# What is the entropy of a classical rigid rotor?

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

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

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

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

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

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

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

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

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

• 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? Commented May 16, 2017 at 16:01