 Deterministic alignment of an initially isotropic orientation distribution

Without rotational diffusion, $x=p_z$ follows $\dot{x}=(1-x^2)/B$, hence $x=(x_0+\tanh (t/B))/(1+x_0\tanh (t/B))$. Pushing forward the initial uniform orientation distribution gives $f=\alpha /[\pi (1+\alpha -(1-\alpha )x{)}^2]$, $\alpha =e^{-2t/B}$, per unit solid angle. Integrating $2\pi xf$ yields the displayed mean, interpreted continuously as zero at $t=0$. Its late deficit is $(4t/B-2)e^{-2t/B}+O((t/B)e^{-4t/B})$. At each finite time the density is smooth and normalized; the infinitely late distribution is a point mass at the upward pole.