# How to think about thermodynamic equilibrium in terms of virtual displacements?

I am currently taking thermodynamics and am finding the concept of thermodynamic equilibrium rather difficult to grasp. My textbook gives a definition in terms of the virtual displacement of with respect to virtual displacements of variable $$x_1,x_2,\dots,x_n$$ and writes

$$\delta S = \left(\frac{\partial S}{\partial x_1}\right)_{x_2,x_3,\dots,x_n} \delta x_1 + \left(\frac{\partial S}{\partial x_2}\right)_{x_1,x_3,\dots,x_n} \delta x_2 + \dots + \left(\frac{\partial S}{\partial x_n}\right)_{x_1,x_2,\dots,x_{n-1}} \delta x_n.$$

From a purely mathematical standpoint, this definition makes sense: essentially it is the differential change in $$S$$ we would expect for a set of mathematically conceivable (but not necessarily physically possible) displacements $$\delta x_1,\dots,\delta x_n$$. The book then states that the equilibrium condition for an isolated system is that the virtual displacement of entropy at constant internal energy and volume is nonpositive: $$(\delta S)_{U,V}\leq 0.$$ I am, however, unsure of what exactly constitutes a virtual displacement. For example, are we allowed to consider a virtual displacement in temperature $$\delta T$$? Physically speaking, this is not possible at constant $$V$$ and $$U$$, but it seems like a mathematically conceivable displacement. One answer I saw referenced dividing the system into two parts and considering displacements where the subsystems exchanged energy so that the total energy of the system would remain constant. I do feel, however, that there should be more displacements than just those that are obtained from exchanges between parts of the system. I am hoping someone could explain what exactly constitutes (and does not constitute) a virtual displacement.

Furthermore, my second point of confusion is the following: suppose our system is in equilibrium and we have a set of virtual displacements $$\delta x_1, \delta x_2, \dots, \delta x_n$$ such that $$(\delta S)_{U,V}\leq 0$$ (this is allowed by our equilibrium condition). Then, if we instead considered the displacements $$-\delta x_1, -\delta x_2, \dots, -\delta x_n$$, shouldn't we have $$(\delta S)_{U,V} >0$$ by definition. This seems like a contradiction to the equilibrium condition that I don't see how to resolve.

I apologize for the length of this question and thank you in advance for your help!

• Thermodynamics is a field plagued with bad textbooks. I would recommend that you try H.B. Callen's book instead of trying to disentangle what your current book is saying. If you are an engineer then you may like the one by Van Wylen and Sontagg.
– Deep
Oct 7, 2023 at 5:47
• Oct 7, 2023 at 15:09

What you are asking is what does it mean that in an isolated system in equilibrium the entropy $$S=S(X_0, X_1,X_2, ..X_n)$$ is "maximum" where $$X_0, X_1,X_2,...X_n$$ are the extensive variables, say, internal energy $$U=X_0$$, volume $$V=X_1$$, mass of chemical species "1" $$m_1=X_2$$, species "2" $$m_2=X_3$$, etc., characterizing the conserved quantities the thermodynamic body may exchange with its environment. How can $$S$$ be varying when we already have demanded that the system be isolated, that is all the conserved quantities are the same?

The only reasonable way to interpret the principle of maximum is that we start from an isolated but non-equilibrium state and follow the evolution of entropy as we keep the variables $$X_0,X_1, X_n$$ constant. But that just means that either nothing happens or we acknowledge that the system when not in equilibrium cannot be described by this set of numbers alone. If the isolated body internally is homogeneous, where homogeneity refers to the spatial distribution of all the intensive parameters together with the local distribution of the corresponding extensive parameters, then nothing will happen and it stays in equilibrium. If, on the other hand, there are internal inhomogeneities, then the evolution of the entropy will be such that at no instant may the total entropy decrease. For this statement to make sense we must require that entropy be definable and be defined for a system that is not in equilibrium be it isolated or not.

One way to achieve that is by assuming so-called local equilibrium. That means dividing the system into sufficiently small but homogeneous pieces over which the equilibrium entropy function can be applied and then use the gradient of the intensive parameters between these homogeneous pieces that are not equilibrium with each other to calculate entropy production. Whether this is a valid assumption or not depends on the system itself and cannot be declared to be valid always.

But when the assumption of local equilibrium is valid then it makes sense to talk about a total entropy at any given instant during the system's evolution. This just means that the local equilibrium assumption is a sufficient condition for a global entropy function but it is not necessary. (There are systems where local equilibrium does not hold and one can still assign a total entropy, but that is another story.)

Now when when we start from dis-equilibrium and we have a total entropy function then the entropy principle states that the spontaneous process of the isolated system terminates in an equilibrium when the total entropy reaches its maximum.

The simplest way I know of formalizing this statement is to say that in all processes during which the system or a piece of the system $$\delta X_k$$ extensive quantity is transported through a gradient of the conjugate intensive $$\delta Y_k$$, it performs an element of work $$\delta Y_k \delta X_k$$ so that the total work performed by that piece of the system whose temperature is $$T\pm \delta T$$ is $$T\sigma =\delta T \delta S + \sum_{k=1}^n \delta Y_k \delta X_k$$ with $$\sigma > 0$$ representing the irreversibly entropy generated in that homogenous piece. If $$\sigma = 0$$ then the local process is reversible.

When the transported pieces are all zero $$\delta S=0$$ and $$\delta X_k=0, k=1,2,..n$$ then of course $$\sigma =0$$ and nothing happens, and if that holds for all pieces then, per force, the total entropy change is zero and also you have equilibrium.

If on the other hand a variation leads to an increase in the total entropy, $$\sum \sigma >0$$, since in all localities $$\sigma \ge 0$$ then it is an allowed process and it will reach equilibrium until it can increase the total entropy, at which point it reaches its maximum, the equilibrium value as represented by the internally homogeneous distribution of its extensive and intensive parameters.

Since, by definition, in an equilibrium nothing happens, when we say that for all virtual "displacement" $$\delta S_{tot} <0$$ we mean that these virtual displacement in $$X_1, X_2, ...$$ while they must be compatible with all the external constraints but may allow for arbitrary unconstrained local infinitesimal variations. For example, while in an isolated system we constrain the total mass, we may allow for local spontaneous density (concentration) variation (fluctuation). Every such virtual variation must be the such that reduces the total entropy if it started from equilibrium that is represented by the maximum of the total entropy.

This, of course, implies a kind of stability of the equilibrium state against small, infinitesimal, fluctuations, for the system is forced to go back to the total entropy maximum. And it is also at the heart of the minimum entropy production principle, as long as we have local equilibrium, because the evolution towards maximum tends to reduce the gradients that themselves are the causes of the entropy production.