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What's Wrong With Black Hole Thermodynamics?

by juanrga
Tags: black hole, thermodynamics
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twofish-quant
#37
Feb20-12, 09:01 PM
P: 6,863
Quote Quote by juanrga View Post
Already taking a look to the first reference, one can find many basic mistakes. The author (astrophysicist?) does not even know what equilibrium is, or more correctly, he confounds the concept of mechanical equilibrium with the concept of thermodynamical equilibrium.
Is anyone else reading this? I don't think I can convince you, but if someone else is reading this, they might be learning some astrophysics, so it's useful. If no one else is following this thread, then it's a waste of my time.

No people aren't confusing mechanical equilibrium with thermodynamic equilibrium.

What happens is that in astrophysical situations, the time scales for hydrodynamic equilibrium are considerable smaller than the thermodynamic equilibrium timescales. Typically in a star, time it takes to reach hydro equilibrium is in minutes, whereas it takes several thousand of years to reach thermo equlibrium. Therefore in modelling stars, thermodynamic equilibrium *is an incorrect assumption*, and the dynamics is driven by hydro rather than by thermo. You can assume (at least in stars) assume *local* thermo equilibrium, which allows you to use equations of state, but that's it, and that's not even true when you are talking about stellar atmospheres which are wildly out of equilibrium.

Because of time scales, you can't take thermo equations and add potentials. You have to start with hydro equations, then add in local thermo equilibrium. Stars are wildly out of thermo equilibrium because gravity pushes them out of thermo equlibrium. If they were in thermo equilibrium, the sun wouldn't shine.

Because gravity affects the behavior of atoms, you often have to rederive the equations from the atomic level using statistical mechanics. Gravity fundamentally changes the thermo behavior of gasses, so that you have to think about things at the atomic level. Equations and relationships that work in the laboratory, just don't work in self-gravitating systems.

This is important for stars. If you dump energy into ordinary gas, it will just expand. Gravity changes the thermo properties of gases so that if you dump energy into them, they will contract. This means that the extra energy has to go somewhere, which is why stars shine. If gravity *didn't* change the thermo property of gases, then stars would not exist.

I repeat, by confounding well-understood thermodynamic stuff you can obtain anything that you want, negative or even imaginary heat capacities... all is possible.
You keep saying that, but it's not true. You can derive the laws of thermodynamics from statistical mechanics, so once you have to rederive the equations from basic principles, those statistical rules still hold, so that you can't get anything. So when you rewrite all of the equations to take into account of gravity, you end up with different equations, but the statistical microphysics makes sure that energy is conserved, entropy increases, and entropy goes to zero as T->0.

You keep saying that people don't understand themodynamics, but personally I think people understand it a lot better than you do. You just can't take equations out of a book and assume they are universally true. You have to understand the *principles* behind those equations, and in stars, they are very different than what you see in the laboratory, and you can get weird stuff.

If you are not willing to learn, then there is no point in me teaching it to you, but if anyone else is interested, I'll keep talking.
jambaugh
#38
Feb21-12, 06:01 AM
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Pardon the long delay in replying I'm behind grading papers.
Quote Quote by juanrga View Post
I was said in my thermo course that residual entropies are the result of ignoring some interaction that breaks the degeneracy. I.e. that those degeneracies are fictitious. In any case I cannot see how substituting the strong form by the weak form [itex]\lim_{T\to 0}S = S_0[/itex] changes anything for BH thermo.
The assertion that there exists some ignored interaction is ad hoc and not part of the theory. It is most especially invalid in the BH case. Where it makes a difference is that with a BH the entropy is increasing with decreasing temperature. Not a violation of 3rd law if you allow it to increase to a "constant" (diverge to infinity) rather than require it be zero.

The zero value clearly ignores entropy of mixing. Yes you can resolve that by asserting the system has a boundary breaking spatial symmetry and so some mixtures have higher energy than others but that's just the thing about a black hole. Its interior has no spatial boundary. The event horizon and the singularity are null surfaces.

How do you adjust the 'exceptional behaviour' of evaporating BHs with the thermodynamic properties of the supposedly emitted thermal radiation?
No adjustment is needed. You can have a BH in thermal equilibrium with its environment. [BIG box with black hole and thermal photon gas] Inject energy into the system and the system will reach a new higher entropy equilibrium with a larger black hole and a colder environment.

I spent some time this weekend seeing if I could integrate the exterior volume to replicate some quantitative figures but its a busy time in the semester and I have other pressing priorities. If I have time and come up with anything worth posting I will.

The most exceptional aspect of a BH is that it must reach infinite size to achieve zero temperature. This is not outside the practical meaning of the 3rd law which is that absolute zero is an asymptotic limit one cannot achieve via finite processes.

If the Bekenstein-Hawking formula is wrong then one should be able to generate a contradiction via some though experiment. I see nothing like this in the OP reference.
PeterDonis
#39
Feb21-12, 10:07 AM
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Quote Quote by jambaugh View Post
The event horizon and the singularity are null surfaces.
The EH is null but the r = 0 singularity (for a Schwarzschild BH) is spacelike. It's true that neither of them are a "spatial boundary"; a spatial boundary would have to have a timelike component (more precisely, it would have to be a family of timelike worldlines, one for each point of the boundary).
PAllen
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Feb21-12, 10:32 AM
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I've been following with interest, but for me you (Twofish-quant) don't need to add anything. I've reading through references you've already provided.
juanrga
#41
Feb21-12, 12:29 PM
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Quote Quote by twofish-quant View Post
No people aren't confusing mechanical equilibrium with thermodynamic equilibrium.

What happens is that in astrophysical situations, the time scales for hydrodynamic equilibrium are considerable smaller than the thermodynamic equilibrium timescales. Typically in a star, time it takes to reach hydro equilibrium is in minutes, whereas it takes several thousand of years to reach thermo equlibrium. Therefore in modelling stars, thermodynamic equilibrium *is an incorrect assumption*, and the dynamics is driven by hydro rather than by thermo. You can assume (at least in stars) assume *local* thermo equilibrium, which allows you to use equations of state, but that's it, and that's not even true when you are talking about stellar atmospheres which are wildly out of equilibrium.

Because of time scales, you can't take thermo equations and add potentials. You have to start with hydro equations, then add in local thermo equilibrium. Stars are wildly out of thermo equilibrium because gravity pushes them out of thermo equlibrium. If they were in thermo equilibrium, the sun wouldn't shine.
His confusion has nothing to see with time scales.
zonde
#42
Feb21-12, 01:28 PM
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Quote Quote by twofish-quant View Post
If you dump energy into ordinary gas, it will just expand. Gravity changes the thermo properties of gases so that if you dump energy into them, they will contract. This means that the extra energy has to go somewhere, which is why stars shine. If gravity *didn't* change the thermo property of gases, then stars would not exist.
This does not make sense. Take your example with satellite. Take away energy and satellite goes into lower orbit (system contracts), put energy into the system and satellite goes into higher orbit (system expands).

It seems that your misunderstanding starts here:
Quote Quote by twofish-quant View Post
The virial theorem says that the internal energy is going to be -1/2 the potential energy. You can get a similar relationship through the equipartition.

This poses an interesting problem. If you just use classical mechanics, then you could in principle collapse the star to zero radius, and extract infinite energy which then violates all sorts of thermodynamic principles. What keeps this from happening is that if you collapse the star enough, it becomes a black hole that that sets a bound for the amount of energy you can extract. With some rather simple arguments you can convert those bounds into entropy bounds, which is what Hawking and Berkenstein have done.
Solution to this problem is simple. Energy that you can extract from gravitationally collapsing system is limited by mass of the system. As system is collapsing it develops bigger and bigger mass defect and this mass defect can not exceed initial mass.

So basically you can ignore gravity in thermodynamical treatment of selfgravitating object because the only thing it does is rearranges structure of energy inside object.
juanrga
#43
Feb21-12, 01:57 PM
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Quote Quote by jambaugh View Post
No adjustment is needed. You can have a BH in thermal equilibrium with its environment.
String theorists need to invoke such tricks as a five-dimensional space to adjust the cubic T-dependence of entropy of radiation with the quadratic T-dependence of the BH entropy.
juanrga
#44
Feb21-12, 03:13 PM
P: 476
Quote Quote by jambaugh View Post
Pardon the long delay in replying I'm behind grading papers.

The assertion that there exists some ignored interaction is ad hoc and not part of the theory. It is most especially invalid in the BH case. Where it makes a difference is that with a BH the entropy is increasing with decreasing temperature. Not a violation of 3rd law if you allow it to increase to a "constant" (diverge to infinity) rather than require it be zero.
As I recall, all the degeneracies are shown to vanish, when more realistic interaction models are considered. That is why Planck stated the third law in that precise form: [itex]\lim_{T\to 0}S = 0[/itex].

Moreover, I do not agree on that the BH model is not ignoring interactions. For instance, the GR model is clearly ignoring the self-interaction of the graviton field.

The real problem here is not that Planck third law of thermo is violated in BHs, but that no generalized third law as [itex]\lim_{T\to 0}S_{gen} = 0[/itex] exists for [itex]S_{gen} = S + S_{BH}[/itex].

Similarly, there is not generalized zeroth law in BH 'thermodynamics', because of the area law (as the OP article remarks).

The only generalized law in BH 'thermodynamics' is the GSL (Generalized Second Law) [itex]\Delta S_{gen} \geq 0[/itex]. But even in this I have my doubts, because for an evaporating BH [itex]\Delta S_{BH} \leq 0[/itex], which means that evaporation cannot be considered a dissipative process at the BH level!
twofish-quant
#45
Feb22-12, 07:41 AM
P: 6,863
Quote Quote by zonde View Post
This does not make sense. Take your example with satellite. Take away energy and satellite goes into lower orbit (system contracts), put energy into the system and satellite goes into higher orbit (system expands).
Exactly, so if you take away energy from a gas, then the system contracts and gets hotter. Put in energy and the system expands and gets cooler. If you take away energy, you are adding to the internal energy, but this comes from the potential energy that you lose by contracting the system.

Solution to this problem is simple. Energy that you can extract from gravitationally collapsing system is limited by mass of the system. As system is collapsing it develops bigger and bigger mass defect and this mass defect can not exceed initial mass.
No. Mass doesn't change. System gets smaller and denser. Now if you could collapse a system to point, then you would have a big problem since you would be able to extract infinite energy from a gravitationally bound system.

But somehow nature creates these darn black holes that keep us from doing that.

So basically you can ignore gravity in thermodynamical treatment of self gravitating object because the only thing it does is rearranges structure of energy inside object.
But we are talking about the energy structure.
twofish-quant
#46
Feb22-12, 08:19 AM
P: 6,863
Quote Quote by jambaugh View Post
If the Bekenstein-Hawking formula is wrong then one should be able to generate a contradiction via some though experiment. I see nothing like this in the OP reference.
Conversely if you assume that black holes are not thermodynamic objects, then it's trivially easy to generate contradictions.

If you assume that naked singularities can exist, then it's trivial to violate the first law of thermodynamics.

If you assume that black holes don't have entropy, then it's trivial to violate the second law of thermodynamics. You just use the black hole as a vacuum to dump waste heat, and it's not hard to make a perpetual motion machine.
zonde
#47
Feb23-12, 10:25 PM
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Quote Quote by twofish-quant View Post
Exactly, so if you take away energy from a gas, then the system contracts and gets hotter. Put in energy and the system expands and gets cooler. If you take away energy, you are adding to the internal energy, but this comes from the potential energy that you lose by contracting the system.
Sorry but it still does not make sense. Basically you are saying that "if you take away energy you are adding energy". That is contradictory statement.

Let's say it differently. System is loosing energy to environment (energy is crossing some closed surface around the body that we consider border between the system and environment). When this happens system gets hotter. Mechanism how it gets hotter is internal to system and is not related to anything additional crossing that border between system and environment. Do you agree so far?

Quote Quote by twofish-quant View Post
No. Mass doesn't change. System gets smaller and denser. Now if you could collapse a system to point, then you would have a big problem since you would be able to extract infinite energy from a gravitationally bound system.
How do you argument that mass doesn't change?

From my side argumentation for change of mass is that mass/energy of the system is conserved (E=mc^2) and so if some energy is crossing border between system and environment (closed arbitrary surface around the system) mass/energy of the system remaining inside the border gets smaller by the amount of energy that left the system.
That certainly works for microscopic systems and is experimentally confirmed (as I believe, I will check if you will doubt that) as mass defect.
twofish-quant
#48
Feb23-12, 11:49 PM
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Quote Quote by zonde View Post
Sorry but it still does not make sense. Basically you are saying that "if you take away energy you are adding energy". That is contradictory statement.
Energy is conserved. You have E = E_star + E_environment and then E_star = E_gravity + E_thermal

For our purposes E is constant. E_star goes down. E_thermal goes up but E_gravity goes down even more.

When this happens system gets hotter. Mechanism how it gets hotter is internal to system and is not related to anything additional crossing that border between system and environment. Do you agree so far?
Well yes.

From my side argumentation for change of mass is that mass/energy of the system is conserved (E=mc^2) and so if some energy is crossing border between system and environment (closed arbitrary surface around the system) mass/energy of the system remaining inside the border gets smaller by the amount of energy that left the system.
Rest mass is conserved and doesn't change.
zonde
#49
Feb25-12, 01:04 AM
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Quote Quote by twofish-quant View Post
Energy is conserved. You have E = E_star + E_environment and then E_star = E_gravity + E_thermal

For our purposes E is constant. E_star goes down. E_thermal goes up but E_gravity goes down even more.
So it's E_gravity that should supply infinite energy, right?

Quote Quote by twofish-quant View Post
Rest mass is conserved and doesn't change.
Is it supposed to say something more than "Mass doesn't change"?

Here is what wikipedia says in binding energy:
"In bound systems, if the binding energy is removed from the system, it must be subtracted from the mass of the unbound system, simply because this energy has mass, and if subtracted from the system at the time it is bound, will result in removal of mass from the system.[5] System mass is not conserved in this process because the system is not closed during the binding process."
Reference 5 turns out to be hyperphysics page about nuclear binding energy.
But wikipedia page has reference to this (pay per view) article as well World Year of Physics: A direct test of E=mc2


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