How to prove using only statistical mechanics and canonical probabilities that $F=U-TS$ I am a little confused, I would like to prove that :
$$ F=U-TS $$
Using only the fact that :
$$ F=-k_b T ln(Z) $$
$$P(\sigma)=\frac{e^{-\beta} E(\sigma)}{Z}$$
$$S(E)=k_b ln(\Omega(E))$$
What I mean by using only stat mech is that I don't want to use the classical thermodynamic relation saying $S=-\frac{\partial F}{\partial T}$.
Is it possible ?
Actually I ended up with either :
$$F=-k_bTln(\sum_E e^{-\beta(E-TS(E))})$$
Or : $$<E>=F+\beta \frac{\partial F}{\partial \beta} $$
But I don't know how to end up with $F=U-TS$ where $U=<E>$ without using the classical thermodynamic relation $S=-\frac{\partial F}{\partial T}$
Is what I want possible or we are forced to use classical thermodynamic to conclude ?
 A: There is a misunderstanding regarding entropy. Given a probability distribution $P(\sigma)$, the statistical entropy is defined as
     $$S[P]=-k_B\sum_\sigma P(\sigma)\ln P(\sigma)$$
The extremum of this quantity with the only constrain that $\sum_\sigma P(\sigma)=1$ leads to the microcanonical ensemble
      $$P_{\rm micro}[\sigma]={1\over \Omega(E)}$$
Plugging this into the statistical entropy gives the microcanonical entropy
     $$S_{\rm micro}=k_B\ln\Omega(E)$$
The extremum of the statistical entropy with the constrains that $\sum_\sigma P(\sigma)=1$ and that the average energy is $\langle E\rangle$ leads to the canonical ensemble
     $$P_{\rm can.}[\sigma]={1\over{\cal Z}}e^{-\beta E}$$
Plugging this into the statistical entropy gives the canonical entropy
     $$S_{\rm can.}=-k_B\sum_\sigma P[\sigma]
 \big(-\ln{\cal Z}-\beta E\big)$$
You can recognize canonical averages so
      $$S_{\rm can.}=k_B\ln{\cal Z}+k_B\beta\langle E\rangle$$
from which follows
      $$-k_BT\ln{\cal Z}=F=\langle E\rangle-TS_{\rm can.}$$
A: For me the most satisfying "proof" is using Liouville's Theorem. Simplifying, using Poisson Brackets, and for a system in statistical equilibrium (recall that Liouville's theorem is nothing more than partial differential chain rule on H, a function of the phase space),
$$\frac{\partial \rho}{\partial t}=-\{\rho, H\}=0$$
One such solution (on the interval $E=H$),
$$\rho(E) = Ae^{-E/kT}=e^{(F-E)/kT}$$
Gives,
$$kT\ln \rho=F-E$$
And since $S=-k\ln \rho$ (easy to show this from microcanonical ensemble),
$$-ST=F-E$$
