This question came to me after a discussion on another post on the site, which you can find here.

I will restrict the following discussion to Hamiltonian systems. Also, I will use the abbreviated notation $\partial_t$ for the partial derivative with respect to time, $\partial/\partial t$.

Let $\rho(p,q,t)$ be the ensemble probability density; this function satisfies in general Liouville's equation:

$$\frac{d\rho}{dt} =\partial_t \rho + \{\rho,H\}=0 \tag{1}\label{1}$$

where $H$ is the Hamiltonian and $\{\cdot\}$ are Poisson brakets.

We can get the average of some thermodynamic quantity $a(p,q)$ by computing

$$A(t) = \langle a(p,q)\rangle = \int \rho(p,q,t) a(p,q) dp dq$$

We know that a necessary condition for thermodynamic equilibrium is that the average of every thermodynamic quantity is independent of time, i.e. $A(t) = A(t') \ \forall t,t'$. A necessary condition to achieve this is that $\rho$ has no explicit time dependence, i.e.

$$\text{Equilibrium} \Rightarrow \ \partial_t \rho = 0$$

In this case, it follows from Eq. \ref{1} that $\{\rho,H\}=0$, and therefore $\rho$ is a function of the Hamiltonian: $\rho=\rho(H)$.

My question is: is this also a sufficient condition? In other words, is

$$\text{Equilibrium} \iff \ \partial_t \rho = 0$$

true? If not, can we give some examples of systems where $\partial_t\rho=0$ which are not at thermodynamic equilibrium, showing explicitly that there is no thermodynamic equilibrium?

An attempt to clarify

What I am thinking about is some kind of metastable state with infinite lifetime, in such a way that the ensemble is "stuck" in this non-equilibrium state ($\partial_t \rho =0$) for an infinite time.

This probably has to do with non-ergodicity; however, in most physical system non-ergodic behavior shows up only for a set of initial conditions which has measure zero. Since $\rho$ is a probability density over an ensemble, i.e. over an infinite number of imaginary copies of the system, even if a discrete subset of the ensemble has a pathological non-ergodic behavior, the whole ensemble will behave just fine and reach equilibrium.

So I guess that in some sense what I am asking is whether there are systems where non-ergodic behavior shows up for a non-zero measure set of initial conditions.

  • $\begingroup$ To clarify, are you assuming $\frac{\mathrm{d}\rho}{\mathrm{d}t} = 0$ throughout? i.e. your question is not about achieving a steady state but about whether that steady state, combined with a lack of explicit time dependence, is enough for thermodynamic equilibrium? $\endgroup$ Jan 30 '18 at 18:14
  • 1
    $\begingroup$ @BySymmetry I am assuming Liouville's equation, so $d\rho/dt=0$. This is valid for systems both at and out of equilibrium. Then, I want to know if $\partial_t \rho$ (no explicit time dependence in $\rho$), which is for sure a necessary condition for equilibrium, is also a sufficient condition. $\endgroup$
    – valerio
    Jan 30 '18 at 18:48
  • $\begingroup$ If $\rho$ is independent of time, your definition of $A(t)$ is clearly independent of time. $\endgroup$
    – Martino
    Jan 31 '18 at 11:54
  • $\begingroup$ @Bzazz Sure. Therefore, as I wrote, it's clear that $\partial_t \rho$ is a necessary condition for equilibrium. My question is, is it also a sufficient condition? $\endgroup$
    – valerio
    Jan 31 '18 at 11:58
  • $\begingroup$ I thought you defined thermodynamic equilibrium as all quantities have averages independent of time, but now I see you call it a necessary condition. Then what's your definition of thermodynamic equilibrium? $\endgroup$
    – Martino
    Jan 31 '18 at 12:01

The short answer to your question is 'no'. There are steady-states that are not thermal.

Before I start, let me clarify: For me a thermal state is, by definition, at state with a density matrix that corresponds to one of the canonical ensembles of statistical physics. For example,

$$ \rho(p,q) = \frac{e^{-\beta H(p,q)}}{\int_{p,q} e^{-\beta H(p,q)}} \, $$

is thermal. See my answer to this question for more on this definition. Then it is easy to construct a steady state that is not thermal. Pick $\rho$ as any other function of $H(p,q)$. This definition is rooted in the idea that the system evolves towards a state where the entropy of the system is maximised (modulo constraints related to conserved quantities).

But, I guess that you really want to know is: 'Are there systems that do not evolve towards a thermal state at long times?'. There, the answer is 'yes'. As you suggest though, these systems are not the most common. I give a few examples:

  • Integrable systems are (without going into details) systems with as many conservation laws as degrees of freedom. These conservation laws prevent the system from evolving to a traditional thermal state. In a 'normal' system, the thermal state is characterised by it's conserved quantities. In the microcanonical ensemble, the thermodynamic coordinates are the total energy, the number of particles, the volume, etc. If there is an infinite (in the thermodynamic limit) number of conserved quantities, then an infinite number of parameters are necessary to describe the equilibrium state. The system remembers it's initial conditions. A 'dumb' example of an integrable system is the ideal gas, $$ H = \sum_i \frac{p_i^2}{2m} \, .$$ For this system, the momenta of each particle is conserved independently (instead of $P = \sum_i p_i$ for an interacting system). Integrable systems are often thought as mathematical exceptions, but they are actually quite common in $1d$ and have been realised experimentally. The go-to theoretical tool here is called generalised Gibbs ensemble.

  • Disordered systems are systems where the microscopic physics appears random. Think of a collection of particles moving around in an erratic potential landscape. There are many local minima of the potential energy and particles can get stuck there for very long times. The 'canonical' example of such systems are spin glasses. These are Ising models with random parameters. You can also look into the phenomena of frustration. The point here is that there is a huge number of local minima of the energy of the system. The dynamics is super-slow because it can easily get stuck for long times in a local (but not global) minimum of the energy. Such relaxation dynamics is some times called ageing. It is still an open question (for theory as well as for experiments) to determine if these system thermalise at long times and how. A super cool theoretical tool here is the replica trick.

  • Everybody knows about closed systems. There energy, particle number, etc. are conserved. The dynamics is described by a Hamiltonian and Liouville theorem applies. Although the phase space typically gets completely jumbled up, the initial conditions are never really forgotten. After all Hamilton's equations are reversible. Then thermalisation is a complicated process. Actually, it never fully happens for finite systems. When there are a finite number of degrees of freedom, Poincaré's recurrence theorem applies and the system eventually comes back very close to its initial conditions. This implies that the limits of infinite time (since the initial conditions) and infinite number of particles do not commute. In thermodynamic ($=$ infinite number of degrees of freedom) systems, thermalisation only sets in for 'normal' observables. If you look at the the phase space density or the the kinetic energy of the system, everything will look thermal pretty quickly. However, strongly non-local observables or high order correlation functions will retain a memory of the initial conditions for a longer time. What happens is that the information on the initial conditions gets mixed up as time goes on and increasingly strange observables are necessary to pull them up. But, strictly speaking, this information is always there and thermal equilibrium is only asymptotically reached.

  • The above example is a bit academic. You can however look into non-thermal fixed point dynamics. The idea here, is to prepare a system really far from equilibrium (but without fine-tuning any parameter). It is a numerical observation that instead of directly thermalising, the dynamics gets caught in a quasi-stationary non-thermal state. What happens is that the system needs to move energy and particles around (because of the initial conditions) in order to thermalise. Since these are conserved quantities, large fluxes appear. It is like a turbulent state. However the turbulence is not maintained by sink and sources of energy (as usual) but the system acts as it's own reservoir and sink. If the system is big enough, this can go on for a very long time. A 'dumb' analogy would be a bathtub with a hole in the bottom. It is not at equilibrium until all the water has flowed out. If you take the thermodynamic limit however, then it takes an infinite amount of time for this to happen and equilibrium is never reached.

  • $\begingroup$ @MassimoOrtolano Thanks for the comment. I rephrased slightly. Do you agree with my statement now? $\endgroup$ Feb 1 '18 at 21:11

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