# The Bekenstein bound : Area versus Volume

## Main Question or Discussion Point

Dear all ---

This question raises concerns already expressed in
but in a more specific form --- so that, hopefully, a more specific answer may be given.

With the Bekenstein-bound-saturated-by-BH argument, we have that a sphere of radius R can have at most a.k.R^2/4 information inside.

Now, let me imagine the following scenario.

Let us say I have some very light physical device able to store one bit of data. It could be anything, perhaps a piece of mirror oriented in one direction or another --- something stable, localized in empty space, optically readable.

Now, very, very far away, perhaps a light year away, I place another one. And then another one. And so on, arranging all of this bits of information in an evenly spaced, infinite, _cubic_ grid, laid in almost flat space.

It seems to me that the attraction between each of these one-bit storage devices is perfectly negligible. Moreover, each device is evenly attracted by the others, so it won't "move". Therefore, we run no risk of them colliding into a BH.

Now, consider taking a subset of these physical devices. For instance, center on one of them, and draw an imaginary ball of radius R. Clearly, the information storage capacity of this ball grow in R^3. This information storage is accessible: I can always send a lightray, or even a spaceship if that is necessary, go and read some device inside: it will take a while but this is doable.

My main question is: does this contradict the Bekenstein-bound-saturated-by-BH argument?
Subquestions:
- do the agree that this will not collapse into a BH?
- do we agree that the information storage capacity grows in R^3, and so even if the information density is rather low, it will eventually exceed the a.k.R^2/4 bound?
- has this been discussed and fixed in any way that someone could explain?

Many thanks.

Last edited:

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@RayLouvreur

How can you be sure the lattice will not collapse ? Unless the mirrors are massless I can't see why they shouldn't.

Clearly, the information storage capacity of this ball grow in R^3.
No.
You have a hidden assumption - matter in these areas is not correlated. You can say: "But I put a matter there is a way so there is no correlation". But Bekenstein bound tells you that it is NOT possible, because you can't "inject" all this stuff into Universe at some arbitrary moment of time, you have past history, past lightcones, and they intersect, so these areas MUST have a common path, and they are correlated.

Bekenstein bound tells us about one important property of the Universe: when you take bigger and bigger volumes, information density becomes lower and lower. Ultimately going to 0, which makes a perfect sense (you can calculate Universe wavefunction from the equations of TOE).

PeterDonis
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With the Bekenstein-bound-saturated-by-BH argument, we have that a sphere of radius R can have at most a.k.R^2/4 information inside.
Basically, yes. For any sphere of radius ##R## that is not the horizon of a black hole, the information inside will be less, probably much less, than the bound.

Clearly, the information storage capacity of this ball grow in R^3.
Initially, yes. But things change as ##R## grows. See below.

does this contradict the Bekenstein-bound-saturated-by-BH argument?
No. In addition to the point tzimie raised, that you can't assume all the information storage devices are uncorrelated, you have overlooked another key point; see below.

do the agree that this will not collapse into a BH?
No. This is the other key point you have overlooked.

Suppose that each information storage device has mass ##m##, and that a sphere of radius ##R## encloses ##N## such devices. Then the mass enclosed inside radius ##R## is ##N m##. Since ##N## increases, basically, with volume enclosed, ##N## should be proportional to ##R^3##, so the total mass ##M## enclosed inside radius ##R## should be proportional to ##m R^3##. We'll call the constant of proportionality ##k##; we don't have to know its numerical value.

For a given total mass ##M## enclosed inside radius ##R## to be a black hole, we must have ##R \le 2M## (we are using "geometric units" in which ##G = c = 1## for simplicity). Given that ##M## is proportional to ##m R^3##, we have ##R \le 2 k m R^3## as the condition for a black hole to be formed. Dividing through and inverting and taking the square root gives the inequality

$$R^2 \ge \sqrt{\frac{1}{2km}}$$

So as soon as ##R## gets large enough, the assembly you describe will indeed be a black hole. Note that the individual devices don't need to "fall together" for this to happen; a large enough sphere, given constant lattice spacing, will automatically enclose enough mass, per the above inequality, to be a black hole.

Dale
Mentor
Let us say I have some very light physical device able to store one bit of data. It could be anything, perhaps a piece of mirror oriented in one direction or another --- something stable, localized in empty space, optically readable.

Now, very, very far away, perhaps a light year away, I place another one. And then another one. And so on, arranging all of this bits of information in an evenly spaced, infinite, _cubic_ grid, laid in almost flat space.
Regardless of how low the density is, there is some limit where there is just so much of it that it will collapse into a black hole.

naima
Gold Member
The main thing is that all the bits have to be visible from the outside of the sphere. or that there is room enough on the boundary to describe them.

naima
Gold Member
I will ask Ray's question differently.
Is a flat Minkowski spacetime with an infinite cubic lattice of motionless dust test particles
an approximate solution of Einstein equation?
Each test particle is the center of a great cube with a problem with Beckentein.
So BH or no BH everywhere?

PeterDonis
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Is a flat Minkowski spacetime with an infinite cubic lattice of motionless dust test particles
an approximate solution of Einstein equation?
Minkowski spacetime by itself is a valid solution. Adding "dust test particles" makes no difference to the solution at all if they are only test particles, because test particles by definition have no stress-energy and do not produce any spacetime curvature. But such objects also cannot contain any information, so they are irrelevant if we're discussing the Bekenstein bound.

If what you really mean is "Minkowski spacetime with an infinite cubic lattice of motionless objects that can each hold at least one bit of information", then no, that is not even an approximate solution of the field equations, because storing even one bit of information requires nonzero stress-energy and therefore must produce nonzero spacetime curvature. Locally, the objects might contribute only a very small amount of spacetime curvature, so the deviation of the solution from Minkowski spacetime might be undetectable. But globally, the solution will have to be very different; it won't even be close to flat Minkowski spacetime.

A technical point: the condition you give, "an infinite cubic lattice of motionless objects", is actually not physically possible. A finite lattice of motionless objects is, at least for an instant, provided the lattice isn't too large. (The next instant, the objects will start falling together by their mutual gravity; they might start falling very slowly, so "motionless" might be a decent approximation for quite a while, but it can't be forever.) But an infinite lattice is not, not even for an instant, for the same reason that an infinite static universe is not a solution of the EFE.

A further technical point: if you include a positive cosmological constant, you can make a finite lattice remain motionless for more than an instant. (It will be an unstable equilibrium, like a pencil balanced on its point, but we don't need to delve into that here.) But it still has to be a finite lattice (compare with the Einstein static universe, which has a finite spatial volume). An infinite lattice can't be made static in this way.

naima
Gold Member
So what about when with this constant you have a big finite cubic lattice where we have the beckenstein problem?

PeterDonis
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So what about when with this constant you have a big finite cubic lattice where we have the beckenstein problem?
Posts #3 and #4 have already answered that, by showing which of the OP's assumptions are not valid. Adding a cosmological constant doesn't change any of that.

Also, note the qualification I gave to the finite static lattice being a solution: "as long as the lattice is not too large". If the lattice gets large enough, a static solution is no longer possible because the lattice must be inside a black hole, and it is not possible for a finite lattice inside a black hole to be motionless, even for an instant; everything inside a black hole must be falling inward. Adding a cosmological constant doesn't change that either. (It might change the details of exactly how large is "too large"; I haven't checked the exact numbers.)

PeterDonis
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If the lattice gets large enough, a static solution is no longer possible because the lattice must be inside a black hole
Actually, there is a wrinkle here. If we assume that we are talking about an (unstable) Einstein static universe type solution, then there is a different problem with the OP's assumptions. Once we specify the density of the lattice (how many information units there are per unit volume, which tells how much stress-energy there is per unit volume), then we have fixed the entire solution, including its (finite) total spatial volume. And as the density of the lattice goes up, the spatial volume goes down as the 3/2 power of the density--i.e., volume goes down faster than density goes up.

So the OP's assumption that you can take a lattice of fixed density and keep extending it indefinitely is not true for this kind of solution. A lattice of a given density can only store a fixed amount of information, which goes down as the density goes up.

Demystifier
Gold Member
Bekenstein bound seems to be valid for classical processes which do not involve negative energy carrying positive entropy. But quantum effects, like Hawking radiation, do involve negative energy carrying positive entropy. As argued in
http://lanl.arxiv.org/abs/1507.00591
it violates Bekenstein bound and suggests that maximal black-hole entropy is proportional to the volume, rather than area.

naima
Gold Member
I read that Jakobson derived the Eistein equation from entropy proportional to area.

Demystifier
Gold Member
I read that Jakobson derived the Eistein equation from entropy proportional to area.
Einstein equation is a classical equation, so this is not inconsistent with the idea that in full quantum gravity entropy does not necessarily need to be proportional to area.

PeterDonis
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quantum effects, like Hawking radiation, do involve negative energy carrying positive entropy.
I'm not sure I understand (and I haven't read through the paper you link to in detail, so it may be that the answer to the question I'm about to ask is "it's explained in the paper"). A black hole emitting Hawking radiation loses energy (mass), and also entropy, as I understand it, at least according to the standard view (i.e., the hole's entropy, taken by itself, still obeys the Bekenstein bound): but the entropy in the emitted radiation is larger than the entropy lost by the hole in the process of emitting it, so the total entropy still increases and the generalized second law holds.

Demystifier
Gold Member
A black hole emitting Hawking radiation loses energy (mass), and also entropy, as I understand it, at least according to the standard view
According to the standard view, black hole loses energy by absorbing negative energy. This negative energy is entangled with positive energy of Hawking radiation, implying that the absorbed negative energy carries positive entropy. So black hole absorbs positive entropy, which contradicts Bekenstein bound and constitutes the infamous black-hole information paradox. A possible way out of the paradox is to deny the Bekenstein bound.

PeterDonis
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According to the standard view, black hole loses energy by absorbing negative energy.
This is one way of looking at it, but it's not the only way. I'll defer further comment until I've read through the paper.

Demystifier
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This is one way of looking at it, but it's not the only way.
I agree that this is not the only way, but that's what semi-classical gravity predicts.

PeterDonis
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that's what semi-classical gravity predicts.
Semi-classical gravity predicts that a black hole will emit Hawking radiation and lose mass. It does not "predict" that the hole loses mass by absorbing negative energy; that's a particular interpretation of what the math underlying the prediction says, and not the only possible one.

naima
Gold Member
Could you explain what is this infamous loss information paradox? and why it is a paradox?

Why can't an infinite motionless lattice be stable or even possible?
I don't know what the constituents of this lattice are but it seems admisable to talk of it as one would talk of a test particle, but with information.
If we set the stage with an infinite lattice and an exactly zero cosmo-constant ( which I gather results in infinite Minkowski space ) universe, what happens? Can't a lattice be stable if it is balanced with just the right negative pressure from a vacuum or some other agent? I thought this was stasis in the Gen.Rel. context.
I gather here that the lattice objects are not mere dust, but carry information requiring nonzero stress-energy. If the vacuum has nonzero stress-energy, does it also carry information? Is there vacuum information and can it stabilize our lattice?

PeterDonis
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Why can't an infinite motionless lattice be stable or even possible?
The exact reason depends on whether there is a cosmological constant or not.

(1) If there is no cosmological constant, the stress-energy of the objects in the lattice will produce attractive gravity, meaning the lattice can't remain motionless for more than an instant.

(2) If there is a cosmological constant, the lattice can be motionless for more than an instant if the repulsive gravity due to the cosmological constant exactly cancels the attractive gravity of the objects in the lattice; but this can only be the case for a finite lattice.

If you want more detail on the above, you should start a separate thread in the relativity forum, because it's really off topic for this discussion; what I've said above has nothing in particular to do with the Bekenstein bound or even with quantum gravity; it's purely classical GR.

Demystifier