# Different definitions of equilibrium in statistical mechanics

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?

• 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). Commented Jun 21, 2023 at 23:25

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: $$\rho(I,\phi)=\delta(I-I_0)$$