I know of two major extremum principles:

Entropy Maximization: for a specified $(U,V)$, the equilibrium state has a higher $S$ than all non-equilibrium states, i.e., equilibrium occurs at the maximum possible $S$

Energy Minimization: for a specified $(S,V)$, the equilibrium state has a lower $U$ than all non-equilibrium states, i.e., equilibrium occurs at the minimum possible $U$.

I have also derived:

Volume Minimization: for a specified $(S,U)$, the equilibrium has a lower $V$ than all non-equilibrium states, i.e., equilibrium occurs at the minimum possible $V$.

Why haven't I heard of this before? It seems rather academic, but no more so than the energy minimization principle...

For reference, my derivation is as follows. I can rearrange the fundamental relation: \begin{align} \text{d}U &= T \text{d}S - P \text{d} V \\ P \text{d}V &= T \text{d}S - \text{d}U \\ \text{d} V &= \frac{T}{P} \text{d} S - \frac{1}{P} \text{d} U \end{align}

Then I can show that $\left( \frac{\partial V}{\partial S}\right)_U = \frac{T}{P}$, which must be positive. This means that, for a given $U$, a plot of $V$ vs $S$ for equilibrium states (the line separating "non-equilibrium" from "impossible" states) has a positive slope. We know that points left of a given equilibrium are non-equilibrium points from the entropy maximization principle. If the slope is positive, points left of one equilibrium point are also above a different equilibrium point. This means that an equilibrium point also has the lowest possible $V$ for a given $U$ and $S$ (a diagram was helpful here, but I don't have one in digital form).

  • $\begingroup$ Soap bubbles and balloons... you have heard it before, you just can't seem to recall it at the moment. $\endgroup$ – CuriousOne Jan 31 '16 at 4:24
  • $\begingroup$ Bubbles and balloons would do boundary work if they expanded - wouldn't that mean they don't have constant U? $\endgroup$ – user1476176 Feb 1 '16 at 5:53
  • $\begingroup$ If you have a boundary it has to have a boundary term. I don't see a problem with that. $\endgroup$ – CuriousOne Feb 1 '16 at 8:40

I think that is ok; only problem, V is usually a constraint from outside, while S cannot be controlled and floats freely to set itself to optimal value. In practice, it is difficult to keep constant U and specially S, while letting V find the equilibrium! A realization of your discovery may be this example:

Take a rarefied gas in a rectangular box, with no collision between molecules; they only bounces against the wall, perfectly as billiard balls. Say they have a preferential direction, they move faster in the $z$ direction than in $x, y$. Thus, there is more pressure on the walls perpendicular to $z$ than those against $x,y$. We can expand the box in the $z$ direction while compressing it in the $x,y$, such that no net work is done. We are stealing energy to the $z$ movement of molecules and putting it in the $x,y$, until $x, y, z$ components became equivalent.

We should have ended up with same energy $U$ (no external work) and same entropy $S$ (since there are no collisions between molecules, no relaxation mechanism). However the velocity distribution is now uniform in $x, y, z$, which has a larger entropy contribution than the initial $z$-oriented. There has been a decrease in entropy contribution from the volume, meaning the volume has shrinked.

Is the status reached an equilibrium? I don't know. Since at $U, S$ constant we must have $PdV=0 $, and we want to move $V$, I tried to invent an example that start with $P$ not uniform (molecules moving around $z$).


All 3 corollaries are valid. I think the reason why it is not mentioned even is the fact of its limited applicability compared to the maximum entropy principle. Most books of physical chemistry don't even mention the energy minimum principle. Suppose that a system undergoes a change at constant U like the isotermal expansion of a ideal gas inside an isolated vessel at constant V. How can You establish if the system has attained the equilibrium using the energy/volume corollaries? U and V are at a minimum since the beginning of the expansion... Same thing for all isochoric/isoenergetic transformations. It is the entropy the true indicator that allows to follow the evolution of a system toward equilibrium.

The proof of the equivalence of the two extremum criteria for S and U can be formulated either as a physical or mathematical argument.

  1. $dS_{U,V , n_i} = 0$
  2. $d^2 S_{U,V, n_i} \le 0$

the first equation says that the first derivative vanishes at an extremum, whereas the second one says that, if the second derivative is negative at the extremum, then the extremum is a maximum.

it can be shown that

\begin{equation} \left(\frac{\partial U} {\partial X} \right)_{S}= -T \left(\frac{\partial U} {\partial X} \right)_{U} \end{equation}

if $(\partial S/\partial X)_U = 0$ vanishes, the so does $(\partial U/\partial X)_U = S$

where $X$ can be any of the variable $V,n_i$

Thus an extremeum for entropy at constant internal energy is also an extremum for internal energy at constant entropy and by calculating the second derivative one finds that:

\begin{equation} \left(\frac{\partial U} {\partial X} \right)_{S}= -T \left(\frac{\partial^2 S} {\partial X^2} \right)_{U} > 0 \end{equation}

The Second Law of Thermodynamics therefore leads to a minimum energy principle: at equilibrium in a constant entropy system the internal energy takes it minimum possible value. In the notation of thermodynamics we write the minimum energy as

  1. $dU_{S,V , n_i} =0$

  2. $d^2 U_{S,V,n_i} \ge 0$

from a mathematical perspective considering now the volume the situation is analogous as can be seen by the correspondence of the differentials of U and V

  1. $dU = -P dV + T dS$

  2. $dV = -\frac{1}{P} dU +\frac{T}{P} dS$

we could graph both functions for the minimum (here shown for U) as

enter image description here

The equilibrium state A as a point of minimum U for constant S.

And write

$dV_{S,U , n_i} =0$

$d^2V_{S,U,n_i} \ge 0$

Now I will show the consistency of the energy minimum principle and of the minimum volume principle. Starting from the energy minimum principle; But I want to make a consideration first.

The entropy maximum and the energy minimum principles are both valid definitions of thermodynamic equilibrium. It is necessary to be extremely careful here, though. The equilibrium state defined by equations (3) and (4) and (5) (6) is not the same as the one defined by equations (1) and (2). It cannot possibly be, as in the first case we are dealing with an isolated system and in the second case we are not. If entropy were kept constant in an isolated system then the system would not change and the minimum energy principle would be meaningless, because internal energy would not be able to change. In order for constant entropy minimization of the internal energy to be possible, the system of interest must be part of a larger isolated system in which entropy does increase.

Consider the example of the two bodies forming the system A at different temperatures, $T_2 > T_1$, shown now in the Fig. below; we want to achieve thermal equilibrium at constant entropy and volume, then system A cannot be isolated. It must be able to exchange energy with its surroundings, which conform an isolated system, labeled B in the figure, and the entropy of B must increase. In order for the entropy of B to increase there must be heat transfer between A and the rest of B. Because equilibrium of A at constant entropy requires that its internal energy be minimized, heat must be transferred from A to its surroundings. Therefore, the final temperature of A if it is allowed to reach equilibrium at constant entropy must be lower than its final temperature if it reaches equilibrium at constant internal energy.

enter image description here

To show the consistency of the volume minimum principle from a physical perspective, consider the same system A now isoentropic and also isoenergetic and also compressible, initially not in mechanical equilibrium ($P^A < P^{\text{ext}}$) since it has not reached the minimum possible volume. In order to attain the equilibrium work should be done on the system compressing it and heat should be transferred from A to its surroundings to keep the system at S and U constant;

  • $\begingroup$ Thank you for pointing out the algebra error. I have edited the question to correct it. I think that you have misunderstood my question though; I mentioned the cases where $U$ and $V$ or $S$ and $V$ are fixed for reference, but I am asking about the case where $U$ and $S$ are fixed. It seems that this case would complete a triad, but I have never seen this discussed before and my question is why. $\endgroup$ – user1476176 Apr 14 '19 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.