# Helmholtz free energy related query.

1. Oct 28, 2012

### A Dhingra

hello

This is a section from Callen, Herbert B - Thermodynamics and an Introduction to Thermostatistics
"Any equilibrium state can be characterized either as a state of maximum entropy for given energy or as a state of minimum energy for given entropy. But these two criteria nevertheless suggest two different ways of attaining equilibrium. Let us consider a piston originally fixed at some point in a closed cylinder. We are interested in bringing the system to equilibrium without the constraint on the position of the piston. We can simply remove the constraint and allow the equilibrium to establish itself spontaneously. Here the entropy increases and the energy is maintained constant by the closure condition. This process is suggested by the maximum entropy principle. Alternatively, we can permit the piston to move very slowly, reversibly doing work on an external agent until it has moved to the position that equalises the pressure on the two sides. During this process energy is withdrawn from the system and entropy remains a constant (the process is reversible and no heat flows).This is the process suggested by the minimum energy principle.
Independent of whether the equilibrium is brought about by either of these two processes, or by any other process, the final equilibrium state in each case satisfies both extrema conditions."

According to the two processes described in the text,
the first process refers to this till the equilibrium state is reached: (∂S/∂V)T
and the second process refers to (∂P/∂T)V

And according to the Maxwell's relations (the Helmholtz free energy minimized ) these two are equal. But according to the last statement : both of these satisfy the extremal condition, but need not be necessarily equal.
So please tell me what is to be added in the theory so that the equivalence seems visible for a process. (i am not asking for a mathematical proof, the one based on the definition of exact differentials)
Or someone please help me visualize that these two processes are actually same (also in what respect as per the Maxwell's relations?)

Thank you for any help

(I was confused in this section hence had to put it here, but i had no intention of copyright violation in any way)

2. Oct 28, 2012

### Studiot

I have never heard of the text you refer to, but is the author really suggesting that both the value of entropy and energy can be specified at the same time?

Are you thinking that these two criteria can be satisfied simultaneously?

3. Oct 29, 2012

### A Dhingra

Last edited by a moderator: May 6, 2017
4. Oct 30, 2012

### pabloenigma

The text you are referring to is the ultimate book of thermodynamics,in my opinion,and also the most frequently cited source in research literature.

However,I dont really get where you bring in the Helmoltz function.The author says,that minimizing the energy holding the entropy constant gets you to the equilibrium state,the same equilibrium state that you would have reached if you had constrained the energy to a constant value(by isolating etc) and maximizing the entropy.

Callen himself has mentioned the logical similarity of this thing with the isoperimetric problem in geometry.The Circle(analogous to equilibrium state) can be characterized either as the 2 dimensional figure of maximum area for a given perimeter,or of a minimum perimeter for a given area..Ref section 5-1

However I dont know where the Helmholtz function comes into this paragraph,and could you please throw more light on the same

5. Oct 30, 2012

### A Dhingra

After reading that section from the book,i just tried to put his words in equation, i mean the first process he has described is allowing the piston move, so for sure letting the volume occupied by the gas change, and (since volume change for sure affects the entropy of the gas) there should be respective change in S. Also it is said that "Here the entropy increases and the energy is maintained constant by the closure condition" , i suppose that the energy they are talking about is the total energy or the internal energy as we call it in thermodynamics. At least for an ideal gas, this can be a constant with the temperature kept constant. So lead me to (∂S/∂V)T
For the second process he said change in pressure is allowed so i jumped on to (∂P/∂T)V.
(I think i used the Maxwell's relation somewhere in my brain trying to fit them here, as i was trying to see these equations being equivalent for two processes, i guess i jumped on conclusions.....)
So can you now help me with two processes (of the Maxwell's relations) and prove them to be equivalent in non mathematical way..

6. Oct 30, 2012

### A Dhingra

Well thanks for making me see that i used the very relations I wanted to see proved!

7. Oct 30, 2012

### Studiot

That book is frighteningly expensive.

I hope it actually makes better statements about equilibrium.

Is a system for which entropy is a maximum or the internal energy is a minimum, but the volume is not constant in equilibrium?

The answer to this has significant implications for the piston in a cylinder described in post#1

If the piston is a diathermal piston then the condition of maximum entropy corresponds to equilibrium. But if the piston is an adiabatic one then this is not the case, since

dStotal = dSleft + dSright = 0

for any position of the piston when the constraint is removed and the piston is free to move under the influence of the substances in the chambers.

Where left and right refer to the chambers either side of the piston and the total refers to the whole system.

Last edited: Oct 30, 2012
8. Oct 30, 2012

### pabloenigma

Actually,in the second process,volume is not constant,its expanding slowly,so you cant use that derivative.And trust me,this has got nothing to do with Maxwells relations.Also,this paragraph tries to explain the logical consistency of the fundamental postulates of theory.Its not right to try to justify the fundamental postulates in term of Maxwells relations,which is just a consequence of postulates.

Eitherway,even if the derivatives were right,there is no reason that they are equal.It just says,that both processes lead to the same final state,theres no reason that two random derivatives of the changing variables should be equal.

9. Oct 31, 2012

### pabloenigma

Well its very cheap in India.Indian edition costs about 300 bucks in Indian currency,which will be about 6 dollars,hahha

It is downright silly to debate the authenticity or correctness of HB Callen.Obviously Volume is constant.I didnt make a good job of expounding what Callen says.I didnt quote the book.I just assumed A Dhingra was reading the book,so he will get what Im saying.

Callen says this : There exists Entropy defined for all equilibrium states,such that The values assumed by the extensive parameters in absence of internal constraints(diathermaic,adiabatic whatever) are those that maximize the entropy over the manifold of constrained equilibrium states.

Now,if I understand your point correctly,you are talking about a movable adiabatic wall. This a subtle from the point of view of Entropy maximum formalism,but can be done nonetheless.Writing dS=(1/T)dU+PdV,it can be shown that maximizing Entropy leads to the Mechanical equilibrium condition P1=P2.

10. Oct 31, 2012

### Studiot

Pabloenigma

You did indeed understand my model, but not my point.

Whilst I agree with you that the equilibrium condition is equalty of pressures, it is very easy to glibly state "It can be shown that................."

I think you cannot prove it using entropy maximization since dS = 0.

Let us consider a sealed pipe containing a frictionless, adiabtic piston which divides the pipe into to chambers A and B and inviscid gasses in each chamber.

Now let the volume of chamber A at equilibrium be Va and consider a small change of state from equilibrium, to Va + dVa

$$d{S_a} = \frac{{d{U_{}}}}{{{T_a}}} + \frac{{{P_a}}}{{{T_a}}}d{V_a} = 0$$

since

$$d{U_a} = - {P_a}d{V_a}$$

Similarly dSb = 0.

So you cannot use the maximum entropy condition to specify equilibrium.

Last edited: Oct 31, 2012
11. Oct 31, 2012

### Studiot

A Dhingra

The thread carries the name Helmholtz free energy in the title and suggests some confusion embodied in the textbook concerning equilibrium.

The following may be of use when reading the textbook.

Starting with the following second law condition

$$\begin{array}{l} \Delta S \ge \int {\frac{{\delta q}}{T}} \\ \delta q \le TdS \\ \end{array}$$

Thus

$$dU - TdS + PdV \le 0$$

With the equality sign holding at equilibrium,
From which can be derived a shedload of equilibrium conditions:-

Substituting H=U+PV

$$dH - TdS - VdP \le 0$$

Thus at constant entropy and pressure

$${\left( {\Delta H} \right)_{SP}} \le 0$$

A system at constant entropy and pressure is in equilibrium when its enthalpy is at a minimum.

Substituting A=U-TS

$$dA + SdT + PdV \le 0$$

Or

$${\left( {\Delta A} \right)_{TV}} \le 0$$

A system at constant T and V is in equilibrium when its Helmholtz free energy is a minimum

Substituting G=U+PV-TS

$$dG + SdT - VdP \le 0$$

A system at constant T and P is in equilibrium when its Gibbs free energy is a minimum

$${\left( {\Delta G} \right)_{TP}} \le 0$$

Thus we have the following set of 5 alternative conditions with equality holding for equilibrium. They may not all be available as shown in post#10

$$\begin{array}{l} {\left( {\Delta S} \right)_{UV}} \ge 0 \\ {\left( {\Delta U} \right)_{SV}} \le 0 \\ {\left( {\Delta H} \right)_{SP}} \le 0 \\ {\left( {\Delta A} \right)_{TV}} \le 0 \\ {\left( {\Delta G} \right)_{TP}} \le 0 \\ \end{array}$$

Note you always need to specify two variables to define equilibrium. This was one of my original points.

Last edited by a moderator: Nov 1, 2012
12. Oct 31, 2012

### pabloenigma

My bad.I apologize.I think I screwed up. But P1=P2 is a result of energy conservation for isolated system,so that holds.The entropy maximum formalism encompasses that,since dU1=-dU2 is part of constraint imposed by isolation of the system. You can say that the entropy maximizing yields nothing NEW,since dS identically vanishes.But as far as I can see,the equilibrium position is still indeterminate in terms of temperature of subsystems,and it is therefore,not a shortcoming of the theory.We cannot do any better from anywhere.Physically,without any damping forces,the piston will perpetually oscillate.And in presence of damping ,the resulting temperatures of the chambers will depend on damping forces. Correspondingly,this formalism is indeterminate with respect to temperatures.

My point is,the theory says(as i quoted above post 9) the entropy is maximized over the manifold of constrained equilibrium states.That is,if we remove the constraint(that of the piston being fixed here),the piston reverts to the state which maximizes the entropy.and this reverting is ofcourse subject to the conditon that U1+U2=0,from where P1=P2 follows.The theory doesnt exclude that

Last edited: Oct 31, 2012
13. Oct 31, 2012

### Studiot

Pabloenigma, you now have it right. You need one of the other (energy) criteria I listed in post#11 to establish equilibrium.

14. Oct 31, 2012

### pabloenigma

All said and done,you should really try to have a look at the book.Its probably available at every university Library.You will love it
http://en.wikipedia.org/wiki/Herbert_Callen
I insist,because, I think, in the above discussion I havent really done justice to this book. In my country, a theoritical physicist Palas Pal(http://www.saha.ac.in/theory/palashbaran.pal/) once said that there are 2 kinds of books in thermodynamics :good and bad.If Callen is of the first kind,all the rest are of the second kind.

15. Nov 1, 2012

### Studiot

Unfortunately a script error crept into the collection of formulae at the end of my post#11 so the last four have the wrong direction to their inequalities.

The correct list is

$$\begin{array}{l} {\left( {dS} \right)_{UV}} \ge 0 \\ {\left( {dU} \right)_{SV}} \le 0 \\ {\left( {dH} \right)_{SP}} \le 0 \\ {\left( {dA} \right)_{TV}} \le 0 \\ {\left( {dG} \right)_{SP}} \le 0 \\ \end{array}$$

16. Nov 1, 2012

### A Dhingra

This is exactly what i wanted to know...
What would make such two random derivatives equal, the final states(which you said does not) or it is something to do with the equivalence principle of maximum entropy and minimum energy? How does this happen exactly, please tell me where to read about it (if in Callen) or anywhere else? Or please explain this through two processes....The same thing I was trying to do, making equations of the processes and then checking if they will be equal or not, but my derivatives turned out to be wrong,I guess.

17. Nov 1, 2012

### A Dhingra

Now i think I got what you were saying in this. With continuous change in volume, the entropy of an isolated system tends to increase till it reaches its maximum for the given internal energy, So the derivative (∂S/∂V)T is definitely incorrect for the equilibrium state in process one..

In fact it might not be possible to form any such derivatives at the equilibrium because there won't be any change in teh states any further. Such derivatives can be used only during the process if the the equation of state exists.
So the derivatives used by Maxwell have actually nothing to do with the equivalence principle of entropy maximum and energy minimum, right?

18. Nov 2, 2012

### Studiot

Since you did not respond to the bulk of my post #11 I'm not really quite sure where you are having difficulty. You asked several questions mixed together.

Do you fully appreciate the distinction between differences (Δx) , differentials, dx, partial differentials ∂x and small variations δx ?

The distinction comes into play when we derive Maxwells relations. Have you seen the derivation?
Are you aware that whilst each partial specifies one constant variable, two are in total specified as I said before?

Can you apply say

$${\left( {\frac{{\partial S}}{{\partial V}}} \right)_T} = {\left( {\frac{{\partial P}}{{\partial T}}} \right)_V}$$

to the piston in cylinder example?

I do not want to waste time writing out a lot of stuff if you are not going to read it, but we can go through it if you like.

19. Nov 2, 2012

### pabloenigma

Yes.The Maxwell derivatives dont have anything to do with minimum energy or maximum entropy principle.Also,do realize that this section of the text doesnt want to say anything about maxwell derivatives or them being equal. It tries to explain that " The energy minimum principle with entropy held constant,or the entropy maximum principle with energy held constant,both lead to the same equilibrium point".I think Callen chapter 5,the initial sections elucidate this points very well.

20. Nov 3, 2012

### A Dhingra

I did not mean to offend you in any may. Sorry that i did not say a word about your earlier replies because I thought i knew the formulas for the potentials; After i read the portion again i had a few queries so asked.

Yes. I have seen the proof using the definition of the potentials and using the idea of exact differential.

I know how to write the Maxwell relations, but i think the process described by Callen can't be written in any such form. This relation being not equal to the situation is what both of you have been mentioning.
I just want to see the equality of the two sides of this or any one of the relation of maxwell through a process (actually two processes). Can you help me with that? I thought that each process which should brings about this equality will hold the minimum energy or maximum entropy principle. But that's not the case like pabloenigma said.
Probably this is not what the text said, but this is what i wanted to see and hence interpreted this way (as in the first post).