Collapse and unitary evolution

[PeterDonis]

What I claim is that photon's energy at infinity is relevant for Bekenstein-Hawking entropy

Which means, for how much mass is added to the hole.

while its wavelength near the horizon is relevant for the absorption

And this is the part I don't understand.

The first photon will get absorbed by the black hole

How much mass will it add to the hole?

The first photon will get absorbed by the black hole, due to the fact that $\tilde{\lambda}\ll R$.

How does the photon "know" that $\tilde{\lambda}\ll R$?

 

[Demystifier]

Which means, for how much mass is added to the hole.

Yes.

And this is the part I don't understand.

The absorption happens around the horizon, so the size when the wave is around the horizon is what matters.

How much mass will it add to the hole?

$\delta M=E=\sqrt{g_{00}(r)}\tilde{E}$

How does the photon "know" that $\tilde{\lambda} \ll R$?

The photon is a wave that responds to the local geometry of spacetime. The response is described by wave equation in curved spacetime.

 

[PeterDonis]

The photon is a wave that responds to the local geometry of spacetime.

But the horizon and its size are not local features of the spacetime geometry; they are global ones.

 

[Demystifier]

But the horizon and its size are not local features of the spacetime geometry; they are global ones.

The event horizon is global, but the apparent horizon is local. Perhaps I was not explicit in the paper, but I had the apparent horizon in mind. Indeed, if one considers a black hole with a non-constant mass $M$, then it is quite obvious that the horizon at $R=2M$ cannot be the event horizon.

 

[martinbn]

The event horizon is global, but the apparent horizon is local. Perhaps I was not explicit in the paper, but I had the apparent horizon in mind. Indeed, if one considers a black hole with a non-constant mass $M$, then it is quite obvious that the horizon at $R=2M$ cannot be the event horizon.

But it isn't obvious that any horizon is at $R=2M$.

 

[PeterDonis]

the apparent horizon is local

The presence of it is, yes, but its surface area and therefore its size is not.

 

[rubi]

There seems to be some kind of misunderstanding about the definition of an event horizon. An event horizon is not a 2-dimensional surface such as $R=2M$, but a 3-dimensional object in spacetime. It's the boundary $\partial J^-(\mathscr S^+)$ of the past of future null infinity.

 

[PeterDonis]

An event horizon is not a 2-dimensional surface

The term "event horizon" is sometimes used, not to refer to the 3-surface you describe, but to one of the 2-sphere surfaces that make it up; the event horizon is a 3-surface made up of an infinite series of 2-spheres. Quantities like the "area" of the horizon (which is what @Demystifier is implicitly depending on to specify the "size" of the horizon relative to an ingoing photon's wavelength) only make sense in reference to one of the 2-spheres.

 

[rubi]

The term "event horizon" is sometimes used, not to refer to the 3-surface you describe, but to one of the 2-sphere surfaces that make it up; the event horizon is a 3-surface made up of an infinite series of 2-spheres. Quantities like the "area" of the horizon (which is what @Demystifier is implicitly depending on to specify the "size" of the horizon relative to an ingoing photon's wavelength) only make sense in reference to one of the 2-spheres.

But those 2-surfaces are not an invariant concept. They depend on the choice of a foliation of spacetime. Each 2-surface is the intersection of the event horizon with one leaf of the foliation. Different foliations may intersect the event horizon differently, much like a plane that intersects a cylinder at different angles. Of course no physics will eventually depend on the choice, because GR doesn't require us to choose a foliation in the first place. One should be very careful when trying to make physical conclusions from frame dependent quantities like $R=2M$.

 

[Demystifier]

The presence of it is, yes, but its surface area and therefore its size is not.

How is it relevant to the question whether the photon can fit into the black hole? What you say looks to me like saying the size of a garage is not local and concluding that therefore a car cannot know whether it is small enough to fit into the garage. Which, of course, would be a nonsense.

 

[PeterDonis]

those 2-surfaces are not an invariant concept. They depend on the choice of a foliation of spacetime

No, they don't. The choice of foliation only affects what coordinate values you use to label each 2-surface. They don't affect which infinite set of 2-surfaces gets combined to form the 3-surface that is the boundary of the causal past of future null infinity, or the surface area of each of the 2-surfaces.

frame dependent quantities like $R=2M$.

$R$ is not frame-dependent; it's the areal radius of the 2-sphere, which is a geometric invariant.

$M$ is also not frame-dependent, if you ask which value of $M$ (we're assuming $M$ is not the same everywhere) applies on a given 2-surface that forms a portion of the event horizon. $M$ as a function of coordinates is of course frame-dependent, since coordinates themselves are.

 

[PeterDonis]

How is it relevant to the question whether the photon can fit into the black hole?

Because you are claiming that the photon's blueshifted wavelength allows it to fit inside the hole based on local quantities only:

The photon is a wave that responds to the local geometry of spacetime.

And I am saying that the area of the horizon, and therefore the areal radius $R$ which you are comparing to the photon's blueshifted wavelength, is not a feature of "the local geometry of spacetime". You can't tell what the horizon's area is just from local measurements. It's a global quantity. So either you are claiming that the photon does not respond to the local geometry of spacetime, in contradiction to what you said in what I just quoted, or you have not justified your claim that the photon can fit inside the hole if its blueshifted wavelength is smaller than $R$.

 

[PeterDonis]

What you say looks to me like saying the size of a garage is not local and concluding that therefore a car cannot know whether it is small enough to fit into the garage. Which, of course, would be a nonsense.

It would be nonsense because the size of the garage and the size of the car are both local in the sense you're using the term. But you have not convinced me that the size of the black hole is local in this sense.

 

[Demystifier]

And I am saying that the area of the horizon, and therefore the areal radius $R$ which you are comparing to the photon's blueshifted wavelength, is not a feature of "the local geometry of spacetime". You can't tell what the horizon's area is just from local measurements. It's a global quantity. So either you are claiming that the photon does not respond to the local geometry of spacetime, in contradiction to what you said in what I just quoted, or you have not justified your claim that the photon can fit inside the hole if its blueshifted wavelength is smaller than $R$.

Suppose that we want to drop a photon the wavelength of which is much smaller than $R$ even at infinity. In this way the issue of blueshift becomes unimportant, because the photon is now small enough even without the blueshift. 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?

 

[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?

No, because the wavelength of the photon at infinity is not local either; it's just the energy at infinity of the photon, which is a global constant of its motion, put into different units by way of Planck's constant.

 

[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?