Follow up to

Intuitively, why is a reversible process one in which the system is always at equilibrium?


How slow is a reversible adiabatic expansion of an ideal gas?

Suppose you have a piston with some air in it and you perform a slow, reversible expansion. The air in the piston must be in an equilibrium state the entire time.

Now suppose you do the expansion quickly. During the expansion, the air is not in an equilibrium state. My question is: should it have well-defined state variables? Should the pressure be well-defined, for example?

Presumably there is air moving around in bulk and a pressure gauge would give different readings depending on where you put it. Similarly, there is a mean kinetic energy of the molecules that might be used to define $T$, but there is no $\beta$ exponential factor because the kinetic energies of the molecules will not follow a simple, single-parameter distribution. This would indicate that concepts like pressure and temperature are not well-defined when out of equilibrium.

Is that right, and is it always the case? Can I have a process where I know what the pressure and temperature are the entire time, but the system is not in equilibrium?

  • 1
    $\begingroup$ This is the realm of nonequilibrium thermodynamics--- the pressure and temperature follow local laws, and if the gradient is not too big compared to the mean free path, you expect the continuum approximation to hold. I don't know if you want the detailed local laws for relaxation in a gas, or if you just want a pointer to the field. $\endgroup$
    – Ron Maimon
    Dec 12, 2011 at 11:05
  • $\begingroup$ @Ron A detailed description would be beyond what I'm looking for. I'm just trying to understand the scope of equilibrium thermodynamics clearly. $\endgroup$ Dec 12, 2011 at 11:55
  • $\begingroup$ Modern thermodynamics can deal with changes in time and, possibly, far-from-equilibrium states: what is needed is a system that can be described in terms of a small number of DOFs, or "state variables" (the evolution is slow enough that only a few, possibly far from equilibrium, DOFs are not relaxed). See the references given in Is thermodynamics only applicable to systems in equilibrium? and Assumptions of thermodynamic & chemical equilibrium in fluid-dynamics. $\endgroup$
    – Quillo
    Sep 13, 2023 at 18:20

4 Answers 4


Strictly speaking there are no reversible processes in Nature; it is an idealization that enables one to get bounds on efficiency of nonequilibrium processes by using techniques of equilibrium thermodynamics only.

A reversible process is therefore primarily a theoretical concept for discussing what would happen in a process if dissipation were absent. It is defined as a motion in equilibrium state space, and hence presupposes that the system always remains in equilibrium.

However, empirically, fast processes generate far more excess entropy (the source of the dissipation) than slow ones, so one can treat a slow process approximately as a reversible one.

In a nonequilibrium state, the extensive variables are still well-defined, while the intensive variables (temperature, pressure, chemical potential) typically aren't. On the other hand, most nonequilibrium processes in ordinary life are well-described by local equilibrium; which means that every small region is approximately in equilibrium, and then temperature, pressure, and chemical potential can be assigned definite values. As a result one gets a temperature field, a pressure field, etc. (This is what you feel when you move through a room from the heating to a open window.)


I've started to take Jaynes' wacky point on this subject more seriously --- the key is experimental reproducibility, of which equilibration is a helpful but neither necessary nor sufficient condition. The point is that you know you have enough "macroscopic" degrees of description when you find that it is sufficient to reproduce the phenomena you are interested in.

So for instance, with gases, if all you're interested in is the state of play from equilibrium to equilibrium, then you find that pressure, temperature, volume, entropy (and partial versions if you have a mixture) are sufficient, and one may sensibly have a theory of it. On the other hand, if you're interested in what happens in between, then you have to be more specific (a rather prosaic example of how coarse graining intellectually drives all condensed matter) --- are you interested in the fluctuation spectrum of things like pressure, or maybe localised versions of the macroscopic variables (as Ron suggested)? Or maybe you just care about the case where you set up a shock wave (in which case I can recommend the excellent book on the subject by Zel'dovich).

In the base case and extensions, the logical starting point is experiment: to reproducibly observe this phenomenon, what are things I have to control? This is not a theoretical problem, and can only be answered with experiment.


For a non equilibrium process, pressure and average kinetic energy are still defined (force/surface) but as you have noticed they no longer necesseraly depend on one parameter. For instance, in atomic physics, it is quite common to have a two temperature system to describe the electron populations around an heavy element. Now the non equilibrium state you mention is in fact a non uniform state where the temperature and pressure varies from point to point. Basic thermodynamics introduces T, P constant within the system. Fluid thermodynamics allows T,P,n to be fonction of space and time. Fluid equations are only valid if the mean free path is shorter than the gradient scale length. One assumes that a fluid cell is a microcosm where the distribution function is a Maxwellian. Kinetic theory is the fundamental theory and is always valid but one can not simply define state variables (assuming they have a meaning).


In thermodynamics, entropy (differences) is (are) defined, at least at first, only for two states which can be connected by a reversible transformation. Hence, those two states have to be equilibrium states. Yet, sooner or later, one always runs into the example of two boxes of gasses, at different temperatures, separated by an insulating wall. The entropy of this joint system is calculated by calculating the entropy of each box separately (each box is indeed an equilibrium state) and adding the results (and forgetting to specify the arbitrary additive constants needed, but never mind...). Then the wall is removed, and an irreversible path (this had better be studied using a ppVV diagram, one p and one V for each box, so that each one can have its own $T$...this is in line with the use of a pressure field) is followed from the now non-equilibrium state of each box's being at a different temperature to the equilibrium state of a temperature for the new joint system. Then its entropy is calculated and shown to be higher than before. Yet the entropy difference between the non-equilibrium state and this state is undefined according to the usual difference.

I was just looking at a paper by someone in a math dept. about the thermodynamics of power plants in which he considered a power plant in thermal contact with two atmospheres: the terrestrial one and the martian one. (No I am not making this up.) This inspired the following observation: what is an irreversible transformation in one setting can be considered a reversible transformation in a different setting.

Change the external conditions of the above systems. Each box is now in contact with a separate heat reservoir, at the different initial temperatures we considered. Insulating walls can be present or absent here, too. Now our thermodynamic state space has to have two p variables and two T variables, one for each box. Initially we have each box in contact with its reservoir. We break that contact and put the boxes in contact with each other infinitesimally short period of time, then break that contact (this is analogous to the infinitesimal addition of a weight to the piston). We have moved, reversibly, from one equilibrium state to another. (This is reversible because we could restore contact with the corresponding reservoirs and get back to where we started.) We do this again and again until finally the boxes have equalised their temperatures. This is a reversible transformation. On the ppVV diagram it looks identical to the irreversible transformation we discussed above. But the physics is completely different, because the latter was in an isolated system and this one is not in an isolated system.

Well, this is part of the answer to your question, too. The notions of equilibrium and reversibility depend on the contact with the environment which is supposed.

Now suppose an isolated system of the simplest kind, a perfect gas. Obviously $T$ cannot change. Can $V$ change? Suppose the mass is finite. If the gas is not enclosed in perfectly fixed, insulating walls, then obviously it will dissipate to infinity and we have $p=0$. There are no equilibrium states... So suppose $V$ is fixed. The first law of thermo still implies $T$ cannot change..not even irreversibly. But then the gas law implies $p$ cannot change, so there are no paths at all, neither reversible nor irreversible. All states are equilibrium states, but none of their entropies can be compared. The same holds if the mass is infinite but there is a constant density.

These two extreme examples show that the notions of equilibrium and reversibility depend on the external conditions. Since the science of Thermodynamics was inspired by heat engines (um, terrestrial ones), the usual analytical framework involves putting it in contact with a heat bath, breaking contact, putting it in contact again, etc., as in the Carnot cycle. Fundamental Theoretical Physics finds the closed conservative system more congenial, but such an isolated system looks really strange to Thermodynamics. Perhaps this is what is responsible for the pedagogical confusion students have with entropy.


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