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 =∞ &gt; 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 =∞ &gt; 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.

 

[PeterDonis]

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.

Yes, this is a good point. If Krauss et al. are right and quantum corrections prevent a horizon from ever forming, then none of the stuff we've been talking about applies anyway. For example, the far-away observer will not see the infalling observer slow down more and more and finally appear to "freeze" at the horizon, because there isn't any horizon. And the far-away observer will be able to assign a finite time coordinate to *every* event in the spacetime (at least in principle this should be true--I don't know that Krauss et al. actually give an explicit example of such a coordinate chart), so there are no events where $t = \infty$ and therefore no issues with what that means physically.

 

[DKS]

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.

Is that really so? It seems to me the quantum correction are significant only at timescales of the order of the evaporation time in the distant observer frame which we are in.

Yes, this is a good point. If Krauss et al. are right and quantum corrections prevent a horizon from ever forming, then none of the stuff we've been talking about applies anyway. For example, the far-away observer will not see the infalling observer slow down more and more and finally appear to "freeze" at the horizon, because there isn't any horizon. And the far-away observer will be able to assign a finite time coordinate to *every* event in the spacetime (at least in principle this should be true--I don't know that Krauss et al. actually give an explicit example of such a coordinate chart), so there are no events where $t = \infty$ and therefore no issues with what that means physically.

That is clear and I thought the topic of this thread was to figure out what actually happens according to current theory.

I guess I'll have to read both papers to figure it out. Are there any references for the "generally accepted view"?

 

[PeterDonis]

It seems to me the quantum correction are significant only at timescales of the order of the evaporation time in the distant observer frame which we are in.

Not if they're going to prevent the horizon from forming. For assessing whether that happens, the distant observer's time is not a good time coordinate, for the reasons I've already given; you have to use something more like the natural timescale of the infalling observer that I described before (which basically equates to Painleve coordinate time). On that timescale, gravitational collapse to form a horizon takes place *much* faster (as in, many, many orders of magnitude faster) than black hole evaporation, so quantum corrections would also have to act on a much faster timescale than black hole evaporation in order to prevent a horizon from forming. We have been observing binary pulsars for times much longer than the time it would take for a system of comparable mass to collapse to a black hole (the collapse timescale in Painleve coordinates for a stellar-mass object is hours, and we have been observing binary pulsars for years), so again, I would expect the effects of quantum corrections to have shown up by now if they were large enough to prevent a horizon from forming.

 

[DKS]

Not if they're going to prevent the horizon from forming. For assessing whether that happens, the distant observer's time is not a good time coordinate, for the reasons I've already given; you have to use something more like the natural timescale of the infalling observer that I described before (which basically equates to Painleve coordinate time). On that timescale, gravitational collapse to form a horizon takes place *much* faster (as in, many, many orders of magnitude faster) than black hole evaporation, so quantum corrections would also have to act on a much faster timescale than black hole evaporation in order to prevent a horizon from forming. We have been observing binary pulsars for times much longer than the time it would take for a system of comparable mass to collapse to a black hole (the collapse timescale in Painleve coordinates for a stellar-mass object is hours, and we have been observing binary pulsars for years), so again, I would expect the effects of quantum corrections to have shown up by now if they were large enough to prevent a horizon from forming.

I don't get that. I though a binary pulsar was a system of two neutron stars, not black holes?

 

[PeterDonis]

I don't get that. I though a binary pulsar was a system of two neutron stars, not black holes?

It is. But the system is tightly bound enough, gravitationally, that one would expect to see some effects from quantum corrections if those corrections were large enough to come into play before a horizon formed if a system of similar mass collapsed.

 

[DKS]

It is. But the system is tightly bound enough, gravitationally, that one would expect to see some effects from quantum corrections if those corrections were large enough to come into play before a horizon formed if a system of similar mass collapsed.

I don't believe that is so.

But coming back to the original question, let me try to rephrase what I think is the "paradox".

Let's take a classical BH and plot the radial coordinate $r_g$ of the horizon and the radial coordinate $r_o$ of an infalling observer versus time. The radial coordinate of a space-time point I define as $(R/M^2)^{-1/6}$ with $R$ the curvature scalar and $M$ the BH mass. It is independent of the coordinate system and BH mass, and a measurable quantity. Time can be whatever you want in any coordinate system.

$r_g$ will be just a horizontal line and $r_o$ is some curve starting above $r_g$, passing through it, and terminating at $r_o=0$. The precise shape of the curve depends on your coordinate system.

Now let's add Hawking radiation to the model (but keep the rest of the universe empty). Now the plot of $r_g$ versus time is a decreasing function of time (as the mass $M$ of the BH decreases), terminating at $r_g=0$ at some time $T_e$ which depends on your coordinate system. Now the plot of $r_o$ will start above $r_g$ and will then decrease and the question is will it ever cross (or overtake) the $r_g$ curve?

Now $M(t)$ can be computed from Hawking's formula but you can't just compute the orbit in a Schwarzschild geometry with a time dependent $M$ term as that is not a solution of Einstein's equations. So let's assume $r_g$ changes much slower than $r_o$ and compute the decrease of $r_o$ over some time segment over which $r_g$ can be assumed constant. Since everything I am calculating is coordinate invariant I might as well use Schwarzschild coordinates while $r_o>r_g$. The result is that $r_g$ asymptotically approaches $r_o$ over this time segment, until that time interval becomes large enough that $r_g$ decreases. We then decrease $r_g$ and repeat the calculation. The end result will be that $r_g$ will become $0$ at some time while $r_o>0$ and the observer has not crossed the horizon before the hole disappears.

Problem here is that it is not really consistent to splice segments with different values of $M$ together like this and to do it properly requires you to compute the stress-energy tensor of the Hawking radiation and include it in the right hand side as a source term in Einsteins equations. How to do that seems to be unclear and those two papers mentioned here before are two attempts to do that calculation (up to where $r_g$ becomes comparable to the Planck length), which apparently lead to different conclusions.

Is that an accurate summary?

 

[PeterDonis]

I don't believe that is so.

Well, it's hard to know for sure when we don't know any details about the purported quantum corrections.

Let's take a classical BH and plot the radial coordinate $r_g$ of the horizon and the radial coordinate $r_o$ of an infalling observer versus time.

Whose time? It makes a huge difference. See below.

The radial coordinate of a space-time point I define as $(R/M^2)^{-1/6}$ with $R$ the curvature scalar

By "the curvature scalar", I assume you mean the Kretschmann scalar:

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

If so, your formula is a bit off: it should be (using $K$ for the Kretschmann scalar to avoid confusion, and using units where $G = c = 1$)

$$
r = \left( \frac{K}{48 M^2} \right)^{-1/6}
$$

Just to note, the standard Schwarzschild radial coordinate $r$ is defined such that the surface area of a 2-sphere at $r$ is $4 \pi r^2$. That's the radial coordinate used in most of the common charts for a black hole spacetime. That definition matches the one above, but its geometric meaning is easier to see.

Time can be whatever you want in any coordinate system.

No, it can't; which time coordinate you use makes a huge difference. See below.

$r_g$ will be just a horizontal line

Are you assuming that time is horizontal and space (i.e,. $r$) is vertical? If so, be aware that this is not the usual convention; the usual convention is for time to be vertical and space to be horizontal. For this post I'll adopt the "time is horizontal" convention since it seems to be the one you're using. I'll also assume that the future is to the right (it would be upward in the usual convention).

Given that, the statement above is true for some charts, but not for others. For the standard Schwarzschild exterior chart, which is the "natural" one for the far-away observer to use, the statement is true with some caveats, which are important in this connection: see below.

$r_o$ is some curve starting above $r_g$, passing through it, and terminating at $r_o=0$.

So you are also using a convention that "up" means "radially outward" in the spatial dimension? Again, the usual convention is for radially outward to be rightward. For this post, I'll adopt the "radially outward = up" convention.

Given that, the statement above is true in the charts in which the previous statement (about $r_g$ being horizontal) was true, but there's a key caveat for the Schwarzschild exterior chart: in this chart, the $r_o$ curve goes off to the right to infinity at $r = r_g$; then it comes back in from infinity at the right and goes back to the left (i.e., $t$ is now *decreasing*) as it goes down to $r = 0$. This is true no matter where at the top of the chart we start the $r_o$ curve, i.e., no matter what coordinate time by the far-away observer's clock the infaller starts falling in. That's a reflection of the fact, which I mentioned before, that the Schwarzschild exterior chart maps an infinite line (the horizon) to a single point ($t = + \infty$); an infinite number of possible $r_o$ curves all go to $t = + \infty$ as they go to $r = r_g$. (Note that this also means that the $r = r_g$ horizontal line is really a "phantom" line in the Schwarzschild chart; no worldlines actually cross it at any finite value in this chart.)

In other charts, such as the Painleve chart, this is *not* the case: $r = r_g$ is a horizontal line, but each distinct possible $r_o$ curve crosses that line at a different, finite point. That's why the Painleve chart is a much better chart for charting events at or near the horizon.

Now let's add Hawking radiation to the model (but keep the rest of the universe empty). Now the plot of $r_g$ versus time is a decreasing function of time (as the mass $M$ of the BH decreases), terminating at $r_g=0$ at some time $T_e$ which depends on your coordinate system.

In the Painleve chart, yes, this works fine. (Note that I here am interpreting "decreasing" as "decreasing to the right", since the right is the future direction, as above.) However, in the Schwarzschild chart, it doesn't work the way you are thinking. See below.

Now the plot of $r_o$ will start above $r_g$ and will then decrease and the question is will it ever cross (or overtake) the $r_g$ curve?

In the Painleve chart, yes, it will; it will cross the $r_g$ curve somewhere well to the left (i.e., to the past) of where $r_g = 0$ is reached. They will then continue down to $r = 0$, again reaching that point somewhere well to the left of where $r_g$ reaches $r = 0$.

In the Schwarzschild chart, all of the different possible $r_o$ curves, that all went up to $t = + \infty$ at $r = r_g$ before, now reach $r = 0$ at $t = T_e$. However, the $r_o$ curves do not end there; they each have a second segment that arcs back up from $r = 0$, $t = T_e$, curving to the left (increasing $r$, decreasing $t$), and then curving back down to $r = 0$ at some value of $t$ that is well to the left (i.e., less than, to the past of) $T_e$. (These second segments correspond to the segments between $r = r_g$ and $r = 0$ in the Painleve chart, described above.)

This is, once again, because the Schwarzschild exterior chart bunches together an infinite line of events (all the events on the horizon) at one time, which is now $t = T_e$ instead of $t = + \infty$. That also means that, in the evaporating hole case, there's not really any line you can draw that correctly captures the horizon, $r_g$. If you draw a "decreasing" line as you describe, it will be a "phantom" line, just as the horizontal $r_g$ line was when the black hole was eternal, as above; no worldlines will actually cross it anywhere except at $t = T_e, r = 0$. But that does *not* mean nothing falls through the horizon; it's clear by looking at the Painleve chart that objects *can* fall through the horizon. The Schwarzschild chart is simply too distorted at the horizon to represent it correctly. That's why it *does* make a big difference which coordinate chart you use.

Now $M(t)$ can be computed from Hawking's formula but you can't just compute the orbit in a Schwarzschild geometry with a time dependent $M$ term as that is not a solution of Einstein's equations.

It is if you include the outgoing radiation. Check out the Vaidya metric:

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

So let's assume $r_g$ changes much slower than $r_o$ and compute the decrease of $r_o$ over some time segment over which $r_g$ can be assumed constant.

That's essentially what I was doing to derive the Painleve chart results that I quoted above. See further comments below.

Since everything I am calculating is coordinate invariant I might as well use Schwarzschild coordinates while $r_o>r_g$.

You can even take the limit as $r \rightarrow r_g$. However, you have to be careful to correctly define what it is you are calculating. See below.

The result is that $r_g$ asymptotically approaches $r_o$ over this time segment

"Asymptotically" in what sense? In the sense of coordinate time $t$, yes; but *not* in the sense of proper time of the $r_o$ worldline. The proper time for $r_o$ to reach $r_g$ is finite, and much, much less than the time it takes for $r_g$ to decrease. And the proper time is the physical invariant; the coordinate time $t$ is *not*. So if you are trying to analyze the physics, you need to use the proper time.

(Btw, the reason Painleve coordinates are so much better adapted to this problem is that Painleve coordinate time is the *same* as the proper time of the infalling observer. Schwarzschild coordinate time is only the same as proper time for the far-away observer, not for the infalling observer. The more precise way of stating the third sentence in the previous paragraph is that the Painleve coordinate time it takes for $r_g$ to decrease is much, much larger than the Painleve coordinate time it takes for $r_o$ to reach $r_g$. So for purposes of analyzing the infalling observer's trajectory, we can assume that $r_g$ is constant.)

Problem here is that it is not really consistent to splice segments with different values of $M$ together like this and to do it properly requires you to compute the stress-energy tensor of the Hawking radiation and include it in the right hand side as a source term in Einsteins equations.

That's what the Vaidya metric does.

 

[DKS]

Thanks for the extensive response.

I meant the double contraction of the Riemann tensor for my $R$ and don't care about the factor $48$.

I thought since I formulated everything in terms of observables my reasoning was coordinate independent. It seems you are right though that it is not really. It seems GR is locally coordinate independent but not globally as the Schwarzschild coordinates cover only part of the spacetime that other coordinate systems cover.

Still, as long as nothing has crossed the horizon the Schwarzschild coordinates are as good as any labeling of space-time.

You write:

"The proper time for $r_o$ to reach $r_g$ is finite, and much, much less than the time it takes for $r_g$ to decrease."

How is that possible? In Schwarzschild coordinates it takes a finite time for $r_g$ to reach $0$, and an infinite time for $r_o$ to reach $r_g$, so in proper time of the infalling observer it will take much less time for $r_g$ to decrease than for $r_o$ to reach $r_g$. In other words, in terms of proper time of the infalling observer, $r_g$ remains constant only for a short time.

 

[PeterDonis]

I thought since I formulated everything in terms of observables my reasoning was coordinate independent.

Only if you're careful about what "observable" means. The Schwarzschild time coordinate $t$ is only an "observable" along the worldline of the far-away observer. Close to the horizon $t$ is not an observable, since there's no observer for whom $t$ is even close to being their proper time.

It seems GR is locally coordinate independent but not globally as the Schwarzschild coordinates cover only part of the spacetime that other coordinate systems cover.

Correct. Although I'm not sure I would phrase this as being "globally coordinate dependent"; in a region of spacetime covered by multiple charts, you can compute any observable in any of the charts and get the same answer. I think it's more a matter of being careful about the actual physical meaning of observables, as above.

Still, as long as nothing has crossed the horizon the Schwarzschild coordinates are as good as any labeling of space-time.

In principle, yes; but in practice, no, not if they lead you into incorrect reasoning. There's a reason why other charts were invented to cover the region at or near the horizon.

Also, if you're trying to assess whether or not something crosses the horizon, and if so, when, you can't use a chart that only works as long as nothing crosses the horizon, because that's precisely the assumption you can't make in advance. See below.

"The proper time for $r_o$ to reach $r_g$ is finite, and much, much less than the time it takes for $r_g$ to decrease."

How is that possible?

If you look at the Penrose diagram I linked to in an earlier post, it should be obvious: an $r_o$ worldline can cross the horizon line anywhere along its length, and the crossing only takes up an infinitesimal portion of that length, whereas the decrease in $r_g$ can only be seen over a significant fraction of the total length of the horizon line. But you can confirm this by doing the actual calculation, as long as you do it in a chart that allows it; this is one case where Schwarzschild coordinates simply don't work. See below.

In Schwarzschild coordinates it takes a finite time for $r_g$ to reach $0$

No: in Schwarzschild coordinates the *entire length of the horizon* is labeled with a single time coordinate, $T_e$ (of course that's if the hole evaporates; for an "eternal" hole, the entire length of the horizon is labeled with $t = + \infty$). So there is no way to tell "how long" the horizon line actually is in Schwarzschild coordinates; to do that, different parts of the line would need to be labeled with different time coordinates, and in that chart they aren't.

and an infinite time for $r_o$ to reach $r_g$

Not if the hole is evaporating: if it's evaporating, $r_o$ reaches $r_g$ at time $T_e$. But as we just saw, that tells you nothing about when $r_o$ actually reaches the horizon, because the entire horizon is labeled with the time $T_e$. So the calculation you are trying to do simply can't be done in Schwarzschild coordinates; you can't even use the workaround of taking limits as $r \rightarrow r_g$ in this case (you can do that to calculate the proper time it takes for $r_o$ to reach the horizon, but you can't to calculate the time it takes for $r_g$ to decrease).

so in proper time of the infalling observer it will take much less time for $r_g$ to decrease than for $r_o$ to reach $r_g$. In other words, in terms of proper time of the infalling observer, $r_g$ remains constant only for a short time.

Incorrect; see above. To correctly do this calculation, you have to use a chart that assigns different time coordinates to different events on the horizon, such as the Painleve chart. In that chart, when you do the computation, it comes out the way I said.

 

[DKS]

It's hard to reply without quoting what you are replying too. A limitation of this forum, I guess.

Only if you're careful about what "observable" means. The Schwarzschild time coordinate $t$ is only an "observable" along the worldline of the far-away observer. Close to the horizon $t$ is not an observable, since there's no observer for whom $t$ is even close to being their proper time.

Nowhere did I claim the Schwarzschild time coordinate $t$ is an "observable" for the infalling observer.

"Me: observables are coordinate independent."
In principle, yes; but in practice, no, not if they lead you into incorrect reasoning. There's a reason why other charts were invented to cover the region at or near the horizon.

Observables are coordinate independent in GR, not only when they lead to conclusions you want.
The only issue is that some coordinates may not cover the whole of space-time (e.g., Schwarzschild coordinates don't not cover the interior of a BH, which I re-address below).

Also, if you're trying to assess whether or not something crosses the horizon, and if so, when, you can't use a chart that only works as long as nothing crosses the horizon, because that's precisely the assumption you can't make in advance. See below.

Of course you can *as long as nothing crosses the horizon* which would be reflected in the behavior if $t_{Schwarzschild} \rightarrow \infty$. But the analysis in Schwarzschild coordinates stops at $t= T_e$ and we don't need the other patch.

No: in Schwarzschild coordinates the *entire length of the horizon* is labeled with a single time coordinate, $T_e$ (of course that's if the hole evaporates; for an "eternal" hole, the entire length of the horizon is labeled with $t = + \infty$). So there is no way to tell "how long" the horizon line actually is in Schwarzschild coordinates; to do that, different parts of the line would need to be labeled with different time coordinates, and in that chart they aren't.

I don't understand what you mean by "the *entire length of the horizon* is labeled with a single time coordinate, $T_e$". The horizon is a 3-dimensional region of spacetime. It's time coordinate in the Schwarzschild frame is labeled by the interval $(-\infty\ T_e)$.

 

[Nugatory]

I don't understand what you mean by "the *entire length of the horizon* is labeled with a single time coordinate, $T_e$". The horizon is a 3-dimensional region of spacetime. It's time coordinate in the Schwarzschild frame is labeled by the interval $(-\infty\ T_e)$.

It is not. An easy way to see this is to look at curves of constant Schwarzschild $t$ on a graph whose vertical and horizontal axes are the Kruskal $u$ and $v$ coordinates. You will see that one curve (actually, this one is a straight line, $u=v$) of constant $t$ is the exact same set of points in spacetime as the event horizon.

(yes, I know that I'm describing the event horizon of an non-decaying black hole here, not a decaying one. It seems like a good idea to understand the behavior of the Schwarzschild $t$ in this simpler case before moving on to the decaying case)

 

[PeterDonis]

Nowhere did I claim the Schwarzschild time coordinate $t$ is an "observable" for the infalling observer.

Not explicitly, but you appear to be implicitly assuming it is, without realizing it. However, that's a minor point compared to the others.

Observables are coordinate independent in GR, not only when they lead to conclusions you want.

But you have to be careful about interpreting what observables mean, physically. That was my point.

Of course you can *as long as nothing crosses the horizon*

You're missing the point. The question at issue is, *does* $r_o$ cross the horizon, by reaching $r_g$ before the hole evaporates, or does it only reach $r_g$ at the moment of final disintegration, so that there is no horizon to cross any more? If you are trying to answer that question, you can't use a chart that only works "as long as nothing crosses the horizon"; you have to use a chart that can correctly represent the *possibility* of something crossing the horizon, since it's precisely that possibility that you are trying to evaluate.

But the analysis in Schwarzschild coordinates stops at $t= T_e$ and we don't need the other patch.

Incorrect; if the hole evaporates, $t = T_e$ has the same coordinate singularity, and therefore the same issues, as $t = + \infty$ does in the case of an eternal black hole. The behavior as the horizon is approached is seen as $t \rightarrow T_e$, not as $t \rightarrow \infty$. See below.

I don't understand what you mean by "the *entire length of the horizon* is labeled with a single time coordinate, $T_e$".

I mean just what I said; the entire region of spacetime that comprises the horizon (which is actually a null 3-surface; see below) all has the same Schwarzschild time coordinate. In the case of an eternal black hole, that time coordinate is $t = + \infty$; in the case of an evaporating black hole, that time coordinate is $t = T_e$.

The horizon is a 3-dimensional region of spacetime.

Yes, strictly speaking I should have referred to it as a "null 3-surface", not a line. However, since the spacetime is spherically symmetric, we can suppress the two angular coordinates and just consider the $t - r$ plane; in that plane, the horizon is a line. (You do the same thing when you talk about the $r_g$ line in previous posts.)

It's time coordinate in the Schwarzschild frame is labeled by the interval $(-\infty\ T_e)$.

No, it isn't. Nugatory's response is correct for the case of an eternal black hole; in the case of an evaporating black hole you just substitute $t = T_e$ for $t = + \infty$, as above.

Yes, I know that if you draw a spacetime diagram in Schwarzschild coordinates, it seems like there is an $r_g$ curve running from $t = - \infty$ to $t = T_e$ (in the evaporating case, or $t = + \infty$ in the eternal case). But that line *labels no events*; it is, as I pointed out in previous posts, a "phantom" line that doesn't actually correspond to any part of spacetime. This is because of the infinite distortion of the Schwarzschild chart at the horizon.

 

[Wade888]

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.

Are you implying both types of black holes exist in nature, or are we still unsure which type actually exists in nature?


I was under the impression that an actual black hole would need to "unify" Relativity and QM.

 

[PeterDonis]

Are you implying both types of black holes exist in nature, or are we still unsure which type actually exists in nature?

Neither model actually quite matches real black holes in nature. We expect that real black holes will emit Hawking radiation, but for a real hole to actually evaporate, meaning lose mass with time, it would have to lose mass via Hawking radiation faster than it gained mass from matter and energy falling in. All of the real black hole candidates we know of in the universe have matter and energy falling in at rate many orders of magnitude greater than the rate at which they are emitting Hawking radiation.

In fact, even the cosmic microwave background radiation, all by itself, is enough to add matter and energy to a black hole of stellar mass or larger at a rate many orders of magnitude greater than it emits Hawking radiation. This will continue to be true for a long, long time, and for black holes with large enough masses, it may even be true forever; see this article:

http://math.ucr.edu/home/baez/end.html

 

[Wade888]

Neither model actually quite matches real black holes in nature. We expect that real black holes will emit Hawking radiation, but for a real hole to actually evaporate, meaning lose mass with time, it would have to lose mass via Hawking radiation faster than it gained mass from matter and energy falling in. All of the real black hole candidates we know of in the universe have matter and energy falling in at rate many orders of magnitude greater than the rate at which they are emitting Hawking radiation.

In fact, even the cosmic microwave background radiation, all by itself, is enough to add matter and energy to a black hole of stellar mass or larger at a rate many orders of magnitude greater than it emits Hawking radiation. This will continue to be true for a long, long time, and for black holes with large enough masses, it may even be true forever; see this article:

http://math.ucr.edu/home/baez/end.html

Thank you. I had suspected something similar in the past (r.e. hawking radiation,) at least in terms of theory, but had no proof of it.

So in order for a stellar mass black hole to actually decay, the universe would need to age to a point where all CMB radiation will have passed the BH by from every direction, assuming a finite universe, but if the universe were infinite this could never happen anyway, because there would always be more CMB "out there" somewhere to keep coming and falling in. Right?