In terms of simplicity, purely mechanical systems have an advantage over thermodynamic systems in that stability and instability can be defined solely in terms of potential energy. For example the center of mass of the tower at Pisa, in its present state, must be higher than in some infinitely near positions, so we can conclude that the structure is not in stable equilibrium. This will only be the case if the tower attains the condition of metastability by returning to a vertical position or absolute stability by exceeding the tipping point and falling over.

Thermodynamic systems lack this simplicity, but in common with purely mechanical systems, thermodynamic equilibria are always metastable or stable, and never unstable. This is equivalent to saying that *every spontaneous (observable) process proceeds towards an equilibrium state, never away from it*.

If we restrict our attention to a thermodynamic system of unchanging composition and apply the further constraint that the system is closed, i.e. no mass transfer takes place between system and surroundings, the fundamental relation of thermodynamics can be written as an exact differential expression exhibiting two independent variables: dU = TdS–PdV. There are therefore 2^{2}–1 possible Legendre transformations, each representing a new function. We thus arrive at a set of four state functions

where the independent variables, shown in parentheses, are the natural (canonical) variables for that function [S = entropy, V = volume, T = temperature, P = pressure].

If the above pairs of natural variables are held constant, the energy function is predictably a minimum at equilibrium. But what if the energy function and one of its natural variables are instead held constant? Which extremum will the other natural variable reach at equilibrium – minimum or maximum? There are eight permutations, giving a grand total of twelve sets of equilibrium conditions.

Of these twelve, only four have straightforward proofs based on the fundamental criteria [derived in my previous blogpost *Reversible and Irreversible Change*] that

The proofs of the other eight involve a lengthier train of logical argument which is now rarely seen either in textbooks or on the internet. To redress this issue, CarnotCycle herewith presents * all twelve proofs in their entirety*.

**– – – –**

**The Twelve Conditions of Equilibrium and Stability**

Here are the four straightforward proofs

**1. For given U and V that S is a maximum**

For a closed system, the First Law may be written

Consider an adiabatic change: dq = 0 so the above criteria for equilibrium (1) and observable change (2) become dS = 0 and dS > 0 respectively. If the volume is also kept constant during the change, it follows from the First Law that dU = 0. The criteria for these conditions may be expressed as follows

Since observable change always proceeds towards equilibrium, S will increase towards a maximum at equilibrium when U and V are held constant.

**2. For given H and P that S is a maximum**

For a closed system, the First Law may be written

Consider an adiabatic change: dq = 0 so the above criteria for equilibrium (1) and observable change (2) become dS = 0 and dS > 0 respectively. If the pressure P of the system is kept constant during the change, and is made to differ only infinitesimally from that of the surroundings, equation (3) becomes

By the definition of enthalpy, we find

In this case therefore, H and P are constant, and the criteria for these conditions may be expressed as follows

Since observable change always proceeds towards equilibrium, S will increase towards a maximum at equilibrium when H and P are held constant.

**3. For given T and V that A is a minimum**

For a closed system undergoing a change at constant temperature, we may write the differential expression

But by the first law

so that

Suppose the system is capable only of PV work. If the volume is kept constant, no work can be done and therefore

Applying criteria (1) and (2) for equilibrium (dq = TdS) and observable change (dq < TdS) respectively, we obtain the criteria in terms of A:

Since observable change always proceeds towards equilibrium, A will decrease towards a minimum at equilibrium when T and V are held constant.

**4. For given T and P that G is a minimum**

For a closed system undergoing a change at constant temperature, we may write the differential expression (cf. equation 4 above)

Suppose the system is capable only of PV work. If the pressure P of the surroundings is kept constant during the change, and is made to differ only infinitesimally from that of the system, then

Applying criteria (1) and (2) for equilibrium (dq = TdS) and observable change (dq < TdS) respectively, we obtain the criteria in terms of G:

Since observable change always proceeds towards equilibrium, G will decrease towards a minimum at equilibrium when T and P are held constant.

**– – – –**

* A word of introduction *to the remaining eight not-so-straightforward proofs.

On page 2 of his masterwork *On the Equilibrium of Heterogeneous Substances*, under the heading *Criteria of Equilibrium and Stability*, J. Willard Gibbs constructs the foundations of chemical thermodynamics on the equivalence of two propositions relating to an isolated system – that for given U and V that S is a maximum, and for given S and V that U is a minimum.

Gibbs shows by neat argument that the truth (according to Clausius) of the first proposition necessitates the truth of the second proposition. In parallel fashion, the proof already given for the first proposition (condition 1 above) will be used to prove that for given S and V that U is a minimum. This sequitur argument repeats throughout, with condition 2 used to prove condition 6, condition 3 used to prove condition 7 etc.

The train of logical argument for these remaining proofs is somewhat labyrinthine, but the method is same in each case. It gets easier to follow as you go from one to the next.

**– – – –**

**5. For given S and V that U is a minimum**

Imagine a closed system in some state of equilibrium for which the values of the entropy, volume and internal energy are S1, V1 and U1. Then in any non-equilibrium state with the same values V1 and U1, the entropy S2 must be less than S1 because observable change always proceeds towards equilibrium, and the condition (1) of observable change at constant U and V requires entropy to increase.

Now imagine a second equilibrium state for which the values of the entropy, volume and internal energy are S2, V1 and U2. Compared with the first equilibrium state the volume is unaltered but S2 < S1. Now U and S always change in the same direction since dU = TdS – PdV implies (∂U/∂S)_{V} = T which is always positive. Hence in the second equilibrium state U2 < U1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the entropy and volume are the same but U2 < U1. Since observable change always proceeds towards equilibrium, U will decrease towards a minimum at equilibrium when S and V are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

**6. For given S and P that H is a minimum**

Imagine a closed system in some state of equilibrium for which the values of the entropy, pressure and enthalpy are S1, P1 and H1. Then in any non-equilibrium state with the same values P1 and H1, the entropy S2 must be less than S1 because observable change always proceeds towards equilibrium, and the condition (2) of observable change at constant H and P requires entropy to increase.

Now imagine a second equilibrium state for which the values of the entropy, pressure and enthalpy are S2, P1 and H2. Compared with the first equilibrium state the pressure is unaltered but S2 < S1. Now H and S always change in the same direction since dH = TdS + VdP implies (∂H/∂S)_{P} = T which is always positive. Hence in the second equilibrium state H2 < H1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the entropy and pressure are the same but H2 < H1. Since observable change always proceeds towards equilibrium, H will decrease towards a minimum at equilibrium when S and P are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

**7. For given A and V that T is a minimum**

Imagine a closed system in some state of equilibrium for which the values of the Helmholtz free energy, volume and temperature are A1, V1 and T1. Then in any non-equilibrium state with the same values V1 and T1, the Helmholtz free energy A2 must be greater than A1 because observable change always proceeds towards equilibrium, and the condition (3) of observable change at constant T and V requires the Helmholtz free energy to decrease.

Now imagine a second equilibrium state for which the values of the Helmholtz free energy, volume and temperature are A2, V1 and T2. Compared with the first equilibrium state the volume is unaltered but A2 > A1. Now A and T always change in opposite directions since dA = –SdT – PdV implies (∂A/∂T)_{V} = –S which is always negative. Hence in the second equilibrium state T2 < T1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Helmholtz free energy and volume are the same but T2 < T1. Since observable change always proceeds towards equilibrium, T will decrease towards a minimum at equilibrium when A and V are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

**8. For given G and P that T is a minimum**

Imagine a closed system in some state of equilibrium for which the values of the Gibbs free energy, pressure and temperature are G1, P1 and T1. Then in any non-equilibrium state with the same values P1 and T1, the Gibbs free energy G2 must be greater than G1 because observable change always proceeds towards equilibrium, and the condition (4) of observable change at constant T and P requires the Gibbs free energy to decrease.

Now imagine a second equilibrium state for which the values of the Gibbs free energy, pressure and temperature are G2, P1 and T2. Compared with the first equilibrium state the pressure is unaltered but G2 > G1. Now G and T always change in opposite directions since dG = VdP – SdT implies (∂G/∂T)_{P} = –S which is always negative. Hence in the second equilibrium state T2 < T1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Gibbs free energy and pressure are the same but T2 < T1. Since observable change always proceeds towards equilibrium, T will decrease towards a minimum at equilibrium when G and P are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

**9. For given U and S that V is a minimum**

Imagine a closed system in some state of equilibrium for which the values of the internal energy, entropy and volume are U1, S1 and V1. Then in any non-equilibrium state with the same values S1 and V1, the internal energy U2 must be greater than U1 because observable change always proceeds towards equilibrium, and the condition (5) of observable change at constant S and V requires the internal energy to decrease.

Now imagine a second equilibrium state for which the values of the internal energy, entropy and volume are U2, S1 and V2. Compared with the first equilibrium state the entropy is unaltered but U2>U1. Now U and V always change in opposite directions since dU = TdS – PdV implies (∂U/∂V)_{P} = –P which is always negative. Hence in the second equilibrium state V2 < V1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the internal energy and entropy are the same but V2 < V1. Since observable change always proceeds towards equilibrium, V will decrease towards a minimum at equilibrium when U and S are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

**10. For given H and S that P is a maximum**

Imagine a closed system in some state of equilibrium for which the values of the enthalpy, entropy and pressure are H1, S1 and P1. Then in any non-equilibrium state with the same values S1 and P1, the enthalpy H2 must be greater than H1 because observable change always proceeds towards equilibrium, and the condition (6) of observable change at constant S and P requires enthalpy to decrease.

Now imagine a second equilibrium state for which the values of the enthalpy, entropy and pressure are H2, S1 and P2. Compared with the first equilibrium state the entropy is unaltered but H2 > H1. Now H and P always change in the same direction since dH = TdS + VdP implies (∂H/∂P)_{S} = V which is always positive. Hence in the second equilibrium state P2 > P1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the enthalpy and entropy are the same but P2 > P1. Since observable change always proceeds towards equilibrium, P will increase towards a maximum at equilibrium when H and S are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

**11. For given A and T that V is a minimum**

Imagine a closed system in some state of equilibrium for which the values of the Helmholtz free energy, temperature and volume are A1, T1 and V1. Then in any non-equilibrium state with the same values T1 and V1, the Helmholtz free energy A2 must be greater than A1 because observable change always proceeds towards equilibrium, and the condition (3) of observable change at constant T and V requires the Helmholtz free energy to decrease.

Now imagine a second equilibrium state for which the values of the Helmholtz free energy, temperature and volume are A2, T1 and V2. Compared with the first equilibrium state the temperature is unaltered but A2 > A1. Now A and V always change in opposite directions since dA = –SdT – PdV implies (∂A/∂V)_{T} = –P which is always negative. Hence in the second equilibrium state V2 < V1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Helmholtz free energy and temperature are the same but V2 < V1. Since observable change always proceeds towards equilibrium, V will decrease towards a minimum at equilibrium when A and T are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

**12. For given G and T that P is a maximum**

Imagine a closed system in some state of equilibrium for which the values of the Gibbs free energy, temperature and pressure are G1, T1 and P1. Then in any non-equilibrium state with the same values T1 and P1, the Gibbs free energy G2 must be greater than G1 because observable change always proceeds towards equilibrium, and the condition (4) of observable change at constant T and P requires the Gibbs free energy to decrease.

Now imagine a second equilibrium state for which the values of the Gibbs free energy, temperature and pressure are G2, T1 and P2. Compared with the first equilibrium state the temperature is unaltered but G2 > G1. Now G and P always change in the same direction since dG = VdP – SdT implies (∂G/∂P)_{T} = V which is always positive. Hence in the second equilibrium state P2 > P1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Gibbs free energy and temperature are the same but P2 > P1. Since observable change always proceeds towards equilibrium, P will increase towards a maximum at equilibrium when G and T are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

**– – – –**

Rather over my head, but interesting to me nonetheless. :)

Reblogged this on nebusresearch and commented:

This month Peter Mander’s CarnotCycle blog talks about the interesting world of statistical equilibriums. And particularly it talks about stable equilibriums. A system’s in equilibrium if it isn’t going to change over time. It’s in a stable equilibrium if being pushed a little bit out of equilibrium isn’t going to make the system unpredictable.

For simple physical problems these are easy to understand. For example, a marble resting at the bottom of a spherical bowl is in a stable equilibrium. At the exact bottom of the bowl, the marble won’t roll away. If you give the marble a little nudge, it’ll roll around, but it’ll stay near where it started. A marble sitting on the top of a sphere is in an equilibrium — if it’s perfectly balanced it’ll stay where it is — but it’s not a stable one. Give the marble a nudge and it’ll roll away, never to come back.

In statistical mechanics we look at complicated physical systems, ones with thousands or millions or even really huge numbers of particles interacting. But there are still equilibriums, some stable, some not. In these, stuff will still happen, but the kind of behavior doesn’t change. Think of a steadily-flowing river: none of the water is staying still, or close to it, but the river isn’t changing.

CarnotCycle describes how to tell, from properties like temperature and pressure and entropy, when systems are in a stable equilibrium. These are properties that don’t tell us a lot about what any particular particle is doing, but they can describe the whole system well. The essay is higher-level than usual for my blog. But if you’re taking a statistical mechanics or thermodynamics course this is just the sort of essay you’ll find useful.