Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 333 / 1

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.