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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.

Choose time $t_c$ when the passing sphere centres are at the same height, and let $S$ reflect positions in that horizontal plane. For each instantaneous configuration, Stokes flow determines particle velocities uniquely and linearly from the fixed external forces. Time reversal reverses all velocities and the dense sphere's applied force; reflecting in the horizontal plane reverses that force again. The neutral sphere remains force-free under both operations. Thus the reflected time-reversed motion obeys the same original forcing.

At $t_c$, reflection fixes both particle centres. Since the spheres have no orientational shape variables and there is no contact, uniqueness of the particle-trajectory initial-value problem makes the reflected backward motion coincide with the forward motion:This reflection symmetry of a two-sphere passing trajectory fixes the transverse coordinates and reverses the relative axial separation. Consequently, at matching large separations above and below, both spheres return to their original distances from the centreline. Their intermediate lateral excursions are undone. This uses kinematic reversibility of Stokes flow, geometric reflection symmetry and the noncontact result, not a dissipation-minimizing trajectory. It assumes the spheres pass one another; a perfectly collinear approach would not supply the side-by-side symmetry event.

Let $\delta \ll a$ be the minimum gap as the spheres approach, and $w=-\dot{\delta }>0$ their normal relative approach speed. For smooth equal-radius spheres the gap has local form $\delta +O(s^2/a)$, so the thin-gap region has lateral scale $\ell \sim (a\delta {)}^{1/2}$. Incompressibility drives a radial escape speed of order $w\ell /\delta$. The lubrication pressure therefore scales asThe available gravitational forcing remains finite. Thus lubrication resistance gives $w=O(W\delta /(\mu a^2))$: the normal approach becomes proportionally slower as the gap closes. In particular, for a bounded coefficient $C$, $\dot{\delta }\ge -C\delta$, so $\delta (t)\ge \delta (t_1)e^{-C(t-t_1)}>0$ at every finite time. Equivalently the estimated time to reach zero contains ${\int }_0^{{\delta }_1}d\delta /\delta$, which diverges. This is how lubrication prevents finite-time collision of smooth spheres.

The small nonzero initial lateral offset breaks the exactly collinear configuration; the spheres can deflect sideways and pass with a positive gap rather than touch. The condition $R>2a$ leaves space for a side-by-side passage. Smooth surfaces, continuum hydrodynamics and the no-slip boundary condition are essential to the noncontact conclusion; the argument is not a model of surface roughness or molecular contact.

One normalization of the Papkovich–Neuber representation uses a harmonic vector $\mathbf{\Phi }$ and harmonic scalar $\psi$:These formulae give $\nabla \cdot \mathbf{u}=0$ because ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, and give $\mu {\nabla }^2\mathbf{u}=\nabla p$. Rescaling the potentials gives the other common normalizations of the representation.

A point force requires a decaying, rotationally covariant field linear in $\mathbf{F}$, with no length scale. Thus choose the harmonic monopole $\mathbf{\Phi }=c\mathbf{F}/r$, $\psi =0$, for $r>0$. Differentiating givesThe Newtonian fluid stress tensor is ${\sigma }_{ij}=-p{\delta }_{ij}+\mu ({\partial }_i{u}_j+{\partial }_j{u}_i)$, which simplifies to ${\sigma }_{ij}=-3\mu c(\mathbf{F}\cdot \mathbf{x})x_i{x}_j/r^5$. Its traction resultant on a sphere, using an outward normal from the origin, is $-4\pi \mu c\mathbf{F}$. Momentum balance with the force $\mathbf{F}\delta (\mathbf{x})$ therefore fixes $c=1/(4\pi \mu )$. The resulting Stokeslet isAn arbitrary constant pressure may be added. The stress has purely radial traction on concentric spheres, and its integrated traction $-\mathbf{F}$ supplies the required distributional point-force normalization.