The floor that gives the drift back

There are now three answers to the question of what happens to the mass a swell carries, and each has depended on what surrounds the water.

In a wave tank, the drift a closed box will not allow found a return current carrying exactly the opposite transport, because the far wall will not accept water and mass conservation leaves no alternative. In the open ocean, the drift a rotating planet takes back found the same cancellation with no wall at all: the Coriolis force acting on the drift drives an Eulerian current whose transport is exactly minus the Stokes transport, at every eddy viscosity. And in the drift that turns a current into rolls the drift stopped being a transport and became a force that overturns the surface layer.

The ocean’s cancellation has a condition hidden in it, and it is the depth. The calculation that produced it put the floor so far down that nothing reached it. Over a continental shelf — where most of the world’s coastal pollution, fish larvae and beached plastic spend their time — the floor is thirty, fifty, a hundred metres down, and the Ekman layer that did the cancelling is ten or twenty metres thick.

The one line that says what must happen #

The steady Eulerian current $W = u + iv$ under a swell on an $f$-plane, with an eddy viscosity $\nu$, obeys

$$\nu W'' - i f W = i f\,U_s(z),$$

where the right-hand side is the Coriolis force acting on the drift. Integrating it from the floor to the surface, with no wind stress at the surface, gives

$$i f\,(M_E + M_S) = -\,\tau_b/\rho,$$

where $M_E$ and $M_S$ are the Eulerian and Stokes transports and $\tau_b$ is the stress the current exerts on the floor.

That is the whole argument in one line. The Coriolis force on the net Lagrangian transport is exactly the bottom stress. In the open ocean there is no floor within reach, the stress is zero, and the transport cancels. Over a shelf the floor holds a stress, and whatever it holds is transport that survives.

The line does not say how large the stress is. For that the layer has to be solved.

The layer over a floor #

The model is the deep calculation with a floor put under it: water of depth $H$, the current held at zero on the floor and free of stress at the surface, a constant eddy viscosity, and linear waves of one period in water of that depth, whose drift is

$$U_s(z) = \frac{a^2\omega k\,\cosh 2k(z + H)}{2\sinh^2 kH}.$$

That drift is larger at the surface and reaches further down than the deep-water drift does, because a wave in finite depth has flattened orbits that still move at the floor. The wave’s amplitude is held at one metre as the depth changes, so that every comparison is between seas carrying the same wave.

The equation has a closed-form solution: a part that follows the drift’s own cosh profile, and two Ekman exponentials that fit the floor and the surface. Written as $e^{qz}$ and $e^{-q(z+H)}$, with $q = (1+i)/\delta$ and the Ekman depth $\delta = \sqrt{2\nu/f}$, neither overflows however deep the sea.

The three profiles show the cancellation being built and being prevented. At 60 metres, more than four Ekman depths, the Eulerian current has room to develop its full spiral, and below a thin surface layer the Lagrangian current reverses, carrying water back against the waves: the transport it carries back is nearly the whole of the drift’s. At 8 metres the floor is inside the Ekman layer. The current the Coriolis force tries to drive is held back by the floor’s friction, it cannot build the return flow, and the drift’s forward motion survives at every depth — including at the floor itself, where the wave’s orbits still slide.

That last detail is one of the model’s approximations showing: the drift of inviscid wave theory does not vanish at a no-slip floor, while the Eulerian current does. A real wave has a thin boundary layer of its own at the floor, with its own streaming, and that is left out.

How much is kept, measured in the one length that matters #

The collapse is the finding. Four seas with viscosities spanning a factor of seventeen and waves of eight and twelve seconds keep fractions of their Stokes transport that agree to within a few per cent at every depth, once the depth is measured in Ekman depths. At half an Ekman depth every one keeps 98 per cent; at one, 77 or 78; at two, between 27 and 30; at three, between 9 and 12.

It is a collapse the problem was not obviously going to make. The wave brings a second length, the drift’s own decay depth, and the ratio of the drift’s depth to the Ekman depth varies by a factor of three across these cases. It barely matters. The fraction kept is set by how much of the water column the Ekman spiral can occupy before the floor stops it, and that is a question about $H/\delta$ alone to within a few per cent.

Which way the kept transport points #

The magnitude is only half of the answer, and the other half is more surprising. The kept transport does not simply shrink as the sea deepens; it turns. At half an Ekman depth it is ten degrees to the right of the waves; at one, about 35 to 45; near two it is crossing the waves at a right angle; and beyond that it is pointing partly back against them while it winds towards zero. The trace is a spiral with the open ocean’s cancellation at its centre.

The direction is where the collapse fails.

The four curves agree in shallow water and separate as the sea deepens. The difference is the drift’s depth. A drift concentrated close to the surface, compared with the Ekman depth — the short-wave, high-viscosity cases — acts like a surface stress, and its kept transport turns the way the transport of a wind stress would. A drift that reaches deep acts more like a body force spread through the column, and its transport turns less. How much survives is set by the depth in Ekman depths; where it goes is set by the wave.

Why shallow water turns the transport by two-thirds of a square #

The shallow end of the spiral has a closed form, and it explains both the collapse and where the collapse stops.

When the floor is well inside the Ekman layer, two things simplify. The drift hardly varies over so short a column, so it can be treated as a constant $U_s$. And the current the rotation drives is small, so the Coriolis force on the current itself is negligible beside the Coriolis force on the drift. The equation becomes $\nu W'' = i f U_s$, with the current held at zero on the floor and free of stress at the surface, and its solution is a parabola:

$$W = \frac{i U_s}{\delta^2}\,(z^2 - H^2).$$

That current points at right angles to the waves — it is purely imaginary — and integrating it over the column gives an Eulerian transport of $-\tfrac{2}{3}\,i\,U_s H^3/\delta^2$. Divided by the Stokes transport $U_s H$:

$$\frac{M_L}{M_S} = 1 - \tfrac{2}{3}\,i\left(\frac{H}{\delta}\right)^2.$$

In shallow water the kept transport turns to the right by $\arctan\big(\tfrac23 (H/\delta)^2\big)$ and scarcely shortens at all. The turn is second order in the depth ratio and the shortening fourth order. Both statements check against the full layer: for 12-second waves the turn is 2.43 degrees at a quarter of an Ekman depth against the formula’s 2.39, 9.79 against 9.46 at a half, and 21.2 against 20.6 at three-quarters; and halving the depth from a half to a quarter of an Ekman depth divides the shortening by 15.7, close to the sixteen a fourth-order term predicts.

The formula has no wave in it, which is the collapse. It has no wave in it because at this order the drift’s depth structure has not yet been felt, and that is also why the collapse holds for the magnitude and fails for the direction further out: the direction picks up the drift’s shape at the next order, while the magnitude’s leading correction is still the parabola’s.

The same result, eighty years apart #

This is a problem that has been solved before with a different forcing. When Walfrid Ekman worked out the wind-driven current in 1905 he did the deep-ocean case first, found the transport at right angles to the wind, and then did what this essay does: put a floor under it. In shallow water his transport turned less than ninety degrees, towards the wind’s direction, and approached the wind’s direction as the sea became shallower than the Ekman depth.

The drift’s problem has exactly that structure, and the kept transport above is Ekman’s shallow-water result with the drift playing the part of the wind. It is not quite the same function — the wind acts only at the surface and the drift acts through the depth of the wave — but the controlling ratio is the same one and the spiralling approach to the deep answer is the same geometry. The layer that stops at a depth found the Ekman depth as the length rotation and diffusion make between them; here it is also the length that decides how much of a wave’s mass transport a coastal sea keeps.

The floor holds exactly what the planet leaves #

The identity is a check on the closed form because its two sides come from different parts of the solution. The net transport is an integral of the whole current profile; the bottom stress is a derivative at one point. If the particular solution or either exponential were wrong, the two would disagree, and they agree to machine precision in every sea tried.

The same closed form was also checked against a solution that shares none of its algebra: second-order finite differences on four thousand levels, with the finite-depth drift sampled at every level rather than integrated analytically. Across seas from 12 to 200 metres deep and viscosities from 0.003 to 0.05 m²/s, the two agree to $1.6 \times 10^{-6}$ of the surface drift in the current and $3 \times 10^{-6}$ of the Stokes transport in the net transport. A sea sixty Ekman depths deep keeps $6 \times 10^{-16}$ of its Stokes transport, which is the open ocean’s exact cancellation recovered as a limit; a sea a sixth of an Ekman depth deep keeps 99.97 per cent of it, pointing within about a degree of the waves.

The check that matters most is the last group, because it is the only one that could have failed for a reason other than an algebra slip: a closed form derived from a simplified equation agreeing with the full one is evidence that the simplification captured what the full solution is doing, and not only that both were typed correctly.

One swell over one shelf #

On a North Sea shelf, with a viscosity of 0.01 m²/s and a swell of eight seconds, the numbers are concrete. Over 10 metres of water the sea keeps 90 per cent of a Stokes transport of 0.55 m² per second, pointing 24 degrees to the right of the waves — a depth-mean drift of 4.3 kilometres a day. Over 20 metres it keeps 45 per cent at 66 degrees, 0.85 kilometres a day. Over 40 metres it keeps 8 per cent, pointing back behind the across-wave direction. By 80 metres there is nothing left worth measuring.

Every one of those numbers moves with the eddy viscosity, which is borrowed and genuinely uncertain on a shelf, where tidal currents often stir the whole column far harder than the waves do. A viscosity of 0.05 m²/s puts the Ekman depth at 29 metres and moves the same percentages to depths more than twice as great. What does not move is the shape of the answer: a sea keeps its drift in proportion to how little room its floor leaves the Ekman spiral.

A cancelled transport is not a still surface #

The transport is only the depth integral, and it is worth separating from what a floating object does, because the two can disagree completely.

A piece of plastic at the surface moves with the Lagrangian current at the surface, not with the depth-mean. Under the same 8-second swell, the surface current over 80 metres of water — where the net transport has fallen to 0.3 per cent of the Stokes transport — is still 58 per cent of the surface drift, 2.9 centimetres a second at 21 degrees to the right of the waves, or 2.5 kilometres a day. The deep sea has taken back the transport by driving a return flow a few metres down, as the profiles showed, and the surface layer above that return flow keeps moving.

Over 10 metres of water the surface current is 91 per cent of the surface drift, 9.4 centimetres a second at 20 degrees — 8.1 kilometres a day, nearly twice the depth-mean. Over 20 metres it is 61 per cent, 3.3 kilometres a day at 27 degrees. So a shallow sea moves floating material faster than a deep one, and in both it moves at an angle to the waves; but the difference between them is a factor of three at the surface against a factor of three hundred in the transport. A calculation of where water goes and a calculation of where floating things go are different calculations, and the second is the one a beach-cleaning survey needs.

What the model cannot show #

The wave’s own bottom boundary layer. The drift used here comes from inviscid wave theory and does not vanish at the floor. Real waves in shallow water have a thin oscillatory boundary layer at the floor whose own streaming drives a mean flow in the direction of wave travel, the effect Longuet-Higgins computed in 1953. It adds transport in shallow water and is not in the calculation.

Wind. There is no wind stress here, deliberately, so that the drift’s own transport can be followed. A real shelf sea has a wind blowing over the swell, and the depth-integrated balance then contains the wind stress too; the drift’s kept transport adds to the wind’s.

A constant eddy viscosity and a quiet floor. Shelf seas are stirred by tides, and the mixing that controls $\delta$ varies through every tidal cycle and with height above the floor. A linear stress on a no-slip floor is the simplest closure that holds a stress, not a model of a turbulent bottom boundary layer.

Horizontal uniformity, and no coast. The swell covers the whole sea and nothing interrupts the kept transport. A real shelf ends at a coast, and a transport that points towards it cannot continue.

Shoaling. The wave’s amplitude was held fixed as the depth changed. A real swell crossing a shelf shoals, steepens and eventually breaks, and its drift changes for that reason as well as for the one computed here.

Still open: what a coast does with the transport it is sent #

The kept transport over a shelf points somewhere, and near a coast it points partly at the coast. Water cannot accumulate indefinitely against a shoreline, so the sea surface rises until the pressure gradient drives a current that carries the transport back or along the coast — a set-up, like a wind’s, rather than a flow into the land.

The next calculation takes the same layer, bounds it with a straight coast, and solves for the surface slope and the coastal current the kept transport produces: how large the set-up is for a given swell, whether the return is carried in a bottom layer or turned into a current along the shore, and what fraction of the drift’s transport ends up delivering floating material to the beach rather than circulating back out. That is the question behind every model of where coastal plastic lands, and the depth-mean speeds above are the first number it needs.