A wake that sheds its fluid sinks further, and past a point does not stop

Stable air takes back the momentum a wing pushes down followed the downward momentum a wing gives the air into a stably stratified atmosphere. Behind the wing it is carried by the trailing vortex pair, which sinks at its own induced speed carrying a body of air with it — Lamb’s oval, two spacings wide and $\sqrt3$ tall. That air keeps the density it had at the start, so as the pair descends through air that grows denser with depth the oval becomes lighter than its surroundings, and buoyancy pushes it back. In Saffman’s model the push returns the pair’s impulse as a spring would, and the pair stops a quarter-period later, $2.33/N$ after the wing passed, at a depth that scales with the span: for an airliner in air of buoyancy frequency 0.01 per second, 319 metres down after four minutes.

That model keeps every parcel of the oval. The essay ended on what a real oval does: it sheds its outer fluid as it descends. The parcels at its edge, lighter than the air around them, are peeled off and left behind in a vertical sheet — the curtain of light air seen in tracer pictures of sinking wakes — and the oval shrinks. It asked whether the stop then moves earlier or later, the depth smaller or larger, and whether the shed sheet, which carries some of the momentum out of the pair, keeps the face behind the wing from emptying.

Shedding, and what it takes with it #

The model keeps Saffman’s pair and lets its oval lose area. As the pair sinks a distance $z$, the oval is lighter than its surroundings by a buoyancy $N^2 z$ per unit mass, and that buoyancy jump across its edge is what drives fluid off it: baroclinic vorticity of the opposite sign forms on the oval’s boundary and carries the light edge fluid away. Greene’s description of vortex decay in stable air took the shedding rate to be set by the jump, and the model here does the same in its simplest form — the area falls at a rate proportional to the oval’s perimeter times the velocity scale of the jump,

$$\frac{dA}{dt} = -\alpha\,P_o\,\sqrt{N^2 z},$$

with $P_o$ the perimeter, traced from Lamb’s outline at 5.97 spacings and shrinking as the square root of the area, and $\alpha$ a detrainment coefficient of the order of a tenth, the size entrainment coefficients take in plumes and jets. With $\alpha = 0$ it is Saffman’s model.

Shedding acts on the pair’s momentum in two ways. The parcels that leave were moving down with the pair, so they take their share of its momentum with them, $w\,|dA/dt|$ per unit time. And they take their share of the buoyancy: the force pushing the pair back is the oval’s remaining area times $N^2 z$, and a smaller oval is pushed less. The first effect stops the pair sooner; the second lets it run on. Which wins is the question, and it is not obvious: the momentum loss is proportional to the speed, which is largest early, and the buoyancy loss to the depth, which is largest late.

The brake goes faster than the load #

The oval empties fastest late in the descent, because that is when the buoyancy jump that drives the shedding is largest — the rate grows as the square root of the depth — and that is also when the buoyancy was doing most of its braking. With $\alpha = 0.05$ the oval has lost half its area by the time the pair stops; with $\alpha = 0.08$, nearly nine tenths.

So the brake goes faster than the load. Every pair that sheds stops later and deeper than Saffman’s, and the stop moves later much faster than it moves deeper: at $\alpha = 0.06$ the pair stops a quarter later in time and only four per cent deeper. The reason is that the late part of the descent, when the shedding has weakened the brake, is slow: the pair has by then given up most of its speed, and the extra time it takes to give up the rest is spent creeping. For an airliner, whose time unit is 22 seconds and whose spacing is 47 metres, Saffman’s stop at four minutes and 319 metres becomes, with $\alpha = 0.05$, four and a half minutes and 326 metres.

The refuted intuition — a pair that loses fluid is weaker and so stops sooner and higher — has the momentum half right and the brake half missing. Losing fluid does weaken the pair, but the pair is not stopped by drag that scales with its strength; it is stopped by the buoyancy of what it carries, and losing what it carries loses that.

Why the brake loses, in one balance #

The two effects can be weighed without the march. The momentum the shed parcels carry off is the speed times the shedding rate, and the shedding rate grows as the square root of the depth; the buoyancy the shedding removes is the depth times the area lost, and the area lost grows with the time integral of the shedding rate. Early in the descent the pair is fast and shallow: the first effect dominates, and shedding does slow the pair a little more than Saffman’s spring would. Late in the descent the pair is slow and deep: the shedding rate is at its largest and the speed at its smallest, so almost nothing is carried off with the parcels, while the buoyancy that leaves with them is at its largest. The pair spends its last stretch being braked by an oval that is shrinking fastest at exactly the time it matters most.

The computed stop times bear the balance out. For small shedding the delay grows in proportion to $\alpha$, about 2.3 times Saffman’s stop time for each unit of $\alpha$ near zero, and then faster than in proportion as the oval’s loss compounds: the delay at $\alpha = 0.08$ is more than ten times the delay at 0.02, though $\alpha$ is only four times larger. The depth moves far less, because the late, slow stretch adds time without adding much distance.

Past a threshold the pair does not stop #

Push the detrainment further and the brake can go altogether. Above $\alpha = 0.085$, in air of $N = 0.22$, the oval is empty before the pair has given up its momentum, and with no oval there is nothing light for the buoyancy to act on. In the model the pair then keeps sinking, slowly, with whatever impulse it has left, and nothing in the model stops it.

The threshold rises with the stratification, roughly as the square root of the buoyancy frequency: a stronger stratification stops the pair sooner, in a time proportional to $1/N$, and leaves the oval less time to empty, though the shedding rate itself grows with $N$. A pair in weakly stratified air has a long descent and can be stripped by modest shedding; a pair in strongly stratified air stops before the shedding matters.

In the atmosphere’s own numbers, an aeroplane’s wake in air of buoyancy frequency 0.005 per second — weakly stable, half a typical tropospheric value, so that the buoyancy a given descent builds is a quarter as large — sits at $N = 0.11$ in these units and is stripped above $\alpha = 0.06$; in the stratosphere just above the tropopause, at 0.02 per second, it sits at 0.44 and survives shedding up to $\alpha = 0.12$. Cruise altitude is often within a kilometre or two of the tropopause, so the same aeroplane’s wake can sit on either side of the threshold for the same shedding, depending on whether it is flying just below the tropopause or just above it. Below, a wake that sheds moderately loses its buoyant brake and descends until its cores decay; above, the same wake is stopped by buoyancy within a few minutes. The tropopause is then not only where contrails persist but where the physics that stops a wake changes hands.

What a real stripped pair does is not this model’s business. Without its oval it is two vortex cores and the thin layers of fluid round them, and their own decay — the cores’ diffusion, the mutual instability that links them into rings — takes over from buoyancy as what ends the descent. That the model’s pair falls on without limit is the statement that buoyancy has stopped being the relevant physics, not a prediction of an endless descent. The threshold is where the model’s own explanation of the stop runs out.

The sheet keeps a quarter of the impulse swinging #

The shed parcels are not lost. Each leaves the oval at the pair’s depth and speed, lighter than the air around it, and is pushed back by its own buoyancy as the oval was; on its own each oscillates up and down at the buoyancy frequency about its neutral level. The sheet’s total downward momentum is the sum over parcels, and because the parcels were shed over a time shorter than a buoyancy period they move nearly in step: when the pair stops, the sheet’s net momentum is near zero, and it then swings between about a quarter of the pair’s original impulse downward and a quarter upward, every 28.6 time units.

That answers the last part of the question, about the face drawn across the wake far behind the wing. In Saffman’s model, once the pair has stopped, the face carries none of the wing’s weight as downward momentum, and the whole weight crosses the box’s upper and lower faces as pressure. With shedding, the face behind keeps an oscillating share — up to a quarter of the weight, alternately down and up, at the buoyancy period. In a real atmosphere the oscillating sheet radiates internal gravity waves, which carry its momentum away at an angle, and the swing decays over a few periods; the face then empties, and the weight is carried to the ground as the earlier essays said, but by a longer route.

The sheet is the contrail’s second curtain #

The shed sheet is not a construction of the model; it is visible. Contrails behind airliners in stable air commonly split into two parts within a minute or two: a primary wake that sinks with the vortex pair, carrying ice crystals down and warming them by compression, and a secondary wake — a vertical curtain that stays near flight level or rises back towards it. Photographs from the ground and lidar cross-sections from below show the curtain connected to the descending pair by a thin sheet, exactly where the detrained parcels would be.

The model says why the curtain sits where it does. Each shed parcel, lighter than the air at the depth where it left the oval, is pushed back up towards its own neutral level, which is the level it started from — flight level — and oscillates about it with the buoyancy period. A parcel shed late, from deep in the descent, has the farthest to return and the largest swing. The curtain is the envelope of those swings, and its top is near flight level because that is every parcel’s neutral level. In this model it is the shed sheet that holds the swinging quarter of the impulse; in the atmosphere it is the secondary wake, and its slow decay by wave radiation is part of why contrails in stable air persist as two layers rather than one.

For an airliner that also says how much of the wing’s downward push is left near flight level behind it. When the primary wake stops, about four and a half minutes and seventy kilometres behind, up to a quarter of the impulse the wing gave the air is still moving — not down with the pair, but up and down in the curtain, alternately, as the curtain’s parcels settle.

Why the airliner and the light single still agree #

The earlier essay found the stop’s time the same for every aeroplane, because Saffman’s quarter-period depends only on $N$, and the depth proportional to the span. In the units used here — spacings and spacing-times — every aeroplane in air of 0.01 per second sits at the same $N$, 0.22, because the descent speed scales with the spacing: an airliner’s pair is 47 metres apart and sinks at 2.15 metres a second, a light single’s 8.6 metres apart and sinks at 0.39. So the shedding’s effect is the same fraction for both, if $\alpha$ is. Whether $\alpha$ is the same is a question about the oval’s edge, which depends on the cores’ structure and the turbulence in the oval, and the large and small wakes need not share it. That is where the universality the earlier essay found could fail.

The convention, and the model’s weak point #

Lengths are in the pair’s spacing $b_0 = \pi b/4$, times in $b_0/w_0$ with $w_0$ the pair’s initial descent speed, and impulses over the pair’s initial impulse per unit density, $\Gamma b_0$. The buoyancy frequency is in units of $w_0/b_0$. The stop is where the pair’s impulse first reaches zero. The detrainment law is the model’s weak point and is stated as such: its form — perimeter times the buoyancy jump’s velocity — follows Greene’s description, and its coefficient is left free, because measurements of shedding from sinking vortex ovals vary by a factor of several between laboratory and atmosphere. Every result is given as a function of $\alpha$ for that reason. The shed parcels move only vertically and do not interact; the pair’s spacing is held fixed, which Saffman’s model does and which real pairs in stratified air do not quite do — their spacing shrinks a little as their impulse falls, which is another way the stop is made earlier.

Who sank vortices #

Scorer and Davenport described the detrainment from sinking vortex pairs in stratified fluid in 1970, from tank experiments, and Saffman’s model of the pair as a spring, with its quarter-period stop, is of 1972. Greene gave a practical model of vortex descent and decay in the atmosphere in 1986, with buoyancy-driven detrainment among its terms, and Spalart’s computations of 1996 followed laminar wakes in stratified fluid and found their spacing and their descent departing from Saffman’s. The threshold for stripping, and the sheet’s oscillating share, are this model’s.

How it was checked #

With $\alpha = 0$ the march reproduces Saffman’s closed form: the stop at $\pi/2\omega$ to eight parts in ten thousand, set by the time step, and the depth $1/\omega$ to a part in a million. Lamb’s oval’s perimeter, traced along its outline from the pair’s stream function, is 5.968 spacings, within a per cent of an ellipse of the oval’s half-width and half-height, the oval’s flatter sides making it a little shorter. Each shed parcel’s buoyancy oscillation is advanced exactly over the time step. The model refuses a buoyancy frequency of zero, which has no stop to move, and a negative detrainment coefficient.

Still open: a pair whose spacing shrinks as it sheds #

The model holds the spacing fixed, as Saffman’s does, and a pair that loses its oval does not: the vorticity that the shedding’s baroclinic layers carry off is of the opposite sign at each core’s outer side, and its departure moves the cores together. A pair whose spacing shrinks sinks faster, since its induced speed is its circulation over its spacing, and stops in a different place. The next calculation would carry the pair’s spacing as a variable, with its rate set by the shed vorticity’s impulse, and ask whether the shrinking spacing, which speeds the descent, outruns the shedding’s loss of brake — and so whether a wake that sinks faster than Saffman’s model before it stops is doing so because of its spacing or its shedding.