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  Does Bose-Einstein distribution work for micro-canonical ensemble?

+ 4 like - 0 dislike

Dear all, we are considering an isolated finite gas system consisting of boson particles. So the total energy and volume should never change. Is the Bose-Einstein distribution still working in this case? How are the chemical potential determined then? 

If we now consider the formation of black holes resulting from particle collision. Also the black holes can evaporate, feeding the same boson particles back into the system. Then the total number of particles won't be conserved. Does this mean we can take the chemical potential to be zero? Here, we have assumed equilibrium is always able to be maintained for the gas.

asked Mar 15, 2016 in Theoretical Physics by ruifeng14 (65 points) [ no revision ]
recategorized Mar 16, 2016 by dimension10

In equilibrium and in the thermodynamic limit all ensembles are quivalent. This applies in a good approximation whenever the system is large enough and there is no macroscopic change. The te Bose-Einstein formulas remain valid.

Formation of a black hole is a dynamical process, and equilibrium considerations do not apply, except locally. But in local equilibrium, there is an exchange of everything, and the grand canonical ensemble must be used.

@ArnoldNeumaier Thank you for your reply. If grand ensemble is used for the gas, could the total energy (gas + black hole) still be conserved? In other words, could the black  holes be the only energy and particle source for the gas system? 

For a macroscopic system, only the total mean energy makes sense. It is conserved if the system is isolated. Thus if you assume that the system of gas + black hole is isolated and that it lives in an asymptotically flat universe (so that total energy is well-defined), the answer should be yes. ( I say 'should' since I am not an expert in general relativity and therefore not completely sure.)

2 Answers

+ 4 like - 0 dislike

Yes, as noted by Arnold Neumaier, all ensembles are equivalent in the thermodynamical limit and the Bose-Einstein distribution applies in all of them. I would just like to anticipate what you are thinking about and point out a few facts

  1. What most people know about the behaviour of a Bose-Einstein gas applies to an ideal, collisionless gas. Once you introduce collisions, there is no thing such as a one-particle ground state to fall to and the argumentation becomes much more difficult.
  2. Finite size of the system (finite volume, number of particles) introduces, at least in an ideal gas of free particles, a finite gap above the ground state but smoothens out phase transitions.
  3. If you want to create black holes by collisions of microscopic (quantum) particles, you probably need $k_B T \approx E_{planck}$. In most thinkable situations, both Bose-Einstein and Fermi-Dirac are indiscernible from the Maxwell distribution at these temperatures.

Anyways, the second part of the question asks about black holes created by collisions. This is hard to comment on, because we do not know how would that actually happen because we have no quantum theory of gravity.

The general thing to remember about gravity is that it has many "antithermodynamic" properties. For instance, increasing the kinetic energy of objects in a gravitating system decreases it's temperature. The question of thermodynamic equilibrium of a general gravitating system is unresolved. Etc. etc. Thus, what you might find is that the black hole creation by collisions might be a runaway process, because every collision with a black hole creates just a larger black hole and larger black holes Hawking-radiate less. I.e., for a certain choice of parameters, there might be no thermodynamic equilibrium, just collapse into a large black hole (which is kinda sorta an equilibrium).

However, you might have some luck with special choices of parameters, the radiation might be strong enough for the black holes to evaporate almost instantaneously (before colliding with more particles) and then the problem is basically the computation of a grandcanonical equilibrium with the chemical potential being set to the energy required for the black hole creation. Any point in space corresponds to the reservoir where the particles can leave at the cost of the black hole creation energy and the evaporation is the spontaneous route back. Of course, this idealization will make sense only if the chemical potential of the black hole creation is lower than the chemical potential of the gas without black hole creation. (Otherwise the mean black hole fraction just tends to zero in the thermodynamical limit.)

answered Mar 16, 2016 by Void (1,645 points) [ no revision ]

Instead of "anti-thermodynamic" properties I'd only say "anti-equilibrium" properties. The reason is that gravitation is very well compatible with hydromechnaics including all its thermodynamic aspects.

@Void Thank you for your comprehensive answer.

+ 3 like - 0 dislike

A bit late for an answer, but I just wanted to clarify a point: The ideal Bose gas is rather peculiar in that the standard ensembles (micro-canonical, canonical and grand canonical) are NOT equivalent. While it is safe to use the micro-canonical or canonical ensemble, the grand canonical ensemble can and should not be used to discuss the condensed phase. The reason is that there is zero energy cost for adding even a macroscopic number of particles to the ground state which results in macroscopic fluctuations of particles in the condensate (so, in a sense, there is no sensible thermodynamic limit for this ensemble). Details can be found in the link
below (Physics Report paper from 1977)

The ideal Bose-Einstein gas, revisited; Ziff, Uhlenbeck, and Zac


A citation from their abstract:

"Particular attention is focused on the difference between the canonical and grand canonical ensembles and a case is made that the latter does not represent any physical system in the condensed region."

answered Nov 17, 2017 by Nils Hasselmann [ no revision ]

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