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I'm trying to calculate a few properties of a classical rigid rotor. Its hamiltonian is given by

\begin{equation} H=\frac{1}{2I} \left( p_{\theta}^{2}+\frac{p_{\phi}^{2}}{\sin^{2}\theta} \right) \end{equation}

where $I$ is the moment of inertia and $p$ is the momentum associated with the angle of rotation in its subindex. $\theta$ is the angle with the $z$-axis and $\phi$ with the $x$-axis. The partition function for this system is simply

\begin{equation} Z=\frac{1}{h^2}\iiiint d\theta \,d \phi \,dp_{\theta}\,dp_{\phi} \exp\left(-\beta H\right) = \frac{8\pi^2}{h^2} I \, k_{B}\,T \end{equation}

where $\beta = (k_{B}\,T)^{-1}$, being $k_{B}$ the Boltzmann constant and $T$ the temperature. Now, the entropy is given by

\begin{equation} S=-k_{B}\frac{1}{h^2}\iiiint d\theta \,d \phi \,dp_{\theta}\,dp_{\phi} \left[\frac{\exp\left(-\beta H\right)}{Z}\mathrm{Ln}\left(\frac{\exp\left(-\beta H\right)}{Z}\right)\right] \end{equation}

Is this expression for the entropy correct? Because if it is then it would be

\begin{equation} S=\frac{\langle E \rangle}{T}+k_{B}\,\mathrm{Ln}Z \end{equation}

where $\langle E \rangle$ is the average energy of the system. I understand the first term but the second one I was not expecting.

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    $\begingroup$ The final equation is fine, as it's just the definition of the free energy $F = -kT \log Z$. Did you have any other questions? $\endgroup$
    – knzhou
    Commented May 16, 2017 at 16:01

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