I'm asked to establish the following relations:

$\left( \frac{\partial C_V}{\partial V} \right)_{T,N} = \frac{T}{N} \left( \frac{\partial^2 P}{\partial T^2} \right)_{V,N} $

$\left( \frac{\partial (\beta F)}{\partial \beta} \right)_{V,N} = U $ where $\beta = \frac{1}{k_B T}$

I'm a bit puzzled by the first one since I can't seem to find a relation between the heat capacity $C_V$ and the other term in the equation, though I do know that $C_V = (\frac{\partial E}{\partial T})_V$

Note that F and U are the Helmholtz free energy and internal energy, respectively.

Also how does the fact that certain quantity are maintained constant throughout the differentiation should be accounted for when expanding say the left hand sides?

Any help would be appreciated.


1 Answer 1


Since this appears to be homework I can only give you a hint. Start with $dU = TdS -pdV$ and $C_v = \left( \frac{\delta Q}{\delta T} \right)_v = T \left(\frac {\partial S}{\partial T}\right)_v$ Now differentiate both sides with respect to $V$ and use the Maxwell relation $\left(\frac {\partial p}{\partial T}\right)_v = \left( \frac {\partial S}{\partial V}\right)_T$.

  • $\begingroup$ what does the term $\left(\frac {\partial S}{dT}\right)_v$ exactly mean? Would it be a regular derivative with respect to T (treating V as a constant)? In which case wouldn't $\left(\frac {d S}{dT}\right)_v$ suffice? A bit confused with how to deal with these derivative, especially from the fundamental relation you wrote previously. $\endgroup$
    – user44212
    Apr 9, 2014 at 0:20
  • $\begingroup$ fixed the typo, it should be clear now $\endgroup$
    – hyportnex
    Apr 9, 2014 at 13:33
  • $\begingroup$ Thanks. One more thing. I'm having trouble figuring out what $\frac{\partial}{\partial V} \left( T (\frac{\partial S}{\partial T})_V\right)$ would be because of that constant V in that temperature derivative. Would it be: $\frac{\partial T}{\partial V} \left( \frac{\partial S}{\partial T} \right)_V + T \left( \frac{\partial^2 S}{\partial V \partial T} \right)_V$ ? $\endgroup$
    – user44212
    Apr 9, 2014 at 22:53
  • $\begingroup$ That is correct, and now you have to ask yourself what are the ${independent}$ variables of your problem, and then you will get the answer. $\endgroup$
    – hyportnex
    Apr 9, 2014 at 23:00

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