Two spheres settling at a slant keep their slant for ever, and three do not

A force without the flow that makes it used the reciprocal theorem of Stokes flow to get Faxén’s law — a sphere in a flow moves with the flow’s average over its surface — without solving for the flow round the sphere at all. In a paragraph near its end it noted the theorem’s other immediate consequence: two spheres settling near each other fall faster than either would alone, because each sinks through the downwash of the other, and the interaction falls off only as one over the distance, the slowest decay in fluid mechanics. That was said and not computed.

This essay computes it, and finds that the speed-up is the less interesting half. A pair settling at a slant also drifts sideways, and it does so for ever, never changing its separation, its slant or its drift. Three spheres released together do none of that. The difference between two and three is a symmetry argument, and what the reciprocal theorem contributes to it is the symmetry.

How much faster #

A sphere of radius $a$ settling alone through a liquid of viscosity $\mu$ under its weight less its buoyancy, $F$, reaches Stokes’s speed $U_0 = F/6\pi\mu a$ in the time it takes viscosity to cross its own radius. Put a second, identical sphere at a distance $r$ and each moves in the flow the other makes. The leading term of that flow at a distance is the point force’s, the Stokeslet, whose velocity is $F/8\pi\mu r$ along the force and half that across it, so the pair falls faster by a factor $1 + 3a/2r$ when it is in line and $1 + 3a/4r$ when it is side by side. The factor of two between them is the Stokeslet’s own anisotropy: a point force drags fluid along its line twice as hard as across it.

Closer in, the finite size of each sphere matters, and the corrections come in powers of $a/r$. Batchelor wrote them out in 1976 for spheres of any two sizes; for equal spheres the speed along the line of centres is $1 + 3/2\rho - 1/\rho^3 - 15/4\rho^4$ and across it $1 + 3/4\rho + 1/2\rho^3 - 17/16\rho^6$, with $\rho = r/a$. For the pair in line there is something better than a series. Stimson and Jeffery solved it exactly in 1926 in bispherical coordinates, as an infinite series that converges at any separation, and it says that two touching spheres falling one above the other settle 1.550 times as fast as one alone. At four radii the exact answer is 1.347 and Batchelor’s series 1.345; at twenty, the two agree to a part in ten million.

The interaction is long-ranged in a way that is easy to underrate. At thirty radii — two spheres a millimetre across, fifteen millimetres apart — the pair still falls five per cent faster in line. Two grains of sand settling through honey a few centimetres apart are not independent, and a sedimentation experiment that drops particles one after another too quickly is not measuring single particles.

The drift at a slant #

When the line of centres is neither vertical nor horizontal, the pair’s speed along that line and across it differ, and the gravitational pull, resolved into those two directions, comes back out of the mobility pointing slightly off the vertical. The pair’s path leans. Write the slant $\theta$ as the angle between the line of centres and the vertical; the velocity has a component $A\cos\theta$ along the line and $B\sin\theta$ across it, with $A$ and $B$ the in-line and side-by-side factors, and the path’s angle from the vertical is $\arctan\big((A - B)\sin\theta\cos\theta / (A\cos^2\theta + B\sin^2\theta)\big)$.

The lean is towards the side the lower sphere is on, because the in-line speed is the larger: the pair moves more readily along its line of centres than across it, so it slides down that line a little as it falls. At five radii the largest lean is 3.1 degrees, at a slant just over 45 degrees. It is small, and it never stops: a pair five radii apart released with its line of centres at forty-five degrees drifts sideways by one part in eighteen of every distance it falls, for as long as it falls; through a column of liquid a metre tall that is five centimetres of sideways travel.

The lean is largest a few radii apart, 3.4 degrees near 3.7 radii, because it depends on the difference between the in-line and side-by-side speeds relative to their size. Far apart both are near one and the difference falls as $3/4\rho$; close together, finite size pulls the side-by-side speed up towards the in-line one, and inside a few radii the lean falls again.

Why the pair keeps its shape #

The striking thing is not the lean but its permanence. Two equal spheres under equal forces move with exactly the same velocity, whatever their slant and separation, so their separation and their slant never change. A pair is a rigid object in Stokes flow, though nothing holds it together.

The argument needs two things, and both are short. The first is a symmetry of the pair: turning every position inside out through the point midway between the spheres swaps the two spheres and leaves the pair looking exactly as it did. The same operation turns every vector into its opposite, so it turns the two downward forces into two upward ones, and it says that sphere one’s velocity in the real problem is minus sphere two’s velocity in the problem with the forces reversed. The second is the absence of inertia. In Stokes flow the velocities are linear in the forces, so reversing every force reverses every velocity: sphere two’s velocity with the forces reversed is minus its velocity with them as they are. Put together, sphere one’s velocity is sphere two’s. No relative motion is possible, whatever the slant.

What the reciprocal theorem adds is that the coupling between the spheres is symmetric: the velocity sphere one acquires from a force on sphere two equals, component by component and transposed, the velocity sphere two acquires from the same force on sphere one. For identical spheres the geometry gives that anyway; for a pair of different sizes it does not, and the theorem is what guarantees that a small sphere settling beside a large one is dragged exactly as hard by the large one’s flow as the large one is by an equal force on the small. The theorem forbids a pair from turning its gravity into a torque on itself — no Stokes-flow pair spins up like a propeller, however it is arranged.

A pair that cannot swim #

The same two facts — linearity and the absence of inertia — are what make a swimmer that cannot go backwards at low Reynolds number: a body that changes its shape by a sequence of strokes and then undoes them in reverse order ends where it began, because every velocity during the return exactly cancels one during the outward stroke. A settling pair is the passive version of the same statement. It has no strokes to make and only one force to obey, and the symmetry hands both spheres that force’s velocity; there is nothing in the problem that could make the shape evolve, so it does not. The reciprocal theorem’s symmetry of the coupling and the minimum-dissipation theorem behind the cheapest shape the walls allow are the two other consequences of the same linearity, and each fails at the same place, the first appearance of inertia.

A pair in a laboratory #

The numbers are easy to see. A steel ball a millimetre across falling through glycerol settles at about two and a half millimetres a second, at a particle Reynolds number of about a thousandth, which is Stokes flow by any standard. Two such balls released five radii apart — two and a half millimetres between centres — one above the other cross a 30-centimetre column in 92 seconds instead of the single ball’s 118; side by side, in 102. Released at a slant of forty-five degrees they arrive at the bottom about one and a half centimetres to the side of where they were dropped, still at the same slant and the same separation, which is the demonstration that makes the rigidity vivid. It is also the demonstration that is easiest to spoil: a few per cent difference in the balls’ sizes, a wall within a few centimetres, or a slight temperature gradient that drives a convection current is enough to make a real pair drift apart over that distance.

Three is the number where the shape goes #

Add a third sphere and the symmetry that protected the pair is gone. Three spheres released ten radii apart in a horizontal row are symmetric under reflection through the middle one, so the middle stays on the axis and the outer two move as mirror images; but nothing makes the middle one move with the outer ones. It sits in the downwash of both and they each sit in the downwash of one, so the middle falls faster. At the moment of release, with the spheres ten radii apart, the middle one falls at 1.151 times a single sphere’s speed — two side-by-side downwashes of 0.0755 each — and each outer one at 1.113, one neighbour at ten radii and the other at twenty. The difference, four per cent of the settling speed, is the whole of what starts the rearrangement, and it never has to grow large: over five hundred radii of fall it accumulates into the middle sphere’s four radii of lead. The outer spheres, falling slower, drift behind it and inward, drawn by the lateral component of its downwash.

Within 520 radii of fall — half a metre for millimetre spheres — the outer two have closed from twenty radii apart to two and a half, and the middle sphere has drawn more than four radii ahead of the trio’s centre. Hocking found this in 1964 and Jánosi, Tél and co-workers in 1997 showed that three point-like particles settling together scatter chaotically: their configuration after a long fall depends on their starting arrangement so sensitively that it cannot be predicted from any finite precision of it. The pair cannot do any of that; it has no internal dynamics at all. So three is the most that can be predicted for point vortices, and in settling it is the fewest that can rearrange.

Where the lean would show #

The sideways drift is the kind of effect that hides in a measurement rather than announcing itself. A settling experiment that tracks particles released together and averages their horizontal positions sees a spreading that looks like dispersion, and attributes it to noise, when part of it is pairs leaning deterministically by a few degrees and keeping it. In a dilute cloud, pairs form and break as third particles pass, so the lean of any one pair is interrupted; but the time between those encounters is long when the cloud is dilute, and a pair can drift several of its own separations sideways before a third sphere arrives to scatter it. The lean is also why two droplets settling in a stagnant layer — rain forming in still air at the smallest sizes, or emulsion droplets creaming in a cosmetic — collide less often in line than a vertical-only picture of their settling suggests: a pair approaching at a slant is carried sideways, round the other, rather than straight into it. That is the beginning of the calculation of a collision efficiency, and in Stokes flow a pair of identical spheres, which cannot change its separation at all, has a collision efficiency of zero.

What it means for a suspension #

A settling suspension is many spheres, and the pair’s speed-up suggests it should fall faster than a single sphere. It falls slower. Batchelor’s calculation of 1972, the one that made sense of this, found the mean settling speed of a dilute random suspension to be the single sphere’s times $1 - 6.55\phi$, $\phi$ the volume fraction, and the reason is the one the slow decay of the pair interaction warns about: summed over all the spheres in a suspension, the $1/r$ downwash of each diverges, and what makes the sum converge is the backflow — the liquid the sinking spheres displace must come up somewhere, and it comes up past them. The pair falls faster because it is alone in an infinite liquid; a suspension falls slower because it is not. That is the same divergence a viscosity made of particles needed an argument about the suspension as a whole to tame. The particle’s own inertia, which makes a particle a low-pass filter of the flow it moves in, is the first thing that breaks every symmetry above, long before the fluid’s does.

The convention, and what the picture leaves out #

Speeds are over the single sphere’s Stokes velocity and distances are between centres in radii. Every number is for Stokes flow, at zero particle Reynolds number; a sphere settling at a Reynolds number of even a tenth has an Oseen wake behind it that is not fore-and-aft symmetric, and the reversal argument that keeps the pair rigid fails at the order of that Reynolds number — how small is small enough is the question of when. Then a pair in line is drawn together, the trailing sphere caught in the leading one’s wake, and a pair side by side is pushed apart. The far-field series is used at three radii and beyond, where it agrees with the exact in-line solution to a few parts in a thousand; inside that, only the in-line exact solution is drawn. The three spheres are marched with the Rotne–Prager tensor, the far-field interaction to third order in $a/r$, and the march is stopped at 2.5 radii, where near-field lubrication forces it does not contain take over. The liquid is unbounded: in a container, the walls’ images add their own drift, and a pair near a wall does change its shape.

Who computed it #

Stimson and Jeffery solved two spheres falling along their line of centres in 1926, among the first exact solutions of Stokes flow for more than one body, and Smoluchowski had estimated the interaction of many particles in 1911. Batchelor’s 1976 paper gave the full mobility functions for two spheres of any sizes at large separation, and Jeffrey and Onishi in 1984 the exact functions at every separation. The rigidity of the settling pair was known to Happel and Brenner and is stated in Guazzelli and Morris’s text. Hocking computed three settling spheres in 1964; the chaos of three point particles is Jánosi, Tél, Wolf and Gallas’s of 1997.

How it was checked #

The exact in-line pair, summed in bispherical coordinates until the terms fall below a part in $10^{16}$, and Batchelor’s series agree to $1.9 \times 10^{-3}$ at four radii, $1.5 \times 10^{-4}$ at six, $6.5 \times 10^{-6}$ at ten and $9 \times 10^{-8}$ at twenty, the agreement improving as the series’ first omitted power says it should. Touching, the exact solution gives 1.550. The Rotne–Prager tensor gives the two spheres of a pair identical velocities at slants of 0°, 30°, 60° and 90° to rounding, which is the rigidity, and its coupling matrix equals its transpose with the spheres exchanged to rounding, which is the reciprocity. The lean is exactly zero in line and side by side. And the model refuses overlapping spheres, a touching pair in bispherical coordinates, where they are singular, and two spheres placed in the same place.

Still open: a pair that is not quite identical #

Every pair here is two identical spheres, and identity is what made the pair rigid. A pair of spheres that differ slightly — one a few per cent larger, or denser — settle at different speeds alone, and together each is carried by the other’s downwash and pushed by its own weight differently. The next calculation takes Batchelor’s mobility functions for unequal spheres and asks what a small mismatch does to the pair’s orbit: whether the faster sphere overtakes the slower one and passes on, or whether the interaction captures them into a pair that tumbles round itself as it falls — and how small a mismatch has to be, against the separation, for the slanted pair’s sideways drift to outlast it.