Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 68 2 ii Solution Created 2026-10-03 Updated 2026-10-14
Keep the background magnetic field, thermal gradient and wavevector fixed as . In particular, is nonzero and of order one. Put and . Multiplying the nonzero-frequency dispersion relation by givesthat is,The sum and product of its two roots areFor large rotation, the sum is of order while the product is independent of . 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 away from the zero-frequency boundary, the rapid-rotation slow magneto-Coriolis branch hasIts frequency is of order , 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 , the slow wave has real nonzero frequency whenStable thermal stratification has and automatically satisfies this condition. With adverse stratification, , sufficiently strong magnetic tension along the wavevector is needed. If the inequality is reversed, and the slow branch has imaginary frequencies, including an exponentially growing mode. Equality gives a stationary limit, rather than a nonzero oscillation. If , the proposed slow wave degenerates into zero-frequency disturbances: there is no magnetic restoring coupling. The remaining nonzero branch has . These qualifications also explain why zero roots introduced by clearing denominators must not be interpreted as ordinary propagating waves.