Extending the ergodic theorem to non-equilibrium systems

Question

I try to make this as short and concise as possible. For equilibrium systems in statistical mechanics, we have the Liouville's theorem which says that the volume in phase space is conserved when the system evolves in time. Thus formally, one could rephrase this as: time evolution of equilibrium/stationary systems is a measure invariant transformation (where the measure here is the volume in phase space).

Now for stationary systems, we have a density function $f(\mathbf{q},\mathbf{p})$ that fulfills Liouville's theorem, using which we can write an ensemble average of some phase space function $A$ as follows: $$ \langle A \rangle = \int f(\mathbf{q},\mathbf{p})A(\mathbf{q},\mathbf{p}) d\mathbf{q} d\mathbf{p} \tag{1} $$ Similarly the time average of the same function $A$ is defined by: $$ \langle A \rangle_{time} = \lim_{t\to \infty} \frac{1}{t}\int_0^t A(t)dt \tag{2} $$ The most important, physically relevant, statement of the ergodic theorem is that (1) and (2) are equal, i.e. the ensemble average and time average of phase space functions are the same. This leads to an important interpretation regarding the time evolution of an ergodic system, which is that all the regions of the accessible part of phase space (i.e. consistent with the system's energy) are visited by the system regardless of the initial condition at $t=0$, and that the system spends an equal amount of time in all of them. This also intuitively explains why the averages (1) and (2) ought to be equal.

From a mathematical point of view, we know that the time average (2) converges for all trajectories taken by the system, since we started with the equilibrium assumption, which allowed us to treat the time evolution as a measure preserving transformation. The question is, how is the ergodic theorem generalized to also account for non-equilibrium systems, i.e. systems with dissipative dynamics and with sources/sinks of particles?

For one thing we no longer have the volume preserving transformation, thus Liouville's theorem is violated and the phase space volume is no longer incompressible under time transformations. Thus evaluating the convergence of (1) and (2) becomes non-trivial. Admittedly, the mathematical challenges aside, I am more interested in the physical aspect of the generalization of the ergodic theory. Does this concept even extend to non-equilibrium? If yes, how is it interpreted?

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user user929304

Comments

@YvanVelenik Thanks for the references, I will look into them asap.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user user929304

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I wrote an answer at: http://physics.stackexchange.com/a/177972/59023. The applicable part is in relation to when $df/dt = 0$ in my answer.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user honeste_vivere

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@honeste_vivere Thanks for the link, very interesting read. Unfortunately in that discussion you don't touch on the ergodicity aspect which is of interest here, would you be interested to take a stab at this question?

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user user929304

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@user929304 - If I had more time, yes. In lieu of that, I would recommend a great review paper by Oliver Penrose from 1979 entitled "Foundations of statistical mechanics" in Rep. Prog. Phys.. The first section (18 pages long) focuses almost entirely on the answer to your question. It's a good read and he provides physically intuitive explanations, not just math, which is very helpful.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user honeste_vivere

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@user929304 - After reading your question and many of the comments posted here in more detail, I strongly suggest you take a look at Penrose's 1979 review paper. He spends an entire section on non-equilibrium systems and he explains, quite clearly, the difference between the ensemble average and the time-average (they are only equal under rather limited circumstances and should not be confused).

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user honeste_vivere

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Liouville's theorem is valid irrespective of the assumption of equilibrium. The Hamiltonian evolution is sufficient.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user Ján Lalinský

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@JánLalinský : I think that he's interested in "open" Hamiltonian systems, for which Liouville's theorem indeed does not hold.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user Yvan Velenik

Answer 1

There are quite a few conceptual confusions in this question.

A system is either closed or open. A system is not "equilibrium" or "non-equilibrium". Also, a system is either conservative or dissipative. The ergodic theorem does not apply to open systems, neither to dissipative systems, since they tend to tend to a fixed point or something like that.

A state of a system can be an equilibrium state, or not, depending on whether it is invariant with the passing of time. A system has two concepts of "state": the one relevant for statistical mechanics is a macrostate, which means not a point in the phase space, but a probability distribution on the phase space. Usually the energy is fixed. If Liouville measure were a probability distribution, which it is not, it would be an equlibrium state since it is invariant under the passing of time. If a fixed-energy surface has finite volume, which it usually does, then if one restricts Liouville measure to that surface, one gets an equilibrium state. This can be done for any closed Hamiltionian conservative system. This has nothing to do with ergodicity.

The ergodic theorem does not apply to every dynamical system, yet the above remarks do. A dynamical system could have equilibrium states whether or not the system is ergodic.

Very few of the dynamical systems are known to be ergodic. Even if a system is ergodic, it is a fallacy that that means the paths nearly always enter into every region. That is the concept of "mixing", which is even harder to prove, and rarer. Ergodic just means the time averages are almost always equal to the phase averages, nothing more nor less. And this has to be true for non-equilibrium states too, so it has nothing to do with equilibrium.

The question of open systems cannot use Liouville's theorem since it is false for open or dissipative systems. And so is the ergodic theorem.

Interestingly, if a system is composed of a very large number of similar components which interact somewhat weakly, one can prove that some of the important time averages are approximately equal to their phase averages even though the ergodic theorem is not applicable to that system. (Khinchin, The Mathematical Foundations of Statistical Mechanics.)

A good reference, a little old but easy to understand, for non-equilibrium statistical mechanics, is the book by de Groot and Mazur, recently reprinted by Dover. It studies fluctuations near equilibrium, which can be related to the amount of dissipation present in the system.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user joseph f. johnson

Comments

Thanks for your answer, I really appreciate that took the time to also clarify some of the poorly expressed parts (or potentially confused), very helpful. I think I grapsed the core of your answer, namely that if out of equilibrium, then the system tends to evolve towards some fixed point, thus ergodicity becomes ill-defined there. But unless i ve misunderstood your answer, which is very likely, there s a potential contradiction: where you say that "ergodic just means time... And this has to be true for non-equilibrium states too..." But didnt we just say it is ill defined out of equil.?

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user user929304

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Let us make precise your meaning: when you said "non-equilibrium system" did you mean an open system? Or a dissipative system? (This was suggested by one of the people who commented on your question).

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user joseph f. johnson

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To be precise I meant "open"(particle sinks, absorbing boundaries, heat exchange and so on) in general. But now that you put it like this I'm confused as to why dissipative also doesn't imply openness!

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user user929304

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It does in a way. but dissipative includes the idea of describing a system in a reduced way, ignoring internal degrees of freedom: e.g., energy gets dissipated into internal heat, not necessarily to an outside system across a boundary. Still, with an incomplete description, it's not a closed system in the sense of classical mechanics.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user joseph f. johnson

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A dissipative system doesn't need to tend to a fixed point. An important example is the case of turbulence in fluids satisfying the Navier-Stokes equations.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user Arnold Neumaier

Answer 2

Non-equilibrium systems are most often considered in the approximation where local equilibrium is valid, yielding a hydrodynamic or elasticity description. Local equilibrium means that equilibrium is assumed to hold on a scale large compared to the microscopic scale but small compared with the scale where observations are made. In this case, one considers a partition of the macroscopic system into cells of this intermediate scale and assumes that each of these cells is in equilibrium, but with possibly different values of the thermodynamic variables.

From a macroscopic point of view, these cells are still infinitesimally small - in the sense that a continuum limit can be taken that disregards the discrete nature of the cells, without introducing too much error. Therefore the thermodynamic variables that vary form cell to cell become fields, tractable with the techniques of continuum mechanics.

On the other hand, from a microscopic point of view, these cells are already infinitely large - in the sense that the ideal thermodynamic limit, that strictly speaking requires an infinite volume, already hold to a sufficient approximation. (The errors in bulk scale with $N^{-1/2}$ for $N$ particles, which is small already for macroscopically very tiny cells.) Thus one can apply all arguments from statistical mechanics to the cells.

To the extent that one believes that an ergodic argument applies to the cell, it will justify (subjectively) the statistical mechanics approximation. However, the ergodic argument is theoretically supported only in few situations, and should be regarded more as a pedagogical aid for one's intuition rather than as a valid tool for deriving results.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user Arnold Neumaier

Comments

I gather the O.P. is not actually asking about non-equilibrium closed systems, but about open systems. The question used incorrect terminology. An open system is (almost never) ergodic and Liouville's theorem does not apply either.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user joseph f. johnson

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@josephf.johnson: If an open or closed system is in local equilibrium, ergodic arguments can (with caution) be applied to the individual cells. This is a way - and the only way - his question can be made sense of. He asked for ''[...] ergodic theory. Does this concept even extend to non-equilibrium? If yes, how is it interpreted?'', and my answer gives the only feasible interpretation.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user Arnold Neumaier

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@user929304: (i) For local equilibrium, one bins space-time (a short time average is also needed to account for contributions of very high frequencies). One can also bin in (macroscopic) phase space, but then gets Boltzmann-like kinetic equations rather than hydrodynamic equations, as all the fields then depend on position and momentum. Each cell has its own set of thermodynamic variables defining its macrostate. (ii) roughly, yes.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user Arnold Neumaier

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@user929304: In principle, the cell size is a free parameter on which the coarse-grained model depends. But for a macroscopic system, the result is nearly independent of it, if it is far away from both microscopic and macroscopic scales. In practice, its choice is therefore not critical, taking the geometric mean of microscopic and macroscopic scales works well. In phase space, things are more delicate as there is no canonical metric on it.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user Arnold Neumaier

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''did you mean that a turbulent flow would stay turbulent indefinitely?'' It depends on the boundary condition. Running water can well be turbulent indefinitely. A closed and isolated turbulent system with enough friction at the container will ultimately settle in an equilibrium state. Dissipation is represented by the parabolic terms in the Navier-Stokes equations. The corresponding conservative system would satisfy instead the Euler equations.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user Arnold Neumaier

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Thanks a lot for your patience with me. If there are references you'd recommend for further reading (on these binning discussions of system and its phase space and ergodicity) I would be most interested of course. By the way you may also be interested in this discussion on mixing and ergodicity that arose recently in this post.

This post imported from StackExchange Physics at 2016-01-02 13:14 (UTC), posted by SE-user user929304