What does equilibrium probability density mean in statistical mechanics?

Question

I know that according to the book Theory of Simple Liquids by Hansen & McDonald:

"The distribution of phase points of systems of the ensemble is described by a phase space probability density $f^{[N]}(\textbf{r}^N, \textbf{p}^N; t)$"

And,

"The quantity $f^{[N]}d\textbf{r}^N d\textbf{p}^N$ is the probability that at time $t$ the physical system is in a microscopic state represented by a phase point lying in the infinitesimal, 6N -dimensional phasespace element $d\textbf{r}^N d\textbf{p}^N$"

But for a system in thermodynamical equilibrium the book says:

"The phase space probability density of a system in thermodynamic equilibrium is a function of the time-varying coordinates and momenta, but is independent of t at each point in phase space. We shall use the symbol $f_0^{[N]}(\textbf{r}^N, \textbf{p}^N)$ to denote the equilibrium probability density."

My question is: What does this equilibrium probability density $f_0^{[N]}$ mean? I'm guessing it is a measure of the probability of a system being in a particular state at equilibrium. I'm trying to find a source I can cite, preferably a book, that proves this statement.

Answer 1

$f^{[N]}(\mathbf{r}^N, \mathbf{p}^N; t)$ assigns a probability (technically a probability density) to each microstate. A microstate is defined by a list of the positions and momenta of each of the $N$ particles of the system. This distribution, in general, depends on time.

When the distribution does not depend on time, then it is a stationary distribution. This may also be an equilibrium distribution in the sense that the system will approach $f^{[N]}$ at large enough times.

In thermal equilbrium, you should find $f^{[N]}(\mathbf{r}^N, \mathbf{p}^N) \propto e^{-E(\mathbf{r}^N, \mathbf{p}^N)/kT}$, ie, the Boltzmann distribution.