Infinite Time Dilation at the Surface of a Black Hole?

[bhagwad]

I've never fully understood how anything can actually fall into a black hole without the black hole evaporating first. Since time dilates exponentially as I fall into a black hole, a point will come where a few seconds for me will be millions of years in the outside world...trillions in fact. Perhaps long enough for the black hole to evaporate?

Let me set up an experiment as precisely as a I can to clarify my question.

Assumptions:

1) I am indestructible. Nothing short of the utter dismantling of space itself can destroy me
2) I have an almost infinitely powerful rocket capable of generating insane amounts of thrust and with a practically unlimited energy supply.

Here's the scenario:

I first set up a clock at a safe distance away from the black hole. I engineer it so that it has just one hand and each complete circle of the hand measures what I experience as one year. For each rotation, a counter increases and there is no limit to the number this counter can reach as long as the clock keeps working. And the clock has an inexhaustible energy supply.

Now, I turn around and start heading towards the black hole. I go closer...closer...closer. The gravitational pull keeps increasing. But because of (1), I don't die. I come closer to the event horizon never actually crossing it. In fact, I come as close to it as theoretically possible. Because of (2), my rocket is capable of counteracting the gravitational force since it's still finite no matter how great.

I hold my position for 5 years of my local time. After 5 years have expired (for me), I give my rocket an extra boost and escape the clutches of the black hole (technically I suppose I was never "in" it) at all.

I now go back to my clock which is still ticking away happily at a safe distance.

Question: Does the clock counter show that millions of years have passed? Or billions? Or is the above experiment moot because the black hole itself would have evaporated out from under me but I still survive because of (1)?

Note: I'm not proposing any explanation or putting forward a theory. I freely admit that my knowledge of general relativity has a lot to be desired. I'm merely asking the question - what does the counter on my clock show that I set up before my trip. Of course, this is dependent on the size of the black hole etc, but I want to know whether the counter can be in the millions or billions.

 

[PeterDonis]

Since time dilates exponentially as I fall into a black hole, a point will come where a few seconds for me will be millions of years in the outside world...trillions in fact. Perhaps long enough for the black hole to evaporate?

If you were going to just fall into the hole, no. For the experiment you describe, perhaps, but see below.

I first set up a clock at a safe distance away from the black hole. I engineer it so that it has just one hand and each complete circle of the hand measures what I experience as one year.

I assume you mean, what you experience as one year while you are co-located with the clock.

I come closer to the event horizon never actually crossing it. In fact, I come as close to it as theoretically possible.

Classically, there's no limit to how close you could come, because classically, spacetime is a continuum, so there is no limit to how small a distance there can be.

When we include quantum mechanics, there should be a limit on how small a distance there can be, because spacetime should no longer act like a continuum on small enough scales. Typically the scale on which these effects become important is taken to be the Planck length, which is about 10^-35 meters. So you could not come within a Planck length or so of the horizon without falling inside.

Because of (2), my rocket is capable of counteracting the gravitational force since it's still finite no matter how great.

If the hole is a quantum hole, so that it emits Hawking radiation, then you will also have to withstand the radiation, which becomes arbitrarily high frequency as you get close to the hole's horizon. You assumed you were indestructible, so you can withstand it.

I hold my position for 5 years of my local time. After 5 years have expired (for me), I give my rocket an extra boost and escape the clutches of the black hole (technically I suppose I was never "in" it) at all.

That's right; you never crossed the horizon, so you never got to see what was "inside" the hole.

Question: Does the clock counter show that millions of years have passed? Or billions?

If the black hole was classical, it could show an arbitrarily large amount of elapsed time, depending on how close you got to the horizon. If it was a quantum hole, so that it was emitting Hawking radiation, it depends on the mass of the hole; see below.

Or is the above experiment moot because the black hole itself would have evaporated out from under me but I still survive because of (1)?

If you got close enough to the hole's horizon and stayed long enough, you would find that the mass of the hole was gradually decreasing during the time you spent close to the horizon; you would see this as a reduction in the rocket thrust you had to keep up to maintain altitude. It's possible, if the mass of the hole were small enough, that it could indeed evaporate completely; but as it did so, the relationship between your local clock time and the time on the clock you left behind at a safe distance would change. I haven't run the numbers to see if there is a maximum elapsed time you could see on the clock you left behind (which would depend on the mass the hole started with), but I think there could be one.

 

[Bill_K]

If the hole is a quantum hole, so that it emits Hawking radiation, then you will also have to withstand the radiation, which becomes arbitrarily high frequency as you get close to the hole's horizon.

Don't get this. Hawking radiation does not even come from the horizon.

 

[Nugatory]

The time dilation will behave as you expect: a clock that is moved very close to the event horizon, kept there for a time, and then accelerated back out to where you left the safe-distance clock will show less time having passed than the safe-distance one. Other time-dependent processes will be consistent with these readings as well; for example, if we have two identical twins, one of them stays with the safe-distance clock, and the other takes the journey almost to the event horizon and back, the traveler will have aged less when they get back together.

However, this doesn't tell us anything about falling into a black hole, because the traveler isn't falling - he has a powerful rocket that carries him back out of the gravity well before he crosses the event horizon. Indeed, this situation is just the general relativistic version of the well-known "twin paradox" of special relativity; you will want to be sure that that you completely understand the SR version before you take on the GR version.

But you started your post with a different question: Given the gravitational time dilation effect, would it be possible for something to actually fall into the black hole if it didn't have that powerful rocket to rescue it before it crossed the event horizon?

There are a bunch of threads on this question already, and the short answer is "yes, you fall in".

The longer answer is:
1) If you consider the eventual evaporation of the black hole : The safe-distance observer will eventually see you falling into the black hole, and then at some later time will see the final flash from the evaporation of the black hole. (The evaporation time of an astronomical black hole is easily fifty orders of magnitude greater than the age of the universe, so "eventually" is a long time).
2) If you don't consider the evaporation of the black hole, safe-distance observer will never see you falling into the black hole, because light from that event will never make it out to his eyes - but that doesn't mean that it didn't happen.

 

[bhagwad]

If you were going to just fall into the hole, no. For the experiment you describe, perhaps, but see below.



I assume you mean, what you experience as one year while you are co-located with the clock.



Classically, there's no limit to how close you could come, because classically, spacetime is a continuum, so there is no limit to how small a distance there can be.

When we include quantum mechanics, there should be a limit on how small a distance there can be, because spacetime should no longer act like a continuum on small enough scales. Typically the scale on which these effects become important is taken to be the Planck length, which is about 10^-35 meters. So you could not come within a Planck length or so of the horizon without falling inside.



If the hole is a quantum hole, so that it emits Hawking radiation, then you will also have to withstand the radiation, which becomes arbitrarily high frequency as you get close to the hole's horizon. You assumed you were indestructible, so you can withstand it.



That's right; you never crossed the horizon, so you never got to see what was "inside" the hole.



If the black hole was classical, it could show an arbitrarily large amount of elapsed time, depending on how close you got to the horizon. If it was a quantum hole, so that it was emitting Hawking radiation, it depends on the mass of the hole; see below.



If you got close enough to the hole's horizon and stayed long enough, you would find that the mass of the hole was gradually decreasing during the time you spent close to the horizon; you would see this as a reduction in the rocket thrust you had to keep up to maintain altitude. It's possible, if the mass of the hole were small enough, that it could indeed evaporate completely; but as it did so, the relationship between your local clock time and the time on the clock you left behind at a safe distance would change. I haven't run the numbers to see if there is a maximum elapsed time you could see on the clock you left behind (which would depend on the mass the hole started with), but I think there could be one.

Thank you for that. So if I understand you correctly, it's possible for my clock to have measured trillions of years if the black hole is reasonably sized is that correct?

 

[PeterDonis]

Don't get this. Hawking radiation does not even come from the horizon.

But it does get redshifted as it goes out to infinity, yes? So if you are close to the horizon, you will see the Hawking radiation as much higher frequency than an observer at infinity does.

 

[PeterDonis]

So if I understand you correctly, it's possible for my clock to have measured trillions of years if the black hole is reasonably sized is that correct?

For the clock you left at a safe distance, yes.

 

[Bill_K]

But it does get redshifted as it goes out to infinity, yes? So if you are close to the horizon, you will see the Hawking radiation as much higher frequency than an observer at infinity does.

Not if it originates above you. Remember that the predominant wavelength of the Hawking radiation is comparable to the size of the hole. In fact I believe that the amplitude of the radiation goes to zero as you approach the hole.

 

[PeterDonis]

Remember that the predominant wavelength of the Hawking radiation is comparable to the size of the hole.

But this is the wavelength as seen by an observer at infinity, correct? The wavelength as seen by an observer close to the horizon would be blueshifted. At least, that's what various physicists who talk about a hot "membrane" close to the horizon (or "stretched horizon", which is what Susskind, for example, calls it) seem to be saying.

 

[Nugatory]

Thank you for that. So if I understand you correctly, it's possible for my clock to have measured trillions of years if the black hole is reasonably sized is that correct?

Sure, but you don't need a black hole to get a clock to measure trillions of years - you just have to wait trillions of years, and that's what the safe-distance observer is doing. The black hole is just one way of creating a situation in which some other clock following a different path through space-time (in this case, passing very close to the event horizon so experiencing extreme time dilation) will correctly measure much less time on its path through space-time.

I said above that you really want to nail down the special relativity version of the twin paradox before you take on this general relativistic version... There's a pretty decent summary here: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

 

[yuiop]

Not if it originates above you. Remember that the predominant wavelength of the Hawking radiation is comparable to the size of the hole. In fact I believe that the amplitude of the radiation goes to zero as you approach the hole.

For Unruh radiation, the temperature is is proportional to the proper acceleration. The closer you get to a black hole the greater the proper acceleration required to hold station, so it would seem by the equivalence principle, that the temperature increases the nearer you are to a black hole. The temperature of a black hole due to Hawking radiation is usually quoted for the measurement at infinity.

While thinking about this I came up with this simple thought experiment that might prove interesting. Consider a 12V car battery with a 120 amphour rating that is connected to a 5 amp lamp rated at 60 watts. Anne is lowered with the battery and lamp to a level in a Schwarzschild metric where the gravitational time dilation is 10 time greater than that experienced by Bob higher up. Anne runs the lamp for 24 hours and runs the battery flat and so uses up approximately 1440 Watt Hours of energy.

The light is focused up to Bob in a tight beam so that he collects all the transmitted photons. Due to time dilation, Bob sees the energy arrive over a period of 240 hours at a rate of 6 watts per hour. The total energy received by Bob is 1440 Watt Hours. The interesting aspect is that photons are said to lose energy as they climb out of a gravitational well, but the number of photons received by Bob is identical to the number of photons sent by Anne and the energy sent and received is identical, so the energy carried by each photon is unchanged.<EDIT>Ooops, shooting from the hip again. Upon reflection, the energy carried by each photon must be reduced by the redshift factor, so the total energy E received by Bob is E'/gamma where E' is the energy sent by Anne. The power P or energy received per unit time by Bob is P'/gamma² where P' is the power as seen by Anne. This implies the power of the Hawking radiation increases by a factor of gamma² where gamma = √(1-2m/r), as you get closer to the black hole.

 

[nickb123]

This comment/question does not directly relate to the current topic. Someone I was talking to mentioned that 95% of the matter in the universe is unaccounted for. If this was the leading matter that was expelled in the big bang, it could explain why the universe is expanding at an accelerating rate (due to the gravitational force of this expanding spherical mass encompassing the universe).
If that is the case, I was wondering if the acceleration rate is increasing or decreasing (of course we can only determine if it was way back in time that we can currently see). Like a black hole, light could not escape inward (towards the visible universe).
If it is increasing, the expansion will eventually reach the point that the black hole mass dilutes to the point that can no longer sustain increased expansion acceleration.
On the other hand, if the acceleration rate is already decreasing, we may already be headed back to a singularity, once the acceleration reaches zero and contraction begins.

 

[dauto]

This comment/question does not directly relate to the current topic. Someone I was talking to mentioned that 95% of the matter in the universe is unaccounted for. If this was the leading matter that was expelled in the big bang, it could explain why the universe is expanding at an accelerating rate (due to the gravitational force of this expanding spherical mass encompassing the universe).
If that is the case, I was wondering if the acceleration rate is increasing or decreasing (of course we can only determine if it was way back in time that we can currently see). Like a black hole, light could not escape inward (towards the visible universe).
If it is increasing, the expansion will eventually reach the point that the black hole mass dilutes to the point that can no longer sustain increased expansion acceleration.
On the other hand, if the acceleration rate is already decreasing, we may already be headed back to a singularity, once the acceleration reaches zero and contraction begins.

There is no leading matter. Big Bang happened everywhere. There is no center for it to expand from and there is no edge for it to expand into.

 

[pervect]

Not if it originates above you. Remember that the predominant wavelength of the Hawking radiation is comparable to the size of the hole. In fact I believe that the amplitude of the radiation goes to zero as you approach the hole.

If we take the limit where the black hole horizon approaches a Rindler horizon (I'd call it the flat space limit, as it corresponds to a very large black hole with no appreciable tidal force), shouldn't the temperature of the local vacuum be the Unruh temperature associated with the very large acceleration the observer needs to hold station?

I'm going to snip some stuff from Wiki, I think it maybe is too distracting in this thread and I'm not sure if it's accurate.

 

[nickb123]

There is no leading matter. Big Bang happened everywhere. There is no center for it to expand from and there is no edge for it to expand into.

Is it known (or can it even be measured) if the universe expansion acceleration rate is increasing or decreasing? I understand that even if known, it would reflect a distant past condition.

 

[dauto]

Is it known (or can it even be measured) if the universe expansion acceleration rate is increasing or decreasing? I understand that even if known, it would reflect a distant past condition.

We know the universe is expanding.
We know the expansion rate is accelerating

That's all that's known for sure for now, based on direct observations

Models can be built in order to try to explain those facts
Within those models it is possible to calculate other variables and try to answer questions such as: "Is the acceleration of the expansion increasing or decreasing over time?". But the answers you get are model dependent and should be taken with a grain of salt. Until we have a better theoretical understanding about what causes the expansion acceleration, I wouldn't bet the farm on any of those models. Scientific progress is often slower than what we wish it would be. Gotta have patience.

 

[DKS]

But you started your post with a different question: Given the gravitational time dilation effect, would it be possible for something to actually fall into the black hole if it didn't have that powerful rocket to rescue it before it crossed the event horizon?

There are a bunch of threads on this question already, and the short answer is "yes, you fall in".

The longer answer is:
1) If you consider the eventual evaporation of the black hole : The safe-distance observer will eventually see you falling into the black hole, and then at some later time will see the final flash from the evaporation of the black hole. (The evaporation time of an astronomical black hole is easily fifty orders of magnitude greater than the age of the universe, so "eventually" is a long time).

I don't understand how the safe-distance observer can observe you falling through the event horizon (assuming that's what you meant), with or without Hawking radiation. At what time (say some fraction of the evaporation time) will the safe-distance observer observe this happening?
As you mention it is not observed classically, how does the Hawking radiation cause you to observe the falling in?

2) If you don't consider the evaporation of the black hole, safe-distance observer will never see you falling into the black hole, because light from that event will never make it out to his eyes - but that doesn't mean that it didn't happen.

Not sure what "did (not) happen" means exactly here. Should I interpret it as for the outside observer it never happens but for the falling observer it does? Since they can never compare notes there is no contradiction?

Kees

 

[PAllen]

Not sure what "did (not) happen" means exactly here. Should I interpret it as for the outside observer it never happens but for the falling observer it does? Since they can never compare notes there is no contradiction?

Kees

An observer never seeing something is a statement about light not 'reality', whatever that is. In the complete description of the universe in classical infall, crossing the horizon and approaching the singularity are part of the description. These are invariant facts of the complete description, not observer dependent.

That light = reality is a false concept is seen readily in SR. Consider a rocket uniformly accelerating away from earth, watching a someone drop a rock on earth. They would see the Earth fade to black, and the rock never reach the ground. Is this a feature of reality or a feature of light not being able to catch the rocket? A scientist on the rocket can readily model dropping a rock on Earth and conclude that it fell like any other rock, irrespective of light being able to catch the rocket.

The BH horizon behavior is completely equivalent.

 

[DKS]

An observer never seeing something is a statement about light not 'reality', whatever that is. In the complete description of the universe in classical infall, crossing the horizon and approaching the singularity are part of the description. These are invariant facts of the complete description, not observer dependent.

That light = reality is a false concept is seen readily in SR. Consider a rocket uniformly accelerating away from earth, watching a someone drop a rock on earth. They would see the Earth fade to black, and the rock never reach the ground. Is this a feature of reality or a feature of light not being able to catch the rocket? A scientist on the rocket can readily model dropping a rock on Earth and conclude that it fell like any other rock, irrespective of light being able to catch the rocket.

The BH horizon behavior is completely equivalent.

I don't get that. The rocket would see the rock falling and hitting the ground at some finite time, perhaps substantially redshifted, but you'd see it.

 

[Nugatory]

I don't get that. The rocket would see the rock falling and hitting the ground at some finite time, perhaps substantially redshifted, but you'd see it.

Google for "Rindler horizon" - as long as the rocket is uniformly accelerating, there is a region of spacetime from which a light signal will never catch up with the rocket.

(Of course if the rocket stops accelerating, starts coasting at a constant velocity, then the light will eventually reach it)

 

[DKS]

Google for "Rindler horizon" - as long as the rocket is uniformly accelerating, there is a region of spacetime from which a light signal will never catch up with the rocket.

(Of course if the rocket stops accelerating, starts coasting at a constant velocity, then the light will eventually reach it)

Thank you, now I understand what I didn't understand. Subtle stuff!

 

[Nugatory]

I don't understand how the safe-distance observer can observe you falling through the event horizon (assuming that's what you meant), with or without Hawking radiation. At what time (say some fraction of the evaporation time) will the safe-distance observer observe this happening?
As you mention it is not observed classically, how does the Hawking radiation cause you to observe the falling in?

The Hawking radiation causes the black hole to eventually evaporate. When it does, the light from me falling through the horizon (this light is moving radially outwards all along but doesn't escape as long as the black hole exists) finally escapes, makes it to the safe-distance observer's eyes.

Conversely, if the black hole never evaporates then the light from me falling through the horizon never gets out, never makes it to the safe-distance observer.

Not sure what "did (not) happen" means exactly here. Should I interpret it as for the outside observer it never happens but for the falling observer it does? Since they can never compare notes there is no contradiction?

It's much more straightforward than that. I do fall through the horizon and light from that event does not reach the safe-distance observer's eyes as long as the black hole is there. There's no contradiction between these two statements.

 

[DKS]

The Hawking radiation causes the black hole to eventually evaporate. When it does, the light from me falling through the horizon (this light is moving radially outwards all along but doesn't escape as long as the black hole exists) finally escapes, makes it to the safe-distance observer's eyes.

What I still don't understand is that for the far-away observer you fall through the horizon at a time $t =∞ > T_e$ where $t$ is the Schwarzschild coordinate, which coincides (almost) with proper time for the distant observer, and $T_e$ is the evaporation time (in Schwarzschild coordinates). So it seems to me that no light from you falling through the horizon will ever reach me, as you will not have fallen through the horizon yet at $t = T_e$. If you survive the explosion you'd tell the distant observer that you saw the black hole explode before you could fall in.

What am I doing wrong?

 

[PeterDonis]

What I still don't understand is that for the far-away observer you fall through the horizon at a time $t =∞ > T_e$ where $t$ is the Schwarzschild coordinate, which coincides (almost) with proper time for the distant observer, and $T_e$ is the evaporation time (in Schwarzschild coordinates).

No, this is not correct. You only fall through the horizon at $t = \infty$ if the black hole never evaporates. If the hole evaporates, you fall through the horizon at $t = T_e$, the evaporation time, according to the far-away observer. This is basically another way of saying what Nugatory was saying.

 

[DKS]

What I still don't understand is that for the far-away observer you fall through the horizon at a time $t=∞>T_e$ where $t$ is the Schwarzschild coordinate, which coincides (almost) with proper time for the distant observer, and $T_e$ is the evaporation time (in Schwarzschild coordinates).

__________________

No, this is not correct. You only fall through the horizon at $t = \infty$ if the black hole never evaporates. If the hole evaporates, you fall through the horizon at $t = T_e$, the evaporation time, according to the far-away observer. This is basically another way of saying what Nugatory was saying.

But at $t = T_e$ there is no longer an event horizon and Hawking's calculation breaks down shortly before $T_e$ (say at time $t_q=T_e - \Delta T$ with $\Delta T$ something like the Planck time) as quantum gravity will kick in.

Would it be correct to say you don't fall through the horizon till time $t_q$ and we don't know what happens after that?

 

[PeterDonis]

But at $t = T_e$ there is no longer an event horizon

Yes, but that just means the light from events that happened on what used to be the horizon can now escape.

Hawking's calculation breaks down shortly before $T_e$ (say at time $t_q=T_e - \Delta T$ with $\Delta T$ something like the Planck time) as quantum gravity will kick in.

Hawking's calculation itself requires at least some aspects of quantum gravity; but I think what you mean to say here is "the *full* theory of quantum gravity, rather than the approximation that Hawking used for his calculation". This is true, but I'm not sure how it affects the light emitted by you at the event of you falling through the horizon, or the time assigned to that event by the far-away observer. See below.

Would it be correct to say you don't fall through the horizon till time $t_q$ and we don't know what happens after that?

I don't think so, because whatever the final fate of the hole is, it will be confined to a tiny region of spacetime--basically one Planck length in spatial extent and one Planck time in temporal extent. That should have negligible effect on the spacetime as a whole, including what the far-away observer sees and how he assigns time coordinates to events like you falling through the horizon.

In other words, to the far-away observer, there is negligible difference between assigning time coordinate $t_q$ or $T_e$ to the event of you falling through the horizon. Either way he's going to see a flash of light containing images of you falling through the horizon and (presumably--see below) the hole evaporating, emitted at $T_e$ and reaching him one light travel-time later.

It's true that some theorists believe that, when we have a full theory of quantum gravity, it will say that the hole doesn't finally evaporate: either it leaves a Planck-size "remnant" or it turns into something else (like a baby universe). I don't think any of these speculative possibilities affect the fact that, as far as the rest of our universe is concerned, whatever happens is confined to a Planck-size region of spacetime, as I said above. It might affect exactly what is contained in the flash of light that is seen by the far-away observer; but I don't think it affects the time the far-away observer ends up assigning to the event of you falling through the horizon.

 

[DKS]

In other words, to the far-away observer, there is negligible difference between assigning time coordinate $t_q$ or $T_e$ to the event of you falling through the horizon. Either way he's going to see a flash of light containing images of you falling through the horizon and (presumably--see below) the hole evaporating, emitted at $T_e$ and reaching him one light travel-time later.

Why? At time $t_q$ you haven't passed the horizon yet and after that no one knows what'll happen. At any time $t < t_q$ you can turn on your rocket and arrive back at the distant observer before the flash of the evaporation.

 

[PeterDonis]

At time $t_q$ you haven't passed the horizon yet

No, at time $t_q$ according to the far-away observer's coordinates you haven't passed the horizon yet. This is a big difference; $t_q$ is not an "absolute" time when you haven't crossed the horizon. It's just a time coordinate assigned by the far-away observer to an event where you haven't yet crossed the horizon.

Furthermore, it's a time coordinate which is highly distorted in the vicinity of the horizon, to the point of being infinitely distorted *at* the horizon. Here's what that means: the time coordinate $T_e$ (the time the far-away observer assigns to the black hole's final evaporation) does *not* label a single event. It labels an infinite sequence of events, all of which take place on the horizon, and all of which are distinct.

For example, suppose you fall through the horizon, and then someone else falls in after you. You will each cross the horizon at distinct events: i.e., you will be spatially separated, and this will be obvious to both of you, and the someone else will see you falling through the horizon when he himself falls through, so it will be clear to him that you fell in first, i.e., the events of your two crossings of the horizon are separated in time. However, *both* of those events will be labeled with the time coordinate $T_e$ by the far-away observer! (And these are just two of the infinite number of events all of which are labeled with time $T_e$ by the far-away observer.) And the flash of light that the far-away observer sees when the black hole finally evaporates will contain images of *both* of you falling through the horizon. (And at time $t_q$ according to the far-away observer, both of you will be close to the horizon but not yet have fallen through.)

I could quote more highly counterintuitive facts about events on or near the horizon, but the above should be enough to show that you have to be very careful basing arguments on the time coordinate that the far-away observer assigns to events on or close to the horizon.

after that no one knows what'll happen.

This statement is much too strong, IMO. We don't have a full theory of quantum gravity, but that doesn't mean we know nothing about what will happen in this scenario. As I said in my previous post, whatever happens in the full quantum gravity regime will be confined to a Planck-sized piece of spacetime; that leaves a lot that *is* known about what will happen.

At any time $t < t_q$ you can turn on your rocket and arrive back at the distant observer before the flash of the evaporation.

According to the distant observer, yes; but if you actually do this, then you never cross the horizon at all, and this whole discussion is irrelevant.

 

[DKS]

>>>At any time $t<t_q$ you can turn on your rocket and arrive back at the distant observer before the flash of the evaporation.

According to the distant observer, yes; but if you actually do this, then you never cross the horizon at all, and this whole discussion is irrelevant.

My point is if you can do it, you can not have crossed the horizon at any time $t<t_q$, per definition of the horizon.
After $t_q$ unknown quantum gravity effects kick in. If we call $\tau(t)$ the proper time of the infalling observer, then he has not experienced crossing the horizon at $\tau(t)$ for $t<t_q$, and after that will experience unknown quantum gravity effects.

It still seems to me you can't cross the horizon before quantum gravity kicks in, from both perspectives.

 

[PeterDonis]

My point is if you can do it, you can not have crossed the horizon at any time $t<t_q$, per definition of the horizon.

This is incorrect as you state it, because time is relative. You can't have crossed the horizon at any time $t < t_q$ according to the far-away observer. But the far-away observer's time coordinate is highly distorted near the horizon. There are other time coordinates which are *not* distorted that way, and according to such a time coordinate, the infaller will have fallen through the horizon long before the hole finally evaporates. See further comments below.

If we call $\tau(t)$ the proper time of the infalling observer, then he has not experienced crossing the horizon at $\tau(t)$ for $t<t_q$

This is technically correct, but it is highly misleading as you state it. You apparently did not read carefully this portion of my previous post:

For example, suppose you fall through the horizon, and then someone else falls in after you. You will each cross the horizon at distinct events: i.e., you will be spatially separated, and this will be obvious to both of you, and the someone else will see you falling through the horizon when he himself falls through, so it will be clear to him that you fell in first, i.e., the events of your two crossings of the horizon are separated in time. However, *both* of those events will be labeled with the time coordinate $T_e$ by the far-away observer! (And these are just two of the infinite number of events all of which are labeled with time $T_e$ by the far-away observer.) And the flash of light that the far-away observer sees when the black hole finally evaporates will contain images of *both* of you falling through the horizon. (And at time $t_q$ according to the far-away observer, both of you will be close to the horizon but not yet have fallen through.)

In other words, there are an infinite number of possible functions $\tau(t)$, each describing the proper time of a *different* observer that falls through the horizon at a *different* event, and *all* of these events are *different* from the event of the hole's final evaporation. Furthermore, if we adopt a different time coordinate, one better suited to describing events at or near the horizon, then all of those "infaller" events will happen *before* the hole's final evaporation; they will happen at times (*different* times for each one) when the hole is still large and there are no quantum gravity effects at the horizon. (And we can verify that this different time coordinate is better suited by computing invariants along each of the infalling worldlines, such as the area of the horizon when each infaller falls through it, and verifying that those invariants confirm that the hole *is* large when each infaller falls through.)

But *all* of those different events will be labeled with the time coordinate $T_e$ by the far-away observer; and all of the different events one Planck length above the horizon along each of those different infalling worldlines will be labeled with the time coordinate $t_q$ by the far-away observer. The far-away observer's time coordinate is so distorted at the horizon that it makes an infinite family of different, distinct events look like one single event. So the far-away observer's time coordinate is *not* a good basis for understanding how things work at or near the horizon.

and after that will experience unknown quantum gravity effects.

No, he won't. See above.

 

[PeterDonis]

The following Wikipedia page contains a spacetime diagram of an evaporating black hole, which may help to illustrate what I've been saying in this thread:

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

The horizon is the 45-degree line going up and to the right from the middle of the left edge (marked r = 0) to the right end of the singularity (the jagged horizontal line). At the point where the horizon meets the singularity, the BH finally disintegrates; if we magnified this tiny area greatly, we would see the Planck-sized region of spacetime (one Planck length wide and one Planck time long) where quantum gravity effects come into play. But for the whole length of the horizon below that, the hole is large and there is plenty of room for lots of different observers to fall in without encountering any quantum gravity effects at the horizon.

A properly adapted time coordinate (such as the vertical direction in this diagram) distinguishes all these different possible infall points from the point of the hole's final evaporation; but the "natural" time coordinate of a far-away observer labels the *entire* horizon with the time coordinate $T_e$, and an entire 45-degree line one Planck length outside the horizon--which would appear just below it in this diagram--with the time coordinate $t_q$. That's why the far-away observer's time coordinate is not a good one to use for understanding what happens at the horizon or how the hole evaporates.

 

[DKS]

The following Wikipedia page contains a spacetime diagram of an evaporating black hole, which may help to illustrate what I've been saying in this thread:

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

The horizon is the 45-degree line going up and to the right from the middle of the left edge (marked r = 0) to the right end of the singularity (the jagged horizontal line). At the point where the horizon meets the singularity, the BH finally disintegrates; if we magnified this tiny area greatly, we would see the Planck-sized region of spacetime (one Planck length wide and one Planck time long) where quantum gravity effects come into play. But for the whole length of the horizon below that, the hole is large and there is plenty of room for lots of different observers to fall in without encountering any quantum gravity effects at the horizon.

A properly adapted time coordinate (such as the vertical direction in this diagram) distinguishes all these different possible infall points from the point of the hole's final evaporation; but the "natural" time coordinate of a far-away observer labels the *entire* horizon with the time coordinate $T_e$, and an entire 45-degree line one Planck length outside the horizon--which would appear just below it in this diagram--with the time coordinate $t_q$. That's why the far-away observer's time coordinate is not a good one to use for understanding what happens at the horizon or how the hole evaporates.

Thanks for the link.

Wikipedia is unreliable for this kind of stuff in my experience, is that really the correct Penrose diagram?

Recent literature I found on the topic appears not as confident as you are. For example http://arxiv.org/abs/gr-qc/0609024 Phys.Rev.D76:024005,2007 concludes that the hole evaporates before anything can fall in. This article is mentioned in a special wikipedia page where it is claimed it has been "refuted" without giving any reference. A citation search yields no refutation or severe criticism in any paper that cites it.

Do you have an opinion on that paper?

 

[Nugatory]

Do you have an opinion on that paper?

Krauss and company are arguing that collapse won't lead to the traditional black hole event horizon. I don't see anything in it that disagrees with what PeterDonis and others have been saying about observations of an object falling through the event horizon if one were to have formed somehow.

 

[PAllen]

Thanks for the link.

Wikipedia is unreliable for this kind of stuff in my experience, is that really the correct Penrose diagram?

Recent literature I found on the topic appears not as confident as you are. For example http://arxiv.org/abs/gr-qc/0609024 Phys.Rev.D76:024005,2007 concludes that the hole evaporates before anything can fall in. This article is mentioned in a special wikipedia page where it is claimed it has been "refuted" without giving any reference. A citation search yields no refutation or severe criticism in any paper that cites it.

Do you have an opinion on that paper?

The majority view is that this paper is refuted, in that the horizon does form before evaporation completes. The main paper considered to refute Krauss et. al. is:

http://arxiv.org/abs/0906.1768

 

[PeterDonis]

Wikipedia is unreliable for this kind of stuff in my experience, is that really the correct Penrose diagram?

"Correct" is a strong word given that we don't have a full theory of quantum gravity. However, that diagram, or something similar to it, appears in a lot of references and seems to be commonly accepted by physicists working in the field. But not by all, as the paper you linked to shows. See below.

Do you have an opinion on that paper?

As PAllen said, the generally accepted view is that that paper is wrong. That's my opinion too, for the same reason as many physicists think that: for a black hole of stellar mass or larger (which basically covers all of the black hole candidates we know of in our universe), the spacetime curvature at the horizon is small enough to be well within the regime where GR should be a good classical approximation to whatever the correct quantum-level physics is. If there really were quantum corrections large enough to keep a horizon from forming when an object of stellar mass or larger collapses, we would expect to see the effects of such corrections in other observations, for example the binary pulsar observations, where GR has been confirmed to a good enough accuracy to rule out quantum corrections of the necessary size.