I got a question I'm not sure how to state precisely or is it even valid. Any help is most welcomed.

I stripped the question of all details because I wanted to emphasize my problem, but should someone think they would bring any clarity (it is a solid state problem) I'll present them.

Ok, let say I have two interacting systems. One of them is a system (S1) in a thermodynamical equilibrium and the other is a well defined classical system (S2). I know how to derive S2 from microcanonical state of S1 and how surrounding of a S1 depends on S2. It is very unclear how to combine these two mathematically but here is a kick - I THINK that S2 is changing much more slowly than S1. So, I was thinking of a iterative approach: to run a Monte Carlo to solve S1, then derive S2, then adjust conditions of S1 based on the new state of S2 and rerun MC etc. So my questions would be: is this approach valid if I assume that S1 is changing adiabatically? Is there a practical way to verify adiabatic change? Is there any circumstance where calculation like this is valid? It feels that if the S2 can't kick S1 out of equilibrium, then I got a powerful edge to clear this problem up - but is this true?


1 Answer 1


I'm not exactly an expert on this, but here goes.

When you say "One of them is a system (S1) in a thermodynamical equilibrium", you are already assuming that the equilibration time of S1 is much faster than the time required for S2 to change significantly. I don't know what this assumption is based on, but the equilibration time of S1 (by itself) can certainly be found out. For example, if S1 is the gas in a room and S2 is a door opening, then how long does it take for the gas to respond to the motion of the door? Well, at the very least, it takes as long as the propagation time of a pressure wave travelig across the room. That's just a lower bound of course. In a simulation, you can try, for example, moving all the gas particles out of some area in the simulation, and see how long it takes for the system to bring the local density back up to normal.

If it is true that S1 is in thermodynamic equilibrium with the instantaneous configuration of S2, then it is possible (as you propose) to do a time-evolution of S2, where in each time-step, S1 is in a random configuration drawn from its equilibrium distribution. To get the right answer, you would need to make sure that in the time required for S2 to do anything significant, you have gone through so many time-steps that S1 has thoroughly sampled its ensemble of configurations. I do not believe this is a wise approach. Instead, I think it's better to calculate the free energy of S1 as a function of the configuration of S2. Take the derivative of that function in order to model the effects of S1 as a deterministic classical force acting on S2.

  • $\begingroup$ thank you for your answer. I'm still thinking my strategy on this, so your input is welcomed. Would you care to elaborate why don't you think my original approach is wise? Thanks!! $\endgroup$ Apr 21, 2012 at 12:35
  • 1
    $\begingroup$ Calculating the free energy of a system is a very common and well-understood thing which can be done efficiently and accurately with procedures described in textbooks. In the original approach, you are calculating something unusual, and therefore you need to figure out for yourself how many timesteps you need, what are the possible sources of systematic error, etc. etc. It is "wise"---simpler in practice and less prone to error---to not reinvent the wheel by trying to calculate unusual things. :-) $\endgroup$ Apr 22, 2012 at 2:24
  • $\begingroup$ Thank your for your answers. I did some reading and noticed that situation that I described is analog of Fermi's golden rule: like interaction of atoms and photon gas - photons thermalise in much shorter times than what it takes to change the number of excited atoms significantly and such change can't kick photon gas out of equilibrium. It is much harder to see how to devise analog of FGR in my situation, but at least this conversation started me thinking. Thank you, sir, you deserve this bounty. $\endgroup$ Apr 22, 2012 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.