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What's Wrong With Black Hole Thermodynamics? 
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#37
Feb2012, 09:01 PM

P: 6,863

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 selfgravitating 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. 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. 


#38
Feb2112, 06:01 AM

Sci Advisor
PF Gold
P: 1,776

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


#39
Feb2112, 10:07 AM

Physics
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PF Gold
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#40
Feb2112, 10:32 AM

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PF Gold
P: 5,059

I've been following with interest, but for me you (Twofishquant) don't need to add anything. I've reading through references you've already provided.



#41
Feb2112, 12:29 PM

P: 476




#42
Feb2112, 01:28 PM

PF Gold
P: 1,376

It seems that your misunderstanding starts here: 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. 


#43
Feb2112, 01:57 PM

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#44
Feb2112, 03:13 PM

P: 476

Moreover, I do not agree on that the BH model is not ignoring interactions. For instance, the GR model is clearly ignoring the selfinteraction 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! 


#45
Feb2212, 07:41 AM

P: 6,863

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


#46
Feb2212, 08:19 AM

P: 6,863

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. 


#47
Feb2312, 10:25 PM

PF Gold
P: 1,376

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? 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. 


#48
Feb2312, 11:49 PM

P: 6,863

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


#49
Feb2512, 01:04 AM

PF Gold
P: 1,376

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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