Sufficient, and not necessary

The number that stops the mixing introduces the Richardson number and its quarter, and five numbers, one name is about which of the several quantities called by that name the quarter belongs to. This essay is about the logical shape of the statement, which is a sufficient condition and is almost universally quoted as a criterion.

The distinction is not pedantic. It changes what a measurement of the Richardson number is evidence for.

What Miles proved #

The theorem, from 1961: if the gradient Richardson number

$$Ri(z) = \frac{N^2(z)}{\left(dU/dz\right)^2}$$

exceeds $\tfrac14$ everywhere in an inviscid, non-diffusive stratified parallel flow, then no normal-mode instability exists.

Three qualifications are in that sentence and all three matter. Everywhere — a single station below a quarter voids it. Inviscid and non-diffusive — viscosity and diffusion of the stratifying agent both break it, and both are present in the ocean. And normal-mode — it bounds eigenvalues and says nothing about transient growth.

The threshold, computed #

The test profile is the standard one: $U = \tanh z$ with $N^2 = J\,\mathrm{sech}^2 z$, chosen because the gradient Richardson number is then exactly $J$ at every height, so “everywhere” and “somewhere” are the same condition and the theorem’s hypothesis is unambiguous.

Solving the Taylor–Goldstein equation by the same complex shooting used for Howard’s semicircle, the fastest growth rate falls smoothly to zero at the quarter. At $J = 0.2$ it is 0.0470; at 0.24, 0.00981; at 0.2499, 0.00106; at 0.26, the solver finds no unstable mode at all.

Stratification does not shift the unstable band so much as squeeze it. The neutral wavenumber comes down, the maximum falls, and both reach zero together — so approaching the threshold there is neither a sudden loss of instability nor a long tail of very slow growth.

On logarithmic axes the approach is clean: the growth goes continuously to zero as the quarter is neared, so the boundary is a genuine crossing rather than a place where the instability becomes too weak to find. That distinction matters for a numerical result, because “no mode found” and “no mode exists” are different claims and only the second is a theorem.

There is a prediction here beyond the threshold. The fastest-growing wavenumber falls with $J$, so a more strongly stratified layer rolls up into longer billows — which is what the spacing of a Kelvin–Helmholtz cloud pattern is a photograph of, and is a testable consequence the threshold alone does not provide.

The phase speed is zero at every Richardson number, because the profile is antisymmetric and the mode travels with the mean of the two streams. Howard’s circle is centred there, so the eigenvalue sits on its vertical axis and rides down it as the stratification rises. Stratification does not move the billow; an asymmetric profile does.

The theorem’s own shape, and why the qualifications are load-bearing #

It is worth taking the three qualifications one at a time, because each of them fails somewhere real and the failures are different in kind.

“Everywhere” is a pointwise condition on a quantity that varies. The proof works by an integral inequality that has to hold at every height, so a single station where $Ri$ dips below a quarter breaks it — and the dip need not be deep or wide. A profile whose Richardson number is 5 over most of its depth and 0.2 across one thin sheet is not covered.

“Inviscid and non-diffusive” is not a small-correction assumption. Viscosity does not merely damp things here; it changes the problem’s character, because the Taylor–Goldstein equation is second order and the viscous one is fourth, with two more boundary conditions and two more families of modes. Diffusion of the stratifying agent is worse, because it lets a fluid parcel exchange buoyancy with its surroundings and so escape the restoring force the theorem’s energy argument depends on. That is the mechanism behind double-diffusive instability, which occurs at Richardson numbers of any size whatever.

And “normal mode” excludes the transient. A superposition of individually decaying modes can grow for a while before it decays, sometimes by factors of hundreds, and a flow that grows a disturbance by two hundred before decaying it is not, operationally, a stable flow. That is the same objection that makes pipe flow’s transition so difficult, and it applies here word for word.

None of that makes the theorem less exact. It makes the set of flows it applies to smaller than the set a reader assumes.

And what it does not say #

Now the logical half.

Take plane Couette flow between rigid walls — $U = z$ — with a uniform stratification $N^2 = 0.05$. The gradient Richardson number is 0.05 at every height, a fifth of the threshold, so Miles’ theorem does not apply and says nothing.

Searching over twenty-five wavenumbers from 0.1 to 4.1 and five starting guesses each — a hundred and twenty-five Newton solves — finds no unstable mode. The flow is stable, comfortably, at a Richardson number five times below the value that is routinely quoted as the onset of instability.

The reason is not subtle and it is not about stratification at all. Couette flow has no inflection point, so Rayleigh’s criterion already forbids inviscid instability with no stratification whatever, and adding some cannot help.

Laid out as a table the logic is plain. Above a quarter the theorem forbids and nothing is found. Below it the theorem is silent — and one of the two profiles below a quarter is unstable while the other is not. Silence is not a prediction.

What the billow actually does, once it has started #

The theorem is about the moment of onset, and it is worth marking where its jurisdiction ends, because the interesting physics is on the other side of that line.

A Kelvin–Helmholtz billow that starts growing does not grow for ever. It rolls up, overturns, and mixes the two layers — and in doing so it raises the local Richardson number, because it has flattened the shear and steepened the density gradient. The instability eats its own cause.

Where it stops is not a linear question and nothing above bears on it, but the observed answer is suggestive: mixing events in the ocean and in the laboratory tend to leave the layer at a Richardson number near a quarter rather than well above it, which is why the number keeps turning up in observational data as a sort of attractor. That is a different claim from the theorem’s, arrived at by a different route, and the two get conflated constantly.

The site’s own account of what the mixing costs is the number that stops the mixing; the point here is only that a measured distribution of Richardson numbers piling up near a quarter is evidence about the end of mixing events, not about their onset.

Below a quarter there is more than one thing that can happen #

The silence is worse than it looks, because below the threshold the Richardson number does not even settle which instability is on offer.

The profile used above matches the density interface to the shear layer, which is convenient and is not what a thermocline or a salt interface looks like: those are usually far thinner than the shear that straddles them. Make the density interface thin and a second mode appears — a pair of them, with equal and opposite phase speeds, travelling in both directions relative to the mean flow rather than sitting still in it.

That is the Holmboe instability, and it looks nothing like a Kelvin–Helmholtz billow. Instead of a roller that overturns and mixes the two layers, it produces cusped waves that ride along the interface and eject thin wisps from their crests. It mixes considerably less for the same growth, and it persists to Richardson numbers at which the stationary mode has already been extinguished — still below a quarter, so no theorem is troubled.

So the phase speed of zero in the figure above is a property of the matched profile and not of stratified shear layers, and below the threshold the outcome is governed by a second parameter — the ratio of the two thicknesses — that the Richardson number does not contain at all.

Why it is so often read the wrong way #

Two reasons, and both are respectable.

The first is that the standard test case obeys the converse. A tanh layer with a matched density profile is unstable at every Richardson number below a quarter, so on the flow everybody uses to illustrate the theorem the condition happens to be necessary as well as sufficient. Generalising from it is natural and wrong.

The second is that the quarter has an independent heuristic behind it that sounds like a criterion.

Overturning a layer of thickness $h$ costs potential energy of order $N^2h^2$; the shear has kinetic energy of order $(dU/dz)^2h^2$ to pay with; the ratio is the Richardson number and the exchange breaks even at a quarter, with the quarter coming from the factors of a half in each energy and from the mixing being to a mean. It is a plausibility argument that happens to give the right number, and because it is framed as an availability of energy it reads as a two-way condition. It is not one: having enough energy to do something is not a reason to expect it to happen.

A quarter is not the only threshold in the picture #

There is a second exact statement in the same problem, and putting the two side by side makes the logical shapes clearer.

Howard’s semicircle theorem — the subject of every unstable wave inside one circle — bounds where an unstable eigenvalue can be. It is also sufficient and not necessary, in the sense that being inside the circle does not make a mode unstable. But nobody misreads it that way, because nobody expects a region to be a prediction.

Miles’ theorem has the same logical form and gets misread constantly, and the difference is that its statement is a number. A number invites the reading “below this, the thing happens”, in a way a region does not, and the invitation is strengthened by the number being memorable and by the standard example obeying the converse.

That is worth generalising, because this collection has several thresholds of each kind. A critical Reynolds number is not a theorem at all and is not even a number. A choking condition is a theorem and is two-way — the flow is sonic at the throat if and only if the pressure ratio is below the critical value. Miles’ quarter is a theorem and is one-way. Knowing which of the three a threshold is settles what a measurement of it is evidence for, and the three are routinely quoted in the same sentence.

What this means for a measurement #

An oceanographer measuring a Richardson number of 0.15 in a shear layer has found that Miles’ theorem does not apply. That is all. Whether the layer is mixing depends on its profile shape, on whether the shear and the stratification are co-located, on the layer’s history, and on the finite-amplitude disturbances already present.

Two practical consequences follow.

A Richardson number is a screening test, not a diagnosis. Above a quarter everywhere, a layer can be excluded. Below it, it goes on the list. Turbulence-closure schemes that switch mixing on below a quarter are making a modelling choice, not applying a theorem — and the choice is defensible on other grounds and is often calibrated to a different number entirely, which is the closure problem in another guise.

And the “everywhere” is the part that fails first. Real profiles have $Ri$ varying by orders of magnitude over a few metres, so a bulk Richardson number computed over a layer — which is what a measurement usually delivers — can be above a quarter while a thin sublayer inside it is far below. The theorem needs the pointwise minimum, and the pointwise minimum is what the measurement resolves worst.

Two ways of getting the number wrong in practice #

The quarter is quoted in two forms in the field and they are not the same number.

The gradient Richardson number is the pointwise one the theorem is about: $N^2$ over the square of the local shear, evaluated at a height. The bulk Richardson number is $g\,\Delta\rho\,h / (\rho\,\Delta U^2)$ over a layer, and it is what a mooring with instruments a few metres apart actually delivers.

For a tanh layer with a matched density profile the two differ by a factor that depends on how the layer thickness is defined — commonly around four — so a bulk value of unity can correspond to a gradient minimum of a quarter. Quoting a threshold of “a quarter” against a bulk measurement is therefore wrong by roughly the factor that decides the answer, and this collection has a whole essay on that particular hazard in five numbers, one name.

The second error is subtler. A profile’s gradient Richardson number has a minimum, and the theorem needs that minimum. An average of $Ri(z)$ over the layer is not the minimum and is systematically larger, because $Ri$ is a ratio whose denominator is largest exactly where the theorem is closest to being violated. Averaging a Richardson number is averaging the wrong quantity, and the direction of the error is always towards reassurance.

What the theorem is genuinely good for #

Having spent most of the essay on what it does not say, it is worth being clear that it says something valuable and rare.

It is an exact result about a nonlinear system’s linear stability with no adjustable content. There are very few of those. The quarter is not fitted, not measured, and not a leading term of anything: it comes out of an integral inequality, and if the hypotheses hold it holds.

It rules things out cheaply. A profile can be excluded from consideration with two derivatives and a division, without solving an eigenvalue problem — and eigenvalue problems for stratified shear flows are unpleasant enough that a cheap exclusion is worth a great deal. Most of the ocean, most of the time, is above a quarter, and the theorem says so without anybody computing anything.

And it identifies the right quantity. That the competition between shear and stratification is governed by their ratio rather than by either separately is the substantive content, and the fact that the ratio has a threshold at all is more surprising than where the threshold is. A criterion that depended on the shear and the stratification separately would be a much less useful thing to have found.

The complaint in this essay is about a direction of inference, not about the result. Read as “above a quarter, safe”, it is exact, cheap and true. Read as “below a quarter, unstable”, it is a guess wearing a theorem’s clothes.

What is not claimed #

Inviscid and non-diffusive. Both fail in every real stratified flow. Viscosity and the diffusion of salt or heat produce their own instabilities — double diffusion is the standard example — and none is covered here or by the theorem.

Normal modes only. A stratified shear flow with $Ri > 1/4$ can still grow transiently, sometimes by large factors, from a superposition of decaying modes. Miles’ theorem is about eigenvalues and transient growth is not an eigenvalue.

The profile was chosen to make “everywhere” unambiguous. $N^2 = J\,\mathrm{sech}^2 z$ against $U = \tanh z$ gives a constant gradient Richardson number, which is convenient and not typical: a real layer’s $Ri$ varies by orders of magnitude across it, and the theorem’s hypothesis then depends on a minimum that is hard to resolve. Nothing here measures how much that matters.

Two-dimensional disturbances. Squire’s theorem extends to the stratified case with a modification, and the three-dimensional problem is not solved here.

The solver’s “no mode found” is bounded by the search. At $J = 0.26$ and above, a hundred and twenty-five Newton solves from different starting guesses returned nothing, and a hundred and twenty-five failures are strong evidence rather than a proof. The proof is Miles’, and the computation is a check on the solver as much as on the theorem — which is the right way round, since a solver that found an unstable mode above a quarter would be reporting its own bug.

And the Couette counterexample is a clean case, not a common one. It works because the profile has no inflection point at all, which is a strong condition. It demonstrates that the converse is false; it does not measure how often the converse fails in flows anybody cares about, and on the evidence of the tanh layer the converse is often true in practice.