Past exam of the geophysical fluid dynamics course of the University of Oxford

Solutions to all parts of the held paper, including the four inline Roman parts of 2(c). Source PDF: Oxford, A15274H1, 9 June 2025. Numbered headings retain the source numbering; question statements are omitted.

The annual, zonal-mean absorbed sunlight falls from the tropics towards the poles much more strongly than outgoing thermal radiation. Thus Earth's top-of-atmosphere radiation budget has a tropical surplus and a polar deficit, with changes of sign around $35^{\circ }$ latitude in a basic sketch. The curves below are illustrative, rather than measured data; their area-weighted global means agree.

In a statistically stationary climate, local conservation of energy requires the surplus to be exported. If $R(\varphi )$ is the net downward radiative heat flux and $\mathcal{H}(\varphi )$ the northward total power carried across the latitude circle, thenHere $a$ is Earth's radius. The terrestrial atmosphere and ocean circulation carry this energy poleward by moving warm and cold fluid, including atmospheric transport of latent heat. Without this transport the assumed steady radiation budget could not persist: the tropics would warm and polar regions cool until the budget changed. The tropical atmospheric contribution includes the Hadley cell: rising warm air near the equator, poleward flow aloft, subtropical descent, and an equatorward lower return.

For a parcel moving without an axial torque, conservation of angular momentum applies in an inertial frame. The specific axial angular momentum at latitude $\varphi$ is“Initially at rest” means $u=0$ relative to the rotating surface at the equator, so $M=\Omega a^2$. Neglecting atmospheric height compared with $a$ givesWith $\Omega =7.292\times 10^{-5}{\mathrm{s}}^{-1}$ and $a=6.371\times 10^6\mathrm{m}$, the eastward velocity at $30^{\circ }\mathrm{N}$ isThis is the ideal angular-momentum-conserving upper branch of the Hadley cell. Real jet streams exchange angular momentum through eddies and friction, so this estimate is not an observational claim about the actual wind speed.

Take $x$ eastward, $y$ northward and $z$ upward, and let $u$ be the eastward geostrophic wind. With Coriolis parameter $f$ independent of $z$, geostrophic balance, hydrostatic balance, and the ideal gas law giveHere $R_d$ is the specific gas constant for dry air. Taking a vertical derivative of the first equation and interchanging mixed partial derivatives givesIt matters that $\rho$ is not constant in the pressure gradient force. Substitutingcancels the two terms proportional to $gfu/(R_{d}T)$ and yields the height-coordinate thermal wind relationIf $|u{T}_z|\ll |T{u}_z|$, this reduces toThe discarded term has relative scale $H|T_z|/T$ when $u_z\sim u/H$. Taking $H=10\mathrm{km}$, $T=250\mathrm{K}$ and a atmospheric lapse rate of $6.5\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ gives about $0.26$. Thus this height-coordinate formula is a useful estimate, with potentially tens-of-percent compressibility-related correction through the vertical temperature gradient; it is not an exact consequence of the two balances. If the temperature gradient vanishes vertically, the displayed approximation is exact within those balances.

The hydrostatic approximation is very good on the broad, shallow atmospheric scales considered. The geostrophic balance is normally justified by a small Rossby number and fails as a general approximation near the equator. At $30^{\circ }\mathrm{N}$, $f=7.292\times 10^{-5}{\mathrm{s}}^{-1}$. With the ideal $134\mathrm{m}{\mathrm{s}}^{-1}$ wind and a $1000$--$2000\mathrm{km}$ horizontal scale, $\mathrm{Ro}=U/(fL)\simeq 0.9$--$1.8$: quantitative accuracy is limited for this especially strong ideal jet. Small vertical aspect ratio alone does not ensure geostrophic balance, although a straight, steady zonal flow can satisfy it even when this scale estimate of the Rossby number is not small.

For the requested estimate, suppose the upper jet stream is $H=10\mathrm{km}$ above a weak surface wind, has temperature $T\simeq 250\mathrm{K}$, and spans a meridional width $L$. Integrating the approximate thermal wind relation in scale form givesThe northward temperature gradient is negative because the eastward wind strengthens upward. A $1000$--$2000\mathrm{km}$ jet width gives roughly $25$--$50\mathrm{K}$. No unique numerical difference follows without specifying the jet width, height and representative temperature; these are explicitly estimated quantities.

Write the velocity field as $u=U+u\prime$, $v=v\prime$. The uniform background has zero relative vorticity, and the perturbation has $\xi \prime =v{\prime }_x-u{\prime }_y$. Keeping only terms linear in the perturbations givesThe incompressibility condition permits a streamfunction $\psi$ with $u\prime =-{\psi }_y$, $v\prime ={\psi }_x$, hence $\xi \prime ={\nabla }_h^2\psi$. Thus the Linear Rossby-wave equation here isChoose a plane wave $\psi =\mathrm{Re}\{\hat{\psi }{e}^{i(kx+ly-\omega t)}\}$, with $K^2=k^2+l^2>0$. Substitution givesand therefore the local Rossby-wave dispersion relation in a zonal jetThe $Uk$ term is the background Doppler shift; the intrinsic phase velocity is westward. A real physical wave can always be represented using the conjugate wavevector if necessary to choose $\omega \ge 0$.

The relevant ground-frame zonal phase velocity and group velocity areTime increases downward in the source diagram. The individual phase contours lean towards increasing longitude as time increases, so their phase velocity is eastward. The envelope moves eastward faster: successive crests grow at the downstream edge and decay upstream. This illustrates Rossby-wave downstream development.

For the intended purely zonal wave packet, $l=0$, andConsequently, with $U,\beta >0$ and the positive-frequency choice $k>0$,There is a small interpretation qualification: if “zonally propagating” specified only the plotted direction while allowing an unknown nonzero meridional wavenumber, eastward phase velocity alone would imply $k^2+l^2>\beta /U$, not the stronger bound on $k$ alone. In this uniform model a strictly zonal group velocity with $\beta k\ne 0$ does imply $l=0$, which supplies the intended interpretation.

For a zonal background $U(y)$, its relative vorticity is $\overline{\xi }=-U_y$. The meridional absolute vorticity gradient is thereforeAt $30^{\circ }\mathrm{N}$,A localized jet stream of velocity scale $U=134\mathrm{m}{\mathrm{s}}^{-1}$ and width $L$ hasFor $L=1000$--$2000\mathrm{km}$ this is about $1.7$--$6.8$ times the planetary contribution, so wind curvature cannot generally be neglected. The coefficient in this scaling depends on the chosen jet profile and width convention.

Linearizing the vorticity equation now gives the Rossby-wave equation for a sheared zonal currentFor a qualitative estimate at the jet core, freeze the local coefficients. At a smooth maximum $U_{yy}<0$, so $Q_y>\beta$: the restoring absolute vorticity gradient is stronger. The local Rossby-wave dispersion relation in a zonal jet gives, for the same local $U$ and zonal wavenumber $k$ with $l=0$,soThe envelope in the diagram would advance eastward faster, while individual phase contours would advance more slowly eastward, become stationary if $Q_y=U{k}^2$, or move westward if $Q_y>U{k}^2$. The separation between envelope and phase speeds increases. On a plot with time downward, compare horizontal displacement at fixed time: the envelope moves farther right, while a given phase moves less far right, or left after reversal. If the frequency changes sign for a fixed signed $k$, conjugate the mode to restore the convention $\omega \ge 0$; the physical propagation is unchanged.

This conclusion concerns the concave-down jet core. On a flank with $U_{yy}>0$ the correction has the opposite sign; the value at the edge of the Hadley cell alone does not specify the curvature everywhere. Indeed the angular-momentum-conserving profile in 1(a) is a rising inner branch, not a complete smooth jet maximum. A global sheared-current normal mode requires its meridional structure and boundary conditions; the local calculation is the qualitative estimate requested.

Define the salinity contrast $\Delta S=S_1-S_2$, positive when the warm box is saltier. The linear equation of state gives the cold-minus-warm density contrast and signed volume fluxHere $q>0$ represents warm upper water travelling poleward and cold deep water returning equatorward. The $\beta$ in this question is the haline density coefficient, unrelated to the planetary vorticity gradient in Question 1. The imposed heating and cooling are taken to maintain $\Delta T$.

For the positive circulation shown in the source diagram, the exact stationary salt budget of the cold box isThe precipitation adds no salt. Replacing the salinity multiplying the small compensating volume flux by a reference $S_0$ gives $q\Delta S\simeq E{S}_0$. When circulation reverses, the upstream box for the exchange reverses too: the homogenizing exchange is $|q|$, not $q$. In the same reference-salinity approximation the two stationary salt budgets reduce toFor example, when $q=-r<0$ and $r>E$, the exact cold-box balance is $(r-E)S_1-r{S}_2=0$, giving $r\Delta S=E{S}_1$ and the same leading approximation.

Substitute $\Delta S=E{S}_0/|q|$ into the density-driven volume flux and multiply by $|q|$:Both $A$ and $B$ have the units of volume flux, so every term has units of squared volume flux. The absolute value expresses the fact that transport in either direction reduces the salinity contrast, while the density contrast sets its direction.

The requested equation is thus the reference-salinity, or virtual salt flux, version of the two-box overturning model. If the finite compensating flow is retained exactly with unequal $S_1,S_2$, one has the appropriate upstream salinity on the right-hand side instead of exactly $S_0$. The approximation assumes $|\Delta S|/S_0\ll 1$ and $E/|q|\ll 1$. Also, for $E>0$, $q=0$ cannot be a steady solution: the imposed freshwater transport would build an uncompensated salinity contrast.

Using the constants as printed, with volume flux measured in Sverdrups, givesIn particular the two small expansion coefficients should not be silently replaced by different oceanographic values.

For a positive steady volume flux, the equation in 2(a) is $q^2-Aq+BE=0$. For a negative steady volume flux, it is $q^2-Aq-BE=0$, with only its negative root admissible. Therefore for $E>0$ the branches areThe upper positive branch $q_{+}$ is stable, the lower positive branch $q_{-}$ is unstable, and the reversed branch $q_{\mathrm{rev}}$ is stable under the usual salt budget dynamics with fixed $\Delta T$. Thus there is bistability below the positive-branch fold. At the fold the two positive equilibria merge with a zero restoring rate; the point is not an ordinary asymptotically stable interior point of either branch. The negative algebraic root of the positive-$q$ quadratic, or positive root of the negative-$q$ quadratic, must never be added as extra solutions.

At $E=0.3\mathrm{Sv}$, the three steady volume fluxes areThe present thermally dominated, northward-upper-flow Atlantic meridional overturning circulation corresponds to $q_{+}\simeq 17.7\mathrm{Sv}$. These algebraic data alone also allow a stable reversed state, so selecting the present branch uses the stated circulation direction. On the large positive branch, the predicted salinity contrast is $\Delta S=E{S}_0/q_{+}\simeq 0.594\mathrm{g}\mathrm{k}{\mathrm{g}}^{-1}$, consistent with the reference-salinity approximation. The weaker branches give larger contrasts, approximately $3.16$ and $4.20\mathrm{g}\mathrm{k}{\mathrm{g}}^{-1}$, so the approximation is less accurate there.

The saddle-node bifurcation occurs atFor $E>E_c$, only the reversed steady volume flux remains. At $E=E_c$ there are still two distinct values, the double positive root and the negative root; “single-valued” holds strictly above this threshold.

Slowly increasing $E$ from the present value causes the stable positive volume flux to weaken. Its salinity contrast then increases because salt exchange is weaker, reducing the cold-box density advantage further: this is the salt-advection feedback. At the fold the positive steady state disappears, producing an abrupt transition to the reversed branch in the quasistatic limit. At the same $E_c$ that branch hasFor a finite slow ramp there can be a small dynamical delay near the fold. In this model subsequently reducing $E$ while keeping it positive follows the stable reversed branch instead of automatically restoring the positive branch, an example of hysteresis. The two small branches approach $q=0$ as $E\to 0^{+}$. At $E=0$ this is an equal-density, no-exchange equilibrium with $\Delta S=\alpha \Delta T/\beta =3.75\mathrm{g}\mathrm{k}{\mathrm{g}}^{-1}$, whereas the positive branch tends to $q=A$ with $\Delta S=0$. One must use the original density relation at this endpoint, rather than divide by $|q|=0$.

The reversal is thermally opposed but salinity driven: freshwater makes the cold box sufficiently fresh that the warm, salty box becomes denser. A real weakening would reduce the northward ocean heat flux and alter regional climate and ocean ventilation. The fold and full reversal here describe the ideal two-box overturning model; its numerical threshold is not a forecast for the real ocean, whose other controls appear in 2(c).

Northern high-latitude cooling removes heat from surface water and raises its density. If the surface water becomes denser than the water below, convection mixes it downward and forms dense water that can be exported in the deep southward branch of the Atlantic meridional overturning circulation. This water-mass transformation helps maintain the contrast represented by $\alpha \Delta T$ in 2(a). Freshening can oppose it by reducing density and increasing stable stratification. Local convection and the basin-integrated sinking volume flux are related but are not the same process: export and the return path also depend on the large-scale dynamics.

Dense water formed in marginal seas spills over sills and descends the continental slope as a marginal-sea overflow. Strong shear flow generates turbulent mixing and entrainment of surrounding, lighter water. The descending current's volume flux grows while its density anomaly is diluted, modifying the depth at which it joins the deep circulation and the temperature and salinity of the exported water. Thus the deep volume flux can substantially exceed the source overflow volume flux; it cannot be inferred from northern surface cooling alone.^[ref]

Southern Ocean westerly wind stress drives northward surface Ekman transport in the Southern Hemisphere. Its divergence and the circumpolar geometry allow deep water to rise towards the surface along tilted isopycnals, providing an important return route for the deep branch of the Atlantic meridional overturning circulation. Surface heat fluxes and freshwater fluxes then transform the upwelled water before it enters shallower return pathways. The winds supply mechanical energy; the overturning cannot be represented entirely by a local northern density difference. Eddy-induced overturning opposes part of the wind-driven contribution, so the net water-mass transformation and transport need not increase in direct proportion to the wind stress.^[ref]

Deep currents and tides moving across rough topography, including the Mid-Atlantic Ridge, generate internal waves. Their breaking produces turbulent mixing across isopycnals, called diapycnal mixing. This transfers buoyancy between deep and overlying waters, transforms dense water into lighter water, and contributes to the upward return needed to close the overturning. The required mechanical energy ultimately comes from winds and tides; roughness is a mechanism for converting this input into mixing, rather than an independent source of energy. Mixing is spatially concentrated, and local vertical motion depends on the divergence of the turbulent buoyancy flux, so enhanced mixing does not imply uniform local upwelling everywhere over a ridge.^[ref]

With the same prescribed $\Delta T$, $S_0$, $\alpha$ and $\beta$, replacing $k$ by $\lambda k$ changes both transport coefficients:Hence the positive steady volume fluxes becomeand the negative steady volume flux isThe bifurcation diagram undergoes a particularly simple rescaling:Thus it contracts horizontally and vertically by the same factor. Its branch stability and saddle-node bifurcation structure persist, but the range of freshwater forcing that admits the stable positive circulation shrinks. At a fixed subcritical $E$, the upper stable positive branch weakens, the lower unstable positive branch moves upward, and the reversed branch decreases in magnitude. These changes reduce the separation of the positive equilibria and the margin against an overturning transition. The right panel of the figure in 2(b) shows this for $\lambda =0.7$ and $0.45$.

For the present $E=0.3\mathrm{Sv}$, a positive steady state exists only ifAt equality the positive branches merge at $q=\lambda A/2=5.6\mathrm{Sv}$; below it only the reversed steady state remains. If a discovery merely revises our estimate of $k$, it revises the inferred branch strengths and threshold, rather than itself causing an instantaneous physical transition. If the real effective $k$ were to decrease slowly through the critical value while $E$ stayed fixed, this model predicts loss of the positive steady circulation and transition to the reversed branch. Uncertainty in $k$ therefore translates directly into uncertainty in the model's freshwater threshold and its inferred proximity to a transition.

Solutions to all eight labelled parts. Source PDF: Oxford, A15274W1, 5 June 2024. The numbered headings preserve the source numbering; question statements are omitted.

Take $x$ eastward, $y$ northward and $z$ upward. In a rotating reference frame, the horizontal Cauchy momentum equation isHere ${\mathbf{F}}_h$ represents other accelerations, such as friction, and $f$ is the Coriolis parameter. When the Rossby number $U/(|f|L)$ and the ratio $1/(|f|\tau )$ are small, and friction is weak, both advective and local acceleration can be neglected. The Coriolis force then balances the pressure gradient force: this is geostrophic balance, with $fv=p_x/\rho$ and $fu=-p_y/\rho$. The geostrophic wind runs along atmospheric isobars, with low pressure on its left in the northern hemisphere.

A rapid change of pressure need not bring the corresponding geostrophic wind into existence at once. An initially weak wind therefore leaves an inward pressure gradient force around the new low. For a smoothed anomaly of horizontal scale $L=5\times 10^5\mathrm{m}$, geostrophic balance would requireFor the illustrative midlatitude value $|f|=10^{-4}{\mathrm{s}}^{-1}$, the implied Rossby number is already about $0.67$, so curved-flow acceleration matters. With a small tropical value $|f|=10^{-5}{\mathrm{s}}^{-1}$, the inferred speed is $333\mathrm{m}{\mathrm{s}}^{-1}$ and the Rossby number about $67$: the assumed geostrophic balance is untenable. Geostrophic adjustment redistributes the anomaly and emits internal gravity waves, leaving a weaker balanced component.

For a target speed $U_{\ast }=10\mathrm{m}{\mathrm{s}}^{-1}$ on the same scale, the required scale estimate isThus $6\mathrm{hPa}$, or $0.30$ of the initial magnitude, is the midlatitude estimate; at the illustrative tropical $|f|=10^{-5}{\mathrm{s}}^{-1}$ it is $0.6\mathrm{hPa}$, or $0.030$ of the initial magnitude. The paper gives no latitude, so there is no unique numerical reduction. At the equator itself one must use equatorial dynamics rather than divide by $f=0$. Even the tropical $10\mathrm{m}{\mathrm{s}}^{-1}$ estimate has Rossby number $2$, so it describes the scale permitted by the stated balance, rather than a controlled tropical geostrophic balance.

There are two qualifications to “unstable”. A literal top-hat pressure anomaly has a discontinuous edge, where no finite classical geostrophic wind can balance its distributional pressure gradient. The $500\mathrm{km}$ estimate assumes a smoothed edge; a thinner edge requires stronger winds. More generally a large low is not, by its amplitude alone, proved to have an exponentially growing instability. For a concrete counterexample to the claim that such an amplitude cannot be stationary, set $p\prime (s)=-P\exp (-s^2/a^2)$, with cylindrical radius $s$, constant $\rho$ and $f>0$. The cyclonic gradient windis real and satisfies $v_{\theta }^2/s+f{v}_{\theta }=p_s/\rho$ for every $P>0$, with a regular limit at the centre. This is a stationary axisymmetric flow; no claim about its perturbation stability is needed. Here the physically justified conclusion is that a rapidly imposed anomaly without its balancing wind is unbalanced.

Small tropical Coriolis parameter allows only weak horizontal pressure gradients to coexist with ordinary large-scale winds. Hydrostatic balance connects the vertical variation of these pressure gradients to horizontal temperature gradients; equivalently the approximate thermal wind relation gives $f{u}_z\simeq -(g/T)T_y$. For comparable vertical wind shear, small $|f|$ means smaller $|T_y|$. Internal gravity waves can spread heating-induced buoyancy anomalies efficiently before rotation arrests the adjustment. This motivates the weak temperature gradient approximation for the tropical free troposphere; it does not require the land surface or atmospheric boundary layer to have uniform temperature.

Use a plane wave with phase $kx+mz-\omega t$, and take $N>0$, $k,m\ne 0$. The positive-frequency branch of the given hydrostatic dispersion relation is $\omega =N|k|/|m|$. The wavevector is $\mathbf{K}=(k,m)$. Differentiating with respect to its components gives the group velocity, which carries the wave packet and its energy:The normal phase velocity, which follows a crest in its normal direction, isConsequently,Thus the group velocity lies along the constant-phase lines $z=-(k/m)x+\mathrm{constant}$, perpendicular to the phase velocity. For $k,m>0$, crests move upward and rightward, whereas energy moves downward and rightward. Choosing $m<0$ gives upward and rightward energy propagation and downward and rightward normal phase propagation. Conjugate or oppositely directed components supply the other outgoing directions from a localized source.

The speeds areThe hydrostatic limit is $|k/m|\ll 1$. In that limit energy propagates almost horizontally, with $|c_{g,x}|=N/|m|$ and $|c_{g,z}|=N|k|/m^2$, while normal phase propagation is almost vertical and has speed approximately $N|k|/m^2$. The horizontal speed at which a crest intersects a fixed height is $\omega /k=c_{g,x}$; this is not the horizontal component of the normal phase velocity vector. For example, $N=10^{-2}{\mathrm{s}}^{-1}$, vertical wavelength $2\pi /|m|=10\mathrm{km}$ and horizontal wavelength $2\pi /|k|=1000\mathrm{km}$ give $|c_{g,x}|\simeq 15.9\mathrm{m}{\mathrm{s}}^{-1}$ and $|c_{g,z}|\simeq 0.159\mathrm{m}{\mathrm{s}}^{-1}$.

These are local speeds in a uniform medium, without a background current. The non-rotating approximation requires $|f|\ll \omega$, while the hydrostatic approximation requires $\omega \ll N$. At lower frequencies, rotation changes the dispersion relation to that of inertia-gravity waves.

The equations describe the linear reduced gravity shallow water equations for a uniform upper layer over a deep, resting lower layer, or equivalently one separated internal vertical mode. The layer is shallow compared with its horizontal scales, permitting hydrostatic balance, and its horizontal velocity is independent of depth. The Boussinesq approximation retains the small density contrast through $g\prime =g({\rho }_2-{\rho }_1)/{\rho }_0>0$. We neglect friction, external forcing, background currents, and quadratic perturbation terms. On an equatorial beta plane, $f=\beta y$ with $\beta >0$, and the equatorial Kelvin wave branch imposes $v=0$. The meridional Cauchy momentum equation then reduces exactly within this linear model to a balance between the Coriolis force and pressure gradient force.

Strictly, the printed $h$ must denote the thickness perturbation $\eta$, with total thickness $h_0+\eta$ and $|\eta |\ll h_0$. A signed oscillation around zero cannot represent the total positive layer thickness. Define $c=\sqrt{g\prime h_0}$. For a nonzero zonal wavenumber $k$, the zonal Cauchy momentum equation and conservation of mass giveOn the eastward branch $\omega =ck$, the wave polarization is $u=(c/h_0)\eta$. Meridional geostrophic balance givesThe branch $\omega =-ck$ instead grows as $\exp (+\beta y^2/(2c))$ and is excluded by decay away from the equator. Matching the paper's Gaussian function therefore givesThe usual equatorial deformation radius is $R_{\mathrm{eq}}=\sqrt{c/\beta }$, with the envelope written $e^{-y^2/(2R_{\mathrm{eq}}^2)}$. Thus the printed exponent uses $L_{\mathrm{eq}}=\sqrt{2}{R}_{\mathrm{eq}}$. Calling that width a deformation radius is a convention; using the standard radius without the factor $1/2$ would fail the printed meridional equation.

At an instant, alternating positive and negative thickness anomalies have their largest magnitudes on the equator and decay symmetrically away from it. Eastward arrows occupy the positive anomalies, westward arrows the negative anomalies; the velocity vanishes at the nodes and weakens away from the equator. The whole pattern travels eastward at speed $c$. Meridionally, $fu=-g\prime {\eta }_y$ is geostrophic balance. For a positive anomaly north of the equator, the northward pressure gradient force balances the southward Coriolis force; both directions reverse south of the equator. Zonally, $fv=0$ and $u_t=-g\prime {\eta }_x$: the zonal pressure gradient force accelerates the flow, so it is not in zonal geostrophic balance. On the equator the meridional terms both vanish smoothly.

Hold $l,m,f_0,N$ fixed when differentiating the dispersion relation of Rossby waves, so that ${\alpha }^2\ge 0$ is independent of $k$. For $k\ne 0$, stationary zonal phase requiresThe zonal group velocity measured relative to the ground isUsing the stationarity condition,Here $\beta =2\Omega \cos \varphi /a>0$ on the ordinary midlatitude beta plane, so the stationary Rossby wave requires an eastward background current and carries its wave packet eastward. If $\alpha >0$, a real nonzero stationary $k$ exists only for $0<U<\beta /{\alpha }^2$. The restriction $k\ne 0$ matters: a zonally uniform disturbance with $k=0$ has $\omega =0$ identically, but its zonal phase velocity is undefined, so it is not a counterexample within the stated meaning of stationary waves.

Kelvin waves: weakened trade winds, or a westerly wind burst, excite a downwelling equatorial Kelvin wave. Its positive upper-layer thickness anomaly travels eastward and deepens the eastern Pacific thermocline. Upwelling then entrains warmer water, favouring eastern Pacific surface warming and the development of El Niño. This is a propagating adjustment of the thermocline, rather than transport of an unchanged parcel of warm water across the entire basin.^[ref]

Gravity waves: changed tropical convection and its latent heat release force atmospheric internal gravity waves. Their rapid propagation spreads the associated pressure and buoyancy disturbances, allowing the tropical terrestrial atmosphere to adjust to the shifted heating and contributing to the divergent circulation and surface wind response coupled to the ocean. This is one atmospheric adjustment mechanism; it does not assert that all wind variability during a particular El Niño is caused by this wave family.

Rossby waves: wind forcing during El Niño also launches westward equatorial Rossby waves with a shoaling thermocline signal. Reflection at the western Pacific boundary can produce an eastward upwelling equatorial Kelvin wave, which shoals the eastern thermocline and promotes cooling. This supplies a delayed negative feedback that can help terminate the event.^[ref] These examples describe mechanisms relevant to events such as 2023/24, rather than claiming an event-specific attribution from the exam data.

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.

Source PDF: mmathphys.web.ox.ac.uk/sitefiles/geophysical-fluid-dynamics.pdf

Choose $y$ northward and $z$ upward in the Northern Hemisphere. In a locally linear Boussinesq approximation, write the background buoyancy as $\bar{b}=b_{\ast }+By+N^{2}z$, with horizontal buoyancy gradient $B<0$ and positive squared buoyancy frequency $N^2$. The stable density stratification resists a purely vertical displacement, but the isopycnals slope upward poleward:The geopotential surfaces are horizontal in this local approximation. Their inclination differs from that of the isopycnals, which is the relevant baroclinicity.