The drift a rotating planet takes back

The drift in a wave that has none found that every parcel under a wave creeps forward while the velocity at any fixed point averages to zero. The drift a closed box will not allow put the wave in a tank and found that the tank cannot hold the result: the drift carries mass towards the far wall, the wall will not take it, and a return current appears carrying exactly the opposite transport — exactly, because mass conservation says so and nothing else is involved. A drift made of two things that average to zero then wrote the drift down as a correlation, and ended its list of limits with a sentence about the ocean: no rotation, and a real ocean adds a Coriolis turning comparable in size to the drift itself.

This essay takes that sentence seriously. The open ocean has no far wall, so the tank’s argument does not apply, and the natural expectation is that the drift survives and a floating object is carried downwave at the drift speed for as long as the swell lasts. It is not. The planet’s rotation cancels the depth-integrated transport as exactly as the tank’s wall did, by a different mechanism and with a different consequence for anything that floats.

The Coriolis force acts on the motion the water actually has #

Four conventions come first. The swell is linear and in deep water, uniform over a wide area, travelling along x, with an 8 second period and an amplitude of 1 metre: wavelength 99.9 m, steepness ak = 0.063, phase speed 12.5 m/s. Its surface drift is $U_s = a^2\omega k$ = 0.0494 m/s, falling with depth as $e^{2kz}$ over an e-folding depth of 7.95 m, and its depth-integrated Stokes transport is $U_s/2k$ = 0.393 m² of water per second per metre of crest. The ocean is an f-plane at latitude 45°, Coriolis parameter $f = 1.031 \times 10^{-4}\,\mathrm{s^{-1}}$, northern hemisphere, so “to the right” means clockwise seen from above. Horizontal velocities are written as complex numbers $W = u + iv$, which turns rotation into multiplication by $i$.

The central physical point is short. The Coriolis force is a force on moving water, and the water’s actual mean motion is the Lagrangian one — the Eulerian mean current $W_E$ that a moored meter would record, plus the Stokes drift $W_S$ that only a float sees. The moored meter cannot see the drift, but the planet can. So the mean momentum balance for a horizontally uniform wave field reads

$$\frac{\partial W_E}{\partial t} + i f\,(W_E + W_S) = \frac{\partial}{\partial z}\!\left(\nu \frac{\partial W_E}{\partial z}\right) - r W_E,$$

with an eddy viscosity $\nu$ and a linear drag $r$ standing in for whatever friction the upper ocean has. The term $i f W_S$ is the Coriolis–Stokes force: a steady sideways push on the Eulerian current, made by a transport no fixed instrument can detect. The Eulerian mean current is zero under a wave in a non-rotating ocean. In a rotating one it is driven, and it grows until it answers the drift.

Without friction, the drift turns instead of stopping #

Start with no friction at all, and with the ocean at rest when the swell arrives. The equation then has a closed solution: $W_E = -W_S(1 - e^{-ift})$, so the Lagrangian mean is $W_L = W_E + W_S = W_S\,e^{-ift}$.

The Lagrangian mean velocity never shrinks. It keeps its full 0.0494 m/s and turns clockwise at the inertial frequency, once every 16.92 hours. The Eulerian current, starting from nothing, swings between zero and twice the drift backwards along the waves, and its average over an inertial period is exactly $-U_s$: a return current, without a wall. Over each inertial period the Lagrangian velocity averages to zero, and the drift the swell promised has been converted into an inertial oscillation.

That is the resolution of a long-standing puzzle in the theory of swell. An argument of Ursell’s in 1950 held that a steady wave could produce no mean mass transport on a rotating planet at all, which seemed to contradict every float ever watched drifting under waves. Hasselmann showed in 1970 that both pictures are right in their place: the transport is there, and rotation turns it into an oscillation rather than a displacement. The figure above is that statement drawn.

The clock is worth comparing with the one How long a fluid takes to forget it was not rotating found for a spinning container. There, water reaches solid-body rotation through thin layers on the end walls, and the time it takes is set by the walls: a number of rotation periods equal to the inverse square root of the Ekman number. Here there is no wall and no layer. The Coriolis–Stokes force acts at every depth at once, and the adjustment takes one inertial period regardless of how deep the water is or how viscous.

For a float the consequence is striking. Integrating the Lagrangian velocity gives a circle of radius $U_s/f$ = 479 m, and after 16.92 hours the float is back exactly where it started. In a non-rotating ocean the same float would have gone 3.0 km straight downwave in the same time. The waves have been working on it the whole while, every orbit has missed closing by the same amount the first essay measured, and the planet has steered the accumulated misses into a loop.

A drag turns the circle into a veered drift #

No real ocean is frictionless, and the circle does not survive friction. A linear drag $r$ on the Eulerian current damps the inertial oscillation, and the Lagrangian mean settles to $W_S\,r/(r + if)$: smaller than the drift and turned to its right by $\arctan(f/r)$.

With a drag of a quarter of $f$ — a friction time of about eleven hours — the float ends up drifting at 24.3 per cent of the Stokes drift, 76.0° to the right of the waves, and after one inertial period it is 0.35 km downwave and 1.02 km to the side. With a drag equal to $f$ it settles at 70.7 per cent of the drift at exactly 45°, and has moved 1.74 km downwave and 1.50 km to the right. With three times $f$ — strong friction, a relaxation time under an hour — it keeps 94.9 per cent of the drift at 18.4°. The drift survives in proportion to how fast friction can destroy the current the Coriolis force is building, which is the sense in which the result is “one number decides”: the ratio of the friction rate to the inertial frequency.

The steady answer is a spiral, and its transport is fixed #

A linear drag is a crude stand-in for turbulence. The classical alternative is an eddy viscosity, which lets the Eulerian current vary with depth and gives the upper ocean a boundary layer — the Ekman layer that The layer that stops at a depth derived for a wind stress. Driven by the Coriolis–Stokes force instead of by wind, with a stress-free surface, the steady equation has a closed solution: a part that follows the drift’s own $e^{2kz}$ shape, and a part that spirals down over the Ekman depth $\delta = \sqrt{2\nu/f}$.

With $\nu$ = 0.01 m²/s the Ekman depth is 13.9 m, somewhat deeper than the drift’s 7.95 m. At the surface the Eulerian current is −0.438 of the drift along the waves and −0.204 across, so the Lagrangian current a float follows is 0.562 of the drift at 20.0° to the right. The Lagrangian current along the waves falls through zero near 8 m and is backwards below it: −0.115 of the surface drift at 15 m and −0.072 at 25 m, while the across-wave current changes sign and reaches +0.067 at 25 m. Near the surface the water goes downwave and slightly right; below the drift’s reach it goes back upwave.

That backward water is the return current, and it is exactly as large as it must be. Integrating the steady equation over depth, the viscous term becomes the stress at the surface minus the stress at great depth, and both are zero. What is left is

$$i f\,(M_E + M_S) = 0,$$

where $M_E$ and $M_S$ are the depth-integrated Eulerian and Stokes transports. The Eulerian transport is minus the Stokes transport, exactly, whatever the viscosity. The Coriolis force has done what the tank’s far wall did: fixed the total transport at zero, and said nothing at all about how the return is distributed.

The spirals make the division of labour visible. With $\nu$ = 0.001 m²/s the Ekman depth is 4.40 m, shallower than the drift, and the surface Eulerian current is −0.748 of the drift along the waves and −0.162 across; the return is concentrated high in the water column, where it nearly cancels the drift locally, and the surface Lagrangian current is only 0.252 of the drift at 32.8°. With $\nu$ = 0.1 m²/s the Ekman depth is 44.0 m, the return is spread thin over a deep layer, the surface Eulerian current is −0.172 along and −0.127 across, and a surface float keeps 0.828 of the drift at 8.7°. The shapes could hardly be more different. Integrated over depth, each carries exactly −0.393 m²/s.

Across four decades of eddy viscosity the transport ratio reads −1.000000000000 with a cross-wave part below 10⁻¹⁶, while the surface Lagrangian speed runs from 0.113 of the drift at $10^{-4}\,\mathrm{m^2/s}$ through 0.598 at 10⁻² to 0.945 at 1. This is the same structure as the tank’s, and the same moral: a conservation argument fixes a total and leaves the answer anybody wants entirely open. What an oil slick or a larva at the surface does depends on the surface value, and that depends on a viscosity that is really a turbulence closure — and what the ocean’s volume budget does depends only on the total, which is exact. The eddy viscosity is a closure of exactly the kind The mean is not the flow introduces for the Reynolds stress: a guess at a correlation nobody has computed, and the surface current inherits the guess while the transport does not.

When the wind blows too, the Lagrangian transport is Ekman’s alone #

Swell far from its storm has no wind over it, but a sea still being raised by the wind has both. Adding a wind stress $\tau$ at the surface changes the depth integral to $i f (M_E + M_S) = \tau/\rho$: the total Lagrangian transport is the Ekman transport, $\tau/\rho f$ directed 90° to the right of the wind, and the Stokes transport contributes nothing to it.

For a 10 m/s wind along the waves the stress is about 0.144 N/m², and the Ekman transport is 1.36 m²/s. The Stokes transport of the swell here is 0.393 m²/s, 28.8 per cent of it — not small. Yet the Lagrangian transport comes out as exactly 1.36 m²/s at right angles to the wind, with nothing downwind, and it is the Eulerian transport that carries the adjustment: 0.393 m²/s upwind and 1.36 m²/s to the right, a current pointing 106° to the right of the wind. A current meter moored in that sea would report an Eulerian transport that does not match Ekman’s classical answer, and a drifter would report one that does. Two averages of one flow warned that quoting one average where the other is meant is an error of order one rather than a rounding; this is a case where the two differ by a quarter of the answer and by a direction.

Towards the equator the circle grows until rotation stops mattering #

Everything above scales with the Coriolis parameter, and that falls to zero at the equator.

At the pole the circle’s radius is 339 m and the inertial period 12.0 hours. At 45° they are 479 m and 16.9 hours; at 30°, 677 m and 23.9 hours; at 10°, 1.95 km and 68.9 hours; at 5°, 3.89 km and 137 hours — nearly six days. Both grow as one over the sine of the latitude. Near the equator the inertial period becomes longer than a swell field lasts or than any friction allows the oscillation to persist, and then the drift behaves almost as it would on a non-rotating planet: the float simply goes downwave. Rotation cancels the Stokes transport only on timescales longer than the inertial period, and within a few degrees of the equator that timescale runs out.

The same comparison is the one The bath does not know the hemisphere made for a draining bath: the Coriolis term is always present, and whether it decides anything is a ratio of its time scale to the others. Here the other time scale is the duration of the swell and the friction time, not the drain; at mid-latitudes both are comparable with a day, which is why ocean drift under swell is neither the straight line of a non-rotating calculation nor the clean circle of a frictionless one.

Every closed form checked by a route that shares none of its algebra #

The results above lean on closed-form solutions, and each is checked independently.

The steady layer was solved a second time by finite differences on a grid reaching 14 times the larger of the Ekman and drift depths, with the surface stress condition applied through a ghost node and a complex tridiagonal sweep. The largest difference from the closed form, as a fraction of the surface drift, is 8.8 × 10⁻⁴ at 250 points, 2.2 × 10⁻⁴ at 500, 5.5 × 10⁻⁵ at 1,000, 1.4 × 10⁻⁵ at 2,000 and 3.5 × 10⁻⁶ at 4,000: slope exactly −2.00, the second-order rate the scheme should have. The transport by trapezoidal quadrature converges the same way, to 2.5 × 10⁻⁶ of the Stokes transport. The same check passes with the 0.144 N/m² wind, where the surface condition is no longer stress-free and a wrong sign in it would show as a solve that converges to the wrong current.

The frictionless spin-up marched by fourth-order Runge–Kutta agrees with $W_S e^{-ift}$ to 9 × 10⁻¹⁶ of the drift after 0.13 inertial periods, 7 × 10⁻¹⁵ after half a period and 9 × 10⁻¹² after 2.7 periods; the float returns to its starting point after one inertial period to 10⁻⁹ of the radius, and its greatest distance is the circle’s diameter to the same precision. The damped march settles on $W_S r/(r + if)$ to 10⁻⁶ at three drag strengths. The surface drift itself is the collection’s own closed form $(ak)^2c$ from the first essay’s calculation, reproduced to twelve figures. And the calculation refuses the equator, a wave too steep for second-order theory, a layer with no viscosity and a negative period.

What the uniform swell cannot show #

Horizontal structure. A uniform wave field has no horizontal gradients, so no pressure gradient appears and no geostrophic current can be set up. A real swell field has edges, and the Coriolis–Stokes force varies across them; the resulting divergences drive upwelling, pressure gradients and currents that are not in this calculation at all. How well such a current would be balanced is one number: its Rossby number, which The balance that is its own error showed is also the fractional error of the geostrophic answer. For a drift of 0.049 m/s varying over a swell edge 200 km wide at 45° it is 0.049/(1.03 × 10⁻⁴ × 2 × 10⁵) = 0.0024, so the response to the edges would be geostrophic to a quarter of a per cent — and computing it needs the horizontal dimension this calculation removed.

Time. The spin-up takes an inertial period and the steady spiral assumes the swell has lasted many. A swell that arrives, stays for a day and leaves drives a transient that ends with the water somewhere else, and where depends on the arrival and departure in a way the steady answer cannot say.

Turbulence. A constant eddy viscosity and a linear drag are placeholders. Beneath them sits a laminar wave boundary layer at the surface itself, a few millimetres thick, whose own nonlinear term drives a steady streaming of the kind An oscillation with somewhere to go computed at a wall — another second-order current of the drift’s own size, added at the top of the column. The real upper ocean’s mixing varies with depth and is itself driven by the waves, and the surface value of the Lagrangian current — the one quantity a float cares about — was shown above to depend on exactly that placeholder, from 11 per cent of the drift to 94.

Stratification, a real spectrum, breaking. A mixed layer of finite depth caps the spiral; a sea has many wave frequencies with drifts of different depths; and breaking waves hand their momentum to the current directly, which is a separate term in the budget. None is here.

Still open: the vortex force that makes rolls #

The drift does one more thing in a rotating or sheared ocean that this calculation cannot see because it is uniform in the horizontal. Where the current has vorticity, the Stokes drift tilts it — the product of the drift and the vorticity acts on the current as a force of its own, first written down by Craik and Leibovich — and in a wind-driven surface layer that force is unstable. It organises the upper ocean into counter-rotating rolls aligned with the wind, the Langmuir circulation that lines floating debris up in windrows. Computing that instability from the same drift profile, and finding the roll spacing and growth rate it predicts against the drift-to-shear ratio, is the next calculation.

Beside it is the stratified version of this one. With a mixed layer of finite depth above a stable thermocline, the Eulerian return flow has a floor, the spiral is truncated, and some of the cancelling transport must be carried somewhere else. Whether the Lagrangian transport still vanishes, and what happens to the part that cannot fit, is the question a mixed-layer calculation with the same Coriolis–Stokes force would answer — and it is the case the real ocean is actually in.