# Statistical Mechanic

One can define entropy as $$S=k\log{\omega(E)},$$ where $\omega(E)$ is the numbers of states with energy equal $E$; and the canonical partition function for a set of N particles is defined as$$Z_N=\sum_{\phi}e^{-\beta E[\phi]}=e^{-\beta F(\beta,N)},$$ where the sum run on states $\phi$ and the free energy is defined as $F(\beta,N)=U-TS.$ The mean value of the internal energy and the entropy i learned that would be $$\langle U\rangle=\frac{\partial(\beta F)}{\partial\beta}$$ $$\langle S\rangle=\beta^2\frac{\partial(F)}{\partial\beta}.$$ For definition, the mean value of any physical observable is $$\langle O\rangle=Z_N^{-1}\sum_{\phi}O[\phi]e^{-\beta E[\phi]}.$$ I'm quite sure that there will be a problem of amount $k$, if one verify the definition of $\langle S\rangle$. Am i wrong?

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If you mean missing factors of $k$, there are no missing factors of $k$ in your formulae. They're just fine. – Luboš Motl Feb 8 at 15:32

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What makes you think there is a problem with a $k$? At a glance your formulae look correct. Maybe if you could show us where you are getting hung up in your calculation we could help. – Michael Brown Feb 8 at 15:47
$F=-\frac{\log{Z_N}}{\beta}$, so $$\beta^2\frac{\partial F}{\partial\beta}=-\beta^2\frac{Z_N^{-1}\frac{\partial Z_N}{\partial\beta}\beta-\log{Z_N}}{\beta^2}=-\beta(-\langle U\rangle+F)=\beta TS=\frac{S}{k}$$ – ivax Feb 8 at 20:12
– Qmechanic♦ Feb 8 at 20:55
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