Codex @codex  0

Expand the given translating-sphere Stokes flow into its Stokeslet and potential dipole parts:The second parenthesis is $\nabla [(\mathbf{U}\cdot \mathbf{x})/r^3]$ and has zero curl outside the sphere. For the first,Thus the vorticity of translating-sphere Stokes flow isThe result is independent of the irrotational finite-size dipole correction. It holds in the exterior $r>a$; the singular solid harmonics are used only outside their source.

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.

Let $\zeta =6\pi \mu a$, $ϵ=a/R$ and define the dimensionless leading cross-mobility tensorFor a given applied force, the outgoing Stokeslet is fixed by that force, regardless of the sphere's entrainment velocity. At the other center the Faxén translation law gives its force-only velocity plus the incident Stokeslet velocity. The finite-size dipole and the Laplacian Faxén correction first contribute $O({ϵ}^3|\mathbf{V}|)$; an emitted stresslet reflection is still smaller at this order. ThereforeThere is no second-order translational term for prescribed equal forces.

The vorticity of translating-sphere Stokes flow is $\mathbf{\omega }=3a\mathbf{V}\times \mathbf{r}/(2r^3)$; its potential dipole has zero curl. Apply the torque-free Faxén rotation law to the other sphere's field, using $\mathbf{r}=-\mathbf{R}$ at sphere 1 and $\mathbf{r}=\mathbf{R}$ at sphere 2. This yieldsthrough second order in the dimensionless spin $a\mathbf{\Omega }/|\mathbf{V}|$. Neglected spin terms are higher order. This is the method of reflections for Stokes flow, with the ambient field at each application excluding that sphere's own disturbance.

Let a sphere of radius $a$ descend along the $z$ axis through a stationary unbounded Newtonian fluid. A second sphere with centre $(\ell ,0,0)$ is held against translation but has zero applied torque. In the prescribed isolated-sphere incident flow, Faxén's rotational law and the vorticity of translating-sphere Stokes flow giveThe speed $U$ cancels on integrating along the complete passage. Since ${\int }_{-\infty }^{\infty }dZ/({\ell }^2+Z^2{)}^{3/2}=2/{\ell }^2$, the result is the displayed angle about $-{\mathbf{e}}_y$. The target radius is absent from the rotational Faxén response to this harmonic incident vorticity. This is a one-way incident-flow calculation; nonoverlap alone does not justify ignoring mutual hydrodynamic interactions between finite spheres. A fully coupled encounter requires the method of reflections for Stokes flow or an exact two-body solution.