, Sagittarius A* has "now" formed an event horizon.

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At the end of this blog post http://backreaction.blogspot.com/2009/11/causal-diagram-of-black-hole.html there's a Penrose diagram for a black hole that forms by collapse and then later evaporates. I don't know if this is right and noncontroversial, but let's assume it is. The reasoning leading up to it seems OK on casual inspection by a non-expert (me).

Based on this diagram, it seems like we have kind of a cute resolution to this whole question. A distant observer sees a relatively short time-scale (years, maybe) for the horizon to be effectively established, where "effectively" means that in practical terms, there are infalling particles that we no longer have any hope of gathering information about. However, the completion of evaporation doesn't lie in the interior of the future light cone of any point on the horizon, nor does it lie in the interior of the light cone of points near the singularity. It's lightlike in relation to the horizon, so in that sense, there is no time delay between formation of the horizon and the completion of evaporation.

However, there is a delay in the following sense. We have the last infalling light ray that, in the Schwarzschild version, would have been able to escape to infinity. This ray moves through some nonzero (in fact, very large) affine parameter between the time when it joins the horizon and the time when it's liberated at evaporation.

audioloop
At the end of this post http://backreaction..com/2009/11/causal-diagram-of-black-hole.html [Broken] there's a Penrose diagram for a that forms by collapse and then later evaporates. I don't know if this is right and noncontroversial, but let's assume it is. The reasoning leading up to it seems OK on casual inspection by a non-expert (me).

Based on this diagram, it seems like we have kind of a resolution to this whole question. A distant observer sees a relatively short time-scale (years, maybe) for the to be effectively established, where "effectively" means that in practical terms, there are infalling particles that we no longer have any hope of gathering information about. However, the completion of evaporation doesn't lie in the interior of the cone of any point on the , nor does it lie in the interior of the light cone of points near the . It's lightlike in relation to the , so in that sense, there is no time delay between formation of the horizon and the completion of evaporation.

However, there is a delay in the following sense. We have the last infalling light ray that, in the Schwarzschild version, would have been able to escape to infinity. This ray moves through some nonzero (in fact, very large) affine parameter between the time when it joins the horizon and the time when it's liberated at evaporation.

penrose solution involves a loss of unitarity,
for a non-unitary black hole evaporation you need a nonlinear quantum mechanics.

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Mentor
At the end of this blog post http://backreaction.blogspot.com/2009/11/causal-diagram-of-black-hole.html there's a Penrose diagram for a black hole that forms by collapse and then later evaporates. I don't know if this is right and noncontroversial, but let's assume it is.

I've seen diagrams like this one before; I think one of Penrose's popular books has a similar one (maybe The Road To Reality?). The things I've been saying about a spacetime with an evaporating BH in it are based on this type of causal structure for the spacetime, which AFAIK is the currently accepted picture. (For example, it's easy to see from this diagram that any observer who finds himself inside the horizon as it currently exists when he falls in, regardless of how much the BH has evaporated by that point, is doomed to be destroyed in the singularity.)

I agree with your follow-on comments about why and in what sense there is a long time delay between the BH forming and its final evaporation.

Mentor
penrose solution involves a loss of unitarity

I've just been re-reading Susskind's The Black Hole War, where he discusses this in some detail. He doesn't give an explicit Penrose diagram for his proposed solution (black hole complementarity), but from what he says in the book (which is, admittedly, not a scientific paper), it seems like such a diagram would look like the one in the blog post bcrowell linked to.

BH complementarity, as I understand it, entails that any quantum information that comes within the stretched horizon (basically within a Planck length above the horizon) gets transferred to the quantum field modes that eventually emerge in the Hawking radiation and thereby get back out to infinity, so unitarity is preserved. Susskind's answer to the question of why this doesn't violate the quantum no-cloning theorem--since there is also a copy of the infalling information that continues to fall in past the horizon and into the singularity--is that the copy that goes down the hole can't send any signals back out, so it's OK. Which sounds hand-wavy, but presumably there's a lot more technical detail in the actual papers.

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Here is my rendition of the a Penrose diagram for a collapsing and evaporating black hole, with all the bells and whistles. This is basically the one given here http://backreaction.blogspot.com/2009/11/causal-diagram-of-black-hole.html and here http://www.math.unb.ca/~seahra/resources/notes/black_holes.pdf , plus my idea of what apparent horizon should be like, based on the discussion in the second reference. I don't think it's consistent with the one at http://en.wikipedia.org/wiki/File:PENROSE2.PNG , which shows a three-way intersection of the infalling matter, singularity, and horizon, and has the singularity being initially timelike. The bow represents the infalling matter.

http://www.lightandmatter.com/penrose_diagram_for_black_hole.jpg [Broken]

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The process of formation of the horizon also takes finite time in the following sense. At sufficiently early times, the spacetime is static and flat to an arbitrarily good approximation, so we can define a family of observers, all at rest relative to the dilute cloud of slowly collapsing matter, and let them synchronize their clocks in the usual way. These observers know which one of them is positioned at the center of the cloud, and can predict the reading on his clock at the event E when he crosses the horizon. Event E is, however, never in the past light cone of any observer who is outside the event horizon.

lukesfn
Yes. You've left out a third possibility: that the region of spacetime inside the horizon, even though it exists, can't send light signals out to the distant observer. The reason it can't is that the curvature of spacetime in that region is great enough to force even outgoing light to fall inward towards the singularity.
No, I don't think that is a third possibility. I am trying to ask about a discontinuity in the particles path. If the outsider can't observe anything once the particle reaches the horizon, the outside observer must observe a discontinuity, of the path of the particle at the time the observer *sees* the particle reach the horizon, no? If there is a part of the particles trip that cannot be observed, it is very hard to understand how there could not be a discontinuity.

Grom doing some further reading, it does sounds like evaporating BHs are not that well understood, and some points are an open question. I am wondering how much of what you are stating is actually an open to question and far from certain.

Yes. Light from the horizon appears to be coming from in front of you, but the curvature of spacetime inside the horizon is extreme enough that you can't trust your intuitions about what that light is telling you.

No. You can never see the singularity, because it's in your future; you can't see it for the same reason you can't see tomorrow.
I think you are thinking of a different coordinate system to me. Perhaps I am thinking of Rindler space... although, I don't really know much about anything.

If you are observing light that originated from behind your current position, it still got deeper into the BH then you at some point right?. Is it true that if you are inside the event horizon of a BH, then there is point between you and the singularity where on one side, light can reach you, and on the other side, light can't reach you. Couldn't this be considered a horizon that you can never cross? Would you expect the Unruh effect to take place here?