Boltzmann distribution for angles? Consider a system whose sole degree of freedom is an angle $\theta$ that goes from $0$ to $2\pi$. Let $E(\theta)$ be its energy function. Obviously, $E(\theta)$ is $2\pi$-periodic. What's the general form for the Boltzmann distribution for $\theta$? Is it just: $P(\theta)\propto e^{-E(\theta)/kT}$? Or is there some issue stemming from the fact that $\theta$ is an angle?
 A: There's no issue with the energy having an angular dependency. This is similar to the case of a spin in a magnetic field, in which the energy is
$$E = -\mathbf{\mu \bullet B}$$
or 
$$E (\theta) = -\mu B cos(\theta)$$
This poses no problem. As you say, the Boltzmann factor is $e^{\mu B cos(\theta)/kT}$, and the partition function is found by integrating the Boltzmann factor w.r.t. $\theta$ from $0$ to $2 \pi$. 
Others are correct in suggesting that in cases of a continuous energy spectrum, you're most often integrating the probability density over a range of interest to find what the probability is that, say, the spin is aligned with the field to within 0.1 radians. But this is common in statistical physics, and there's nothing special about the case of an angular dependence.
A: It sounds like you're looking for an analog to the density of states. The density of states $D\left(E\right)$ tells you about the number of states in a given energy range $\left[E,E+dE\right]$. So the probability of ending up in a state with energy $E$ is $P\left(E\right) \propto e^{E/k_bT} D\left(E\right)$.
Here, you would have an angular density of states $D\left(\theta\right)$, the number of states in the range $\left[\theta,\theta+d\theta\right]$ -- leading to $P\left(\theta\right) \propto e^{E\left(\theta\right)/k_bT} D\left(\theta\right)$. If $D\left(\theta\right)$ is constant, then it drops out and $P\left(\theta\right) \propto e^{E\left(\theta\right)/k_bT}$. (I think that $E\left(\theta\right)$ would have to be continuous for that to hold, but I can't think of any other restrictions.)
