Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 333 (Fluid Dynamics of Climate)

The basic state has velocity fieldand buoyancy $B=N^{2}z$. For disturbances independent of $y$, the linearized equations areSubstituting a plane wave proportional to $\exp [i(kx+mz-\omega t)]$ and eliminating $p$, $v$, and $b$ gives the dispersion relationThus the requested coefficients are

An instability exists precisely when some wavenumber pair makes ${\omega }^2<0$. Since the Brunt–Väisälä frequency satisfies $N^2>0$, this is possible exactly whenThe basic relative vorticity is $\Lambda {\mathbf{e}}_z$, so its absolute vorticity is $(f+\Lambda ){\mathbf{e}}_z$. Its Ertel potential vorticity is thereforeThe instability criterion can consequently be written as $fQ<0$: the vertical absolute vorticity has the opposite sign to the planetary vorticity. This is inertial instability, approached most directly by disturbances with $|m/k|\gg 1$.

Now let the basic buoyancy beThermal-wind balance requires the basic along-front velocity to have vertical shear $V_z=M^2/f$, soThe basic absolute vorticity and buoyancy gradient areand hence

For the prescribed disturbance the buoyancy perturbation vanishes, so the linearized buoyancy equation and incompressibility giveThe wavevector must therefore satisfythe disturbance velocity lies along a basic isopycnal. The along-front momentum equation becomesProjecting the remaining momentum equations onto the divergence-free direction eliminates the pressure and yieldsThus this isopycnal disturbance grows if and only if $fQ<0$. Geometrically, $Q$ is the component of the absolute vorticity along the buoyancy gradient, multiplied by $|\nabla B|$; the horizontal buoyancy gradient reduces that component through the term $-M^4/f$. When $M=0$, the result reduces to the most unstable, nearly horizontal-wavevector limit of part i.

Represent a horizontal vector $(u,v)$ by the complex number $W=u+iv$, let $s=\mathrm{sgn}f$, and define the two Ekman layer depths and drag coefficients byThe atmospheric and oceanic departures from their respective geostrophic flow satisfyThe solutions that decay away from the ice areandHere $W_a^{\infty }$ and $W_o^{\infty }$ denote the atmospheric and oceanic geostrophic velocities.

The viscous stress exerted on the ice by the atmosphere and ocean is, respectively,Because the ice is an infinitesimally thin, freely moving sheet, its horizontal force balance is ${\tau }_a+{\tau }_o=0$. The common Ekman turning factor cancels, leavingThus the ice moves along the weighted mean of the two geostrophic currents. In particular, as ${\nu }_a\to 0$ one has $r_a\to 0$ and $W_i\to W_o^{\infty }$: an atmosphere with vanishing viscosity transmits no finite stress to the ice.

The atmospheric Ekman transport relative to its geostrophic current isThe atmospheric stress isIf the two geostrophic currents are parallel but unequal, both directions are obtained by rotating their velocity difference: in the Northern Hemisphere the stress lies $45^{\circ }$ anticlockwise from $W_a^{\infty }-W_o^{\infty }$, while the atmospheric transport lies $45^{\circ }$ clockwise from its negative. Both rotations reverse in the Southern Hemisphere. If the two currents are identical, the shear, stress, and relative Ekman transport all vanish.

At the displaced free surface, constant atmospheric pressure and the hydrostatic approximation give the dynamic conditionThe linearized kinematic condition is $w=D_g\eta /Dt$. Combining these relations withgives

Linearization about rest replaces the material derivative by ${\partial }_t$, so the quasi-geostrophic potential-vorticity equation becomesInsert the separation of variables $\psi =\varphi (x,y,t)P(z)$ and imposeThe horizontal amplitude then satisfiesThe Linear Rossby-wave equation therefore governs each vertical normal mode, whose Rossby deformation radius is $c/f_0$.

The rigid lower boundary has $w=0$. For a time-dependent normal mode this givesThe free-surface condition from part i givesWriting $q=N/c$, the lower condition selectsThe upper condition therefore requiresEquivalently, with $\mu =NH/c$ and $ϵ=N^{2}H/g$,Each positive root ${\mu }_n$ determines the barotropic mode or a baroclinic mode through $c_n=NH/{\mu }_n$.

Suppose $ϵ=N^{2}H/g\ll 1$. The smallest root satisfiesso the leading external gravity wave eigenfunction is depth-independent:For $n\ge 1$, the roots lie just above $n\pi$:and henceIn particular, up to an arbitrary normalization and sign,The hierarchy $c_0>c_1>c_2>\cdots$ separates the fast, nearly barotropic free-surface mode from the internal baroclinic modes.

Expand both the forcing and response in the orthogonal vertical normal modes. Projection onto $P_n$ givesIf the horizontal forcing scale obeys $L_h\gg c_n/f_0$, the relative vorticity term ${\nabla }_h^2{\varphi }_n$ is small compared with the stretching term. The resulting long-wave equation isThus each mode communicates the forcing westward at its long Rossby wave speed $C_n$, with the barotropic mode fastest because $c_0$ is largest.

The method of characteristics shows that a point is affected only after a westward characteristic from the forcing region reaches it. At a western observation point $X_1$, during the intervalthe fast barotropic signal has arrived but the first and higher baroclinic signals have not. Consequentlywhich is nearly independent of depth. At the eastern point $X_2$, no westward Rossby-wave characteristic arrives from the forcing region, so the disturbance remains zero under the stated initial and radiation conditions.

The propagation speed derived in part v is $C_n=\beta c_n^2/f_0^2$. Thus the question's notation $c_{0}t>L_X>c_{1}t$ must be read as a comparison with these modal Rossby signal speeds; using the gravity-wave speeds $c_n$ literally would not describe the long-wave equation derived above.

In a steady state the time derivative of quasi-geostrophic potential vorticity vanishes, leaving the local balanceChoose the undisturbed state immediately east of the compact forcing as the integration condition. The solution is thenThe discontinuity in this ideal expression reflects the discontinuous prescribed forcing at $z=-D$; a vertically smooth forcing gives the corresponding smooth vertical profile.

The adjustment begins with the rapidly propagating barotropic mode and is therefore initially almost depth-independent. Successively slower baroclinic modes then arrive from the forcing region. Their Fourier series in the vertical normal modes progressively reconstructs the forcing's upper-layer profile, tending to the stated steady solution while points to the east remain unaltered.

Let every perturbation be proportional to $\exp [i(kx+mz-\omega t)]$ and define the intrinsic frequencyEliminating pressure with incompressibility, and then eliminating buoyancy, gives the internal gravity wave dispersion relationThe vertical group velocity isThus energy propagates upward when $\hat{\omega }m<0$ and downward when $\hat{\omega }m>0$. Equivalently, on the positive-intrinsic-frequency branch upward propagation requires $m<0$, while on the negative branch it requires $m>0$.

The exact no-penetration condition on the stationary boundary is that the velocity be tangent to it. Linearized at $z=0$, it isThe terrain is stationary, so $\omega =0$ and $\hat{\omega }=-kU$. The dispersion relation givesA vertically propagating wave therefore exists if and only ifThe upward radiation condition $\hat{\omega }m<0$ selects $km>0$ for $U>0$. Taking $k>0$ gives $m>0$. Lines of constant phase satisfyso they tilt upstream as height increases. The discarded sign of $m$ gives the mirror-image phase tilt and downward energy propagation.

For $k>0$, choose the upward-radiating rootThe boundary condition gives a vertical-velocity amplitude of magnitude $Uk{h}_0$. Incompressibility gives $u^{\prime }=-(m/k)w^{\prime }$, so horizontal averaging over one wavelength yields the wave momentum fluxIf the wave is completely absorbed in a remote dissipation layer, this negative flux increases to zero across the layer. The integrated force on the mean flow is thereforeIt acts opposite to the positive basic flow.

The sloping boundary exerts the pressure dragon the fluid. At generation this equals ${\rho }_0\overline{u^{\prime }{w}^{\prime }}$: the mountain supplies a negative vertical flux of horizontal momentum to the wave. Where the wave dissipates, convergence of that flux transfers the same westward momentum to the mean flow. A forcing that maintains the prescribed uniform basic current must supply the compensating eastward momentum.

Define the perturbation vorticity componentTaking the curl of the linear momentum equation and using the buoyancy equation givesMultiply the first equation by $b^{\prime }$, the second by ${\zeta }^{\prime }$, and horizontally average. Periodicity makes the averaged $x$ derivatives vanish, whileby incompressibility and integration by parts. It follows thatTherefore the wave activity conservation law is

The force on the mean flow and the wave-activity tendency obeyHence growth of positive wave activity produces a negative mean-flow force, while decay deposits negative momentum into the mean flow. Since part iii found a negative force for a growing, upward-radiating wave with $U>0$, its wave activity is positive.

If $U<0$, stationarity still gives $\hat{\omega }=-kU$, but upward propagation now selects $km<0$. For $k>0$ one takes $m<0$, and thereforeThe integrated force reverses and accelerates the mean flow in the positive $x$ direction. The activity of a growing wave correspondingly has the opposite sign and is negative.