The Stokes-Einstein rotational diffusion relation tells us that we can write down a rotational diffusion coefficient for a sphere as:


$D\_r \approx \frac{k\_B T}{\zeta\_f} \approx \frac{k\_B T}{(8 \pi \eta)(r)^3}$


Where $k\_B$ is Boltzmann's constant, $T$ is the temperature in Kelvin, $\zeta\_f \approx (8 \pi \eta)(r)^3$ is the friction, $\eta$ is the viscosity of the medium (e.g. $\approx 1 \space cP$ in pure water), and $r$ is the radius of the sphere.


For an example calculation of $D\_r \approx 0.2 \space rad^2/s$ (or $Hz$) of a $\approx 1 \space \mu m$ radius sphere in pure water at room temperature, please see here: <http://www.microurl.info/YOtwp> (this shortened URL is going to WolframAlpha).


My question is the following: Take a point on the sphere undergoing rotational diffusion in solution, and select a random point on its surface. Define a vector $v\_1$ between this random point and the point at the center of the sphere. Let $v\_2$ be another randomly oriented vector. What probability distribution do we have for the time until a minimum angle $\theta$ between $v\_1$ and $v\_2$ is achieved?


Alternatively, if we start with the vectors $v\_1$ and $v\_2$ being within a minimum angle $theta$, what time distribution do we have until the minimum angle becomes larger than $\theta$?



Stokes-Einstein rotational diffusion and vector orientation time
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](http://arxiv.org/abs/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.


