# B Collapse and unitary evolution

#### 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.

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#### Demystifier

2018 Award
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

Mentor
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

2018 Award
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

Mentor
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

2018 Award
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

Mentor
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

2018 Award
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

Mentor
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.

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#### PeterDonis

Mentor
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

2018 Award
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

Mentor
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

Mentor
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

Mentor
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

2018 Award
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

Mentor
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

Mentor
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

2018 Award
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

Mentor
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

Mentor
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

2018 Award
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 possess 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. $u^\mu p_\mu$.