Codex @codex  0

Hold the two-sphere object fixed in a uniform stream $\mathbf{V}$. In the free-draining approximation, its drag forces are ${\mathbf{f}}_i={\zeta }_i\mathbf{V}$, with ${\zeta }_i=6\pi \mu a_i$. The hydrodynamic torque about ${\mathbf{r}}_c$ isTo make it zero for every incoming-stream direction, choose the hydrodynamic centre of a two-sphere dumbbell:It lies closer to the larger sphere, whose larger drag requires a shorter lever arm. Uniform translation does not spin an isolated sphere, so no additional isolated-sphere spin torque enters this leading approximation.

Finite-separation hydrodynamic interactions and connector drag can change the displayed location; the calculation has the same approximation as part c.

For rotation about the hydrodynamic centre of a two-sphere dumbbell, the sphere centres have velocities ${\mathbf{v}}_i=\mathbf{\Omega }\times ({\mathbf{r}}_i-{\mathbf{r}}_c)$. Their resultant drag force in the free-draining approximation isEach sphere also experiences a spin drag torque, but its isolated rotation causes no net force. Those spin contributions therefore do not alter the cancellation.

This illustrates translation-rotation reciprocal coupling. About any fixed origin, write the fluid's rigid-body force and torque asThe transposed cross-blocks follow from the Lorentz reciprocal theorem. If pure translation causes zero torque about an origin for every $\mathbf{U}$, then $B^T=0$, hence pure rotation about that origin causes zero force. This implication is general for linear Stokes flow; the existence and leading location of such an origin here use the two-sphere geometry and the approximation in part c.