{c}
{title2=$w_E$}
{wiki}

Spatial convergence or divergence of <Ekman transport> drives vertical motion at the base of the Ekman layer. With upward vertical coordinate,
$$
w_E=\widehat{\mathbf z}\mathbin\cdot
\nabla\times\left(\frac{\boldsymbol\tau}{\rho f}\right).
$$

Ekman pumping

Assume a steady, small-<Rossby number> surface boundary layer, neglect horizontal viscosity and nonlinear acceleration, take the pressure gradient to be independent of depth, impose no normal flow at the surface, and let the viscous stress vanish at $z=-h$. Subtract the depth-independent <geostrophic balance> from horizontal momentum. For the ageostrophic velocity,
$$
-fv_a=\nu u_{zz},
\qquad
fu_a=\nu v_{zz}.
$$
Integrating from $-h$ to $0$ and using
$$
\boldsymbol\tau=\rho\nu(u_z,v_z)|_{z=0}
$$
gives the <Ekman transport>
$$
\boxed{
\mathbf M_E=
\int_{-h}^0(u_a,v_a)\,dz
=\left(\frac{\tau_y}{\rho f},
-\frac{\tau_x}{\rho f}\right)
=\frac{\boldsymbol\tau\times\widehat{\mathbf z}}{\rho f}}.
$$

Depth-integrated <mass conservation>, with $w(0)=0$, gives
$$
\nabla_h\mathbin\cdot\mathbf M_E-w(-h)=0.
$$
Hence the vertical velocity entering the ocean interior is the <Ekman pumping> velocity
$$
\boxed{
w_E\equiv w(-h)
=\widehat{\mathbf z}\mathbin\cdot
\nabla_h\times
\left(\frac{\boldsymbol\tau}{\rho f}\right)}.
$$
For constant $f$ this reduces to
$$
w_E=\frac{\tau_{y,x}-\tau_{x,y}}{\rho f}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 333 / 2 / a / Solution

{c}
{title2=$\mathbf M_E$}
{wiki}

Ekman transport is the depth-integrated ageostrophic flow in a frictional rotating boundary layer. Beneath a surface stress $\boldsymbol\tau$,
$$
\mathbf M_E=\frac{\boldsymbol\tau\times\widehat{\mathbf z}}{\rho f}
$$
with the stated orientation convention.


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