The work done during a process between two equilibrium states can be described by thermodynamics. Even when process itself is out of equilibrium, the thermodynamic laws can still be used, though calculating the work is much more difficult. But if the initial or final states, or both, are not in equilibrium, can the work done in driving the system from one to the other be calculated? How?

Edit: @Roy @genneth @Marek I do mean far out of equilibrium. There wasn't a nonequilibrium tag, and I don't have enough reputation points to create one! I can't seem to post comments today, so I'll write this here instead: I'm familiar with Jarzynski's equality and while it is very useful, it is still only valid between equilibrium initial and final states (though the final state being in equilibrium can be relaxed). I'd like to know if there's any way work (or heat) can be defined or calculated when the initial and final states are out of equilibrium, possibly very out of equilibrium.

Edit2: @Roy I'd like to know if it's possible in a system with an initial state which is not in equilibrium which is then driven to a final value which is still not in equilibrium. I don't want to make any other assumptions if possible. So I don't necessarily expect local equilibrium to hold, although I'd still be interested to know if work between the 2 nonequilibrium states can be found in that case.

  • $\begingroup$ You mentioned when processes are out of equilibrium; did you mean en.wikipedia.org/wiki/Jarzynski_equality ? $\endgroup$
    – genneth
    Feb 10, 2011 at 22:15
  • $\begingroup$ @Jane: thanks for clarifying this. I'd also like to hear the answer to the general case but am actually very skeptical whether anything can be said at all. All the tools that I know about rely to some degree on equilibrium thermodynamics. $\endgroup$
    – Marek
    Feb 11, 2011 at 12:05
  • $\begingroup$ @Jane : You might need to clarify this further as it seems more like a research area, than a question from the textbooks. For example do you expect "Local Equilibrium" to hold? (This would enable T(position), etc and no large position gradients.) Shock waves dont meet this condition. $\endgroup$ Feb 11, 2011 at 17:11
  • $\begingroup$ @Jane: just to add to Roy's comment above: it's usually clearer to think of things in terms of statistical physics; the key statement about equilibrium states is that they are fully described by their macroscopic description, and the microscopic states are equiprobable. To have a chance of describing what you want, we would at least need to have a way of describing these non-equilibrium states --- but in general there is no way to do so. In the most general case, you have a problem of non-linear evolution in a large, non-integrable Hamiltonian system --- about which very little can be said. $\endgroup$
    – genneth
    Feb 12, 2011 at 12:16
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    $\begingroup$ I was never very happy with the state we left this question in; I've since been digging around a little, and there's definitely been work done on this. Sections D and C (read in that order!) of bayes.wustl.edu/etj/articles/stand.on.entropy.pdf is a nice place to start. Perhaps you can chase up the references contained there-in? $\endgroup$
    – genneth
    Feb 21, 2011 at 20:06

2 Answers 2


This is a topic in Non-Equilibrium Thermodynamics. There is a standard concept in Thermodynamics of "Thermodynamic Force", "Thermodynamic Flux" and so on. In the Physical Chemistry context you might be familiar with "Affinity" and "Chemical Potential". These are the mechanisms used to explain chemical reaction directions, etc.

So to summarise this large area with an example two substances S1 (with Temperature T1) and S2 (with Temperature $T2 < T1$) are in thermal contact, the combined system is not in equilibrium. The Thermodynamic Force here is: $F=(1/T2 - 1/T1)$. This is derivable from the formula for change in Entropy in this situation:$dS = -dS1/T1 + dS2/T2$. In general a "thermodynamic force" causes a change in Entropy - a situation that can only arise in non-Equilibrium situations. The thermodynamic force will become zero when T1=T2 and the system S1+S2 is in equilibrium. So the idea is that Thermodynamic force models Entropy change (in a mechanics-like manner).

Associated with this Force is the Flux $J_Q$: the time dependent construct
$dQ/dt = J_Q = \alpha(T1-T2)$ where $\alpha$ is the Fourier coefficient of heat conductivity.

In general the First Law of Thermodynamics holds the key to the constructions.

$dU=TdS - PdV + \Sigma \mu_i dN_i $

(here the $\mu_i$ are the chemical potentials of the ith particle species) - useful if they are created or destroyed (as in a chemical or nuclear reaction.) This equation could be written as:

$dU=TdS - \Sigma X_i dx_i$

where the $X_i$ are the generalised forces and $x_i$ are the generalised conjugate variables.

When the extra variables include electrical potentials we can have thermo-electric equations, etc. These can describe the flows in electric-chemical batteries and the like.

One point of debate is whether and to what extent thermodynamic variables are really "local" as would be required in a true continuum based theory. Thus how valid is "Temperature at a point" etc.

These topics are covered in Thermodynamics texts e.g. Callen: "Introduction to Thermodynamics and Thermostatics" or Prigogine "Modern Thermodynamics".

  • $\begingroup$ I believe these formalisms still describe quasi-equilibrium states and their evolutions? Surely in the most general case, nothing can be said about truly out of equilibrium systems? In which case I would have thought that if a statistical approach is possible, there needs to be some restraints applied? (By truly out of equilibrium I mean we're so far from the bottom of a free energy potential that non-quadratic effects are important.) $\endgroup$
    – genneth
    Feb 10, 2011 at 22:11
  • $\begingroup$ @genneth: this question is labeled as thermodynamics. So I guess one is free to assume quasi-equilibrium states. Otherwise you are of course correct, it's hard to say anything about an arbitrary state. $\endgroup$
    – Marek
    Feb 10, 2011 at 22:31
  • $\begingroup$ @genneth : I didnt take this question as really about far-out-of-equilibrium theory - just the basics of non-equilibrium TD. Having said that the basic theory here is just 1st Law. Prigogine does make the point that "local equilibrium" is a key condition for much of his later non-equilibrium theory. Also in these extreme cases the relations between variables become non-linear, but they are assumed linear in near-equilibrium theory: that difference is important but too much to discuss in one answer! $\endgroup$ Feb 10, 2011 at 22:37
  • $\begingroup$ @Marek: the Jarzynski equality is notably about highly out of equilibrium processes (though admitted it requires the starting state to be in equilibrium), and it is purely a statement of thermodynamics. Perhaps I gave the impression of being too pessimistic --- I actually believe that there are some very general thermodynamic-esque statements possible for out-of-equilibrium processes, but it will involve some (possible mild) restrictions on what states are allowed. $\endgroup$
    – genneth
    Feb 11, 2011 at 9:35
  • $\begingroup$ @genneth: the processes are out of equilibrium but initial and final states are still equilibrium states. So while I agree that the picture is not completely black (thanks to all those fluctuation theorems) we still can't say anything about processes between general (i.e. not equilibrium or quasi-equilibrium) initial and final states. Right? $\endgroup$
    – Marek
    Feb 11, 2011 at 11:13

If you want to treat a non equilibrium problem, you need to introduce a distribution function for each component and to solve the corresponding kinetic equations. That is currently done in fusion plasma physics. However, keep in mind that it is mathematically involved and that the thermodynamic temperature is not even defined for non equilibrium processes. The thermodynamical temperature can indeed be identified with a parameter which appears in the equilibrium solution the kinetic equation (Boltzmann's distribution), out of equilibrium the distribution function is no longer Maxwellian and the temperature becomes undefined. It is usual to keep identifying the temperature as the average particle kinetic energy, but to compute this average you need the distribution function that may depend on more than one parameter... And of course one can compute the work done calculate the work done between two non equilibrium states, assuming you have solved the kinetic equations.


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