What is the maximum entropy a system can have based on its energy and size?

[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

 

[michael879]

anybody?

 

[PeterDonis]

The way I understand the derivation of the bekenstein bound is

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

 

[michael879]

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

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

 

[PeterDonis]

I'm getting this from a number of sources, I've been reading up on the topic recently.

Can you give any specific references?

Any black hole absorption process is going to radiate.

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

 

[michael879]

Can you give any specific references?

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

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

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...

 

[PeterDonis]

First of all there would be gravitational radiation.

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.

Then there would also be electromagnetic radiation near the event horizon as the rock is ripped apart.

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

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...

Why? How would it get generated?

Plus, this bound applies to all systems not just uncharged ones.

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.

 

[michael879]

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

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.

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

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

Why? How would it get generated?

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

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.

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...

 

[PeterDonis]

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.

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.

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!

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.

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.

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.

However there is gravitational radiation and I have seen at least one person mention it and decide it was negligible.

Reference?

 

[michael879]

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.

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.

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.

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!

If tidal gravity is negligible then the effect you're talking about is also negligible.

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!

Reference?

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

 

[PeterDonis]

The Bekenstein bound is for bounded system in the absense of a black hole.

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.

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.

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.

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!

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.

 

[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)

 

[PeterDonis]

The black holes I was referring to were the black holes you drop the system into to derive the Bekenstein bound.

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.

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)!

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.

However, black holes do not satisfy the assumptions made in deriving the bound

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

 

[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?

 

[PeterDonis]

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.

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.

 

[PeterDonis]

How do you show that a black hole has maximal entropy for its radius and mass?

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.

 

[michael879]

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.

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

 

[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)²When Bekenstein and other folks say "lowered" they really mean lowered, not dropped, I think.

 

[PeterDonis]

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

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.

 

[PeterDonis]

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

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

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

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.

 

[michael879]

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.

There is though, because if you actually go through the derivation you will see that the "R" you quoted is actually the radius of the hypothetical black hole, NOT the radius of the system in question! The way you are able to swap the two Rs is by assuming the black hole's radius is "close" to the system's radius up to some factor. It turns out this factor needs to be 1/2 in order to arrive at the correct bound. However, your original assumption that the black hole needs to be much more massive than the system can't be satisfied simultaneously with 1/2 radius if the system in question is (or is close to becoming) a black hole!

 

[PeterDonis]

There is though, because if you actually go through the derivation you will see that the "R" you quoted is actually the radius of the hypothetical black hole, NOT the radius of the system in question!

Huh? R is defined (in the Bekenstein paper) such that the smallest 2-sphere that entirely encloses the "system" has area $4 \pi R^2$. If the "system" is a black hole, then R is determined by the area of the hole's horizon. Note that the "system", if it is a black hole, is *not* the same as the (much larger) black hole that the "system" falls into. Bekenstein's formula for the bound doesn't include *any* parameters of the large black hole that the "system" gets dropped into; the "R" in his formula is the "radius" (actually the surface area--see below) of the smallest 2-sphere that can enclose the "system".

The way you are able to swap the two Rs is by assuming the black hole's radius is "close" to the system's radius up to some factor.

No, I'm not assuming this at all. The black hole (the small one) *is* the "system". See above.

It turns out this factor needs to be 1/2 in order to arrive at the correct bound.

Why do you say this? Bekenstein's paper doesn't mention any factor of 1/2. I think you're confusing Bekenstein's derivation with another one.

However, your original assumption that the black hole needs to be much more massive than the system can't be satisfied simultaneously with 1/2 radius if the system in question is (or is close to becoming) a black hole!

As above, I think you're confusing the "system" being a black hole with the (much larger) black hole that the "system" gets dropped into. They're not the same. Bekenstein's formula for the bound doesn't include *any* parameters of the (much larger) black hole; it only includes parameters of the "system", which if it is a black hole will be much smaller than the hole the "system" gets dropped into.

 

[michael879]

Ok so I read through 2 of Bekenstein's papers on the subject, and now I'm a little confused. In his paper specifically on this bound, he references his paper titled "Black Holes and Entropy" in which he shows that the area increase of a black hole after absorbing the system must be at least $8\pi ER$. From this he makes the simple deduction that the initial entropy of the system must be less than $2\pi ER$. Now I have 2 questions about this

1) He still doesn't address the entropy of the radiation emitted during the "fall", which would increase the upper bound on the system's entropy

2) I can not understand how he found a lower bound on the area change of the black hole..

2a) In his derivation of this he uses a general Kerr-Newman black hole. If you drop a particle with charge and/or angular momentum and no mass into a Kerr-Newman black hole, doesn't the area actually decrease??

2b) I follow his derivation up until the point where he suddenly ends up with this bound. For the life of me, I can't figure out how he arrived at this area bound which is independent of ALL of the parameters of the black hole!

2c) The radius R of the system is a completely adjustable parameter... There is a lower bound on R, the minimal radius to bound the system, but no maximum! So I really don't understand why for example, doubling R would double the lower bound on the area increase of the final state black hole... Even if you define R as the minimal radius to bound the system, you could still take some infinitesimal piece of the system and move it 3R away from the center of mass to make R go up to 2R! So in essence you can altar the minimal R by making infinitesimal changes to the system! When talking about an upper bound on the entropy of the system this freedom in choose R is fine, because you only care about the minimal upper bound. However when you're talking about a lower bound on the area increase, you care about the maximal lower bound, which can be raised to infinity!

 

[PeterDonis]

1) He still doesn't address the entropy of the radiation emitted during the "fall", which would increase the upper bound on the system's entropy

Yes, I don't see radiation discussed at all. However, the entropy contained in the radiation does not fall into the hole, so it would still be counted as part of the total entropy of the "universe" (i.e., idealizing the situation so there's nothing else present); therefore it drops out of the calculation of the *change* in entropy when the object falls into the hole, which is what Bekenstein is focusing on.

2b) I follow his derivation up until the point where he suddenly ends up with this bound. For the life of me, I can't figure out how he arrived at this area bound which is independent of ALL of the parameters of the black hole!

The critical step seems to me to be the assertion that when the object with energy E enclosed in a 2-sphere of radius R falls into the (much larger) hole, the area of the hole's horizon increases by $8 \pi E R$. The paper referenced at that point appears to be behind a paywall, so I can't read it to see the details of how this formula is arrived at.

2a) In his derivation of this he uses a general Kerr-Newman black hole.

Does he? He considers the case of a Kerr-Newman hole to check if it satisfies the bound when treated as the "object" of energy E and radius R (which it does; a Schwarzschild hole saturates the bound); but as far as I can tell he doesn't make any assumptions about the object of energy E and radius R in actually deriving the bound.

If you drop a particle with charge and/or angular momentum and no mass into a Kerr-Newman black hole, doesn't the area actually decrease??

How can a particle have charge and/or angular momentum but no mass? (Assuming that by "mass" you really mean "energy"; obviously particles with zero rest mass can still have angular momentum, since photons do, but they will also have energy.)

Also, I don't think he's considering the (much larger) hole that the object with energy E and radius R falls into to be a Kerr-Newman hole; I think he is treating it as a Schwarzschild hole. But I can't be sure because, as I noted above, the paper referenced at the critical step is behind a paywall.

2c) The radius R of the system is a completely adjustable parameter... There is a lower bound on R, the minimal radius to bound the system, but no maximum!

I think he means "R" to stand for the lower bound, i.e., that the area of the *smallest* 2-sphere that encloses the system is $4 \pi R^2$. But it's true that he doesn't make that explicit. Also, as you note, one could still technically increase R drastically while leaving E essentially the same by rearranging very small pieces of the system. However, I don't think this is a real issue (see below).

When talking about an upper bound on the entropy of the system this freedom in choose R is fine, because you only care about the minimal upper bound. However when you're talking about a lower bound on the area increase, you care about the maximal lower bound, which can be raised to infinity!

I'm not sure this is a real issue. Consider two systems of energy E, one where the pieces are "normally" arranged, with radius R, and one with a very small piece taken out to radius 10^6R (or some very large number). If we drop the second "system" into a much larger black hole (M >> E), there will be *two* distinguishable "infall" processes instead of one, and we can treat each one separately. The first one will be the same as the one for the first system with all pieces "normally" arranged; the second one will basically be a non-event since the piece of the system we took out to 10^6R was infinitesimal.

In short, I think the only cases of real interest for Bekenstein's derivation are cases where the "object" being dropped into the much larger black hole can't be divided up into separate parts that can be detected as falling into the hole separately. But it's true, once again, that Bekenstein doesn't consider this issue explicitly.

 

[PeterDonis]

the entropy contained in the radiation does not fall into the hole, so it would still be counted as part of the total entropy of the "universe" (i.e., idealizing the situation so there's nothing else present); therefore it drops out of the calculation of the *change* in entropy when the object falls into the hole, which is what Bekenstein is focusing on.

On thinking this over some more, there's another issue as well with trying to include radiation: you now have to determine the entropy increase involved with the radiation emission process. There has to be an entropy increase because the radiation emission is irreversible; and since the degrees of freedom in the radiation are spatially unbounded (because the radiation escapes to infinity), the entropy increase can potentially be large. None of that entropy can be counted as part of the "original" entropy that was confined within radius R in the original system.

 

[michael879]

On thinking this over some more, there's another issue as well with trying to include radiation: you now have to determine the entropy increase involved with the radiation emission process. There has to be an entropy increase because the radiation emission is irreversible; and since the degrees of freedom in the radiation are spatially unbounded (because the radiation escapes to infinity), the entropy increase can potentially be large. None of that entropy can be counted as part of the "original" entropy that was confined within radius R in the original system.

Yes I was about to make that same comment. That is the entire point of this thread, I don't see why the radiation can be ruled out as negligible. The radiation exists in the final state but not the initial state, so any entropy it carries away changes the entropy bound on the system! Yes, the energy of the radiation can be made infinitesimal, but the entropy is potentially huge! As far as I can tell the radiation emitted by the system as it falls into the whole could ENTIRELY account account for the entropy lost to the black hole, effectively leaving entropy unbounded!

The critical step seems to me to be the assertion that when the object with energy E enclosed in a 2-sphere of radius R falls into the (much larger) hole, the area of the hole's horizon increases by $8 \pi E R$. The paper referenced at that point appears to be behind a paywall, so I can't read it to see the details of how this formula is arrived at.

http://www.webcitation.org/5pvpyakNu

off the references on wikipedia's page on the subject. The paper is called "Black Holes and Entropy". There is a critical step in which he goes from an equation highly dependent on the parameters of the large black hole, to an inequality completely independent of these parameters, and only dependent on E and R of the system. I got completely lost in how this step was done, and I can't figure it out.

Also, I don't think he's considering the (much larger) hole that the object with energy E and radius R falls into to be a Kerr-Newman hole; I think he is treating it as a Schwarzschild hole. But I can't be sure because, as I noted above, the paper referenced at the critical step is behind a paywall.

How can a particle have charge and/or angular momentum but no mass? (Assuming that by "mass" you really mean "energy"; obviously particles with zero rest mass can still have angular momentum, since photons do, but they will also have energy.)

Does he? He considers the case of a Kerr-Newman hole to check if it satisfies the bound when treated as the "object" of energy E and radius R (which it does; a Schwarzschild hole saturates the bound); but as far as I can tell he doesn't make any assumptions about the object of energy E and radius R in actually deriving the bound.

You will notice that he actually uses a Kerr-Newman black hole with arbitrary mass, charge, and angular momentum as the "large" black hole used to derive this bound. And sorry, yes you're right a particle can't have 0 energy and charge/angular momentum. HOWEVER, fundamental particles have significantly more charge and angular momentum than mass. So it would be possible to hypothetically increase the charge and angular momentum orders of magnitude more than the mass of a black hole (J/M ~ (planck mass)/(electron mass) for an electron). This should lower the area!

 

[michael879]

I'm not sure this is a real issue. Consider two systems of energy E, one where the pieces are "normally" arranged, with radius R, and one with a very small piece taken out to radius 10^6R (or some very large number). If we drop the second "system" into a much larger black hole (M >> E), there will be *two* distinguishable "infall" processes instead of one, and we can treat each one separately. The first one will be the same as the one for the first system with all pieces "normally" arranged; the second one will basically be a non-event since the piece of the system we took out to 10^6R was infinitesimal.

In short, I think the only cases of real interest for Bekenstein's derivation are cases where the "object" being dropped into the much larger black hole can't be divided up into separate parts that can be detected as falling into the hole separately. But it's true, once again, that Bekenstein doesn't consider this issue explicitly.

That is a good point, but there is still the issue of a blob of uniform density (possible in GR, not nature of course) with an "arm" extending very far out that is infinitesimal in energy/radius. You can't split that object up into multiple systems!

 

[PeterDonis]

As far as I can tell the radiation emitted by the system as it falls into the whole could ENTIRELY account account for the entropy lost to the black hole, effectively leaving entropy unbounded!

I don't think the radiation entropy can be unbounded; that would require the radiation to have arbitrarily long wavelength, which isn't possible if it is to be emitted by an object confined within a finite radius R. But I think I agree that there should be some argument given for why the entropy, as well as the energy, of the radiation should be negligible. There is an argument about a system of massless particles confined within a finite cavity satisfying the bound, but I'm not sure that argument would apply to radiation escaping to infinity.

I could see an argument possibly being made based on the maximum wavelength of the emitted radiation being of order R, and then arguing that radiation of that wavelength and total energy very small compared to E would have negligible entropy compared to $2 \pi E R$. But I haven't tried to actually work through such an argument, and I haven't seen one anywhere in any of the papers or textbooks I've read on this subject.

There is a critical step in which he goes from an equation highly dependent on the parameters of the large black hole, to an inequality completely independent of these parameters, and only dependent on E and R of the system. I got completely lost in how this step was done, and I can't figure it out.

I'll take a look, I haven't gone through this paper in detail.

You will notice that he actually uses a Kerr-Newman black hole with arbitrary mass, charge, and angular momentum as the "large" black hole used to derive this bound.

I must have missed this; where is it specified?

 

[PeterDonis]

That is a good point, but there is still the issue of a blob of uniform density (possible in GR, not nature of course) with an "arm" extending very far out that is infinitesimal in energy/radius. You can't split that object up into multiple systems!

Not if the "arm" was continous, true; but I'm also not sure that the dynamics of this object falling into a much larger black hole would satisfy the assumptions used to derive the bound. In particular, I'm not sure you could assume that tidal gravity would be negligible as the object falls through the horizon.

 

[michael879]

I must have missed this; where is it specified?

It's in the paper you couldn't access that I referred you to. The entropy bound he sets comes from the area bound he derives in that paper.

Not if the "arm" was continous, true; but I'm also not sure that the dynamics of this object falling into a much larger black hole would satisfy the assumptions used to derive the bound. In particular, I'm not sure you could assume that tidal gravity would be negligible as the object falls through the horizon.

True, that's a good point, the black hole's mass must grow with the system's R to account for the tidal gravity. But there is still the issue that the lower bound on the acquired area can grow substantially by making negligible changes to the system! R doesn't have to be unbounded for there to be a problem.

Imagine a ball of size R and mass M, and another ball of size 2R and mass M (1/4 the density). Surely the area acquired by the black hole should be almost identical in both cases. However for whatever reason the larger ball has twice as large a lower bound on the area increase!

 

[PeterDonis]

But there is still the issue that the lower bound on the acquired area can grow substantially by making negligible changes to the system!

Looking at the "Black Holes and Entropy", Appendix A, it looks like the derivation there assumes a spherical particle, so it would not apply to configurations with a long "arm" sticking out. However, I can't see that any assumption is made about the density of the spherical particle that falls in; see below.

Imagine a ball of size R and mass M, and another ball of size 2R and mass M (1/4 the density). Surely the area acquired by the black hole should be almost identical in both cases. However for whatever reason the larger ball has twice as large a lower bound on the area increase!

Yes, given that no assumption is made about the object's density, it looks like the area increase goes up with R if M is held constant, implying that an object with the same M but larger R has more entropy. This does make a kind of sense: a mass M confined in a larger radius R should have more degrees of freedom since the position of its parts is less constrained, and therefore should have more entropy (assuming we know nothing about its state other than M and R).

I'm still digesting the paper, and Appendix A in particular; I may have further comments later on.

 

[michael879]

That's a good point, and I see what you're saying about the larger system having more entropy in general. However, when you think about it in terms of area it makes no sense! Why would simply expanding the system make the final state black hole's area any larger?? I look forward to any other comments, I thought I understand the normal bound pretty well but apparently I don't... I mean, the heuristic argument makes perfect sense, but you're left with an unspecified factor and the question of radiation remains...

 

[jartsa]

This article I find unconvincing but understandable:

http://www.scholarpedia.org/article/Bekenstein_bound

First we derive a bound that has the radius of the black hole as one parameter.

Then we replace "the radius of the black hole" with "couple of times the radius of the dropped object".

Then we find the exact value of "couple of times".

 

[jartsa]

So, if I have a system with very small mass, very large radius, and very large entropy, then it seems that I can violate the second law of thermodynamics by dumping the system into a small black hole.

(If the system is a gas cloud, just put the system and the black hole in a hermetic container, and wait)

 

[PeterDonis]

Why would simply expanding the system make the final state black hole's area any larger??

I'm still digesting the papers, but I think this is because expanding the system changes the details of the process of it falling into the black hole, in such a way as to make the lower bound on the area increase larger.