Infinite Time Dilation at the Surface of a Black Hole?

[PeterDonis]

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

This won't happen even in a finite universe; the CMB radiation fills the entire universe, so it is always present at every spatial location. Individual CMB photons basically "circle around" the universe.

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?

This happens even in a finite universe (see above), because CMB radiation keeps circling around the universe. But in any case, whether or not there is CMB radiation present is not the critical factor. The critical factor is the temperature of the CMB radiation, compared to the Hawking temperature of the black hole. As long as the latter is less than the former, the hole will gain energy. The CMB temperature decreases as the universe expands, but it will take a long, long time for the universe to expand enough for the CMB temperature to be lower than the Hawking temperature of a stellar mass black hole; and for a large enough black hole, the Hawking temperature is less than the minimum temperature the CMB radiation will ever reach (about $10^{-30}$ Kelvin, according to the article I linked to), so the hole will *never* be able to lose energy via Hawking radiation.

 

[DKS]

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.

I'm beginning to see it. It is a "phantom line" because the metric is not defined (singular) on the horizon, correct?

In non-singular coordinates (any of those you recommended) what would a plot of $r_g$ and $r_o$ versus time look like? Is there some definite computable time (coordinate and/or proper time) at which $r_o$ crosses $r_g$ while $r_g>0$?

I ordered Penrose's book "cycles of time" in the hope of mastering these Penrose diagrams and should probably shut up till I've read it. If you have other recommondations for books on the subject, I'm all ears. Thanks again for your patience, I learned a lot.

 

[PeterDonis]

I'm beginning to see it. It is a "phantom line" because the metric is not defined (singular) on the horizon, correct?

Correct.

In non-singular coordinates (any of those you recommended) what would a plot of $r_g$ and $r_o$ versus time look like?

The easiest to describe is Painleve coordinates, which look like this:

For the case of an eternal black hole, $r_g$ would be a horizontal line (using your "time is horizontal" convention), and $r_o$ would be a line starting at the top (i.e., at some large value of $r$) and gradually sloping more and more downward as it went to the right (using the "future is to the right" convention), crossing $r_g$ at some finite point, and at some later finite point reaching $r = 0$.

For the case of an evaporating black hole, $r_g$ would be a line slowly moving downward as it moves to the right; but its slope would be gentle enough that its height would not change appreciably during the entire span of an $r_o$ curve, which would look the same as I described above.

Eddington-Finkelstein coordinates look more or less the same as the above, but the precise scaling of the time coordinate is somewhat different because those coordinates are adapted to ingoing light rays instead of ingoing free-falling observers.

Kruskal coordinates look similar to the Penrose diagram I posted earlier, but with somewhat different scaling, and extending out to infinity instead of having "infinity" as a finite point on the diagram--the latter was one of the chief motivations for Penrose diagrams, so that rigorous theorems about "what happens at infinity" could be more easily formulated and proven. In these charts, $r_g$ is the 45-degree line going up and to the right, and $r_o$ curves are more or less vertical (the time dimension is normally presented as vertical in these charts) and cross the $r_g$ line somewhere below and to the left of where it terminates at the moment of final disintegration of the evaporating hole.

Is there some definite computable time (coordinate and/or proper time) at which $r_o$ crosses $r_g$ while $r_g>0$?

Yes to both. The proper time is the same no matter which chart you use to compute it (since it's a direct observable); the coordinate time depends on the chart you use but is always finite.

I ordered Penrose's book "cycles of time" in the hope of mastering these Penrose diagrams and should probably shut up till I've read it. If you have other recommondations for books on the subject, I'm all ears.

Kip Thorne's Black Holes and Time Warps is a good non-technical presentation of GR, although IIRC he doesn't say much about black hole evaporation.

Thanks again for your patience, I learned a lot.

You're welcome!

 

[DKS]

Kip Thorne's Black Holes and Time Warps is a good non-technical presentation of GR, although IIRC he doesn't say much about black hole evaporation.

I have that book, I'm looking for something more technical.
Can you recommend me your favorite grad-level textbook on these issues?

 

[Akashks001]

Hi, it was a nice question...

As far as i know, the time in your clock after you come back to the safe distance would be around five years (the time what you spent before the event horizon of the black hole).

For the time being, ignore the standard definition of the event horizon of the black hole. Just take black hole as a super massive star with stupendous gravitational field. I think this assumption does not contradict anywhere as long as you are outside the event horizon. Now, if you say you are just outside a event horizon, you are just in a strong gravitational field and clocks do run slower in stronger fields. If you spend 5 years in strong gravitation, it would mean millions of years to the outer world. But, when you come to the outer world, your clock will still show that 5 years and odd, but the rest of the world would be millions of years ahead of your clock... You can get the clear idea if you compare this situation to the twin paradox of Einstein. Except, time dilation occurs due to gravitation in this case, whereas time dilation occurs due to velocity in twin paradox.

And for the evaporation of black hole... No, the black hole may not evaporate. Because it is also in the same gravitational field (or in an even stronger field) as you are and both your and the black hole's time would run at approximately same speed...

Note: Since this is a forum, my answer is of the nature of forum discussion. Please think over it again. Thanks.

 

[PeterDonis]

I have that book, I'm looking for something more technical.
Can you recommend me your favorite grad-level textbook on these issues?

To be honest, I don't know of a good textbook that covers the issues we've been discussing in this thread adequately. (If any of the other experts here know of one, please post!) It would have to be pretty recent, since our understanding of black hole evaporation has evolved a lot over the last few decades. Most of the useful information on the topic at this point appears to be in papers, not textbooks.

Wald's General Relativity has some discussion of black hole evaporation, but mostly to do with the reasons why Hawking radiation is a robust expectation based on semiclassical quantum gravity arguments. (Wald also has a monograph, Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, which covers this in much more detail, but it's hard to find.) However, he doesn't really discuss the details of things like how an $r_o$ trajectory would work in a spacetime with an evaporating black hole. (I suspect, though of course I don't know for sure, that a key reason for this was that it seemed obvious to Wald that an $r_o$ trajectory would work the same in the evaporating case as it does in the eternal case for any black hole large enough to be of practical interest, for the reasons I explained in this thread. But I admit it would have been nice to see something more explicit on this score.)

For the general properties of the horizon in Schwarzschild spacetime, Misner, Thorne & Wheeler does a good, thorough job, but since it was published in 1973 it only covers the "eternal" black hole case. However, it's still the best treatment I know of regarding the limitations of Schwarzschild coordinates, and how the other coordinate charts I've mentioned work. (If an updated edition of MTW had been published, which is unfortunately impossible now since Wheeler died in 2008, I'm sure it would have covered things like an infalling $r_o$ trajectory in an evaporating black hole spacetime, the same way it covers similar trajectories in an eternal black hole spacetime. But most textbooks don't strive to be as comprehensive as MTW did.)

 

[PeterDonis]

And for the evaporation of black hole... No, the black hole may not evaporate.

Do you have any references to back up this claim? As you will be aware if you've read through this thread, this is *not* the mainstream view.

Because it is also in the same gravitational field (or in an even stronger field) as you are and both your and the black hole's time would run at approximately same speed...

I'm not sure what this even means. The black hole *is* the gravitational field; it's not "in" it. And I don't know what you mean by "the black hole's time".

 

[PAllen]

There is a view advocated by Ellis of Hawking and Ellis fame that, while clearly not the majority view, can hardly be called crackpot. In his latest work on this he argues, in contrast to majority view, that:

- Hawking radiation originates mainly inside the horizon
- a realistic black hole will never evaporate, even given infinite time

http://arxiv.org/abs/1310.4771

However, as far as the main issues of this thread, this paper is much more in the mainstream camp than Krauss et.al. :

- Hawking radiation cannot prevent either horizon or singularity formation

 

[TumblingDice]

There is a view advocated by Ellis of Hawking and Ellis fame that, while clearly not the majority view, can hardly be called crackpot.

I was curious and followed the link to review the abstract. I'm not casting any opinion on the value of the paper or the qualifications of the author, however something drew my attention that I've been chewing on:

...and most of the Hawking radiation ends up at this singularity rather than at infinity. Whether any Hawking radiation reaches infinity depends on...

My first reaction was that, if a new member used wording like this in any context what-so-ever, veteran members would consider the statement was fraught with misconceptions.

But perhaps I need to learn more. To me, infinity is a concept, not a destination. To suggest anything "ends up" or "reaches" infinity sounds incorrect - naive at best. I also prefer thinking of a singularity as a concept, associating the terminology with the uniqueness of where the math breaks down and the equation becomes meaningless, as opposed to a 'point of infinite density' that is often a point of confusion for new learners. Like infinity, I feel uncomfortable reading that something "ends up" at "this singularity".

Don't want to get off topic, just wondering if statements like these can be considered 'crackpot' when they're made by someone at the author's level of understanding...

 

[PAllen]

I was curious and followed the link to review the abstract. I'm not casting any opinion on the value of the paper or the qualifications of the author, however something drew my attention that I've been chewing on:My first reaction was that, if a new member used wording like this in any context what-so-ever, veteran members would consider the statement was fraught with misconceptions.

But perhaps I need to learn more. To me, infinity is a concept, not a destination. To suggest anything "ends up" or "reaches" infinity sounds incorrect - naive at best. I also prefer thinking of a singularity as a concept, associating the terminology with the uniqueness of where the math breaks down and the equation becomes meaningless, as opposed to a 'point of infinite density' that is often a point of confusion for new learners. Like infinity, I feel uncomfortable reading that something "ends up" at "this singularity".

Don't want to get off topic, just wondering if statements like these can be considered 'crackpot' when they're made by someone at the author's level of understanding...

Conformal infinity is a well defined concept in GR, and is used in Penrose diagrams, for example. The author is referring to it in the technical sense. There are different types of infinity in GR:

Bondi Mass measures total mass of the universe measured at null-infinity, which means it doesn't include radiation that escapes forever.

ADM mass measures total mass of the universe at spatial infinity, including radiation.

The difference between the two, for example, can be used to model the energy lost to gravitational radiation from a binary system.

[edit: as for the singularity, this author, along with Hawking and Penrose, helped produce our modern understanding of singularities as a rigorous concept. Again, they come in different varieties - this paper focuses only on spacelike singularities (which effectively mean a moment in time rather than a location). This is interesting because the Kerr singularity is timelike, but it is unknown what the nature of the singularity is - classically - for a realistic collapse with rotation. ]

 

[DKS]

The easiest to describe is Painleve coordinates, which look like this:

For the case of an eternal black hole, $r_g$ would be a horizontal line (using your "time is horizontal" convention), and $r_o$ would be a line starting at the top (i.e., at some large value of $r$) and gradually sloping more and more downward as it went to the right (using the "future is to the right" convention), crossing $r_g$ at some finite point, and at some later finite point reaching $r = 0$.

I got that.

For the case of an evaporating black hole, $r_g$ would be a line slowly moving downward as it moves to the right; but its slope would be gentle enough that its height would not change appreciably during the entire span of an $r_o$ curve, which would look the same as I described above.

But not this. Maybe you can point out where I go wrong?

The Hawking radiation power goes like $1/M^2$ so in Schwarzschild time $t$ we have $dM/dt = -a/3M^2$ with $a$ a constant, so $M(t)=(M_0^3- at)^{1/3}$ where $M_0$ is the mass at time 0.

According to http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates#GP_coordinates the Painleve time $t_r$ is related to the Schwarzschild coordinates by $t_r = t - f(r,M)$ with $f(r,M)$ given on that webpage.

If we now substitute the formula for $M(t)$ we get $t_r = t - f(r,t)$ (I mean $f(r,t) = f(r,M(t))$). Solving $t_r = t - f(r,t)$ for $t$ gives $t = g(r,t_r)$ with $g$ some complicated function which has no explicit formula, but I can compute it numerically if I want. Now if I substitute this I get in Painleve coordinates $r_g(t_r) = 2M = 2(M_0^3- ag(r,t_r))^{1/3}$ which depends on $t_r$ and on $r$.

How can I now get a plot of $r_g$ versus $t_r$? It also depends on $r$?!

Perhaps the next step would be to compute the orbit of the falling observer, $r_0(t_r)$, and use $r_o(t_r)$ in the equation for $r_g$?

 

[PeterDonis]

Maybe you can point out where I go wrong?

The equations you're using for Hawking radiation are only valid at infinity; at infinity, Painleve coordinate time and Schwarzschild coordinate time are the same (because the function $f$ giving the difference between them goes to zero as $r \rightarrow \infty$). So the method you're using won't work to find $r_g$ as a function of Painleve coordinate time, because for that you need the behavior of the spacetime at finite $r$.

I'm not sure if I've mentioned it before in this thread, but we do not have a full analytical solution for an evaporating black hole spacetime. The best model we have right now is a "quasi-static" decrease in $M$; i.e., we model the spacetime as a succession of slices with gradually decreasing $M$, where each slice looks just like the ordinary static geometry of an eternal black hole with that $M$. (Note that we have to use Painleve coordinate time, or some other time coordinate that's not singular at the horizon, to label the slices.) That means that $r_g$ on each slice is just $2M$, twice the mass $M$ for that slice. I don't know that anyone has actually derived a detailed equation for the "shape" of the $r_g$ curve with respect to, say, Painleve coordinate time; you'll notice I didn't commit to any details about that shape, I just said the curve was decreasing.

As far as the decrease in $r_g$ being negligible while an object is falling through the horizon, the Painleve coordinate time $T$ of infall is given by

$$
T = 2M_0 \left( \frac{R_0}{2M_0} \right)^{\frac{3}{2}}
$$

where $R_0$ is the radius from which the object starts falling in and $M_0$ is the mass of the hole at the time the object starts falling in. So the infall time is linear in $M_0$, while the evaporation time is cubic in $M_0$; that means the ratio of the times is proportional to the inverse square of $M_0$. If you plug in numbers, you find that the proportionality constant is such that for a stellar-mass black hole, the ratio is something like $10^{-70}$.

(Note that the ratio does depend on $R_0$; but the order of magnitude I just gave is for $R_0 \approx 10^6 M_0$, i.e., already a huge multiple of $M_0$. But you could still increase it by many orders of magnitude and the ratio of infall time to evaporation time would still be miniscule.)

 

[DKS]

The equations you're using for Hawking radiation are only valid at infinity; at infinity, Painleve coordinate time and Schwarzschild coordinate time are the same (because the function $f$ giving the difference between them goes to zero as $r \rightarrow \infty$). So the method you're using won't work to find $r_g$ as a function of Painleve coordinate time, because for that you need the behavior of the spacetime at finite $r$.

According to that wiki page $f(r) = O(\sqrt{r})$ and goes to infinity as $r \rightarrow \infty$. Is the wiki page wrong? I also thought it should go to 0.

How do you mean "you need the behavior of the spacetime at finite $r$"? I have the metric in Paileve coordinates, what more do I need?

I'm not sure if I've mentioned it before in this thread, but we do not have a full analytical solution for an evaporating black hole spacetime. The best model we have right now is a "quasi-static" decrease in $M$; i.e., we model the spacetime as a succession of slices with gradually decreasing $M$, where each slice looks just like the ordinary static geometry of an eternal black hole with that $M$. (Note that we have to use Painleve coordinate time, or some other time coordinate that's not singular at the horizon, to label the slices.) That means that $r_g$ on each slice is just $2M$, twice the mass $M$ for that slice. I don't know that anyone has actually derived a detailed equation for the "shape" of the $r_g$ curve with respect to, say, Painleve coordinate time; you'll notice I didn't commit to any details about that shape, I just said the curve was decreasing.

Yes $r_g$ obviously decreases as does $r_o$. I don't see how, if no equation for $r_g(t_r)$ is known, you can then claim so confidently that $r_o$ must cross $r_g$.
I can see it is true if I assume, as you do, that $r_g=2M(t_r)$ decreases slower than $r_o$, but that is assuming what needs to be calculated. And I should be able to compute $M(t_r)$, I know it explicitly in Schwarzschild coordinates!

Kees

 

[PeterDonis]

According to that wiki page $f(r) = O(\sqrt{r})$ and goes to infinity as $r \rightarrow \infty$. Is the wiki page wrong? I also thought it should go to 0.

I'm not sure; I'll have to look at some other references when I get a chance. I'm quite sure that the "evaporation time" in terms of Painleve coordinate time is similar, if not exactly the same, as the formula you quoted; some of the constants may be slightly different, but it still goes like $M_0^3$ and is very, very, very large for a stellar mass black hole.

How do you mean "you need the behavior of the spacetime at finite $r$"?

I mean you need the actual metric for an evaporating black hole spacetime (see below) in order to derive a specific equation for $r_g$ as a function of Painleve coordinate time. But you do *not* need that metric to derive the formula for Hawking radiation that you quoted.

I have the metric in Painleve coordinates, what more do I need?

We have the metric in Painleve coordinates *for an eternal black hole*. Nobody has a specific expression for the metric, in Painleve or any other coordinates, for an evaporating black hole. All we have are "semiclassical" heuristic arguments and approximations.

For example, the Hawking radiation formula you quoted is derived using the metric for an "eternal" black hole; it basically amounts to analyzing what happens if you "scatter" quantum fields off that metric. More precisely, it analyzes what happens if you take a quantum field state containing zero particles (i.e., a vacuum state) coming in from past infinity, "scatter" that state off the black hole, and look at what the scattered quantum field state looks like at future infinity. It turns out that the future infinity state is a thermal state at the Hawking temperature for the black hole mass $M$. Yes, that means $M$ has to change over time, and the derivation of the formula assumed a constant $M$; so at the very least, there have to be some missing pieces that we haven't filled in yet. But in any case, the derivation certainly does not amount to finding an actual expression for an evaporating black hole metric, which is what you would need to derive a specific function for $r_g$ vs. coordinate time.

(There are other metrics, such as the Vaidya metric, which I think I alluded to before, that *do* model a spacetime containing a "central mass" that decreases with time as it emits radiation; but AFAIK this metric doesn't work "as is" if the central mass is a black hole, so at best it's another source of heuristic approximations.)

I don't see how, if no equation for $r_g(t_r)$ is known, you can then claim so confidently that $r_o$ must cross $r_g$.

Because, even if we don't know the detailed equation for $r_g$ vs. $t_r$, we know that at time $t_r = 0$, when $r_o$ starts falling in from radius $R_0$, $r_g$ has the value $2 M_0$, because that is the hole's mass at that Painleve time. And we know that $r_g$ has the value $0$ at a Painleve time equal to the "evaporation time" from the formula you gave, which for a stellar mass black hole works out to something like $10^{67}$ years. In between those two Painleve times, therefore, we must have $0 < r_g < 2M_0$. The Painleve time at which $r_o = 0$ lies in that range of Painleve times; note that, while the formula I gave before was for the time for $r_o$ to reach the horizon, there is a similar formula for the Painleve time for $r_o$ to reach $r = 0$ (in fact, now that I think of it, the formula I gave before may actually have been for the time to reach $r = 0$, not the horizon--I'll check when I get a chance). For a stellar mass black hole, this formula gives a time of a few hours for $r_o$ to reach $r = 0$, and at that point it has certainly crossed $r_g$, since $r_g > 0$ at that time.

I can see it is true if I assume, as you do, that $r_g=2M(t_r)$ decreases slower than $r_o$, but that is assuming what needs to be calculated.

No, it's not an assumption, it's a calculation; see above.

And I should be able to compute $M(t_r)$, I know it explicitly in Schwarzschild coordinates!

No, you don't; the formulas you gave for Hawking radiation do *not* tell you this. They are heuristic approximations; they certainly are *not* exact functions for $M$ as a function of any time coordinate. See above.

 

[PeterDonis]

A quick follow-up to clarify two points from my previous post:

I'm not sure; I'll have to look at some other references when I get a chance.

I'm still not sure exactly where the Wiki page is going wrong, though I think it's in evaluating the integral for $f(r)$. But to see that $f(r) \rightarrow 0$ as $r \rightarrow \infty$, you don't actually have to evaluate the integral (i.e., you don't need to write down a closed-form expression for $f(r)$ that's valid at any radius). First, define

$$
\beta(r) = \sqrt{\frac{2M}{r}}
$$

It is evident that $\beta \rightarrow 0$ as $r \rightarrow \infty$. Now we can write $f'(r) = \beta / \left( 1 - \beta^2 \right)$, or

$$
t_f (t, R) = t - \int_R^{\infty} \frac{\beta}{1 - \beta^2} dr
$$

where $R$ is the specific radius (i.e., value of $r$) at which we are evaluating $t_f$. But now it's obvious that, as $R \rightarrow \infty$, the integral goes to zero (because $\beta$ in the numerator of the integrand goes to zero, and the denominator goes to $1$), so $t_f \rightarrow t$ as $R \rightarrow \infty$.

the formula I gave before may actually have been for the time to reach $r = 0$, not the horizon--I'll check when I get a chance

I checked, and the formula I gave *is* for the time to fall to $r = 0$, not the horizon. The time to fall to the horizon is the time I gave, minus $2M$.

 

[DKS]

Because, even if we don't know the detailed equation for $r_g$ vs. $t_r$, we know that at time $t_r = 0$, when $r_o$ starts falling in from radius $R_0$, $r_g$ has the value $2 M_0$, because that is the hole's mass at that Painleve time. And we know that $r_g$ has the value $0$ at a Painleve time equal to the "evaporation time" from the formula you gave, which for a stellar mass black hole works out to something like $10^{67}$ years.

It seems you are mixing two coordinate systems. The formula I gave for $M$ is in Schwartzschild time. In Schwartzschild time the evaporation time is indeed something like $10^{67}$ years, but the infall time is even larger, infinity. In Painleve coordinates I don't know what the evaporation times is. The calculation I showed in my previous post gives that in P coordinates $M$ depends on $t_r$ as well as $r$, which sounds reasonable as time and space are mixed up.

About the approximation used. I just substitute $M(t)$ for $M$ in the Schwarzschild metric. This defines a spacetime with a metric which is an approximate solution of Einsteins equations as long as $M(t)$ varies slow enough in some appropriate sense. I then have a metric tensor in hand and I can do anything I want including changing coordinates and compute anything I want.
Computing the worldline of the infalling observer is easy, just solve the geodesic equation. Computing $r_g$, or equivalently, $M(t_r,r)$ is what's not clear to me as it depends on time and $r$. I think if I compute it on the orbit $(t_r(\tau) r(\tau))$ of the infalling observer I should get what he sees.

Is that the right approach?

 

[pervect]

At one time, we had a discussion of the Krauss paper, and a rather technical rebuttal of it. I can't find the related thread, I recall that the authors noted the calculation for the evaporating hole was "non trivial".

 

[DKS]

I'm still not sure exactly where the Wiki page is going wrong, though I think it's in evaluating the integral for $f(r)$. But to see that $f(r) \rightarrow 0$ as $r \rightarrow \infty$, you don't actually have to evaluate the integral (i.e., you don't need to write down a closed-form expression for $f(r)$ that's valid at any radius). First, define

$$
\beta(r) = \sqrt{\frac{2M}{r}}
$$

It is evident that $\beta \rightarrow 0$ as $r \rightarrow \infty$. Now we can write $f'(r) = \beta / \left( 1 - \beta^2 \right)$, or

$$
t_f (t, R) = t - \int_R^{\infty} \frac{\beta}{1 - \beta^2} dr
$$

where $R$ is the specific radius (i.e., value of $r$) at which we are evaluating $t_f$. But now it's obvious that, as $R \rightarrow \infty$, the integral goes to zero (because $\beta$ in the numerator of the integrand goes to zero, and the denominator goes to $1$), so $t_f \rightarrow t$ as $R \rightarrow \infty$.

That can't be right. $f'(r) = O(1/r^{1/2})$ for large $r$ so $f(r) = O(r^{1/2})$, consistent with wikipedia.

Of course if $M$ depends on Schwarzschild-time $f$ will also depend on $t$ and additional terms arise in the Painleve metric, but they are proportional to $M'(t)$ which we assume is small, so perhaps it is still a good approximation.

 

[PeterDonis]

The formula I gave for $M$ is in Schwartzschild time.

As I said before, it's for Schwarzschild time at infinity. Which, as I've shown, is the same as Painleve coordinate time at infinity. [Edit: See my response to your follow-up post on this.] But that may be a confusing way to say it: see below.

In Schwartzschild time the evaporation time is indeed something like $10^{67}$ years, but the infall time is even larger, infinity.

No, it isn't. We went over this before. The Schwarzschild coordinate time for the infall (to the horizon) is also $10^{67}$ years, i.e., it's $T_e$, the same as the evaporation time. But that's because Schwarzschild coordinates label *everything* that happens at the horizon with the same time coordinate; they're infinitely distorted at the horizon.

In Painleve coordinates I don't know what the evaporation times is.

Yes, you do; it's $10^{67}$ years, the same as it is in Schwarzschild coordinates. But perhaps the following is a better way to present how this conclusion is arrived at:

Consider an observer at a very, very large $r$ coordinate (call it $R$), where $R$ is large enough that the metric at $R$ is basically flat (i.e., $g_{tt}$ and $g_{rr}$ are $1$). This observer is static, i.e., he stays at $r = R$ forever. At some time $\tau = 0$ by this observer's clock, he drops an object towards the black hole. At this time, he measures the hole's mass (by putting a test object in orbit about the hole and measuring its orbital parameters, then applying Kepler's Third Law), and gets a value $M_0$. At some much later time $\tau_e = K M_0^3 + k R$ by his clock, he sees the flash of light from the black hole's final evaporation. Here $K$ is the constant factor in the formula for $M_0$ that you posted, and $k$ is a correction factor for the light travel time (see further comments below); in other words, the flash of light takes time $k R$ to travel from the final evaporation at $r = 0$ to radius $R$. So the observer can correct for light travel time and assign a time $T_e = K M_0^3$ to the hole's final evaporation.

Now: what coordinate chart is $T_e$ a coordinate time in? The answer is, it is formally a valid coordinate time for *both* the Schwarzschild chart *and* the Painleve chart; the only thing that might vary is the light travel time correction factor $k$, and that factor won't be much different between the two. Why? First, the observer is at a large enough $R$ that the two charts both assign essentially the same time coordinate to events on his worldline, which will just be the observer's proper time; so the event of the observer seeing the flash of light will have coordinate time $\tau_e$ in both charts. (Also, the event of dropping the object into the hole will have coordinate time $t = 0$ in both charts.)

Second, spacetime is flat everywhere to the future of the outgoing flash of light from the hole's final evaporation; and in flat spacetime (i.e., $M = 0$), the Painleve and Schwarzschild charts are identical. So on the actual worldline of the outgoing flash of light, the two charts can differ only by a small amount (since the charts are continuous), and that means the factor $k$ can only differ by a small amount from $1$ for both charts (because spacetime is "almost flat" on the worldline of the flash of light), meaning that the coordinate time values that each chart assigns for $T_e$ can only differ by a small amount (because the charts are almost identical).

The calculation I showed in my previous post

Which is wrong because the Wikipedia page formula for $t_r$ in terms of $t$ is wrong; as I posted before, I don't know exactly where the Wiki calculation goes wrong, but the formula has to be wrong because, as I showed, $t_r \rightarrow t$ as $R \rightarrow \infty$. [Edit: See my response to your follow-up post on this.]

About the approximation used. I just substitute $M(t)$ for $M$ in the Schwarzschild metric. This defines a spacetime with a metric which is an approximate solution of Einsteins equations as long as $M(t)$ varies slow enough in some appropriate sense.

No, it doesn't, because you can't define "slow enough" using Schwarzschild coordinates, because of the coordinate singularity at the horizon. You have to be able to assign different time coordinates to different events on the horizon in order to do what you're trying to do, and you can't do that in Schwarzschild coordinates.

Is that the right approach?

Not using Schwarzschild coordinates; see above. You could do it in Painleve coordinates, making $M$ a function of Painleve coordinate time; that's more or less what I've been doing.

 

[PeterDonis]

$f'(r) = O(1/r^{1/2})$ for large $r$ so $f(r) = O(r^{1/2})$, consistent with wikipedia.

You're missing the key point I was making: the integral that appears in the formula for $t_r$ is a *definite* integral: it's taken from some finite value $R$ of the $r$ coordinate to infinity. So if you're trying to evaluate how $t_r$ relates to $t$ at infinity, you need to let the lower limit of the integral, $R$, tend to infinity itself. In *that* limit, the integral does vanish; heuristically, this is because you're adding together fewer and fewer terms, and the "number of terms" decreases linearly while the "size of the terms" increases only sub-linearly (I'm not sure it's as simple as just a square root dependence, but it's certainly sub-linear). I don't think the Wiki page is taking this into account.

 

[DKS]

You're missing the key point I was making: the integral that appears in the formula for $t_r$ is a *definite* integral: it's taken from some finite value $R$ of the $r$ coordinate to infinity. So if you're trying to evaluate how $t_r$ relates to $t$ at infinity, you need to let the lower limit of the integral, $R$, tend to infinity itself. In *that* limit, the integral does vanish; heuristically, this is because you're adding together fewer and fewer terms, and the "number of terms" decreases linearly while the "size of the terms" increases only sub-linearly (I'm not sure it's as simple as just a square root dependence, but it's certainly sub-linear). I don't think the Wiki page is taking this into account.

You are wrong. You can't integrate $1/r^{1/2}$ to infinity.

 

[PeterDonis]

You are wrong. You can't integrate $1/r^{1/2}$ to infinity.

The Wikipedia page is implicitly doing the same thing; it's just not telling you what the limits of integration are.

On reading what you quoted, however, I admit it's garbled. I need to look at references again before trying to un-garble it.

 

[DKS]

The Wikipedia page is implicitly doing the same thing; it's just not telling you what the limits of integration are.

On reading what you quoted, however, I admit it's garbled. I need to look at references again before trying to un-garble it.

Wikipedia is fine. $t_r$ is undefined for $r \rightarrow \infty$. You may want to reconsider your reasoning that $t_r(evaporation) = t(evaporation)$.
E.g., http://www.math.unb.ca/~sseahra/resources/notes/black_holes.pdf eq. 108.

 

[PeterDonis]

http://www.math.unb.ca/~sseahra/resources/notes/black_holes.pdf eq. 108.

That equation does match what I'm finding in other references. However, it doesn't make sense to me, because, as you note, it makes $t_f \rightarrow \infty$ for $r \rightarrow \infty$. I had always understood that the Painleve chart "matched up" with the Schwarzschild chart as $r \rightarrow \infty$, but of course that can't be true if these formulas are right. I need to consider this some more.

 

[PeterDonis]

You may want to reconsider your reasoning that $t_r(evaporation) = t(evaporation)$.

No, the reasoning I gave in post #68 is still valid, though perhaps I didn't make it clear enough, hence the following.

First, that reasoning only involved events at finite values of $r$, and at any finite value of $r$, the difference between Painleve and Schwarzschild coordinate time is finite. Second, the reasoning was about the *difference* in coordinate time values between two events, the event where the far-away observer measures the mass of the hole to be $M_0$ (call this event M), and the event of the hole's final evaporation (call this event E). When taking differences in coordinate time values, the finite constant offset (at the finite value of $r$ where the far-away observer is) between Painleve and Schwarzschild coordinate time cancels out.

In other words, the formula you posted for the evaporation time as a function of $M_0$ is a formula for the *difference* in time between events M and E. The only wrinkle is that the far-away observer seeing the hole's final evaporation is delayed due to light travel time; the final flash of light has to get out to the far-away observer's $r$ from $r = 0$, and that takes time, so the coordinate time the far-away observer assigns to event E will be earlier than the time he observes it (call the event where he actually observes the flash event O). But I showed in my post how the "adjustment" in coordinate time to get the time of event E from the time of event O must be almost the same for Schwarzschild and Painleve coordinates; so the *difference* in Painleve coordinate time between event M and event E will be almost the same as the difference in Schwarzschild coordinate time between those two events. So in practical terms, the formula you gave can be interpreted as giving the coordinate time between events M and E in either chart.

 

[DKS]

No, the reasoning I gave in post #68 is still valid, though perhaps I didn't make it clear enough, hence the following.

First, that reasoning only involved events at finite values of $r$, and at any finite value of $r$, the difference between Painleve and Schwarzschild coordinate time is finite. Second, the reasoning was about the *difference* in coordinate time values between two events, the event where the far-away observer measures the mass of the hole to be $M_0$ (call this event M), and the event of the hole's final evaporation (call this event E). When taking differences in coordinate time values, the finite constant offset (at the finite value of $r$ where the far-away observer is) between Painleve and Schwarzschild coordinate time cancels out.

In other words, the formula you posted for the evaporation time as a function of $M_0$ is a formula for the *difference* in time between events M and E. The only wrinkle is that the far-away observer seeing the hole's final evaporation is delayed due to light travel time; the final flash of light has to get out to the far-away observer's $r$ from $r = 0$, and that takes time, so the coordinate time the far-away observer assigns to event E will be earlier than the time he observes it (call the event where he actually observes the flash event O). But I showed in my post how the "adjustment" in coordinate time to get the time of event E from the time of event O must be almost the same for Schwarzschild and Painleve coordinates; so the *difference* in Painleve coordinate time between event M and event E will be almost the same as the difference in Schwarzschild coordinate time between those two events. So in practical terms, the formula you gave can be interpreted as giving the coordinate time between events M and E in either chart.

That does not seem to be so. The coordinates for event M are $(t,r)=(0,R)$ in Schw. and $(0-f(R,M_0),R)$ in Painl. with $f$ the formula from wiki or the literature I quoted. For $r>>M$ $f \approx 2(2Mr)^{1/2}$, it goes to $-\infty$ at $r=2M$, then comes back up to 0 at $r=0$. At event E $M=0$ and $f=0$ so in both charts E is labeled $(T_e,R)$.

So in P. coordinates it takes $T_e+f(R,M_0)$ to evaporate, in S. coord. $T_e$. But that means nothing, a coordinate is just a bookkeeping device. The S. evaporation time does have a physical meaning: it is the proper time for the far observer.

If I understand you correctly you say that if you now also would calculate the time in P. coordinates that it takes for a free falling observer to fall into the singularity, this will be less than $T_e+f(R,M_0)$. This is clearly false on physical grounds if you just drop from $r=R$ for sufficiently large $R$ as he will never get even near the hole before it evaporates, so you clearly have to drop from sufficiently nearby but then I don't know how to compare coordinates at different $r$ for the dropper and the far observer. I know that if $M$ is constant the proper time to fall in is tiny compared to $T_e$ but since your time slows down if you approach the horizon, you should see $M$ decreasing faster and it's not clear to me who wins the race.

Can't we use Schwarzschild coordinates to compute at least the proper time it takes to reach the horizon, for the case of a slowly decreasing $M(t)$? If $M$ is constant this is very easy: you find that $r_o$ reaches $2M$ in the limit $t \rightarrow \infty$, but the limit of the proper time on the geodesic is finite. To compute what happens after that you have to use better coordinates, but if you don't care S. coordinates are just fine. So why can't we do this same calculation of the proper time to reach the horizon if $M = M(t) = M_0(1 - t/T_e)^{1/3}$ for the case of Hawking radiation?

 

[PAllen]

At one time, we had a discussion of the Krauss paper, and a rather technical rebuttal of it. I can't find the related thread, I recall that the authors noted the calculation for the evaporating hole was "non trivial".

See post #34 of this thread. I was the one who posted it in that other thread, and linked to it here again.

 

[PeterDonis]

So in P. coordinates it takes $T_e+f(R,M_0)$ to evaporate, in S. coord. $T_e$.

Yes, this is fine; and the $f(R, M_0)$ "correction term" is much, much smaller than the $T_e$ term for a black hole of stellar mass or larger and any reasonable value for $R$. For example, for the stellar mass black hole we've been considering, $T_e \approx 10^{67}$ years, and if we take a huge value for $R$, something like the radius of the observable universe (which is about $10^{23} M_0$), we get $f(R, M_0) \approx 1$ year.

But that means nothing, a coordinate is just a bookkeeping device. The . evaporation time does have a physical meaning: it is the proper time for the far observer.

True (at least to a very, very good approximation for the above value of $R$). But Painleve coordinate time is the proper time for an infalling observer, so if it turns out that the evaporation time calculated above (which, as I've just shown, is essentially the same in both charts for any practical value of $R$) is much, much larger than the Painleve coordinate time for the infalling observer to reach the horizon (which is the same as his proper time to reach the horizon and therefore has a direct physical meaning), then the infalling observer can indeed reach the horizon before the hole evaporates. See below.

This is clearly false on physical grounds if you just drop from $r=R$ for sufficiently large $R$

But what *is* a "sufficiently large" $R$? Have you calculated it? Obviously not, as we will see.

We have, from what I posted before, the infaller's proper time time for infall (to the singularity, since that formula is the one I actually posted, as I noted in some post or other recently), which is the same as the Painleve coordinate time for infall, $T = 2M_0 \left( R / 2M_0 \right)^{3/2}$. For the same numbers I gave above, $R \approx 10^{23} M_0$ and a stellar mass black hole, this gives $T \approx 10^{22}$ years, which is indeed much, much less than the evaporation time.

So while it is, technically, true that for "large enough" $R$, the infaller can't reach the hole before it evaporates, no value of $R$ within the observable universe is even close, by many, many orders of magnitude, to being "large enough". Put another way, there is clearly a *huge* range of $R$ values for which the infaller *can* reach the hole before it evaporates. (See further comment below on why time variation in $M$ does not affect the above.)

I don't know how to compare coordinates at different $r$ for the dropper and the far observer.

You don't have to. The infaller's proper time can be calculated in terms of the radius $R$ at which he starts falling, which can be taken to be the same $R$ at which the far-away observer is permanently located. That's what I just did above.

I know that if $M$ is constant the proper time to fall in is tiny compared to $T_e$ but since your time slows down if you approach the horizon, you should see $M$ decreasing faster

Only if you start falling at a large enough $R$ for the variation in $M$ to be significant during the time of infall. But as the above numbers show, the time of infall into a stellar mass black hole even for an $R$ the size of the observable universe is many, many orders of magnitude less than the evaporation time, so the variation in $M$ during the infall is negligible, as I've been saying all along.

Can't we use Schwarzschild coordinates to compute at least the proper time it takes to reach the horizon, for the case of a slowly decreasing $M(t)$?

You don't have to; see above. But also see further comment below.

why can't we do this same calculation of the proper time to reach the horizon if $M = M(t) = M_0(1 - t/T_e)^{1/3}$ for the case of Hawking radiation?

I don't think this will work, even taking limits, in Schwarzschild coordinates, because $t \rightarrow T_e$ in those coordinates as the horizon is approached for *any* infalling worldline, so $M(t) \rightarrow 0$ is going to happen because of the distortion in the coordinates.

Also, as I commented before, I don't think this equation for $M(t)$ is an exact equation, even in Schwarzschild coordinates; I think it's just somebody's heuristic guess based on the formula for $T_e$, which is also not an exact formula but just based on heuristic approximations. The heuristic approximations should work OK at very large $r$, where the far-away observer is, but I don't know that they still work close to the horizon, and doing the sort of integral you are talking about requires relying on that formula for events close to the horizon.

 

[PeterDonis]

I need to consider this some more.

Ok, after considering and getting input from experts, I agree that the formula given earlier is correct, which means that the Painleve coordinate time $T \rightarrow \infty$ as $r \rightarrow \infty$. So I was incorrect in using the $r \rightarrow \infty$ limit as I was using it before. (As I noted in a previous post, this means that there will be some sufficiently large $r$ for which an infaller indeed cannot reach an evaporating black hole before it evaporates; however, "sufficiently large" turns out to mean "much, much greater than the size of the observable universe" for a hole of stellar mass or larger, so in practical terms this possibility can be ignored.)

It's worth noting that there is still a sense in which Painleve coordinates "approach" Schwarzschild coordinates as $r \rightarrow \infty$: the surfaces of constant Painleve coordinate time get closer and closer to being parallel to the surfaces of constant Schwarzschild coordinate time as $r$ gets larger and larger.

 

[PeterDonis]

The heuristic approximations should work OK at very large $r$, where the far-away observer is, but I don't know that they still work close to the horizon, and doing the sort of integral you are talking about requires relying on that formula for events close to the horizon.

On thinking this over, there's a dramatic way of seeing that the heuristic approximations can't possibly work close to the horizon. Consider a curve of constant $t = T_e$ in Schwarzschild coordinates; as we've seen, this curve *is* the horizon. But now consider this: $M$ is *not constant* along this curve of constant $t$! It can't be, because $M$ can't be constant along the horizon, since the hole is evaporating. So looking at $M$ as a function of $t$ in Schwarzschild coordinates can't possibly be right. At the very least, in Schwarzschild coordinates, $M$ would have to be a function of $t$ and $r$; but even that requires thinking carefully about how the $r$ coordinate is defined: the $r$ coordinate of the horizon has to change as you move along the horizon, but this change itself can't be a function of $t$, since $t$ is constant along the horizon. (This, btw, is a better way of putting the objection I made earlier to considering $r_g$ as a function of $t$ in Schwarzschild coordinates.)

In principle the same thing applies to Painleve coordinates; it's quite possible that $M$ is not constant along a curve of constant Painleve coordinate time $T$ either. But it should be a lot closer to being constant along curves of constant $T$ close to the horizon, because close to the horizon curves of constant $T$ are close to being orthogonal to curves of constant $t$. Certainly events along the horizon have different $T$ coordinates, so in principle $M$ along the horizon could just be a function of $T$. However, I'll need to do some computations to investigate this further.

Of course this raises the question, is there a coordinate chart in which $M$ *is* always constant along curves of constant time? My initial guess is that there should be such a chart, but it might not be analogous to any of the "standard" charts. However, I also would like to take a look at Eddington-Finkelstein coordinates, since those are closer to the coordinates used in the Vaidya metric, which, as I've noted, explicitly includes a time-varying $M$.

 

[PeterDonis]

But it should be a lot closer to being constant along curves of constant $T$ close to the horizon

Thinking this over, it can't be right. Here's why: idealize the hole as spherically symmetric, and the outgoing Hawking radiation as successive spherically symmetric shells that move outward at the speed of light. At any given radius $r$, an observer sees $M$ decrease slightly when each successive shell of outgoing Hawking radiation passes him. So a surface of constant $M$ must be a null surface "in between" two successive shells of outgoing radiation. But a null surface can't be a surface of "constant time", which must be spacelike, not null. So $M$ must change along any surface of "constant time" in any chart.

In fact, the above analysis suggests that $M$ should be a function of a coordinate that is constant along the outgoing null surfaces described above. This is basically what the Vaidya metric does, and the coordinates are basically outgoing Eddington-Finkelstein coordinates (the double-null version of that chart). So finding $M$ as a function of $T$ and $r$ in Painleve coordinates (or $t$ and $r$ in Schwarzschild coordinates) can be done by finding $M$ as a function of $u$, the "outgoing" null Eddington Finkelstein coordinate, and then transforming between charts. I would expect that somewhere in the literature there is a paper where someone has done this, but I have not found one.

 

[DKS]

On thinking this over, there's a dramatic way of seeing that the heuristic approximations can't possibly work close to the horizon. Consider a curve of constant $t = T_e$ in Schwarzschild coordinates; as we've seen, this curve *is* the horizon. But now consider this: $M$ is *not constant* along this curve of constant $t$! It can't be, because $M$ can't be constant along the horizon, since the hole is evaporating. So looking at $M$ as a function of $t$ in Schwarzschild coordinates can't possibly be right. At the very least, in Schwarzschild coordinates, $M$ would have to be a function of $t$ and $r$; but even that requires thinking carefully about how the $r$ coordinate is defined: the $r$ coordinate of the horizon has to change as you move along the horizon, but this change itself can't be a function of $t$, since $t$ is constant along the horizon. (This, btw, is a better way of putting the objection I made earlier to considering $r_g$ as a function of $t$ in Schwarzschild coordinates.)

Aha! I think you hit the nail on the head. To check that I understand let me say it in my own words.
For a far away observer a static BH is described naturally by the Schwarzschild metric, and you can measure the mass parameter $M$ in the manner you described in an earlier post. If the BH is static you can prove that the Schw. solution describes the geometry everywhere outside $r=2M$.

Now I want to approximate evaporation and take Hawking formula to be correct. The radiated power is then computable and the far away observer can in principle measure $M$ every $10^60$ year and discover it slowly decreases. He can then model his region of spacetime by the S. metric with a slowly varying parameter $M = M_o(t)$. But unlike the static case he can't use Einsteins equations to compute the rest of the geometry outside his immediate vicinity, because when you make $M$ time dependent you have a well-defined metric but is does not solve Einsteins equations. So all we can conclude is that the true geometry will have some function $M(t,r)$ and all we can say is that $M(t,r) \approx M_o(t)$ for large $r$, but it's dependence on $r$ is not determined. To determine it you'd have to compute the backreaction of the Hawking radiation, i.e., the effective stress-energy tensor of the particle field and include that as a source term in Einsteins equations, a subject of several papers without definite consensus.

Before you do that calculation I don't believe you can show that an infalling observer must fall in. You argue that the $t_r$ in Painleve coordinates is proper time for the infalling observers, but this is only true if $M$ is constant.

I found a paper that claims to prove that for any rotationally symmetric backreaction stress-energy tensor, the horizon can never be crossed before it evaporates, more or less along the line of my reasoning: http://arxiv.org/abs/1102.2609.

It appeared in a peer-reviewed journal, so I assume it is not crackpottery. I did not find any refutation anywhere though.

I looked Ward's book on GR and Penrose's "Road to reality". Frustratingly they talk a lot about BH evaporation without showing that an observer can still fall in and hit the singularity before it evaporates. I'm starting to think that no-one really knows. If it was so obviously provable as you seem to believe, why would they not prove it?

 

[PeterDonis]

Now I want to approximate evaporation and take Hawking formula to be correct. The radiated power is then computable and the far away observer can in principle measure $M$ every $10^{60}$ year and discover it slowly decreases. He can then model his region of spacetime by the S. metric with a slowly varying parameter $M = M_o(t)$. But unlike the static case he can't use Einsteins equations to compute the rest of the geometry outside his immediate vicinity, because when you make $M$ time dependent you have a well-defined metric but is does not solve Einsteins equations. So all we can conclude is that the true geometry will have some function $M(t,r)$ and all we can say is that $M(t,r) \approx M_o(t)$ for large $r$

Yes, I agree with all of this (with the caveat that the metric with $M(t)$ slowly varying is an approximate solution of the EFE at large $r$; the problem comes when you try to extend the solution to small values of $r$).

but it's dependence on $r$ is not determined.

Not just from the above, no. But there might be other ways of figuring out, if not the complete function $M(t, r)$, at least enough about its behavior to show that there are plenty of infalling trajectories that reach the singularity before the hole finally evaporates. See below.

To determine it you'd have to compute the backreaction of the Hawking radiation, i.e., the effective stress-energy tensor of the particle field and include that as a source term in Einsteins equations, a subject of several papers without definite consensus.

This is what you'd have to do to get a complete analytical solution, yes. (Although the solution would still be semi-classical, not a full quantum gravity solution.) However, I don't think you need to do that to show that an infalling observer must fall in. But I'll defer further comment on that until I've looked at the paper you linked to and some other references.

 

[PeterDonis]

I found a paper that claims to prove that for any rotationally symmetric backreaction stress-energy tensor, the horizon can never be crossed before it evaporates, more or less along the line of my reasoning: http://arxiv.org/abs/1102.2609.

On a quick read-through, I don't think that's what this paper is saying. What it is saying is that the BH can never form in the first place; more precisely (I'm rephrasing somewhat to put it in terms that seem to me to be clearer), in compact regions of spacetime where current mainstream theory says there is a BH, there isn't actually a BH; instead, there is a "quasi-BH" and a "quasi-WH" ("WH" for "white hole"), joined together so that there is never an actual horizon anywhere. (Fig. 7 in the paper shows what this means.) If there is no horizon, then of course no observer can ever cross one. (There is also no singularity, so no observer can ever hit one.)

Note that even if this paper is correct, the *reason* it is correct is not that I have been making incorrect deductions from the model I've been using; it's because the model I've been using is (if the paper is correct) not the right model. I.e., the mathematical description of spacetime I have been using (based on the Penrose diagram I referred to lo, these many posts ago--a similar one is given on the right of Fig. 2 in the paper) is not the one that is actually physically realized; what is physically realized is a *different* mathematical description, the one in Fig. 7 of the paper. Everything I have been saying would still be correct, as a statement about the mathematical model I have been working from; that model would just not be physically applicable.

I'm still working through the details of the paper's computation purporting to show the above, but one thing struck me at the outset: the author makes a point of noting that the "standard" BH model including Hawking radiation (which is the one I've been using, the one with the Penrose diagram I referred to before) violates the dominant energy condition. This is true, and is a well-known property of an evaporating BH, which is known to be necessary for the BH (i.e., an "actual" BH with a horizon) to evaporate. The author, however, views this as a reason to invalidate the standard model; he basically believes that no physically reasonable model can violate the dominant energy condition. So his model does not (and that's why it can't have any horizons in it; his model has "Hawking radiation" observed at infinity, but the only way a horizon can radiate is if the dominant energy condition is violated--at least, assuming that the semiclassical approximation is valid.)

It's worth noting, though, that the author's belief that the dominant energy condition can't be violated is *not* a mainstream belief. It is well known that quantum fields can fairly easily produce effective stress-energy tensors, in the semiclassical approximation, that violate this condition. (Indeed, the quantum fields used in the standard derivation of Hawking radiation from a BH horizon have this property, as they must given the above.)

As I said, I'm still working through the details in the paper, but it may well turn out that our disagreement in this thread is, as I noted above, a disagreement about which mathematical model is physically realized, not about the properties of a particular mathematical model; we can both be right about the properties of the particular models we are using, we are just using different models. So we may have to just leave it at that.

 

[PeterDonis]

After reading through the paper some more, I have a couple of additional comments:

(1) The argument in section B, that for a "classical" evaporating BH, an infalling observer (indeed, an ingoing light ray) can never reach the horizon before it evaporates, seems inconsistent to me. The metric being used is derived assuming the dominant energy condition; but a BH can only evaporate in the first place if the dominant energy condition is violated. This is not just a heuristic result; it's one of the classic theorems in BH physics, proved first (I believe) by Hawking & Ellis in their classic monograph. The theorem says that, if the dominant energy condition holds, the area of a classical BH's event horizon can never decrease. So this section of the paper appears to be talking about a scenario that is not possible. (The same would go for the brief section C that talks about generalizations of the section B argument to non-spherically-symmetric BHs; the theorem referred to above makes no assumptions about symmetry.)

(2) In the "quasi-BH plus quasi-WH" model in section D, a key item gets almost no discussion: the "Hawking-Unruh mechanism" that makes the quasi-BH "bounce" and turn into a quasi-WH. I haven't looked at any of the paper's references, so I don't know what sort of detailed mechanism is being considered here: but whatever it is, it must be a *large* correction to the standard model of gravitational collapse. The general nature of this correction should be to change the effective stress-energy tensor of the collapsing matter to something that can successfully resist collapse.

However, if the effective stress-energy tensor still obeys the dominant energy condition, it's extremely hard to see how it could stop the collapse, for a simple reason: the dominant energy condition basically says that $\rho + p > 0$, i.e., the sum of energy density and pressure is positive, and that condition *increases* the chance of collapse. To decrease the chance of collapse, you need something like "dark energy" that can create a repulsive effect to counteract the attractive effect of the collapsing matter; and that violates the dominant energy condition.

 

[DKS]

After reading through the paper some more, I have a couple of additional comments:

Thanks, I printed that and attached it to my copy of the paper.

 

[PeterDonis]

$M$ should be a function of a coordinate that is constant along the outgoing null surfaces described above.

Just to clean up this point, PAllen reminded me of two interesting references that bear on this. One is the paper refuting Krauss et al. by showing that semiclassical back-reaction effects cannot prevent horizon formation given some fairly general assumptions:

http://arxiv.org/abs/0906.1768

The other is the paper by Ellis (of Hawking & Ellis) arguing that black holes cannot completely evaporate when the effects of external energy falling in from the rest of the universe (such as the CMBR) are taken into account:

http://arxiv.org/abs/1310.4771

(Both of these papers, btw, present a picture of how Hawking radiation is generated that is quite different from the usual heuristic picture. They essentially argue that Hawking radiation is generated when matter collapses sufficiently rapidly to form a trapped surface, and it is generated just outside the collapsing matter, i.e., at an apparent horizon, *not* at the event horizon of the spacetime as a whole. This makes a *big* difference; as Ellis argues, it essentially means that most of the Hawking radiation emitted by a black hole goes into the singularity rather than escaping to infinity. This is a very interesting subject, but it's mostly not relevant to the question of how to find a function $M(t, r)$ that captures how the total mass varies in an evaporating black hole spacetime, so I won't comment further on it here except in one particular respect where it helps--see below.)

The key point that I take away from both of these papers is that even when using Vaidya metric-type coordinates, there is no way to make $M$ a function of just one coordinate throughout the spacetime. The argument I gave for why $M$ should be a function only of the outgoing null coordinate $u$ (which labels outgoing null rays) applies at large $r$ (which also means at "late times", since it takes time for outgoing null rays to reach large $r$), but breaks down close to the horizon, because the horizon itself is an outgoing null ray, yet $M$ has to vary along it.

So what to do? The answer, close to and inside the horizon, is to instead look at *ingoing* null rays, which are labeled by the other Vaidya-type coordinate $v$ (each ingoing null ray is a curve of constant $v$, just as each outgoing null ray is a curve of constant $u$). Using the $v$ coordinate, events along the trajectory of the collapsing matter can be labeled, as well as events along the horizon (if a horizon forms, as we'll see it does), so we can construct a function $M(v, r)$ that works in the region of small $r$ where trying to construct a function $M(t)$ based on Schwarzschild coordinate time (or $M(u)$ based on outgoing null rays) breaks down. (If we wanted to construct a single function for $M$ valid throughout the spacetime, the easiest way would probably be to use double-null, i.e., Kruskal-type, coordinates, and make $M$ a function of $u$ and $v$ by finding a function $r(u, v)$ that transforms the $r$ coordinate into a function of the null coordinates.)

The specific form of $M(v, r)$ will depend on the details of the scenario, of course; but as the Padmanabhan paper shows, under fairly general conditions, even though $M$ may decrease during the collapse process (i.e., as $v$ increases from the value $v = 0$ that is assigned to the start of the collapse), it can't decrease fast enough to stop a horizon from forming. (The basic reason is that the rate of emission of Hawking radiation--here "rate" means "rate of change with respect to $v$"--depends on the *speed* of collapse, so there is a tradeoff: decreasing $M$ slows the collapse, but that also slows the rate of emission of radiation and thereby the rate of decrease of $M$, which prevents the collapse from slowing down enough to actually stop.) So there will be some $v$ (which will, in general, be of order $M(v = 0, R_0)$, the original mass of the collapsing object, if I am reading the papers right) at which an event horizon has formed and we have a black hole. (One way of physically interpreting this is to say that, if we have an observer who stays behind at $R_0$ when the object starts collapsing and sends null rays inward and watches for their reflections back from the surface of the collapsing matter, there will be some proper time of order $M(v = 0, R_0)$ by his clock at which the null ray he sends inward catches up with the collapsing matter just as the horizon forms, so he never sees that ray reflected back to him. This works because the transformation from $v$ to the observer's proper time is linear, so if $v$ is of order $M$, so is the observer's proper time.)

Once the hole forms, we can use $v$ as a time coordinate along the horizon, and we find that in terms of $v$, things work pretty much as I described them before in terms of Painleve coordinates. An object that falls in from some radius $R_0$ at some $v = v_0$ will reach the horizon at a $v$ of order $v_0 + M_0 \left( R_0 / M_0 \right)^{3/2}$, where $M_0$ is the value of $M$ that solves $M_0 = M(v_0, r = 2M_0)$; whereas, if the hole does evaporate completely (which it might never do if Ellis' arguments are correct), it will do so in a $v$ of order $M_0^3$, i.e., a much, much, much larger $v$ for a black hole of stellar mass or larger.