Infinite Time Dilation at the Surface of a Black Hole?

[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?

 

[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.