Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 81 / 4 / ii / Solution

With $x=\cos \theta$ and $D_r=0$, the Fokker-Planck equation becomes the probability-transport equationIts characteristic obeys $\dot{x}=(1-x^2)/B$. Writing $s=t/B$ and $q=\tanh s$ givesConservation of orientation probability pushes the initially uniform density forward by this Jacobian. ThusSince $q=(1-\alpha )/(1+\alpha )$ with $\alpha =e^{-2t/B}$, the deterministic alignment of an initially isotropic orientation distribution isAt $t=0$, $\alpha =1$ and $f=1/(4\pi )$; the characteristic Jacobian preserves $2\pi {\int }_{-1}^{1}fdx=1$ at every finite time. Its reciprocal is quadratic in $x$, as in the suggested substitution.

To compute the mean, integrate explicitly:ConsequentlyThe apparent singularity at $t=0$ is removable: $⟨x⟩=2t/(3B)+O((t/B{)}^3)$, so the initial mean is zero. At late times,Except for the exactly downward initial orientation, which has zero probability in the initial smooth distribution, every characteristic aligns upward. The infinite-time orientation distribution is a point mass at the upward pole.