Hawking radiation prevents event horizon crossing?

[Asher Weinerman]

No one seems to be bothered by this except me:

Black holes have a finite lifetime measured in Schwartzchild time due to Hawking radiation. Similarly, the universe probably has a finite lifetime measured in Schwartzchild time. In that case, nothing ever falls through the event horizon of a black hole because it takes infinite Schwartzchild time so the black hole would disappear and the universe would come to an end before you crossed the event horizon. You could even calculate dr/dt of the event horizon of a black hole shrinking due to Hawking radiation which would presumably be non-zero and presumably negative. Since an infalling observer's dr/dt approaches zero at the event horizon, as you fall towards the event horizon it will shrink away from you as you approach and you will never cross it. So why does everyone spend time analyzing what happens after you cross the event horizon, lured by the fact that proper time is finite? I really just don't understand - I'm not trying to be clever - I just want to know what I am missing here.

 

[phinds]

No one seems to be bothered by this except me:

Black holes have a finite lifetime measured in Schwartzchild time due to Hawking radiation. Similarly, the universe probably has a finite lifetime measured in Schwartzchild time. In that case, nothing ever falls through the event horizon of a black hole because it takes infinite Schwartzchild time so the black hole would disappear and the universe would come to an end before you crossed the event horizon. You could even calculate dr/dt of the event horizon of a black hole shrinking due to Hawking radiation which would presumably be non-zero and presumably negative. Since an infalling observer's dr/dt approaches zero at the event horizon, as you fall towards the event horizon it will shrink away from you as you approach and you will never cross it. So why does everyone spend time analyzing what happens after you cross the event horizon, lured by the fact that proper time is finite? I really just don't understand - I'm not trying to be clever - I just want to know what I am missing here.

This is all wrong. You should read some more (and not pop-sci stuff) about black holes.

Things always fall through the EH of a BH, it just doesn't look that way to a remote observer. The in-falling object never notices that there even IS such a thing as an EH.

 

[PeterDonis]

nothing ever falls through the event horizon of a black hole because it takes infinite Schwartzchild time so the black hole would disappear and the universe would come to an end before you crossed the event horizon.

No, this is not correct. First of all, Schwarzschild coordinates are not good ones to use to describe an object falling through the event horizon even for a purely classical black hole, which does not emit Hawking radiation, because those coordinates are singular at the horizon. Painleve or Eddington-Finkelstein coordinates, which are nonsingular at the horizon, are much better. In these coordinates, it only takes a finite time for an object to fall through the horizon.

If we add Hawking radiation to the mix, and use coordinates that are nonsingular at the horizon, like Painleve or Eddington-Finkelstein (more precisely, if you use the analogues of these coordinates for a spacetime with an evaporating black hole, which is different from a spacetime with a purely classical black hole), then we find that, as before, it only takes a finite time for an object to fall through the horizon, and this time will be shorter than the time (in the same coordinates) that it takes the hole to evaporate via Hawking radiation. If, however, you insist on using Schwarzschild coordinates (more precisely, coordinates that are analogous to them), then you find that, while the coordinate time it takes for an object to reach the horizon is infinite, the time it takes for the hole to evaporate is also infinite. So these coordinates can't even represent the evaporation process properly, and you can't use them to answer the question of whether objects can reach the horizon before the hole evaporates.

 

[Asher Weinerman]

But it takes finite time in Schwartzchild coordinates for a black hole to evaporate if the universe is open. If the universe is closed I guess we all fall into the same black hole. I say finite time if the universe is open because the Hawking radiation will dissolve the black hole until it is in thermal equilibrium with the cosmic background radiation. This cosmic background radiation will become less and less powerful until it disappears entirely if the universe is open, which means all black holes will eventually evaporate. Apparently, according to wiki, at the moment only a black hole the size of the moon or less will evaporate before it is in thermal equilibrium with the current state of the cosmic background radiation.

So again, if the universe is open, all black holes will evaporate before anything falls into them. Right?

 

[PeterDonis]

it takes finite time in Schwartzchild coordinates for a black hole to evaporate

No, it doesn't. See my previous post. Please become familiar with the actual math and theory before making claims about it.

If the universe is closed I guess we all fall into the same black hole.

If the universe is closed, then strictly speaking there is no such thing as a black hole, because there can't be any event horizons. Event horizons can only exist in a spacetime that has a future null infinity, and the spacetime of a closed universe does not have one.

the Hawking radiation will dissolve the black hole until it is in thermal equilibrium with the cosmic background radiation. This cosmic background radiation will become less and less powerful until it disappears entirely if the universe is open, which means all black holes will eventually evaporate

Yes, this is true, but the "time" being used here cannot be Schwarzschild coordinate time (see my previous post), even though it will be very close to Schwarzschild coordinate time far away from the black hole (this is true of both of the alternate coordinate charts I mentioned in my previous post). Close to the horizon, the "time" implicitly being used here will differ greatly from Schwarzschild coordinate time; at the horizon, it will differ infinitely greatly, since Schwarzschild coordinates assign an infinite time coordinate to the horizon.

if the universe is open, all black holes will evaporate before anything falls into them. Right?

Wrong. See above.

 

[PAllen]

But it takes finite time in Schwartzchild coordinates for a black hole to evaporate if the universe is open. If the universe is closed I guess we all fall into the same black hole. I say finite time if the universe is open because the Hawking radiation will dissolve the black hole until it is in thermal equilibrium with the cosmic background radiation. This cosmic background radiation will become less and less powerful until it disappears entirely if the universe is open, which means all black holes will eventually evaporate. Apparently, according to wiki, at the moment only a black hole the size of the moon or less will evaporate before it is in thermal equilibrium with the current state of the cosmic background radiation.

So again, if the universe is open, all black holes will evaporate before anything falls into them. Right?

Wrong. Let's describe the physics observed by two different observers, and forget coordinates which are just labels attached to events.

Bob freely falls into BH, reaching singularity in quite finite time on their watch. Later, the BH completes evaporating, including the energy contributed by Bob.

Distant observer watches Bob disappear, progressively, on approach to horizon. 'Disappear' means that any light from Bob is zillions of orders of magnitude less than either CMB or the prior Hawking radiation from the hole (which, for a stellar mass BH order of magnitude less than CMB). The BH continues to radiate and shrink until a final burst. The total emitted includes radiation equivalent to all matter that has fallen into the BH - including that of Bob and all else that has reached the singularity. There is complete consistency between this and Bob's observation. It is true that distant observer cannot see Bob crossing horizon but since when has what some observers sees been taken as the definition of the universe?

As for coordinates, SC coordinates simply don't cover events at or inside the horizon, any more than Rindler coordinates cover the boundary or beyond the Rindler wedge. Thus, the complete picture of matter forming the BH, more matter falling in, and all evaporating is simply not covered by SC coordinates. Instead, all events at or inside the (evolving) horizon are simply not labeled by these coordinates. However, the subset of events covered by SC coordinates are completely consistent with the more complete description provided by other coordinates - it is just that some part of spacetime is not labeled.

 

[PAllen]

It is worth noting that your overall argument is common one that has been refuted repeatedly.

 

[Asher Weinerman]

I appreciate you helping me to understand things, since I'm not an expert in this subject like you are.
Every reference I've checked says that black holes will evaporate via Hawking radiation in finite far-away time on the order of 10^60 times the age of the universe (wiki for example). What reference are you using - can you give me a link to it so I can become familiar with the actual math and theory?

 

[Asher Weinerman]

Hi PAllen and PeterDonis - thank you for helping me understand.
PAllen - my question is that doesn't BH evaporate before Bob falls through the horizon in quite finite proper time, since I'm saying black holes evaporate in finite far-away time, which occurs before Bob enters BH.
Maybe you could give me a link that states black holes take infinite far-away time to evaporate by Hawking radiation if the universe is open?

 

[Asher Weinerman]

Oops - I meant to say above "before Bob enters EH, not BH"

 

[pervect]

Ted Bunn's cosmology FAQ is one resource on this topic, see for instance http://cosmology.berkeley.edu/Education/BHfaq.html#q9
The sci.physics.faq at http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html also covers the issue. Both FAQ's come to the same conclusion, the black hole will not evaporate before you fall in.

On the more techincal side, T Vachaspati, D Stojkovic and L M Krauss, Phys. Rev. D76, 024005 (2007) [arXiv:gr-qc/0609024] do make claims which could be regarded as saying that black holes evaporate before you fall in. Rebutting this paper we have:

"Radiation from collapsing shells, semiclassical backreaction and black hole formation" which is a response to the Krauss paper that can be found on arxiv at http://arxiv.org/abs/0906.1768

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

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

3) I haven't seen a published response by Krauss, et al, to P&P's comments. (It's possible I could have missed it).

I recall some other comments on the topic that were much simpler than Paranjape and Padmanababhn's, and came to the same conclusion, but not exactly where they were or who made them. It might have been Unruh. It was a rather simple argument about the implausibility of any terms in the stress-energy tensor due to evaporation that could halt collapse, as I recall.

 

[PeterDonis]

Every reference I've checked says that black holes will evaporate via Hawking radiation in finite far-away time

"Far-away time" is not the same as "Schwarzschild time". Do any of those references actually say that the hole will evaporate before anyone falls in? If so, please give an explicit quote and link. It looks to me like you are misinterpreting what the references are telling you.

finite far-away time, which occurs before Bob enters BH.

No, it does not. This has already been addressed repeatedly. There is no point in continuing to ask a question that has already been answered.

Maybe you could give me a link that states black holes take infinite far-away time to evaporate by Hawking radiation if the universe is open?

We can't because that statement is false. The hole does evaporate in finite "far-away time". But people can fall into the hole in a finite "far-away time" that is much shorter than the far-away time it takes for the hole to evaporate. Again, this has already been stated.

The question you should be asking is, how can someone fall into the hole in finite "far-away time"? The answer is that you need to use appropriate coordinates, such as the ones I mentioned previously (Painleve or Eddington-Finkelstein), which are not singular at the horizon, so they can be used to assign a well-defined "far-away time" to events near or at or inside the horizon. Schwarzschild coordinates cannot do this because they are singular at the horizon, so you can't use Schwarzschild coordinates as your "far-away time" if you are trying to analyze events near or at or inside the horizon.

 

[stevendaryl]

No one seems to be bothered by this except me:

Black holes have a finite lifetime measured in Schwartzchild time due to Hawking radiation. Similarly, the universe probably has a finite lifetime measured in Schwartzchild time. In that case, nothing ever falls through the event horizon of a black hole because it takes infinite Schwartzchild time so the black hole would disappear and the universe would come to an end before you crossed the event horizon. You could even calculate dr/dt of the event horizon of a black hole shrinking due to Hawking radiation which would presumably be non-zero and presumably negative. Since an infalling observer's dr/dt approaches zero at the event horizon, as you fall towards the event horizon it will shrink away from you as you approach and you will never cross it. So why does everyone spend time analyzing what happens after you cross the event horizon, lured by the fact that proper time is finite? I really just don't understand - I'm not trying to be clever - I just want to know what I am missing here.

You have two events: (1) The infalling observer crosses the event horizon. (2) The black hole evaporates. Obviously, event (1) must happen before event (2), if it happens at all.

If there were no Hawking radiation, then for events taking place outside the event horizon, you can use Schwarzschild coordinates, and event (1) never happens. But "never happens" just means that there is no [itex]t[/itex] coordinate for that event. The Schwarzschild coordinates just don't cover that event. It happens, but just not at a finite [itex]t[/itex] coordinate. To see that this doesn't really mean that it doesn't happen, you can compare the Schwarzshild coordinates with the Rindler coordinates:

[itex]X = \sqrt{x^2 - c^2 t^2}[/itex]
[itex]T = tanh^{-1}(\frac{gt}{cx})[/itex]

where [itex]x[/itex] and [itex]t[/itex] are the usual inertial coordinates.

An object dropped from the point [itex]X_0, T_0[/itex] will approach the point [itex]X=0[/itex] as [itex]T \rightarrow \infty[/itex], but it will never reach it in Rindler coordinates. But it will reach it at time [itex]t=X_0/c[/itex] in inertial coordinates.

So getting back to black holes, event (1) does happen, it's just that it isn't covered by Schwarzschild coordinates. What about event (2)? Well, Schwarzschild coordinates are appropriate for eternal black holes. For evaporating black holes, you have to use some other coordinate system. If you pick a coordinate system such that the evaporation takes place at a finite time, then event (1) will also take place at a finite time. What'll happen is that the event horizon will shrink with time, and the infalling observer will hover just outside. The black hole will disappear when the event horizon shrinks to nothing, and at that moment, the infalling observer will be squashed into zero volume, also.

 

[PeterDonis]

What'll happen is that the event horizon will shrink with time, and the infalling observer will hover just outside.

If you mean this is what the infalling observer's worldline will actually be, this is not correct. The infalling observer will fall through the horizon long before the hole evaporates, and will be destroyed in the singularity. (At least, this is what will happen in the original model of a hole evaporating by Hawking radiation, the one Hawking originally worked with. There are lots of other proposed models where different things happen, but I don't think we need to open that can of worms in this thread. ) If we use appropriate coordinates, such as ingoing Eddington-Finkelstein coordinates, the infalling observer will fall through the horizon at an Eddington-Finkelstein time that is long before the Eddington-Finkelstein time at which the hole evaporates, and will hit the singularity at an E-F time that is somewhat later, but still long before the E-F time at which the hole evaporates.

One could, however, set up a sort of "Schwarzschild-like" coordinate chart in which the "time" coordinate was assigned based on when light signals are seen by a far-away observer. (This would actually be more like outgoing Eddington-Finkelstein coordinates.) In such a chart, the infalling observer would appear to "hover" outside the horizon until the hole evaporated, because the light signals he emitted just before reaching the horizon would be extremely delayed getting out to the far-away observer; they would reach that observer just before the light from the hole's final evaporation does. Light emitted by the infalling observer at the instant he crossed the horizon would reach the far-away observer at the same instant as the light from the hole's final evaporation, and so would be assigned the same "time" coordinate in this chart.

However, in this latter chart, no time could be assigned to any event inside the horizon, including the event at which the infalling observer hits the singularity. The far-away observer will never see any of those events, because the light from them will have hit the singularity. The event horizon, while it exists, is still an event horizon; light from events inside it can never get out. When the far-away observer sees the light emitted by the infalling observer as he crosses the horizon, those signals will show that observer in the (presumably) normal condition he was in when he crossed the horizon; the far-away observer will never see him being crushed in the singularity. It will just look like the infalling observer vanished at the instant he crossed the horizon.

 

[stevendaryl]

If you mean this is what the infalling observer's worldline will actually be, this is not correct. The infalling observer will fall through the horizon long before the hole evaporates, and will be destroyed in the singularity.

I am talking about from the point of view of someone observing the saga from a safe distance. I realize that that doesn't uniquely determine a coordinate system. However, during the time when the black hole is shrinking slowly, I would think that a distant observer could use, throughout most of the lifetime of the black hole, use a coordinate system that is Schwarzschild coordinates with a slow time-dependence in the [itex]M[/itex] parameter. In such a coordinate system, the infalling observer would be hovering outside the black hole.

If we use appropriate coordinates, such as ingoing Eddington-Finkelstein coordinates, the infalling observer will fall through the horizon at an Eddington-Finkelstein time that is long before the Eddington-Finkelstein time at which the hole evaporates, and will hit the singularity at an E-F time that is somewhat later, but still long before the E-F time at which the hole evaporates.

One could, however, set up a sort of "Schwarzschild-like" coordinate chart in which the "time" coordinate was assigned based on when light signals are seen by a far-away observer. (This would actually be more like outgoing Eddington-Finkelstein coordinates.) In such a chart, the infalling observer would appear to "hover" outside the horizon until the hole evaporated, because the light signals he emitted just before reaching the horizon would be extremely delayed getting out to the far-away observer; they would reach that observer just before the light from the hole's final evaporation does. Light emitted by the infalling observer at the instant he crossed the horizon would reach the far-away observer at the same instant as the light from the hole's final evaporation, and so would be assigned the same "time" coordinate in this chart.

That's what I mean.

However, in this latter chart, no time could be assigned to any event inside the horizon, including the event at which the infalling observer hits the singularity. The far-away observer will never see any of those events, because the light from them will have hit the singularity. The event horizon, while it exists, is still an event horizon; light from events inside it can never get out. When the far-away observer sees the light emitted by the infalling observer as he crosses the horizon, those signals will show that observer in the (presumably) normal condition he was in when he crossed the horizon; the far-away observer will never see him being crushed in the singularity. It will just look like the infalling observer vanished at the instant he crossed the horizon.

Yes, the importance of the "Schwarzschild-like" coordinates is just to understand that there is no contradiction between the black hole evaporating and the infalling observer never crossing the event horizon. From the point of view of this coordinate system, the evaporation and the crushing of the infalling observer are simultaneous.

 

[PAllen]

It is not so clear to me what 'SC equivalent coordinates' would mean for an evaporating BH. Classical models are usually based on the Vaidya metric, and the P&P paper Pervect mentions states that despite quantum issues near the horizon, the distant region must be described by a Vaidya metric.

SC coordinates for the SC spacetime are similar to radar coordinates, and cover the same spacetime region. In radar coordinates defined by a distant observer, it seems to me that something interesting happens to the labeling of an infaller world line in the case of an evaporating BH. Light trapped on the horizon would be released simultaneous to the final BH evaporation. This means (following the way radar assign coordinate time), the actual event of infaller reaching the horizon would be given a finite coordinate time before the event final evaporation (because the inward light reaching infaller at horizon would start much earlier than an inward light path reaching the horizon at moment of evaporation). There would still be no labeling events inside the horizon. The world line of the infaller would just cease at some finite coordinate time corresponding to horizon crossing, at the horizon radius at the time of crossing. The energy associated with the infaller would then come out later in the Hawking radiation.

 

[PeterDonis]

during the time when the black hole is shrinking slowly, I would think that a distant observer could use, throughout most of the lifetime of the black hole, use a coordinate system that is Schwarzschild coordinates with a slow time-dependence in the ##M## parameter.

Far away from the hole, you can do this; but if you try to extend such a chart close to the horizon, you will find that it doesn't work; you can't have ##M## depend on just the ##t## coordinate. It has to depend on both ##t## and ##r##. If you want ##M## to just be a function of the "time" coordinate, you have to use something like Eddington-Finkelstein coordinates, as I said before.

If you're ok with ##M## depending on ##r## as well as ##t##, then you can use these "Schwarzschild-type" coordinates, but in these coordinates, the ##r## coordinate of the horizon changes with time, which makes a difference. See below.

Also, remember that this spacetime is not a vacuum solution; the Hawking radiation has a nonzero stress-energy tensor. This sort of spacetime is actually more like the outgoing Vaidya solution than the Schwarzschild black hole solution. (PAllen has already mentioned this.)

In such a coordinate system, the infalling observer would be hovering outside the black hole.

Not quite. In these coordinates, as I said above, the ##r## coordinate of the horizon decreases with time; so while the infalling observer might appear to "slow down" as he falls, he won't "hover", because, roughly speaking, the horizon is "falling" too.

From the point of view of this coordinate system, the evaporation and the crushing of the infalling observer are simultaneous.

No, they aren't. From the point of view of this coordinate system, the evaporation and the falling through the horizon of the infalling observer are simultaneous. This coordinate system cannot assign a time to the event where the infalling observer hits the singularity and is crushed (or to any event inside the horizon).

 

[PAllen]

It is not so clear to me what 'SC equivalent coordinates' would mean for an evaporating BH. Classical models are usually based on the Vaidya metric, and the P&P paper Pervect mentions states that despite quantum issues near the horizon, the distant region must be described by a Vaidya metric.

SC coordinates for the SC spacetime are similar to radar coordinates, and cover the same spacetime region. In radar coordinates defined by a distant observer, it seems to me that something interesting happens to the labeling of an infaller world line in the case of an evaporating BH. Light trapped on the horizon would be released simultaneous to the final BH evaporation. This means (following the way radar assign coordinate time), the actual event of infaller reaching the horizon would be given a finite coordinate time before the event final evaporation (because the inward light reaching infaller at horizon would start much earlier than an inward light path reaching the horizon at moment of evaporation). There would still be no labeling events inside the horizon. The world line of the infaller would just cease at some finite coordinate time corresponding to horizon crossing, at the horizon radius at the time of crossing. The energy associated with the infaller would then come out later in the Hawking radiation.

Following this up, the key point I intended is that contrary to the SC spacetime, radar coordinates for a distant observer of an evaporating BH do not give infinite coordinate time to horizon crossing or horizon formation. Both coordinate times are finite [due to final evaporation 'releasing' frozen outward null geodesics], and causal ordering is as expected: horizon formation precedes later infaller crossing horizon, which precedes final evaporation.

 

[stevendaryl]

No, they aren't. From the point of view of this coordinate system, the evaporation and the falling through the horizon of the infalling observer are simultaneous. This coordinate system cannot assign a time to the event where the infalling observer hits the singularity and is crushed (or to any event inside the horizon).

Right, but after evaporation, the infalling observer is effectively gone from the universe of the external observer. Whether or not you assign a time to his demise, there is a time after which he's beyond interacting with.

 

[stevendaryl]

Following this up, the key point I intended is that contrary to the SC spacetime, radar coordinates for a distant observer of an evaporating BH do not give infinite coordinate time to horizon crossing or horizon formation. Both coordinate times are finite [due to final evaporation 'releasing' frozen outward null geodesics], and causal ordering is as expected: horizon formation precedes later infaller crossing horizon, which precedes final evaporation.

Okay, here's a question. If we temporarily forget about Hawking radiation, we can use the Schwarzschild metric to compute two numbers of an infalling observer:

1. The time [itex]\tau_1[/itex] on his watch when he crosses the event horizon.
2. The time [itex]\tau_2[/itex] on his watch when he smashes into the singularity. (Ignoring the fact that his watch will be smashed before he reaches the singularity)

So if we assume that Hawking radiation is a tiny perturbation, then those two numbers are not changed much by evaporation.

The question for a distant observer is this: Does evaporation allow the light from these two events reach a distant observer?

 

[PeterDonis]

Does evaporation allow the light from these two events reach a distant observer?

For event 1, yes. For event 2, no.

 

[Asher Weinerman]

Thank you everybody - especially to pervect who gave to me the links I requested, and to stevendaryl who answered in a respectful tone.

I have a master's degree in particle physics but have always been interested in GR - I'm half finished the excellent introductory workbook by Thomas Moore. For the record I'm not convinced by the arguments presented here, and I'm sorry if I angered some of you with my persistence and naivete. Maybe after a few years of experience I will be as confident as you are about the theory of black holes.

 

[PeterDonis]

contrary to the SC spacetime, radar coordinates for a distant observer of an evaporating BH do not give infinite coordinate time to horizon crossing or horizon formation. Both coordinate times are finite [due to final evaporation 'releasing' frozen outward null geodesics], and causal ordering is as expected: horizon formation precedes later infaller crossing horizon, which precedes final evaporation.

Hmm. I see what you are saying, but I'm still not sure that these coordinates are "what you get if you adapt standard Schwarzschild coordinates to an evaporating black hole spacetime". More precisely, I'm not sure these coordinates are the unique result of doing that.

The key feature of standard Schwarzschild spacetime (more precisely, of the region outside the horizon) that standard Schwarzschild coordinates "match up" with is the timelike Killing vector field. Each surface of constant Schwarzschild coordinate time is orthogonal to the integral curves of this KVF. The reason Schwarzschild coordinates become singular at the event horizon is that the KVF there becomes null.

It also happens to be true that Schwarzschild coordinates in this case match up with "radar coordinates" for a faraway observer, because since the spacetime is stationary, round-trip light paths are time-symmetric; the event at which such a path is reflected, at some finite radius above the horizon, is "halfway between" the events of emission and reception far away because the ingoing path is just the time reverse of the outgoing path.

In an evaporating black hole spacetime, we no longer have a timelike KVF because the spacetime is no longer stationary. Also, ingoing and outgoing light paths are no longer time-symmetric. But I'm not sure that these two failures of time symmetry match up, so to speak; that is, if I do my best to pick out surfaces of "constant time" that are orthogonal to the closest thing I have to integral curves of a timelike KVF, I'm not sure those will be the same as surfaces of constant radar time. An obvious example is events on the horizon: the horizon is a null surface and so is itself orthogonal to the "timelike KVF" pointing along it, so all events on the horizon should have the same "orthogonal time", just as they do in standard Schwarzschild spacetime (the only difference is that this time is now finite instead of plus infinity, because there is now a region of the spacetime to the future of the null surface I get when I extend the horizon out past the hole's final evaporation); but as you point out, events on the horizon will have different radar times.

None of this possible ambiguity about coordinates affects the actual physics, so I'm not sure how much it's worth belaboring it. My personal preference would be to use one of the other coordinate charts I mentioned, where the ambiguity does not arise.

 

[PAllen]

Hmm. I see what you are saying, but I'm still not sure that these coordinates are "what you get if you adapt standard Schwarzschild coordinates to an evaporating black hole spacetime". More precisely, I'm not sure these coordinates are the unique result of doing that.

I never implied they were. They are not the same as SC coordinates even for SC geometry (since they are normally built around one chosen timelike world line treated as a time axis). However, they have the virtue of being defined by simple physical procedure (that couples to two way causal connection), and have similar global characteristics as SC coordinate for SC geometry. They cover the same spacetime region, and they give time coordinate approaching infinity as an infaller approaches the horizon. Thus, they have just as good a claim to describing distant experience as SC coordinates. Then, given that, I ask what they say for an evaporating BH.

 

[Khashishi]

I know very little on this subject, but, if the Unruh effect exists, then I think it is the apparent acceleration of the black hole's horizon that causes it to emit radiation. The Unruh effect is a hypothesis that an accelerating observer sees radiation coming from the vacuum. If you are free-falling into a black hole, then I figure you don't see Hawking radiation. Is this logically consistent?

 

[PAllen]

I know very little on this subject, but, if the Unruh effect exists, then I think it is the apparent acceleration of the black hole's horizon that causes it to emit radiation. The Unruh effect is a hypothesis that an accelerating observer sees radiation coming from the vacuum. If you are free-falling into a black hole, then I figure you don't see Hawking radiation. Is this logically consistent?

There is some debate about this, but certainly many GR/QG experts have argued that a free faller does not detect Hawking radiation.

 

[m4r35n357]

Maybe it's a bit late to add anything here, but I think there might be an elephant in the room (c.f. Hawking radiation is really really tiny). The Schwarzschild solution (SC) is static, and is also a single body solution. Both these aspects prevent proper consideration of "horizon crossing", and we should not be surprised to see "paradoxes" like this appear. In numerical simulations the infalling particle eventually "merges" with the black hole and the event horizon expands to engulf both. I am assuming here for the sake of science that this effect is detectable externally, at least in theory, since "horizon crossing" is something that we can only theorize about from the outside.
But is seems maybe that inferring "horizon crossing" times from the SC is not really useful and has had its day . . . . ?

 

[PeterDonis]

The Schwarzschild solution (SC) is static, and is also a single body solution. Both these aspects prevent proper consideration of "horizon crossing"

How so?

In numerical simulations the infalling particle eventually "merges" with the black hole and the event horizon expands to engulf both.

No, the infalling particle just falls in; it doesn't "merge" with anything, because there's nothing to "merge" with--the black hole is vacuum everywhere. (The particle does eventually reach the singularity and is destroyed there, but this is not a "merger" in any useful sense.)

The event horizon, if you insist on thinking of it as a "place", starts expanding outward before the infalling particle reaches it; the horizon just reaches its new radius (the radius derived from the mass of the original hole plus the mass of the infalling particle) at the instant the infalling particle reaches that radius (and therefore reaches the horizon).

It is true that you can't model these effects in Schwarzschild coordinates, but that doesn't mean you can't model them; it just means you need to pick better coordinates. In Painleve or Eddington-Finkelstein coordinates, there is no coordinate singularity at the horizon and all these phenomena can be modeled just fine.

I am assuming here for the sake of science that this effect is detectable externally, at least in theory, since "horizon crossing" is something that we can only theorize about from the outside.

Of course the effect is detectable externally; the mass of the hole increases.

 

[m4r35n357]

How so?

I mean that it is static, in that neither it's mass nor its horizon can grow; it cannot even form, it just is. There is only one body so the falling particle does not exist either. I think that's all I meant ;)

No, the infalling particle just falls in; it doesn't "merge" with anything, because there's nothing to "merge" with--the black hole is vacuum everywhere. (The particle does eventually reach the singularity and is destroyed there, but this is not a "merger" in any useful sense.)

I try not to agonize too much about what happens inside a horizon, I treat it as a mathematical issue not a physical one, since no measurements can be made from the outside.

The event horizon, if you insist on thinking of it as a "place", starts expanding outward before the infalling particle reaches it; the horizon just reaches its new radius (the radius derived from the mass of the original hole plus the mass of the infalling particle) at the instant the infalling particle reaches that radius (and therefore reaches the horizon).

I agree totally completely 100% with that, I thought I was clear about it but maybe not.

It is true that you can't model these effects in Schwarzschild coordinates, but that doesn't mean you can't model them; it just means you need to pick better coordinates. In Painleve or Eddington-Finkelstein coordinates, there is no coordinate singularity at the horizon and all these phenomena can be modeled just fine.
Of course the effect is detectable externally; the mass of the hole increases.

I tend to treat Gullstrand-Painleve, Eddington-Finklestein and Kruskal–Szekeres etc. as transformations of the SC, and therefore equally static (no explicit dependence on time), but maybe that's my mistake, or one of them. If so, I have misunderstood the meaning of Birkhoff's theorem, which is also quite possible.

BTW, I seem to have messed up the quoting a bit, but I think it's possible to follow . . . (Mentor note: fixed.)

 

[PeterDonis]

I mean that it is static, in that neither it's mass nor its horizon can grow; it cannot even form, it just is. There is only one body so the falling particle does not exist either.

Ah, ok. Yes, this is true; strictly speaking, Schwarzschild spacetime is a model of a single black hole that is "eternal", it never changes. So you can't use it as a model of something falling into a black hole and increasing its mass. (You also can't use it as a model of a hole evaporating by Hawking radiation, as has been pointed out already.)

However, you can model holes with slowly increasing mass due to infalling matter, or holes evaporating very slowly due to Hawking radiation, using spacetimes that are "very close" to Schwarzschild spacetime, so that the structure of the models is very close to being a series of slices of Schwarzschild spacetimes with slowly increasing or decreasing mass. The description I gave of what happens to the event horizon when an object falls into a black hole is taken from such a model.

I treat it as a mathematical issue not a physical one, since no measurements can be made from the outside.

If you're outside the horizon, you can't receive signals from inside, yes. But you can still see indirect effects, such as the mass of the hole, and the fact that objects fall into that region of spacetime and never come out.

I tend to treat Gullstrand-Painleve, Eddington-Finklestein and Kruskal–Szekeres etc. as transformations of the SC, and therefore equally static (no explicit dependence on time)

"Static" is a property of the spacetime, not the coordinates; the spacetime is static even if we choose coordinates where the metric coefficients do depend on the "time" coordinate (which they do, for example, in Kruskal coordinates).

 

[m4r35n357]

OK, thanks for clearing up a couple of points. Where my mind is leading me is along these lines. I was watching some numerical simulations of merging masses (black holes) and it occurred that this model of merging "horizons" is much nicer to visualize than infinite red-shifting/smearing at a static "horizon". I'm quoting the word horizon because I've always associated it with the infinite future, and I'm not sure how or even whether the concept of future infinity arises in a simulation. I suppose it must because they plotted it. So I've replaced one uncertainty with another!
I suppose I'll have to look into collapsing dust equations . . . feel free to close this if I've said nothing of interest!

 

[Asher Weinerman]

The Schwarzschild solution (SC) is static, and is also a single body solution. Both these aspects prevent proper consideration of "horizon crossing", and we should not be surprised to see "paradoxes" like this appear. In numerical simulations the infalling particle eventually "merges" with the black hole and the event horizon expands to engulf both.

Maybe the above quote contains the answer I'm looking for. In numerical simulations, will the infalling particle merge with the black hole and the event horizon expand to engulf both IN A FINITE AMOUNT OF FAR_AWAY TIME?

 

[m4r35n357]

Maybe the above quote contains the answer I'm looking for. In numerical simulations, will the infalling particle merge with the black hole and the event horizon expand to engulf both IN A FINITE AMOUNT OF FAR_AWAY TIME?

According to these simulations, yes. The relevance/accuracy of the computer models and initial/boundary conditions is another matter though.

 

[Asher Weinerman]

According to these simulations, yes.

Thank-you. I now understand that the infinite (far-away) time calculation is for a single body, and produces a paradoxical result. A correct two body treatment requires a computer simulation and results in a FINITE (far-away) time for the horizon crossing because the horizon itself expands to engulf the infalling particle.

I knew the answer should be simple, but I couldn't find it anywhere!

Thanks again,
Asher

 

[PeterDonis]

In numerical simulations, will the infalling particle merge with the black hole and the event horizon expand to engulf both IN A FINITE AMOUNT OF FAR_AWAY TIME?

The ingoing Vaidya metric is a closed form solution that shows the same effect; it models spherically symmetric radiation falling into a black hole, and the hole's horizon radius increases as the radiation falls in.

(The outgoing Vaidya metric, which has been mentioned in this thread as being a suitable idealized model for a black hole emitting Hawking radiation, is just the time reverse of the above; as the radiation goes out, the horizon radius of the hole decreases.)