I have a question about a characterization of thermodynamic equilibrium as given on German wiki article: https://de.wikipedia.org/wiki/Thermodynamisches_Gleichgewicht#Abgeschlossenes_System

What does it state? Let $S=S(U, V, N)$ the entropy and the expression $S=S(U, V, N)$ make sense as well a state with macroscopical parameters $U,V,N$ is in thermodynamical equilibrium. The claim is

"Ein abgeschlossenes System befindet sich im thermodynamischen Gleichgewicht, wenn seine Entropie $S$ maximal ist. Entsprechend gilt, für das Differential


Translated that means

A closed system is in thermodynamic equilibrium when its entropy $S$ is at a maximum. For the differential this means $dS=0$.

And the last condition I not understand. By definition writing $S= S(U,V,N)$ (or for any other arbitrary thermodynamic potential e.g. internal energy $U(S,V,N)$ or free Helmholtz $F(T,V,N)$)

makes only sense if the given state parametrized by macro parameters $U,V,N$ is already in thermodynamic equilibrium. Otherwise, $S= S(U,V,N)$ don't make any sense as a thermodynamical system that is not in a thermodynamic equilibrium is too complex to be described by only three independent macro parameters $U, V,N$ and futhermore the concept of associating a thermodynamical potential to a state only works if one consider a state beeing already in equilibrium.

Therefore what is $dS=0$ (or more precisely how to interpret it in this context?) and why does it make sense to use it as characterization of equilibrium?

The only way I'm familar how the differential $dS$ is used in thermodynamics is explaned in following setting. We start with a state parametrized by $(U_1,V_1,N_1)$ that is already in thermodynamic equilibrium and then we start a certain thermodynamical process (reversible or irreversible) that finally brings the system in another state $(U_2,V_2,N_2)$ that is also in a new thermodynamical equilibrium after a long time.

The point is that how we pass from $(U_1,V_1,N_1)$ to $(U_2,V_2,N_2)$ we don't know precisely as in nature we pass during the process non equilibrium states that we can't describe with our formalism. One possible ansatz is to consider it a sequence of quastatical processes such that each intermediate state is assumed to be also in thermodynamical equilibrium. This is of course a strong idealisation.

With this idealization passing from $(U_1,V_1,N_1)$ to $(U_2,V_2,N_2)$ can be indeed visualized as a curve in Phase space as long as we consider it as a quasistatic process.

then indeed also the expression $dS= S(U_2,V_2,N_2)-S(U_1,V_1,N_1) $ make sense.

But in this case it makes no sense to use $dS=0$ as characterization of a state beeing in themodynamic equilibrium as by $dS$ we consider allways diferences of enropies of states that are already in equilibriums.

Does anybody have an idea how the "characterization thermodynamic of quilibrium" $dS=0$ should here be understood and what is the error in my reasonings above?


Let us imagine a box of gas in equilibrium with volume $V$ and energy $U$. And let there be a heat conducting piston that divides the volume into two parts, $V_1 = αV$ and $V_2 = (1 − α)V$. They are at some common temperature $T_1 = T_2 = T$, but not necessarily at a common pressure. The system is forced into equilibrium despite this due to a constraint, holding the piston in place. Since entropy is additive, the entropy of the combined system is just the sum: $$S=S_1\left(V_1\right)+S_2\left(V_2\right)=S(αV)+S\left((1 − α)V\right)$$ Now we let the piston move. It may no longer stay in place and α could change. But where will it settle down? Here is where the maximum entropy comes in. We are told it will settle down at the value that maximizes S: $$0=dS =dS_1+dS_2\\ \frac{\partial S_1}{\partial V_1}dV_1+ \frac{\partial S_2}{\partial V_2}dV_2\\ =\left(\frac{P_1}{T_1}-\frac{P_2}{T_2}\right) Vd\alpha\\ 0=\left(\frac{P_1-P_2}{T}\right)d\alpha $$ which is the correct physical answer: in equilibrium, when S is maximized, the pressures will be equal. So the principle of maximum entropy means that when a system held in equilibrium by a constraint becomes free to explore new equilibrium states due to the removal of the constraint, it will pick the one which maximizes S. In this example, where its options are parametrized by α, it will pick$$\frac{\partial S}{\partial\alpha}=0$$

  • $\begingroup$ Yes, this example reveals exactly the point that confuses me. As you said we start with a system (in our case box of gas) that is in equilibrium with fixed $V,U$ consisting of two subsystems with volumes $V_1= \alpha V, V_2 = (1- \alpha) V$ where at first $\alpha$ is fixed as the piston is fixed at begin; thus it's mathematically a constrain. Thus the entropy $S(U,V)$ at the beginning is maximal with respect the given constrain $\alpha$. Now we let piston move (= throw away the constrain $\alpha$) and waiting still the new system becomes an equilibrium. $\endgroup$ – katalaveino Feb 19 '20 at 2:53
  • $\begingroup$ Mathematically as you say it's a extremal problem and this is exacly what confuses me: over which argument space lives $S$ while we maximize it with respect $\alpha$ ? The naive phase space $(\{V_1, V_2, U, N) \vert V_1+V_2 =V \}$? If yes, then I don't understand the point, since in the phase space every state hast to be an equilibrium space, otherwise it cannot be expressed in a such few number of macroparameters. $\endgroup$ – katalaveino Feb 19 '20 at 2:55
  • $\begingroup$ More concretly the following point confuses me: on the one hand when we vary $\alpha$ we „fluctuate“ in the phase space and physics tells that the equilibrium state is exactly the state for which $S$ becomes maximal. On the other hnd every state in phase space is an equilibrium state as otherwise it cannot be expressed in a such few number of macroparameters. Since physics tells that only a equilibrium state can be discribed by such „compact“ set of macro paramters like $(U,V,N)$. Thus this extremal problem sounds like „we maximize a function in order to find unique equilibrium $\endgroup$ – katalaveino Feb 19 '20 at 2:55
  • $\begingroup$ state although every state in phase space is an equilibrium by argument above“. This of course absurd and that’s exactly what confuses me. Do you see my thinking error at this point? $\endgroup$ – katalaveino Feb 19 '20 at 2:57
  • $\begingroup$ Possibly I miss the whole point but would like to explain my unterstanding of the whole story with respect the $dS=0$ conditions as possibly it will reveal my thinking error that I have in order to understand your answer. Firstly the entropy $S$ (or any other thermodynamic potential) can generally be associated to every state (independent if it is in equilibrium or not). The difference is that if we have such a state in non equilibrium then it can only be described not by tree macroparameters $(U,V,N)$ but a huge number of further parameters, $\endgroup$ – katalaveino Feb 19 '20 at 3:00

The criterion $dS=0$ characterizes equilibrium, as by assumption equilibrium is the maximum entropy state, thus any fluctuation from the state will result in decrease in entropy. Thus, from Fermat's theorem (stationary points) you get that the derivative of $S$ at this state is zero.

This rule is used differently for static cases (to find equilibrium) and dynamic cases (move from one to another). For static cases you assume perfect knowledge of your single macro state $(U,V,N)$ and calculate the entropy consistent with this macroscopic configuration. Notice the assumption of closed system in the wiki definition, it means setting specific $(U,V,N)$ and finding appropriate equilibrium. For dynamic cases you change your macro state $(U,V,N)$. Notice that $S(U,V,N)$ is single valued function, that is - single value for any $(U,V,N)$, but it doesn't mean that many different states $(U,V,N)$ cannot have the same value. By proper adjustment of the macroscopic parameters you can pass through equilibrium states. That means that if you stop the process at any time - the system remains there (as opposed to perturbing the system out of equilibrium, which starts the time dependent process of equilibration)

Modern definition of equilibrium is by property of detailed balance, that is, the phase space being static, without probability currents*. Proper modern treatment of the topic comes from perspective of non-equilibrium statistical mechanics, with tools such as Fokker-Planck equation and stochastic processes.

*Notice that non-vanishing probability currents can still result in static distribution. In this case the state is not called "equilibrium" but a "non-equilibrium steady state".

  • $\begingroup$ The usage for static case I probably undrstand. Do you refer here to such kind of problem where $U,V,N$ are fixed, considered system is isolated/closed and in equilbrium and we asking if for example if this already imply eg that everywhere in the system the presure or temperature most be equal. That is we split the system in two parts $(U_1, V_1,N_1)$ and $(U_2, V_2,N_2)$ with conditions $U=U_1 +U_2$ ans the same for $V$ & $N$ and then derive $T_1=T_2$ and $p_1=p_2$ by the static ansatz & aditivity $0=dS= dS_1(U_1, V_1,N_1)+dS_2(U_2, V_2,N_2)=...$ That's what you mean by static case, right? $\endgroup$ – katalaveino Feb 18 '20 at 1:21
  • $\begingroup$ Your explanation on dynamic case I not fully understand. As you said we start with macro state $(U,V,N)$ and want to change it. Do I understand you correctly that the motivation in dynamic case is to obtain informations about $S$ by going throught the phase space configured by variables $U,V,N$ from starting "point" $(U_1, V_1,N_1)$ to another one *along a path in phase space (thus the whole time quasistatical; otherwise this "path" not make any sense) which not change $S$, thus only along equipotential paths? $\endgroup$ – katalaveino Feb 18 '20 at 1:21
  • $\begingroup$ And the motivation is that if certain paths are "alowed" we gain new information about the shape of $S$? Is this what you mean or did I misunderstood you? $\endgroup$ – katalaveino Feb 18 '20 at 1:22

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