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If I have understood you correctly, your rotational diffusion problem is equivalent to the problem of a particle diffusing on the surface of a sphere (angular diffusion constant $D_r$), starting at some polar angle $\theta_0\in(0,\pi)$ and first crossing the angle $\theta\in(0,\theta_0)$ at time $t$. This is a classic problem in first-passage-times, see for example S. Redner, A guide to first-passage processes, or arXiv:1101.5043.

The mean first passage time is $$\langle t\rangle=\frac{2}{D_{r}}\ln\left(\frac{\sin(\theta_0/2)}{\sin(\theta/2)}\right).$$

If you choose the starting point at random, uniformly on the sphere, then you would perform one more average over $\theta_0$ (with weight $\sin\theta_0$), producing the average

$$\langle t\rangle=-\ln[\sin(\theta/2)]-\frac{1+\cos\theta}{4}.$$

The full probability distribution of the first passage time seems more problematic, I have not found it in the literature.

If I have understood you correctly, your rotational diffusion problem is equivalent to the problem of a particle diffusing on the surface of a sphere (angular diffusion constant $D_r$), starting at some polar angle $\theta_0\in(0,\pi)$ and first crossing the angle $\theta\in(0,\theta_0)$ at time $t$. This is a classic problem in first-passage-times, see for example S. Redner, A guide to first-passage processes, or arXiv:1101.5043.

The mean first passage time is $$\langle t\rangle=\frac{2}{D_{r}}\ln\left(\frac{\sin(\theta_0/2)}{\sin(\theta/2)}\right).$$

If you choose the starting point at random, uniformly on the sphere, then you would perform one more average over $\theta_0$ (with weight $\sin\theta_0$), producing the average

$$\langle t\rangle=-\frac{2}{D_{r}}\left[\ln[\sin(\theta/2)]+\frac{1+\cos\theta}{4}\right].$$

The full probability distribution of the first passage time seems more problematic, I have not found it in the literature.

If I have understood you correctly, your rotational diffusion problem is equivalent to the problem of a particle diffusing on the surface of a sphere (angular diffusion constant $D_r$), starting at some polar angle $\theta_0\in(0,\pi)$ and first crossing the angle $\theta\in(0,\theta_0)$ at time $t$. This is a classic problem in first-passage-times, see for example S. Redner, A guide to first-passage processes, or arXiv:1101.5043.

The mean first passage time is $$\langle t\rangle=\frac{2}{D_{r}}\ln\left(\frac{\sin(\theta_0/2)}{\sin(\theta/2)}\right).$$

If you choose the starting point at random, uniformly on the sphere, then you would perform one more average over $\theta_0$ (with weight $\sin\theta$).

The full probability distribution of the first passage time seems more problematic, I have not found it in the literature.

If I have understood you correctly, your rotational diffusion problem is equivalent to the problem of a particle diffusing on the surface of a sphere (angular diffusion constant $D_r$), starting at some polar angle $\theta_0\in(0,\pi)$ and first crossing the angle $\theta\in(0,\theta_0)$ at time $t$. This is a classic problem in first-passage-times, see for example S. Redner, A guide to first-passage processes, or arXiv:1101.5043.

The mean first passage time is $$\langle t\rangle=\frac{2}{D_{r}}\ln\left(\frac{\sin(\theta_0/2)}{\sin(\theta/2)}\right).$$

If you choose the starting point at random, uniformly on the sphere, then you would perform one more average over $\theta_0$ (with weight $\sin\theta_0$), producing the average

$$\langle t\rangle=-\ln[\sin(\theta/2)]-\frac{1+\cos\theta}{4}.$$

The full probability distribution of the first passage time seems more problematic, I have not found it in the literature.

If I have understood you correctly, your rotational diffusion problem is equivalent to the problem of a particle diffusing on the surface of a sphere (angular diffusion constant $D_r$), starting at some polar angle $\theta_0\in(0,\pi)$ and first crossing the angle $\theta\in(0,\theta_0)$ at time $t$. This is a classic problem in first-passage-times, see for example S. Redner, A guide to first-passage processes.

The mean first passage time is $$\langle t\rangle=\frac{1}{D_{r}}(\theta_0-\theta)(2\pi-\theta_0-\theta)$$.

If I have understood you correctly, your rotational diffusion problem is equivalent to the problem of a particle diffusing on the surface of a sphere (angular diffusion constant $D_r$), starting at some polar angle $\theta_0\in(0,\pi)$ and first crossing the angle $\theta\in(0,\theta_0)$ at time $t$. This is a classic problem in first-passage-times, see for example S. Redner, A guide to first-passage processes, or arXiv:1101.5043.

The mean first passage time is $$\langle t\rangle=\frac{2}{D_{r}}\ln\left(\frac{\sin(\theta_0/2)}{\sin(\theta/2)}\right).$$

If you choose the starting point at random, uniformly on the sphere, then you would perform one more average over $\theta_0$ (with weight $\sin\theta$).

The full probability distribution of the first passage time seems more problematic, I have not found it in the literature.