Wobbling saddles keep the quarter, and raise it

A lattice of vortices folds a cloud at the saddles’ quarter followed a cloud of heavy particles — droplets, dust — through the steady cellular flow $\psi = \sin x\,\sin y$, a chessboard of vortices joined at saddle points. The particles relax towards the air’s velocity over a response time $\tau$, and in units of the saddles’ strain rate that time is the Stokes number. The vortices fling the particles outwards to the cell walls, and the saddles where the walls meet decide whether they cross: below a Stokes number of a quarter nothing folds — no two particles ever arrive at the same place with different velocities — and above it the cloud folds onto the walls in finite time. The quarter is the converging saddle’s own threshold, the overshoot of a damped oscillator driven by a strain of one.

That essay ended with a doubt about its own result. Particles in turbulence fold at every Stokes number, at a rate that is exponentially small when the Stokes number is small but never zero, and a steady lattice is about as far from turbulence as a flow can be. Was the quarter a property of saddles, or of saddles that hold still? The simplest flow between the two is a lattice that moves, and this essay computes it.

A lattice that wobbles #

The stream function $\psi = \sin(x + \varepsilon\sin\omega t)\,\sin y$ is the same lattice shifted sideways by $\varepsilon\sin\omega t$, so its cells and saddles slide back and forth with amplitude $\varepsilon$ and frequency $\omega$. The air is still incompressible at every instant and still has saddles of strain one; what changes is that they move, and that fluid near the walls, which in the steady lattice circulates for ever in its own cell, is now handed back and forth between neighbouring cells. Solomon and Gollub built exactly this flow in 1988 — a row of convection rolls whose pattern was made to oscillate — and measured tracers diffusing along the row through those exchanges.

The particles’ calculation is the steady one with a moving field. Each particle carries the gradient of the particles’ own velocity field, $Z$, by its Riccati equation, $\dot Z = (\nabla u - Z)/\tau - Z^2$, and the cloud folds where the trace of $Z$, the rate at which the cloud is converging, runs to minus infinity in finite time — the particles’ version of following the parcel, with the parcel replaced by a particle that lags it. A fold is declared when it falls through $-50$. A cloud is a hundred or more particles seeded over one cell with the air’s own velocity, followed for sixty time units unless stated.

The threshold #

The answer is in the first figure, and it is the opposite of the suspicion. With no oscillation the calculation recovers the steady lattice’s quarter, 0.251 over this run length. Moving the lattice never lowers the threshold. At every amplitude up to one radian of the cell and every frequency from one to ten times the saddles’ strain rate it stays at or above a quarter, and at the largest amplitude it rises: to 0.283 at $\omega = 1$, 0.311 at 3 and 0.324 at 10. A cloud at $\mathrm{St} = 0.24$ in a lattice wobbling with $\varepsilon = 1$ and $\omega = 3$, followed for three hundred time units, does not fold at all.

So the quarter is not a property of steady saddles. It belongs to something that moving the lattice does not change, and the next figure shows what.

The floor under the convergence #

Along a single path the trace of $Z$ swings as the particle passes saddles and vortices, and at $\mathrm{St} = 0.24$ it never falls below −0.54. At 0.4 it falls through −50 at sixteen time units: a fold. The difference is what the Riccati equation does in a strain it cannot escape. Along the compressing direction of a saddle of strain $s$ the equation is, in one dimension, $\dot z = (-s - z)/\tau - z^2$. Its right-hand side has real zeros — a floor the convergence settles on — exactly when $1 \geq 4s\tau$; above that the zeros are complex, nothing holds $z$, and it runs away. With $s = 1$, the strongest strain anywhere in this flow, the floor exists for $\tau \leq \tfrac14$.

The floor has a value. At $\tau = 0.24$ in a strain of one the two zeros are at $-1.67$ and $-2.5$: the convergence can settle at $-1.67$ and no lower, and a particle that started gentler than that approaches it from above. At $\tau = 0.25$ the two zeros merge at $-2$, and for any longer response there is no zero to settle on. The measured path never comes near $-1.67$ — its least, $-0.54$, is reached in passing — because a particle in this lattice does not sit on a saddle’s compressing axis for long; it is swept past the saddle, as a parcel goes straight while the streamlines curve, and the compression it feels is a pulse rather than a steady push. The floor is the worst case, and the worst case is still bounded.

Moving the lattice changes where the strain is and when a particle meets it. It does not change its size: the oscillating flow is the steady one evaluated at a moving point, and its strain is never more than one. So a particle can be carried from saddle to saddle as often as the wobble likes, and at each it meets a strain the drag can hold. That is why the quarter survives. And it explains, in the same terms, why turbulence folds below any threshold: a turbulent strain is not bounded, and a rare fluctuation of strain $s$ larger than $1/4\tau$ held for long enough folds the cloud there, at a rate set by how rare such fluctuations are.

Above the threshold, the wobble saves particles #

The threshold rises with amplitude, and the share of the cloud folded shows why from the other side. Each curve switches from nothing to everything over a narrow band of Stokes numbers, and the band moves up with the wobble. At $\mathrm{St} = 0.28$ the steady lattice has folded all of its cloud, the lattice wobbling with $\varepsilon = 0.5$ has folded 81 per cent, and the one with $\varepsilon = 1$ none. By 0.36 all three have folded everything.

The reason is time. A fold needs a particle to sit in the compressing direction of a saddle long enough for its overshoot to develop — a few response times. In the steady lattice the particles flung to a wall slide along it into a saddle and stay in its neighbourhood as long as they like. In the moving lattice the saddle moves away from them, or they are carried across the wall into the next cell, before the overshoot has finished; only a particle with a longer response, and so a larger margin over the threshold, gets there in the time it is given. The wobble does not weaken the strain. It shortens the dwell.

That margin depends on the run length and on how finely the cloud samples the cell, which is a caution the steady calculation already found: its threshold fell from 0.2565 over twenty time units to 0.2501 over eighty. The moving lattice’s raised thresholds are measured over sixty time units with a hundred particles; at $\varepsilon = 1$ they hold over three hundred, but at smaller amplitudes a longer run with a finer cloud finds rare folds closer to the quarter — a cloud at 0.27 in the lattice wobbling with $\varepsilon = 0.5$ folds eleven per cent of its particles by three hundred time units. What no run finds is a fold below a quarter.

What the frequency does #

The three frequencies in the threshold figure tell the same story about dwell. A slow wobble, at a third of the strain rate, moves the saddles so gently that a particle sitting in one barely notices: the threshold at amplitude 0.3 is 0.253 and at amplitude one 0.257, hardly above the steady quarter. At the strain rate itself the threshold at amplitude one is 0.283. At three and ten times it, 0.311 and 0.324, and the curves for those two lie close together.

The particles’ response explains the order. A particle relaxes to the air over a time $\tau$, so it follows a wobble whose period is long compared with $\tau$ and cannot follow one whose period is short — a particle is a low-pass filter, and the wobble is a signal it filters. When $\omega\tau$ is small the particle moves with the pattern, sits in its saddle as the saddle moves, and the steady result holds. When $\omega\tau$ is large the particle barely moves while the pattern sweeps back and forth past it, so the saddle’s compression is applied to it in short bursts rather than continuously, and a particle needs a longer response — a larger margin — to accumulate the overshoot that folds it. At $\omega = 10$ and $\mathrm{St} = 0.3$, $\omega\tau$ is three: the particle is already mostly deaf to the wobble, which is why the threshold stops rising much beyond $\omega = 3$. The rise saturates at the amplitude that sets how far the saddles sweep, not at the frequency.

That is the same lag that makes a droplet turn or not at a bend in the flow, read in time rather than in space: the particle’s own time decides which features of the air it sees.

The particles travel and still do not fold #

None of this means the wobble leaves the particles where they were. The steady lattice keeps every particle in its own cell for ever. Wobbling with amplitude 0.3, it hands twelve per cent of a cloud of near-tracers to other cells within forty time units — the Solomon–Gollub transport — and seventy per cent of a cloud of particles at $\mathrm{St} = 0.2$. The inertial particles move far more because their inertia flings them out of the vortex cores to the walls, which is exactly where the exchange happens; a tracer that is not one drifts from the streamlines in this way whenever the flow curves.

So the moving lattice separates two things the steady one kept together. In the steady lattice, being gathered onto the walls and being folded onto them happen to the same particles above the same threshold. In the moving one, below the threshold, the particles are gathered onto the walls, handed from cell to cell, and spread through the lattice — and the cloud never once has two particles at the same place with different velocities.

The paths show the difference at a glance. In the steady lattice each particle spirals out to its own cell’s walls and circulates there, at a Stokes number at which it will be gathered onto the wall without limit but never fold. In the moving lattice the same particles leave their cells within a few circuits and wander through the lattice, and their cloud does the same thing without folding.

A droplet’s view #

For a cloud droplet the Stokes number is the droplet’s response time divided by the strain time of the air around it. A forty-micrometre droplet responds in about five milliseconds; at the dissipation scale of a vigorous cumulus the strain time is ten to forty milliseconds, so its Stokes number is between about a tenth and a half, and droplets of that size sit around the quarter. The calculation here says that the unsteadiness of that air does not by itself fold the droplets below a quarter, however much it moves them around; what folds them is strain stronger than the typical — the intermittent, rare, intense strain events that a cloud’s own turbulence carries. A droplet collision rate that depends on folding is therefore a statement about the tail of the strain distribution, not about the flow’s general unsteadiness.

What an experiment would see #

The prediction is testable in the apparatus that made the flow famous. A row of electromagnetically driven vortices in a thin layer of salt water — the descendant of Solomon and Gollub’s rolls, with the forcing made to oscillate — seeded with heavy particles of a known response time, photographed from above. Below a quarter, at any wobble, the particles should gather on the cell walls and travel along the row without ever forming the sharp crossing lines, the caustics, that appear when a cloud folds; just above a quarter in a steady forcing they should form caustics at once, and with the forcing oscillating they should not form them until the Stokes number has risen by the margin in the first figure. The caustics are visible directly as bright lines of concentrated particles, which is how they have been seen in turbulent sprays.

How it was checked #

The moving-lattice code reduces to the steady one at $\varepsilon = 0$ and returns its threshold, 0.2518 over forty time units, against the steady calculation’s 0.2519 over the same run. A cloud at $\mathrm{St} = 0.01$ in a lattice wobbling with $\varepsilon = 0.6$ never folds, as a cloud of near- tracers cannot. The flow’s divergence is zero to the last digit at every point and time tested, as it must be for a pattern that is only shifted. The Riccati form itself was checked in the steady calculation against finite differences of neighbouring paths, and is unchanged here except that the field it differentiates now carries a time.

The convention: Stokes number against the strongest strain #

The Stokes number is the particle’s response time in units of the inverse of the saddles’ strain rate, which is one in this lattice whether it moves or not. The oscillation’s amplitude $\varepsilon$ is in radians of the cell — a quarter-period of the pattern is $\pi/2$ — and its frequency $\omega$ in units of the strain rate. A fold is the trace of the particle-velocity gradient passing through −50, which the steady calculation showed is a definite crossing rather than a threshold on noise.

What the picture cannot show #

The flow is two-dimensional, periodic in space and in time, and has one frequency; turbulence has none of those properties. The calculation shows what unsteadiness alone does — it moves particles without folding them below a quarter — and the argument it gives for turbulence, that folding below a quarter needs strain stronger than one, is a reading of the Riccati equation’s one-dimensional form, not a computed turbulent rate. Particles here feel only drag: no gravity, which would make them settle through the lattice and fold them sooner, no collisions, and no effect on the air.

Who threw dust into moving cells #

Maxey and Riley wrote down the equation of a small heavy particle in 1983, and Maxey and Corrsin followed particles through a steady cellular flow in 1986, finding them gathered on the walls. Solomon and Gollub measured the transport of tracers between oscillating convection rolls in 1988, the flow used here. Falkovich, Fouxon and Stepanov in 2002 and Wilkinson and Mehlig in 2005 explained the folds of particle clouds in turbulence — caustics — as the Riccati equation’s runaway in a random strain, at a rate that is exponentially small in the inverse Stokes number. The observation that a bounded strain cannot fold particles whose response is quicker than a quarter of its time is the deterministic counterpart of their result.

Still open: a strain that is sometimes stronger #

The lattice here has its strain fixed at one, so the only way it could fold below a quarter would be through a strain it does not have. The next calculation gives it one: a lattice whose amplitude, and so whose strain, fluctuates — $\psi = A(t)\sin x\sin y$ with $A$ a random signal of mean one and a stated spread — and asks how the folding rate below a quarter depends on how often the strain exceeds $1/4\tau$, and whether it reproduces the exponential dependence on the inverse Stokes number that random flows give, which would put the turbulent caustic rate on the same footing as the quarter found here.