The drift a closed box will not allow

The drift in a wave that has none is the collection’s account of Stokes drift: the velocity at a fixed point under a passing wave averages to exactly zero, and every parcel of water moves steadily forward anyway. This essay is about what happens when the water has nowhere to go — which produces an exact constraint and, immediately afterwards, a question the constraint cannot answer.

The transport #

For a linear wave in water of depth $h$,

$$u_S(z) = \frac{a^2\omega k \cosh 2k(z+h)}{2\sinh^2 kh},$$

which is forward everywhere and largest at the surface. Its depth integral — the mass the wave carries per unit width — comes out in closed form as

$$M_S = \frac{a^2\omega}{2\tanh kh},$$

which in deep water is $a^2\omega/2$ with no depth in it at all.

Checked by integrating the drift profile rather than by trusting the algebra, at three grid resolutions, the two agree to eight figures at the finest.

Shallow water carries more: at $kh = 0.3$ the transport is 3.43 times its deep-water value at the same amplitude and frequency. It is finite and non-zero at every depth, which is what makes the constraint below bite.

Why the drift exists at all, in one paragraph #

The transport above is a second-order quantity and it is worth being clear where it comes from, because the whole essay hangs on it being real rather than an artefact of the expansion.

Under a linear wave a parcel moves on a closed ellipse — to first order. To second order the orbit does not close, and the reason is that the parcel spends slightly longer in the part of its orbit where the wave-induced velocity is forward, because it is higher there and the orbital velocity decays with depth. The mismatch per orbit is small and it never reverses, so it accumulates: the drift is a bias rather than an oscillation, and integrating it over depth gives a mass flux.

The consequence worth carrying is that Stokes drift is not a separate current added to the wave. It is the wave, described in the coordinates a fluid parcel uses rather than the coordinates a fixed instrument uses — which is the difference the drift in a wave that has none is entirely about, and is why the Eulerian mean velocity can be exactly zero while every parcel advances.

The constraint #

Now put the wave in a closed channel — a wave tank, a lake, a laboratory flume. The total volume flux through any cross-section must be exactly zero, because the water has nowhere else to be.

That is a mass-conservation statement and nothing else. No viscosity, no turbulence model, no assumption about anything. So an Eulerian return current appears whose depth-integrated transport is exactly $-M_S$.

Three candidate return currents, each normalised to carry exactly the required transport, and each verified to do so at the quadrature’s own accuracy.

A uniform slab moving back at $M_S/h$; a parabola that vanishes at the bed and has no shear at the surface; and a profile concentrated near the surface where the wave is. All three are admissible in the only sense the constraint defines, and they are completely different flows.

What “admissible” means here, and how weak it is #

The comparison class is worth stating precisely, because the essay’s whole point is how large it is.

A return current is admissible if it is a horizontal flow whose depth integral is $-M_S$. That is one scalar condition on a whole function of depth. Nothing requires it to be smooth, to vanish anywhere, to be single-signed, or to have any particular shape — and the three profiles used here are a deliberately tame sample of what satisfies it.

Compare that with the comparison class in how much more than the least, where the admissible fields have to be solenoidal and have the right normal component on every boundary, and the theorem still leaves an infinite family. Here there is one condition rather than two families of them, and the freedom is correspondingly larger.

The general point is that a constraint’s usefulness is measured by how small it makes the space of possibilities, not by how exactly it holds. This one holds exactly and leaves a space of functions. Kelvin’s minimum energy theorem holds exactly and picks out a single member of its space — because it is a variational principle rather than a single integral condition, and a variational principle is a condition at every point.

That is the difference between an exact statement that determines something and an exact statement that does not, and both are common.

And what it does not fix #

Adding the Stokes drift to each return current gives the net Lagrangian drift, which is the quantity a float measures and the quantity a sediment grain feels. All three agree that the surface moves forward, because the Stokes drift there swamps any return current of the right size — but they agree about nothing else.

The uniform return current reverses the drift at 23 per cent of the depth below the surface, so three-quarters of the water column is moving upstream. The two that satisfy no slip at the bed reverse it at 98 and 100 per cent — which is to say they barely reverse it at all, and essentially the whole column drifts forward.

That is a spread of 77 per cent of the water column, between three flows that carry exactly the same transport.

At the bed the disagreement is not quantitative but categorical. The uniform current takes the water back upstream; the two that satisfy no slip cannot, because they vanish there, so what is left is the Stokes drift — forward.

The sign of the near-bed drift decides which way sediment moves, which is the question the whole calculation is usually being done to answer, and the exact constraint has no opinion about it. The answer needs the dynamics: the viscous boundary layers at the bed and at the free surface, the turbulence in between, and the wave’s own second-order stress — which is the streaming an oscillation produces, computed properly rather than assumed.

The same shape of problem, twice before #

An exact integral constraint that fixes a total and not a distribution is a shape this collection has met twice already, and it is worth putting the three side by side.

A body’s lift is exactly $\rho U\Gamma$ and its pitching moment is not fixed at all — which is one formula that does not ask the shape, where six bodies at one circulation carry identical lifts and moments spanning a factor the theorem cannot see.

The dissipation’s two forms have exactly equal averages and are uncorrelated pointwise — which is equal on average and nothing else.

And here, the transport is exactly cancelled and the profile is free.

In all three the exact statement is an integral over a region and the interesting question is about the integrand. That is not a coincidence: conservation laws are statements about regions, and a conservation law can only ever constrain the total of what it conserves. Reading one as a statement about a distribution is the commonest way to over-read an exact result, and this essay is the cleanest example of it in the collection because the constraint is so simple and the spread it permits is so wide.

What the constraint is good for anyway #

Three things, and none of them is the profile.

It bounds the error of ignoring it. A calculation that leaves out the return flow has the whole Stokes transport as a systematic error in its mass budget, and that error is exactly computable from the wave alone. In a flume with a ten-centimetre wave in half a metre of water it is not small.

It says the mean Eulerian current is not zero. A moored current meter under a wave field measures the Eulerian velocity, which in a closed channel is negative — upstream — while a surface drifter measures the Lagrangian one, which is positive. The two instruments disagree by construction, and the difference is exactly the Stokes drift.

And it identifies where the constraint applies. In the open ocean nothing forces the transport to cancel, and it does not: the Stokes transport is a real mass flux that has to appear in any budget, and it is large enough to matter for oil spills, for larval dispersal and for the momentum balance of the surface layer. The exactness of the cancellation is a property of the box, not of the wave, and carrying a tank result into open water is carrying a boundary condition that is not there.

What pushes the water back #

Mass conservation says a return transport must exist. It says nothing about what drives it, and the answer is worth having because it explains which of the three profiles is the natural one.

The water piles up. Waves running along a closed flume carry mass towards the far end, and it arrives faster than it can be carried back, so the mean surface tilts — standing higher downwave — until the hydrostatic pressure gradient that tilt produces drives a return flow of exactly the required transport, and no more. The slope is tiny and it is not optional: it is the only thing in a closed channel capable of pushing water the wrong way.

So the return current is a pressure-driven flow, which is why the parabola is the natural laminar candidate — a uniform pressure gradient between a no-slip bed and a stress-free surface produces exactly that shape — and why the uniform slab, which needs no gradient at all, is the least physical of the three.

The same balance in open water is what makes the mean sea level fall slightly beneath a shoaling wave field and rise inside the surf zone. There it is not a wall doing the pushing but the along-shore change in the wave’s own momentum flux, and the tilt adjusts against it in the same way. Closed box and open beach are one equation with two boundary conditions.

What actually decides the profile #

Since the constraint does not, it is worth naming what does, even though none of it is computed here.

The viscous boundary layer at the bed. A no-slip bottom forces the Eulerian velocity to zero there, and within a Stokes layer of thickness $\sqrt{2\nu/\omega}$ — millimetres in water at wave frequencies — the wave’s own second-order stress drives a steady streaming that is forward, towards the wave. That is Longuet-Higgins’ bottom streaming, and it is why sediment under a shoaling wave tends to move shorewards even where the return current above is seawards.

The free surface. A clean surface and a contaminated one behave completely differently: a surfactant film makes the surface behave as a no-slip boundary rather than a stress-free one, which reverses the sign of the streaming just below it. Laboratory measurements of return flow are notoriously sensitive to how clean the water is, and that is the reason.

And the turbulence, in any real sea. Above laboratory scales the return flow’s shape is set by a turbulent eddy viscosity whose profile is itself modelled, so the answer inherits a closure’s own guess.

Three mechanisms, all of which act at scales far smaller than the wave, all of which change the answer to the question the constraint could not answer. That is the usual arrangement: the exact result is cheap and general, and everything expensive is in what it leaves out.

What a wave tank measurement is measuring #

The practical form of all this is a warning about laboratory data, and it is worth spelling out because wave-tank results are the source of a good deal of what is believed about waves.

A flume with waves running along it is a closed box, so its return flow is present, is exactly determined in total, and is entirely undetermined in shape by anything the experimenter controls. Its shape depends on the flume’s depth, its length, whether the beach at the far end absorbs or reflects, how clean the surface is, and whether the flow has had time to set up.

So a velocity profile measured in a flume is the wave’s drift plus a return current that is an artefact of the flume — and the artefact is comparable in size to the thing being measured. At the depths where the two nearly cancel, the measured net drift is a small difference of two larger quantities, one of which is the apparatus.

The standard defence is to measure the Eulerian mean and add the theoretical Stokes drift, which works if the theory is right about the drift and does not need it to be right about the return flow. The alternative — measuring the Lagrangian drift directly with floats — measures the sum, artefact included, and is the one that does not transfer to the open sea.

Neither approach is wrong. What is wrong is quoting a measured drift profile as a property of waves, and this collection’s usual name for that is the instrument in the answer.

One number that is not a convention #

Amid all the freedom there is a second exact statement worth extracting, and it is about the surface.

Every one of the three profiles has the surface drifting forward, and the values are within 28 per cent of each other — 0.0282, 0.0266 and 0.0219 in the units used. That is not an accident of the three shapes chosen. Near the surface the Stokes drift is at its largest and the return current, whatever its shape, has to average to a much smaller number over the depth, so the return current cannot be comparable with the drift there without violating its own integral.

So the surface drift is nearly determined, and its value is close to the Stokes drift itself minus the mean return, which is $u_S(0) - M_S/h$. That is a formula containing only the wave and the depth, and it is what a surface drifter in a flume should read to within tens of per cent whatever the return profile turns out to be.

That is worth having as a matter of method: when an integral constraint leaves a profile free, the places where the answer is nevertheless nearly determined are the places where one term dominates. The surface is one; the bed is emphatically not, which is why the two ends of the same water column are respectively the best and the worst thing to predict here.

What is not claimed #

Linear waves, to second order. The Stokes drift above is the second-order term of a small-amplitude expansion, and at a steepness of 0.1 the third-order correction is a few per cent. Nothing here is valid for a breaking wave, where the transport is much larger and is not given by any expansion.

The three return profiles are shapes, not solutions. They are chosen to be admissible and different, which is what the argument needs. Longuet-Higgins’ conduction solution — the profile a laminar flume actually develops, from the viscous boundary layers at both surfaces — is a fourth possibility, and it is not computed here because this module has no viscosity in it.

One shape had to be rejected as inadmissible. The first candidate for the third profile was $3s^2 - 2s$, whose depth integral is exactly zero — so normalising it to carry a stated transport divides by nothing. A shape that carries no net transport cannot be scaled to carry one, and it is a recirculation rather than a return flow.

The channel is two-dimensional and the flow is steady in the mean. A real basin has ends, sides and a spin-up time; the constraint applies to a cross-section of an infinitely long channel with the flow established, and none of those is a wave tank.

The wave’s own momentum flux is described and not computed. The section on what drives the return flow says what it does; no figure here contains it, and the momentum balance of the surface layer has a term for it that this essay does not evaluate — the same distinction between what is carried and what is stored that the momentum with no value is about in a completely different setting.

And the reversal depths are for one wave in one depth. The 23 per cent and the 77 per cent spread are for $kh = 5$ at a steepness of 0.1. The qualitative conclusion — that the constraint fixes an integral and leaves the profile open — does not depend on that, and the numbers do.