The Bekenstein bound : Area versus Volume

  • #26
PeterDonis
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See the standard textbook Birrel Davies "Quantum Fields in Curved Space", the un-numbered equation at page 269. Is it an interpretation or a prediction?
I don't have this textbook, and unfortunately Google Books won't show me pg. 269 in preview. I have read Wald's monograph "Quantum Fields in Curved Spacetime and Black Hole Thermodynamics". From what I read there, it seems to me that one could just as easily interpret what is happening with Hawking radiation as "incoming quantum field modes corresponding to a vacuum state scatter off the spacetime curvature of the hole in such a way as to transfer energy from the hole's curvature to quantum field modes corresponding to outgoing particles". No "negative energy" anywhere, just a perfectly ordinary scattering process.
 
  • #27
naima
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I read in the second link that QM gives probabilities for outputs that sum to one. Then the author writes: in QM when particles collide information is not really loss. Everybody knows that given an output you have a probability for preparations giving this output. He only says as a proof look at fig 1 !
 
  • #28
PeterDonis
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in QM when particles collide information is not really loss.
There's a hidden qualifier to this statement (other treatments that I've seen make it explicit): information is not lost in QM provided there is no wave function collapse. This is a simple consequence of unitarity. Many physicists prefer the MWI, in which there is never any collapse at all, so they talk about complex processes taking place, like particles colliding and falling into black holes, but still assume that unitarity holds.

Everybody knows that given an output you have a probability for preparations giving this output.
This is true if there is wave function collapse. In no collapse interpretations, like the MWI, initial states map uniquely to final states. Again, this is a simple consequence of unitarity.
 
  • #29
naima
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In this war (Susskind vs Hawking) have we collapsers vs anticollapsers?
 
  • #30
PeterDonis
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In this war (Susskind vs Hawking) have we collapsers vs anticollapsers?
No. Pretty much all of the physicists involved prefer the MWI, so they would be "anticollapsers". Certainly Susskind and Hawking both prefer the MWI.

Roger Penrose has speculated that wave function collapse might be a quantum gravity effect, but AFAIK none of those speculations ever led to anything testable. I don't think he's involved in the "black hole war".
 
  • #31
naima
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In QM things that cannot be measured are not worthwhile.
Maybe MWIsts like to think that there is state vector that keeps all the information as time flows. But information for who? not for peculiar instrumentalists in one of the worlds. Maybe for a deity above physicists?
Susskind likes the image of strings out of the BH with their ends on the horizon. it happens that a string C twists to OC and separate in O and C
O is then a particle which escapes.
 
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  • #32
PeterDonis
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In QM things that cannot be measured are not worthwhile.
Maybe MWIsts like to think that there is state vector that keeps all the information as time flows. But information for who? not for peculiar instrumentalists in one of the worlds. Maybe for a deity above physicists?
This is getting off topic (and also into territory that's philosophy, not physics). The MWI is a mainstream interpretation of QM. It has open issues, but so does every interpretation of QM. We're not going to resolve them here.
 
  • #33
ohwilleke
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RIP Bekenstein
 
  • #35
Dear All,

Thank you for your replies.

They come in three kinds, two I find unconvincing, one I find surprising.

1 The first kind of reply says: "You could have all theses storage devices, but you won't use them." I find that unconvincing. --- if the above setup was a stable, non-collapsing solution of the EFEs, I could just assume that the mirrors are independent from one another at t=-infinity. So, this can only be a consequence of the next argument.

2 The second kind of reply says: "You could put all this information right there, but you wouldn't be able so access it, because every large enough ball would be a non-collapsing BH." Again I find that unconvincing. --- if the above setup was a stable, non-collapsing solution of the EFEs, I could just travel through it --- or just send lightrays to read out my mirrors --- at will. The contrary would be absurd: it would mean I can't access even a nearby mirror, just because it belongs to some very large ball having some far away center.

3 The third kind of reply says: "You believe your arrangement is stable, but it is not." I find that surprising, let me argue why.

According the schwarzschild solution, an isolated mass induces a pertubation with respect to the minkowski metric which is in ~m/r, and so by taking the mass m to be negligible compared to the empty ball of radius r that surrounds it ---- so that the ratio m/r is lower than the some planck length and the metric perturbation becomes "unphysical" ---- we can safely say that this little mass as no impact on stability of the rest, whatsoever.

Well this is what I used to believe, at least, and its seems to follow quite clearly from the hypothesis of a cut-off Planck length.

Now, say we forget about such a cut-off hypothesis.
Then what you say is that the all these negligible curvatures, induced by these little masses, somehow cumulate non-vanishingly in order to form a BH. I still find that surprising. Indeed:

Let us say I start at t=0 from this lovely, perfectly arranged, evenly-spaced cubic lattice. How will it evolve and deform? Why should it collapse into a BH here and not there, given the perfect initial arrangement?

Or is it that you are saying that the infinite cubic lattice is not even an acceptable initial condition? But then is it not too much to ask of partial differential equations (be it the EFEs) to forbid certain initial conditions based on rather global conditions?


Regards.
 
  • #36
Demystifier
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Or is it that you are saying that the infinite cubic lattice is not even an acceptable initial condition? But then is it not too much to ask of partial differential equations (be it the EFEs) to forbid certain initial conditions based on rather global conditions?
A perfect infinite cubic lattice is an acceptable initial condition and it will not collapse. However, a perfect lattice is essentially unique; it cannot be constructed in many different ways and hence cannot contain much information. To put a lot of information into the lattice, you need some (at least small) imperfections.

But imperfect lattice will not be stable. The imperfect lattice will undergone a gravitational collapse, eventually forming black holes or other stable compact objects (stars, planets, ...). These stable compact objects will obey the Bekenstein bound.
 
  • #38
PeterDonis
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According the schwarzschild solution
The solution you are describing is not the Schwarzschild solution, so reasoning based on that solution does not apply to the case you are considering. If you're going to make claims about your solution, you need to first be clear about which solution it is. You can't just wave your hands; you have to actually do the math.

an isolated mass
An infinite lattice of small objects is not "an isolated mass". Such a solution is not asymptotically flat, and that is what "an isolated mass" means.

the ratio m/r is lower than the some planck length and the metric perturbation becomes "unphysical"
Even leaving out the fact that we don't have a theory of quantum gravity and so you are speculating here in an area where there is no theory to back it up: assuming that the objects were that small, the same tentative speculations about quantum gravity that talk about perturbations smaller than the Planck length being unphysical also say that an object that small cannot store any information. So a lattice of such objects would not meet your requirements.
what you say is that the all these negligible curvatures, induced by these little masses, somehow cumulate non-vanishingly in order to form a BH.
No, I didn't say that. I said that all those small masses add up to a non-negligible global gravitational effect. Whether or not they will collapse to form a BH is a separate question that is irrelevant to refuting your claim. You are claiming that such a lattice is stable--that it can persist indefinitely in a static state, neither expanding nor contracting. When you actually do the math, you find that this claim is false.

is it that you are saying that the infinite cubic lattice is not even an acceptable initial condition?
Yes. To expand on (and somewhat correct) what I said in post #23: there are two possible cases, zero cosmological constant and nonzero cosmological constant.

If the CC is zero, then you can have an infinite lattice, but it can't be motionless, even for an instant. (In post #23 I thought it could, but I was wrong.) An infinite lattice corresponds to either the spatially flat or spatially open FRW solutions, and neither of those have even a single instant of time where they are "motionless" (neither expanding nor contracting). They either start out from a "Big Bang" singularity and expand forever into the infinite future, or (the time reverse of that) they contract forever from the infinite past to a "Big Crunch" singularity. (A finite lattice can be motionless for an instant if the CC is zero, but only for an instant.)

If the CC is nonzero, then you can have a lattice that is motionless, but it can only be finite. (This solution is basically the Einstein static universe, and it is unstable, like a pencil balanced on its point, but as an idealization it does exist.) There is no way to have an infinite lattice with the CC exactly balancing the density of matter to keep the solution static.

To see why the above is true, look at the Friedmann equations:

$$
\frac{\dot{a}^2 + k}{a^2} = \frac{8 \pi \rho + \Lambda}{3}
$$

$$
\frac{\ddot{a}}{a} = \frac{- 4 \pi \left( \rho + 3 p \right) + \Lambda}{3}
$$

"Static" (for more than a single instant) means ##\dot{a} = 0## and ##\ddot{a} = 0##, which gives

$$
\frac{k}{a^2} = \frac{8 \pi \rho + \Lambda}{3}
$$

$$
\Lambda = 4 \pi \left( \rho + 3 p \right)
$$

Substitute the second into the first to obtain

$$
\frac{k}{a^2} = 4 \pi \left( \rho + p \right)
$$

Since ##\rho + p## is positive for any ordinary matter and energy, we must have ##k## positive, which means a closed, spatially finite universe.

If, instead, we assume ##\Lambda = 0## and we want our universe to be static for a single instant, then we still have to have ##\dot{a} = 0## for that instant; so at that instant, the first equation would read:

$$
\frac{k}{a^2} = \frac{8 \pi \rho}{3}
$$

If we want an infinite universe, we must have ##k \le 0##; but that would require ##\rho \le 0## by the first equation, which is impossible (negative energy density). So an infinite universe can't be static even for an instant if ##\Lambda = 0##. A finite universe can, since it has ##k > 0##. But for a finite lattice, the second equation with ##\Lambda = 0## says:

$$
\frac{\ddot{a}}{a} = - \frac{4 \pi}{3} \left( \rho + 3 p \right)
$$

For ordinary matter and energy, ##\rho + 3p## is positive, so a finite universe with zero CC that is motionless at a single instant will start contracting; it won't stay motionless.
 
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  • #39
PeterDonis
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A perfect infinite cubic lattice is an acceptable initial condition and it will not collapse.
I don't think this is correct; see my previous post. What solution do you have in mind for a perfect infinite cubic lattice that is static and stable?
 
  • #40
Demystifier
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I don't think this is correct; see my previous post. What solution do you have in mind for a perfect infinite cubic lattice that is static and stable?
You are right that it will collapse. (Actually, I liked your post for correcting my mistake). However, even though such a Universe may collapse, it will not collapse into a black hole. It will collapse into a naked singularity, similar to the Big Bang singularity.
 
  • #41
PeterDonis
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even though such a Universe may collapse, it will not collapse into a black hole. It will collapse into a naked singularity, similar to the Big Bang singularity.
Yes, agreed.
 

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