Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2017 / 5

Take an isothermal hydrostatic reference atmosphere, ${\rho }_0(z)={\rho }_s{e}^{-z/H}$, and linearize about rest. The horizontal acceleration and pressure-gradient force retain the equatorial Coriolis parameter $f\simeq \beta y$, with $\beta =2\Omega /a$. The traditional approximation retains only its vertical component. Quadratic perturbation advection is neglected. The vertical acceleration is balanced by perturbation buoyancy and pressure gradient after removing the reference hydrostatic balance.

The continuity constraint is the anelastic approximation, $\nabla \cdot ({\rho }_0\mathbf{u})=0$, giving $u_x+v_y+w_z-w/H=0$. It retains stratification-driven expansion while filtering acoustic density fluctuations. It is therefore not the full compressible continuity equation. Adiabatic displacement in a stably stratified background gives $b_t+N^{2}w=0$, with $N$ the Brunt–Väisälä frequency; upward displacement develops restoring negative buoyancy.

Pressure convention matters. With the entropy-buoyancy equation just described, a consistent reduced-pressure form writes the pressure acceleration as $-\nabla \Phi$, where $\Phi =p^{\prime }/{\rho }_0$. Its vertical term is $-{\partial }_z(p^{\prime }/{\rho }_0)$, not simply $-{\rho }_0^{-1}{p}_z^{\prime }$. The original source explicitly specifies variable ${\rho }_0(z)$ but prints the latter. Those differ by $p^{\prime }/({\rho }_{0}H)$, so the two cannot silently be equated. Parts b and c display the consequence and then solve the explicitly corrected reduced-pressure model that yields the requested wave equation. This is a mathematical inconsistency in the printed set, not missing source material.

First check the printed equations literally. Set $v=0$ and use ${\rho }_0^{\prime }/{\rho }_0=-1/H$. Differentiating continuity in time and using horizontal momentum givesVertical momentum and buoyancy give $w_{tt}+N^{2}w=-p_{zt}/{\rho }_0$. Combining them therefore yieldswhich differs from the requested equation. The factor two arises when differentiating both ${\rho }_0$ and $w_z-w/H$. Keeping variable density in continuity while treating it as constant during that differentiation would be inconsistent.

For the intended anelastic reduced-pressure Kelvin wave, make the explicit correction $w_t-b+{\Phi }_z=0$, while $u_t+{\Phi }_x=0$ and $\Phi =p^{\prime }/{\rho }_0$. Continuity then gives ${\Phi }_{xx}=w_{zt}-w_t/H$, and vertical momentum gives $w_{tt}+N^{2}w=-{\Phi }_{zt}$. Cross differentiation now producesThis proves the advertised equation for the corrected model, without claiming it follows from the incompatible printed set.

A concrete counterexample makes the distinction testable. In nondimensional units $H=N=k=1$, take $m=0$, $\omega =2/\sqrt{5}$ and $w=e^{z/2}{e}^{i(x-\omega t)}$, multiplied by its Gaussian meridional envelope. This satisfies the last boxed wave equation, but substitution into the literal printed-system wave operator gives $-2w/5\ne 0$. No choice of accompanying pressure and horizontal velocity can then make it a solution of all five printed equations. Conversely, the literal operator has vertical factors $e^{z/H+imz}$ and dispersion ${\omega }^2=N^2{k}^2/(k^2+m^2)$; its scaling is different from the requested half-scale-height factor.

Work with the explicitly corrected reduced-pressure system in b. Write $d=1/(2H)$ and use the complex wave amplitude, taking real parts at the end. The vertical factor satisfies $({\partial }_z^2-H^{-1}{\partial }_z)w=-(m^2+d^2)w$, so the wave equation givesFor a Kelvin wave, continuity, zonal momentum and the buoyancy equation then determine all polarizations:In particular $u,w,b,\Phi$ have amplitude proportional to $e^{z/(2H)}$, whereas pressure $p^{\prime }$ has amplitude proportional to $e^{-z/(2H)}$. Directly substituting these expressions into vertical momentum gives ${\Phi }_z=-i\omega (m^2+d^2)w/k^2=i(\omega -N^2/\omega )w$, verifying both the sign and the dispersion relation.

Meridional momentum is $\beta yu+{\Phi }_y=0$. Since $\Phi =cu$, it integrates to a Gaussian $\exp [-\beta y^2/(2c)]$. ThusThe eastward intrinsic branch $c>0$ is selected by decay away from the equator; choosing $c<0$ would make the meridional amplitude grow without bound. Although the squared dispersion relation has both frequency signs, a meridionally trapped Kelvin wave has this one phase-propagation direction.

For short vertical wavelengths, $m^2\gg k^2+d^2$, so ${\omega }^2\simeq N^2{k}^2/m^2$, $c\simeq N/|m|$ and $L_D^2\simeq N/(\beta |m|)$. The wave is a slowly oscillating, hydrostatic internal gravity wave with strong vertical variation and relatively narrow trapping. Here $u\simeq -(m/k)w$ and $p^{\prime }/{\rho }_0\simeq -(\omega m/k^2)w$; the buoyancy expression above remains valid.

For long vertical wavelengths, $|m|\ll \sqrt{k^2+d^2}$, the exact limiting speed is $c=N/\sqrt{k^2+d^2}$. To obtain the stated height-controlled limit also requires long horizontal wavelength, $|k|H\ll 1$. With both $|m|H\ll 1$ and $|k|H\ll 1$,The finite density scale height prevents the hydrostatic $1/|m|$ phase speed from diverging as $m\to 0$. The polarizations tend to $u\simeq -iw/(2Hk)$, $\Phi \simeq -iNw/k$ and $b\simeq -iNw/(2Hk)$. Long vertical wavelength alone at fixed large $kH$ would not justify $c=2NH$. In the opposite horizontally short limit the frequency approaches $N$ instead. The growth $e^{z/(2H)}$ compensates decreasing reference density in wave-energy density; a radiating wave train is a local modal description, not a finite-total-energy eigenfunction over an infinite vertical domain.

Apply the intended, corrected dispersion relation in c. A uniform mean zonal wind gives ${\omega }_{\mathrm{ground}}=\hat{\omega }+kU$. Stationarity with $U=-20\mathrm{m}{\mathrm{s}}^{-1}$ therefore requires eastward intrinsic phase speed $\hat{\omega }/k=-U=20\mathrm{m}{\mathrm{s}}^{-1}$. With $k=1/a$,The period determines $\Omega =2\pi /(24.4\times 3600\mathrm{s})$ and $\beta =2\Omega /a\simeq 4.21\times 10^{-11}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$. Hence

A storm within roughly $12^{\circ }$ of the equator has substantial projection on the Gaussian mode. There is no sharp latitude cutoff: $|y|=L_D$ gives amplitude $e^{-1/2}$ and the $1/e$ amplitude latitude is about $16.4^{\circ }$. Forcing outside that core has progressively weaker overlap.

The intrinsic wave is eastward even though its observed ground pattern is stationary; westward advection cancels its phase propagation. Its intrinsic period is about $12.4$ Earth days. The inferred $|m|\gg 1/(2H)$ puts it in the short-vertical-wavelength branch, well below the long-wave speed $2NH=200\mathrm{m}{\mathrm{s}}^{-1}$. For upward energy radiation, $\partial \hat{\omega }/\partial m$ must be positive, which selects $m<0$ for positive $k,\hat{\omega }$: vertical phase propagation is then downward. Dust absorption supplies thermal forcing with a zonally asymmetric component; the response does not imply that dust itself moves eastward at the intrinsic phase speed.

Southern spring to summer, the warmer perihelion-side dust season, is a reasonable climatological window for such a regional storm. This seasonal inference uses observed dust climatology, not the dispersion relation: the specified wind and wave parameters alone do not determine a unique season, and individual regional storms can occur outside that window.^[ref]