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What's Wrong With Black Hole Thermodynamics?

by juanrga
Tags: black hole, thermodynamics
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twofish-quant
#19
Feb18-12, 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 self-gravitating 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).
PAllen
#20
Feb18-12, 01:58 AM
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Quote Quote by twofish-quant View Post
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 self-gravitating 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).
Right, I completely agree.
juanrga
#21
Feb18-12, 06:32 AM
P: 476
Quote Quote by twofish-quant View Post
Also, a non-radiating star or black hole is a closed system.
Reading again stuff that was corrected before is one thing, but your claim that a black hole is a closed system is too nonsensical to continue.
juanrga
#22
Feb18-12, 06:56 AM
P: 476
Quote Quote by PAllen View Post
Of interest to discussion of entropy of self gravitating systems is the following essay. Of particular interest is the discussion of gravothermal catastrophe - that self gravitating systems really have no equilibrium point short of a black hole.

This essay is not relevant to the core question of the validity of Bekenstein-Hawking entropy. It takes that as a given, and argues that this is exceptional and not some natural limit of ordinary gravitational collapse processes.

http://philsci-archive.pitt.edu/4744...nt_archive.pdf
The entropy of gas ideal is function of its internal energy Eint.

Its equation 5 reads E=K+U, but the thermodynamic expression in presence of fields is E=U+K+Eint.
juanrga
#23
Feb18-12, 07:11 AM
P: 476
Quote Quote by twofish-quant View Post
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.

[...]
The problems with your "thought experiment" already start here. What energy? The total energy is E=K+V+U, where K is kinetic energy, V is potential energy due to gravitational field and U is internal energy.

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).
PAllen
#24
Feb18-12, 09:34 AM
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Quote Quote by juanrga View Post
The entropy of gas ideal is function of its internal energy Eint.

Its equation 5 reads E=K+U, but the thermodynamic expression in presence of fields is E=U+K+Eint.
In the context of discussion in that section, internal energy is held constant and can be ignored. For that section the author is explicitly treating the gas particles has incapable of changes of state. Later sections broaden that to discuss fusion and other process that do change internal energy.

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,"
juanrga
#25
Feb18-12, 01:31 PM
P: 476
Quote Quote by PAllen View Post
In the context of discussion in that section, internal energy is held constant and can be ignored. For that section the author is explicitly treating the gas particles has incapable of changes of state. Later sections broaden that to discuss fusion and other process that do change internal energy.

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
systemís current temperatures,"
The internal degrees of freedom he is referring are essentially atomic/molecular variables (an example is molecular dipole orientation). This has nothing to see with the concept of internal energy Eint. The internal energy of an ideal gas is computed assuming that atoms/molecules do not have internal degrees of freedom.

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).
twofish-quant
#26
Feb18-12, 11:38 PM
P: 6,863
Quote Quote by juanrga View Post
Reading again stuff that was corrected before is one thing, but your claim that a black hole is a closed system is too nonsensical to continue.
Why? You can treat stars and black holes are something very close to closed thermodynamic systems. The only thing that makes them non-closed is radiation and 1) you can include radiative transfer in the book keeping, and 2) it doesn't make a huge difference. The total energy that is radiated at a given point is much, much smaller than the total internal energy.

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 self-gravitating objects" it's not going to end well nor should it.

And yes this works both ways..... If a condensed matter physicist nicely e-mailed 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.
twofish-quant
#27
Feb18-12, 11:43 PM
P: 6,863
Quote Quote by juanrga View Post
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).
It has *everything* to do with thermodynamics and negative heat capacities. Once you have enough satellites, it becomes a gas, and once you have a gas, you can do thermodynamics on it. The system of earth and satellites is a closed thermodynamic system and you can define temperatures and heat capacities.

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.
juanrga
#28
Feb19-12, 08:35 AM
P: 476
Quote Quote by twofish-quant View Post
It has *everything* to do with thermodynamics and negative heat capacities.
If internal energy U of the satellite is held constant and only the mechanical energy (K+V) is varied, then you do not need thermodynamics, but only mechanics. Indeed, the laws of mechanics alone explains, perfectly, why the satteliite modifies its orbit and falles when you take away kinetic energy.

It is evident that you do not need thermodynamics for this. Gibbs and heat capacities are not even mentioned.

Quote Quote by twofish-quant View Post
Once you have enough satellites, it becomes a gas, and once you have a gas, you can do thermodynamics on it. The system of earth and satellites is a closed thermodynamic system and you can define temperatures and heat capacities.
Taking an enough collection of satellites does not magically convert a mechanical problem into a thermodynamic one. Moreover, even assuming that the system of satellites can be considered a thermodynamic system, still there is none need to confound internal energy U with total energy E (including potential energy due to Earth) as done in previous posts. Similar remarks about stars.

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 well-know, 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.
twofish-quant
#29
Feb19-12, 09:50 AM
P: 6,863
Quote Quote by juanrga View Post
If internal energy U of the satellite is held constant and only the mechanical energy (K+V) is varied, then you do not need thermodynamics, but only mechanics. Indeed, the laws of mechanics alone explains, perfectly, why the satteliite modifies its orbit and falles when you take away kinetic energy.
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.

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 self-gravitating objects is that you *can't* separate mechanics and thermodynamics. Every thermodynamic equation in astrophysics has to include mechanical interactions.

Taking an enough collection of satellites does not magically convert a mechanical problem into a thermodynamic one.
Yes it does. You change those satellites into atoms and then have several gazillion of them. At which point you have a gas.

Moreover, even assuming that the system of satellites can be considered a thermodynamic system, still there is none need to confound internal energy U with total energy E (including potential energy due to Earth) as done in previous posts. Similar remarks about stars.
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.)

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.

[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]

the heat capacity is the same, as is well-known.
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.

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.



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.
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.

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.
I don't get the problem with considering a static black hole (or star) a closed thermodynamic system. It's just sitting there in empty space. The only thermodynamic interaction it has with the rest of the universe is radiation and you can add that to the bookkeeping.

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 self-gravitating 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.
twofish-quant
#30
Feb19-12, 10:23 AM
P: 6,863
To summarize the special thermodynamic properties of astrophysical self-gravitating 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.
juanrga
#31
Feb19-12, 04:39 PM
P: 476
Quote Quote by twofish-quant View Post
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.

Quote Quote by twofish-quant View Post
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, [itex]\left\langle T \right\rangle[/itex], of a stable system consisting of N particles, bound by potential forces, with that of the total potential energy, [itex]\left\langle V_\text{TOT} \right\rangle[/itex], 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.

Quote Quote by twofish-quant View Post
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 .

Quote Quote by twofish-quant View Post
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.

Quote Quote by twofish-quant View Post
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, [itex]\tau_k[/itex] is the coupling constant, not the gravitational potential.

Second, as is well-known, 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.

Quote Quote by twofish-quant View Post
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 [itex]\pi:\nabla v[/itex], where [itex]\pi[/itex], is the dissipative pressure tensor. q and [itex]\pi[/itex] are non-hydrodynamical quantity, unless you want to reinvent hydrodynamics also .

Quote Quote by twofish-quant View Post
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
#32
Feb19-12, 10:25 PM
P: 6,863
Quote Quote by juanrga View Post
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 [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.
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
#33
Feb19-12, 11:02 PM
P: 6,863
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.
juanrga
#35
Feb20-12, 12:16 PM
P: 476
Quote Quote by twofish-quant View Post
Except that the kinetic energy in this situation is microscopic.
You understood nothing.
juanrga
#36
Feb20-12, 12:36 PM
P: 476
Quote Quote by twofish-quant View Post
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.


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