A stronger strain folds a cloud only if it lasts

A lattice of vortices folds a cloud at the saddles’ quarter found that heavy particles in a steady cellular flow fold onto its walls — their paths cross and their density becomes infinite on a curve — only above a Stokes number of a quarter, the overshoot threshold of the converging saddles. Wobbling saddles keep the quarter, and raise it moved the saddles back and forth without changing their strength and found the quarter intact: the threshold belongs to the strongest strain the flow has. It ended on the case that test could not reach. Random flows fold particles at every Stokes number, at a small rate that falls exponentially with the inverse Stokes number, and a random flow’s strain is sometimes stronger than its average. A lattice whose strain is itself sometimes stronger would say whether that is the whole of the explanation.

The answer has a time scale in it that neither of the earlier essays needed: a stronger strain folds particles only if it lasts.

A lattice whose strength changes #

The flow is the steady lattice multiplied by a time-varying amplitude, $\psi = A(t)\sin x\sin y$. Its saddles stay where they are and its strain is $A(t)$ at every instant, so the steady lattice’s fold condition — a Stokes number above a quarter of the strain — can be applied to the instantaneous strain and compared with what the particles actually do. Each particle relaxes towards the air’s velocity over its own time $\tau$, and carries the gradient of the particle velocity field, $Z$, by the same Riccati equation as before; it folds when the trace of $Z$ falls through $-50$, the signal that neighbouring paths are about to cross. A cloud is seeded over one cell, and the fold threshold is the least Stokes number at which any of its particles folds, found by bisection.

Two amplitudes are used. A pulse, $A = 1 + a\sin\omega t$, isolates the role of time: its strongest strain, $1 + a$, recurs with period $2\pi/\omega$ and lasts for a stated fraction of it. A random signal of mean one, standard deviation $s$ and correlation time $T_c$ — an Ornstein–Uhlenbeck process, stepped exactly and seeded so that a run can be repeated — is the case the earlier essay asked about.

Slow pulses fold at the peak, fast ones at the mean #

With a pulse of amplitude 0.3 the strain swings between 0.7 and 1.3. If the steady rule applied to the instantaneous strain, the cloud would fold whenever the Stokes number exceeded a quarter of the peak strain, $1/(4 \times 1.3) = 0.192$, at any frequency. It does at a slow pulse: at $\omega = 0.05$ the threshold is 0.194. As the pulse quickens the threshold climbs — 0.197 at $\omega = 0.2$, 0.206 at 0.5, 0.220 at 1, 0.240 at 2 — and at $\omega = 5$ it is 0.249, the quarter of the mean strain. The larger and smaller pulses do the same between their own limits: 0.168 to 0.246 at amplitude 0.5, 0.219 to 0.250 at 0.15.

A fast pulse, in other words, is invisible to the particles, and a slow one is felt at its strongest. The two limits are not symmetric in what they cost: a particle between the quarters is folded by any slow pulse whose peak exceeds its own quarter, however brief the pulse’s strong phase is in absolute terms, as long as it outlasts the window; and it is spared by any fast pulse, however strong its peak, whose strong phases are each shorter than the window. The steady rule was never a statement about the strain at an instant; it is a statement about a strain that has acted for long enough.

The same particle, two pulses #

One particle at a Stokes number of 0.21 — between the two quarters — shows the difference directly. Under the slow pulse its $\mathrm{tr}\,Z$ dives and it folds at $t = 10.7$, as the strain spends a long stretch above $1/(4 \times 0.21) = 1.19$. Under the fast pulse the strain passes 1.19 for the same share of each cycle, but each pass lasts a fifth of a time unit, and the particle’s gradient swings with the pulse and recovers each time; it is still unfolded at $t = 80$. The fold is a runaway of the Riccati equation, and a runaway needs time to run.

Why the runaway needs time #

The fold is the end of a runaway, and the runaway has a speed that sets the window. Near a converging saddle of strain $A$, a particle’s velocity gradient along the converging direction obeys, in the steady lattice, $\dot z = -(z + A)/\tau - z^2$: it relaxes towards the air’s gradient $-A$ over the time $\tau$, and the quadratic term pulls it further down once it is negative. Below a quarter, $\tau A < 1/4$, the equation has a fixed point and $z$ settles on it; above, there is none and $z$ falls to minus infinity in a finite time. That finite time is the runaway’s duration, and it is not short just above the threshold: it grows without limit as $\tau A$ approaches a quarter from above, because the trajectory lingers near the ghost of the vanished fixed point.

So a strain that rises just past $1/4\tau$ for a moment starts a runaway that has nowhere near enough time to finish before the strain falls back and the fixed point reappears to catch it. A strain well past $1/4\tau$ finishes the runaway in a strain time or two. The measured window of about four strain times is the duration of the runaway at the strains the pulses provide — a few tenths above threshold — and that is why it fits all the cases roughly rather than exactly: a more strongly over-threshold pulse needs a shorter stay. Drag decides whether a cloud folds set up the same equation for a single strain; the pulse makes its ghost the part that matters.

How far along the threshold has moved #

Scaled between its two limits, each amplitude’s threshold slides across over two decades of frequency, from about $\omega = 0.05$, where it is at the peak’s quarter, to about 5, where it is at the mean’s. The three amplitudes do not slide at the same rate. At $\omega = 1$ the smallest pulse is seven-tenths of the way to the mean’s quarter and the largest a third: a larger pulse spends a longer stretch of each cycle above any given strain, so its strong stretches last longer and are felt longer.

What the particle remembers #

The pattern has a simple reading. Suppose a particle feels not the strain at an instant but its average over a window of fixed length $w$, and folds by the steady rule applied to the largest such average. For a sinusoidal pulse the largest windowed average is $1 + a\,|\sin(\omega w/2)|/(\omega w/2)$, and the threshold is a quarter of its inverse. One window fits all eighteen thresholds — three amplitudes, six frequencies — to within 3.7 per cent: 3.83 time units, about four strain times of the steady saddle. That is the time the Riccati runaway takes to develop from a cloud’s ordinary state to a fold at a strain near the threshold, and it is not a property of the particle alone: it is the saddle’s own time scale, multiplied by the few e-folds the runaway needs.

The reading is a description, not a derivation, and its 3.7 per cent is the measure of how much it leaves out — the window is not sharp and the runaway’s speed depends on how far above threshold the strain is. But it says which quantity a fluctuating flow has to supply to fold a cloud: a strain above $1/4\tau$ held for several strain times. A particle is a low-pass filter is the same idea for the particle’s velocity, and a parcel goes straight while the streamlines curve its origin in a fluid parcel’s own inertia — a particle cannot follow fluctuations faster than its own response time — applied one level up, to the gradient.

A random strain folds in bursts #

The random lattice shows what that does to a flow without a period. At a Stokes number of 0.17 the steady rule would require a strain above 1.47 — 1.6 standard deviations above the mean of a signal of deviation 0.3 — and the signal spends a few per cent of its time there. The cloud does not fold at a steady trickle. It folds in steps: long stretches with no folds at all, then a large share of the cloud at once, each step at an excursion that is both strong and long. Every particle feels the same strain at the same time, so an excursion that is long enough folds every particle that happens to be near the saddles when it comes.

That is a warning about reading a rate. A per-particle fold rate measured from such a cloud depends almost entirely on how many long excursions the run happened to contain, and the exponential dependence on the inverse Stokes number that random flows show is a statement about the statistics of their long, strong strain events, not about a property of folding that a single-strain lattice could reveal by itself. In a turbulent flow different regions see independent strain histories and the bursts average out into a rate; in a lattice every particle sees the same history and they do not.

A droplet cloud near a pulsing vortex street #

The pulsing lattice is a cartoon of a real arrangement: droplets carried past a bluff body whose wake sheds vortices at a fixed frequency, so that the strain at each saddle between vortices rises and falls with the shedding. A wake shedding at the Strouhal number a wake chooses, near 0.2, behind a body of diameter $d$ in a stream $U$ has a shedding frequency of $0.2U/d$ and saddle strains of order $U/d$, so in the lattice’s units its pulse frequency is of order one or a little more — squarely in the range where the threshold is between the two quarters. Droplets in such a wake should fold at Stokes numbers below the steady quarter of the mean strain only by about half the gap to the peak strain’s quarter, and a calculation that applied the steady rule to the peak strain would overpredict the folding, and with it the collision and coalescence rate that folding drives.

That is the practical content of the window: whenever a flow’s strain varies on a time comparable to a few of its own strain times, neither the mean strain nor the peak strain is the right one to put into the fold condition, and the lattice says by how much each is wrong.

What this says about the quarter #

The essays on this lattice together put the quarter where it belongs: on the strongest strain the flow sustains. This one adds the qualifier that matters for any real flow — sustains for long enough. A wobbling lattice keeps the quarter because its strain never changes; a pulsing one moves it because its strain does, and only by as much as the strain’s strong stretches last. Whether the droplet turns is the same competition in a steady flow round a body, where the strain a droplet meets lasts as long as it takes to pass the body, and the tracer that is not one changes the response time itself. The folding threshold in a fluctuating flow is a property of the strain’s history as much as of its size.

What a tank experiment would show #

The pulsing lattice can be built: electromagnetically driven cells in a thin layer of salt water, with a current through the layer that can be modulated, give a cellular flow whose strength follows the drive. Heavy particles — small glass beads — dropped into it and photographed from above show where they gather. At a fixed mean drive and a bead size between the two quarters, a slowly modulated drive should collect the beads into sharp lines along the cell walls once per modulation cycle, the lines forming at the strong phase and dispersing at the weak one; a quickly modulated drive at the same mean and the same amplitude should leave them spread, as the steady drive at the mean would. Sweeping the modulation frequency across the window — a cycle of four or five strain times — is the whole of the experiment, and the frequency at which the lines stop forming is the measurement of the window.

The bursts of the random case would show as well, and would be the more striking sight: lines that form suddenly across the whole tank at an excursion of the drive and nowhere in between, because every cell sees the same drive. A tank with independently driven cells would be needed to see them average out.

The steady lattice and the two pulse limits, as checks #

With no pulse the lattice is the steady one, and a cloud at a Stokes number of 0.22 does not fold in sixty time units while one at 0.30 folds entirely. The slowest pulse’s threshold is within one per cent of the peak strain’s quarter and the fastest’s within $10^{-3}$ of the mean’s. The random signal, built by the exact update of an Ornstein–Uhlenbeck process, has a mean of one and a deviation of 0.3 to a few parts in a thousand over twenty thousand time units, and a correlation at its correlation time within $3\times10^{-3}$ of $1/e$. Thresholds are bisections to about $4\times10^{-4}$ in Stokes number, on clouds of 64 particles; the same step control as the earlier essays keeps the integration accurate through a fold.

The convention: the Stokes number on the mean strain #

The Stokes number is the particle’s response time over the steady lattice’s strain time, so a pulse of mean one leaves its scale unchanged. A fold is $\mathrm{tr}\,Z < -50$. Frequencies and windows are in the same time unit. The pulse’s limits are the quarter of the mean strain, 0.25, and the quarter of the peak, $1/(4(1 + a))$.

What a lattice of pulsing saddles leaves out #

A particle’s own history before the run. Every cloud starts from the air’s velocity and gradient at $t = 0$, which for a slow pulse is a choice of phase; a cloud started at the pulse’s peak folds sooner than one started at its trough, and the threshold here is for clouds that see at least one full peak. Gravity, which drives particles across the cells and changes which ones reach the saddles. Spatial variation in the strain. Every saddle here pulses together; in a turbulent flow neighbouring saddles have independent histories, which is what turns this lattice’s bursts into a turbulent flow’s rate. Collisions once paths cross, which the fold signals and the model does not follow.

Dust, cells and caustics #

Maxey’s equation for small heavy particles is from 1987, and the Riccati form for their velocity gradient, which makes a fold a finite-time singularity, came with the study of caustics in random flows in the 2000s by Wilkinson, Mehlig and others, who found the fold rate’s exponential dependence on the inverse Stokes number. Cellular flows with pulsing or random strength are standard tests of mixing; using one to separate the size of a strain from its duration in the fold condition is the step taken here.

Still open: saddles that pulse independently #

The bursts come from every saddle pulsing together. The next calculation gives each saddle of a larger lattice its own independent random amplitude, so that different regions see different strain histories, and asks whether the fold rate then becomes steady, whether it depends on the inverse Stokes number exponentially or as the Gaussian tail of the strain’s distribution, and how the answer depends on the strain’s correlation time measured against the four strain times a particle needs — which would say whether the turbulent fold rate is set by how often the strain is strong or by how often it is strong for long enough.