# Bekenstein bound

1. Feb 28, 2013

### michael879

ok so my main question here is about the normal bekenstein bound, but I will go into why I'm asking too in case anyone has any comments on that.

The way I understand the derivation of the bekenstein bound is:
If you have some closed system with energy E bounded by radius R, you can derive the bound by "lowering" it into a black hole with mass M (radius 2M) and enforcing the 2nd law of thermodynamics: $S_{bh}(M) + S_{system}(E,r) \leq S_{bh}(M+E-E_{rad}) + S_{rad}$. Neglecting the radiation if M >> E, you get: $S_{system}(E,r) \leq S_{bh}(M+E) - S_{bh}(M)$. If you reduce M to the same order as R, while keeping M >> E you arrive at the bekenstein bound up to some constant. As far as I can tell this constant is determined by requiring a black hole bounded by its event horizon to have maximum entropy.

Now my question is why and how you eliminate the entropy of the radiation from this equation?? Clearly any system falling into a black hole will produce radiation which will carry away energy and entropy. I understand that if M>>E, the energy of the radiation is negligible. Why though is the entropy contribution negligible??

The reason I'm asking this question is because I am trying to derive a generalized Bekenstein bound for a system with parameters (m,q,j) by "lowering" it into a black hole with parameters (M,Q,J). By adjusting M, Q, and J you should be able to come up with a lower maximal entropy of the system as it takes more into account. The issue is that a charged system falling into a black hole emits much more radiation and an uncharged one, and if the black hole is rotating the situation is even worse! So I would like to know what restrictions I need to place to remove the radiative terms from the equation

Last edited: Feb 28, 2013
2. Mar 7, 2013

anybody?

3. Mar 7, 2013

### Staff: Mentor

Where are you getting this understanding from? In particular, why do you think the derivation assumes that radiation is emitted?

4. Mar 7, 2013

### michael879

I'm getting this from a number of sources, I've been reading up on the topic recently. As for the radiation, the derivation would be SEVERELY flawed if it didnt assume radiation! Any black hole absorption process is going to radiate.. You need to show that it is negligible

5. Mar 7, 2013

### Staff: Mentor

Can you give any specific references?

Really? What if I take a rock and drop it into a black hole? Why would it radiate?

6. Mar 7, 2013

### michael879

Not off the top of my head but I'll go back and find them

Of course it would... First of all there would be gravitational radiation. Then there would also be electromagnetic radiation near the event horizon as the rock is ripped apart. If the black hole is so big that the rock isn't ripped apart, then the electromagnetic radiation would negligible but it would still be there...

Plus, this bound applies to all systems not just uncharged ones. Charged systems will radiate as they fall into any black hole, so this radiation must at least be addressed even if its thrown out as negligible...

7. Mar 7, 2013

### Staff: Mentor

This is true, since the rock breaks spherical symmetry. But it would be negligible if the rock is small enough compared to the hole.

Also, we could eliminate the gravitational radiation entirely by making the rock a spherically symmetric shell of matter that falls into the hole.

Not if the hole were large enough compared to the rock, as you note.

Why? How would it get generated?

That's the claim, yes, but I haven't actually seen a derivation that explicitly deals with charged objects or radiation. That's why I asked you for specific references; if you have some, I'd be interested to read them.

8. Mar 7, 2013

### michael879

I agree that the energy of the radiation would be negligible, but we're talking about an ENTROPY bound here. My question is why the entropy of the radiation is ignored.

I agree with you, but this bound is supposed to apply to ALL systems (that aren't close to becoming black holes at least), not just spherically symmetric ones!! It isn't topologically possible to drop a rock into a black hole with a comparable radius without breaking spherical symmetry

even if the atoms aren't being ripped apart there would be a very small strain on each electron pushing it away from the nucleus. Like I said, this effect would be incredibly small, but there would be SOME radiation. Its pretty hard to get 0 radiation when you're talking about charged particles in a non-symmetric, arbitrary configuration

Sorry I misunderstood you, I have no reference where they explicitly deal with charge, that's what I'm trying to do here. However there is gravitational radiation and I have seen at least one person mention it and decide it was negligible. Any derivation that doesn't even consider radiation is doing something wrong...

9. Mar 7, 2013

### Staff: Mentor

Again, it's hard to answer this question without seeing a reference that actually addresses the question; none of the references I've seen have. They all deal with an idealized case in which there is zero radiation. If nothing else, that should be obvious from the assumption that the system as a whole can be confined within a finite radius R; any emitted radiation would escape to infinity and so would not be confined within any finite radius. So any analysis of the Bekenstein bound that included radiation would not be able to make that assumption; but all of the analyses that I have seen do make it.

Yes, I know, but the derivation you're referring to is heuristic, so I'm trying to distill it down to its simplest heuristic form. As I noted above, my understanding of the simplest heuristic form is that it deals with an idealized case in which there is no radiation emitted.

If tidal gravity were significant, yes. But the whole point of making the rock small enough compared to the hole is to make tidal gravity negligible. If tidal gravity is negligible then the effect you're talking about is also negligible.

Reference?

10. Mar 7, 2013

### michael879

The Bekenstein bound is for bounded system in the absense of a black hole. The black hole is only introduced to find the entropy bound, so the bounded system doesn't need to, and shouldn't be expected to stay bounded once you add the black hole in! Like I said, it is impossible to lower an arbitrary system into a black hole without any kind of radiation.

The thing is, in order to derive the bound you actually need to make the black hole approximately the same size as the system in question (spatial size, not massive). I think to get the correct bound the black hole ends up with half the radius of the system, making it smaller!

Ignoring the above, like I mentioned earlier you're only considering the energy of the radiation. The entropy of the radiation is a totally different beast and I see no obvious reason why it would be negligible. For all I know the entropy of the radiation could exactly cancel out any violations to the 2nd law, resulting in no entropy bound!

I will find you references, I'm looking for them

11. Mar 7, 2013

### Staff: Mentor

No, it applies to black holes too. Black holes saturate the bound--i.e., they have more entropy than any other object with the same mass enclosed within a given radius--more precisely, with the same mass enclosed within a 2-sphere of a given area. In fact, that's an alternate way of stating the bound: of all objects with mass M enclosed in a 2-sphere of surface area A, the black hole of mass M has the most entropy. See further comments below.

Um, black holes are bounded systems. Also, an object that emits radiation as it's falling into a black hole has to emit it *before* it enters the hole, i.e., before it crosses the horizon. Once it's crossed the horizon, any radiation it emits is also trapped within the hole and so doesn't count as "escaped".

So an alternate way of describing the idealization I think the heuristic argument for the bound is using is that the "object" that falls into the hole only consists of what actually crosses the horizon; any radiation emitted before that escapes to infinity and isn't counted in either the energy or the entropy calculation.

However, I think an even better way to look at it is to rephrase the bound the way I did above: of all objects of mass M that can be enclosed in a 2-sphere of some radius R (or, better, some surface area A), the one with the maximal entropy is the black hole of mass M.

It can be much smaller than half the enclosing radius, depending on the "system" in question. For example, take the Earth and enclose it in a 2-sphere just large enough to contain all of its matter. The radius of that 2-sphere is something like a million times the radius of a black hole with the mass of the Earth, and the bound says that that black hole has more entropy than the Earth does.

12. Mar 7, 2013

### michael879

I think you misunderstood most of what I said in the last post. The black holes I was referring to were the black holes you drop the system into to derive the Bekenstein bound. In deriving the Bekenstein bound you need to assume that:
1) The black hole's mass is much larger than the system's
2) The black hole's radius is of the same order as the system's, so you can interchange the radii in the resulting formula. To derive the Bekenstein bound quoted as "correct" you must use a black hole with a radius of 1/2 the radius of the system

Yes, black hole's saturate the Bekenstein bound, but I'm pretty sure this is by construction (i.e. the 1/2 factor is introduced in order to make black holes saturate the bound)! However, black holes do not satisfy the assumptions made in deriving the bound, so they really have no place in this discussion when we're talking about the test system.

Of course the system doesn't radiate once it falls into the black hole, but in order to derive the bound you must do the following:
1) Consider a bound system (in empty space if you wish) far from becoming a black hole. If it is close to becoming a black hole you can't apply this heuristic!
2) Introduce a black hole, isolated from the system
3) Recognize that the entropy of the black hole and the entropy of the system you've just placed near the black hole are your "initial" entropy. Clearly to keep the entropy of the system near the black hole equal to the entropy of the system in empty space, you would need to drop the system from infinity (so it's not actually near the even horizon). The closer you place it to the black hole the less accurate your bound is on the system in question!
4) Let the system fall into the black hole, and recognize the entropy of the resulting black hole as the "final" entropy. However, since you necessarily need to drop the system from infinity, there is absolutely no way to avoid some kind of radiation in the process. Therefore you need to make sure the entropy of this radiation is negligible, or you will have a serious problem.

So the black holes I'm talking about are mathematical tools used to place an upper bound on the entropy of systems nowhere near black holes! My question involves the derivation of the Bekenstein bound, so we really shouldn't be talking about any other black holes (i.e. black holes as the initial system, or near the initial system)

13. Mar 7, 2013

### Staff: Mentor

I understand that that is one heuristic way to derive the bound, yes. I'm not sure it's the only way. That's why I'm suggesting other possible ways of deriving the bound, as well as of stating what it says.

In the heuristic version you are describing, yes, you are left with an undetermined constant, which is determined by assuming that the black hole saturates the bound. However, in the alternate derivation I'm describing, where you consider all possible objects with mass M that can be confined within a 2-sphere of radius R, I think the fact that the black hole is the one with maximal entropy is not "by construction", it's actually derived. However, I need to check references myself to confirm that.

Perhaps not in the derivation you are describing, but again, I'm not sure that's the only derivation.

14. Mar 7, 2013

### michael879

Ah I see, I don't know that derivation. How do you show that a black hole has maximal entropy for its radius and mass?

15. Mar 7, 2013

### Staff: Mentor

It turns out that Bekenstein's original papers are available online; I found links to them on the Wikipedia page on the Bekenstein bound:

http://en.wikipedia.org/wiki/Bekenstein_bound

The two papers that seem to give the most straightforward discussion of his heuristic derivation of the bound are the ones on the generalized second law and on the universal upper bound of the entropy-energy ratio; the one on black holes and the second law also has useful background on the black hole entropy formula.

His heuristic derivation goes like this (from the universal upper bound paper, Sec. II): take an object with energy E which can be contained within a 2-sphere of radius R and drop it into a black hole with mass M >> E. The area of the hole's horizon will increase by $8 \pi E R$ (strictly speaking, this is the minimum possible increase of the horizon area, but he assumes one can arrange the process so the minimum possible increase is attained); the hole's entropy increases by 1/4 of this, or $2 \pi E R$. But that means that, unless the original entropy of the object is less than $2 \pi E R$, the generalized second law (here he references his previous paper on that) will be violated. So $2 \pi E R$ must be an upper bound on the entropy of any object with energy E contained within a 2-sphere of radius R (strictly speaking, with area $4 \pi R^2$--he makes that point later on in the section).

Notice that there is nothing in there about the object emitting radiation; he assumes that the energy E of the object is the energy that is added to the hole. However, later on in the same paper he talks about the entropy to energy ratio of a system of massless particles, i.e., radiation, and concludes that it must also satisfy the bound.

16. Mar 7, 2013

### Staff: Mentor

This is simple given the derivation I just posted from Bekenstein's original paper. A black hole of mass M has an entropy of 1/4 its horizon area, or $4 \pi M^2$; since $R = 2M$ for a black hole, that gives entropy $2 \pi M R$, which saturates the bound.

17. Mar 7, 2013

### michael879

now THAT is circular! I was asking how you derive a bound that predicts black holes saturate the bound

18. Mar 8, 2013

### jartsa

When a small black hole is slowly and carefully lowered into a large black hole, extracting mass-energy from the small black hole, then:

The mass of the resultant black hole = m1+m2 - extracted mass
The entropy of the resultant black hole is proportional to (m1+m2-extracted mass)²

When the black hole that is lowered is small, the extracted mass is about the same as the mass of the small black hole, the mass increase of the large black hole is about zero, and the area increase of the large black hole is about the area of the small black hole.

If we drop a black hole into a black hole then:

the mass of the resultant black hole = sum of masses of the original black holes
the entropy of the resultant black hole is proportional to (m1+m2)²

When Bekenstein and other folks say "lowered" they really mean lowered, not dropped, I think.

Last edited: Mar 8, 2013
19. Mar 8, 2013

### Staff: Mentor

I'm not sure I understand. The $2 \pi E R$ bound is derived without making any assumptions about what kind of object falls into the large black hole; just that the object has energy $E$ and is enclosed in a 2-sphere of area $4 \pi R^2$. Showing that if the object is a black hole (i.e., a *second* black hole, much smaller than the large black hole it falls into), it saturates the bound, is just a matter of plugging in the relationship between $E$ and $R$ for a black hole. That relationship wasn't assumed anywhere, so there's no circularity that I can see.

20. Mar 8, 2013

### Staff: Mentor

This isn't possible; you can't extract mass from a black hole in a controlled fashion.

It gets a bit confusing because Bekenstein refers to other thought experiments that do talk about lowering objects (*not* black holes) and extracting work from the process; for example, Bekenstein refers to a thought experiment by Geroch along these lines. But I'm not sure Bekenstein's own heuristic derivation requires the object to be slowly lowered; I think his works fine if the object is just dropped.