Collapse and unitary evolution

In summary, Susskind's book "The Black Hole War" discusses the concept of unitarity in quantum mechanics and how it relates to the issue of information loss in black holes. He argues that information cannot be lost in quantum mechanics because of unitarity, but this raises questions about the collapse interpretation of quantum mechanics. Susskind favors the many-worlds interpretation and believes that collapse never occurs. However, the issue of information loss in black holes remains even in no collapse interpretations like the many-worlds interpretation. Susskind also discusses the AdS/CFT correspondence and its role in solving the paradox of information in black holes. There is some bias in his conclusion, as the AdS/CFT correspondence is still unproven and may
  • #106
Demystifier said:
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.

Demystifier said:
while its wavelength near the horizon is relevant for the absorption

And this is the part I don't understand.

Demystifier said:
The first photon will get absorbed by the black hole

How much mass will it add to the hole?

Demystifier said:
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##?
 
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  • #107
PeterDonis said:
Which means, for how much mass is added to the hole.
Yes.

PeterDonis said:
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.

PeterDonis said:
How much mass will it add to the hole?
##\delta M=E=\sqrt{g_{00}(r)}\tilde{E}##

PeterDonis said:
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.
 
  • #108
Demystifier said:
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.
 
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  • #109
PeterDonis said:
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.
 
  • #110
Demystifier said:
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##.
 
  • #111
Demystifier said:
the apparent horizon is local

The presence of it is, yes, but its surface area and therefore its size is not.
 
  • #112
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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  • #113
rubi said:
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.
 
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  • #114
PeterDonis said:
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##.
 
  • #115
PeterDonis said:
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.
 
  • #116
rubi said:
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.

rubi said:
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.
 
  • #117
Demystifier said:
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:

Demystifier said:
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##.
 
  • #118
Demystifier said:
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.
 
  • #119
PeterDonis said:
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?
 
  • #120
Demystifier said:
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.
 
  • #121
Demystifier said:
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?
 
  • #122
PeterDonis said:
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?
 
  • #123
Demystifier said:
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.

Demystifier said:
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.
 
  • #124
PeterDonis said:
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).
 
  • #125
PeterDonis said:
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##.
 
  • #126
Demystifier said:
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.
 
  • #127
PeterDonis said:
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".)
 
  • #128
Demystifier said:
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.
 
  • #129
Demystifier said:
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.

Demystifier said:
(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.

martinbn said:
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.
 
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  • #130
PeterDonis said:
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... ?
 
  • #131
Boing3000 said:
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.
 
  • #132
PeterDonis said:
"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 ?
 
  • #133
Boing3000 said:
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.
 
  • #134
Boing3000 said:
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.
 
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  • #135
PeterDonis said:
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##.
 
  • #136
Demystifier said:
For practical purposes, "infinity" can be interpreted as a large radius ##r\gg R=2M##.
And how is radius ##r## defined?
 
  • #137
martinbn said:
And how is radius ##r## defined?
It's the usual radial coordinate in the Schwarzshild metric. Is that precise enough?
 
  • #138
Demystifier said:
It's the usual radial coordinate in the Schwarzshild metric. Is that precise enough?
Wait, the whole discution is about the Schwarztschild solution only?
 
  • #139
martinbn said:
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". o0)
 
  • #140
Demystifier said:
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". o0)
No, I'll ask if there is such a result about the approximation, or is it just expected. What about stubility?
 
<h2>1. What is the difference between collapse and unitary evolution?</h2><p>Collapse refers to the collapse of a quantum state into a definite state when it is observed or measured. Unitary evolution, on the other hand, is the continuous and deterministic evolution of a quantum state according to the Schrödinger equation.</p><h2>2. How does collapse occur in quantum systems?</h2><p>Collapse occurs when a quantum system interacts with a classical measuring apparatus, causing the superposition of states to collapse into a definite state. This is known as the measurement problem in quantum mechanics.</p><h2>3. Can unitary evolution and collapse coexist?</h2><p>Yes, they can coexist in the Copenhagen interpretation of quantum mechanics. In this interpretation, quantum systems evolve unitarily until they are observed, at which point collapse occurs. However, there are alternative interpretations that do not involve collapse, such as the Many-Worlds interpretation.</p><h2>4. How does the concept of entanglement relate to collapse and unitary evolution?</h2><p>Entanglement is a phenomenon where two or more quantum systems become correlated in such a way that their individual states cannot be described independently. This can lead to collapse when one of the entangled systems is observed, causing the other system to also collapse. Unitary evolution can also preserve entanglement between systems.</p><h2>5. Can collapse and unitary evolution be tested experimentally?</h2><p>Yes, there have been numerous experiments that have tested the predictions of quantum mechanics, including the concepts of collapse and unitary evolution. These experiments have confirmed the validity of quantum mechanics, but the exact mechanism of collapse is still a subject of debate and ongoing research.</p>

1. What is the difference between collapse and unitary evolution?

Collapse refers to the collapse of a quantum state into a definite state when it is observed or measured. Unitary evolution, on the other hand, is the continuous and deterministic evolution of a quantum state according to the Schrödinger equation.

2. How does collapse occur in quantum systems?

Collapse occurs when a quantum system interacts with a classical measuring apparatus, causing the superposition of states to collapse into a definite state. This is known as the measurement problem in quantum mechanics.

3. Can unitary evolution and collapse coexist?

Yes, they can coexist in the Copenhagen interpretation of quantum mechanics. In this interpretation, quantum systems evolve unitarily until they are observed, at which point collapse occurs. However, there are alternative interpretations that do not involve collapse, such as the Many-Worlds interpretation.

4. How does the concept of entanglement relate to collapse and unitary evolution?

Entanglement is a phenomenon where two or more quantum systems become correlated in such a way that their individual states cannot be described independently. This can lead to collapse when one of the entangled systems is observed, causing the other system to also collapse. Unitary evolution can also preserve entanglement between systems.

5. Can collapse and unitary evolution be tested experimentally?

Yes, there have been numerous experiments that have tested the predictions of quantum mechanics, including the concepts of collapse and unitary evolution. These experiments have confirmed the validity of quantum mechanics, but the exact mechanism of collapse is still a subject of debate and ongoing research.

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