Kelvin's sheet asks for six metres a second, and the water starts at two

A sheet that cannot stay a sheet took a vortex sheet with nothing to hold it back: every wavelength grows, the shortest fastest, and the problem is so badly posed that a computer’s rounding error decides the answer. Every wavelength at once stated the rescue that a real interface carries. Put the sheet at the surface of water, and gravity holds down its long waves and surface tension its short ones, so that a band of wavelengths survives only above a critical speed. Air over water comes out at 6.67 metres a second.

Then that essay added a sentence and moved on: water is observably disturbed by winds well below that. It is the most visible failure of an exact calculation that exists. Anyone can stand beside a pond in a light breeze, about two metres a second on a good anemometer, and watch cat’s paws run across the surface. Beaufort’s scale, which is a list of what sailors could see, puts “small wavelets” between 1.6 and 3.3 metres a second. This essay asks what the sheet got wrong, and finds that nothing about its arithmetic is at fault. It answers a different question, about a wind that does not exist, and the wind that does exist cannot reproduce its answer by any adjustment, because the two have different physics in them. One of them has a threshold in the air. The other has its threshold in the water.

Why a sheet has a threshold at all #

A sheet is two uniform streams and an interface. Suppose the interface is rippled. The air has to go over the crests, it speeds up there, and by Bernoulli its pressure drops; in the troughs the opposite. So the pressure the air exerts is lowest exactly where the surface is highest, and it pulls the crests up. That suction is proportional to the density of the air, the wavenumber and the square of the speed, and against it stand gravity and surface tension, which try to flatten the wave. For a density ratio like air’s over water, the suction is a small correction to the water’s own stiffness. It softens the wave: as the speed rises, the frequency of the wave falls, in the frame of the water. At Kelvin’s speed the frequency reaches zero, the stiffness has been cancelled and the crest keeps going.

In the formula, the wave’s speed is

$$c = \frac{\rho_a U}{\rho+\rho_a} \pm \sqrt{c_0^2 - \frac{\rho\rho_a U^2}{(\rho+\rho_a)^2}},$$

where $c_0$ is the capillary-gravity speed on still water. Everything happens through the sign under the root. The minimum of $c_0$ over wavelengths is 0.231 metres a second, and that fixes the whole answer: the wind needs to be about $c_0$ multiplied by $\sqrt{\rho/\rho_a}$, which is twenty-nine for air on water, so the sheet needs 6.67 metres a second. The wavelength is $2\pi\sqrt{\sigma/g\Delta\rho}$, where gravity’s stiffness equals tension’s, 1.71 centimetres.

The essential point is what happens below that speed. The two roots are real. The wave moves, a little altered, and its pressure stays exactly in phase with its elevation: lowest over the crest, highest over the trough. A pressure in phase with elevation pushes down where the water is moving neither up nor down, and it does no work at all, over a whole cycle, on the wave. Energy can reach a wave only through a pressure in phase with its slope. So a sheet below Kelvin’s speed is not weakly transferring energy. It is transferring none, to the digit, however close to the threshold it is.

Above the threshold the growth is violent. At 8 metres a second the amplitude multiplies by $e$ in thirteen milliseconds. At 10 metres a second, in five. That alone should give the theory away, because nobody has seen a pond do this. A real ripple under a real ten-metre-a-second wind takes something of the order of a second to double its energy. The sheet is wrong on both sides of its threshold: too quiet by an infinite factor below it, and too violent by a factor of a hundred above it.

The air has a height at which it travels with the wave #

A real wind is not a sheet. It rises from nothing at the water, roughly as the logarithm of height, and its friction velocity $u_*$ sets its scale:

$$U(z) = \frac{u_*}{\kappa}\,\ln\frac{z}{z_0},$$

with $\kappa = 0.41$ and $z_0$ the small roughness height the law of the wall fits. And that innocent change introduces something the sheet cannot contain: there is a height at which the air moves at exactly the wave’s speed. Call it the critical level. Below it the air is slower than the wave, and above it faster. In the sheet that height has been crushed into the interface. The speed jumps from zero to $U$, passing through $c$ in zero thickness, and nothing can happen there.

In a continuous wind a parcel of air at the critical level travels with the wave and sees a steady pressure pattern, so it does not average the wave’s push away. That is resonance, and Miles found in 1957 what it does. Rayleigh’s equation for the air’s disturbance,

$$(U - c)\left(\phi'' - k^2 \phi\right) - U'' \phi = 0,$$

is singular where $U=c$, and the singularity is exactly the transfer. Going round it on the correct side, the side a slightly growing wave or a slightly viscous fluid picks out, makes the solution pick up an imaginary part. Above the critical level the air’s disturbance carries a momentum flux into the level, and below it there is less, so the difference is delivered to the water. The size of the jump is proportional to $-U''/U'$ and to the square of the disturbance at that level. A logarithmic profile has $U''<0$ everywhere, so the sign is always the wave-growing one. That is the part of the physics that depends on the profile’s curvature rather than on its speed, and the sheet had no way of having it.

The calculation here integrates Rayleigh’s equation downward from high in the air, where the disturbance decays as $e^{-kz}$, along a path in the complex height plane that passes beneath the critical level. No viscosity is used in the air, no turbulence and no eddy model. The result is a pressure at the surface with a part in phase with the slope, written as Miles wrote it:

$$\frac{1}{E}\frac{dE}{dt} = \frac{\rho_a}{\rho}\,\beta\,\omega\left(\frac{u_*}{\kappa c}\right)^{2}.$$

The coefficient $\beta$ depends on the wave’s age, $\Omega = \kappa c/u_*$, and the roughness.

Nothing in that curve is a threshold. It holds steady near three and a half while the wave is younger than the wind, then falls away smoothly as the wave outruns it, because the critical level is climbing to a height $z_0 e^{\Omega}$, and a wave can only feel air within about one wavelength. A level one wavelength up holds a disturbance $e^{-1}$ of the surface’s, and the air there hardly knows the wave exists. So the payment does not stop. It fades, and by age 8 it is a thousandth of its peak, and the arithmetic has never had a reason to reach zero. The most important thing about this result is the shape of the statement it permits: at any wind speed at all, however slight, a wave slow enough to be overtaken by that wind receives some energy.

Where a threshold comes back from #

Then why do ponds not ruffle in any breath of air? Because the water charges a fee. A small wave loses energy to its own viscosity at a rate Lamb calculated, $4\nu k^2$, independent of how strongly anything is driving it. For the 1.71 centimetre ripple that is 0.54 per second, and its energy would last less than two seconds on its own. For a 30 centimetre wave it is 0.00175 per second, three hundred times less.

So the crossing is a threshold, and it looks like one on a pond, which makes it a number that is not a property of the fluid in much the same way a transition Reynolds number is not. But it is a threshold of an entirely different kind from Kelvin’s. It has a viscosity in it. With inviscid water, the sheet’s answer stays what it was, 6.67 metres a second, and the layer’s threshold disappears: any wind pays any slower wave, and with no fee the wave grows however slowly it is paid. With a more viscous liquid, the sheet’s answer does not change, because viscosity is not in it, and the layer’s threshold rises. That experiment has been done: Keulegan’s wind tanks of 1951, filled with liquids of raised viscosity, needed stronger winds to raise any waves at all, and Kelvin’s formula, with no viscosity, has no way of saying so.

Read across wavelengths and the second difference appears. Kelvin’s threshold picks out one ripple, since his minimum is where stiffness is least. The layer picks differently. At the threshold the ripple is the most expensive thing on the surface, because of that $k^2$, and the wave that starts first is longer and lower in frequency than the sheet’s.

The curve has a shallow minimum of 1.8 metres a second at about twenty centimetres, and that is Beaufort’s force two. Short waves cost too much in viscosity, and long waves run faster than the wind can catch them. A metre-long wave moves at 1.25 metres a second, so a wind of 2.1 metres at ten metres is barely overtaking it near the surface. It is rather a good account of why a light breeze on open water shows wavelets rather than a sudden burst of fine ripple.

What a growth rate this small looks like #

The sheet’s growth is measured in milliseconds, and the layer’s in seconds and minutes. A 30 centimetre wave at a friction velocity of 0.1 metres a second gains energy with an $e$-folding time of 134 seconds. At 0.2 it takes 34 seconds; at 0.3, 14.5. At these rates a wave near its threshold needs minutes to gather anything, longer than a light gust lasts, which is what a light wind on a pond looks like: patches of scale-like ripples that come and go, with no sudden overturning of the surface anywhere.

The order is right. Plant in 1982 gathered every measurement of wave growth available at the time and fitted them with one line, $0.04\,(u_*/c)^2$ per radian, across three orders of magnitude of $u_*/c$, with a scatter of about a factor of two either way.

The computed rate sits about 1.6 times below that line at the friction velocities where the measurements are densest: about 0.6 of the fit for a 30 centimetre wave at a friction velocity of 0.1 metres a second. Neither the shape nor the value was fitted. The only input apart from the profile itself is $\kappa$. It is a surprisingly good result for a calculation with no turbulence in it at all, and the fact that a real wind is turbulent and the model’s is not is what keeps anyone from treating that as more than an order-of-magnitude agreement. The measured line is not this mechanism alone either, and there is an argument, still open after sixty years, about how much of the observed growth comes from turbulent stresses and from the air separating off steep waves. Jeffreys’s sheltering argument from 1925 was the first attempt at a theory, and it remains a live contribution near breaking.

Where the sheet is the right theory #

The mistake is thinking of the sheet as a crude model of a wind. It is an accurate model of something else. Kelvin’s threshold scales with the square root of the density ratio, and at a density ratio near one it says something sensible: the critical speed for 800 kilograms a cubic metre of light oil over water, at an interfacial tension of 30 millinewtons a metre, is 19 centimetres a second. For two water layers of a slightly different salinity it is four centimetres a second, at a tension of a millinewton a metre, and the wave is slow and the layers comparable in density. There the suction is not a tiny correction to a large stiffness, and the interface is the thinnest thing in the problem compared with the waves it carries. In that limit the critical-level mechanism has little to do, because the wave is scarcely slower than the flow. Thorpe’s tilted-tube experiments and the billows in a stratified estuary are the Kelvin–Helmholtz instability, and a layer with a kink in it and its neighbours belong to that world.

Air over water is the opposite extreme. The density ratio is eight hundred. The waves are slow compared with any wind that could matter, so the air sees a nearly solid, wavy wall, and the only question is how the wall’s motion extracts anything from a flow that is mostly going over it. The answer has to be a small, resonant, delicate one, and the sheet has none of those qualities.

So the surprising connection is between this and the problem of the instability a viscosity creates in a boundary layer. Both are the same equation. Rayleigh’s equation with a critical level is what governs waves in a shear flow everywhere: in a Tollmien–Schlichting wave in a wall layer, in a Rossby wave absorbed in the stratosphere. Everywhere, the critical level is the place where a wave exchanges momentum with a mean flow. It happens that the first problem anyone tried with waves on a shear did not have one, and it set the question’s terms for eighty-six years, in the wrong direction.

What the picture cannot show #

The figures show a growth rate. They do not show a sea, and three things about the calculation limit how much it can claim.

The critical level of a ripple sits inside the viscous sublayer. At the threshold, a 1.71 centimetre ripple’s critical level is about 0.04 millimetres from the water, and at that friction velocity the air’s viscous sublayer is a millimetre and a half thick, where the logarithm is simply not the profile. So for the shortest waves the calculation is using a law where the law does not hold, which is the same defect, for the same reason, as using the sheet. The proper treatment puts in a linear viscous profile and the air’s viscosity with it, as Valenzuela did in 1976; the true level then lies several times higher, near a quarter of a millimetre, and the viscous stress in the air takes over much of the transfer. At their thresholds a 30 centimetre wave’s level is about 3.6 millimetres up, just above a sublayer of 2.8, and a metre wave’s about 30 millimetres, so the logarithm is on firm ground only from about 30 centimetres upward. Left of that, the curve is the inviscid mechanism carried beyond its range, and its numbers are a statement of size and not a prediction.

Nothing here is nonlinear, and a pond’s waves are. Energy passes upward and downward between wavelengths; once a band exists, ripples form on the crests of longer waves by a parasitic mechanism at a rate that has nothing to do with the wind. Phillips’s 1957 resonance with the air’s pressure fluctuations, which needs no wave to start with and so explains how a wave begins from a flat surface at all, is absent too. Those are separate contributions and the calculation has no room for them.

The profile is held fixed. The wave takes momentum out of the air, so the wind just above it must adjust, as it must over any change of surface, and once the waves are big enough to matter, that feedback is a dominant part of a real sea’s stress. Here the wave takes energy from a wind it cannot affect. And the roughness model is too simple where it matters: above a friction velocity of about 0.3 metres a second, a Charnock roughness puts the short ripples inside their own roughness, and the integration is no longer about anything, so the budget figure stops there.

The conventions, and the values #

Water at 20 degrees Celsius, of density 998 kilograms a cubic metre, kinematic viscosity $10^{-6}$ square metres a second, surface tension 0.0728 newtons a metre, under air of density 1.2. Rates of change are of the wave’s energy, not its amplitude, except in the sheet’s rate figure, which plots amplitude because that is what the formula gives; the two differ by a factor of two, which at a factor of a hundred’s disagreement is not what matters. The roughness is the larger of the smooth-wall value $0.11\nu_a/u_*$ and Charnock’s $0.012u_*^2/g$. The wind at ten metres is read from the same logarithm, and in very light airs the logarithm’s reach to ten metres is itself a stretch. Kelvin’s speed is a relative speed across a sheet, and the wind at ten metres is taken as the fairest comparison with it, since a sheet has only one speed.

How the numbers were checked #

The sheet’s minimum from the closed form agrees with a direct search of the speed against wavelength. In the integration, doubling the path’s steps moves $\beta$ by less than one part in a thousand. The admittance’s imaginary part has the sign a growing wave requires, and the size of $\beta$ falls in the range Conte and Miles tabulated in 1959. The model refuses a density ratio the wrong way up, a negative wave age and a zero wavenumber.

Who found it, and when #

Helmholtz stated the instability of a sheet in 1868 and Kelvin added gravity and tension in 1871, deriving his minimum as a statement about the first ripples a wind can raise; Lamb’s textbook later recorded that real water disagreed with it. Jeffreys tried a sheltering mechanism in 1925. Miles’s critical-layer theory appeared in 1957, the same year and the same journal as Phillips’s resonance. Lighthill gave the physical account of the critical level in 1962, Snyder and Cox first measured the growth at sea in 1966, Valenzuela treated the viscous sublayer in 1976, and Plant’s fit dates from 1982.

Still open: when the wave pays back #

The wave’s own effect on the air was left out. Each wave that takes momentum from the wind lowers the wind just above it, and for young waves that loss is a sizeable part of the whole stress; it changes the very curvature at the critical level on which the payment depends. The next calculation would take a spectrum of waves growing together, feed the momentum they gain back into the mean profile at each one’s critical level, and ask what the coupled system settles at: whether the profile’s curvature is flattened until the waves’ growth stops, as it is when electrons plateau a plasma’s velocity distribution, and so whether a young sea’s roughness is set by this quasi-linear balance and not by a constant of Charnock’s.