I have just encountered this formula for obtaining the Helmholtz free energy $F(T,V)$ of a system from the internal energy $U(T,V)$:


The book I got it from says you can prove it knowing that

$$U(T,V)=-T^2\left[ \dfrac{\partial}{\partial T} \left( \dfrac{F(T,V)}{T} \right) \right]_V$$ and $$F(T,V)=U(T,V)+T\left(\dfrac{\partial F}{\partial T}\right)_V$$

by employing the third law of thermodynamics.

I already knew the formulas above, but I could not prove the formula. If I try to integrate the second formula I don't know how to deal with having to integrate from zero to $T$ and the temperature being at the denominator. Can someone help me?


1 Answer 1


I have just encountered this formula...$$F(T,V)=U(0,V)-T\int_0^TdT'\dfrac{U(T',V)-U(0,V)}{T'^2}.\tag{A}$$

The book I got it from says you can prove it knowing that

$$U(T,V)=-T^2\left[ \dfrac{\partial}{\partial T} \left( \dfrac{F(T,V)}{T} \right) \right]_V\tag{B}$$ and $$F(T,V)=U(T,V)+T\left(\dfrac{\partial F}{\partial T}\right)_V\tag{C}$$

Can someone help me?

Instead of considering $F(T, V)$, you can consider: $$ \Delta F(T, V) \equiv F(T, V) - F(0, V) = F(T,V) - U(0,V)\;. $$

The utility of considering $\Delta F$ instead of $F$ is that it is easy to show that $$ \lim_{T\to 0}\frac{\Delta F}{T} \to -S(T=0)\;, $$ which is finite by the third law of thermodynamics. The above limit is calculated using the usual rule for an indefinite ratio (where both the numerator and denominator go to zero): $$ \lim_{T\to 0}\frac{\Delta F}{T} = \lim_{T\to 0}\frac{\frac{\partial \Delta F}{\partial T}}{\frac{\partial T}{\partial T}} $$ $$ =\lim_{T\to 0}\frac{\frac{\partial \Delta F}{\partial T}}{1} =\lim_{T\to 0}\frac{\frac{\partial F}{\partial T}}{1} =\lim_{T\to 0}\frac{\partial F}{\partial T} = -S(0)\;. $$

The above limit can also be evaluated by expanding in a power series like: $$ \Delta F \equiv F(T, V) - F(0, V) \approx T\frac{\partial F}{\partial T} + O(T^2) $$

The above limit can also be evaluated by recognizing that it is just the definition of the derivative of $F$ at $T=0$: $$ \lim_{T\to 0}\frac{\left(F(T,V) - F(0, V)\right)}{\left(T - 0\right)}\equiv \left.\frac{\partial F}{\partial T}\right|_0 $$

So, anyways, using $\lim_{T\to 0}\frac{\Delta F}{T} \to -S(T=0)$, the lower limit at $T=0$ is not a problem.

You can further show, using Eq (B) above, that: $$ U(T, V) - U(0, V) = -T^2\frac{\partial}{\partial T} \left(\frac{\Delta F}{T}\right)\;. \tag{1} $$

Rearranging my Eq (1) above gives: $$ \frac{-1}{T^2}\left(U(T, V) - U(0, V)\right) = \frac{\partial}{\partial T} \left(\frac{\Delta F}{T}\right)\;.\tag{2} $$

Integrating both sides from $0$ to $T$, I find: $$ F(T,V) = U(0, V) - TS(0,V) - T\int_{0}^T dT'\frac{U(T',V) - U(0,V)}{T'^2}\;, $$ which is the same expression as OP's Eq. (A) except for the addition of the term $TS(0, V)$.

If there is a unique ground state then $S(0, V) = 0$ and the two expressions agree, otherwise $S(0, V) = \text{log}(g_0)$, where $g_0$ is the degeneracy of the ground state.

Not sure if OP needs to use $S(0, V) = 0$, or if there is an error in my calculation somewhere, or if there is some other issue.

  • $\begingroup$ OK, I updated my answer to explain why the lower limit is not a problem when you consider $\Delta F/T$ rather than $F/T$. I am not giving all the details, since I think OP should provide the details. $\endgroup$
    – hft
    Mar 1 at 18:56
  • $\begingroup$ I updated my answer. $\endgroup$
    – hft
    Mar 1 at 20:50
  • 1
    $\begingroup$ @TobiasFünke Sure, in hindsight this is trivially just the definition of the derivative at $T=0$. But, it was apparently not obvious on initial inspection, so some explanatory text and symbolic manipulation is probably helpful. $\endgroup$
    – hft
    Mar 1 at 21:05
  • $\begingroup$ I agree. Actually, just to check your last result: Have you tried to take the derivative of it (with respect to $T$)? If it is correct, then it should yield $-S(T)$. You can then compare with the derivative of the first equation of the question; there should be a difference due to the presence of the $S(0)$ term (which I suspect however is correct). I guess one can take the derivative with respect to $V$ and check if it is related to the pressure in the well-known way. Then, I guess, one can use these arguments to show that your eq. is correct (i.e. turn the necessary into suff. conditions). $\endgroup$ Mar 1 at 21:09
  • $\begingroup$ @TobiasFünke I did try taking the derivative as a sanity check, but it doesn't immediately seem very helpful, at least to me. $\endgroup$
    – hft
    Mar 1 at 21:19

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