A thin interface keeps its waves past a quarter

Sufficient, and not necessary computed Miles’ theorem on the layer it is always illustrated with: a hyperbolic-tangent shear whose density changes over exactly the depth its velocity does. There the quarter is sharp. The billow grows at every Richardson number below it, its growth rate falls to a thousandth at 0.2499, and at 0.26 the search finds nothing at all. The essay then warned that the matched profile is a convenience — a thermocline or a salt interface is usually far thinner than the shear across it — and that a thin interface opens a second kind of instability that the Richardson number does not even name. It described Holmboe’s travelling waves and did not compute them.

This essay computes them, and the result is sharper than the warning. A thin interface does not merely add a second mode below the quarter. It removes the quarter. The stationary billow dies sooner than on the matched layer, but a pair of waves travelling in opposite directions replaces it and is still growing at a bulk Richardson number of 1.3, more than five times the threshold, with no sign of a value at which it stops. Miles’ theorem is not contradicted anywhere, and seeing why it is not is the most useful thing the calculation shows.

Thinning the interface without changing the jump #

The velocity is the familiar one, $U = \tanh z$, in units of the half-velocity difference and the half-thickness of the shear. The stratification is moved, not strengthened. Its buoyancy frequency squared is

$$N^2(z) = J\,R\,\operatorname{sech}^2(Rz),$$

so that the whole buoyancy jump across the layer, $\int N^2\,dz = 2J$, is the same at every thickness ratio $R$. At $R = 1$ this is the matched layer, and $J$ is then its gradient Richardson number at every height. At $R = 3$ the same density difference is squeezed into a third of the depth, and $J$ is the bulk Richardson number: the jump times the shear’s thickness over the square of the velocity difference, the quantity a measurement across a thermocline reports. Five numbers, one name spent its length on the difference between those two, and here the difference is the whole story.

The question asked of each layer is the linear one. Disturbances of wavenumber $k$ and complex phase speed $c$ obey the Taylor–Goldstein equation,

$$(U - c)(\phi'' - k^2\phi) - U''\phi + \frac{N^2}{U - c}\,\phi = 0,$$

and a mode grows if $c$ has a positive imaginary part. The equation is shot from below, starting on the solution that grows away from the layer, and Newton’s method on the complex $c$ asks that the solution decay above. The search is the same eigenvalue search the threshold the walls decide runs for convection, asked of a shear rather than a heated layer. A mode with $c_r = 0$ is stationary in the frame of the layer’s mean velocity — a billow. A mode with $c_r \neq 0$ travels, and reflecting the layer through its centre turns it into a second mode with $-c_r$ and the same growth, so travelling modes come in pairs.

The billow dies, and the waves do not #

The faint curve is the matched layer, and it does what the previous essay found: 0.1897 with no stratification, 0.126 at a tenth, a hundredth at 0.24, and nothing past the quarter. Two times thinner, the billow dies sooner. Its growth falls from 0.072 at a tenth to 0.024 at 0.12, and by 0.125 there is no growing mode of either kind. Concentrating the density jump has made the layer more stable at small Richardson numbers, which is the opposite of what a reader of the warning might expect, and it is because the billow lives at the centre of the layer — on the inflection point that a layer with a kink in it showed every inviscid instability needs — — where the thin interface now puts all of its stratification. At the centre the gradient Richardson number is $JR$, twice or three times $J$, and the billow feels that.

Three times thinner, the billow is gone by a Richardson number of about 0.1. But by then something else is growing. At a bulk number of a tenth there is a travelling wave with a growth rate of 0.034, and as $J$ rises its growth barely moves at first — 0.0347 at 0.15, 0.0335 at the quarter — and then falls steadily: 0.0252 at a half, 0.0109 at one, 0.0063 at 1.3. On the logarithmic axis the fall is close to a straight line, an $e$-fold for every 0.63 of Richardson number, and nothing in it suggests a value at which it reaches zero. At two and a half times thinner the waves are weaker by a factor of two to six and fall the same way, to a thousandth at 1.3.

Those numbers need one caution, and it is about the instrument. Near neutral the critical level of a mode — the height where the flow moves at the wave’s speed — is a feature as narrow as the imaginary part of $c$, and an integrator that does not resolve it invents modes. A first scan of this problem found a family of them: growth rates of a few thousandths at wavenumbers above one on the matched layer, where none exists. Every mode reported here was solved twice, on a grid and on one twice as fine, and kept only if its phase speed moved by less than a ten-thousandth. That filter is why the curves stop at 1.3 rather than continuing. Beyond it the fastest wave’s $c_i$ falls below a few thousandths, a second branch at shorter wavelengths starts to compete, and the continuation that follows the curve can no longer be trusted to have stayed on the fastest one.

Where the quarter went #

Miles’ theorem says that if the gradient Richardson number $N^2/U'^2$ is at least a quarter at every height, no mode grows. It is a statement about the smallest value in the profile. On the matched layer every height carries the same number, so the smallest is $J$ and the theorem’s quarter is a quarter of $J$’s. On a thin interface the profile is

$$\mathrm{Ri}_g(z) = \frac{J R\,\operatorname{sech}^2(Rz)}{\operatorname{sech}^4 z},$$

which is $JR$ at the centre and, far from it, behaves as $\tfrac14 JR\,e^{(4-2R)|z|}$. The density gradient decays as $e^{-2R|z|}$ and the shear squared as $e^{-4|z|}$, and whichever decays faster wins. For $R < 2$ the shear dies first and the Richardson number climbs without limit away from the centre. At $R = 2$ the two decay together and it tends to $J/2$. For $R > 2$ the density gradient dies first, and the Richardson number falls to zero at the edges of the shear.

The figure draws it at $J = 0.5$. The matched layer carries 0.5 everywhere. Three times thinner, the centre carries 1.5 — three times as stable as the matched layer — and by three half-thicknesses out the number has fallen below a thousandth. The quarter is crossed two-thirds of a half-thickness above and below the centre. So for any $R$ above two the smallest gradient Richardson number is zero, whatever $J$ is, and Miles’ hypothesis is never met. The theorem is silent on every one of these layers, not because it is weak but because it was never asked.

This is the precise version of the warning sufficient, and not necessary gave. A bulk Richardson number is one number standing in for a profile — exactly the kind of number what “of order one” is worth warns is only as good as the length put into it — and the profile’s minimum is what the theorem reads. A thermocline that reports a bulk number of one may contain no height at which the gradient number is a quarter, and the reader who checks the bulk number against Miles has checked the wrong quantity.

A shorter, slower, travelling disturbance #

The waves are a different disturbance, and the band shows it. The matched layer at a tenth grows a billow fastest at $k = 0.45$, a wavelength of fourteen half-thicknesses, at a rate of 0.126. The thin interface at the same Richardson number grows waves at wavenumbers from 0.1 to 0.65, fastest at 0.034 — less than a third of the billow’s rate. At a half, where the matched layer is stable, the thin interface’s band has moved up to wavenumbers 0.4 to 1.1 and the fastest wave has a length of eight half-thicknesses, a little over half the billow’s, growing a fifth as fast as the billow did at a tenth. As the stratification strengthens the waves shorten — the fastest wavenumber rises from 0.37 at a tenth to 1.49 at 1.3 — and slow down.

A growth rate of 0.03 in these units is slow. For a thermocline with a velocity difference of twenty centimetres a second across a shear layer two metres thick, the unit of time is ten seconds, and the wave e-folds in about five minutes where a billow on the matched layer would e-fold in eighty seconds. That is consistent with what Holmboe waves are known for in the laboratory: they mix much less than a billow for the same shear, because they grow slowly and do not overturn the interface. The linear calculation cannot say how much less, but it can say why the mixing is not zero. Something is growing.

Two families in one plane #

Every unstable wave is inside one circle proved the bound this figure tests: for any stratified shear flow, the complex phase speed of a growing mode lies inside the semicircle whose diameter is the range of the velocity. Here that range is $-1$ to $1$, and every mode found lies inside it, the farthest at 0.917 of its radius. The two families separate cleanly. The matched layer’s billows sit on the vertical axis, with no phase speed, high up because their growth per unit wavenumber is large. The thin interface’s waves sit off it, with phase speeds from 0.18 to 0.92 of the half-velocity and imaginary parts below a tenth, and each one has its mirror travelling the other way. On a symmetric layer a wave travelling right and one travelling left grow equally, and a real interface shows both, as a pattern of cusps leaning first one way and then the other.

The semicircle explains one detail of the growth curve. As $J$ rises the waves’ phase speed approaches the edge of the velocity range, 0.83 at a bulk number of 1.3, and inside a semicircle a mode near the rim has room for only a small imaginary part. The waves are pressed against the boundary of what any mode can do.

Where the wave draws its energy #

A travelling wave has a critical level, the height $z_c$ at which $U(z_c) = c_r$, and for the tanh layer that height is $\tanh^{-1} c_r$. It is where a neutral wave’s equation is singular — the same singular level that decides which waves a vortex used as a waveguide carries and which it absorbs — and where a growing wave exchanges most of its energy with the mean shear. For the fastest wave on the interface three times thinner it lies 0.58 half-thicknesses from the centre at a bulk number of a quarter, and climbs to 1.2 at 1.3. The interface’s own half-thickness is a third. The wave’s critical level is therefore not in the interface at all. It sits in the shear above it, where the velocity is still changing and the density has almost stopped.

That is the physical reading of the Holmboe mode that the wave-interaction account gives it: a gravity wave carried on the thin interface, whose speed is set by the density jump, locked in phase with a vorticity wave carried on the edge of the shear, whose speed is set by the velocity profile. Neither grows alone. Together, at a wavenumber where their speeds match, each pushes the other. The density jump decides the speed, and the shear outside the interface supplies the energy.

The quarter, read at the right height #

Put the two previous figures together and the theorem comes back, in a form that is not a theorem but is surprisingly exact. Evaluate the gradient Richardson number at each wave’s critical level. At every bulk Richardson number solved, for both interfaces, it is below a quarter. On the interface three times thinner it never exceeds 0.232, although the bulk number runs to 1.3. As the stratification strengthens the wave moves its critical level outwards, keeping it in shear weak enough in buoyancy to pay for growth.

This is not Miles’ theorem, which is a statement about the smallest value anywhere and gives no reason why the value at a critical level should obey a quarter. It is an observation from the calculation, and it is the one worth carrying. The quarter is local. A mode grows where the shear, at the height the mode actually lives, outruns the stratification there, and a single bulk number cannot say whether such a height exists. On a matched layer the answer is the same at every height, which is why the bulk number seemed to be enough.

The number that stops the mixing derived the quarter from an energy exchange between two parcels. That derivation is local too — the parcels are at one height, with one shear and one density gradient between them. A thin interface makes the local nature of both the heuristic and the theorem impossible to miss.

Two codes and a doubled grid #

The shooting code here is a second implementation of the equation solved in the previous essays, written in scalar complex arithmetic for speed, and the first check is that the two codes give the same residual for the same layer and the same trial $c$. They agree to $10^{-14}$. With no stratification the fastest mode is Michalke’s: a growth rate of 0.18970 at $k = 0.4446$, against his 0.1897. On the matched layer at 0.26 and 0.4 no mode of either kind grows, as Miles requires — a test the solver could fail, since it is the same search that finds the waves at $R = 3$. A travelling wave’s mirror, with phase speed $-c_r$, is a root of the same equation to a residual of $10^{-10}$. Every mode survives the doubled grid. The tests also refuse a wavenumber of zero and a negative one.

What an inviscid wave cannot say #

Viscosity and diffusion. The calculation is inviscid and non-diffusive. A real thin interface is thin because heat or salt diffuses slowly, and it thickens as it is stirred, so $R$ is not a fixed property of a real flow but a state it passes through. Diffusion also damps the shortest waves, as viscosity reorders the verdicts in the profile Rayleigh cleared and viscosity did not, and since the waves shorten as $J$ rises, a viscous calculation would probably find a finite Richardson number at which they stop. Where that is depends on the Reynolds and Prandtl numbers, and no inviscid curve can say.

Amplitude. These are growth rates of infinitesimal disturbances. Whether a Holmboe wave saturates as a train of cusped waves or breaks into turbulence, and how much it mixes when it does, are nonlinear questions. The laboratory says the mixing is weaker than a billow’s; the calculation says only that it is not zero.

Asymmetry. The layer here is symmetric: the interface sits exactly at the centre of the shear. A real interface is offset, and then the two travelling waves are no longer equal — one grows faster, and the pattern drifts in one direction.

The far end of the curve. The waves are followed to a bulk Richardson number of 1.3 and are still growing there. The calculation does not show that they grow at every $J$. The argument from the profile shows only that nothing forbids it.

The convention: which Richardson number is $J$ #

$J$ is the bulk Richardson number of the whole layer: half the buoyancy jump, times the shear’s half-thickness, over the square of half the velocity difference. On the matched layer that equals the gradient number at every height, which is why the two have so often been treated as the same quantity. On a thin interface the gradient number is $JR$ at the centre and falls to zero at the edges, so a statement about “the Richardson number” of a thermocline must say which one. The results here are in units of the half-velocity and the half-thickness of the shear, so a growth rate of 0.03 means an e-folding time of about thirty-three of those time units.

Holmboe’s two sheets, and Hazel’s smooth layers #

Holmboe proposed the travelling instability in 1962 from a piecewise-linear model, two vortex sheets and a density step, and showed it persisted at every Richardson number in that idealisation. Hazel solved smooth profiles numerically in 1972, the calculation repeated here, and found the waves for interfaces thinner than about half the shear. Browand and Wang photographed them in 1972. The wave-interaction account is due to Baines and Mitsudera, 1994, and Carpenter and colleagues made it quantitative. Alexakis, in 2005, studied the smooth profiles at large Richardson number and the role of the gradient number at the edges of the shear.

Still open: the interface that diffuses while it waves #

The calculation fixes $R$. A real interface between warm and cold water thickens by diffusion while it carries waves, and the waves themselves stir it. The next calculation lets the interface’s thickness grow at a molecular diffusivity, follows $R$ down through two over the life of a thermocline, and asks how long a Holmboe-unstable interface stays thin enough to keep its waves — at the Prandtl number of heat in water, seven, and of salt, seven hundred — and whether a salt interface is thin long enough for its waves to grow by a factor that matters before the quarter rules it again. Beside it is the viscous version of the same eigenvalue problem, which would put a number on the Richardson number at which the shortening waves are finally damped.