# Homework Help: Stability, Helmholtz free energy mathematical relation proof

1. May 26, 2013

### fluidistic

Stability, Helmholtz free energy mathematical relation "proof"

1. The problem statement, all variables and given/known data
I must show that $$\left ( \frac{\partial ^2 F}{\partial V ^2} \right ) _T=\frac{\frac{\partial ^2 U}{\partial S^2} \frac{\partial ^2 U}{\partial V ^2} - \left ( \frac{\partial ^2 U}{\partial S \partial V} \right ) ^2}{\frac{\partial ^2 U}{\partial S^2}}$$
Where F is the Helmholtz free energy defined as $F=U-TS$.
Hint: See that $\left ( \frac{\partial ^2 F}{\partial V ^2} \right ) _T=-\left ( \frac{\partial P}{\partial V } \right ) _T$ and consider P as a function of S and V.
2. Relevant equations
As you can see, it is not clear at all what variables are kept constant when he does (the book) the partial derivatives. Furthermore, I don't find any sense to considering $-\left ( \frac{\partial P}{\partial V } \right ) _T$ if P is a function of S and V rather than V and T.

3. The attempt at a solution
I can't seem to make sense of the hint given. So I started with the definition of the Helmholtz free energy and performed twice the derivative with respect to V with T fixed to reach $\left ( \frac{\partial ^2 F}{\partial V ^2} \right ) _T=\left ( \frac{\partial ^2 U}{\partial V ^2} \right ) _T-T\left ( \frac{\partial ^2 S}{\partial V ^2} \right ) _T$.
Now $T=\left ( \frac{\partial U}{\partial S} \right ) _V$ so that $\left ( \frac{\partial ^2 F}{\partial V ^2} \right ) _T=\left ( \frac{\partial ^2 U}{\partial V ^2} \right ) _T-\left ( \frac{\partial U}{\partial S} \right ) _V \left ( \frac{\partial ^2 S}{\partial V ^2} \right ) _T$.
Now I multiply and divide by $\left ( \frac{\partial ^2 U}{\partial S ^2} \right ) _T$ to reach $$\left ( \frac{\partial ^2 F}{\partial V ^2} \right ) _T=\frac{\left ( \frac{\partial ^2 U}{\partial V ^2} \right ) _T\left ( \frac{\partial ^2 U}{\partial S ^2} \right ) _T- \left ( \frac{\partial U}{\partial S} \right ) _V \left ( \frac{\partial ^2 U}{\partial V ^2} \right ) _T}{\left ( \frac{\partial ^2 U}{\partial S ^2} \right ) _T}$$ so I've almost the right expression but not quite. I don't know how to proceed further. Nor do I know how to use the hint given, nor do I understand it.
Any help is appreciated. Thank you!

2. May 26, 2013

### TSny

In arriving at your last expression, it looks like you set $\left ( \frac{\partial ^2 S}{\partial V ^2} \right ) _T \cdot \left ( \frac{\partial ^2 U}{\partial S ^2} \right ) _T = \left ( \frac{\partial ^2 U}{\partial V ^2} \right ) _T$. But this isn't correct. Note that the dimensions of the left and right sides don't match. (The entropy dimension does not cancel out on the left).

In the original identity that you want to show, I believe that the energy U is considered a function of V and S. So, the quantities that are kept constant in the partial derivatives on the right side are either V or S rather than V or T.

The hint that is given will work. Start with dU = -PdV + TdS to find an expression for P as a partial derivative of U. Then use this to try to express $-\left ( \frac{\partial P}{\partial V } \right ) _T$ in terms of partials of U.

3. May 26, 2013

### fluidistic

I see, you are right I have been sloppy with the partial differentials. I didn't know what I did was wrong, so thanks a lot.
$dU=-PdV+TdS+\mu \underbrace {dn}_{=0} \Rightarrow P = - \left ( \frac{\partial U}{\partial V} \right ) _S$.
So that $- \left ( \frac{\partial P}{\partial V} \right ) _T = \frac{\partial }{\partial V} \left [ \left ( \frac{\partial U}{\partial V} \right ) _S \right ] _T$.
Is it okay so far? I still don't see where to use the hint but I'll think about it.

4. May 26, 2013

### dextercioby

Ok so far.

5. May 26, 2013

### fluidistic

I'm lost. I've tried some non sensical stuff but that lead me to nowhere. Like the Euler relation U=TS-PV, then $P=\frac{TS-U}{V}$ and if I think P as a function of S and V like the hint says, then I reach that $\left ( \frac{\partial P}{\partial V} \right ) _T=\frac{U-ST}{V^2}$. Of course this doesn't make any sense but I've reached the point where I don't have any physical intuition but I only see a bunch of equations of several variables.

6. May 26, 2013

### TSny

Welcome to thermodynamics!

You need to find an expression for $\left( \frac{\partial P}{\partial V} \right )_T$.
We are taking P to be a function of V and S. So, for any infinitesimal process $dP = \left ( \frac{\partial P}{\partial V} \right )_S dV + \left ( \frac{\partial P}{\partial S} \right )_V dS$

Apply this general relation to the specific case where P, V and S change while T is held fixed. Denote the infinitesimal changes in P, V and S while T is held fixed by $dP_T$ ,$dV_T$ and $dS_T$.

So, $dP_T = \left ( \frac{\partial P}{\partial V} \right )_S dV_T + \left ( \frac{\partial P}{\partial S} \right )_V dS_T$

Divide through by $dV_T$ and interpret.

7. May 26, 2013

### fluidistic

I see. I reach a "famous" formula for partial derivatives that I demonstrated a few months ago. Namely that $\left ( \frac{\partial x}{\partial y} \right ) _z=\left ( \frac{\partial x}{\partial y} \right ) _w+\left ( \frac{\partial x}{\partial w} \right ) _y\left ( \frac{\partial w}{\partial y} \right ) _z$. So here I reach $\left ( \frac{\partial P}{\partial V} \right ) _T=\left ( \frac{\partial P}{\partial V} \right ) _S+\left ( \frac{\partial P}{\partial S} \right ) _V\left ( \frac{\partial S}{\partial V} \right ) _T$. I didn't know it would haunt me up to here...
I don't really know what to interpret here. I've rewritten $\left( \frac{\partial P}{\partial V} \right )_T$ into a more complicated form. I know that minus $\left( \frac{\partial P}{\partial V} \right )_T$ must equal $\frac{\left ( \frac{\partial ^2 U}{\partial V ^2} \right ) _T\left ( \frac{\partial ^2 U}{\partial S ^2} \right ) _T- \left ( \frac{\partial U}{\partial S} \right ) _V \left ( \frac{\partial ^2 U}{\partial V ^2} \right ) _T}{\left ( \frac{\partial ^2 U}{\partial S ^2} \right ) _T}$. I guess I must use another relation between partial derivatives to show that equality?

8. May 26, 2013

### TSny

Good. Now you want to write the right-hand side in terms of partials of U. You found earlier that $P = -\left( \frac{\partial U}{ \partial V} \right)_S$.

Use that in the right-hand side of your equation for $\left( \frac{ \partial P}{ \partial V} \right)_T$.

9. May 27, 2013

### fluidistic

$\left( \frac{ \partial P}{ \partial V} \right)_T=-\left( \frac{ \partial ^2 U}{ \partial V^2} \right)_S- \frac{\partial }{\partial S} \left [ \left( \frac{ \partial U}{ \partial V} \right)_S \right ] _V \left( \frac{ \partial S}{ \partial V} \right)_T$. Which is equal to $-\left( \frac{ \partial ^2 U}{ \partial V^2} \right)_S-\left( \frac{ \partial T}{ \partial V} \right)_S \left( \frac{ \partial S}{ \partial V} \right)_T$.
Here I used the cyclic relation $\left( \frac{ \partial T}{ \partial V} \right)_S \left( \frac{ \partial V}{ \partial S} \right)_T \left( \frac{ \partial S}{ \partial T} \right)_V=-1$ to get another expression for $\left( \frac{ \partial T}{ \partial V} \right)_S$ but after replacing it in the expression of $\left( \frac{ \partial P}{ \partial V} \right)_T$, it just looks more complicated.
I'm sure I'm missing something, once again.

10. May 27, 2013

### TSny

Don't do anything to the $\frac{\partial }{\partial S} \left [ \left( \frac{ \partial U}{ \partial V} \right)_S \right ] _V$ since that is part of what you want to end up with.

Use your cyclic relation to find another expression for $\left( \frac{ \partial S}{ \partial V} \right)_T$. Then remember that $T = \left( \frac{ \partial U}{ \partial S} \right)_V$.

11. May 27, 2013

### Staff: Mentor

This is a straight mathematical problem in manipulating partial derivatives. Maybe this will help:
Write P = P(S,V) and T = T(S,V). Then,

$$dP=\left( \frac{\partial P}{\partial S}\right)_VdS+\left(\frac{\partial P}{\partial V}\right)_SdV$$
$$dT=\left( \frac{\partial T}{\partial S}\right)_VdS+\left(\frac{\partial T}{\partial V}\right)_SdV$$

From the first equation,

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

From the second equation, if dT = 0, then
$$\left(\frac{\partial S}{\partial V}\right)_T=-\frac{\left(\frac{\partial T}{\partial V}\right)_S}{\left( \frac{\partial T}{\partial S}\right)_V}$$

Combining these two equations, we get:

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

Now,
$$dU=\left( \frac{\partial U}{\partial S}\right)_VdS+\left(\frac{\partial U}{\partial V}\right)_SdV=TdS-PdV$$
Therefore,
$$\left( \frac{\partial U}{\partial S}\right)_V=T$$
$$\left(\frac{\partial U}{\partial V}\right)_S=-P$$
Therefore,
$$\left( \frac{\partial ^2U}{\partial S \partial V}\right)=\left(\frac{\partial T}{\partial V}\right)_S=-\left(\frac{\partial P}{\partial S}\right)_V$$
So,
$$\left( \frac{\partial ^2U}{\partial S \partial V}\right)^2=-\left(\frac{\partial T}{\partial V}\right)_S\left(\frac{\partial P}{\partial S}\right)_V$$

The rest should be easy.

12. May 29, 2013

### fluidistic

Thanks guys, I had been busy and couldn't spend time on this problem but now I'm back.
So I had $\left( \frac{ \partial P}{ \partial V} \right)_T=-\left( \frac{ \partial ^2 U}{ \partial V^2} \right)_S- \frac{\partial }{\partial S} \left [ \left( \frac{ \partial U}{ \partial V} \right)_S \right ] _V \left( \frac{ \partial S}{ \partial V} \right)_T$. Rewritten: $\left( \frac{ \partial P}{ \partial V} \right)_T=-\left( \frac{ \partial ^2 U}{ \partial V^2} \right )- \frac{\partial ^2 U }{\partial S \partial V} \left( \frac{ \partial S}{ \partial V} \right)_T$.
Using the cyclic relation on the last term (basically equivalent to what Chestermiller did) I get that $\left( \frac{ \partial P}{ \partial V} \right)_T=-\left( \frac{ \partial ^2 U}{ \partial V^2} \right)_S + \frac{\left ( \frac{\partial T}{\partial V} \right ) _S}{\left ( \frac{\partial T}{\partial S} \right ) _V} \frac{\partial ^2 U}{\partial V \partial S}$.
Now using your tip TSny about T, I get that $\left ( \frac{\partial T}{\partial V} \right ) _S= \frac{\partial ^2 U}{\partial V \partial S}$ and $\left ( \frac{\partial T}{\partial S} \right ) _V= \frac{\partial ^2 U }{\partial S ^2}$.
Plugging this back into the equation, I get that $$\left ( \frac{\partial P}{\partial V} \right ) _T=\frac{- \left ( \frac{\partial ^2 U}{\partial V ^2} \right ) \left ( \frac{\partial ^2 U}{\partial S^2} \right ) }{\frac{\partial ^2 U}{\partial S^2}}+\frac{\left ( \frac{\partial ^2 U}{\partial S \partial V} \right ) ^2 \left ( \frac{\partial ^2 U }{\partial V \partial S} \right ) }{\left ( \frac{\partial ^2U}{\partial S ^2} \right ) }$$. So it's almost what I should get, except for the second term where I have an extra multiplicative term $\left ( \frac{\partial ^2 U}{\partial V \partial S} \right )$.
So I've rechecked all the algebra around this second term, but I didn't spot any mistake. I don't see where I went wrong! Can you spot my mistake(s)?

13. May 30, 2013

### Staff: Mentor

Yes. Your mistake is easy to spot.
The equation
$\left( \frac{ \partial P}{ \partial V} \right)_T=-\left( \frac{ \partial ^2 U}{ \partial V^2} \right)_S + \frac{\left ( \frac{\partial T}{\partial V} \right ) _S}{\left ( \frac{\partial T}{\partial S} \right ) _V} \frac{\partial ^2 U}{\partial V \partial S}$
is correct. But then when you substituted $\left ( \frac{\partial T}{\partial V} \right ) _S= \frac{\partial ^2 U}{\partial V \partial S}=\frac{\partial ^2 U}{\partial S \partial V}$ into the equation, you got an extra factor of $\frac{\partial ^2 U}{\partial S \partial V}$.

14. May 30, 2013

### TSny

You have $\left( \frac{ \partial P}{ \partial V} \right)_T=-\left( \frac{ \partial ^2 U}{ \partial V^2} \right)_S + \frac{\left ( \frac{\partial T}{\partial V} \right ) _S}{\left ( \frac{\partial T}{\partial S} \right ) _V} \frac{\partial ^2 U}{\partial V \partial S}$ which looks correct.

The relation $\left ( \frac{\partial T}{\partial V} \right ) _S= \frac{\partial ^2 U}{\partial V \partial S}$ gives you one more factor of $\frac{\partial ^2 U}{\partial V \partial S}$.

So, overall, you should end up with only two factors of $\frac{\partial ^2 U}{\partial V \partial S}$ rather than three factors.

[EDIT: I see Chestermiller already caught the mistake.]

15. May 30, 2013

### fluidistic

Thank you guys! Problem solved then.