 Solution

Use the rotation convention of part (a), so the orientation drift is $B^{-1}({\mathbf{e}}_3-p_3\mathbf{p})-\Omega {\mathbf{e}}_2\times \mathbf{p}$. The steady gyrotactic orientation Fokker-Planck equation on the unit sphere isLet $\eta =(B\Omega {)}^{-1}$, $\lambda =(B{D}_R{)}^{-1}$, and introduce the rotation generator $\mathcal{A}=({\mathbf{e}}_2\times \mathbf{p})\cdot {\nabla }_p$. Dividing by $\Omega$ givesThe leading isotropic density is $f_0=1/(4\pi )$. At first order, writing $f=f_0+\eta f_1+\cdots$, rotational diffusion makes no contribution because ${\Delta }_p{f}_0=0$. The tangential drift is the spherical gradient of $p_3$, whose divergence is ${\Delta }_p{p}_3=-2p_3$. ThereforeIn the polar convention of the paper, $p_1=\sin \theta \cos \varphi$, and $\mathcal{A}{p}_1=p_3$. Thus, in the specified first-harmonic form,The correction integrates to zero, preserving normalization. The first correction has no dependence on $\lambda$ because the isotropic leading density has zero spherical Laplacian; $\lambda$ enters the next-order equations.

Spherical symmetry gives $\int p_i{p}_{j}d\Omega =4\pi {\delta }_{ij}/3$. Consequently the rapid-rotation mean gyrotactic orientation isIt is this small horizontal swimming bias, rather than a mean vertical orientation at leading order, that enters the concentration equation of part (c).