Definition of the number of microstates accessible to a system

[Sam_Goldberg]

Hi guys, I'm going through Reif's book on statistical mechanics and have a question on the parameter omega, the number of microstates accessible to a system. Let's imagine a big box, and divide it into two equal halves. Now place a gas with a macroscopic amount of molecules in the left half, and say that it has an energy between E and E + dE. There is an initial value here of omega characterized by the energy and the volume.

Now remove the division in the box. Reif would say that since the volume is now increased, the number of accessible states of the system instantly increases; however, the probability of the system of being in most of them is zero, at least instantly after the division is removed (after an amount of time this is not the case). He says: since the probabilities of being in most of the accessible states are zero, the system will tend to evolve so that the probabilities become equal (due to the fundamental postulate of statistical mechanics). But this means that when the division is removed, the inverse of the absolute temperature instantly increases, for it is the partial derivative of log(omega) with respect to energy.

Is it correct that the absolute temperature instantly increases? Here is an alternative: maybe the value omega stays the same instantly after the division is removed; it then increases more or less continuously, approaching a maximum, final, value. Then, since the definition of temperature uses a partial derivative, the value would not change instantly after the division is removed. Rather, it would just be the same.

Which view of the amount of microstates is correct? I want to be sure I understand this book and the author's message. Thanks.

 

[LeonhardEuler]

What is important to keep in mind is that when one talks about a number of accessible states, one is always speaking about a specific ensemble of states. In the most straightfoward way of thinking about the situation you describe the number of accessible states instantaneously increases because you instantly change the ensemble you are talking about. What happens immediately after the partition is physically removed is not part of the subject of equilibrium statistical mechanics because the system moves into a non-equilibrium state and does not sample from the equilibrium distribution of states.

You could analyze the situation in different ways and come to different conclusions. You could consider the ensemble of states to include the entire container from the beginning, for instance. In that case, when the gas is confined to half the container by the partition, it is in a non-equilibrium state of this ensemble because it is kinetically trapped by the barrier from reaching certain states. This would not be a particularly useful way of analyzing the system, but my point is that in statistical mechanics, one has to keep in mind the ensemble one is talking about.

As far as the inverse of the temperature goes, I think you're making a mistake. For each particle the number of positions it can be in doubles. If there are N particles, then the number of accessible states increases by a factor of $$2^N$$. Thus
$$\Omega_{new} = 2^N\Omega_{old}$$
$$\rightarrow S_{new} = k\ln(\Omega_{new}) = k\ln(\Omega_{old})+k\ln{2^N} = S_{old} + k\ln(2^N)$$
$$\rightarrow \frac{1}{T_{new}} = \frac{dS_{new}}{dE}=\frac{dS_{old}}{dE}=\frac{1}{T_{old}}$$

If there was a similar change that did bring about a temperature change by passing through non-equilibrium states, then whether there is a gradual change in temperature or not depends on what definition of temperature you use. If you use a definition from the kinetic theory of gases that relies only on the velocities of the particles, then it will probably change gradually. If you use a statistical mechanical definition that depends on a derivative of the entropy, then it doesn't make sense to talk about temperature outside of equilibrium in the first place.

 

[Sam_Goldberg]

Euler, that makes perfect sense. Thanks very much for your help.