Course notes · OCN 620
10 Ekman layers
10.1 Motivation
The horizontal momentum equation as discussed so far includes the following:
$$ \begin{equation} \begin{array}{ccccccc} \frac{\partial u}{\partial t} & + & \vec{u}\cdot\nabla u & - & fv & = & -\frac{1}{\rho}\frac{\partial p}{\partial x} \\[2pt] \frac{\partial v}{\partial t} & + & \vec{u}\cdot\nabla v & + & fu & = & -\frac{1}{\rho}\frac{\partial p}{\partial y} \\[4pt] \text{local accel.} & & \text{nonlinear} & & \text{Coriolis} & & \text{pressure} \\[2pt] \uparrow & & & & \uparrow & & \\[-2pt] & & \text{inertial oscillations} & & & & \\[6pt] & & & & \uparrow & & \uparrow \\[-2pt] & & & & & \text{geostrophy} & \end{array} \label{eq:momsofar} \end{equation} $$Note that the nonlinear terms are not involved in either of the basic balances indicated above. We will continue to neglect them for the most part, except insofar as they lead to Reynolds stress (turbulent stress) that can be parameterized as eddy viscosity. That will leave us with a linear system, simplifying the solution enormously. This approximation will be a good one for the steady, large-scale flow in the interior of the ocean basin, and that is the focus of this and the following two lectures.
Equations $\eqref{eq:momsofar}$ are obviously still missing something important: there is neither forcing nor dissipation. So, now we will add the stress terms through which the wind drives the ocean.
10.2 Horizontal stress
There are two basic types of force on a fluid:
body forces such as gravity, that act on the bulk of a fluid parcel
surface forces such as pressure, that act on the surface of a fluid parcel
Pressure forces act normal to the surface of a parcel; shear stress is tangential to the surface. The part of the shear stress that we are concerned with now is the horizontal force due to the change of horizontal velocity in the vertical. Ignoring some subtleties about the nature of stress, and using the eddy viscosity concept, we can define a horizontal stress vector parameterized as
$$ \begin{equation} \vec{\tau} = \rho A_v\frac{\partial\vec{u}_H}{\partial z} \label{eq:stresspar} \end{equation} $$where $\vec{u}_H = u\hat{\imath} + v\hat{\jmath}$ is the projection of the velocity vector on the horizontal plane. Recall that the shear stress is a momentum flux density, so that the force per unit volume is the divergence of the stress. Divide that by $\rho$ to get force per unit mass, and neglect the minuscule effect of the variation of $\rho$. Then, leaving out the nonlinear terms for compactness, we get
$$ \begin{equation} \begin{array}{ccccccc} \frac{\partial u}{\partial t} & - & fv & = & -\frac{1}{\rho}\frac{\partial p}{\partial x} & + & \frac{\partial}{\partial z}A_v\frac{\partial u}{\partial z} \\[2pt] \frac{\partial v}{\partial t} & + & fu & = & -\frac{1}{\rho}\frac{\partial p}{\partial y} & + & \frac{\partial}{\partial z}A_v\frac{\partial v}{\partial z} \\[4pt] \text{local accel.} & & \text{Coriolis} & & \text{P.G.F.} & & \text{stress div.} \\[2pt] \uparrow & & \uparrow & & & & \\[-2pt] & \text{inertial oscillations} & & & & & \\[6pt] & & \uparrow & & \uparrow & & \\[-2pt] & & & \text{geostrophy} & & & \\[6pt] & & \uparrow & & & & \uparrow \\[-2pt] & & & & \text{Ekman layer} & & \end{array} \label{eq:momstress} \end{equation} $$Now we have stress in the momentum equations, but no explicit wind stress forcing. That is because the wind stress does not force the equations directly, as a body force would, but through the boundary condition. A boundary condition at the sea surface is that the stress in the water equals the wind stress, $\vec{\tau}_w$:
$$ \begin{equation} \rho A_v\frac{\partial\vec{u}_H}{\partial z} = \vec{\tau}_w\ \text{ at } z = 0 \end{equation} $$10.3 A simple ocean model: plywood on ice
Imagine a very simple ocean model: an infinite sheet of plywood sliding without friction on an infinite sheet of ice. Consider $f$ constant.
What happens when a stress is applied to the top of the sheet of plywood? It initially accelerates downwind, but as soon as its velocity is non-zero, it experiences a Coriolis force to the right of that velocity. If the stress were to stop, the plywood would oscillate, with each point moving in a circular trajectory: an inertial oscillation. But if the stress is steady, then in addition to the oscillation resulting from the initial change in stress, there is a steady component of motion in which the Coriolis force balances the wind stress. The Coriolis force per unit surface area is $-\rho Hf\hat{k} \times \vec{u}_H$, where $\rho$ here is the density of the plywood and $H$ is its thickness. Setting this equal to minus the wind stress gives
$$ \begin{align} u &= \frac{\tau_w^y}{\rho Hf} \\ v &= \frac{-\tau_w^x}{\rho Hf} \end{align} $$Hence the steady component of the motion is at right angles to the wind stress, and the stress does no work. Neither does the Coriolis force, of course—it never does. Note how rotation has completely changed the physics of the situation: if there were no rotation, there would be no steady solution at all; the plywood would accelerate forever.
Now, suppose there are two sheets of plywood, with friction between them. If the top sheet is deflected to the right of the wind stress, then the stress that it applies to the lower sheet will also be to the right of the wind stress, and the lower sheet will therefore go even further to the right of the wind stress. Stack up more sheets, and you will find a spiral, with each lower sheet moving somewhat to the right of the sheet above it, and at a slower speed. The sum of all the layers will be just as in the case of a single layer: net transport exactly to the right of the wind stress. (Everywhere I say “right”, it applies to the northern hemisphere; change it to “left” for the southern hemisphere.)
10.4 The ocean, vertically integrated
Because the system in $\eqref{eq:momstress}$ is linear, we can solve it one piece at a time. Let’s look at the balance of terms labeled “Ekman layer”: Coriolis and stress divergence. Note that we will formulate the balance in terms of the stress itself, without using $\eqref{eq:stresspar}$ to parameterize the stress as a function of the velocity.
$$ \begin{align} -fv &= \frac{1}{\rho}\frac{\partial\tau^x}{\partial z} \\ fu &= \frac{1}{\rho}\frac{\partial\tau^y}{\partial z} \end{align} $$with the boundary conditions
$$ \begin{align} \text{at the surface, } z = 0,&\quad \begin{cases} \tau^x = \tau_w^x \\ \tau^y = \tau_w^y \end{cases} \\ \text{at some depth, } z = -H,&\quad \tau^x = \tau^y = 0 \end{align} $$In other words, the stress at the sea surface is just the wind stress, and we assume there is some depth $-H$ at which the stress is negligible. That is, we assume that this Ekman balance is confined to a boundary layer at the surface. Later we will see that this is a good assumption. Integrate (7, 8) over depth, apply the boundary conditions, and rearrange to get the volume transport per unit width:
$$ \begin{align} U \equiv \int_{-H}^0 u\,dz &= \frac{\tau_w^y}{\rho f} \\ V \equiv \int_{-H}^0 v\,dz &= \frac{-\tau_w^x}{\rho f} \label{eq:vtrans} \end{align} $$Note this rather surprising and profound result: the transport per unit width of the Ekman layer—its net effect—depends only on the wind stress and $f$, the Coriolis parameter. It does not depend on any assumption about the eddy diffusivity. Note also that the mass transport per unit width calculated here is identical to that calculated for the sheet of plywood. Again, it is at right angles to the wind stress.
10.5 The ocean, with constant eddy diffusivity: top Ekman layers
Although we don’t usually need to know the exact structure of the flow in an Ekman layer, it is fun to calculate it for a simple case. This will also demonstrate that the balance between Coriolis force and horizontal stress really does lead to a boundary layer. By contrast, the geostrophic balance tends to prevail throughout a fluid volume, not just near boundaries.
The simple case in which we can easily calculate the structure of an Ekman layer is constant diffusivity, $A_v$, which could be eddy or molecular. Then (7, 8) become
$$ \begin{align} -fv &= A_v\frac{\partial^2 u}{\partial z^2} \label{eq:cek1} \\ fu &= A_v\frac{\partial^2 v}{\partial z^2} \label{eq:cek2} \end{align} $$with boundary conditions (9, 10) written as
$$ \begin{align} \text{at the surface, } z = 0,&\quad \begin{cases} A_v\frac{\partial u}{\partial z} = \frac{1}{\rho}\tau_w^x \\ A_v\frac{\partial v}{\partial z} = \frac{1}{\rho}\tau_w^y \end{cases} \\ \text{far from the surface, } z = -\infty,&\quad \tau^x = \tau^y = 0. \end{align} $$The trick for solving this system easily is to add $\eqref{eq:cek1}$ to $i$ times $\eqref{eq:cek2}$, giving a single second-order differential equation for the complex variable $u + iv$. The solution is
$$ \begin{align} u &= \frac{z_e}{\rho A_v\sqrt{2}}\exp\left(\frac{z}{z_e}\right)(\tau^x\cos\phi - \tau^y\sin\phi) \\ v &= \frac{z_e}{\rho A_v\sqrt{2}}\exp\left(\frac{z}{z_e}\right)(\tau^x\sin\phi + \tau^y\cos\phi) \\ z_e &\equiv \left(\frac{2A_v}{f}\right)^{\frac{1}{2}} \\ \phi &\equiv \mathrm{sgn}(f)\left(\frac{z}{z_e} - \frac{\pi}{4}\right) \\ \mathrm{sgn}(f) &= \begin{cases} 1 & \text{if } f > 0 \\ -1 & \text{if } f < 0 \end{cases}. \end{align} $$This solution is the Ekman spiral.
- The velocity at the ocean surface is in a direction 45° to the right (or left in the southern hemisphere) of the wind stress.
- From there the velocity vector decays exponentially and rotates anticyclonically with increasing depth.
- The constant $z_e$ is the depth scale of the Ekman layer. It is the depth at which the velocity has decayed by $e^{-1}$ and has rotated one radian (57.3°). (Different authors use different measures of the “Ekman depth”; the e-folding scale, $z_e$, makes the most sense to me.)
- $z_e$ varies as $f^{-\frac{1}{2}}$; it goes to infinity at the equator, so the Ekman layer does not exist there.
This last point may be surprising at first. After all, if the Ekman layer is due to rotation, should it not get thinner closer to the equator, where the relevant components of the Coriolis force are weaker? No, because the effect of rotation is to deflect the velocity vector. The stronger the rotational effects, the faster the velocity vector is deflected, and the thinner the layer. If there were no rotation there would be no steady frictional boundary layer at all; frictional stress would be important throughout the depth of the ocean. (This is true for a linear, constant-$A_v$ model; nonlinearities and/or variations in $A_v$ can and do additionally limit the depth range of frictional effects.)
10.6 Bottom Ekman layers
An Ekman layer will be found at any non-vertical boundary of a rotating fluid. For example, the frictional boundary layers at the bottom of the ocean, and at the bottom of the atmosphere, are Ekman layers. A “bottom Ekman layer” is usually one in which the boundary condition is not a specified stress but a specified velocity, normally zero for all practical purposes. (The correct boundary condition for the bottom of the atmosphere over the ocean is that the air and water velocities match. The water moves so slowly compared to typical winds, however, that for most purposes it can be approximated as zero.)
Suppose the flow near a bottom boundary consists of the sum of two components: the geostrophic interior, $\vec{u}_g$, and the Ekman layer, $\vec{u}_e$. Then the no-slip boundary condition requires
$$ \begin{equation} \vec{u}_e = -\vec{u}_g \text{ at } z = 0. \end{equation} $$We have shifted the vertical coordinate so that now $z = 0$ at the bottom boundary. Then the solution for the bottom Ekman layer is
$$ \begin{align} u_e &= \exp\left(-\frac{z}{z_e}\right)\left[-u_g\cos\left(\frac{z}{z_e}\right) - v_g\,\mathrm{sgn}(f)\sin\left(\frac{z}{z_e}\right)\right] \\ v_e &= \exp\left(-\frac{z}{z_e}\right)\left[u_g\,\mathrm{sgn}(f)\sin\left(\frac{z}{z_e}\right) - v_g\cos\left(\frac{z}{z_e}\right)\right] \end{align} $$and the total velocity near the bottom is
$$ \begin{align} u &= u_g\left[1 - \exp\left(-\frac{z}{z_e}\right)\cos\left(\frac{z}{z_e}\right)\right] - v_g\,\mathrm{sgn}(f)\exp\left(-\frac{z}{z_e}\right)\sin\left(\frac{z}{z_e}\right) \\ v &= u_g\,\mathrm{sgn}(f)\exp\left(-\frac{z}{z_e}\right)\sin\left(\frac{z}{z_e}\right) + v_g\left[1 - \exp\left(-\frac{z}{z_e}\right)\cos\left(\frac{z}{z_e}\right)\right]. \end{align} $$Characteristics of bottom Ekman layers:
- The net transport in a bottom Ekman layer is directed 135° cyclonically from the geostrophic component of the velocity. Hence, it has a component down the pressure gradient and an equal component opposite to the geostrophic velocity.
- The component of Ekman flow down the pressure gradient is the reason why surface winds do not blow exactly along the isobars on a weather map. Instead there is a component inward toward the center of a low, and outward away from a high.
- Because the boundary condition on a bottom Ekman layer is a velocity condition, not a stress condition, the transport is proportional to $z_e$ and to $\vec{u}_g$; it is not independent of $A_v$.
10.7 Summary
- Because of the rotation of the earth, the direct effect of wind stress is confined to a boundary layer, called the top Ekman layer, of order 10–100 m thick.
- The transport per unit width of the top Ekman layer depends only on the wind stress and on $f$; not on the details of $A_v$ or on the precise structure of the layer.
- The transport per unit width of the top Ekman layer is exactly to the right of the wind stress in the northern hemisphere, to the left in the southern.
- Bottom Ekman layers have a component of transport down the pressure gradient, that is, from high pressure to low.
Appendix: details in the complex plane
It’s easier to see how the Ekman layer works by taking full advantage of the complex plane. Represent the velocity vector as a point in the complex plane, $u_c = u + iv$, and similarly for the stress, $\tau_c = \tau_x + i\tau_y$. Then the momentum balance becomes
$$ \begin{equation} ifu_c = \frac{1}{\rho}\frac{\partial\tau_c}{\partial z} \end{equation} $$and its integral yields
$$ \begin{align} \int_{-H}^0 u_c\,dz &= \frac{-i\tau_{cw}}{\rho f} \\ &= \frac{\tau_{cw}\exp\left(-i\frac{\pi}{2}\right)}{\rho f} \end{align} $$where $\tau_{cw}$ is the wind stress in complex form. This is very similar to 12, again showing that the transport is 90° to the right of the wind stress in the northern hemisphere.
The big advantage of the complex plane becomes clear when calculating the Ekman spiral structure itself—not just its vertical integral—for the special case of a uniform coefficient for the vertical diffusivity of horizontal momentum, $A_v$, which could represent either molecular viscosity in the case of the laboratory tank, or turbulent viscosity in the ocean. (A constant value of $A_v$ is generally a poor approximation in the ocean, but it makes the problem analytically tractable and illustrates the basic concept of the Ekman spiral.)
In this case the Ekman momentum balance becomes
$$ \begin{equation} ifu_c = A_v\frac{\partial^2 u_c}{\partial z^2}. \end{equation} $$Rearranging, and noting that the only spatial dimension directly involved is the vertical, we have a linear second-order ODE with constant (but complex) coefficients:
$$ \begin{equation} \frac{d^2 u_c}{dz^2} - \frac{is|f|}{A_v}u_c = 0 \end{equation} $$where $s = 1$ in the northern hemisphere, and $-1$ in the southern. Substitute a solution of the form $u_c = u_0 e^{\alpha z}$ to find
$$ \begin{equation} \alpha^2 - \frac{is|f|}{A_v} = 0 \end{equation} $$with the two solutions
$$ \begin{equation} \alpha = \pm\left(\frac{|f|}{A_v}\right)^{\frac{1}{2}}(is)^{\frac{1}{2}}. \end{equation} $$The last term on the right is just the square root of $i$, or of $-i$, depending on the hemisphere. If $i$ is a rotation 90°, or $\frac{\pi}{2}$ CCW from the real axis, then $\sqrt{i}$ is just half the rotation:
$$ \begin{align} \sqrt{i} &= e^{i\frac{\pi}{4}} = \frac{1 + i}{\sqrt{2}} \\ \sqrt{-i} &= e^{-i\frac{\pi}{4}} = \frac{1 - i}{\sqrt{2}}. \end{align} $$Using $z_e = \sqrt{\frac{2A_v}{|f|}}$, we get an alternative form for $\alpha$:
$$ \begin{equation} \alpha = \pm\left(\frac{1 + is}{z_e}\right). \end{equation} $$We can then write the general solution for the Ekman balance:
$$ \begin{equation} u_c = u_p e^{z/z_e}e^{isz/z_e} + u_m e^{-z/z_e}e^{-isz/z_e}, \end{equation} $$where $u_p$ and $u_m$ are complex amplitudes to be determined by the boundary conditions.
We see the product of an exponential growth or decay times a rotation, both with the characteristic length scale $z_e$. For the ocean, unless the water is very shallow, we have two separate cases: we can have an Ekman spiral that decays downward from the surface, so $u_m = 0$ and $u_p$ is determined by the surface boundary condition (the wind stress), and we can have a spiral that decays upward from the bottom, so $u_p = 0$ and $u_m$ is determined by the bottom boundary condition (no slip, so the Ekman and geostrophic parts of the velocity field sum to zero at the bottom). In both cases, as one moves away from the boundary, the velocity rotates to the right in the northern hemisphere, and to the left in the southern.
Notes by Eric Firing · Last updated 2019/10/09