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I have a problem and I don't understand how the book solves it. It says I have a bar such that $$F=aT^2(L-L_0)$$ where $F$ is the force or tension of the bar, $T$ is the temperature, and $L$ the longitude. $L_0$ is the longitude when $F=0$. It says that $C_{L_0}=bT$ and it asks me about the general function $C_L(L,T)$, being $C_L$ the specific heat at constant $L$. The book starts solving it with this expression: $$\left(\frac{\partial C_L}{\partial L} \right)_T=-T\left(\frac{\partial^2 F}{\partial T^2}\right)_L$$ From there, I know how to go on, but I have no idea where that comes from. I have tried to get that out of the generalised Mayer relation but if that's the way, I can't do it.

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@tpg2114 Yes, it was a typo – MyUserIsThis Feb 1 '13 at 20:00
I don't have time for a full answer, but look at question 17 on It seems to be doing exactly what you are doing. – tpg2114 Feb 1 '13 at 20:09
up vote 1 down vote accepted

Recall first that the definition of $C_L$ is $$ C_L = \left(\frac{\partial U}{\partial T}\right)_L $$ The first law of thermo for the bar reads $$ dU = T dS + F dL $$ Taking $S=S(L, T)$ gives $$ dS = \left(\frac{\partial S}{\partial T}\right)_L dT + \left(\frac{\partial S}{\partial L}\right)_T dL $$ Combining these two equations gives $$ dU = T\left(\frac{\partial S}{\partial T}\right)_L dT + \left[T\left(\frac{\partial S}{\partial L}\right)_T +F\right]dL $$ From which we see that $$ \left(\frac{\partial U}{\partial L}\right)_T = T\left(\frac{\partial S}{\partial L}\right)_T +F $$ On the other hand, one has the the following Maxwell relation (I'll leave it to you to derive it, but I can show you how if you want): $$ \left(\frac{\partial S}{\partial L}\right)_T = -\left(\frac{\partial F}{\partial T}\right)_L $$ So plugging this into the last equation gives $$ \left(\frac{\partial U}{\partial L}\right)_T = -T\left(\frac{\partial F}{\partial T}\right)_L +F $$ Taking the partial of both sides with respect to $T$, noting that partial derivatives commute, and then using the definition of $C_L$ gives precisely the equation you want! Let me know if you want more detail!


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Thank you very much, that was more than enough, I know how to get that Maxwell relation. You've already solved me a couple of headaches. – MyUserIsThis Feb 1 '13 at 21:22
Great! Glad it helped! – joshphysics Feb 1 '13 at 21:26

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