What holds the shear decides whether the waves last

A thin interface keeps its waves past a quarter took a stratified shear layer and concentrated its density jump into an interface thinner than the shear, by a factor R. Above about R = 2 the stationary billow that Miles’ quarter governs dies early, and a pair of travelling waves — Holmboe’s — takes its place and keeps growing at Richardson numbers several times the quarter, because at the edges of the shear, where those waves draw their energy, the local Richardson number has fallen to nothing. The calculation held R fixed, and the essay ended on the fact that nothing in a real ocean or tank holds it: a thermocline thickens by conduction while it waves, a salt interface by molecular diffusion a hundred times more slowly, and the question was how long either stays thin enough to keep its waves.

The question as asked has an assumption hidden in it — that the interface thickens and the shear layer does not. A laminar shear layer thickens too, by viscosity, and for water viscosity is the faster of the two. Whether the waves last turns out to depend less on how fast the interface diffuses than on what, if anything, is holding the shear.

The profiles diffusion leaves #

A step in velocity or in density spread by diffusion takes the shape of the error function, and — unlike the hyperbolic tangent of the earlier calculation — keeps it exactly: a velocity $U = U_0\,\mathrm{erf}(z/\delta_u)$ diffuses into the same shape with $\delta_u^2 = \delta_{u0}^2 + 4\nu t$, and a buoyancy $b = \Delta b\,\mathrm{erf}(z/\delta_\rho)$ with $\delta_\rho^2 = \delta_{\rho0}^2 + 4\kappa t$, for as long as the flow does not stir them. So the state at any time is two numbers, the thickness ratio $R = \delta_u/\delta_\rho$ and the bulk Richardson number $J = \Delta b\,\delta_u/U_0^2$, and both move:

$$R^2 = \frac{\delta_{u0}^2 + 4\nu t}{\delta_{\rho0}^2 + 4\kappa t} \;\longrightarrow\; \frac{\nu}{\kappa} = \mathrm{Pr}, \qquad J \propto \delta_u.$$

A free interface’s thickness ratio heads to the square root of the Prandtl number, whatever it started at. For heat in water that is 2.64; for salt in water, 25.8; for heat in air, 0.84.

The ratio, free and held #

The case followed throughout is a laboratory one: a shear layer of half-thickness a centimetre with a velocity difference of three centimetres a second across it, so $U_0$ is 1.5, and a density interface a quarter as thick, at a bulk Richardson number of 0.2. Free, the ratio for heat falls from 4 towards 2.64 and stays there; for salt it rises, towards 25.8, because the shear layer outgrows an interface that diffuses six hundred and seventy times more slowly than momentum. Held — a shear layer maintained at a centimetre by whatever drives it, so that only the density diffuses — the ratio falls without limit, through 2 after 50 seconds for heat and 4724 seconds for salt.

The held case is the one the earlier essay’s question had in mind, and it is real: in an exchange flow through a strait, or a tank driven continuously, the shear layer’s thickness is set by the flow rather than by viscosity. The free case is a shear layer left to itself, as in an experiment where two layers are set sliding and allowed to evolve.

Where the waves grow, on diffused profiles #

The waves on these profiles had to be found again, because the earlier essay’s growth rates were for hyperbolic-tangent profiles, and the shape of the profiles’ tails is exactly what Holmboe’s waves depend on. The Taylor–Goldstein equation was solved on the error-function profiles — whose growing modes lie inside one circle, Howard’s, like every other profile’s — with the same complex shooting and the same acceptance rule — a mode counts only if it survives a doubling of the grid — and the fastest travelling mode tabulated over thickness ratio and Richardson number.

At a ratio of 1.7 nothing grows faster than the shooter’s floor of a few thousandths. At 2 the waves are weak, a growth rate of at most 0.011 in units of $U_0/\delta_u$; at 2.5 they reach 0.039, at 3 0.062, at 4 0.091. Every ratio’s waves weaken past a Richardson number of about one and are nearly gone by 1.5. On these profiles the threshold is close to 2, as it was on the tangent profiles, and the waves above it are stronger — at a ratio of 3 and a Richardson number of one, 0.043 here against 0.011 on the tangent profiles.

Gaussian tails, and a threshold that is not the tails’ #

The earlier essay explained Holmboe’s waves by the tails: on tangent profiles the local Richardson number far from the centre goes as $e^{(4-2R)|z|}$, so it falls to nothing for any R above 2, and Miles’ theorem, which needs the smallest Richardson number to exceed a quarter, never applies. On error-function profiles both gradients have Gaussian tails and the local Richardson number goes as $e^{(2-R^2)z^2}$, falling to nothing for any R above $\sqrt2$. If the tails were the whole story, diffused interfaces would carry the waves from a ratio of 1.41.

They do not: at 1.7 the tails’ Richardson number vanishes and still no travelling mode grows. The tails make the waves possible by removing Miles’ bound; whether a wave actually grows needs the interface to be thin enough that a disturbance at the edge of the shear can feel the density jump at all, and that takes a ratio near 2 on both families of profile. The comparison is worth having for its own sake: a sufficient condition for stability failing to hold is not a growing mode, which is the distinction sufficient, and not necessary drew for the quarter itself.

A held shear: heat loses the waves in a minute #

With the shear held, the wave’s growth is integrated along the interface’s path: at each moment the thickness ratio and the Richardson number give a growth rate from the table, multiplied by $U_0/\delta_u$, and the e-folds are its integral. For heat the window closes after 50 seconds and the wave has gained 1.8 e-folds — a factor of six. A disturbance that starts at a thousandth of the interface’s thickness finishes the window still far below anything visible. For salt the window lasts 4724 seconds and the wave gains 168 e-folds, which is more than any disturbance needs to become a finite wave and break.

That is the whole answer to the earlier question for a maintained shear, and it is a sharp one: in a laboratory tank under a sustained shear, a heat-stratified interface cannot show Holmboe’s waves and a salt-stratified one cannot avoid them. In the experiments where they were first photographed the stratification was salt, and the arithmetic says it had to be.

The number that counts the e-folds #

The two held cases differ by a factor of 95 in e-folds and by the same factor in diffusivity, and that is not a coincidence. With the shear held, the window lasts as long as conduction takes to thicken the interface to half the shear layer, $(\delta_u^2/4 - \delta_{\rho0}^2)/4\kappa$, and during it the wave grows at a rate of order $0.05\,U_0/\delta_u$. Their product is a number times $U_0\delta_u/\kappa$, the Péclet number of the interface on its shear layer’s scale — the same competition between carrying and conducting that the other layer sets out for a heated wall. Here it is 1049 for heat and 100,000 for salt, and the e-folds come out at very nearly one for every six hundred of it in both cases.

That turns the laboratory answer into a rule that scales. A held interface keeps its waves for long enough to matter when its Péclet number is in the thousands or more, which at laboratory scales means salt and at geophysical scales means almost anything. A thermocline sheared by a tide, with a shear layer of a metre and a velocity difference of ten centimetres a second, has a thermal Péclet number of about 350,000: its window is four days long and holds some six hundred e-folds, and heat alone is ample. What makes heat-stratified Holmboe waves rare in tanks and ordinary in estuaries is the size of the Péclet number, not the nature of the stratifying agent.

Why viscosity outruns conduction #

The free case turns on a comparison that is easy to get backwards. Heat in water diffuses more slowly than momentum — the Prandtl number is about seven — so a shear layer and a temperature interface of equal thickness do not stay equal: the shear layer thickens $\sqrt7$ times faster in thickness. A centimetre shear layer takes 25 seconds to double its thickness squared by viscosity; a 2.5-millimetre interface takes 11 seconds by conduction, but in those 11 seconds the shear layer has grown more in absolute terms, and the ratio of their thicknesses moves towards $\sqrt7$ from whichever side it starts. Thick interfaces under a free shear get thinner relative to it, which is why a free salt interface’s ratio rises from 4 towards 26: the layer outgrows it. The same arithmetic, applied to a vortex rather than a shear layer, is what viscosity cannot take away — a diffusing shape that keeps its form while its scale grows as the root of time.

A free shear: the waves outlast the interface’s thinning #

Left free, the shear layer thickens by viscosity, and that changes both numbers. The ratio no longer falls through 2 for heat — it settles at 2.64 — so the interface never stops being thin enough. What ends the waves instead is the Richardson number: $J$ grows in proportion to the shear layer’s thickness, from 0.2 to 2 in 2475 seconds, and the waves weaken past one and are gone near two. Along the way heat gains 19.7 e-folds and salt at least 30 — at least, because the table stops at a Richardson number of 2, where salt’s very thin interface still has growing waves. Heat in air, whose ratio heads to 0.84, gains less than a tenth of an e-fold: an air interface’s waves, if it ever had any, are lost in the first seconds.

So the Prandtl number decides the outcome twice over. Below 1 the shear layer outgrows nothing and the interface loses its thinness whatever holds the shear. Between about 4 and the hundreds, a free layer keeps the ratio above 2 indefinitely and the interface’s own diffusivity hardly matters — heat and salt gain e-folds of the same order — while a held layer makes the diffusivity the whole story.

What the tank would show #

The fastest waves on the starting interface, a ratio of 4 at a Richardson number of 0.35, have a wavenumber of 0.95 on the shear layer’s thickness — a wavelength of about 6.6 centimetres in the centimetre layer — and travel at 0.54 of the half-velocity difference, eight millimetres a second, in both directions at once: Holmboe’s pair, one riding above the interface and one below, crossing each other. At a ratio of 3 and a Richardson number of a half they are slightly longer, 6.5 centimetres, and faster, 0.64 of the half-difference. Their growth time, the inverse of the rate in seconds, is about seven seconds on the salt interface at the start, so in a held salt-stratified tank they are visible as cusped crests within a minute or two and the interface between the layers is a moving, interlocking pattern for the rest of the hour. On the heat-stratified interface in the same tank the same waves would have a growth time of seven seconds too, and fifty seconds to use it in.

The billow that Miles’ quarter governs is absent throughout: on these profiles no stationary mode grows past a Richardson number of 0.2 at ratios of 2 and 2.5, and none grows at all at 3 and 4, the same early death the tangent profiles showed. The interface either carries travelling waves or nothing, and which depends on the clock that diffusion runs, which is the thin layer question asked of a density profile rather than a velocity one.

Two doors out of the region #

Drawn in the plane of the two numbers, the cases are two paths out of one region. A held interface moves straight down at fixed Richardson number and leaves through the floor near a ratio of 2 — the door the earlier essay’s question was about. A free one moves to the right as its shear layer thickens and levels off at $\sqrt{\mathrm{Pr}}$, so it leaves through the side, at a Richardson number between 1.5 and 2. The five quantities called the Richardson number are why this needs saying: the bulk number grows here while the gradient number at the interface’s centre, $J R\sqrt\pi/2$, grows with it, and the gradient number at the edges stays at nothing throughout.

The check, and what was held #

The error-function option was added to the existing shooter rather than written as a second one, and the original’s growth rate at a ratio of 3 and a Richardson number of 1 is unchanged to $10^{-6}$. The profiles’ centre Richardson number is $JR\sqrt\pi/2$ exactly, and their tails fall at a ratio of 1.5 and rise at 1.3, either side of $\sqrt2$. The free ratio approaches $\sqrt{\mathrm{Pr}}$ to $10^{-6}$ by $10^7$ seconds. The growth table uses a grid of 1600 steps accepted on 3200, which reproduces the finer grid’s rates to the third figure at the points compared.

The convention: J on the shear layer’s thickness #

The bulk Richardson number here is $J = \Delta b\,\delta_u/U_0^2$, with $\Delta b$ half the buoyancy jump, $U_0$ half the velocity difference and $\delta_u$ the shear layer’s error-function thickness — the same definition as the earlier essay’s, on a different profile, and the one that grows as the shear layer thickens. The thickness ratio is the shear layer’s thickness over the interface’s, both as error-function widths. Growth rates are in units of $U_0/\delta_u$ at the moment they are evaluated.

What a quasi-steady reading cannot show #

Every growth rate is the linear rate of the profiles as they stand at that moment, which is right only while a wave grows faster than the profiles change: true throughout the salt cases and for most of the heat ones, and not for the slowest-growing stretches near the ends of each window. The waves stir what they grow on: a finite Holmboe wave mixes the interface and thickens it faster than diffusion does, so the held salt interface’s 170 e-folds end long before 4724 seconds, in a mixed layer that is a different problem. Viscosity acts on the waves as well as the layer, and at these centimetre scales and speeds the layer Reynolds number is 150, low enough that the viscous version of the eigenvalue problem would trim the weakest rates. And the starting interface is stated, four times thinner than the shear at a Richardson number of 0.2; a different start moves every time, not the doors.

Holmboe’s interface and the tanks that showed it #

Holmboe proposed the travelling-wave instability of a sharp interface in a shear layer in 1962, by an argument with piecewise-linear profiles. Hazel’s numerical study of smooth profiles in 1972 found it on tangent profiles with a thickness ratio above about 2, and Browand and Wang photographed the waves the same year, in salt-stratified water. Later studies of error-function and Gaussian profiles, and of the effect of the Prandtl number on direct simulations of the waves, found them far more robust in salt than in heat, for reasons that included the one computed here. The free-layer asymptote $R \to \sqrt{\mathrm{Pr}}$ is a line of diffusion arithmetic, and it is the reason a free heat interface in water never quite stops being thin enough.

Still open: the interface a wave thickens #

The calculation keeps the interface thickening by molecular diffusion alone, and a growing Holmboe wave does more than that: once finite, its cusps eject wisps of the interface into the layers either side and the interface thickens by mixing, faster as the wave grows. The next calculation lets the interface’s thickness grow at a rate set by the wave’s own amplitude — a mixing efficiency times the wave’s energy — and asks whether a held salt interface reaches a steady thickness at which the waves it carries just maintain it, the state the number that stops the mixing points at for a stratified layer, and how that equilibrium thickness depends on the shear that holds it.