This question is related to this one. I would ask you to read that question and my answer to the question itself before answering this one.

The problem is the following. In his book Thermodynamics, appendix C, Callen introduces the problem of the adiabatic piston: to find the equilibrium state of an isolated system with an internal movable adiabatic wall.

Callen's argument is the following:

Total volume is fixed:


So that

$$dV_1 = - dV_2$$ And the piston is adiabatic, i.e. it allows no heat exchange, so that:

$$dU_{1,2} = -P_{1,2} dV_{1,2}$$

The total entropy of the system is

$$S(U,V)= S_1 (U_1, V_1)+ S_2 (U_2, V_2)$$


$$d S = dS_1 + dS_2 = \frac{dU_1}{T_1} + \frac{P_1}{T_1} dV_1 + \frac{dU_2}{T_2} + \frac{P_2}{T_2} dV_2$$

Since $dU_{1,2} = -P_{1,2} dV_{1,2}$, we see that $dS$ vanishes identically, so that we can say nothing about $P_{1,2}$ and $T_{1,2}$: the entropy maximum principle is thus inconclusive.

On the other hand, there is conservation of energy. We require that $dU=0$ since our system is isolated from the environment, hence

$$dU_1 + dU_2 = 0 \to P_1 d V_1 + P_2 dV_2 = 0$$

But $V=V_1+V_2$ and $V$ is fixed, so that $dV_1 = - dV_2$ and we obtain


I know that this conclusion is correct, i.e. it is a necessary condition for thermodynamic equilibrium. On the other hand, in the article Thermodynamics of systems with internal adiabatic constraints: time evolution of the adiabatic piston by C. Gruber (1998) the author says:

It was then noted that Callen's argument, which was repeated by Leff, could not be correct since the equilibrium condition was derived from the first law, rather than the second law.

My question is: What does the author of this article mean? Why should Callen's argument be flawed?

I stress that I don't want an alternative proof, but just an explanation of what is wrong in Callen's argument, possibly using only thermodynamics and not kinetic theory or fluid dynamics.


An "answer" to this question is actually contained in the article A Thermodynamic Consideration of Mechanical Equilibrium in the Presence of Thermally Insulating Barriers by A. E. Curzon (1968), which is mentioned explicitly in the article I cite in my question. The problem is that to me it in not clear why Curzon argument should be the "right one" while Callen's is the "wrong one"...

This is Curzon's argument:

The system is isolated and its volume is fixed, so that $$\label{1}\tag{1}dU_1+dU_2=0$$ $$\label{2}\tag{2}dV_1+dV_2=0$$ If we assume that the wall separating the two system does not allow exchange of matter, we have $$\label{3}\tag{3}T_{1} dS_{1} = d U_{1} + P_{1} dV_{1}$$ $$\label{4}\tag{4}T_{2} dS_{2} = d U_{2} + P_{2} dV_{2}$$ The entropy of an isolated system with fixed volume at equilibrium is at a maximum: $$dS=dS_1+dS_2=0\tag{5}\label{5}$$ From \ref{1}, \ref{2} and \ref{4} we obtain $$\tag{6}\label{6}T_2 dS_2 = -dU_1 -P_2 dV_1$$ Substituting \ref{6} and \ref{3}in \ref{5}, we obtain $$dS = \left( \frac 1 {T_1} - \frac 1 {T_2} \right) d U_1 + \left(\frac{P_1}{T_1}-\frac{P_2}{T_2} \right) dV_1 = 0 \tag{7}\label{7}$$ Now, since the piston is adiabatic we have for the two subsystems $$\tag{8} \label{8} dS_{1,2} \geq 0$$ (this is a consequence of the Clausius inequality). Equations \ref{5} and \ref{8} can only be consistent if $$dS_1=dS_2=0\label{9}\tag{9}$$ From \ref{3} and \ref{9} we obtain $$dU_1=-P_1 dV_1\tag{10}\label{10}$$ Substituting \ref{10} in \ref{7} we finally obtain $$(P_1-P_2) \frac {dV_1} {T_2} = 0\label{11}\tag{11}$$ Being $dV_1$ arbitrary, we conclude that $$P_1=P_2$$

I am not completely convinced that this can be considered the "correct" proof that $P_1=P_2$ is a necessary condition for equilibrium while Callen's is wrong. Curzon actually addresses the problem of analogous derivations of many authors (for example Kubo), i.e. the fact that they state that the condition $P_1=P_2$ can be derived only if we assume $T_1=T_2$. But Callen does not make such an assumption! In fact, his argument is very similar to Curzon's and it seems to me that Curzon's argument is quite the mathematical trick.

Indeed, from \ref{4} and \ref{9} we would obtain $dU_2 = -P_2 dV_2$ (analogos to \ref{1}): this, together with \ref{1}, \ref{2} and \ref{10}, would give us Callen's argument again!

So, in conclusion, to me it looks like Curzon's argument is quite the same as Callen's: therefore it is not clear what Gruber meant when he wrote that sentence, and I consider the question to be still open.

  • 3
    $\begingroup$ "Wrong" seems like a very strange word to use here. $\endgroup$ Commented May 29, 2016 at 21:33
  • 1
    $\begingroup$ @dmckee "Incorrect" or "flawed" if you prefer. It has been pointed out in more than one article that, although the conclusion is correct, the argument is not. But nobody explains clearly why. $\endgroup$
    – valerio
    Commented May 29, 2016 at 23:11
  • $\begingroup$ As UtilityMaximizer pointed out in his 2nd and 3rd paragraph, piston being adiabatic does not imply the relation $dU_1 = -P_1dV_1$ in general; during the process of approaching the equilibrium, if it is not too violent, we have $dU_1 = T_1 dS_1-P_1dV_1$ with non-zero $dS_1$. In words, piston being adiabatic does not prevent increase of entropy, it only prevents heat transfer. The relation $dU_1 = -P_1dV_1$ is valid when $dS_1 = 0$, which is e.g. for virtual variation in thermodynamic equilibrium, when total $dS$ is zero. In this respect Curzon's argument (as you've written it) is more correct. $\endgroup$ Commented Dec 30, 2023 at 17:24
  • $\begingroup$ This seems to be also the core idea in the untidy paper by Curzon and Leff, Resolution of an entropy maximization controversy, Am. J. Phys. 47, 385-386 (1979) where they explain why Callen's argument is wrong. $\endgroup$ Commented Dec 30, 2023 at 17:42

5 Answers 5


What is meant by

It was then noted that Callen's argument, which was repeated by Leff, could not be correct since the equilibrium condition was derived from the first law, rather than the second law.

is this. The first law is a statement of the conservation of energy. It holds even if the system is not in thermodynamic equilibrium. If the argument were correct it would imply that we never see a pressure imbalance in nature, which is false.

What is the flaw in the argument then? It is this. The movement of the piston into its equilibrium position is irreversible, so the equation $dU_i = dW_i = - P_i dV_i$ is false ($dW_i$ being work done on the system). Instead we have the inequality $dU_i = dW_i \geq - P_i dV_i$ with equality only for reversible processes. If the equilibration process were reversible then as you say "the entropy maximum principle is thus inconclusive" because any movement of the piston leaves total entropy unchanged.

What Curzon's argument shows is that, as the piston moves to its equilibrium position, its final increment of volume is exchanged reversibly. This can be seen because he derives $dU_i = - P_i dV_i$ from the equality $dS = 0$ (it would be false otherwise). This equality of differentials holds only when the piston is in its equilibrium position, and so the second law has been used in an essential way.

  • $\begingroup$ Can you explain why the equation $dU_i = dW_i = - P_i dV_i$ is incorrect in general? If I am anywhere on the "manifold of constrained [equilibrium] states" which Callen talks about in Chapter 1 (equilibrium just meaning "well-defined macrostate", not necessarily the final state of interest here), can I not imagine a (virtual) quasistatic process in which I change the volume of each subsystem slightly? Or are you saying that at such a point, that the change in energy $dU_i$, whatever it is, would nevertheless not equal $dU_i = -P_idV_i$ precisely because it would be pushing against some... $\endgroup$
    – EE18
    Commented May 15, 2023 at 3:00
  • $\begingroup$ ...force $P_jA_j$ which is not equal to its own? $\endgroup$
    – EE18
    Commented May 15, 2023 at 3:01
  • 1
    $\begingroup$ @EE18 It's a corollary of the Clausius inequality. The relation $dU = TdS - P dV$ holds generally, even for irreversible processes, since it is a relation between state functions. By energy conservation we have $dU_i = dQ_i + dW_i$. The Clausius inequality then asserts that $T_i dS_i \geq dQ_i$ which implies that $dW_i \geq -P_i dV_i$. Another way of seeing that is that if each subsystem is not allowed to exchange heat with any other subsystem, then $dS_i \geq 0$ for all $i$. This gives $dU_i = dW_i \geq - P_i dV_i$. $\endgroup$ Commented May 16, 2023 at 10:40

I quote from the abstract of one of the articles which the OP linked in the question preceding this one

The principle of increase of entropy is used to obtain the condition for mechanical equilibrium in an isolated system divided into two parts by a frictionless, weightless piston which is made of a perfectly thermally insulating material. The result emphasizes that the principle can be used to obtain the condition for mechanical equilibrium without the assumption, frequently made in the textbooks, that the mechanical equilibrium is accompanied by thermal equilibrium.

It seems to be explicitly stated here that Callen's "flaw" is this assumption. See also this physics.SE question for an explanation of the difference between mechanical and thermal equilibrium in one of the answers.

To summarise in my own words: $P_1 = P_2$ only makes sure that you are in a state where nothing is moving/accelerating (i.e. that you are in mechanical equilibrium), which would obviously be the case if $P_1 \neq P_2$ (thus it is a necessary condition, as the OP states). However this is not sufficient to conclude that you are in thermodynamic equilibrium, because due to thermal fluctuations you always get $P_1 \neq P_2$ some of the time. Now it could be the case that the $(P_1 = P_2)$-state was metastable and drifts away from the mechanical equilibrium induced by the small fluctuations. As far as I can tell from the articles linked in the other question that is not the case, so $P_1 = P_2$ does turn out to be the correct condition, but Callen did not rigorously show that the "restoring force" accompanying the fluctuations leads back to the mechanical equilibrium.

I am not 100% sure if this answers the question, nor if it is correct. In particular I personally think that Callen's argument can easily be made complete by noting that the mechanical equilibrium $P_1 = P_2$ is unique. I mean if your system leaves mechanical equilibrium due to fluctuations it has to go somewhere. And there is no second mechanical equilibrium for it to go to. I.e. necessary $\Leftrightarrow$ sufficient since it is also unique.


This is to address the update to the question (v6).

Additionally to the original question the update contains an outline of the argument that is supposed to be correct, yielding the same outcome as Callen's argument. The OP then points out that the arguments seem to be the same.

I don't think this is the case. As stated previously by the OP Callen only uses the first law while for a correct proof of the equilibrium the second law should be employed. The reason for this is what I tried to point out above in my original answer. The first law is a mere statement of energy conservation. You can't extract information about thermodynamic equilibrium by simply imposing the constraints of your system. That only shows that mechanical equilibrium is realised, which is a necessary but not sufficient condition for thermodynamic equilibrium (as the OP already stated repeatedly).

The second law provides the sufficiency. Why? Because it is a statement about the nature of thermodynamic equilibrium itself, in particular it tells you how fluctuations behave around the point of equilibrium and that you will bounce back into your original state under fluctuation perturbations, to say it from the statistical mechanics viewpoint.

One of the OPs doubts is

Indeed, from (4) and (9) we would obtain $dU_2 = -P_2 dV_2$ (analogos to (1)): this, together with (1), (2) and (10), would give us Callen's argument again!

Of course your are going to get the same relations as from the first law if you plug the relations obtained from the second law argument into each other, because mechanical equilibrium happens to be thermal equilibrium in the situation considered. However I fail to see how this coincidence means that they are the same argument, since they start from completely different points.

  • $\begingroup$ You are right in saying that that article contains the answer to this question, but I think that the motivation is more formal than physical. Actually the article I cited contains a reference to that article, so this question came from an oversight. I think I will shortly post an answer myself explaining everything (also my oversight). $\endgroup$
    – valerio
    Commented Nov 5, 2016 at 13:32
  • $\begingroup$ @valerio92 yeah, it would be great to see a full resolution of the problem, which is surprisingly complex and interesting. Thank you for bringing it up on SE! Do you think I pointed out the flaw in Callen's argument correctly though? I find the reason it is wrong quite obvious now, although I wouldnt claim to understand the full scope of the papers you linked and how to solve the problem, but I thought that wasnt the question you asked here was it? $\endgroup$ Commented Nov 5, 2016 at 21:27
  • $\begingroup$ I think that what you say is right, but it is not what the author of the article cited was referring to when he wrote that "Callen's argument [...] could not be correct since the equilibrium condition was derived from the first law, rather than the second law." And yes, here I only want to discuss Callen's argument, I don't presume to solve the adiabatic piston problem :-) $\endgroup$
    – valerio
    Commented Nov 5, 2016 at 22:52
  • $\begingroup$ (I wrote that what you said is right, but I have to specify that Callen does not assume that T1=T2T1=T2 so that cannot be the flaw of his argument. But the rest of your post is right, thought it does not really answer the question. ) $\endgroup$
    – valerio
    Commented Nov 5, 2016 at 23:44
  • $\begingroup$ After a more in-depth reading of Curzon's article, I am convinced that it does not contain the answer to my question, and that his argument is the same as Callen's. You can find every detail in the "update" section of my question. Sorry for the many comments, I promise this is the last one! $\endgroup$
    – valerio
    Commented Nov 6, 2016 at 0:18

It goes wrong when conservation of energy is applied to the total internal energy. One has to add the kinetic energy of the piston here. This allows for the two pressures to be different, the piston will then oscillate about its equilibrium position. Dissipation of this energy will lead to entropy increase; at maximum entropy the piston will have settled down at the equilibrium position and the pressures will have equalized.

  • $\begingroup$ I'm not sure this is really a flaw. I mean, it's an argument in equilibrium thermodynamics. Here you claim it is wrong for an out of equilibrium system (true) and that it will be right again when the system achieves equilibrium. In other words, that it is valid for equilibrium thermodynamics, no? $\endgroup$ Commented Jun 1, 2016 at 21:19
  • $\begingroup$ @dmckee If we compare this to e.g. an isolated system consisting of two subsystems that are a internal thermal equilibrium at different temperatures. Then we can see that total entropy being maximal implies that the two temperatures must be equal. If they are not equal, then you don't have thermal equilibrium. But it's possible to treat this case using equilibrium thermodynamic methods as the two subsystems are in internal thermal equilibrium. The case of two different pressures should i.m.o. be considered in an analogous way. $\endgroup$ Commented Jun 1, 2016 at 21:43
  • $\begingroup$ The complication is only that unequal pressure leaves us with two subsystems with oscillating pressures but you can still treat that approximately within the framework of equilibrium thermodynamics. By not taking into account the physical situation where the pressures are not equal, one is imposing the conditions valid in the final state and that obscures the role of the second law in achieving equilibrium between the pressures. $\endgroup$ Commented Jun 1, 2016 at 21:46
  • $\begingroup$ You can also say that the whole point of the second law is to consider how equilibrium is approached, so you have to consider a non-equilibrium situation. That we have tricks available to do that within the framework of equilibrium thermodynamics (dividing up a non-equilibrium system into parts such that each one is in internal equilibrium and changes are happening in a quasi static way) is a nice thing, but we should not get carried away too much with the details of using these tricks. What matters is that there exists a well defined initial state and a well defined final state. $\endgroup$ Commented Jun 1, 2016 at 21:51
  • $\begingroup$ I don't think that anything you've said here is wrong (and it's not my downvote), but I do think they represent philosophical reasons for preferring a different argument rather than some kind of invalidation of Callen's approach. Sure, it doesn't generalize very well. Sure, it doesn't take advantage of a pedagogical opportunity to show off the second law. But so what? Special case analyses are often a big win for a few cases and no good for others. $\endgroup$ Commented Jun 1, 2016 at 22:29

Callen's argument seems like argument of virtual work. If the system is in equilibrium, then a small displacement consistent with constraints existing on the system should end up doing no work. This, as far as I can tell, is a correct argument. However perhaps the objection is that this doesn't count as a thermodynamic argument, which must make explicit use of entropy maximization.


We start with the diferential form of the total entropy S of the isolated composite system of $1$ & $2$.

$dS_{total} = dS_1 + dS_2 = (1/T_1)dU_1 + (P_1/T_1)dV_1 +(1/T_2)dU_2 + (P_2/T_2)dV_2$

We then apply that since the composite system is isolated, $dU_{total} = dV_{total} = 0$, so $dU_2 = -dU_1$ and $dV_2 = -dV_1$

$dS_{total} = (1/T_1 - 1/T_2)dU_1 + (P_1/T_1 - P_2/T_2)dV_1$

The equilibrium condition is that $dS_{total} = 0$, so each term must be $0$.

$(1/T_1 - 1/T_2)dU_1 = 0$

$(P_1/T_1 - P_2/T_2)dV_1 = 0$

Since the internal volume is not fixed, $dV_1$ is not $0$, and thus $(P_1/T_1 - P_2/T_2)$ must be $0$. Since the volume is not fixed, it is possible for $U_1$ to change (through volume work), and thus $dU_1$ is not $0$ either. Therefore $(1/T_1 - 1/T_2)$ must also be $0$.

Combining both equations (and assuming neither $T$ is zero) we obtain two equilibrium conditions:

$T_1 = T_2$ and $P_1 = P_2$

Note that the adiabaticity of the wall was unnecesary for the deduction. In fact it is irrelevant. If the wall is movable the temperatures will equalize at equilibrium, so we could say we proved no wall is adiabatic unless it is completely isolating.

$dS_{total} = 0$ is the equilibrium condition and its formula includes neither 'heat $Q$' nor 'work $W$', only $dU_1$ and $dV_1$. Any proof that involes $Q$ or $W$ will be less fundamental than this one, as it involes adding additional postulates about the relation between $Q$ and $W$ and $U$.

Furthermore, it's my understanding (like it's said in UtilityMaximiser' answer) that the formula $dW_{1,2} = -P_{2,1}dV_{1,2}$ is not always valid. It's only valid if $P_{1,2}$ is well defined. And the only situation in which both $P_1$ and $P_2$ are well defined simultaneously is if the system is in equilibrium.

Therefore, by proposing $dU_{1,2}=−P_{2,1}dV_{1,2}$ one is already imposing that the system be in equilibrium, so any other equilibrium condition becomes impossible to obtain.


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