Leading interaction of two sedimenting spheres ($\dot{\mathbf{r}}=-\delta \hat{\mathbf{z}}-3D(I+\mathbf{nn})\hat{\mathbf{z}}/(4r)$)

For two well-separated spheres with isolated downward speeds $V_i$ and radii $a_i$, let $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $r=|\mathbf{r}|$, $\mathbf{n}=\mathbf{r}/r$, and let $\hat{\mathbf{z}}$ point upwards. Their leading Stokeslet interactions giveEach sphere's Stokes drag law supplies force $-6\pi \mu a_i{V}_i\hat{\mathbf{z}}$ on the fluid. Evaluate the other's Stokeslet at its centre and subtract the two velocities to obtain the equation. Corrections from finite size and potential dipoles are higher order in radius divided by separation. The approximation must remain well separated throughout the segment of trajectory to which it is applied.

If $\delta D<0$, the leading interaction of two sedimenting spheres has a formal vertical equilibrium at $r_{\ast }=-3D/(2\delta )$. Choose labels with $\delta >0,D<0$ and put sphere one above sphere two. On that invariant vertical line, $\dot{r}=-\delta +\delta r_{\ast }/r$, so both sides approach $r_{\ast }$ and the radial eigenvalue is $-\delta /r_{\ast }$. A small tilt instead grows at rate $\delta /(2r_{\ast })$: the full configuration-space equilibrium is a saddle, not an attracting node. A physically valid far-field bound pair requires $r_{\ast }\gg a_1+a_2$, or $|D|\gg (a_1+a_2)|\delta |$. Reversing both driving forces reverses the vector field, interchanging radial attraction and repulsion, consistent with kinematic reversibility of Stokes flow.

In the leading interaction of two sedimenting spheres, put $c=3D/4$ and let $\theta$ be the angle from the upward vertical. The equations are $\dot{r}=-(\delta +2c/r)\cos \theta$ and $r\dot{\theta }=(\delta +c/r)\sin \theta$. Their ratio separates toso $r(\delta r+2c){\sin }^2\theta$ is invariant. When $\delta ,c$ have the same nonzero sign, choose labels so both are positive. For a noncollinear encounter the invariant is positive, ensuring a strictly positive closest separation. Meanwhile $\dot{z}=-\delta -c(1+{\cos }^2\theta )/r<0$, so the vertical order reverses and eventually $r\ge |z|\to \infty$. Exactly collinear contact is outside this passing argument.