6.3 Upwelling & Downwelling

Step 1: Coastal Geometry and Wind Forcing

Consider a north-south oriented coastline on the eastern boundary of an ocean basin in the Northern Hemisphere, with $x$ pointing offshore (eastward) and $y$ along the coast (northward). An equatorward wind stress $\tau_y < 0$ blows along the coast.

$$\boldsymbol{\tau} = (0, \tau_y), \quad \tau_y < 0 \quad \text{(equatorward)}$$

Step 2: Ekman Transport Direction

From the Ekman transport relation derived earlier, the depth-integrated mass transport in the Ekman layer is perpendicular to the wind. In the NH, transport is 90 degrees to the right of the wind:

$$\mathbf{M}_E = \frac{\boldsymbol{\tau} \times \hat{k}}{\rho f} = \left(\frac{\tau_y}{\rho f},\; 0\right)$$

Since $\tau_y < 0$ and $f > 0$ in the NH, $M_{Ex} < 0$ -- the Ekman transport is directed offshore (westward, away from an eastern boundary coast). This removes surface water from the coast.

Step 3: Continuity Requires Vertical Motion

Mass conservation in a control volume adjacent to the coast requires that the offshore Ekman transport be compensated by upward vertical motion from below. The vertically integrated continuity equation gives:

$$\frac{\partial M_{Ex}}{\partial x} + \frac{\partial M_{Ey}}{\partial y} + \rho w_E = 0$$

At the coast ($x = 0$), the offshore transport must start from zero (no flow through the boundary) and reach its full value $M_{Ex}$ over the upwelling zone width $L$.

Step 4: Upwelling Velocity

The divergence of offshore transport occurs over a coastal strip of width $L$ (approximately the Rossby radius). The resulting upwelling velocity at the base of the Ekman layer is:

$$w_E = \frac{|M_{Ex}|}{\rho L} = \frac{|\tau_y|}{\rho f L}$$

For $\tau_y = 0.1$ N/m$^2$, $f = 10^{-4}$ s$^{-1}$, $\rho = 1025$ kg/m$^3$, and $L = 20$ km: $w_E \approx 5 \times 10^{-5}$ m/s $\approx 4$ m/day.

Step 5: Upwelling Front and Geostrophic Jet

The cold upwelled water creates a cross-shore temperature (density) gradient. This pressure gradient drives an equatorward geostrophic jet in thermal wind balance:

$$f\frac{\partial v_g}{\partial z} = -\frac{g}{\rho_0}\frac{\partial\rho}{\partial x} \implies v_g \sim -\frac{g\,\Delta\rho\,D}{\rho_0\,f\,L}$$

This equatorward coastal upwelling jet is observed in all major eastern boundary current systems, with typical speeds of 0.2-0.5 m/s.

Step 1: Ekman Transport Components

The depth-integrated Ekman mass transport per unit horizontal area is given by the wind stress divided by the Coriolis parameter:

$$M_{Ex} = \frac{\tau_y}{\rho f}, \quad M_{Ey} = -\frac{\tau_x}{\rho f}$$

Step 2: Horizontal Divergence of Ekman Transport

The vertical velocity at the base of the Ekman layer equals the horizontal divergence of the Ekman transport (from mass conservation):

$$w_E = \frac{\partial M_{Ex}}{\partial x} + \frac{\partial M_{Ey}}{\partial y} = \frac{\partial}{\partial x}\left(\frac{\tau_y}{\rho f}\right) - \frac{\partial}{\partial y}\left(\frac{\tau_x}{\rho f}\right)$$

Step 3: Expand the Derivatives

Using the product rule and noting that $f$ depends only on $y$ (through latitude), so $\partial f/\partial x = 0$ and $\partial f/\partial y = \beta$:

$$w_E = \frac{1}{\rho f}\left(\frac{\partial\tau_y}{\partial x} - \frac{\partial\tau_x}{\partial y}\right) + \frac{\beta\tau_x}{\rho f^2}$$

Step 4: Identify the Two Terms

The result contains two distinct physical contributions:

$$w_E = \underbrace{\frac{1}{\rho f}\text{curl}_z(\boldsymbol{\tau})}_{\text{wind stress curl term}} + \underbrace{\frac{\beta\tau_x}{\rho f^2}}_{\beta\text{-term}}$$

The first term arises from spatial variations in the wind itself. The second ($\beta$-term) arises because the Coriolis parameter varies with latitude -- even a uniform zonal wind produces Ekman pumping because the Ekman transport varies with $f$.

Step 5: Compact Form and Sign Convention

Both terms combine into the elegant compact form:

$$w_E = \frac{1}{\rho}\,\hat{k}\cdot\nabla\times\!\left(\frac{\boldsymbol{\tau}}{f}\right) = \frac{1}{\rho}\,\text{curl}\!\left(\frac{\boldsymbol{\tau}}{f}\right)$$

Cyclonic wind stress curl ($\text{curl}_z(\boldsymbol{\tau}) > 0$ in the NH) produces $w_E > 0$ (Ekman suction = upwelling). Anticyclonic curl produces $w_E < 0$ (Ekman pumping = downwelling). Typical open-ocean values are $|w_E| \sim 10^{-6}$ m/s or about 30 m/yr.

Step 1: Advection-Diffusion Equation for Nutrients

The concentration of a nutrient $N$ (e.g., nitrate) in a one-dimensional vertical column is governed by the advection-diffusion equation with a biological sink:

$$\frac{\partial N}{\partial t} + w\frac{\partial N}{\partial z} = K_v\frac{\partial^2 N}{\partial z^2} - J(N)$$

where $w$ is the vertical (upwelling) velocity, $K_v$ is the vertical diffusivity, and $J(N)$ is the biological uptake rate.

Step 2: Advective Nutrient Flux

At the base of the euphotic zone (depth $z = -D_E$), the upward advective nutrient flux is the product of upwelling velocity and the nutrient concentration at that depth:

$$F_N^{adv} = w_E \cdot N(z = -D_E)$$

If we approximate the nutrient profile as having a constant gradient $dN/dz$ below the mixed layer, the flux entering the euphotic zone from the nutricline is:

$$F_N = w_E \cdot \frac{dN}{dz}\bigg|_{z=-D_E} \cdot \Delta z$$

Step 3: Combine with Diffusive Flux

The total vertical nutrient flux includes both the advective (upwelling) and turbulent diffusive components:

$$F_N^{total} = w_E \cdot N_{deep} + K_v\frac{dN}{dz}\bigg|_{z=-D_E}$$

In typical upwelling regions, $w_E \sim 5\times10^{-5}$ m/s and $N_{deep} \sim 20$ $\mu$M, giving $F_N^{adv} \sim 1$ $\mu$mol m$^{-2}$ s$^{-1}$. The diffusive flux ($K_v \sim 10^{-4}$ m$^2$/s, $dN/dz \sim 0.1$ $\mu$M/m) gives $F_N^{diff} \sim 10^{-5}$ $\mu$mol m$^{-2}$ s$^{-1}$. Advection dominates by a factor of ~100.

Step 4: New Production and the f-Ratio

The upwelled nitrate fuels "new production" (as opposed to "regenerated production" from recycled ammonium). Using the Redfield C:N ratio of 6.625 mol C per mol N:

$$P_{new} = R_{C:N} \cdot F_N^{adv} = 6.625 \cdot w_E \cdot [NO_3^-]_{deep}$$

Step 5: Estimate for a Typical Upwelling System

For the Peru-Chile upwelling system with $w_E \approx 5$ m/day = $5.8\times10^{-5}$ m/s and $[NO_3^-]_{deep} \approx 25$ $\mu$M:

$$F_N = 5.8\times10^{-5} \times 25\times10^{-3} = 1.45\times10^{-6} \;\text{mol m}^{-2}\text{s}^{-1} \approx 125\;\text{mmol m}^{-2}\text{d}^{-1}$$

$$P_{new} = 6.625 \times 125 \approx 830\;\text{mmol C m}^{-2}\text{d}^{-1} \approx 10\;\text{g C m}^{-2}\text{d}^{-1}$$

This is among the highest primary production rates in the global ocean, explaining why eastern boundary upwelling systems support ~20% of global fish catch from less than 1% of the ocean surface area.