= Solution
Fix the sense of rotation by taking the fluid <velocity> to be $\mathbf u=\Omega(-x_3,0,x_1)$, with $\Omega>0$. Its <vorticity> is $-2\Omega\mathbf e_2$. In the convention where $B$ is the <gyrotactic reorientation time> appearing with aligning rate $1/B$, the <bottom-heavy spherical-cell orientation dynamics> is
$$
\dot{\mathbf p}=\frac1B(\mathbf e_3-p_3\mathbf p)-\Omega\mathbf e_2\times\mathbf p.
$$
For the stable orientation in the transverse plane, write $\mathbf p=(\sin\vartheta,0,\cos\vartheta)$. The angle equation is $\dot\vartheta=-\sin\vartheta/B-\Omega$. If $B\Omega<1$, its stable fixed point satisfies
$$
\sin\vartheta_*=-B\Omega,\qquad\cos\vartheta_*=\sqrt{1-B^2\Omega^2},\qquad
\boxed{\mathbf p_*=(-B\Omega,0,\sqrt{1-B^2\Omega^2}).}
$$
The other planar fixed point has negative $p_3$ and is unstable. The positive-$p_3$ orientation also damps a small out-of-plane component because $\dot p_2=-p_3p_2/B$.
Once the orientation has relaxed to $\mathbf p_*$, the transverse trajectory obeys
$$
\dot x_1=-\Omega x_3-V_sB\Omega,\qquad
\dot x_3=\Omega x_1+V_s\sqrt{1-B^2\Omega^2}.
$$
Define the shifted axis by
$$
\boxed{x_{1A}=-\frac{V_s}{\Omega}\sqrt{1-B^2\Omega^2},\qquad x_{3A}=-V_sB.}
$$
Then $(x_1-x_{1A},x_3-x_{3A})$ rotates rigidly with angular speed $\Omega$, so its squared length is constant. These are the <gyrotactic circular paths in a rotating cylinder>, and
$$
\boxed{|OA|=\frac{V_s}{\Omega},\qquad \alpha=\arcsin(B\Omega).}
$$
Here $\alpha$ is the acute angle between the plane through the two parallel axes and the horizontal plane; the directed displacement $OA$ in the chosen cross-section points to the lower-left. Reversing the sense of rotation reverses the horizontal displacement. Arbitrary initial orientations have an orientation transient before the exact circular trajectories; neglecting randomness alone does not remove that transient. The bulk result also assumes that the trajectory does not intersect the cylinder wall.
Back to article page
