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Two positive numbers $\alpha$ and $\beta$ are given. We are going to describe a process of choosing a random vector on the unit sphere $S$ in $\mathbb R^3$ (given by $x^2+y^2+z^2=1$).

A vector $u\in S$ is chosen uniformly at random in the spherical disc with center $(0,0,1)$ and radius $\alpha$ on the sphere. After that a vector $v\in S$ is chosen uniformly at random on the spherical circle with center $u$ and radius $\beta$.

Can we describe the distribution of $v$ on the sphere (e.g., by specifying its density function or otherwise)? Clearly, $v$ lies in the spherical disc around $(0,0,1)$ of radius $\alpha+\beta$, however its distribution doesn't seem to be uniform. Can we estimate somehow the most frequent angle that $v$ subtends with $(0,0,1)$? The answer should certainly depend on $\alpha$ and $\beta$.

(By a "spherical disc" with center $c$ and radius $r$ we understand the set of points on the sphere whose distance from $c$ is at most $r$. Similarly, a "spherical circle" with center $c$ and radius $r$ is the set of points on the sphere whose distance from $c$ is exactly $r$.)

This question arouse related to the distribution of the polarization vector in a randomly grained polycrystalline material (ceramics).

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I think you can get the answer as a one-dimensional integral using the spherical law of cosines. –  Douglas Zare Sep 30 '12 at 20:44
    
Let $\theta$ be the angle the second radius $uv$ makes with the first radius $pv$ (where $p=(0,0,1)$) and let $\phi$ be the angle $v$ makes with $p$. Then the distribution of $v$ depends only on $\phi$, which in turn depends only on $\theta$, which is uniformly distributed. Calculate $\phi(\theta)$ and take $d\phi/d\theta$ and you have your answer. I get $\sqrt{1 - (\cot \alpha \cot \beta - \cos \phi \csc \alpha \csc \beta)^2} \csc \phi \sin \alpha \sin \beta$ in Mathematica, but I did not check this carefully. –  Yoav Kallus Sep 30 '12 at 21:43
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