One hears occasionally in the popular media that one possible consequence of global warming might be the disappearance of the Gulf Stream. This makes physical oceanographers cringe. The Gulf Stream and its analogs in other ocean basins exist for fundamental physical reasons. Climate change may well bring changes in the Gulf Stream. It may not be in the same place, may not be of the same strength or have the same temperature and salinity characteristics, but as long as the continents bound the great ocean basins, the sun shines, the earth turns toward the east and the wind blows in response, there will be a Gulf Stream. There will also be a Kuroshio, as the analogous current in the north Pacific is called, as well as the other western boundary currents, so called because, like the Gulf Stream, they form on the western boundaries of ocean basins.

The dynamical description of the Gulf Stream can be found in just about any text on physical oceanography or geophysical fluid dynamics. I first learned the general outlines of ocean circulation as a young postdoc, when my mentor **Allan Robinson** walked into my office and dropped a copy of “The Gulf Stream” by Henry Stommel on my desk and said “Read this, cover to cover.” We’ve learned a great deal about western boundary currents in the fifty years since “The Gulf Stream” was published, but it’s still an excellent introduction to the subject and and a good read. “The Gulf Stream” is long out of print, but used copies can occasionally be found. I got mine at Powell’s for six bucks, with original dust jacket. Electronic versions in many formats can be found **here**.

There are lots of places to learn about the wind driven ocean circulation. My purpose here is to present the fundamental picture as an exercise in perturbation technique.

1. The **beta-plane**

We model an ocean basin in Cartesian coordinates:

\begin{eqnarray}

(1)\qquad& x &= R \cos(\phi_0 )(\lambda – \lambda_0)\\

(2)\qquad& y &= R (\phi – \phi_0)\\

(3)\qquad& z &= r-R

\end{eqnarray}

where $(\phi , \lambda , r )$ are the coordinates of a point with latitude $\phi$, longitude $\lambda$ and distance $r$ from the center of the earth, and $R$ is the radius of the earth. $\phi_0$ and $\lambda_0$ are the latitude and longitude of a reference point in the mid-latitudes. We account for the rotation of the earth by the Coriolis parameter $f=2\Omega \sin\phi$, where the angular speed of rotation of the earth is $\Omega = 2\pi /86400s$. We approximate the Coriolis parameter as a linear function of latitude $f = f_0 + \beta y$. This is the only effect we will consider of the fact that the earth is round.

2. The Reduced Gravity Model

We model the density-stratified ocean as being composed of two immiscible fluids of slightly different densities in a stable configuration, i.e., the more dense fluid lying below the less dense one. In such a configuration, there is a family of waves that is much slower than the waves on the surface of a homogeneous fluid of the same depth would be. You’ve probably seen the effect in the clear plastic boxes containing different colored fluids, usually one clear and the other blue, that are sold as toys in stationery stores, or on the internet, e.g., **here**. When you tip the box you see waves propagate slowly along the interface. We simplify the two layer model further by assuming that the deeper layer is motionless, so the thickness of the upper layer adjusts in such a way as to make the pressure gradient vanish in the lower layer. The equations of motion for the upper layer are formally identical to the shallow water equations, but with the

acceleration of gravity $g\approx 10 ms^{-2}$\ reduced by a factor of $\Delta \rho /\rho_0$, where $\rho_0$ is the density of the upper layer.

The equations for steady linear flow on the $\beta$-plane for the reduced gravity model in the rectangle $x\in [0, a], y\in [0,b]$:

\begin{eqnarray}

(4)\qquad& -fhv + g^{\prime}hh_x &= -(\tau_0/\rho_0) \cos(\pi y/b) – A(uh)\\

(5)\qquad& fhu + g^{\prime}hh_y &= -A(vh)\\

(6)\qquad& (hu)_x + (hv)_y &= 0

\end{eqnarray}

$h$ is the thickness of the upper layer, $(u,v)$ are the horizontal velocity components and $g^{\prime}=g\Delta \rho / \rho_0$\ is the reduced gravity. The basin dimensions $a$ and $b$ are typically thousands of kilometers. We assume linear drag with constant drag coefficient $A$, and wind stress with amplitude $\tau_0$ in the $x-$direction only. The model defined by (4)-(6) is intended to be a schematic picture of a mid-latitude ocean basin in the northern hemisphere. The stress pattern of winds from the east in the southern half of the domain and from the west in the northern half is intended as a schematic model of the trade winds south of a relatively calm region at about $30^o N$, the **horse latitudes**, and westerly winds to the north.

From the continuity equation (6), we can define a transport streamfunction:

\begin{eqnarray}

(7)\qquad& \psi_x &= hv; \quad \psi_y = -hu

\end{eqnarray}

The boundaries of our idealized ocean are assumed impermeable, so we choose $\psi = 0$ on the boundaries. Taking the curl of the momentum equations (5)-(6) leads to

\begin{eqnarray}

(8)\qquad& A\nabla^2 \psi + \beta \psi_x &= \frac{-\tau_0 \pi}{b\rho_0} \sin(\pi y/b)

\end{eqnarray}

Look for a separable solution of the form

\begin{eqnarray}

(9)\qquad& \psi &= F(x) \sin(\pi y/b)\\

(10)\qquad& F^{\prime \prime} + \frac{\beta} {A} F^{\prime} – \frac{\pi^2}{b^2}F &= -\frac{\tau_0 \pi}{Ab\rho_0 }

\end{eqnarray}

So $F=\tau_0 b/(\pi \rho_0 A) +G$, and $G$ obeys the homogeneous equation

\begin{eqnarray}

(11)\qquad& G^{\prime \prime} + \frac{\beta}{A}F^{\prime} – \frac{\pi^2}{b^2} G &= 0

\end{eqnarray}

so $G = G_+ \exp(\lambda_+ x) + G_- \exp(\lambda_- x)$ for constants $G_+$ and $G_-$ where $\lambda_\pm = -\beta / (2A)(1 \pm (1+4\pi^2 A^2 /\beta^2)/b^2)^{1/2})$. This could be solved as a boundary value problem but it is more enlightening to rescale the problem.

Scale (8) by $x\rightarrow x/L_x; y\rightarrow y/b$, so (8) becomes

\begin{eqnarray}

(12)\qquad& \frac{A}{\beta L_x}(\psi_{xx} + \frac{L_x^2}{b^2}\psi_{yy}) + \psi_x &= \left ( \frac{L_x}{b}\right ) \left ( \frac{-\tau_0 \pi}{\beta b \rho_0}\right ) \sin(\pi y)

\end{eqnarray}

For basin scale motions we may choose $L_x = b$. Dissipation in the ocean is very weak, so $A/(\beta b) \ll 1$ and the leftmost term in parentheses can be neglected, leaving

\begin{eqnarray}

(13)\qquad& \beta{hv} &= \frac{-\tau_0 \pi}{b\rho_0} \sin(\pi y/b)

\end{eqnarray}

where dimensions have been restored. This is a special case of the **Sverdrup relation**, $hv = \nabla \times (\tau^{(x)},\tau^{(y)})/\beta$, i.e., transport in the north-south direction is proportional to the curl of the wind stress. This is a good approximation in the interior, but (13) is a first-order equation and cannot satisfy the boundary conditions at the eastern and western boundaries, so the Sverdrup balance cannot hold uniformly. (Also, in this example, the transport in the interior of the basin is southward, and there must be a northward return flow) Near at least one of the boundaries the momentum balance must be different.

Choose $\psi=0$ at $x=a$, so the interior solution is

\begin{eqnarray}

(14)\qquad& \psi &= \frac{\tau_0 \pi a}{\beta b \rho_0} \sin(\pi y /b)(1 -x/a)

\end{eqnarray}

We must now find an approximate solution to (8) in a thin strip near $x=0$ with $\psi (0,y) = 0$ and $\psi(x,y) \rightarrow ((\tau_0 \pi a)/(\beta b \rho_0)) \sin(\pi y /b)$ as $x\rightarrow \infty$. If we choose $L_x = A/\beta$, (12) becomes

\begin{eqnarray}

(15)\qquad& (\psi_{xx} + \frac{A^2}{\beta^2 b^2}\psi_{yy}) + \psi_x &= \left ( \frac{A}{\beta b}\right ) \left ( \frac{-\tau_0 \pi}{\beta b \rho_0}\right ) \sin(\pi y)

\end{eqnarray}

$A/(\beta b) \ll 1$ so to leading order, near the boundary, $\psi_{xx}+\psi_x = 0$ and $\psi = C_0 + C_1 \exp(-x)$ for constants $C_0$ and $C_1$. So, in dimension, near the boundary,

\begin{eqnarray}

(16)\qquad& \psi &= \frac{\tau_0 \pi}{\beta b \rho_0} \sin(\pi y)(1-e^{-x\beta /A}),

\end{eqnarray}

very close to Stommel’s solution. This boundary layer at $x=0$ is the analog, within our simple model, of the Gulf Stream.

What would happen if we were to choose the interior solution as

\begin{eqnarray}

(17)\qquad& \psi &= \frac{\tau_0 \pi a}{\beta b \rho_0} \sin(\pi y /b)(-x/a)

\end{eqnarray}

and attempt to fit a boundary condition at $x=a$? It wouldn’t work. We would need to find $C_1$ and $C_2$ such that, in a neighborhood of the boundary, $\psi = C_0 + C_1 \exp(-x\beta /A)$ such that $\psi (x=a,y) = 0$ and $\psi (x,y) \rightarrow -((\tau_0 \pi a)/(\beta b \rho_0)) \sin(\pi y /b)$ as $x\rightarrow -\infty$, which is clearly impossible.

So the fact that western boundary currents occur on the west side of ocean basins is a consequence of the fact that $\beta$ is positive, and western boundary currents form in the southern hemisphere as well.

Robert Miller

College of Earth, Ocean, and Atmospheric Sciences

Oregon State University

miller@coas.oregonstate.edu