# Helmholtz free energy

1. Sep 18, 2009

### LocationX

1. The problem statement, all variables and given/known data

The Helmholtz free energy is written as:

$$dA=-SdT-pdV-VMdH$$

when an incompressible liquid is placed in a magnetic field H. Thus, the free energy can be written as since $$-pdV=0$$:

$$dA=-SdT-VMdH$$

The two heat capacities can be defined as:

$$C_H = T (dS/dT)_H$$ and $$C_M = T (dS/dT)_M$$

Where one is at constant H field and the other is at constant magnetization.

I need to find the relation for $$C_H - C_M$$

I'm really not sure where to begin for this problem, any advice would be really helpful. Thank you.

2. Sep 18, 2009

### Mapes

There's a trick to these problems that's not immediately obvious: express dS as

$$dS=\left(\frac{\partial S}{\partial T}\right)_H\,dT+\left(\frac{\partial S}{\partial H}\right)_T\,dH$$

and take the derivative with respect to T at constant M. Know what I mean?

3. Sep 18, 2009

### LocationX

I'm not sure exactly if I have the entire concept down but here's what I'm thinking:

$$C_M=T \frac{1}{dT} \left( \left(\frac{\partial S}{\partial T}\right)_H dT+\left(\frac{\partial S}{\partial H}\right)_T dH \right)_M$$

$$C_M=T \left( \left(\frac{\partial S}{\partial T}\right)_H +\left(\frac{\partial S}{\partial H}\right)_T \frac{dH}{dT} \right)_M$$

$$C_H=T \frac{1}{dT} \left( \left(\frac{\partial S}{\partial T}\right)_M dT+\left(\frac{\partial S}{\partial M}\right)_T dM \right)_H$$

$$C_H=T \left( \left(\frac{\partial S}{\partial T}\right)_M +\left(\frac{\partial S}{\partial M}\right)_T \frac{dM}{dT} \right)_H$$

thus:

$$C_H-C_M$$
$$C_H-C_M=T \left( \left( \frac{\partial S}{\partial T} \right)_{M,H} - \left( \frac{\partial S}{\partial T} \right)_{M,H} + \left( \frac{\partial S}{\partial M} \right)_{T.H} \left( \frac{\partial M}{\partial T} \right)_H - \left( \frac{\partial S}{\partial H} \right)_{T.H} \left( \frac{\partial H}{\partial T} \right)_M \right)$$
$$C_H-C_M=T \left( \left( \frac{\partial S}{\partial M} \right)_{T.H} \left( \frac{\partial M}{\partial T} \right)_H - \left( \frac{\partial S}{\partial H} \right)_{T.H} \left( \frac{\partial H}{\partial T} \right)_M \right)$$

4. Sep 19, 2009

### Mapes

This is pretty good. Remember, though, that the temperature derivative is $\partial/\partial T$ (rather than $1/\partial T$, and this may have just been a typo), and that the condition of constant H or M is attached to the derivative only, so there's no $((\partial S/\partial T)_H)_M$, there's only $(\partial S/\partial T)_H$, because the M condition vanishes with $(\partial T/\partial T)_M$.

Did you notice that your expression for $C_M$ contains $C_H$?

Keep going!

5. Sep 19, 2009

### LocationX

I think I got it, but there are two answers:

$$C_H-C_M = T \left( \frac{\partial S}{\partial M} \right)_T \left( \frac{ \partial M}{\partial T} \right)_H$$

and also:

$$C_H-C_M = - T \left( \frac{\partial S}{\partial H} \right)_T \left( \frac{ \partial H}{\partial T} \right)_M$$

Are these both correct? Thanks again for that helpful hint :)

The second part of the question asks obtain an expression for the heat capacity difference in terms of the field (H), temperature (T), and the Curie constant (a).

Using Curie's law, we are given the approximation that:

$$\mu_o M/H \approx a/T$$
$$M \approx \frac{aH}{T \mu_o}$$

$$C_H-C_M = T \left( \frac{\partial S}{\partial M} \right)_T \left( \frac{ \partial M}{\partial T} \right)_H = \frac{aTH}{\mu_o} \left( \frac{\partial S}{\partial M} \right)_T$$

I'm not sure how to change the $$\left( \frac{\partial S}{\partial M} \right)_T$$ to the variables stated... am I on the right track?

6. Sep 19, 2009

### Mapes

Pretty much. Your first two expressions are equivalent, which you can show with Maxwell relations. And you can handle $(\partial S/\partial M)_T$ with a Maxwell relation. But I'm not following how you differentiated M with respect to T to get your current answer.

7. Sep 19, 2009

### LocationX

Whoops, this is what i meant:
$$C_H-C_M = T \left( \frac{\partial S}{\partial M} \right)_T \left( \frac{ \partial M}{\partial T} \right)_H = - \frac{aH}{T\mu_o} \left( \frac{\partial S}{\partial M} \right)_T$$

The Maxwell relation I got when using the Helmholtz free energy is:

$$\left( \frac{\partial S}{\partial H} \right)_T = \left( \frac{\partial (VM)}{\partial T} \right)_H$$

Since M is constant, I cant get the relationship with dS/dM?

8. Sep 19, 2009

### Mapes

There's a Maxwell relation for $(\partial S/\partial M)_T$, but it requires a different potential, one that includes dT and dM.

9. Sep 19, 2009

### LocationX

Here's another shot:

$$\left( \frac{\partial S}{\partial M} \right)_T = VM \frac{ \partial ^2 H }{\partial T \partial M} = \frac{V M \mu_o}{a}$$

$$C_H-C_M = - \frac{aH}{T\mu_o} \left( \frac{\partial S}{\partial M} \right)_T = -\frac{HVM}{T}$$

Last edited: Sep 20, 2009
10. Sep 19, 2009

### LocationX

Also, why wouldnt we use the chain rule to evaluate the derivative? That is...

$$H \frac{\partial }{\partial T} \left( \left( \frac{ \partial S}{\partial H} \right)_T \right)_H + \left( \frac{\partial H}{\partial T} \right)_H \left( \frac{ \partial S}{\partial H} \right)_T$$?

11. Sep 20, 2009

### Mapes

I get MVH/T after trying a couple different ways, so I think you might have a sign error somewhere in your Maxwell relation, but in any case you know the general approach now.

We do always use the chain rule to evaluate the derivative of a product, but differential terms like dH will always be negligible compared to derivatives like $(\partial H/\partial T)_M$, so we end up dropping these terms. I'm not sure if this answers your question?