Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 74 / 1 / vii / Solution

When both spheres have the same effective downward weight, the second sphere adds a force-driven disturbance instead of being merely a force-free rigid inclusion. The power now contains both $W{V}_{1z}$ and $W{V}_{2z}$, so the one-forced-sphere bound from part (iv) no longer applies.

For $a\ll R$, place the spheres in a tandem configuration with $a\ll s\ll R$. Locally the flow around the second sphere is dominated by its downward Stokeslet. Its axial incident velocity at the first sphere is downward, so the first sphere falls faster to leading order. By contrast, the zero total flux for localized tube Stokes flow requires compensating upward flow away from the downward-moving particle core in sections cutting the sphere. There are therefore positions in the return flow where the disturbance acts upwards.

One can choose the second sphere in such an upward-flow region generated by the first. The Lorentz reciprocal theorem makes the axial cross coefficients of the hydrodynamic mobility matrix equal, so the downward force on the second sphere then produces an upward contribution to the first sphere's velocity. With small well-separated spheres, this force-driven interaction dominates the higher-order self-resistance change from simply inserting the other sphere. In mobility notation, with downward positive,and the cross coefficient can have either sign in the tube. ThusThe distinction is the sign of the incident flow, not whether adding a second geometrical obstacle lowers a fixed-velocity dissipation.