Drag decides whether a cloud folds, and the air decides at what

A parcel goes straight while the streamlines curve followed a medium in which nothing acts on a parcel, so that every parcel keeps its velocity for ever. An incompressible flow can do that only as a pure shear. A cloud that can compress — spray, sand, a dilute dust — can do it with any starting velocity, and pays for the freedom in finite time: wherever the cloud converges, neighbouring parcels arrive at one place together, the density there becomes infinite, and the paths cross. The first such caustic comes at $t^* = -1/\lambda_{\min}$, where $\lambda_{\min}$ is the most negative eigenvalue of the starting velocity gradient anywhere in the cloud, and in two dimensions it folds the cloud into the thin sheets Zel’dovich called pancakes.

A real cloud of droplets or grains is not free. Each particle drags on the air, relaxing towards the air’s velocity over its own response time τ, which for a small sphere is its density times its diameter squared over eighteen times the air’s viscosity. The earlier essay’s closing question was when that drag is enough to prevent the caustic altogether, and where it is not, how much later the first sheet forms. The answer turns out to depend less on the particles than on the air.

Still air: the drag caps the distance #

Throw a cloud into still air. Each particle’s velocity then decays as $e^{-t/\tau}$, so a particle that started with velocity $\mathbf u_0$ travels at most $\tau\mathbf u_0$, however long it is followed. The flow map is exact and simple:

$$\mathbf x = \mathbf a + \tau\left(1 - e^{-t/\tau}\right)\mathbf u_0(\mathbf a),$$

which is the free cloud’s map with the time $t$ replaced by an effective time $\tau(1 - e^{-t/\tau})$. The effective time grows like $t$ at first and stops at τ. Everything the earlier essay found about the free cloud carries over with that one substitution, and its most important consequence follows at once: the free cloud folds when its effective time reaches $-1/\lambda_{\min}$, and the dragging cloud’s effective time never gets past τ. If τ is shorter than $-1/\lambda_{\min}$, the cloud never folds.

The ratio of the two is the Stokes number of the cloud, $\mathrm{St} = \tau|\lambda_{\min}|$: the particles’ response time over the time the fastest-converging part of the cloud would take to close. The threshold is exactly one.

The figure is the earlier essay’s one-dimensional cloud, started with velocity $-\sin a$ so that its fastest convergence rate is one, with drag added. At a Stokes number of 0.8 every particle is stopped before its neighbours reach it: the paths bend towards each other, bunch, and then run parallel and vertical — particles at rest, closer together than they started, and in the same order. At a Stokes number of 2 the free-streaming distance is enough, the paths cross, and the cloud has three streams over a widening interval, as the free cloud did.

How much later #

Where the cloud does fold, setting the effective time equal to $-1/\lambda_{\min}$ gives the moment:

$$t_c = -\tau\ln\left(1 - \frac{1}{\mathrm{St}}\right) = t^*\,\mathrm{St}\,\ln\frac{\mathrm{St}}{\mathrm{St} - 1}.$$

Heavy particles, with large Stokes numbers, fold almost as soon as free ones: at $\mathrm{St} = 10$ the delay is five per cent. As the Stokes number falls towards one the delay grows without limit: 2.15 times the free time at 1.2, 4.66 at 1.01. The figure puts the closed form beside the time found directly, by bisecting on the first moment at which the map’s own Jacobian, computed by differencing the map at two thousand points, touches zero; the two agree to about a part in a million. The logarithm is worth noticing. A threshold approached from above by a quantity that saturates exponentially produces a delay that diverges only logarithmically, so a cloud just above the threshold folds eventually, and late — which is the sense in which the threshold is sharp.

The sheet that forms and stops #

Below the threshold the cloud still converges, and the question is how far. Mass conservation says a parcel’s density is its starting density over the factor by which the map has stretched it — mass that has nowhere to go piling up where the map squeezes — and in one dimension that factor at the densest point is $1 - \tau(1 - e^{-t/\tau})|\lambda_{\min}|$. It falls towards $1 - \mathrm{St}$ and stops there. The densest point’s density therefore rises and levels off at

$$\frac{\rho_{\max}}{\rho_0} = \frac{1}{1 - \mathrm{St}},$$

twice the starting density at a Stokes number of 0.5, ten times at 0.9, a hundred times at 0.99. The figure draws all three rising and flattening, and a cloud above the threshold, at 1.5, rising through every level and reaching infinity at its caustic.

That is the cloud’s version of a sheet that forms and stops. Below one, drag makes a dense band where the free cloud would have folded, of a density set by how close to one the Stokes number is, and then freezes it in place, because the particles are at rest. In a real cloud the air would not stay still — the particles’ own drag pushes it — and the band would drift and spread; but the ceiling is a clean statement about how much concentration a burst of spray or a puff of dust can achieve before it has to stop.

In two dimensions #

The two-dimensional cloud behaves the same way, with the same threshold, because its map is again the free one with the time capped: the Jacobian $\det(I + T A_0)$ first vanishes where the most negative eigenvalue of the starting gradient is, when the effective time T reaches its inverse. At a Stokes number of 0.9 the cloud gathers into bands along the directions it was converging fastest and stops, its densest point eleven times its starting density — a little more than the one-dimensional ceiling of ten, because at that point the cloud is converging gently across the band as well as fast along it. At 3 the cloud folds into the sheets the free cloud made, and at one and a half times its first caustic it looks much like the free cloud did at one and a half times its own.

Converging air: the drag keeps pushing #

Still air only brakes. Air that is itself moving does something else: it carries the particles, and where the air converges it keeps pushing them towards the convergence even after they have arrived. A particle near a line on which the air converges at a rate σ obeys

$$\tau\ddot x + \dot x + \sigma x = 0,$$

a damped oscillator whose spring is the air’s convergence and whose damping is the drag. With $\sigma\tau < \tfrac14$ it is overdamped: the particle approaches the line and never reaches it, as a parcel of the air itself would not. With $\sigma\tau > \tfrac14$ it overshoots: its own inertia carries it through the line, and particles arriving from the two sides cross.

The figure releases particles with the air’s own velocity into a steady flow converging on a line, $u = -\sin x$, and integrates each one through the whole nonlinear flow. At a Stokes number of 0.2 the particles approach the line and crowd towards it without reaching it, their density growing only as the air’s does. At 0.5 each overshoots, and particles from the two sides cross at 1.6 strain times. Integrated this way, particles at 0.24 never cross and at 0.26 do, at 7.1 strain times, and the crossing times agree with the linear oscillator’s to the integration’s step: the first crossing happens at the convergence line, where the flow is linear.

That quarter is not new to this site. Whether the droplet turns found the same condition deciding whether a droplet in a stream hits a cylinder at all: at the stagnation point the air decelerates linearly, a droplet approaching it is the same damped oscillator, and while the air’s deceleration rate times the droplet’s response time is below a quarter it is turned aside before it arrives, so the cylinder collects nothing. For a cylinder, whose stagnation point decelerates the air at twice the stream’s speed over the radius, that quarter is the critical Stokes number of an eighth quoted there. The quarter that decides whether a droplet hits a wing is the quarter that decides whether a cloud carried by converging air folds. Both are the discriminant of one quadratic: the particle’s characteristic equation $\tau s^2 + s + \sigma = 0$ has real roots, and monotonic approach, exactly when $1 - 4\sigma\tau \ge 0$.

One number, two thresholds #

The figure puts the two cases side by side, each in its own flow’s time. A cloud thrown into still air needs a Stokes number of one to fold, and its crossing time diverges logarithmically at the threshold. A cloud carried by converging air needs a quarter, and its crossing time diverges as the oscillator’s frequency vanishes, as the inverse square root of the distance from the threshold. At large Stokes numbers both tend to the same place — heavy particles cross in about one of the flow’s own times whatever the air does, because the drag barely affects them.

The factor of four between the thresholds is the difference between a push that stops and a push that continues. A particle thrown into still air must carry all the way to the crossing on the velocity it started with, and the drag spends that velocity. A particle in converging air is resupplied by the air at every moment, and needs only enough inertia to overshoot. Which case a real cloud is in is a question about the air, not the particles: spray from a nozzle into a quiet room is the first; dust in a gust, droplets in a turbulent eddy, sediment in a converging current are the second. In turbulence the air’s convergence comes and goes, and the preferential concentration of heavy particles into sheets between vortices — a particle is a low-pass filter is the essay about the particles’ side of that — happens at Stokes numbers near the quarter rather than near one.

A drag is not a viscosity #

The cloud’s two thresholds invite a comparison with the gas the earlier essay set beside it. A converging gas has the same problem a free cloud has — its characteristics cross in finite time — and every compression becomes a shock described what the gas does about it: the pressure intervenes before the crossing, and a thin layer in which viscosity and conduction act turns the would-be fold into a shock. That regularisation works at every strength. However gentle the compression, the gas steepens and a shock forms; viscosity sets how thick it is, never whether it exists.

A drag works the other way round. It does not act in a thin layer where the gradients are steep; it acts everywhere, on every particle, all the time, draining the velocity that would carry a particle into its neighbour’s place. So it can prevent the fold altogether, but only when it drains the velocity faster than the convergence can use it — below a threshold. Above the threshold it prevents nothing, and the cloud folds as a free one would, only later. Viscosity is a regularisation with no threshold and a drag is a threshold with no regularisation: past its critical Stokes number, a cloud of particles has no mechanism of its own to stop the paths crossing, which is why dense sprays and particle-laden flows are described as several interpenetrating streams at once rather than as one fluid with a shock in it.

The same contrast explains why a thrown cloud and a carried one differ by a factor of four. The drag’s effect on a particle depends on the particle’s velocity relative to the air, and streamlines are not paths for exactly this reason: in still air that relative velocity is the particle’s whole velocity, so the drag attacks everything the particle has; in moving air the particle and the air share most of their velocity and the drag acts only on the difference, while the air keeps replacing what the drag takes. The particles in the carried cloud are fighting a much weaker brake. What decides the fold is how much of a particle’s motion is its own, and in converging air most of it is borrowed.

Where the thresholds apply #

Both results need the particles to respond to the air through a drag that is linear in their slip, which is what how small is small enough establishes for a small sphere in slow flow. For a water droplet in air, $\tau = \rho_p d^2/18\mu$ is about a third of a millisecond for a ten-micron droplet and thirty milliseconds for a hundred-micron one. A spray leaving a nozzle at ten metres a second with its velocity varying by a tenth across a centimetre has a convergence rate of about a hundred per second, so its ten-micron droplets are at a Stokes number of 0.03 and gather without folding, and its hundred-micron ones are at three and fold at about 1.2 times the free cloud’s time. A factor of ten in droplet size is a factor of a hundred in Stokes number, and the threshold sits in the middle of the range an ordinary spray contains.

What was checked #

The still-air results are closed forms, and the check is that the map, computed and differenced numerically, agrees: its Jacobian first vanishes at the closed form’s time to within nine parts in ten million at Stokes numbers of 1.2, 2 and 5, and at 0.999 it never vanishes. The converging-air results are integrated, particle by particle through the nonlinear flow by fourth-order Runge–Kutta, and checked against the linear oscillator: no crossing at 0.24, and at 0.3 a crossing at 3.090 strain times against the oscillator’s 3.086. The tests refuse a Stokes number of zero, a negative one, and a tolerance of zero.

What the picture cannot show #

The air’s response. Every particle drags on the air, and a dense cloud drags the air along with it; the still air stops being still, and the converging air converges differently. The calculation is the dilute limit, a cloud whose mass loading is small compared with the air’s.

Stokes drag. The drag is linear in the slip velocity, which holds while the particle’s Reynolds number is below about one. A millimetre raindrop falling at several metres a second is far outside that, and its drag grows faster than linearly, which makes its effective response time depend on how fast it is slipping.

The terms the Stokes number leaves out. The tracer that is not one found that a particle’s added mass and the pressure gradient of the air around it change its response time, by half again for a neutrally buoyant particle, and reverse the drift of a bubble. For heavy particles in gas, the case here, both are negligible.

Gravity. A falling cloud adds a settling velocity to every particle equally, which moves the cloud without changing its map’s Jacobian; in a converging air flow gravity changes where each particle sits relative to the convergence, which can move the crossing.

The convention: a Stokes number of the cloud #

The Stokes number here is the particles’ response time multiplied by the fastest rate at which the cloud, or the air, is converging: $|\lambda_{\min}|$ of the starting velocity gradient for the thrown cloud, the strain rate σ at the convergence line for the carried one. It is a local number, set at the place the fold would first form, and a cloud with a range of convergence rates has a range of Stokes numbers; the first fold, if any, happens where the Stokes number is largest.

Who found it, and when #

The linear drag law is Stokes’s, from 1851, and the response time of a sphere follows from it. The threshold of a quarter for particles approaching a stagnation point is Taylor’s, from the 1940s, in the theory of the collection of droplets by aircraft and cloud particles that Langmuir and Blodgett developed at the same time. Caustics in clouds of inertial particles — the “sling effect” in which fast-converging eddies throw droplets across one another — were analysed by Falkovich, Fouxon and Stepanov in 2002 and by Wilkinson and Mehlig from 2005, as a mechanism that speeds up the collisions that grow raindrops.

Still open: a cloud in air that turns #

The two airs here are the extremes, one that does nothing and one that converges steadily. Turbulent air turns as well as converges, and a region of rotation flings heavy particles outwards while a region of strain gathers them. The next calculation puts the dragging cloud in the cellular flow $\psi = \sin x\sin y$ — a lattice of vortices with converging lines between them — and asks at what Stokes number the cloud first folds there, whether the answer is the strain’s quarter or something between the two thresholds because each particle spends part of its time in a vortex, and how dense the sheets between the cells become below the threshold.