Why is any real process which proceeds through non-equilibrium states necessarily irreversible?

As per the title, why is any real process which proceeds through nonequilibrium states necessarily irreversible?

The question came up when reading Callen's definition of "reversible process" (the whole discussion is in the context of closed/isoaltes systems):

Callen's definition: The limiting case of a quasi-static process in which the increase in the entropy becomes vanishingly small is called a reversible process.

Quasi-static processes were (for the purposes of this question) earlier defined as idealized processes which at all times have the given system in equilibrium. Thus, on the basis of this definition, we see that any (real) process which proceeds through nonequilibrium states cannot be reversible. But I have a hazy sense that this could instead arise as a theorem somewhere down the line. That is, I believe that Callen could have defined a reversible process as

A possible re-definition: The case of a process in which the increase in the entropy is vanishingly small is called a reversible process.

Then, we could recover his old definition as a theorem if we could show that if a process is reversible ($$\Delta S = 0$$) then it necessarily proceeded only through equilibrium states (did not proceed through nonequilibrium states).

• It would be helpful if you included Callen's definition of a reversible process. I say this because a reversible process is, in fact, one that occurs when the entropy generated becomes vanishingly small, which in turn occurs when disequilibrium becomes vanishingly small. So I'm not sure what your issue is. Commented May 26, 2023 at 21:14
• If a system is out of equilibrium (i.e., there's a gradient or spatial difference in some intensive variable, such as pressure, whose conjugate variable, volume, is unconstrained), then the system will move toward equilibrium through some flux or shift. Entropy production scales with the product of the flux and the gradient. Irreversible processes are associated with entropy production. Is this the kind of reasoning you're looking for? Commented May 26, 2023 at 21:32
• @BobD I did provide Callen's definition verbatim, but my apologies as I'm just seeing the formatting did not come out as I intended. I will highlight it to make it clearer.
– EE18
Commented May 26, 2023 at 21:35
• @Chemomechanics Yes that is the sort of thing I had in mind.
– EE18
Commented May 26, 2023 at 21:35
• Your definition seems like circular reasoning because, to determine the entropy change of a system using classical thermodynamics, you need to use a reversible path between the initial and final states of the system. Commented May 27, 2023 at 11:55

A mathematical view of what reversibility means is the following: a process is reversible if the thermodynamic relationships that characterize it are all thermostatic, in other words, the relationships between the various quantities are all function relationships. This definition, for example, excludes time derivatives among others.

The relationships that are allowed in a reversible process are of this kind, say, $$S=f^s(T, p, m_1,m_2, e_1, e_2, ...)$$, where the independent variables are temperature, pressure, mass of chemical compound 1, mass of chemical compound 2, electric charge of chemical compound 1, electric charge of chemical compound 2, etc. Notice that no derivative of, say. mass 1, occurs, just the mass itself.

If a system is not in equilibrium then it will not stay there but will move towards something else, so there is your time dependence creeping in, and then the above definition is not applicable any more, and you get irreversibility in the mathematical sense.

Your redefinition of reversibility leaves out the quasi-static aspect of the process. That quasi-static aspect is the one that is manifested in the function relationships as it has no time aspect in it, you know infinitely slow... If you leave that out then you are implying that a non-static process must always generate some amount of entropy. This is empirically true to the extent it can be observed for arbitrary slow processes and is also implied by the Postulate II but I think it is better to state explicitly.

• I think your last paragraph gets at the heart of my question. You say it is empirically true that nonquasistatic processes generate entropy. Do we take that as an axiom of the theory or is it derivable? You say that it's implied by postulate 2 but I'm not sure why that should be.
– EE18
Commented May 27, 2023 at 1:31
• In what is a bit pompously called "rational thermodynamics" that is the starting postulate. Above, I should have better said "kind of implied" by postulate ii since we cannot have equilibrium unless $S$ is maximum, so starting from equilibrium having removed a constraint a spontaneous process can only increase it otherwise. This is not a problem in practice only in theory. Example: isothermal AND reversible transformation. Commented May 27, 2023 at 1:48
• This really cannot happen because if the reservoir and system are of the same temperature nothing will change. You must have some amount of temperature differential otherwise nothing will happen. When you have a nonzero gradient it is not reversible any more and there is a $\delta S=S\frac{\delta T}{T} > 0$ positive entropy produced as the gradient transports $S$ entropy between them. To overcome this problem one must postulate that $\frac{\delta S }{S}$ is of a higher order infinitesimal than the temperature drop $\delta T$ makes in the other parts of the system. Commented May 27, 2023 at 1:50
• I have never seen this problem analyzed from this point of view. Instead what people do is to postulate that such transport exists as idealized limit of all irreversible processes one can think of. Just as a personal view, I am a bit skeptical about it, though, because I cannot be sure that such limit operation is always allowed but nobody has complained yet. It reminds me of the troubles that "singular perturbation" may cause, unless one is very careful, see the two-capacitor paradox. This is why I would always say it explicitly that if it is irreversible then $\delta S >0$ and vice versa. Commented May 27, 2023 at 2:01

I think this is somewhat incorporated in the postulate that the entropy is maximised.

I cannot remember the exact details, but the book that really made this glaringly obvious is Ian Ford's Statistical Physics: An Entropic Approach. There is a reason why he wins the department teaching award every year that I was studying there.

The general gist is like this: You can study non-quasistatic processes by comparing the initial and final equilibrium states. Say, a temperature difference in the question you recently asked and I just added an answer, or a pressure difference, or both, etc.

The crucial step is to express the spontaneous entropy generation over infinite time, as a function of the ratio of the parameter that is different. One of the graph will look like $$y=x-1,$$ whereas the other will look like $$y=\ln x.$$ The difference between them is the entropy that will be generated if you remove the constraint and allow non-equilibrium motion to occur. The graph $$y=x-1-\ln x$$ is one that increases on both sides of $$x=1,$$ which is saying that entropy is always generated if the ratio of the parameters is different from 1, and it does so almost quadratically near 1. This is important, because it also tells us that if we want the spontaneous entropy generation to be minimal, then we do not have to exactly reach equilibrium, but rather have a small enough difference that the total entropy generation is smaller than whatever tolerance we want to impose. It is also saying that the limit, in which an idealisation of a real process can actually be reversible, exists.

Nota Bene however, that it is the entropy of the universe that you have to take as increasing vanishingly. Or some form of spontaneous entropy generation. Because entropy can be transferred between systems reversibly, and you do not want to exclude reversible entropy transfer in the definition of reversible!