The layer that stops at a depth

A boundary layer on a plate grows without limit. Its thickness is $\sqrt{\nu x/U}$, which contains the distance travelled, and there is nothing in the problem to stop it: diffusion has all the time in the world and keeps reaching further out.

Two things on this site stop it. One is suction through the wall, which balances diffusion against a velocity somebody has to supply. The other needs no apparatus at all — the fluid merely has to be rotating, which on a planet it always is.

Where the depth comes from #

The Coriolis term contributes something no other term in the equations of motion has: a frequency, $f = 2\Omega\sin\phi$, with no length and no velocity in it. Set that against viscous diffusion, whose only parameter is $\nu$, and dimensional analysis has one length to offer:

$$\delta = \sqrt{\frac{2\nu}{f}}.$$

There is no distance travelled in that, no free-stream speed, and no time. The layer does not grow; it exists. At forty-five degrees of latitude $f$ is $1.03\times10^{-4}$ per second, so an ocean with an eddy viscosity of $10^{-2}\,\mathrm{m^2/s}$ has an Ekman depth of fourteen metres, and an atmosphere with an eddy viscosity of ten has one of four hundred and forty. Both are the numbers those layers are observed to have.

The exact solution, and the two angles in it #

The steady linear equations for a fluid driven by a stress at its surface are, in complex form,

$$i f W = \nu W'' ,\qquad W = u + iv,$$

which is solved by $W \propto e^{-(1+i)z/\delta}$ — an exponential decay multiplied by a rotation, in one expression. That is the spiral.

The sign of the exponent is not a convention. The other root, $e^{-(1-i)z/\delta}$, solves the equation with $f$ reversed, which is the southern hemisphere; both give a surface current at forty-five degrees and a transport at ninety, so the picture looks right either way and only the residual test tells them apart. This site’s version is put back into the equations it claims to solve by central differences on the tabulated values, which is the discipline every exact solution here is held to.

Forty-five degrees at the surface is the first exact angle. It comes from the $(1+i)$: the two parts of that complex number are equal, so the velocity at $z = 0$ makes equal components along and across the driving stress. Nothing in the fluid, the forcing or the geometry can change it, because none of them appear.

The transport, which has no viscosity in it #

Integrate the whole spiral over its own depth and the second exact angle appears, along with the result that made Ekman’s name.

$$\mathbf{M} = \frac{\boldsymbol{\tau}}{\rho f}, \qquad \text{at ninety degrees to } \boldsymbol{\tau}.$$

The viscosity has cancelled. It decides how deep the layer is ($\delta \propto \sqrt{\nu}$) and how fast the water at the top moves ($\propto 1/\sqrt{\nu}$), and the product of a depth and a speed is a transport, so the two dependencies annihilate.

That is not a curiosity. The eddy viscosity of the upper ocean is not known to within an order of magnitude — it is a stand-in for turbulence that the model does not contain, and different estimates of it differ by a factor of thirty. Every quantity in this problem that depends on it is therefore uncertain by a factor of five or six.

And the quantity oceanography actually needs is the one that does not depend on it. Ekman transport is what drives coastal upwelling: a wind blowing along a coast drives water at right angles to itself, offshore or onshore depending on the sign, and the water that replaces it comes up from below carrying nutrients. The rate at which that happens is $\tau/\rho f$, computable from a wind measurement and a latitude, with no fluid property in it at all.

Why the boundary condition is the whole story #

This is the phase’s thesis in its cleanest form, so it is worth stating flatly.

The equation contains $\nu$. The solution contains $\nu$. Every profile drawn above contains $\nu$. The integral of the solution contains only the boundary condition — the stress at the surface — and the equation’s own coefficient has dropped out of it.

The reason is a conservation statement rather than an accident. In steady state the Coriolis force on the whole layer must balance the stress applied to it, since there is nothing else acting: integrate $f\,\hat{z}\times\mathbf{M} = \boldsymbol{\tau}/\rho$ and the transport follows without solving anything. The spiral is how the fluid arranges itself to do that; the transport is that it must.

There is one more place the cancellation shows itself, and it is a good test of whether the argument has been understood. Doubling the eddy viscosity doubles the depth’s square, so the layer becomes $\sqrt{2}$ times deeper; the surface current becomes $\sqrt{2}$ times slower, because the same stress is now shared over more fluid; and the transport, being the product, is unchanged. A measurement of the surface current alone therefore says nothing about the transport unless the depth is known, and a measurement of the transport says nothing about the viscosity at all.

The transport before it settles, which overshoots by a factor of two #

Everything above is steady, and the steadiness is doing more work than it looks. The transport was obtained twice — once by integrating the spiral, and once by balancing the Coriolis force on the whole layer against the stress applied to it — and the second derivation used no viscosity and no profile. It is worth running that second argument without dropping the time derivative, because the answer is not a mild transient.

Integrate the momentum equation over the depth of the layer and nothing survives but three terms:

$$\frac{d\mathbf{M}}{dt} + f\,\hat{z}\times\mathbf{M} = \frac{\boldsymbol{\tau}}{\rho}.$$

There is no viscosity in it, no depth, and no profile — the same cancellation as before, now with a clock. The particular solution is Ekman’s $\mathbf{M}_E = \boldsymbol{\tau}/\rho f$ at right angles to the wind. The homogeneous solution is a vector of constant length rotating at the rate $f$, which is an inertial oscillation, and nothing in this equation damps it.

So start a wind blowing over an ocean at rest. The transport is the steady answer plus whatever rotating vector is needed to make the total vanish at $t = 0$, which is $-\mathbf{M}_E$ rotated:

$$\mathbf{M}(t) = \mathbf{M}_E\left[(1 - \cos ft)\,\hat{e}_1 + \sin ft\;\hat{e}_2\right].$$

The tip of that vector traces a circle of radius $|\mathbf{M}_E|$ centred on $\mathbf{M}_E$ and passing through the origin. Half an inertial period after the wind starts, the transport is $2\mathbf{M}_E$ — exactly twice the steady answer — and half a period later it is zero again. It is never the Ekman transport except at two instants per cycle, and its average over a cycle is the Ekman transport exactly.

The period is $2\pi/f$, which at forty-five degrees is 16.9 hours, so the overshoot happens about eight and a half hours after a front passes. That is not a laboratory number: it is why current meters in the upper ocean record a near-circular oscillation dominating everything else for several days after a storm, rotating clockwise in the northern hemisphere, and why the wind-driven signal has to be extracted by averaging over the inertial period rather than read off directly.

Two things are worth carrying out of this. The first is that the quantity with no fluid property in it is also the quantity that arrives last. The spiral needs viscosity and takes an Ekman spin-up time to form; the transport needs neither and is exact from the first instant, but it is exact as an oscillation about the steady value rather than as the steady value. A theory whose one robust result is an integral will describe a real ocean well only after the oscillation about that integral has been averaged out — and the averaging is over the planet’s rotation rather than over anything the fluid does.

The second is that the friction left out of this balance is what eventually kills the oscillation. The Ekman layer’s own dissipation drains it over a spin-down time $H/\sqrt{\nu f}$ — the same time as the teacup’s, arrived at from the same layer — so the two results in this essay are not independent: the mechanism that lets the transport be computed without viscosity is the mechanism that requires viscosity to make the computation stop wobbling.

There is a third consequence, and it is the one that makes the inertial period a design number for anybody predicting an upwelling. The balance above is a linear oscillator driven by the wind stress, undamped except by the layer’s own friction, and its natural frequency is $f$. A wind that varies on a timescale near the inertial period is therefore driving it at resonance, and the transport it produces is not the quasi-steady $\tau/\rho f$ evaluated instant by instant — it is larger, by whatever the damping allows. Weather systems at mid-latitudes pass in one to three days against an inertial period of seventeen hours, which is close enough that the ocean’s response to a moving storm is dominated by what side of the track it is on: a wind that rotates with the current’s own sense pumps it and one that rotates against it does not.

The bottom layer, and the spin-down it causes #

The same solution run at a floor rather than a surface gives the other Ekman layer: a geostrophic current above, brought to rest at the bottom, with the spiral turning the other way.

Its consequence is out of proportion to its size. The layer is thin, but it carries a transport across the geostrophic flow, and where that flow is a vortex the transport is radial — inward at the bottom of a cyclone, and it has to go somewhere. What it does is rise, filling the interior with fluid from the boundary layer, and the whole vortex spins down in a time

$$t \sim \frac{H}{\sqrt{\nu f}}$$

rather than the diffusive $H^2/\nu$. Ekman pumping is faster than diffusion by a factor of the square root of the Ekman number, which for a stirred teacup is a factor of several hundred: the tea stops in tens of seconds rather than in hours, which is diffusion beaten, and the leaves gather in the middle because the bottom layer swept them there.

The number that says whether the layer is thin #

Rotation supplies a frequency and the geometry supplies a depth, so there is a dimensionless group, and it decides everything about whether any of this is a boundary layer at all:

$$E = \frac{\nu}{f H^2},$$

the Ekman number, which is the square of the ratio of the layer’s own depth to the depth of the fluid, divided by two. A small Ekman number means the layer is thin compared with the ocean, the interior is geostrophic and knows nothing about the bottom, and the whole two-region picture holds. A large one means the layer fills the container and there is no interior left to speak of.

For the ocean, with $H$ a few kilometres, $E$ is around $10^{-5}$ or smaller and the picture is excellent. For a teacup, with $H$ a few centimetres and molecular viscosity, it is around $10^{-3}$ — still small, which is why the tea leaves work. For a bathtub it is not the Ekman number that fails but the Rossby number, and that failure is a factor of nearly four hundred.

What the picture cannot show #

Four limits, and the first is the size of the whole edifice.

The eddy viscosity is not a fluid property. It is a stand-in for turbulent momentum transport, constant here because the exact solution requires it, and the real upper ocean’s is neither constant nor isotropic. What survives that substitution is the transport, which is the reason this solution is still used; what does not survive is the spiral’s shape. Measured spirals are flatter than this one, the surface angle is nearer twenty or thirty degrees than forty-five in most observations, and the discrepancy has been argued about for a century.

The layer is linear. The nonlinear terms have been dropped, which requires a small Rossby number, and the surface layer of an ocean under a strong wind is not always in that regime, and a stratified one mixes differently again.

The surface is flat and the stress is steady. A real wind-driven ocean has waves in it, and the momentum flux from wind to water goes partly through wave growth rather than directly into the mean flow — the Stokes drift and the wave-driven circulation are not in this solution at all.

And the two exact angles are properties of the model, not measurements. Forty-five degrees is what a constant-viscosity layer does. It is quoted here as an exact consequence of an assumption, which is a different kind of statement from a fact about the sea.

The pumping that follows, and why it matters more than the layer #

The layer is thin and its consequence is not, and the reason is that its transport can diverge.

Ekman transport is $\tau/\rho f$ at right angles to the wind. Where the wind varies from place to place — as it does across any ocean basin — that transport varies too, and where it converges, water has to go somewhere. It goes down. Where it diverges, water comes up.

The vertical velocity that results is tiny, of order metres per year, and it drives the ocean’s interior circulation. The whole of the classical theory of wind-driven gyres is that statement plus the conservation of potential vorticity: the interior of the ocean moves because the surface layer’s divergence squeezes it, not because the wind reaches down.

The coastal case is more visible. A wind blowing along a coast — a flow with one number deciding its regime — with the land on its left in the northern hemisphere drives water offshore; water rises to replace it, bringing nutrients from below; and the world’s most productive fisheries — Peru, California, Namibia, north-west Africa — are the four coastlines where the prevailing wind does that reliably. The number that predicts the strength of an upwelling is the one with no fluid property in it, computable from a wind measurement and a latitude, which is why it was usable long before anybody could measure an eddy viscosity.

Who found it, and why #

Fridtjof Nansen noticed, during the Fram expedition of 1893–96, that sea ice does not drift downwind: it drifts twenty to forty degrees to the right of the wind, consistently, in the Arctic. He put the question to Vilhelm Bjerknes, who gave it to a student, and Vagn Walfrid Ekman’s answer — published in 1905, in the first volume of a journal Bjerknes founded — is the solution above.

The surprising part of the history is what Ekman got right by being wrong. His constant eddy viscosity is a poor model of the ocean’s turbulence and gives a spiral that is not quite the observed one. But the transport does not depend on it, so the one result that mattered was insensitive to the one assumption that was weakest, and the theory survived a century of measurements that disagreed with its profile while agreeing with its integral.

The connection worth keeping is with the rest of this collection. An exact solution is usually valued for the field it gives. This one is valued for an integral of that field, and the integral is fixed by the boundary condition alone — the same shape of argument as the vorticity flux at a wall, whose rate is set by the pressure gradient and contains no viscosity either. When a quantity turns out not to depend on the coefficient in the equation, it is usually because a conservation law has been hiding in the boundary condition.

Where the ladder goes next #

The rung above is the pumping rather than the layer: the vertical velocity that a horizontal divergence of Ekman transport forces on the interior, which is what drives the ocean’s gyres and which turns a wind field into a circulation without any of the fluid in between being stirred.

The one beside it is the same balance in a duct rather than a half-space, where the layer stops for a completely different reason — because it meets the one growing from the opposite wall. That is how far a duct takes to forget its inlet, and the length comes out proportional to the Reynolds number, which is the opposite of what a first reading suggests.