Ekman Number

Fridtjof Nansen deliberately froze his ship into the Arctic ice in 1893 to drift across the pole, and noticed that the ice did not travel downwind. It went consistently 20 to 40 degrees to the right of the wind. He mentioned it to Vilhelm Bjerknes, who handed the problem to a graduate student named Vagn Walfrid Ekman, and in 1905 Ekman published the answer.

The wind puts a stress on the surface. That stress is transmitted downward by friction, and each layer of water, once moving, is deflected by the Coriolis force. The result is a spiral: the surface layer moves at an angle to the wind, the layer below at a greater angle and more slowly, and so on down, until the motion dies away. Integrate the whole spiral and the net transport is at right angles to the wind — 90 degrees, exactly, in the idealized case. That is Ekman transport, and it is why an alongshore wind causes coastal upwelling, why the great ocean gyres pile water into their centres, and why the productive fisheries of the world sit where they do.

The dimensionless group that governs the whole picture is viscosity against the Coriolis force:

\[\mathrm{Ek} = \frac{\nu}{fL^2}\]

and it is really a depth in disguise. The Ekman layer thickness is $\delta = \sqrt{2\nu/f}$, and substituting gives $\mathrm{Ek} = \tfrac{1}{2}(\delta/L)^2$. A small Ekman number says the frictional layer is thin compared with the flow you care about.

In the atmosphere and ocean it is very small indeed, and that is the single most important structural fact about geophysical fluid dynamics. Friction is confined to thin boundary layers at the top and bottom, and the interior between them behaves as if it were inviscid — which is what licenses geostrophic balance, and what licenses the Taylor–Proudman theorem's remarkable claim that a slowly moving rotating fluid tries to be two-dimensional, with columns of fluid moving together as if rigid.

But the boundary layers are where the interesting behaviour lives. Inside the atmospheric Ekman layer the wind turns with height, and at the surface it crosses the isobars towards low pressure instead of running along them. That inward component is what actually fills a low; without friction a low-pressure system would circulate forever. The same convergence forces air upward out of the top of the boundary layer — Ekman pumping — which is the mechanism by which surface friction spins down a vortex, and by which wind stress curl drives vertical motion in the ocean.

Now the honesty, and on this page it is a large one: $\nu$ is almost never the molecular viscosity.

Air's molecular kinematic viscosity is $1.5\times10^{-5}$ m²/s. Put that into $\delta = \sqrt{2\nu/f}$ at mid-latitude and you get an Ekman layer about half a metre deep. The real atmospheric boundary layer is 500 to 1500 m. The discrepancy is three orders of magnitude, and it is not a small correction — it means the classical solution, taken literally, is wrong about the thing it is most famous for predicting.

What closes the gap is turbulence. Momentum in a real boundary layer is carried by eddies, not by molecules, and representing that transport as though it were viscous requires an eddy viscosity of order 1 to 100 m²/s — six or seven orders above the molecular value. That eddy viscosity is a fitted parameter, not a property of air. It varies with wind speed, with surface roughness, and above all with stability; it is not constant with height even though Ekman's solution assumes it is; and on a clear calm night, when the surface layer stratifies, it collapses towards nothing and the layer becomes tens of metres deep instead of a kilometre. Modern boundary-layer schemes abandon constant eddy viscosity entirely.

So an atmospheric or oceanic Ekman number is an order-of-magnitude statement about regime, and never a precise figure. The saving grace is the square root: a hundredfold uncertainty in $\nu$ is only a tenfold uncertainty in $\delta$. That insensitivity is the reason Ekman's solution has survived a century of knowing that its central parameter is a fiction.

One notational warning. Some texts write $\mathrm{Ek} = \nu/(2\Omega L^2)$, using the full planetary rotation rate rather than $f = 2\Omega\sin\varphi$. The two differ by a factor of $\sin\varphi$, which at 30° latitude is a factor of two. This page uses $f$.