High School Collapse and unitary evolution

[Demystifier]

I understand why the wavelength is relevant. I don't understand how you can consider the photon's wavelength to be blueshifted but not its energy.

Unlike wavelength, the energy is conserved. So when the photon energy is blushifted, one can say that what is increased is the kinetic energy of the photon, while its potential energy in the gravitational field is decreased (by becoming negative), so that the total energy does not change.

 

[PeterDonis]

when the photon energy is blushifted, one can say that what is increased is the kinetic energy of the photon, while its potential energy in the gravitational field is decreased (by becoming negative), so that the total energy does not change.

This is just another way of saying that the energy the photon adds to the hole is its energy at infinity, which I already agree with.

If I understand you correctly, you are basically saying that there is no "wavelength at infinity" corresponding to energy at infinity. But that still doesn't explain why it's justified to use the blueshifted wavelength as the criterion for whether the photon will "fit inside the black hole". The blueshifted wavelength is the wavelength that would be measured by observers hovering close to, but outside, the horizon--but those observers will also measure the photon's energy to be blueshifted (they will measure what you are calling the kinetic energy of the photon above). I'm not aware of any observer who will measure the photon's wavelength to be blueshifted but still measure its energy to be the same as its energy at infinity. So what justifies using the blueshifted wavelength while still using the energy at infinity?

 

[Demystifier]

This is just another way of saying that the energy the photon adds to the hole is its energy at infinity, which I already agree with.

If I understand you correctly, you are basically saying that there is no "wavelength at infinity" corresponding to energy at infinity. But that still doesn't explain why it's justified to use the blueshifted wavelength as the criterion for whether the photon will "fit inside the black hole". The blueshifted wavelength is the wavelength that would be measured by observers hovering close to, but outside, the horizon--but those observers will also measure the photon's energy to be blueshifted (they will measure what you are calling the kinetic energy of the photon above). I'm not aware of any observer who will measure the photon's wavelength to be blueshifted but still measure its energy to be the same as its energy at infinity. So what justifies using the blueshifted wavelength while still using the energy at infinity?

The observer far from the black hole (Alice) cannot measure the wavelength of the photon near the horizon. All what she can is to determine whether the photon was absorbed by the black hole or merely scattered. The blueshift of the wavelength makes sense only from the point of view of the observer near the horizon (Bob). So Bob will see a blueshift in both wavelength and energy. And due to the blueshift in wavelength, he will conclude that near the horizon the wavelength is sufficiently small so that the wave can enter the black hole. And so the photon will be absorbed from the point of view ob Bob. But Alice cannot disagree on the fact that the photon has been absorbed, so she will observe absorption (or more precisely the lack of scattering) too. How will Alice interpret this? She cannot see the shrinking of the wave (because she cannot see the wave at all because she is far from the wave when it gets shrinked), but she will say that the wave shrinked objectively, without her observation.

 

[PaleMoon]

You write that the photon is absorbed from the point of view of Bob.
Do you think that events are observer dependent?

 

[Demystifier]

You write that the photon is absorbed from the point of view of Bob.
Do you think that events are observer dependent?

No.

 

[PeterDonis]

Bob will see a blueshift in both wavelength and energy.

Yes, agreed.

due to the blueshift in wavelength, he will conclude that near the horizon the wavelength is sufficiently small so that the wave can enter the black hole

But if he concludes this, doesn't he also have to conclude that the absorption process adds the photon's blueshifted energy to the hole's mass?

Alice cannot disagree on the fact that the photon has been absorbed

Agreed, whether or not the photon is absorbed must be an invariant. But so must the increase in mass of the black hole as a result, correct? And yet it seems like Bob will see a different mass increase than Alice.

 

[Demystifier]

But if he concludes this, doesn't he also have to conclude that the absorption process adds the photon's blueshifted energy to the hole's mass?

No, because mass, unlike energy, is defined as an invariant, observer independent quantity. See e.g. https://en.wikipedia.org/wiki/Mass_...ndi_masses_in_asymptotically_flat_space-times

 

[PeterDonis]

mass, unlike energy, is defined as an invariant, observer independent quantity

This is quibbling. The mass of a black hole is its energy in the asymptotically flat frame normally used to describe it. And the photon's energy at infinity, which is the energy it adds to the hole, is also an invariant.

If your argument is that we should focus on invariants, then what invariant corresponds to the photon's blueshifted wavelength? Wavelength is no more invariant than the photon's blueshifted energy is, by the argument you are making.

 

[Demystifier]

The mass of a black hole is its energy in the asymptotically flat frame normally used to describe it. And the photon's energy at infinity, which is the energy it adds to the hole, is also an invariant.

If your argument is that we should focus on invariants, then what invariant corresponds to the photon's blueshifted wavelength? Wavelength is no more invariant than the photon's blueshifted energy is, by the argument you are making.

Ah, I think I understand now what bothers you, so now I think I finally have the answer that will satisfy you. One can introduce the observer-dependent black-hole mass $\tilde{M}(r)$, which depends on the observer's position $r$ according to the Tolman's law
$$\tilde{M}(r)=\frac{M}{\sqrt{g_{00}(r)}}$$
where
$$g_{00}(r)=1-\frac{2M}{r}$$
In particular,
$$\tilde{M}(\infty)=M$$
is the usual ADM mass seen by the observer at infinity. So now you can say that the observer at position $r$ sees a blueshifted mass given by the first equation above.

However, the Bekenstein-Hawking entropy is given by the equation
$$S_{BH}=\frac{A}{4}=4\pi M^2$$
and entropy does not depend on the observer. In my paper the mass is only needed to determine the Bekenstein-Hawking entropy, so for this purpose I need $M$, not $\tilde{M}(r)$. Of course, you can argue that the physical mass observed by observer at $r$ is $\tilde{M}(r)$, but then I can say - fine, the entropy can be written in terms of $\tilde{M}(r)$ as
$$S_{BH}=4\pi g_{00}(r) \tilde{M}^2(r)$$
which in fact does not depend on $r$. So you are right that there is a blueshift of mass, but it is not relevant in the context of my paper which is really about entropy. That's why I am allowed to talk only about the invariant mass $M$, and not about the blueshifted mass $\tilde{M}(r)$. On the other hand I keep talking about the blueshifted wavelength because it is relevant for the question whether the wave can fit into the black hole. I could rephrase the arguments in my paper in terms of the blueshifted mass $\tilde{M}(r)$, but that would sound somewhat unusual and would not influence the results.

 

[PeterDonis]

One can introduce the observer-dependent black-hole mass $\tilde{M}(r)$, which depends on the observer's position rrr

Do you have a reference for this? I've never seen it in any GR texts or papers I have read.

Also, what physical measurement does $\tilde{M}(r)$ correspond to? Measuring the mass of the hole by the usual methods--Keplerian orbit parameters--gives what you are calling $\tilde{M}(\infty)$, not $\tilde{M}(r)$.

 

[A. Neumaier]

I'll start a new thread soon, as I think it does change something of substance and I'm not sure of the degree to which "made of quantum fields" is true either.

I am looking forward to this thread. I would say, every object in nature may be regarded as being that part of the collection of quantum fields defined by the standard model plus gravity localized in the region of space-time where the object is located.

 

[pBrane]

The problem is not associated with absorption of matter but with Hawking radiation.

I think I agree with Demystifier, but I only guess.

My view; The information about the particle (label) is left splattered at the event horizon. The information in the particle (value) goes into the non-3D arena for audit history purposes and for the pleasure of those in ((3D)+).

Information is not lost at at all. It just goes beyond our view.

Without ever meeting the guy, my guess is that Dr Susskind would say the the particle's label info reflects the last interaction of the particle and is no particular value to anyone or thing.

The result/effect of that last interaction would have been added to the value in the particle itself, and that is very valuable to the particle.

Hawking can radiate whatever he wants from the horizon surface, nothing of value is coming out any no-time soon.

Does anyone know the bars Susskind hangs out in?

I need a drink and a shower, i need a drink most.

 

[Demystifier]

Do you have a reference for this? I've never seen it in any GR texts or papers I have read.

Also, what physical measurement does $\tilde{M}(r)$ correspond to? Measuring the mass of the hole by the usual methods--Keplerian orbit parameters--gives what you are calling $\tilde{M}(\infty)$, not $\tilde{M}(r)$.

I have never seen before such a formula for mass (which is why I haven't wrote it before), but it is a well known formula for any energy-like quantity that suffers a blueshift. It was you who insisted that mass is just energy and hence must obey a blueshift, which made me to comply with you and say - fine, if you insist that mass must be blushifted, then it can only be blushifted by the formula above. (I called it the "Tolman" law, but strictly speaking the Tolman law is the law for temperature, $\tilde{T}(r)=T/\sqrt{g_{00}(r)}$.)

Now if you ask me how that mass would be measured, my answer is that I don't know. That's why I hesitated to talk about blueshifted mass, until you insisted.

But now it looks as if I can never satisfy you. If I say that mass is not blueshifted because mass is not measured that way, then you object that it is just energy so must be blueshifted. If I try to comply with you and say, fine, mass is also blueshifted, then you object that I haven't specify how to measure this blueshift. Do you have your own strong opinion on that (in which case it would help if you could express it unambiguously), or are you just confused?

My view is that a blushifted mass can be introduced formally, just for the sake of theoretical idea that any energy-like quantity should be blueshifted, but that such a concept of a blueshifted mass is not very useful form a practical experimental point of view. I'm sure someone could contrive some method of measurement of mass that would obey the blueshift formula above, but at the moment nothing simple and natural of that kind comes to my mind.

Or we can work this way. First you give a precise definition of what exactly do you mean by "mass", and then I will tell you whether this mass is blushifted or not, and what, in the context of your definition, that means. Before giving your definition, I want to remind you that it is very tricky and ambiguous https://en.wikipedia.org/wiki/Mass_in_general_relativity

 

[PeterDonis]

It was you who insisted that mass is just energy and hence must obey a blueshift

No, that's not what I said. I have already agreed that the mass that the photon adds to the hole is the mass equivalent of its energy at infinity, not its blueshifted energy. And I have never said that the mass of the black hole should be blueshifted.

What I do not agree with is saying that one can use the blueshifted wavelength to determine whether the photon "fits" into the hole, while still saying that the photon's energy at infinity determines the mass that gets added to the hole. If the photon's energy at infinity is what is relevant, then the photon's wavelength at infinity should also be what is relevant. You keep insisting that the photon's blueshifted wavelength is somehow relevant, and I keep asking for some argument to justify this, because I don't think it is.

 

[Demystifier]

If the photon's energy at infinity is what is relevant, then the photon's wavelength at infinity should also be what is relevant. You keep insisting that the photon's blueshifted wavelength is somehow relevant, and I keep asking for some argument to justify this, because I don't think it is.

Relevant for what? What I claim is that photon's energy at infinity is relevant for Bekenstein-Hawking entropy, while its wavelength near the horizon is relevant for the absorption. There is no contradiction, because different quantities at different positions are relevant for different things.

For a justification, consider the following thought experiment. A quantum-optics laboratory is built at a position $r$ very near the horizon at $R=2M$. In the laboratory two photons are produced, each with energy $\tilde{E}$ as seen in the laboratory. The energy $\tilde{E}$ is sufficiently big so that the wavelength $\tilde{\lambda}=1/\tilde{E}$ satisfies $\tilde{\lambda}\ll R$. One photon is sent to the black-hole interior, while the other is sent to the infinity. The first photon will get absorbed by the black hole, due to the fact that $\tilde{\lambda}\ll R$. The second photon will be eventually observed by observer at infinity, who will see that it has the wavelength $\lambda\gg R$, where $\lambda=1/E$ and $E=\sqrt{g_{00}(r)}\tilde{E}$. So the photons were created with equal energy, yet one gets absorbed and another becomes bigger than the black hole. I don't see any contradiction in that.

 

[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.