Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 68 / 2 / ii

Keep the background magnetic field, thermal gradient and wavevector fixed as $|\Omega |\to \infty$. In particular, $k_z/K$ is nonzero and of order one. Put $H=g\alpha \beta$ and $X={\omega }^2$. Multiplying the nonzero-frequency dispersion relation by ${\omega }^2$ givesthat is,The sum and product of its two roots areFor large rotation, the sum is of order ${\Omega }^2$ while the product is independent of $\Omega$. The quadratic therefore has one large root and one small root. More explicitly,The fast frequencies are consequentlywhich are inertial waves at leading order. For nonzero $a$ away from the zero-frequency boundary, the rapid-rotation slow magneto-Coriolis branch hasIts frequency is of order $|\Omega {|}^{-1}$, much smaller than the fast inertial wave frequency. Coriolis force balances most of the magnetic and buoyant forces in this slow motion, with inertia supplying only a small correction.

For $a^2>0$, the slow wave has real nonzero frequency whenStable thermal stratification has $H<0$ and automatically satisfies this condition. With adverse stratification, $H>0$, sufficiently strong magnetic tension along the wavevector is needed. If the inequality is reversed, $X_s<0$ and the slow branch has imaginary frequencies, including an exponentially growing mode. Equality gives a stationary limit, rather than a nonzero oscillation. If $\mathbf{B}\cdot \mathbf{k}=0$, the proposed slow wave degenerates into zero-frequency disturbances: there is no magnetic restoring coupling. The remaining nonzero branch has ${\omega }^2=(4{\Omega }^2{k}_z^2-H{h}^2)/K^2$. These qualifications also explain why zero roots introduced by clearing denominators must not be interpreted as ordinary propagating waves.