Statistical mechanics tries to predict macroscopic behavior of a thermodynamic system from microscopic considerations. Suppose we are dealing with system with phase space $\Gamma = (\Lambda \times \mathbb{R}^{d})^{N}$ where $N \ge 1$ is large and $\Lambda \subset \mathbb{R}^{d}$. Let me write $x = (q,p) \in \Gamma$ for a point in the phase space.

Boltzmann argued that when we measure an observable $f$ (continuous function on $\Gamma$), the measurament takes some time. Hence, it is natural do expect that the equilibrium value of $f$ is defined by the limit: $$\bar{f}(x) := \lim_{T\to \infty}\frac{1}{T}\int_{0}^{T}f(x_{t})dt \tag{1}\label{1}$$ where $x$ is the state of the system at an initial time $t=0$ and and $x_{t}$ is the state of the system at a posterior time $t$. Physically, we would expect that this measurement does not depend on the initial state of the system, so it becomes a constant, say, $\bar{f}$. This is the ergodic hypothesis and it implies that this time average $\bar{f}$ coincides with the ensemble expectation: $$\langle f\rangle = \int f(x)\rho(x)dx \tag{2}\label{2}$$ where $\rho$ is the probability density (of states) on $\Gamma$.

I am happy with the above definitions. However, my professor took a slightly different route in his classes. He starts with a probability density $\rho_{0}$ at time $t=0$ and $\rho_{t}$ is the probability density at time $t > 0$. By Liouville Theorem, one can show that $\rho_{0}(x) = \rho_{t}(x_{t})$. He then wrote that, at time $t$, the ensemble average of $f$ is given by: $$\langle f\rangle_{t} = \int f(x)\rho_{t}(x)dx, \tag{3}\label{3}$$ to which I agree. This is the tricky part: he defines the equilibrium value of $f$ to be: $$\langle f\rangle_{\text{eq}} = \lim_{T\to \infty}\frac{1}{T}\int_{0}^{T}\langle f\rangle_{t}dt \tag{4}\label{4}$$

Even when (\ref{1}) does not depend on the initial state $x$, it seems that the previous time average $\bar{f}=\langle f\rangle$ and the equilibrium value $\langle f\rangle_{\text{eq}}$ are not equivalent definitions. So, I wonder: what is the correct definition of the equlibrium state of $f$, and how to reconcile these two approaches?

  • $\begingroup$ If I were you I would ask the professor first, then I would argue for or against the answer I'd receive (I am assuming this is not a recorded class). $\endgroup$
    – Themis
    Commented Jun 21, 2023 at 23:25

1 Answer 1


Your professor’s approach is a generalized version of yours. You can recover your definition when $\rho_0$ is a Dirac delta centered on your initial condition.

The corresponding $\rho_{eq}=\lim_{t\to\infty}\rho_t$, defined by the weak limit, will thus correspond to your ensemble in the case of an ergodic system: $$ \rho_{eq}=\rho $$ If your system is not ergodic, then your $\rho$ will depend on the initial condition and similarly $\rho_{eq}$ will depend on $\rho_0$.

For a non ergodic case, take for example the harmonic oscillator: $$ H=\frac{p^2+q^2}{2} $$ Going to action angle coordinates: $$ q=\sqrt{2I}\cos\phi\\ p=\sqrt{2I}\sin\phi $$ This gives: $$ \rho_{eq}(I,\phi)=\int \rho_0(I,\phi’)d\phi’ $$ ie the density is averaged over the energy isolines.

Similarly, your $\rho$ will now depend on the initial energy, i.e. initial $I$. Your $\rho$ will be supported on the initial energy isoline: $$ \rho(I,\phi)=\delta(I-I_0) $$

Hope this helps.


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