Register to reply 
What's Wrong With Black Hole Thermodynamics? 
Share this thread: 
#19
Feb1812, 01:56 AM

P: 6,863

I've been trying to think of a "thought experiment" that illustrates what happens so that we can argue about actual science rather than personalities.
Here's an attempt..... You have a satellite that is in orbit around the earth in a distant orbit. Because it is in a distant orbit, it orbits slowly. Now you take energy away from the satellite. What happens to it? Well, because the satellite has less energy, it's going to start falling into earth. Once it gets to a lower orbit, it's going to orbit faster. Now imagine a cloud of atomic sized satellites in orbit around the earth. You take energy away from them. They'll drop to lower orbits. Once they drop to lower orbits, they will move faster. If you measure the temperature of the cloud, you'll find that it has increased because faster atoms = higher temperature. So if you have a selfgravitating object, pumping energy into the object will boost things into higher orbits which slows things down and makes things cooler. Taking energy out will put things into lower orbits which speeds things up and make things hotter. In other words the object has negative heat capacity. What's wrong with this picture? (My answer is that nothing is wrong with this picture and this is exactly what happens). 


#20
Feb1812, 01:58 AM

Sci Advisor
PF Gold
P: 5,059




#21
Feb1812, 06:32 AM

P: 476




#22
Feb1812, 06:56 AM

P: 476

Its equation 5 reads E=K+U, but the thermodynamic expression in presence of fields is E=U+K+E_{int}. 


#23
Feb1812, 07:11 AM

P: 476

If I take away internal energy dU<0 (example via cooling the satellite), but maintain constant its mechanical energy (K+V), satellite will continue to orbit exactly in the same orbit. If I take away kinetic energy dK<0 (example via decelerating the satellite), but maintain constant its internal energy (U), satellite will start falling into Earth... but this have nothing to see with thermodynamics or with negative heat capacities (indeed in this case Q=0). 


#24
Feb1812, 09:34 AM

Sci Advisor
PF Gold
P: 5,059

For example, just a few lines before the equation you criticize is: "Consider an ideal, uniform gas of total energy E and kinetic energy K, containing N particles whose internal degrees of freedom I will assume to be either nonexistent or irrelevant at the system’s current temperatures," 


#25
Feb1812, 01:31 PM

P: 476

The entropy of an ideal gas is a function S=S(U,V,N) http://en.wikipedia.org/wiki/Sackur_Tetrode Notice that internal energy is usually denoted by U and potential energy by V so that total energy is E=K+V+U The author that you cite use U for potential energy (although in a latter section uses U supposedly for internal energy, which he calls "some energy") and confounds internal energy with total energy. If composition is constant and V=V(R) this reduces to S=S(U,R). He confounds internal energy with total energy, among other mistakes, and writes S(E,R). One does not need to appeal to fusion to change the internal energy U of an ideal gas (and would be ridiculous because if exists fusion then the gas is NOT ideal). 


#26
Feb1812, 11:38 PM

P: 6,863

Considering stars and black holes as something close to closed thermodynamic system is standard basic astrophysics. If you want something to be taken seriously by astrophysicists, and you think that this is ridiculous, you are going to have to argue the point since it's a standard approximation and a very good one. There's also the issue of professional courtesy. I think part of the issue is that astrophysicists may be using terminology in ways that are different and perhaps bizarre to people in other areas of physics, but there are usually reasons behind this. If you go in with the attitude "you idiot astrophysicists don't understand anything about stars or selfgravitating objects" it's not going to end well nor should it. And yes this works both ways..... If a condensed matter physicist nicely emailed me a copy of that paper asking for feedback, and we were talking over lunch, I'd explain the flaws in a very nice way. The context is a paper posted to LANL, and once someone does that, I'm expecting that he wants it to be peer reviewed, at which point I go into "sadistic attack dog" mode, and rip it to shreds with strong language if I think the arguments are weak. I can say with very strong certainty, that I'm making the points that anyone with astrophysics background would make, and that if his goal is to get the paper published in Ap.J., A.A., MNRNAS or Physics Review, that the reviewers are doing to toss it out with perhaps even stronger language that I've been using. If it tries to publish in some journal that is not astrophysics, what is likely to happen is that the editor will send the paper to someone with astrophysics background, and then they'll say the same thing. Now if the author goes back, takes some of the points I've made seriously and tries to rewrite the paper, and then comes up with the same result, *that* would be interesting. 


#27
Feb1812, 11:43 PM

P: 6,863

This is all pretty standard stuff in astrophysics. It's also pretty important. Stars lose energy, and the heat capacity tells you what happens to the temperature of the star as it radiates. As stars radiate, the core temperatures increase which eventually causes fusion reactions that don't occur at lower temperatures. 


#28
Feb1912, 08:35 AM

P: 476

It is evident that you do not need thermodynamics for this. Gibbs and heat capacities are not even mentioned. The equation [tex]\frac{\partial u}{\partial t} = c_V \frac{\partial T}{\partial t} + u_k \frac{\partial n_k}{\partial t}[/tex] in presence of field with potential [itex]\psi[/itex] changes to [tex]\frac{\partial u}{\partial t} = c_V \frac{\partial T}{\partial t} + (u_k + \tau_k \psi) \frac{\partial n_k}{\partial t}[/tex] with [itex]\tau_k[/itex] the coupling factor. The heat capacity [itex]c_V[/itex] is the same, in presence or absence of field, this is wellknow, at least for thermodynamicians. Of course if you want to confound rest energy with internal energy U and with total energy E, if you pretend that a black hole is a closed thermodynamic system, etc. then you can obtain anything that you want, but that is 'astrophysics', not thermodynamics. 


#29
Feb1912, 09:50 AM

P: 6,863

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 The thing about selfgravitating objects is that you *can't* separate mechanics and thermodynamics. Every thermodynamic equation in astrophysics has to include mechanical interactions. 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. 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 selfgravitating systems, it does. 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. And it turns out that you can mostly ignore the radiation when calculating stellar structure. The amount of energy that a star radiates over a year is an insignificant amount of the total internal energy. If you look at nuclear energy, the star can shine for billions of years, and if you look at gravitational energy, you get enough energy to shine for tens of millions. So if you are looking at the dynamics of a star over a day, you can consider the system to be adiabatic. And you can't get anything you want. People (Chandrasekar) worked out how all of these things interact years ago, and it's in every first year astrophysics graduate text. Again it's a problem of tone. If someone asked "it's very odd that selfgravitating bodies have negative heat capacity, why is that?" that's one thing. If the implied message is "astrophysicists are being obvious idiots for ignoring something stupidly obvious" that's not going to provoke a friendly reaction. Also, uploading a paper to LANL implies that you want people to tear it to shreds. 


#30
Feb1912, 10:23 AM

P: 6,863

To summarize the special thermodynamic properties of astrophysical selfgravitating systems:
1) you cannot separate the mechanics from the thermodynamics. If you have a hydrogen gas in orbit around the earth, you have to take into account orbit mechanics in looking at the behavior of said gas. 2) you cannot treat external potentials as a constant addition to the thermo equations. Adding and removing energy from the system will change the interaction of the system with the field and that needs to be taken into account. Even if the field is constant (i.e. hydrogen atoms surround the earth) the fact that the atoms will wind up in different orbits means that you have to add extra terms in the thermo equations. 3) unlike laboratory systems, the hydrodynamic timescales of astrophysical systems are much smaller than the global thermodynamic time scales. This means that arguments based on global thermodynamic equilibrium will not work. If you have a isothermal laboratory system and add energy to it, you can expect the system to quickly move to isothermal. This isn't true with astrophysical systems where the thermal time scales are huge in comparison to the hydro scales, and also long in comparison to the observational time scale (i.e. the thermal time scale of stars are in the thousands of years). There are in fact astrophysical objects (white dwarfs) whose thermal equilibrium time is longer than the age of the universe. For stars you can assume local thermal equilibrium, but for atmospheres, even LTE is not true. This means that if you dump energy into an astrophysical system that's in thermo equilibrium, it will quickly reach hydro equilibrium but that will drive it out of thermal equilibrium, which means that arguments based on thermal equilibrium won't work. 


#31
Feb1912, 04:39 PM

P: 476

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. Second, as is wellknown, the thermodynamic expression [tex]\frac{\partial u}{\partial t} = c_V \frac{\partial T}{\partial t} + (u_k + \tau_k \psi) \frac{\partial n_k}{\partial t}[/tex] can be also rewritten as [tex]\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}[/tex] using [itex]\tilde{u} = u + \tau_k \psi[/itex]. Of course the value of the heat capacity [itex]c_V[/itex] 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 [itex]\tilde{u}[/itex] but maintaining also constant [itex]\psi[/itex] evidently. If you insist on confounding thermodynamic quantities with mechanical quantities then you can obtain any bizarre result that you want even dS<0. 


#32
Feb1912, 10:25 PM

P: 6,863

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


#33
Feb1912, 11:02 PM

P: 6,863

Also one has to be careful about the bookkeepping. In the case of monoatomic 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.



#34
Feb2012, 12:45 AM

P: 6,863

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...cle/jeans.html http://www2.astro.psu.edu/users/rbc/a534/lec10.pdf http://www.vikdhillon.staff.shef.ac....13_virial.html https://casper.berkeley.edu/astrobak...Virial_Theorem http://www.jb.man.ac.uk/~smao/starHt...arEquation.pdf http://www.astro.gla.ac.uk/users/lyn...R/SSE3_05b.pdf http://jila.colorado.edu/~pja/astr3730/lecture15.pdf http://www.astro.caltech.edu/~jrv/Ay...rstructure.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. 


#35
Feb2012, 12:16 PM

P: 476




#36
Feb2012, 12:36 PM

P: 476

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 sonamed 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 wellunderstood thermodynamic stuff you can obtain anything that you want, negative or even imaginary heat capacities... all is possible. 


Register to reply 
Related Discussions  
Black Hole Thermodynamics  Advanced Physics Homework  8  
Black Hole Thermodynamics Resources?  Special & General Relativity  2  
An Introduction to Black Hole Thermodynamics  Astronomy & Astrophysics  21  
Looking for conferences/workshops on Black Hole Thermodynamics / Holography  Beyond the Standard Model  1 