# 7. Basis of wind-driven circulation: Ekman spiral and transports

In Section 6, it was mentioned that the large-scale currents at the ocean surface are all driven by the wind. This seems logical enough at first sight, but the Arctic explorer Fridtjof Nansen noticed something strange: icebergs tend to drift at an angle to the right of the prevailing wind direction. To explain this remarkable observation, Ekman (1905) formulated a theory that is still a cornerstone of physical oceanography. The central assumption is that near the ocean surface, the largest deviations from geostrophic balance occur as a result of the wind stress which leads to momentum diffusion in the vertical direction. This means that to a good approximation, the horizontal momentum balance equations $$(1.2a)$$ and $$(1.2b)$$ in Section 1 become:

$\dfrac{\left(\frac{dp}{dx}\right)}{\rho}=f \times v +K_v\dfrac{d^2u}{dz^2} \tag{7.1a}$

$\dfrac{\left(\frac{dp}{dy}\right)}{\rho}=-f \times u +K_v\dfrac{d^2v}{dz^2} \tag{7.1b}$

We now split the velocity up in a geostrophic part ($$u_g,v_g$$) and an ageostrophic Ekman velocity ($$u_E,v_E$$):

$\dfrac{\left(\frac{dp}{dx}\right)}{\rho}=f \times (v_g+v_E) +K_v\dfrac{d^2(u_g+u_E)}{dz^2} \tag{7.2a}$

$\dfrac{\left(\frac{dp}{dy}\right)}{\rho}=-f \times (u_g+u_E) +K_v\dfrac{d^2(v_g+v_E)}{dz^2} \tag{7.2b}$

From equations $$(5.1a)$$ and $$(5.1b)$$ in Section 5, we can see that the geostrophic velocities cancel against the pressure gradient terms on the lefthand side; the terms $$K_v\dfrac{d^2u_g}{dz^2}$$ and $$K_v\dfrac{d^2v_g}{dz^2}$$ can be neglected. Therefore, the equations simplify to:

$f \times v_E =-K_v\dfrac{d^2 u_E}{dz^2} \tag{7.3a}$

$f \times u_E =K_v\dfrac{d^2 v_E}{dz^2} \tag{7.3b}$

which can be reformulated through substitution into one fourth-order ordinary differential equation:

$u_E =-\left(\dfrac{K_v}{f}\right)^2 \dfrac{d^4 u_E}{dz^4} \tag{7.4}$

with the (real) solution:

$u_E = A_1 \cos\left(\sqrt{\frac{f}{2K_v}}z+\phi_1\right)e^{\sqrt{\frac{f}{2K_v}}z}+A_2 \cos\left(\sqrt{\frac{f}{2K_v}}z+\phi_2\right)e^{-\sqrt{\frac{f}{2K_v}}z} \tag{7.5}$

To determine the different coefficients, we use two boundary conditions:

1) The direct impact of the wind stress disappears in the deep ocean: $$u_E \rightarrow 0$$ for $$z \rightarrow -\infty$$. Therefore, $$A_2$$ must be equal to $$0$$.

2) In Section 6, we argued that close to the ocean-atmosphere interface, the wind stress is linearly proportional to the vertical velocity gradient; this means that if the wind is blowing in the zonal (West-East) direction, $$\dfrac{du_E}{dz}=\dfrac{\tau_w}{\rho K_v}$$ (equation $$6.1$$), $$\dfrac{dv_E}{dz}=0$$ for $$z=0$$. This leads to $$\phi_1=-\dfrac{\pi}{4}$$, $$A_1=\dfrac{\tau_w}{\rho \sqrt{f K_v}}$$.

Overall, we have:

$u_E = \dfrac{\tau_w}{\rho \sqrt{f K_v}} \cos\left(\sqrt{\frac{f}{2K_v}}z-\frac{\pi}{4}\right)e^{\sqrt{\frac{f}{2K_v}}z} \tag{7.6a}$

$v_E = -\dfrac{K_v}{f}\dfrac{d^2 u_E}{dz^2}=\dfrac{\tau_w}{\rho \sqrt{f K_v}} \sin\left(\sqrt{\frac{f}{2K_v}}z-\frac{\pi}{4}\right)e^{\sqrt{\frac{f}{2K_v}}z} \tag{7.6b}$

The Ekman transports per unit area in the zonal and meridional directions are respectively:

$$M_{E,x}=\rho\int_{-\infty}^0 u_E\, dz$$, $$M_{E,y}=\rho\int_{-\infty}^0 v_E\, dz$$

These could be calculated by integrating $$(7.6a)$$ and $$(7.6b)$$, but it is much easier to use $$(7.3a)$$ and $$(7.3b)$$:

$M_{E,x}=\dfrac{K_v \rho}{f}\int_{-\infty}^0 \dfrac{d^2 v_E}{dz^2}\, dz=\dfrac{K_v \rho}{f}\left(\dfrac{dv_E}{dz}(z=0)-\dfrac{dv_E}{dz}(z\rightarrow-\infty)\right) =0 \tag{7.7a}$

$M_{E,y}=-\dfrac{K_v \rho}{f}\int_{-\infty}^0 \dfrac{d^2 u_E}{dz^2}\, dz=-\dfrac{K_v \rho}{f}\left(\dfrac{du_E}{dz}(z=0)-\dfrac{du_E}{dz}(z\rightarrow-\infty)\right) =-\dfrac{\tau_w}{f} \tag{7.7b}$

What does all this mean? At the ocean surface, $$u_E=\dfrac{\tau_w}{\rho \sqrt{2f K_v}}$$ and $$v_E=-\dfrac{\tau_w}{\rho \sqrt{2f K_v}}$$ (from equations $$7.6a$$ and $$7.6b$$), that is, the Ekman velocity is at an angle of $$45^{\circ}$$ to the right of the wind direction in the Northern Hemisphere (and to the left of the wind in the Southern Hemisphere) due to the Coriolis force. Going deeper, the Coriolis force keeps turning the direction of the flow further to the right, while the water speed decreases exponentially with depth. As illustrated in the Figure below (courtesy of NOAA), the overall flow pattern forms a so-called Ekman spiral. Furthermore, $$(7.7a)$$ and $$(7.7b)$$ imply that the net Ekman transport is at $$90^{\circ}$$ to the right of the wind direction in the Northern Hemisphere.