What's Wrong With Black Hole Thermodynamics?

[juanrga]

OK. Suppose the satellite is a hydrogen atom, and you are trying to calculate the properties of a large number of "satellites". At this point, the kinetic energy of the satellites *becomes* the internal energy of gas of satellites.

No. If I assume that atoms can be treated classically as point-like particles then the thermodynamic internal energy of the gas of atoms is not the kinetic energy of the gas. The thermodynamic internal energy of a gas is defined as the total energy minus the average macroscopic kinetic energy.

There are several other ways of calculating this. You can use the virial theorem or the equipartition theorem. See

http://en.wikipedia.org/wiki/Equipartition_theorem
http://en.wikipedia.org/wiki/Virial_theorem

From the wikipedia link:

In mechanics, the virial theorem provides a general equation relating the average over time of the total kinetic energy, \left\langle T \right\rangle, of a stable system consisting of N particles, bound by potential forces, with that of the total potential energy, \left\langle V_\text{TOT} \right\rangle, where angle brackets represent the average over time of the enclosed quantity.

But thermodynamics --which is not mechanics-- deals with internal energy, which is neither total kinetic energy nor total potential energy.

Yes it does. You change those satellites into atoms and then have several gazillion of them. At which point you have a gas.

A gas is not defined as a N-collection of particles with N large. If you were right then virtually every macroscopic piece of matter would be a gas and is not .

Except that you don't. You have the internal energy of the gas which consists of the kinetic energy of the gas molecules (H), and then the potential energy of the gravitational field (U). The relationship between the two are defined by the equipartition theorem and the virial theorem. Because the sign of the potential energy is negative, what happens is that as you remove total energy from the system, the kinetic/internal energy of the gas increases. (It's annoying here, because we are running out of letters.)

Using standard notation (i.e. letters)

E = U + K + V

In mechanics kinetic energy is often denoted by T, but in thermodynamics T is temperature and K is standard symbol for kinetic energy.

If you insist on confound internal energy U with kinetic energy K or with potential energy V, then you can obtain virtually any result. The sum (K+V) is usually named mechanical energy. This mechanical energy is what mechanics is interested in. This mechanical energy verifies virial theorem. Read again the vikipedia link that you give. Pay attention to the first two words.

Thermodynamics is usually interested in U: internal energy. Indeed, thermodynamic textbooks contains a chapter introducing the concept of internal energy and its main properties.

You are missing a term. The problem is that the field is not an external field but the result of self-interactions. What that means is that you have an extra \partial (\tau_k) / \partial t term which includes how the gravitational potential changes in response to external interactions.

This is off the top of my head so don't shoot me if there is a problem (and I'm getting confused with all of the letters meaning different things), but if you add energy into a gravitationally bound system then the \partial (\tau_k) / \partial t is going to be strongly positive which means that once you add that term then c_V is going to turn negative.

I'm not surprised that someone that isn't familiar with stars would make that mistake, since in laboratory experiments interacting with the system doesn't change the potential, but in self-gravitating systems, it does.

First, \tau_k is the coupling constant, not the gravitational potential.

Second, as is well-known, the thermodynamic expression

\frac{\partial u}{\partial t} = c_V \frac{\partial T}{\partial t} + (u_k + \tau_k \psi) \frac{\partial n_k}{\partial t}

can be also rewritten as

\frac{\partial \tilde{u}}{\partial t} = c_V \frac{\partial T}{\partial t} + (u_k + \tau_k \psi) \frac{\partial n_k}{\partial t} + \tau_k \frac{\partial \psi}{\partial t}

using \tilde{u} = u + \tau_k \psi.

Of course the value of the heat capacity c_V is the same, because in the first case one takes partial derivative of internal energy u at constant composition and, in the second case, one takes the partial derivative of \tilde{u} but maintaining also constant \psi evidently.

One other wrinkle is that hydrodynamics beats thermodynamics. A star takes a few minutes to reach hydrodynamic equilibrium, but several thousand years to reach thermodynamic equilibrium. Because of these time scale differences, the hydrodynamics drives the thermodynamics. What that means is that if you add energy to the system, the hydro will cause the change to happen immediately that create temperature gradients that cause the system to go out of general thermodynamic equilibrium.

So you start with an isothermal gas, and add energy to it. Rather than staying in thermodynamic equilibrium, the energy will get distributed hydrodynamically, which will throw the system out of thermo equilibrium.

It seems you are confused about what thermodynamics really is and does. It is well-known that thermodynamics extends the balance laws of hydrodynamics in many ways. For instance, the balance law for internal energy contains the heat flux q and the term \pi:\nabla v, where \pi, is the dissipative pressure tensor. q and \pi are non-hydrodynamical quantity, unless you want to reinvent hydrodynamics also .

No confusion. You have total energy which consists of internal energy + potential energy. Internal energy consists of kinetic energy of the gas molecules + internal degrees of freedom. Equipartition and virial theorems give you very simple relationships between these quantities. So once you have figured out the total energy, the virial theorem gives you the relationship between that and the internal energy and the potential energies.

The total energy E of a gas is E = K + V + U, with K the total kinetic energy, V the total potential energy and U the internal energy. The virial theorem of mechanics (read the wikipedia link that you give) applies to K and V (unsurprisingly K+V is named the mechanical energy). The virial theorem does not apply to U.

If you insist on confounding thermodynamic quantities with mechanical quantities then you can obtain any bizarre result that you want even dS<0.

 

[twofish-quant]

No. If I assume that atoms can be treated classically as point-like particles then the thermodynamic internal energy of the gas of atoms is not the kinetic energy of the gas. The thermodynamic internal energy of a gas is defined as the total energy minus the average macroscopic kinetic energy.

Except that the kinetic energy in this situation is microscopic. I have a hydrogen atom. It goes in orbit around the earth. All of the gravitational rules apply so that if I remove energy from the atom, it will drop to a lower orbit and gain microscopic KE.

I drop a bag of hydrogen gas in high orbit. The net KE is zero, but the atoms in the gas have a KE that is from microscopic motions. The gas molecules go into random orbit around the earth. I suck energy from the molecules. They go into lower orbits. They are now moving faster, but the net KE is still zero.

But thermodynamics --which is not mechanics-- deals with internal energy, which is neither total kinetic energy nor total potential energy.

And in astrophysical gases without internal modes, the microscopic kinetic energy is the internal energy. We aren't talking about the bulk flows, but rather than microphysics. The point is that gravity will change the thermodynamic properties of a fluid. Gravity doesn't know whether it is working on an apple or a hydrogen atom so it has to couple with the kinetic energy. You drop a hydrogen atom, it will go into an elliptical orbit until something stops it.

You can't separate mechanics and thermodynamics when you have a gravitational field.

Thermodynamics is usually interested in U: internal energy. Indeed, thermodynamic textbooks contains a chapter introducing the concept of internal energy and its main properties.

And you can't separate the kinetic properties of a gas from the internal energy when you are dealing with astrophysical fluids. You can ignore this in lab experiments, but you can't when you add gravity. This causes all of the formula to be slightly different.

Of course the value of the heat capacity c_V is the same, because in the first case one takes partial derivative of internal energy u at constant composition and, in the second case, one takes the partial derivative of \tilde{u} but maintaining also constant \psi evidently.

And you can't do this with self-gravitating fluids. What happens is that the internal kinetic energy of the gas changes the potential via the virial theorem. You keep insisting that you can separate out the "internal energy" from the "kinetic energy" but you can't, since the virial theorem doesn't only apply to the *macroscopic kinetic energy* but rather also to the *microscopic kinetic energy* of the gas that comprises the internal energy of gases. (Again, solids are different.)

Gravity applies to all motions. It doesn't know whether the object that is dropping is an apple or a hydrogen atom. So the virial theorem includes not only the *macroscopic* kinetic energy, but also the *microscopic* kinetic energy.

It seems you are confused about what thermodynamics really is and does.

I'm just telling you how things are done in astrophysics. In astrophysics, there isn't a strong boundary between the microscopic motions and the macroscopic ones, and hydrodynamics interacts with the thermodynamics.

This is all very standard stuff. In the laboratory, you can make the distinction between "kinetic energy" and "internal energy". In astrophysics the distinction is much less sharp.

The total energy E of a gas is E = K + V + U, with K the total kinetic energy, V the total potential energy and U the internal energy. The virial theorem of mechanics (read the wikipedia link that you give) applies to K and V (unsurprisingly K+V is named the mechanical energy). The virial theorem does not apply to U.

Yes it does. Look at the derivation of the virial theorem. All it involves are self-gravitating particles. Those particles don't have to be macroscopic. Hydrogen atoms will do. Once you have hydrogen atom as the self-gravitating bits, then the virial theorem applies to the microscopic kinetic energy of the atoms and hence to the internal energy.

Here is another thought experiment. You have to clouds of gas. You say that the virial theorem applies to bulk motions, so you can calculate the gravitational potential. Now you collide the clouds to that the bulk kinetic energy energy goes to zero, and the bulk kinetic energy goes into heat, so the bulk motions disappear, so bulk kinetic energy is now zero.

If your interpretation was correct then the gravity would just disappear. But it doesn't. The reason it doesn't is that the virial theorem applies both to macroscopic KE and microscopic internal energy.

This is very basic astrophysics.

If you insist on confounding thermodynamic quantities with mechanical quantities then you can obtain any bizarre result that you want even dS<0.

Blame God. I didn't make the rules. And it turns out that you can't obtain any bizarre result. Everything works out so that energy is conserved and entropy increases. Somehow the rules work out so that happens.

Once you have self-gravitating objects then mechanics changes the microphysics which changes the thermodynamics. Gravity works on individual hydrogen atoms, which means that the virial theorem must include the microscopic kinetic motions of atoms.

But somehow the bookkeepping works out so that dS > 0 in closed systems.

 

[twofish-quant]

Also one has to be careful about the bookkeepping. In the case of mono-atomic ideal gases, all of the internal energy is in the form of kinetic motions, and so it's subject to the virial theorem. This is less true for gases with internal degrees of freedom, and not true at all for condensed matter. So you not only have to include internal energy, but you have to look at the form of the internal energy.

 

[twofish-quant]

Also if you insist that the virial theorem doesn't apply to thermal internal energy, you not only have to argue with me, but also with...

http://web.njit.edu/~gary/321/Lecture8.html
http://www.astro.utoronto.ca/~mhvk/AST221/L6/L6_4.pdf
http://burro.astr.cwru.edu/Academics/Astr221/LifeCycle/jeans.html
http://www2.astro.psu.edu/users/rbc/a534/lec10.pdf
http://www.vikdhillon.staff.shef.ac.uk/teaching/phy213/phy213_virial.html
https://casper.berkeley.edu/astrobaki/index.php/Virial_Theorem
http://www.jb.man.ac.uk/~smao/starHtml/stellarEquation.pdf
http://www.astro.gla.ac.uk/users/lyndsay/TEACHING/STELLAR/SSE3_05b.pdf
http://jila.colorado.edu/~pja/astr3730/lecture15.pdf
http://www.astro.caltech.edu/~jrv/Ay20/ws/ws_stellarstructure.pdf

whew... That's only after about five minutes of searching on the web. If you want book citations, and I can get you several of those.

 

[juanrga]

Except that the kinetic energy in this situation is microscopic.

You understood nothing.

 

[juanrga]

Also if you insist that the virial theorem doesn't apply to thermal internal energy, you not only have to argue with me, but also with...

http://web.njit.edu/~gary/321/Lecture8.html
http://www.astro.utoronto.ca/~mhvk/AST221/L6/L6_4.pdf
http://burro.astr.cwru.edu/Academics/Astr221/LifeCycle/jeans.html
http://www2.astro.psu.edu/users/rbc/a534/lec10.pdf
http://www.vikdhillon.staff.shef.ac.uk/teaching/phy213/phy213_virial.html
https://casper.berkeley.edu/astrobaki/index.php/Virial_Theorem
http://www.jb.man.ac.uk/~smao/starHtml/stellarEquation.pdf
http://www.astro.gla.ac.uk/users/lyndsay/TEACHING/STELLAR/SSE3_05b.pdf
http://jila.colorado.edu/~pja/astr3730/lecture15.pdf
http://www.astro.caltech.edu/~jrv/Ay20/ws/ws_stellarstructure.pdf

whew... That's only after about five minutes of searching on the web. If you want book citations, and I can get you several of those.

Sure that you can find a hundred of 'astrophysics' references claiming that dA>0 (often named second law of black hole mechanics) was a fundamental law of nature... except that was based in a flawed analogy with real thermodynamics.

If thermodynamics had been studied, and different kind of energies had not been confused, then those hundred of authors had understood that dA<0 was perfectly possible and that their so-named fundamental law (the black hole analogue of the second law of thermodynamics they said to us) was only smoke .

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.

I repeat, by confounding well-understood thermodynamic stuff you can obtain anything that you want, negative or even imaginary heat capacities... all is possible.

 

[twofish-quant]

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]

Pardon the long delay in replying I'm behind grading papers.

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 \lim_{T\to 0}S = S_0 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]

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]

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]

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]

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:

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]

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]

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: \lim_{T\to 0}S = 0.

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 \lim_{T\to 0}S_{gen} = 0 exists for S_{gen} = S + S_{BH}.

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) \Delta S_{gen} \geq 0. But even in this I have my doubts, because for an evaporating BH \Delta S_{BH} \leq 0, which means that evaporation cannot be considered a dissipative process at the BH level!

 

[twofish-quant]

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]

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]

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?

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]

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]

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?

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