# How To COUNT Information in Black Hole?

1. Nov 5, 2012

### referframe

I have not read any formal (mathematical) explanations of black hole thermodynamics, only popular literature on that subject.

I have read that the total amount of information, and therefore its entropy, that a black hole can contain is proportional to the area of its event horizon (measured in Planck areas, etc.).

But the literature is always a little vague regarding how one would COUNT how much information is in any specific black hole. Surely, the number of elementary particles in the black hole, along with their particle types, would contribute to the count of total information (I realize that the entropy would be the LOGARITHM of this total count). The particles various spins would obviously count. I’m not sure if continuous observables like position and momentum would be counted.

MY QUESTION: Would the number and type of COMPOSITE objects, such as atoms, be counted SEPARATELY from the number of their constituent elementary particles?

Thanks in advance.

2. Nov 5, 2012

### Naty1

The only mass [particles] inside the horizon of a black hole are those recently swallowed and falling towards the singularity where presumably they are crushed from existence as we know it...in quantum foam....

The entropy is proportional to the logarithm of the number of microstates.....

just how they count such states is 'above my paygrade'.... see here:

http://en.wikipedia.org/wiki/Microstate_(statistical_mechanics [Broken])

In general, a composite object [say, a particle] has more degrees of freedom than an elementary particle. So an atom has more degrees of freedom, more entropy, than any of it's individual constituents, and that same atom has additional degrees of freedom when part of a larger composite object, like a lattice in a metal.

Last edited by a moderator: May 6, 2017
3. Nov 5, 2012

### Naty1

Here is part of a really good discussion which was referenced in another thread here: [glad I wimped out on explaining 'microstates!!]

[from prof [UCLA] Steve Carlip]....

http://www.2physics.com/2007/06/symmetries-horizons-and-black-hole.html

Last edited: Nov 5, 2012
4. Nov 6, 2012

### Dmitry67

If I witness a collapse of some body, then before the collapse entropy was lower than after the collapse, as Black hole has maximum entropy.

The question is, WHEN (relative to the distant observer) does it happen, because there is no global notion of 'now'. So, when theories talk about entropy of BH, what time is used, or is it a 'timeless' concept?

5. Nov 6, 2012

### Staff: Mentor

That's because, as others have posted, nobody knows for sure how to do it. However...

...one thing we do know is that ordinary "particle" states are *not* the ones that need to be counted. The counting is not done over the interior region; it is done at the horizon, as indicated by the fact that it's the area of the horizon, *not* any measure over the interior, that gives the total entropy. The problem used to be that nobody could find *any* microstates over the horizon that counted up to the right answer; now, as the Carlip quote says, the problem is that we know *too many* ways to count such microstates, but nobody knows why they all give the same answer. But *none* of those candidates for counting microstates over the horizon are counting anything like ordinary "particles".

Technical point: I think what you mean to say here is that if we have, for example, a system of N constituents, each with f degrees of freedom, then the total number of degrees of freedom F of the system is *greater* than Nf. This assumes that there are interactions between the constituents; a "gas" of completely free particles, with no interactions between them, would have exactly Nf degrees of freedom; there would be nothing "extra" over and above the degrees of freedom of the individual constituents. (Of course, in any real object, there *will* be interactions, but there are plenty of idealized theoretical models where there aren't.)

If you are willing to call the various candidates for the microstates on a BH horizon "elementary particles", then yes, this is correct (as it would be in an ordinary object). However, as I said above, whatever these microstates are, they are certainly *not* what we ordinarily think of as elementary particles: they are not electrons, quarks, photons, etc.

For an ordinary object, yes. For a BH, see above.

For an ordinary object, yes, they would. For a BH, see above.

Last edited: Nov 6, 2012
6. Nov 6, 2012

### Staff: Mentor

As I understand it, the BH entropy calculation is done over a stationary state; in other words, it assumes a BH that does not change with time. That means the entropy doesn't change either, so it's independent of what notion of "time" you use.

Of course a real BH would change with time, as objects fell into it (or, sometime very, very far in the future when the universe becomes cold and empty enough, as it evaporates via Hawking radiation). But the change with time is so slow that it can be modeled as quasi-static, i.e., as a succession of stationary states with slightly different masses. So the entropy calculation can just use each of those stationary states, without having to worry about the details of time evolution.

7. Nov 6, 2012

### Dmitry67

Well, but collapse can't be called "quasi-static", right?

8. Nov 6, 2012

### Staff: Mentor

I don't know if anyone has tried to do a "continuous" accounting of entropy change during gravitational collapse. AFAIK the only BH entropy calculations that have been done have been for stationary BHs. AFAIK it has been proven, however, that a BH has the maximum possible entropy of any object with the same mass, and also that a BH has the maximum possible entropy that can be put into its volume. So nobody doubts that the second law of thermodynamics holds when an ordinary object collapses to a BH.

9. Nov 6, 2012

### Naty1

yes....'each with f degrees of freedom, individually'....

I equate the idea with an orbital electron in an atom having the degrees of freedom to absorb a photon while an individual free electron does not....

[edit: I am quite content in accepting that you know what I meant to say better than I know what I meant to say....and even if you did not know what I meant to say, I am still content to accept your explanation. This is also the way it works at my home and family so I am used to it....But here it actually makes sense!!]

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10. Nov 6, 2012

### Staff: Mentor

"Free" electrons *can* emit and absorb photons, so this isn't a good way to distinguish "free" particles from interacting ones in the sense of degrees of freedom. In the real world there are no completely "free" particles; every particle interacts with something, so there are always some extra degrees of freedom present beyond the "free particle" ones. Whenever you see an accounting of degrees of freedom that doesn't include interactions, it means something is being left out of the model (usually because the interactions are too small to matter for the problem under consideration).

The sense of "free" in which an electron in, say, a cathode ray tube is free while an electron in an atom is bound has to do with the *kind* of interaction potential that is present, not in the presence or absence of one. A free electron in this sense has an interaction potential that is not spatially localized, so there is a continuous spectrum of states; that means the electron can interact with photons of any energy. An electron bound in an atom has a spatially confined interaction potential, so its spectrum of states is discrete; that means the electron can only interact with photons that have the right energy to kick it from one of the discrete states to another.

Last edited: Nov 6, 2012
11. Nov 7, 2012

### Dmitry67

So, radius of BH is proportional to M
Amount of information in BH is proportional to R^2
So entropy of BH increases faster, than mass...

If I drop small dust particles into BH, why every new particle adds more entropy then a previous one?

12. Nov 7, 2012

### clamtrox

Certainly the collapsing states are not in thermal equilibrium, so I don't see how you could model it as being quasistatic.

Because the temperature of the black hole decreases as you add more stuff into it. So if you use the 2nd law of thermodynamics, you have dM~TdS, and since T~1/M, integrating that you get S~M2.

13. Nov 7, 2012

### Naty1

PeterDonis:
I found one source which underlies my post, from TomStoer in another thread.....

Maybe these are different views of a 'free electron'??

This view seemed substantiated when I posted and someone confirmed:

[I did not record which thread.]

14. Nov 7, 2012

### Staff: Mentor

I think so. In the quote it talks about extra degrees of freedom, but they're not extra DoF of the electron itself; they're extra DoF that are present because there is another particle present (the proton in the hydrogen atom). In the particular example of the hydrogen atom, the system as a whole happens to be a bound system. But you could also set up a system (such as a free electron laser) where the "extra particles" are just other electrons, and none of them are bound, yet they can still emit and absorb photons (because the Coulomb interactions between the electrons provide the extra DoF required to allow the conservation laws to hold).

If in your original statement you mean by a "free" electron one with absolutely no interactions with anything else, then yes, you are correct that such an electron could not emit or absorb photons--by definition, because photon emission/absorption is an interaction, and you've stipulated that the "free" electron can't interact. But as I noted before, there are no truly "free" particles in this sense--everything interacts with something--so it's not a very useful idea IMO.

For example, even if I want to model a single electron, alone in the entire universe, and all I want to do is calculate how it goes from one place to another, according to quantum field theory I still have to include the possibility of the electron emitting and absorbing *virtual* photons (and in turn those virtual photons can produce virtual electron/positron pairs, which can annihilate into virtual photons, etc., etc.). If I don't include all these possibilities, I will get the wrong answer for the mass of the electron, which appears on the surface to be a property of the "free" electron, not requiring any interactions.

15. Nov 7, 2012

### Naty1

PeterDonis,
as always very good explanation in your prior post....
I especially found the last paragraph helpful: I had not realized an isolated 'free' electron would routinely interact with virtual photons, probably because I never thought about it....

Thanks....

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