Collapse and unitary evolution

[PeterDonis]

If I tell you that the photon will be absorbed by the black hole because it is smaller than the black hole, would you still find it controversial because the size of the horizon is not local?

To answer this another way: I would agree that the photon will be absorbed if its energy at infinity, converted to a wavelength, is much less than $R$, but I might object to the ordinary language phrasing "because it is smaller than the black hole", because neither the Schwarzschild radius $R$ nor the photon wavelength are "sizes of objects" in the usual sense of that ordinary language term. What both of them are are invariant quantities that describe the hole and the photon, respectively. Back in post #98, I asked what invariant corresponds to the photon's blueshifted wavelength. Is there one?

 

[Demystifier]

To answer this another way: I would agree that the photon will be absorbed if its energy at infinity, converted to a wavelength, is much less than $R$, but I might object to the ordinary language phrasing "because it is smaller than the black hole", because neither the Schwarzschild radius $R$ nor the photon wavelength are "sizes of objects" in the usual sense of that ordinary language term. What both of them are are invariant quantities that describe the hole and the photon, respectively. Back in post #98, I asked what invariant corresponds to the photon's blueshifted wavelength. Is there one?

How is the photon's energy or wavelength at infinity an invariant? I don't think that it's an invariant because it depends on the choice of the Lorentz frame.

Hence I do not think that the whole problem can easily be formulated in terms of invariants. Instead, I think it's much easier to formulate it in terms of "preferred" frames, that is frames which are physically natural to the problem at hand.

For instance, in discussing effects related to Lorentz contraction, the "preferred" frame is the one in which the rod is at rest. This is the simplest way to understand e.g. the Bell spaceship paradox, which is not easy to understand in terms of invariants.

Similarly, with issues where black holes are involved, a "preferred" frame is a one in which black hole is at rest. But there is an ambiguity here, because there is a big difference between static observer at infinity and static observer near the horizon, even though they both see that the black hole is at rest. So my argument (which, I admit, is not fully rigorous) is that the photon absorption happens when it crosses the "line" $R=2M$, so for the absorption purposes the "preferred" frame is a frame of an observer close to $R=2M$. It seems physically natural to me, even if I cannot make a fully rigorous argument.

If you think that the "preferred" observer is a static observer at infinity, can you give a rigorous argument for that?

 

[PeterDonis]

How is the photon's energy or wavelength at infinity an invariant?

It's a constant of the motion because the spacetime has a timelike Killing vector field.

I don't think that it's an invariant because it depends on the choice of the Lorentz frame.

The constant of the motion called "energy at infinity" is not frame-dependent; it's just $\xi^\mu p_\mu$, where $\xi$ is the timelike Killing vector field and $p$ is the photon's 4-momentum. This is manifestly invariant.

 

[PeterDonis]

The constant of the motion called "energy at infinity" is not frame-dependent; it's just $\xi^\mu p_\mu$, where $\xi$ is the timelike Killing vector field and $p$ is the photon's 4-momentum. This is manifestly invariant.

And just to be clear, this invariant also tells how much the hole's mass increases when it absorbs the photon (or any other object, for that matter).

 

[Demystifier]

It's a constant of the motion because the spacetime has a timelike Killing vector field.

The constant of the motion called "energy at infinity" is not frame-dependent; it's just $\xi^\mu p_\mu$, where $\xi$ is the timelike Killing vector field and $p$ is the photon's 4-momentum. This is manifestly invariant.

Now it's my turn to be critical. If the mass of the black hole is not constant, then the spacetime does not possesses a Killing vector field. So in general, invariant quantities cannot be defined in terms of Killing fields.

What one can always do is to pick some observer with 4-velocity $u^\mu$ and define invariants in terms of that, e.g. $u^\mu p_\mu$.

 

[PeterDonis]

If the mass of the black hole is not constant, then the spacetime does not possesses 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]

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]

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]

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]

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]

"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]

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]

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]

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]

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]

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]

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]

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]

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 $M\rightarrow M(t)$.

I'll ask if there is such a result about the approximation

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]

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]

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.

Ah, ok, so there is an assumption that the space-time is spherecally symmetric.

 

[Demystifier]

By the way, I have proposed a version in which Hawking radiation is completely irrelevant.
https://arxiv.org/abs/1802.10436

The paper is now published in Universe. It's completely open: everyone can see the paper, the names of reviewers, their reports, and my responses to reviewers.
https://www.mdpi.com/2218-1997/9/1/11

 

[physika]

new interactions does emerge as complexity increases, that was physically impossible at lower complexity.

But this is not a viable strategy for one human interacting with other humans. Instead the complex systems develops behaviour that due to chaos can not be inferred from knowledge of interaction of parts.

Example; altruistic attitudes.

 

[physika]

I'm not sure who said people weren't made of particles.

Indeed.