# Thermodynamic equilibrium with fixed energy/entropy

• Jonk
Jonk
If you take a system with fixed entropy S0 and let it evolve, it reachs equilibrium. Let Ueq be the energy of the system at equilibrium.
Now take the same system with fixed energy U=Ueq (S is not fixed anymore), how do you know that the equilibrium reached is the same as before, that means with entropy S0?

(I'm reading Callen's book on thermodynamics and he uses that fact in chapter 5 to prove the equivalence between the maximum entropy principle and the minimum energy principle.)

Mentor
If you take a system with fixed entropy S0 and let it evolve, it reachs equilibrium. Let Ueq be the energy of the system at equilibrium.
Now take the same system with fixed energy U=Ueq (S is not fixed anymore), how do you know that the equilibrium reached is the same as before, that means with entropy S0?

(I'm reading Callen's book on thermodynamics and he uses that fact in chapter 5 to prove the equivalence between the maximum entropy principle and the minimum energy principle.)
Maybe you can provide more information on exactly what is said. As you post stands, I doesn't make sense (to me).

vanhees71
Jonk
Yes of course.

The goal is to prove the implication Entropy maximum principle => Energy maximum principle. As a reminder,

Entropy maximum principle : The equilibrium value of any unconstrained internal parameter is such as to maximize the entropy for the given value of the total internal energy.
Energy minimum principle : The equilibrium value of any unconstrained internal parameter is such as to minimize the energy for the given value of the total entropy.

Callen shows rigorously that
$$\dfrac{\partial S}{\partial X}=0 \mathrm{~~and~~} \dfrac{\partial^2S}{\partial X^2}<0 \implies \dfrac{\partial U}{\partial X}=0 \mathrm{~~and~~} \dfrac{\partial^2U}{\partial X^2}>0$$

Hence a maximum of S at constant U corresponds to a minimum of U at constant S.

Now it should be enough to show that a system at constant S will be at equilibrium when $$\dfrac{\partial S}{\partial X}=0 \mathrm{~~and~~} \dfrac{\partial^2S}{\partial X^2}<0$$ since it would imply that U is minimum.

We supposed that S is constant (since we want to show the energy minimum principle), but U isn't constant.
The entropy maximum principle requires U constant, so how to use it here?
That's what I don't understand.

Mentor
Can you give an example of a process in which the entropy is constant, and release of an unconstrained parameter results in the internal energy reaching a minimum at equilibrium?

Staff Emeritus
Are you perhaps asking about the difference between a closed system and an isolated system?

https://en.wikipedia.org/wiki/Principle_of_minimum_energy
The principle of minimum energy is essentially a restatement of the second law of thermodynamics. It states that for a closed system, with constant external parameters and entropy, the internal energy will decrease and approach a minimum value at equilibrium. External parameters generally means the volume, but may include other parameters which are specified externally, such as a constant magnetic field.

In contrast, for isolated systems (and fixed external parameters), the second law states that the entropy will increase to a maximum value at equilibrium. An isolated system has a fixed total energy and mass.
A closed system, on the other hand, is a system which is connected to another, and cannot exchange matter (i.e. particles), but other forms of energy (e.g. heat), with the other system. If, rather than an isolated system, we have a closed system, in which the entropy rather than the energy remains constant, then it follows from the first and second laws of thermodynamics that the energy of that system will drop to a minimum value at equilibrium, transferring its energy to the other system. To restate:

• The maximum entropy principle: For a closed system with fixed internal energy (i.e. an isolated system), the entropy is maximized at equilibrium.
• The minimum energy principle: For a closed system with fixed entropy, the total energy is minimized at equilibrium.

Jonk
@Chestermiller an ideal gas in a vertical and adiabatic piston covered with sand will reach a minimum of U if the sand is removed progressively (to ensure that the process is quasistatic).

@anorlunda I got your point about isolated and closed systems. Could you please give a proof of the energy minimum principle using the first and second laws?

Callen's proof is as follow:
Consider a closed system with fixed total entropy and wait for it to be in equilibrium.
The entropy maximum principle states that this equilibrium corresponds to a maximum of S at fixed U.
It has been shown that a maximum of S at fixed U corresponds to a minimum of U at fixed entropy.
Thus this equilibrium corresponds to a minimum of U at fixed S, which completes the proof.

So he uses the entropy maximum principle for an evolution where U isn't constant (and S is constant)!
What am I missing here?