= Solution
With $x=\cos\theta$ and $D_r=0$, the <Fokker-Planck equation> becomes the probability-transport equation
$$
Bf_t+(1-x^2)f_x-2xf=0,
\qquad
\partial_tf+\partial_x\left[\frac{1-x^2}{B}f\right]=0.
$$
Its characteristic obeys $\dot x=(1-x^2)/B$. Writing $s=t/B$ and $q=\tanh s$ gives
$$
x=\frac{x_0+q}{1+qx_0},\qquad x_0=\frac{x-q}{1-qx},\qquad
\frac{dx_0}{dx}=\frac{1-q^2}{(1-qx)^2}.
$$
Conservation of orientation probability pushes the initially uniform density forward by this Jacobian. Thus
$$
f(x,t)=\frac{1-q^2}{4\pi(1-qx)^2}.
$$
Since $q=(1-\alpha)/(1+\alpha)$ with $\alpha=e^{-2t/B}$, the <deterministic alignment of an initially isotropic orientation distribution> is
$$
\boxed{f(\theta,t)=\frac{\alpha}{\pi[1+\alpha-(1-\alpha)\cos\theta]^2}.}
$$
At $t=0$, $\alpha=1$ and $f=1/(4\pi)$; the characteristic Jacobian preserves $2\pi\int_{-1}^1f\,dx=1$ at every finite time. Its reciprocal is quadratic in $x$, as in the suggested substitution.
To compute the mean, integrate explicitly:
$$
\begin{aligned}
\langle x\rangle
&=\frac{1-q^2}{2}\int_{-1}^1\frac{x\,dx}{(1-qx)^2}\\
&=\frac1q+\frac{1-q^2}{2q^2}\log\frac{1-q}{1+q}
=\coth s-s\operatorname{csch}^2s.
\end{aligned}
$$
Consequently
$$
\boxed{\mathbf V_c(t)=V_s\left[\coth(t/B)-\frac{t}{B}\operatorname{csch}^2(t/B)\right]\mathbf k.}
$$
The apparent singularity at $t=0$ is removable: $\langle x\rangle=2t/(3B)+O((t/B)^3)$, so the initial mean is zero. At late times,
$$
1-\langle x\rangle=(4t/B-2)e^{-2t/B}+O((t/B)e^{-4t/B}).
$$
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.
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