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 $f^{[N]}(\textbf{r}^N, \textbf{p}^N; t)$"


"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.


1 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.

  • $\begingroup$ Then is it correct to say that in thermodynamic equilibrium $f^[N]$ is the probability of a system being in a particular state (which is in equilibrium)? And if so, What book would you use to validate this? $\endgroup$
    – felipe
    Commented Dec 14, 2022 at 0:01
  • $\begingroup$ @felipe I wouldn't phrase it like that. I would say the equilibrium distribution $f^{[N]}(\mathbf{r}^N, \mathbf{p}^N)$ is a time-independent probability distribution over the system's microstates, which are labeled by $\{\mathbf{r}^N, \mathbf{p}^N\}$. $\endgroup$
    – Andrew
    Commented Dec 14, 2022 at 1:50
  • $\begingroup$ The phrase "the state of the system" is ambiguous, since it could refer to the microstate of the system, or the macrostate of the system, and possibly other meanings. Without other context, I would tend to think "the state of the system" means a macrostate, so I would make statements like "the state of the system is in equilibrium if its distribution of microstates (given by $f^{[N]})$ is an equilibrium distribution." But again, the phrase "state of the system" can be ambiguous, so best to be more precise if you can. $\endgroup$
    – Andrew
    Commented Dec 14, 2022 at 1:56

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