Average sine of an angle between two rays in a cone

Question

I'm looking for an average value of sine of an angle between two rays, lying within a cone with a certain angle.

Given a cone with an aperture of ${2\chi}$ and two rays lying within the cone. The rays can be represented as vectors in a spherical coordinate system:

$$ {\vec{e_1}=\lbrace1,\phi_1,\theta_1 \rbrace},{\vec{e_2}=\lbrace1,\phi_2,\theta_2 \rbrace} $$

where ${\phi_1,\phi_2\in[0,2\pi], \theta_1,\theta_2\in[0,\chi]}$ (assuming the axis of the cone is aligned with the z axis). The distribution of every angle is uniform. It is needed to find the average value of the sine of the angle between the rays (vectors).

We can get an answer by simply integrating the sine within a needed area: $$ {{\frac{1}{4\pi^2\chi^2}}\int\limits_{0}^{\chi}\int\limits_{0}^{\chi}\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}\sin(\vec{e_1},\vec{e_2})d\phi_1\phi_2\theta_1d\theta_2} $$

We can get cosine of an angle between the vectors using the dot product:

$${\cos(\vec{e_1},\vec{e_2})=\frac{(\vec{e_1},\vec{e_2})}{|\vec{e_1}||\vec{e_2}|}}=\sin\theta_{1}\sin\theta_{2}\cos\left(\phi_{1}-\phi_{2}\right)+\cos\theta_{1}\cos\theta_{2}$$

Using this an average value of cosine can be easily got:

$$ {{\frac{1}{4\pi^2\chi^2}}\int\limits_{0}^{\chi}\int\limits_{0}^{\chi}\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}\cos(\vec{e_1},\vec{e_2})d\phi_1\phi_2\theta_1d\theta_2=\frac{\sin^2\chi}{\chi^2}} $$

But I didn't have much luck trying to get the average value of sine. I can't solve this:

$$ {{\frac{1}{4\pi^2\chi^2}}\int\limits_{0}^{\chi}\int\limits_{0}^{\chi}\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}\sqrt{1-\left(\sin\theta_{1}\sin\theta_{2}\cos\left(\phi_{1}-\phi_{2}\right)+\cos\theta_{1}\cos\theta_{2}\right)^{2}}d\phi_1\phi_2\theta_1d\theta_2} $$

Using the Monte-Carlo simulation I got this curve:

Any help would be appreciated!

Answer 1

This is how I might set up this problem: just a hint at one possible start. (Not an answer!)

If we parametrize the unit sphere $S^2$ by $\mathbf{x}=(\theta,\phi)\in[0,2\pi)\times[0,\pi]$, then the area element is $dS=\sin\phi\,d\theta\,d\phi$, and we are interested in the polar region $R=[0,2\pi)\times[0,\chi]$ which has area $$ A=\int_{R}\,dS=\int_{0}^{\chi}\int_{0}^{2\pi}\sin\phi\,d\theta\,d\phi =2\pi\int_{0}^{\chi}\sin\phi\,d\phi =2\pi\left(1-\cos\chi\right). $$ The average value of the sine of the angle $\alpha=\alpha_{\mathbf{xy}}$ between uniformly random points $\mathbf{x},\mathbf{y}\in R$ is then $$ \eqalign{ \overline{\sin\alpha}& =\frac1{A^2}\iint_{R^2}\sin\alpha\,dS_{\mathbf{x}}\,dS_{\mathbf{y}} =\frac1{A^2}\iint_{[0,2\pi)\times[0,\chi]^2} \sin\alpha \sin\phi_\mathbf{x}\sin\phi_\mathbf{y} \, d\phi_\mathbf{x}\, d\phi_\mathbf{y} \, d\theta } $$ where now, like you, using subscripts $1$ and $2$ in place of $x$ and $y$, $$ \eqalign{ \cos\alpha &=\cos\phi_1\cos\phi_2+\sin\phi_1\sin\phi_2\cos\theta\\ &=\frac{\cos(\phi_1-\phi_2)+\cos(\phi_1+\phi_2)}2 +\frac{\cos(\phi_1-\phi_2)-\cos(\phi_1+\phi_2)}2\cos\theta\\ &=\cos(\phi_1+\phi_2)\frac{1-\cos\theta}2 +\cos(\phi_1-\phi_2)\frac{1+\cos\theta}2\\ &=\cos(\phi_1+\phi_2)\,\sin^2\frac\theta2 +\cos(\phi_1-\phi_2)\,\cos^2\frac\theta2\\ } $$ since $\theta=\theta_\mathbf{x}-\theta_\mathbf{y}$ is uniformly distributed in $[0,2\pi)$ (up to a trivial additive multiple of $2\pi$).