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Two widely separated spheres rigidly joined by a hydrodynamically negligible connector obey the free-draining approximation. Their translational hydrodynamic resistance matrix is $6\pi \mu (a_1+a_2)I$. The displayed point is the centre about which every pure translation has zero hydrodynamic torque: ${\sum }_i{\zeta }_i({\mathbf{r}}_i-{\mathbf{r}}_c)=0$. A pure rotation about that point gives no resultant force at this order. This is an exact cancellation within the approximation, rather than an exact formula at arbitrary finite separation. It does not assert that every chiral rigid body has an origin eliminating its translation-rotation coupling.

Use the free-draining approximation: neglect the hydrodynamic interaction between the widely separated spheres and neglect the connector's viscous drag. Stokes law gives ${\mathbf{f}}_i=-6\pi \mu a_i\mathbf{U}$ for each sphere. Their resultant isThis is the leading two-sphere approximation, not an exact finite-separation hydrodynamic resistance matrix. Writing $d=|{\mathbf{r}}_2-{\mathbf{r}}_1|$ and $\mathbf{n}=({\mathbf{r}}_2-{\mathbf{r}}_1)/d$, the first Stokeslet reflection gives, if the connector remains negligible,The negative correction expresses reduced drag from fluid entrainment by the other sphere. A very thin but arbitrarily long rod need not have negligible total drag: its typical transverse drag scales as $\mu d/\log (d/b)$ for rod radius $b$. Thus the simple boxed result uses the intended negligibly dragging connector limit as well as $a_i/d\ll 1$; the two printed size inequalities alone do not specify a uniform limit in which rod drag vanishes.

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.