Boltzmann entropy, what is $\Omega$, in $S=k_B \ln( \Omega )$?

Question

The Boltzmann's entropy formula $S=k_b \ln \Omega$ is valid both for equilibrium and non-equilibrium systems. On my textbook, it is written that $\Omega$ is the number of accessible microstates of the system. So, i guess that, considering an isolated system, we can write:

• $S_{non-eq}=k_b \ln\Omega_{non-eq}$ entropy of the system, when the system is in a macroscopic state of non-equilibrium.
• $S_{eq}=k_b \ln\Omega_{eq}$ entropy of the system, when the system is at equilibrium.

Now, since $\Omega$ represents the number of accessible microstates of the system, $\Omega(E,N,V)$. And, since for an isolated system $N$, $V$, and $E$ are constant, $\Omega_{non-eq}=\Omega_{eq}$. This imply $S_{non-eq}=S_{eq}$, that, obviously, is wrong.

How it's possible? Maybe the meaning of $\Omega$ is not "the number of accessible microstates", but something else, like "the number of accessible microstates given certain conditions", suggestions?

Answer 1

Usually we say that $\Omega$ is the number of microstates compatible with $E$ which is not really true. In fact, $\Omega$ is the number of microstates (or the volume of phase space) compatible with the macrostate your system is in. Usually, the equilibrium macrostate has a way greater volume (in the phase space, so has more microstates associated to it) than all the other macrostates (the non equilibrium ones). This is the second law.

So effectively, we can just calculate the number of microstates compatible with E. This gives the volume of the sum of all macrostates in phase space. And it will approximately give the volume of the equilibrium macrostate since all other contributions will be negligible.

This is why we can calculate the thermodynamical entropy from statistical mechanics, because of the size of the equilibrium macrostate in the phase space.

Back to your question: $\Omega_{eq}>>>>>\Omega_{non-eq}$. As to whether this definition of entropy is useful for closed out of equilibrium system or not is an other question.

Entropy is a tricky thing and most textbook (in my opinion) just repeat each other and the same misconceptions (often found in this website also..). Here is a good starting point by S. GOLDSTEIN

Answer 2

In the Boltzmann definition $S = k_B\ln \Omega$, $\Omega$ is not necessarily the number of accessible microstates, but it is the number of microstates compatible with the macrostate $\mathbf X$ (e.g. in equilibrium state, the triplet $E,V,N$, or in non-equilibrium state, $V$ and the density fields $e,n$). Only in equilibrium there is sometimes an expectation that all or most compatible microstates are also accessible. But this is not an important question. There may be microstates that are compatible but not accessible, and as long as there is not too many of them, macroscopic entropy is the same.

The difference manifests more in non-equilibrium macrostates $\mathbf X_{noneq}$, where a microstate that is compatible with future equilibrium macrostate $\mathbf X_{eq,2}$ is accessible, but it is not compatible with the macrostate $\mathbf X_{noneq}$. Such a microstate is not a part of $\Omega(\mathbf X_{noneq})$.

System won't get into non-equilibrium state from an equilibrium state on its own. There has to be an intervention, and this intervention changes the macroscopic constraints such as $E$ or $V$.