Past exam of the geophysical fluid dynamics course of the University of Oxford / 2024 (A15274W1)

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.