Past exam of the geophysical fluid dynamics course of the University of Oxford / 2024 / 2

Use eastward $x$, northward $y$, constant layer depth $H$, constant reference density ${\rho }_0$, and a depth-independent horizontal velocity field $(u,v)$. Assume a flat bottom and a rigid upper boundary, giving the two-dimensional incompressibility condition $u_x+v_y=0$. The traditional approximation retains only the vertical component of planetary rotation in the horizontal equations. Take the Coriolis parameter $f=f(y)$, a constant linear drag rate $r$, and no other torques. The component Cauchy momentum equations areDefine relative vorticity $\xi =v_x-u_y$. Apply ${\partial }_x$ to the second equation and subtract ${\partial }_y$ of the first. The pressure gradient force has zero curl because ${\rho }_0$ is constant and mixed partial derivatives commute. Direct expansion of the material-acceleration terms givesThe curl of the rotation terms isThe imposed zonal wind stress contributes $-{\tau }_{s,y}/({\rho }_{0}H)$, and constant linear drag contributes $-r\xi$. HenceSince $Df=v{f}_y$, defining absolute vorticity $\zeta =\xi +f$ givesWithout the incompressibility condition, the left-hand side would contain the stretching term $\zeta {\nabla }_h\cdot \mathbf{u}$. Variable layer depth, variable density or spatially varying linear drag would introduce further terms. No small-Rossby number assumption was needed to obtain this constant-depth vorticity equation.

On a beta plane, $f=f_0+\beta y$, the preceding vorticity equation isIn the broad ocean interior, assume a statistically steady flow, weak advection of relative vorticity, and negligible friction compared with planetary-vorticity advection. The constant-depth, constant-density assumptions of 2(a) exclude topographic and baroclinic vorticity generation contributions. More precisely, require $|{\xi }_t|,|\mathbf{u}\cdot \nabla \xi |,|r\xi |\ll |\beta v|$. Then Sverdrup balance isSmall Rossby number and slow forcing help justify these conditions in the interior; they do not justify them inside an intense western boundary current. For fluctuations, the mean eddy contribution to relative-vorticity advection must also be negligible.

The depth-integrated northward volume flux per unit zonal distance is $V=Hv$. For a basin $0<x<L$ and wind stress independent of $x$, the northward interior volume flux isA negative value denotes southward transport. The cancellation of $H$ is why this interior volume flux is insensitive to the chosen constant depth. In a closed, steady basin the net volume flux across a full latitude section is zero: a boundary return must compensate $T$. Thus the displayed formula cannot represent the total closed-basin transport including that return.

For a concrete circulation sketch, use a northern-hemisphere subtropical basin $0<y<B$ andThis represents westward trade winds at the southern edge and eastward westerlies at the northern edge. The negative wind stress curl drives southward interior flow. Define an ocean transport streamfunction by $Hu=-{\Psi }_y$, $Hv={\Psi }_x$ and impose no normal flow at the eastern wall through $\Psi (L,y)=0$. Integrating Sverdrup balance westward givesSince $V=0$ at $y=0,B$, this also gives no normal flow at the north and south walls. Interior water moves westward in the southern part, southward across the basin, and eastward in the northern part. A northward western boundary current completes the clockwise ocean gyre.

Pure Sverdrup balance cannot satisfy the western impermeability condition as well: ${\Psi }_I(0,y)=-LV(y)$ varies along that wall. Indeed imposing no normal flow on both east and west walls would require their constant streamfunction values to agree in this simply connected closed basin, whereas their difference here is the nonzero $T(y)$. The shaded return in the sketch represents a boundary region where the interior approximation breaks down. It is supplied by friction in 2(c), rather than inferred from Sverdrup balance alone. A southern-hemisphere subtropical ocean gyre has the reflected, anticlockwise circulation.

For an order-of-magnitude estimate, use ${\tau }_0=0.1\mathrm{N}{\mathrm{m}}^{-2}$, ${\rho }_0=10^3\mathrm{kg}{\mathrm{m}}^{-3}$, $\beta =2\times 10^{-11}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$, $L=5\times 10^6\mathrm{m}$ and a wind stress variation scale ${\ell }_{\tau }=10^6\mathrm{m}$. ThenHere $1\mathrm{Sv}=10^6{\mathrm{m}}^3{\mathrm{s}}^{-1}$ is a Sverdrup. The sinusoidal example has maximum $|{\tau }_{s,y}|=\pi {\tau }_0/B$; taking $B=\pi \times 10^6\mathrm{m}$ makes it agree with the chosen variation scale. Other reasonable basin widths and wind profiles change the coefficient, but retain a transport of order tens of Sverdrups.

Let $V_b$ be the northward speed in the western boundary current, whose zonal width $\delta$ is much less than its meridional scale. Its dominant relative vorticity is $\xi \simeq v_x\sim V_b/\delta$. Neglecting local wind stress curl and relative-vorticity advection in the narrow return region, the linear stationary vorticity equation becomesConsequentlyThis decay into the interior from the western wall explains why the Stommel boundary layer lies on the west. The sign also matters: a northward return has $v_x<0$, so its negative relative vorticity supplies a positive drag contribution $-r\xi$ to balance $\beta v$. Here $r$ has units ${\mathrm{s}}^{-1}$; no extra factor of $f_0$ belongs in $r/\beta$ with the paper's definition of linear drag.

For example, at fixed $y$ the leading correction to the interior ocean transport streamfunction isIt cancels the western value of ${\Psi }_I$ and carries ${\int }_0^{\infty }H{v}_{B}dx=-LV=-T$. Exponentially small eastern-wall corrections and the turning regions near the northern and southern walls do not affect the leading width estimate.

Using one year $=3.156\times 10^7\mathrm{s}$, the representative $\beta$ of 2(b), and $H=10^3\mathrm{m}$ givesTo return the $25\mathrm{Sv}$ interior volume flux, conservation of mass requiresFor the exponential profile this is the boundary speed associated with its effective transport width $\delta$. It is much narrower and faster than the familiar large-scale western boundary currents, whose energetic cores span tens of kilometres and have speeds of order $1\mathrm{m}{\mathrm{s}}^{-1}$. The simplified constant-depth linear drag model with a one-year decay time therefore does not give a realistic quantitative boundary structure. Lateral viscosity, eddies, stratification and inertia can all affect a real current.

The corresponding relative vorticity scale isThus relative vorticity cannot be discarded relative to planetary vorticity in this predicted return. Even a more realistic $1\mathrm{m}{\mathrm{s}}^{-1}$ across $50\mathrm{km}$ gives $|\xi |\sim 2\times 10^{-5}{\mathrm{s}}^{-1}$, an appreciable fraction of $|f_0|$.

Magnitude alone does not prove that its material advection is large: a perfectly parallel current $v=v(x)$, $u=0$ has $\mathbf{u}\cdot \nabla \xi =0$ exactly. For a gyre return varying over meridional scale $B$, however, incompressibility gives $u\sim \delta V_b/B$, and both generic contributions to $\mathbf{u}\cdot \nabla \xi$ scale as $V_b^2/(B\delta )$. Relative to $\beta V_b$, this isAbsent special cancellations, the assumed neglect of relative-vorticity advection is therefore not self-consistent for these parameters. Its natural inertial width is instead ${\delta }_I\sim \sqrt{V_I/\beta }$, where $V_I=|T|/(HL)\simeq 5\times 10^{-3}\mathrm{m}{\mathrm{s}}^{-1}$; this gives about $16\mathrm{km}$. That comparison diagnoses the missing inertia rather than supplying a complete nonlinear boundary-current solution.