A lattice of vortices folds a cloud at the saddles' quarter

Drag decides whether a cloud folds, and the air decides at what found two thresholds for a cloud of heavy particles, each relaxing towards the air’s velocity over its own response time. In still air a cloud released with a spread of velocities folds — two particles arrive at the same place from different starts, and the density there is infinite along a curve — only if the response time exceeds the time the spread would take to close the cloud: a Stokes number of one. In air converging at a steady rate, a particle approaching the convergence line is a damped oscillator, and it overshoots, meeting particles from the other side, once the response time exceeds a quarter of the convergence time.

Those were the two simplest airs, and turbulent air is neither. It turns as well as converges: its vortices fling heavy particles outwards and its regions of strain gather them. The essay ended by asking what the threshold is in air that does both — the cellular flow $\psi = \sin x\sin y$, a lattice of vortices with converging lines between them — and whether the answer would be the strain’s quarter, or something between the two thresholds because each particle spends part of its time in a vortex. It is the quarter, exactly, and the reason says something about where in a turbulent flow a cloud first folds.

Cells and saddles #

The cellular flow is a chessboard of square vortices, each turning the opposite way to its four neighbours. Its velocity is $(\sin x\cos y,\ -\cos x\sin y)$. At the centre of each cell the air rotates with vorticity two. At each corner four cells meet in a saddle, where the air arrives along one direction and leaves along the other at a strain rate of one. The cell walls are the streamlines joining the saddles, and the air never crosses them.

A heavy particle obeys $d\mathbf v/dt = (\mathbf u(\mathbf x) - \mathbf v)/\tau$, and with the strain rate one the Stokes number is just $\tau$. The cloud starts as the air: one particle at every point, each moving with the air’s velocity there.

Where a cloud folds, and how to find it #

The paths show the two behaviours. At a Stokes number of 0.15 the rotation flings the particles outwards — they cannot follow the curving streamlines and drift across them — and they approach the cell walls ever more closely without crossing them, so the cloud is compressed towards a set of lines. At 0.4 they reach a wall with enough momentum to overshoot it into the next cell, where particles from that cell are arriving the other way. Wherever two such streams meet, the cloud has folded.

Finding the fold everywhere at once needs more than paths. Each particle carries the gradient of the particle velocity field around it, $Z = \partial\mathbf v/\partial\mathbf x$ along the cloud. It obeys a Riccati equation,

$$\frac{dZ}{dt} = \frac{\nabla\mathbf u - Z}{\tau} - Z^2,$$

and the cloud’s local area changes at the rate $\operatorname{tr} Z$. Where the cloud folds the area goes through zero, and $\operatorname{tr}Z$ runs to minus infinity in finite time. A fold is declared when $\operatorname{tr}Z$ falls below $-50$.

The threshold is a quarter #

Above a quarter the cloud folds, and the first fold’s time falls steeply as the Stokes number rises: 1.21 time units at a Stokes number of one, 6.24 at 0.3, 12.8 at 0.26, 21.5 at 0.255. Approaching a quarter the first fold recedes towards infinity. Below a quarter, in runs of up to 160 time units — eighty turnovers of the cells — nothing folds anywhere.

The threshold found by bisection over a cell’s cloud is 0.2565 when the run is 20 time units, 0.2519 at 40, and 0.2501 at both 80 and 160. It is the quarter. The vortices do not lower it, though the particles spend most of their time inside them, and they do not raise it, though they fling particles apart rather than together.

Why the saddles decide #

The vortices’ part in the story is to deliver. Flung out of the cells, every particle ends up near a wall, and every wall leads to a saddle. A particle travelling into a saddle along its converging direction meets air whose velocity is $-\sin y$, so near the saddle $\tau\ddot y + \dot y + y = 0$ — the converging-air oscillator of the previous essay with the convergence rate one. Below a quarter it is overdamped: the particle creeps into the saddle and never passes it, and is carried out along the diverging direction. Above a quarter it is underdamped: it overshoots the saddle into the next cell, swings back, and meets particles arriving from that side. At exactly a quarter it is critically damped.

So the lattice’s threshold is the saddle’s threshold, because the saddle is the only place the cloud’s particles can be carried through a line along which they are converging. Inside a cell the rotation separates particles; along a wall they approach it asymptotically; at a saddle they either stop short or overshoot. The rotation affects how soon a cloud folds above the threshold, by how quickly it delivers particles to the saddles, and not where the threshold is.

The previous essay drew the same conclusion in a different setting: in a nonlinear converging flow the first crossing happens where the convergence is strongest, and the threshold is the linear one there. In the lattice the strongest convergence is at the saddles, and the strain rate there is the whole of the answer. In turbulence the saddles are the strain-dominated regions between vortices, and the prediction it makes is that clouds fold first there, at a Stokes number of a quarter based on the local strain — a prediction about simulated turbulence that this calculation does not test.

What the vortices contribute #

The vortices are not idle. They decide how fast particles reach the saddles, and so how long a cloud takes to fold above the threshold, and they decide where the cloud goes below it. A particle in a vortex is being accelerated towards the centre all the time, the centripetal acceleration that steady does not mean nothing is happening measured for parcels in a steady flow; a heavy particle cannot supply that acceleration fully from its drag, so it falls outwards, at a speed of order the Stokes number times the centripetal acceleration. Near the centre of a cell, where the rotation is nearly solid-body, that drift is a slow outward spiral; near the walls the flow is mostly strain and the drift becomes an approach to the wall.

What the vortex cannot do is make paths cross. Its rotation is spin, not a convergence: it sends neighbouring particles outwards at nearly the same rate, and a cloud being spread cannot fold. Only a region where the air converges — and in this flow the convergence is concentrated at the saddles, the hyperbolic stagnation points whose index the sign a stagnation point carries in space counted — can bring two particles from different starts to the same place. So the threshold is set where the convergence is, and the vortices only feed it.

A threshold that was rounding #

The first version of this calculation found a different answer, and it is worth showing why. It carried the deformation itself, $X = \partial\mathbf x/ \partial\mathbf a$, along each path, and declared a fold where its determinant changed sign. Over runs of 20 and 40 time units it agreed with the Riccati calculation to four figures. Over longer runs its threshold kept falling — 0.2454 at 80 time units, 0.1435 at 160, and at 320 it had reached 0.10 — as though any cloud would fold if one waited.

It would not. Along the attracting walls the deformation stretches along the wall and compresses across it, both exponentially. Its entries grow like $e^t$ while its determinant, the product, shrinks; after a few tens of time units the determinant is smaller than the rounding error in the entries it is computed from, and its sign is noise. Every spurious fold the direct method found was a sign flip in a number that had no significant digits left. The Riccati form carries the velocity gradient, which stays of order one along the walls and diverges only at a real fold, and it is the right tool for exactly the reason the other one fails.

That trap is general. Any calculation that integrates a deformation gradient through a flow with an attractor — dust in a turbulent flow, ink in a mixing tank, a cloud of droplets in a vortex street — reaches the point where the determinant is lost in rounding within a few multiples of the inverse contraction rate, and from then on it reports folds that are not there.

Below the threshold, the walls gather everything #

Below a quarter the cloud never folds, but it does not stay spread either. The previous essay’s question had a second half: how dense the sheets between the cells become. The answer is that they have no limit. The cloud’s area shrinks at a steady exponential rate, the average of $-\operatorname{tr}Z$ along the paths: 0.056 per time unit at a Stokes number of 0.05, 0.28 at 0.15 and 0.72 at 0.24. Tracers keep their area; heavier particles are concentrated onto the walls faster the closer they are to the threshold, and the density on the walls grows without bound for as long as the flow lasts.

The particles’ distances to the walls show the same thing directly. A cloud spread evenly over a cell would have half its particles within 0.46 of a wall. After thirty time units at a Stokes number of 0.05 the median particle is 0.094 from a wall; at 0.1, 0.013; at 0.15, 0.0013; and at 0.24, less than a millionth. In a steady lattice the cloud becomes a set of sheets of zero thickness, the cell walls, which is the steady-flow form of the preferential concentration that the tracer that is not one described: heavy particles leave the vortices and gather in the strain. In a turbulent flow, where the walls themselves move and break, the sheets reach a finite thickness set by how long a wall lasts.

Checks on the folding cloud #

The Riccati equation is checked against the definition it replaces. Paths started a millionth apart give the deformation and the particle-velocity differences by finite differences, and $V X^{-1}$ from them agrees with the carried $Z$ to $5\cdot10^{-7}$ after three time units at a Stokes number of 0.4. A single particle released on a saddle’s converging axis passes the saddle at 0.26 and not at 0.24, as the oscillator says. At a Stokes number of 0.02 — nearly a tracer — no particle folds in twenty time units. And the threshold over 160 time units with 256 particles is a quarter to four figures. The tests also refuse a Stokes number of zero and a negative one.

A droplet’s view #

The numbers translate into a cloud of water droplets in turbulent air. A droplet’s response time is $2\rho_w a^2/9\mu$, with $a$ its radius and $\mu$ the air’s viscosity — about a thousandth of a second for a ten-micrometre droplet and a hundredth for thirty micrometres. In a cloud whose strain regions turn over in a tenth of a second, the quarter is reached at a response time of $0.025$ seconds, a droplet of about forty-five micrometres. In the most turbulent cumulus, where the smallest eddies turn over in two hundredths of a second, droplets of twenty micrometres reach it. The strain rate of the eddies a droplet meets, not its own size alone, decides which side of the quarter it is on, and a cloud of one droplet size can be below the threshold in its calm parts and above it in its turbulent ones. The saddle threshold therefore marks the size at which droplets start to cross one another’s paths in the strain between eddies rather than merely gathering there — which is the size range at which cloud droplets’ growth by collision is known to be hard to explain.

Whether the droplet turns asked the same question of a droplet meeting a wing, with one strain and one approach; a particle is a low-pass filter gave its response to a fluctuating air in frequency. The lattice adds the geometry: the strain a droplet meets is at the saddles between the eddies, and a parcel goes straight while the streamlines curve is the pressure-free limit of the same caustic, reached at a Stokes number of infinity.

What a steady lattice cannot show #

A flow that changes. The cellular flow is steady and its walls are fixed lines. Turbulent strain regions live for a few of their own strain times and move, so a cloud is delivered to a saddle that may have gone by the time it arrives. The quarter is then a statement about the strain a particle meets on its way through, and a threshold measured in turbulence is spread over the distribution of strains.

Gravity. The particles have no weight. Settling particles cross the lattice’s cells, and in cellular flow are known to be swept along preferred paths that make them fall faster than in still air — a separate effect with its own Stokes number dependence.

The particles’ effect on the air. The cloud is dilute and the air does not feel it. On the walls below the threshold the density grows without bound, and long before that the particles would drag the air with them.

Collisions. A fold is where particle paths cross; in a real cloud, crossing particles can collide. The fold is where collisions become possible, which is why caustics matter for the growth of raindrops — and why the quarter is a number worth pinning down.

The convention: Stokes number against the saddle’s strain #

The Stokes number is the response time times the strain rate at the saddles, which is one in these units. The cells are π wide and the vorticity at their centres is two, so a Stokes number based on the vortices’ rotation rate, one, would read the same; one based on the cell’s turnover time, about 2π, would read six times larger. Lengths are in units of the cell width over π. A fold is where the particle-velocity gradient’s trace passes $-50$, which is within a fiftieth of a time unit of the true blow-up.

Who threw dust into cells #

Maxey studied heavy particles in the cellular flow in 1987 and found them leaving the cells and gathering on the walls; Wang and Maxey carried the question into simulated turbulence in 1993, where preferential concentration became a subject. The Riccati form of the particle-velocity gradient and the identification of caustics with its blow-up are due to Falkovich, Fouxon and Stepanov, and to Wilkinson and Mehlig, in the early 2000s, who showed that in random flows caustics form at a rate exponentially small in the inverse Stokes number. That in the steady lattice the threshold is exactly the saddle’s quarter, and that the direct determinant reports a spurious lower one, is what this calculation adds.

Still open: a lattice that moves #

The steady lattice has saddles that never move and walls that last for ever. The next calculation lets the cells oscillate — $\psi = \sin(x + \epsilon\sin\omega t) \sin y$, the simplest time-dependent cellular flow, whose walls wobble and exchange fluid between cells — and asks whether the threshold stays at a quarter, or whether particles delivered to a saddle that is moving can fold below it, as random flows fold at every Stokes number at a small exponential rate. It would say whether the quarter is a property of saddles or of steady saddles, and so how much of the folding seen in turbulence happens below the local strain’s quarter.