# Collapse and unitary evolution

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PeterDonis
Mentor
If the mass of the black hole is not constant, then the spacetime does not possess a Killing vector field.
Technically that is true, yes. But the process you are considering adds, in the limit, zero mass to the hole, so the hole's mass does not change. If there is a finite lower bound to the mass that can be added to the hole by a photon falling in, so that the hole's mass has to change, then your argument breaks down anyway.

Demystifier
2018 Award
Technically that is true, yes. But the process you are considering adds, in the limit, zero mass to the hole, so the hole's mass does not change. If there is a finite lower bound to the mass that can be added to the hole by a photon falling in, so that the hole's mass has to change, then your argument breaks down anyway.
It doesn't seem that our discussion converges to an agreement. Can we at least agree to disagree and finish with this discussion? (Which, by the way, is an off topic because the thread is entitled "Collapse and unitary evolution".)

martinbn
Can we at least agree to disagree and finish with this discussion?
Before that can someone phrase the qustion (and may be the statement) in a strict GR language.

PeterDonis
Mentor
It doesn't seem that our discussion converges to an agreement. Can we at least agree to disagree and finish with this discussion?
I agree it doesn't seem like we're converging, and I don't see what further progress we can make.

(Which, by the way, is an off topic because the thread is entitled "Collapse and unitary evolution".)
I'll look back and see if it's feasible to spin this discussion off into its own thread.

Before that can someone phrase the qustion (and may be the statement) in a strict GR language.
My understanding of the claim @Demystifier made in the paper of his that he linked to (many posts back) is that the Bekenstein bound can be violated by letting soft photons of arbitrarily low energy at infinity fall into a black hole. Such a process, if it could take place, would add, in the limit, zero energy to the black hole, while still adding a finite positive amount of entropy associated with the additional degrees of freedom of the photons (over and above the degrees of freedom already inside the hole). So one could, in principle, add an unbounded amount of entropy to a black hole while keeping its mass constant.

My objection to the claim is that it requires that photons of arbitrarily low energy at infinity (and therefore arbitrarily long wavelength as seen by an observer very far away) can still "fit" into the hole. @Demystifier argues that, for the purpose of determining whether a photon can "fit" inside the hole, we should look at its wavelength in the limit as the horizon is approached. That wavelength will be highly blueshifted, to the point where a photon of arbitrarily low energy at infinity can still fit inside the hole. But to me, this is trying to have it both ways: to use the energy at infinity (i.e., non-blueshifted) to determine how much mass gets added to the hole (in the limit, zero), while using the wavelength near the horizon (i.e., highly blueshifted) to determine whether the photon can "fit" inside the hole. That doesn't seem consistent to me, but neither of us have been able to convince the other.

Boing3000
Gold Member
by letting soft photons of arbitrarily low energy at infinity fall into a black hole.
That will again sound like a stupid question but if you can mathematically build some arbitrary low energy photon only "at infinity", isn't it the case that those photon simply cannot reach the BH at all... ?

PeterDonis
Mentor
if you can mathematically build some arbitrary low energy photon only "at infinity"
"Energy at infinity" is a technical term; heuristically, you can think of it as what says the same as an object falls in a gravitational field, with its kinetic energy increasing and its potential energy decreasing. It's not a quantity that can only be represented mathematically at infinity or that only has physical meaning at infinity.

Boing3000
Gold Member
"Energy at infinity" is a technical term; heuristically, you can think of it as what says the same as an object falls in a gravitational field, with its kinetic energy increasing and its potential energy decreasing. It's not a quantity that can only be represented mathematically at infinity or that only has physical meaning at infinity.
I understood that. I should have said "compute" not "build". But how can this exact value be relevant in that scenario ? No photon will ever be at infinity, and every BH will be evaporated long before that anyway.

Maybe a more practical value would be at a distance = c * evaporation time ?

PeterDonis
Mentor
how can this exact value be relevant in that scenario ? No photon will ever be at infinity
Because, as I said, "energy at infinity" is a technical term; it does not mean the value of that quantity is only relevant or well-defined at infinity. "Energy at infinity" just happens to be the ordinary language term that physicists have adopted for this quantity.

PeterDonis
Mentor
how can this exact value be relevant in that scenario ?
To answer this another way, if you read through the thread (which admittedly is getting long now), you will see that @Demystifier and I both agree that the energy at infinity of the photon determines the mass that gets added to the hole when the photon falls in--not the energy the photon has as seen by an observer hovering close to the horizon.

Demystifier
2018 Award
Because, as I said, "energy at infinity" is a technical term; it does not mean the value of that quantity is only relevant or well-defined at infinity. "Energy at infinity" just happens to be the ordinary language term that physicists have adopted for this quantity.
For practical purposes, "infinity" can be interpreted as a large radius $r\gg R=2M$.

martinbn
For practical purposes, "infinity" can be interpreted as a large radius $r\gg R=2M$.
And how is radius $r$ defined?

Demystifier
2018 Award
And how is radius $r$ defined?
It's the usual radial coordinate in the Schwarzshild metric. Is that precise enough?

martinbn
It's the usual radial coordinate in the Schwarzshild metric. Is that precise enough?
Wait, the whole discution is about the Schwarztschild solution only?

Demystifier
2018 Award
Wait, the whole discution is about the Schwarztschild solution only?
Almost. It's about uncharged nonrotating black hole, with a possibly non-constant mass. If the mass changes sufficiently slowly, then the metric can be approximated by Schwarztschild metric with $M\rightarrow M(t)$.

Now you will ask me to define "sufficiently slowly".

martinbn
Almost. It's about uncharged nonrotating black hole, with a possibly non-constant mass. If the mass changes sufficiently slowly, then the metric can be approximated by Schwarztschild metric with $M\rightarrow M(t)$.

Now you will ask me to define "sufficiently slowly".
No, I'll ask if there is such a result about the approximation, or is it just expected. What about stubility?

Demystifier
2018 Award
No, I'll ask if there is such a result about the approximation, or is it just expected. What about stubility?
It's expected by my physical intuition, but maybe I should add that I expect it for $r>2M(t)$. I'm sure that someone made explicit calculations and haven't obtained anything very surprising, because otherwise I would already heard about that. After all, the mass of the Sun is not constant, yet the Schwarzschild metric describes the motion of planets around Sun very well.

PeterDonis
Mentor
It's the usual radial coordinate in the Schwarzshild metric.
Or, to make clear that this $r$ labels a geometric invariant, it is the "areal radius" of the 2-sphere on which a given event lies, i.e., if the 2-sphere's area is $A$, then $A = 4 \pi r^2$, so $r = \sqrt{A / 4 \pi}$.

PeterDonis
Mentor
It's about uncharged nonrotating black hole, with a possibly non-constant mass. If the mass changes sufficiently slowly, then the metric can be approximated by Schwarztschild metric with $M\rightarrow M(t)$.
The Vaidya metric is an exact solution describing a non-rotating, uncharged black hole either emitting or absorbing null dust (basically incoherent EM radiation emitted or absorbed isotropically, equally in all directions). See here:

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

Note that the usual way of writing this metric is in the equivalent of Eddington-Finkelstein coordinates, where there is a null coordinate $u$ or $v$ (depending on whether you are looking at the outgoing--emitting radiation--or ingoing--absorbing radiation--case) instead of the Schwarzschild $t$. For this case, $M$ is a function of $u$ or $v$ only, as shown in the article, and this is true regardless of the rate of change of $M$ with respect to $u$ or $v$.

I believe what @Demystifier is referring to is an approximation in which (for the ingoing case) $dM / dv$ is small enough that you can transform into standard Schwarzschild coordinates and still have $M$ be a function of only $t$ (instead of both $t$ and $r$, as it would be in the general case with unrestricted rate of change) for the duration of the process he is analyzing.

PeterDonis
Mentor
the mass of the Sun is not constant, yet the Schwarzschild metric describes the motion of planets around Sun very well
The usual treatment of the Solar System is basically the order by order PPN expansion of the Schwarzschild metric for the case of small $M / r$, yes. But this works well because the mass loss (from radiation, solar wind, etc.) is so small compared to $M$ that its effects are negligible. It's not because the solution explicitly uses an $M$ that is changing with time (at least, that's my understanding).

martinbn
It's expected by my physical intuition, but maybe I should add that I expect it for $r>2M(t)$. I'm sure that someone made explicit calculations and haven't obtained anything very surprising, because otherwise I would already heard about that. After all, the mass of the Sun is not constant, yet the Schwarzschild metric describes the motion of planets around Sun very well.
But the Sun is very far from a black hole. I think it requires more justification than intuition to extrapolate. It would be interesting to see a theorem.
Or, to make clear that this $r$ labels a geometric invariant, it is the "areal radius" of the 2-sphere on which a given event lies, i.e., if the 2-sphere's area is $A$, then $A = 4 \pi r^2$, so $r = \sqrt{A / 4 \pi}$.
Which two sphere?
The Vaidya metric is an exact solution describing a non-rotating...
My exclamation wasn't about which solution describes the scenario but about the fact that it is one solution that he had in mind. The statement seems to be about black holes in general, then it turns out that when physicists say "black hole" they mean the Schwartzschild solution or Vaiday, or something else but just one solution. Demystifier said that it should be a good enough approximation, which is probably true, but it would be interesting to see a precise statement.

PeterDonis
Mentor
Which two sphere?
Whichever 2-sphere the particular event you are interested in (the one labeled with a given value of $r$) lies on. The entire spacetime is spherically symmetric, which means every event in the spacetime lies on some 2-sphere with a definite area.

martinbn
Whichever 2-sphere the particular event you are interested in (the one labeled with a given value of $r$) lies on. The entire spacetime is spherically symmetric, which means every event in the spacetime lies on some 2-sphere with a definite area.