# How to derive Helmholtz free energy in statistical mechanics?

So I was going through a book of statistical mechanics, but the explanation skipped quite a few steps of the derivation. Given the internal energy: $$U=k_B\cdot T^2\cdot \frac{\partial ln(Z)}{\partial T}$$ And the Gibbs-Helmholtz equation from classical thermodynamics: $$U=\left[\frac{\partial\frac{A}{T}}{\partial\frac{1}{T}}\right]_{V,N}$$ How does one arrive to the statistical mechanics definition of the Helmholtz energy (A) given below? $$A = -k_B\cdot T\cdot ln(Z)$$

I thought it was worth adding, I have tried simplifying the RHS of the Gibbs-Helmholtz relation, and it resulted in the following PDE: $$U=A-T\cdot\frac{\partial A}{\partial T}$$ But I'm not too sure that is heading in the right direction, or if so I feel like I may be missing something...

If you just equate the first two lines of your question (I will keep $$k_B=1$$ for convenience and rename $$A=F$$ for the reason of habit):

$$T^2 \frac{\partial \ln Z}{\partial T} = \frac{\partial (F/T)}{\partial (1/T)} = -T^2 \frac{(F/T)}{\partial T}$$

By integrating both sides you can conclude that

$$\ln Z +\mathrm{const} = - F/T$$

Which basically brings you to the sought third line of your question.

• Makes quite a lot of sense... Thanks!
– HWIK
Dec 7, 2022 at 15:36