- #1
nonequilibrium
- 1,439
- 2
With an extensive system, I mean a system where energy is an extensive variable. But no need to state it so generally, since I have a specific system in mind which will make sure we don't get into a battle of semantics: if you double the size of a gravitational system, the energy is not multiplied with a factor of two, not even by approximation, due to the long interaction of gravity.
Now my question is, say we have a big gravitational system that is (on some relevant time-scale) in equilibrium, can we still use the canonical ensemble for a small subset of that system (and thus regarding the rest of the system as a heat bath)?
If we look at the heat bath derivation http://en.wikipedia.org/wiki/Canonical_ensemble, then on first sight everything seems to work out, even for a gravitational system, but I think there is the implicit assumption that [itex]\Omega^*(E^*) = \Omega(E) \Omega'(E')[/itex] (with the notation there, where they use the notation ' to refer to the heat bath, * to refer to the total system, and nothing for the small subsystem) [reason: otherwise [itex]S' = k \ln \Omega '(E')[/itex] won't be true, and then beta doesn't mean the inverse temperature, which you want it to mean].
But such a product doesn't work for gravitational systems, due to lack of extensiveness. More specifically, such a notation doesn't even make sense, because to know what energy the one system has (e.g. the bath) you have to know what the position of the particles/planets in the other system (e.g. the subsystem) is, due to the long interaction, so you can't just regard both segments separately.
So would you agree that
a) the heat bath derivation is not applicable to a gravitational system in momentary/local/... equilibrium;
b) the Boltzmann factor cannot be derived for a gravitational subsystem in contact with a gravitational heat bath (= more general statement than (a))
?
Thank you.
PS: this discussion is not so much about the question if gravitational systems can be regarded as being in some (local) thermal equilibrium; although I'm interested in views on that matter, please keep them separate from my main question, as for what concerns the above post, I'm assuming some local thermal equilibrium, so don't make it about that
Now my question is, say we have a big gravitational system that is (on some relevant time-scale) in equilibrium, can we still use the canonical ensemble for a small subset of that system (and thus regarding the rest of the system as a heat bath)?
If we look at the heat bath derivation http://en.wikipedia.org/wiki/Canonical_ensemble, then on first sight everything seems to work out, even for a gravitational system, but I think there is the implicit assumption that [itex]\Omega^*(E^*) = \Omega(E) \Omega'(E')[/itex] (with the notation there, where they use the notation ' to refer to the heat bath, * to refer to the total system, and nothing for the small subsystem) [reason: otherwise [itex]S' = k \ln \Omega '(E')[/itex] won't be true, and then beta doesn't mean the inverse temperature, which you want it to mean].
But such a product doesn't work for gravitational systems, due to lack of extensiveness. More specifically, such a notation doesn't even make sense, because to know what energy the one system has (e.g. the bath) you have to know what the position of the particles/planets in the other system (e.g. the subsystem) is, due to the long interaction, so you can't just regard both segments separately.
So would you agree that
a) the heat bath derivation is not applicable to a gravitational system in momentary/local/... equilibrium;
b) the Boltzmann factor cannot be derived for a gravitational subsystem in contact with a gravitational heat bath (= more general statement than (a))
?
Thank you.
PS: this discussion is not so much about the question if gravitational systems can be regarded as being in some (local) thermal equilibrium; although I'm interested in views on that matter, please keep them separate from my main question, as for what concerns the above post, I'm assuming some local thermal equilibrium, so don't make it about that