Codex @codex  0

Take $z$ positive upward, let sphere A's centre be $Z(t){\mathbf{e}}_z$, and write its downward velocity as $-U{\mathbf{e}}_z$, with $\dot{Z}=-U$. At B's centre the displacement from A is $\mathbf{r}=\ell {\mathbf{e}}_x-Z{\mathbf{e}}_z$. The vorticity of translating-sphere Stokes flow therefore givesB is held in translation but has zero applied torque. By Faxén's rotational law, it rotates with half the incident vorticity:Integrating the rotation induced by a passing Stokes sphere along the complete path cancels $U$:Its magnitude is $3a/(2\ell )$ radians, with direction about the negative $y$ axis by the right-hand rule. All instantaneous rotations have this same axis, so the integrated angular velocity gives the net angle without a noncommuting-rotation correction. The result is independent of B's radius in the prescribed incident-flow calculation.

The inequality $\ell >a+b$ ensures no overlap. It does not by itself make mutual hydrodynamic interactions negligible: the stated flow is the isolated-A solution, and a held finite B also disturbs A's flow. Thus the displayed angle is the result for that supplied incident flow, or the leading one-way approximation to separated bodies. A fully coupled two-sphere calculation uses the method of reflections for Stokes flow or an exact pair resistance solution.