Thermodynamics - Rods, bubbles and magnets

[MikamiHero]

Homework Statement

For an elastic rod, show that

\[ \frac{\partial C_{L}}{\partial L}_{T} = -T*\frac{\partial^2 f}{\partial T^2}

where C_{L} is the heat capacity at constant length L.

Homework Equations

dU = Tds + fdL
F = U - TS

The Attempt at a Solution

I've used
dF = dU - Tds - SdT = fdL - SdT

Therefore, S = -\frac{\partial F}{\partial T}_{L} and
f = \frac{\partial F}{\partial L}_T
Hence, \frac{\partial S}{\partial L}_T = -\frac{\partial f}{\partial T}_L

I've also found that C_{L} = T*\frac{\partial S}{\partial T}_L

Just not sure what conditions there are when I partially differentiate C_{L}

EDIT: Sorry, I've seen posts here where the TeX code gets translated directly. Is there anyway I can write in TeX language and have it appear as it were compiled in TeX?

 

[omoplata]

If you type within

[tex][/tex]

tags, it's like the equation environment.

If you type within

[itex][/itex]

tags, it's like the typing within $..$ (inline)

You can choose to see the latex source code you typed by right clicking the expression.

I've added tex and itex tags below. Please correct if anything is not as you intended it to be.

Homework Statement

For an elastic rod, show that

$$\[ \frac{\partial C_{L}}{\partial L}_{T} = -T*\frac{\partial^2 f}{\partial T^2}$$

where $C_{L}$ is the heat capacity at constant length $L$.

Homework Equations

$$dU = Tds + fdL$$
$$F = U - TS$$

The Attempt at a Solution

I've used
$$dF = dU - Tds - SdT = fdL - SdT$$

Therefore, $S = -\frac{\partial F}{\partial T}_{L}$ and
$$f = \frac{\partial F}{\partial L}_T$$
Hence, $$\frac{\partial S}{\partial L}_T = -\frac{\partial f}{\partial T}_L$$

I've also found that $C_{L} = T*\frac{\partial S}{\partial T}_L$

Just not sure what conditions there are when I partially differentiate $C_{L}$

 

[omoplata]

OK, what exactly do you mean "not sure what conditions there are when you partially differentiate $C_{L}$"?

I did, $$\frac{\partial C_{L}}{\partial L_{T}} = \frac{\partial}{\partial L_{T}} \left( T \frac{\partial S}{\partial T_{L}} \right) = T \frac{\partial}{\partial L_{T}} \left( \frac{\partial S}{\partial T_{L}} \right)$$ , and from there on had an apparently clear path to the solution.

I can't write down all the steps of the solution. If I do the admins will kick me out.

 

[MikamiHero]

Hello omoplata,

Thank you very much for your reply and help with TeX. I'll re-write it out to make everything clear:

1. Homework Statement
For an elastic rod, show that

$\left( \frac{\partial C_{L}}{\partial L}\right)_{T} = -T \left( \frac{\partial^2 f}{\partial T^2}\right)_{L}$

where $C_{L}$ is the heat capacity at constant length L.

2. Homework Equations
$dU = Tds + fdL$
$F = U - TS$


3. The Attempt at a Solution
I've used
$dF = dU - Tds - SdT = fdL - SdT$

Therefore, $S = -\left( \frac{\partial F}{\partial T}\right) _{L}$ and
$f = \left( \frac{\partial F}{\partial L} \right)_T$
Hence, $\left( \frac{\partial S}{\partial L}\right) _T = -\left( \frac{\partial f}{\partial T}\right)_L$

I've also found that $C_{L} = T\left(\frac{\partial S}{\partial T}\right)_L$

Just not sure what conditions there are when I partially differentiate $C_{L}$
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New comments:

Oh symmetry of partial derivatives, you got me again :(.

Thank you very much, omoplata! I see what to do now.