I am reading Callen's Thermodynamics and introduction to Thermostatics (second edition).

In section $2.1$, Callen states that for a quasi-static process, we have

$$\delta Q = T d S \tag{1} $$

where a quasi-static process is defined as an (ideal) succession of equilibrium states.

Now, I know (see for example Fermi, Thermodynamics) that only for a reversible process between $A$ and $B$

$$\Delta S_{A\rightarrow B} = \int_A^B \frac{\delta Q}{T} \tag{2}$$

Since $(1)$ is basically an infinitesimal version of $(2)$, this would seem to imply that $\delta Q = T dS$ is only true for a reversible process.

Now, I know for a fact that Callen is not using the terms "quasi-static" and "reversible" interchangeably (see chapter $4$ in his book). Therefore, the only plausible conclusion I can think about to avoid a contradiction is that an infinitesimal, quasi-static process must also be reversible. Note that I am aware that in general quasi-static $\nRightarrow$ reversible (whereas reversible $\Rightarrow$ quasi-static), but I am asking about the particular case of an infinitesimal process.

Another possibility is that $(2)$ does not imply that the infinitesimal version $\delta Q= T dS$ is only valid for a reversible process, but I don't see how this could be.

Another way to rephrase my question (though slightly different) is: what is the condition to write $\delta Q = T dS$? That the process is reversible or that it is quasi-static? Or are these two conditions equivalent for an infinitesimal process?

I am aware that this could seem like an artificial problem, but I am trying to re-learn thermodynamics following an axiomatic approach, so it is very important for me to get all these conceptual issues right.

  • $\begingroup$ the thermodynamic process of a hysteretic ferromagnetic material is never reversible no matter how slowly is executed. Another rate independent hysteretic example that is irreversible though quasi-static is strain hardening. $\endgroup$
    – hyportnex
    Feb 20, 2018 at 14:55
  • $\begingroup$ @hyportnex Yeah, I am aware that quasi-static processes are not always reversible, but I am asking about the particular case of an infinitesimal process. Is irreversible=quasi-static in this case? $\endgroup$
    – valerio
    Feb 20, 2018 at 15:02
  • $\begingroup$ if the hysteresis is rate independent then it does not matter how slowly it happens or in how small steps per unit time it is still irreversible. $\endgroup$
    – hyportnex
    Feb 20, 2018 at 15:04
  • $\begingroup$ @hyportnex (In the previous comment I meant to write reversible, not irreversible) So you are saying that also for an infinitesimal process quasi-static and reversible are different concepts? To me, it looks like this means that there is a contradiction, as explained in the post. How would you solve it? $\endgroup$
    – valerio
    Feb 20, 2018 at 15:09
  • $\begingroup$ I can only give you what I think of this but I also believe that quasi-static and reversible are different concepts. Reversible, in practice, is always done slow to some extent, but not the other way around as rate independent hysteresis (memory effects) show. To me reversible means the existence of thermal and caloric functions of state, such that $T=T(p,V, M,...)$ and $S=S(p, U, M,...)$, hold during any instant of the process irrespective of how fast the process is. In practice this means "slow" to avoid inhomogeneities to develop, but would exclude any memory and rate effects. $\endgroup$
    – hyportnex
    Feb 20, 2018 at 15:20

2 Answers 2


The condition to write $\delta Q = T dS$ is that the process be reversible. In general, we have $$\delta Q \leq T dS.$$ A quasi-static process is not generally reversible. Taking infinitesimals makes no difference. Indeed, if you thought that every infinitesimal quasi-static process were reversible, you would have to conclude that all quasi-static processes were reversible, since by definition every quasi-static process is composed of a series of infinitesimal steps.

For a concrete example, consider two gases separated by an insulating piston with friction. One can consider a quasi-static process where the piston slowly moves to the right. Even an infinitesimal motion is not reversible, because heat is produced by friction. The equality $\delta Q = T dS$ is violated, as $dS >0$, while $\delta Q = 0$.

I agree that it's frustrating that sources disagree on this. Part of the issue is that whenever you're doing something concrete, it's easy to see what to do (i.e. you certainly wouldn't forget the friction in the above example). That leads to a degree of sloppiness when formulating general principles, and inconsistency between and within sources. My impression from answering thermodynamics questions on this site is that Callan is sloppier than most textbooks.

  • $\begingroup$ That's what I originally thought, so that's why the claim that $\delta Q = T dS$ is true if the process is quasi-static caught me by surprise. Yeah, you may be right in saying that Callen is just being sloppy, even if his book is considered one of the classics for thermodynamics (even if a bit old, but I would say that it has a status that is equivalent to that of Jackson for electromagnetism). So...strange. $\endgroup$
    – valerio
    Feb 20, 2018 at 16:25
  • $\begingroup$ I have to say though that I am not 100% convinced about your argument to extend the property of an infinitesimal part of the process to the whole process...It feels right, but limits are always tricky. For example consider the harmonic series $\sum_{k=1}^{\infty} (1/k)$: the generic term goes to $0$ as $k \to \infty$, but the whole series diverges. This is a stupid example, but...Do you see where I'm going? $\endgroup$
    – valerio
    Feb 20, 2018 at 16:29
  • $\begingroup$ @valerio Sure, there's a limiting procedure going on. But if you haven't accepted that the limiting procedure is physically valid, you shouldn't even believe in any quasi-static processes at all -- after all, no quasi-static process exists in reality. (There must always be a tiny pressure/temperature difference for anything to happen.) To establish the limit rigorously you would have to do some exhausting microscopic physics involving, e.g. the Navier-Stokes equations. The whole point of thermodynamics is that we don't worry about the microscopic stuff; we postulate it works out. $\endgroup$
    – knzhou
    Feb 20, 2018 at 16:34

I think that for many authors (especially the old one) quasi-static <=> reversible. Fermi that you cited says:

A transformation is said to be reversible when the successive states of the transformation differ by infinitesimals from equilibrium states. A reversible transformation can therefore connect only those initial and final states which are states of equilibrium. A reversible transformation can be realized in practice by changing the external conditions so slowly that the system has time to adjust itself gradually to the altered conditions.

For him the concepts are equivalent. If you think about it the only difference is when friction or similar processes are involved. The equation should be valid for a quasi-static transformation.

  • $\begingroup$ You are right, Fermi does not make the distinction indeed. You should add the section of the book where the quote is taken from (1.1.). $\endgroup$
    – valerio
    Feb 20, 2018 at 16:22
  • $\begingroup$ One could say that a rate independent hysteretic material is never in equilibrium because our process is not slow enough, and then Fermi's $quasistatic=reversible$ definition will hold. Callen by assuming with Gibbs that $dU=TdS+...$ holds is restricted between two nearby equilibrium states but such equation would not hold for infinitesimal magnetising steps. See Bridgman "The Thermodynamics of Plastic Deformation and Generalized Entropy", REVIEWS OF MODERN PHYSICS VOLUME 22. NUMBER 1 JANUARY, 1950. $\endgroup$
    – hyportnex
    Feb 20, 2018 at 16:57

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