# Entropy increase in gravitational collapse.

I want to hear your opinion on this:
Let's say that the universe in time zero consists just of a cloud of matter. Now as the time progresses, and the matter interacts gravitationally, it will gradually collapse into a sphere. Will the entropy of the universe really increase?

Mapes
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The total entropy in the universe tends to increase for every spontaneous process. http://math.ucr.edu/home/baez/entropy.html" [Broken] might interest you.

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atyy
http://math.ucr.edu/home/baez/entropy.html" [Broken] might interest you.

I always thought site that was rather diabolical! Last edited by a moderator:
Mapes
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Thanks for the link, it seems very interesting. So where does the entropy actually go? Since the system is closed, my only tip would be that it increases the "entropy of the gravitational field", but I don't really know what to understand under that.

Gravitational collapse increases spatial order at the expense of increasing velocity disorder. Then the process waits, until the system manages to offload that velocity disorder (which it will do by increasing the temperature of the surrounding universe).

Andy Resnick
The total entropy in the universe tends to increase for every spontaneous process. http://math.ucr.edu/home/baez/entropy.html" [Broken] might interest you.

That's not exactly correct; the change in entropy is positive for all irreversible processes. The change in free energy is negative for spontaneous processes. Also, it is not clear how to define an isolated system in the OP- is the entire universe an isolated system? What are the boundary conditions? Has any work been performed during the process of collapse?

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If we consider the system to be a closed system then the boundary conditions are adiabatic boundary counditions and no work is performed/extracted.

The system is thus kept at some constant energy. Then the total entropy (defined in a suitable way, like minus the sum over p_i Log(P_i) ), can only increase.

Free energy does not apply here, because the system is not kept at constant temperature. Even if the system is approximately at consant temperature, one has to be careful with applying the usual results of thermodynamics, because of the hidden assumptions like there being only short range interactions. That's clearly not the case if gravity is relevant.

So, due to the long range gravitational interaction, the entropy and internal energy won't be the usual extensive functions and thus the standard results like E = T S - P V + mu N are not going to be valid anymore.

Gravitational collapse increases spatial order at the expense of increasing velocity disorder. Then the process waits, until the system manages to offload that velocity disorder (which it will do by increasing the temperature of the surrounding universe).

But this is exactly what Baez calculates in his article - and he shows that even if we take into acount the entropy of the velocity distribution, it will still give us smaller entropy than at the beginning. So where is the remaining entropy?

But [Baez] shows that even if we take into acount the entropy of the velocity distribution, it will still give us smaller entropy than at the beginning. So where is the remaining entropy?
Read the whole article. Distinguish between the initial phase of collapse (that occurs even if the system is isolated) and its continuation (in which the total energy of the system is allowed to decrease). Note Baez is focusing on the latter (by beginning with the virial theorem).

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You may remember from astrophysics class that (loosely speaking) gas clouds can only collapse if they are transparent to their own thermal radiation.

Mapes
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The total entropy in the universe tends to increase for every spontaneous process.

That's not exactly correct; the change in entropy is positive for all irreversible processes. The change in free energy is negative for spontaneous processes.

Minimization of free energy for isothermal systems is equivalent to maximization of entropy for constant-energy systems. (See Callen's Thermodynamics on the extremum principle, for example.) As Count Iblis pointed out, surely it makes more sense to model the universe to have constant energy as opposed to being isothermal.

Andy Resnick
If we consider the system to be a closed system then the boundary conditions are adiabatic boundary counditions and no work is performed/extracted.

The system is thus kept at some constant energy. Then the total entropy (defined in a suitable way, like minus the sum over p_i Log(P_i) ), can only increase.

<snip>

Minimization of free energy for isothermal systems is equivalent to maximization of entropy for constant-energy systems. (See Callen's Thermodynamics on the extremum principle, for example.) As Count Iblis pointed out, surely it makes more sense to model the universe to have constant energy as opposed to being isothermal.

I agree that for a static universe consisting only of a uniform subvolume of dust the results are reasonably well-understood. But reconciling thermodynamics with GR is not as straightforward. Assigning a temperature can be ambiguous, for example: Unruh radiation (http://en.wikipedia.org/wiki/Unruh_radiation). And I don't think we can simply demand that the universe has an adiabatic boundary, see for example black hole thermodynamics (http://en.wikipedia.org/wiki/Black_hole_thermodynamics).

One possible reason for the difficult nature of this problem is that the initial conditions are (perhaps) ill-posed- we did not completely specify the initial state. How can a closed universe filled uniformly with dust come to be? What is the initial total energy, momentum, angular momentum? Can they be unambiguously assigned in a way consistent with GR?

I don't know the answers to these questions- it's way outside my expertise. Tolman's book "Relativity, Thermodynamics and Cosmology" has a chapter on this subject- this could be a good excuse for me to read it.

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Mapes