A free pair turns, and forty-five degrees decides

A pair set free in a stream collides or parts let two cylinders move along their line of centres under the forces of an ideal stream and found one rule behind both outcomes: the pair moves so as to look bigger to the stream. Side by side, the pair’s total doublet grows as the cylinders close, and they close until they collide. In tandem it grows as they part, and they part for ever. The force was the slope of that doublet sum, and the motion was a particle rolling on it.

That essay ended on what its one coordinate could not show. At a fixed spacing the doublet sum is largest side by side and smallest in tandem, so a pair free to move in the plane should turn as well as close. This essay frees the angle and asks three questions: whether every released pair ends side by side, which ones collide and which part, and whether the tumbling that falling spheres show — two spheres touching and swinging round to fall side by side — is this.

The stream’s potential in two coordinates #

The relative motion of the pair has two coordinates: the spacing $D$ between the centres, in radii, and the angle $\varphi$ the line of centres makes with the stream. The pair’s centre drifts with the stream and, by the pair’s symmetry, its motion decouples from the relative one.

The stream’s side of the problem needs one observation. The pair’s doublet responds linearly to the stream, and by symmetry a stream along the line of centres induces a doublet along it and a stream across induces one across. So the pair’s total doublet along the stream at any angle is a mixture of the two arrangements the earlier essay computed,

$$S(D, \varphi) = S_t(D)\cos^2\varphi + S_s(D)\sin^2\varphi,$$

with $S_t$ the tandem sum and $S_s$ the side-by-side one. Computed directly from the successive images at four intermediate angles, the mixture matches to four parts in $10^{16}$ — the first check. The force on each cylinder is $\pi$ times the gradient of $S$, as before, so now it has a component across the line of centres too, $(\pi/D)\,\partial S/\partial\varphi$, and it turns the pair towards the angle at which $S$ is largest: side by side.

Two ways a pair can move relative to itself, and two added masses #

The fluid’s inertia enters through added mass, and a pair moving relative to itself does so in two ways. Closing, each cylinder moves towards the other along the line of centres and squeezes the fluid out of the gap between them; by symmetry that is a circle approaching a wall, the added mass the earlier essay computed. Sliding, each moves past the other the opposite way, across the line of centres. By the same symmetry argument sliding is a circle moving along a plane on which the velocity potential vanishes — a mirror that reverses a moving doublet instead of reflecting it — and its added mass comes from successive images by the same construction. A panel solution of a circle moving along such a plane matches it to two parts in $10^5$ at three gaps.

The first figure shows that they behave oppositely. Closing gets heavy as the gap shrinks — 1.26 displaced masses at a gap of one radius, 2.10 at a hundredth — because the fluid in a narrowing gap has to be driven out ever faster. Sliding gets lighter, to 0.65 at a hundredth of a radius: the two surfaces moving opposite ways past each other barely disturb the fluid between them, which in the limit is at rest in the gap while the bodies slide. A pair near contact is therefore easy to turn and hard to close, which will matter for how it meets.

The two masses differ for the reason the borrowed mass the boundary decides set out: a body near a wall and a body near a free surface borrow different masses from the same fluid, because one boundary stops the flow crossing it and the other stops the pressure changing along it. Each cylinder in the pair is the other’s boundary, and which kind of boundary it acts as depends on how the two are moving. The mass a body has to borrow is not a property of the body alone even in an unbounded fluid; here it is not even a property of the pair’s geometry, but of the direction of its motion.

With masses in the fluid’s own units and $m$ the cylinders’ own mass — one for neutrally buoyant bodies, zero for bubbles — the relative motion follows from

$$\frac{L}{\pi} = \tfrac14\left[(m + m_c)\dot D^2 + (m + m_p)D^2\dot\varphi^2\right] + S(D, \varphi),$$

integrated by fourth-order Runge–Kutta from rest. The energy it conserves is kept to two parts in $10^7$ over a whole collision.

Released at an angle, the pair swings #

The second figure is what released pairs do. Every one turns towards side by side as it moves, which is the rule of the earlier essay extended to the angle. Released near side by side, a pair closes and collides; released near tandem, it turns but parts, moving away along a path that curves across the stream as it goes.

Neither end is tidy. A pair that turns towards side by side arrives there moving, and the torque that brought it there only reverses once it has passed; there is no friction in the fluid and no loss anywhere, so the pair swings through side by side like a pendulum through the bottom of its arc. The paths that end on the contact circle arrive at many angles, not one.

The swing, and the gap closing under it #

The third figure follows one pair. Released six radii apart at 50° from the stream, it turns towards side by side over the first twenty radii of stream travel while its spacing hardly changes; reaches 90°; swings past; and comes back. All the while the spacing falls, slowly at first and then fast, because a pair near side by side is pulled together and the pull grows as the gap closes. The swing and the closing share the time. The pair meets at 77° from the stream, thirteen degrees short of side by side, after 34.8 radii of stream travel — the cylinders’ own diameter seventeen times over.

The pendulum’s period is set by the torque, which is the difference between the two arrangements’ sums, and that difference falls as the inverse square of the spacing. Far apart the pair swings slowly and closes slowly; close in it does both fast. That is why a pair released close to side by side from nearby hardly swings at all before it meets, and one released farther off can swing through side by side more than once.

A colliding pair meets wherever the swing has reached #

The fourth figure answers the first question directly: no. A pair released exactly side by side meets side by side; any other release that collides meets at whatever angle its swing has reached, and across the range of releases that collide the meeting angle runs all the way from side by side to within a few degrees of 45°. From four radii released at 47°, a pair meets at 48° — having swung up to side by side and all the way back.

Side by side is where the stream’s torque vanishes, and it is the arrangement a pair would settle in if anything removed its energy. In an ideal fluid nothing does. The arrangement the stream favours is an equilibrium the pair keeps passing through, not one it reaches.

Forty-five degrees, and why #

The fifth figure answers the second question, and the answer far apart is a number that can be predicted without integrating anything. Released from rest ten radii apart, a pair collides if it starts more than 44.6° from the stream and parts if less. Released from six radii the division is at 43.6°, from four at 41.3°, from three at 36.5° and from 2.2 at 19.5°: close in, even a pair released nearly in tandem is drawn round and collides.

The far-field value is 45° exactly, and the reason is a theorem. Far apart the two arrangements’ doublet sums approach 2 from opposite sides by the same amount, $S_s \approx 2 + 2/D^2$ and $S_t \approx 2 - 2/D^2$, so the potential the pair moves in is

$$S(D, \varphi) \approx 2 - \frac{2\cos 2\varphi}{D^2},$$

and the added masses are both one. That potential is homogeneous of degree $-2$ in the separation, and for such a potential the Lagrange–Jacobi identity says that the second time derivative of $D^2$ is proportional to the total energy and to nothing else. A pair released from rest has only the potential energy of its release, $2\cos 2\varphi_0 / D_0^2$ measured from the far-apart value, which is negative above 45° and positive below it. With negative energy, $D^2$ curves downward from the moment of release and must reach contact; with positive energy it curves upward and must grow without limit. Exactly at 45° the energy is zero, $D^2$ neither curves nor moves, and the pair keeps its spacing for as long as the far field holds while the angle swings.

Near the division the pair hesitates. Released from four radii, a pair at 89° meets after 6.5 radii of stream travel, one at 60° after 8.6, one at 45° after 17.3 and one at 42° — just above that spacing’s 41.3° — after 32.2, having swung round side by side and back more than once. A pair released exactly on the division would take for ever. Below it, a parting pair leaves at whatever angle its swing had reached when the pull faded, 26°, 80° or 40° from the stream for releases at 5°, 20° and 35°, and with a relative speed that the energy fixes: the depth of the potential it started in, $2 - S_0$, turned into kinetic energy with the far-apart added masses, a relative speed of 0.48, 0.42 and 0.26 of the stream’s.

It is the same kind of statement that makes a planet’s orbit bound or unbound by the sign of its energy, but sharper: for an inverse-square potential, rather than the inverse-distance one of gravity, the sign decides even the direction of the very first motion.

Bubbles, with no mass of their own, divide a little farther from tandem at every spacing — 20.4° from 2.2 radii, 44.7° from ten — because their inertia is all borrowed from the fluid, and the borrowed part changes with the gap.

The torque is the pair’s Munk moment #

A pair that has closed to touching behaves like a single body shaped like a figure eight, and the torque that turns it is recognisable. A body with no lift and a moment anyway found that an elongated body in an ideal stream carries no net force but does carry a couple, Munk’s moment, which turns it broadside. The pair’s stream potential is that moment’s source: in tandem the pair is a body aligned with the stream, side by side it is broadside, and $\partial S/\partial\varphi$ is the couple that turns one into the other. The same couple, acting on a single long body, is what makes an airship’s hull unstable in yaw and why its fins are the size they are.

It is worth setting this beside the other way ideal flow makes bodies move each other. Vortices move each other by carrying each other along in their own velocity fields, at speeds proportional to their strengths, with no inertia anywhere in the argument. Cylinders in a stream move each other by forces, through the pressure the stream’s disturbance puts on each surface, and the result is an acceleration divided by a mass that is mostly borrowed from the fluid — the same force of getting going that makes a body accelerating from rest feel heavier than it is. A pair of vortices never collides and never swings; a pair of cylinders does both, and the difference is entirely that one kind of object has inertia and the other has none.

Is this the tumble that falling spheres show? #

Two spheres falling one behind the other in a liquid, at Reynolds numbers of tens to hundreds, draft, kiss and tumble: the trailing sphere sits in the leading one’s wake, catches up, touches, and the touching pair swings round to fall side by side, after which the two drift apart. The ideal calculation predicts the first stage backwards — a tandem pair in an ideal stream parts — because drafting is a wake effect, and an ideal fluid has no wake.

The tumble is a different matter. A kissing pair aligned with the flow is a body in tandem, and the ideal couple turns it broadside, in the direction the spheres go. The calculation also says what an ideal fluid cannot do: it cannot make the pair stop side by side, since nothing damps the swing. Measured tumbles end with the pair side by side and separating, and both of those are viscous: the damping that ends the swing, and the side-by-side repulsion that viscous wakes produce at low Reynolds number. So the answer to the third question is yes and no. The last stage’s turning is this couple; the settling and the separation are not in it.

What was checked #

The ledger holds three checks: the stream potential at an angle against the image sum computed directly at that angle; the sliding added mass against a panel solution of the equivalent single circle; and the integration against its conserved energy. The far-field critical angle, 44.6° from ten radii, sits within half a degree of the theorem’s 45°, and the approach is from below because the near field, which the theorem leaves out, favours collision.

What the ideal pair leaves out #

A wake. Every force here is the ideal stream’s. A real pair at moderate Reynolds number has wakes, and the trailing body’s drag falls in the leading one’s wake — the drafting the ideal theory reverses.

Damping. Nothing removes the swing’s energy, so the swing never ends. In a real fluid viscosity damps it, and the pair would settle towards side by side if it did not collide first.

Three dimensions. Spheres rather than cylinders weaken every interaction, since a sphere’s disturbance falls as the cube of distance rather than the square — three dimensions are kinder in exactly this sense. The pair’s stream potential then falls as the inverse cube of the spacing, a homogeneous potential of degree −3, and the Lagrange–Jacobi identity no longer ties the curvature of the spacing to the energy alone; the critical angle is no longer a pure number, and would have to be computed.

Contact. The calculation stops at a gap of a fiftieth of a radius. What happens at contact — a bounce, a rolling contact, a lubrication film — is outside it.

The convention: angle from the stream #

Lengths are in cylinder radii, the stream’s speed is one and time is in radii of stream travel. The angle $\varphi$ is the line of centres’ angle from the stream: zero in tandem, 90° side by side, and folded into that quadrant because the pair is symmetric. Masses are in the fluid mass a cylinder displaces, $\rho\pi a^2$ per unit length. The potential $S$ is the pair’s total doublet along the stream in units of a single cylinder’s.

Who worked it out #

The forces between bodies moving in an ideal fluid go back to Kelvin, Kirchhoff and Bjerknes in the 1860s–70s, and Lagally’s theorem of 1922 gives the force on a body in terms of the singularities inside it. The Lagrange–Jacobi identity is from Lagrange’s and Jacobi’s work on the many-body problem, in which a homogeneous potential of degree −2 is the case that fixes the sign of the curvature of the moment of inertia by the energy alone. Drafting, kissing and tumbling were named by Fortes, Joseph and Lundgren in 1987.

Still open: a pair whose swing is damped #

The ideal pair keeps its swing because nothing takes the energy away. The next calculation adds the simplest damping that respects the physics — a drag on each cylinder’s motion relative to the stream proportional to its speed, as at low Reynolds number, with a coefficient set by a Reynolds number — and asks how strong it must be for a released pair to settle side by side before it collides, whether the damped pair released beyond 45° from tandem still collides at all or is caught into side by side at a finite spacing, and whether that spacing, where the damped pendulum comes to rest, depends on the Reynolds number in the way the side-by-side separation of falling spheres is measured to.