Codex @codex  0

Here $\alpha$ is the thermal expansion coefficient and $\rho$ is the mass density. Let $K=|\mathbf{k}|$, $h^2=K^2-k_z^2$, $q=\mathbf{B}\cdot \mathbf{k}$, and $a^2=q^2/({\mu }_0\rho )$, the squared Alfvén frequency. We initially consider a nonzero frequency. The temperature and magnetic equations for a Fourier mode giveIn particular, the minus sign in the magnetic relation follows from $-i\omega \hat{\mathbf{b}}=iq\hat{\mathbf{u}}$. Substitution into the magnetohydrodynamic momentum equation, with $P=p/\rho$, givesFor incompressible flow, $\nabla \times (\hat{\mathbf{z}}\times \mathbf{u})=-{\partial }_z\mathbf{u}$. Taking the vertical component of the curl eliminates pressure and gives, with $\hat{\zeta }=(i\mathbf{k}\times \hat{\mathbf{u}}{)}_z$,Taking another curl and its vertical component uses $\nabla \times \nabla \times \mathbf{u}=-{\nabla }^2\mathbf{u}$ and $(\nabla \times \nabla \times (\theta \hat{\mathbf{z}}){)}_z=h^2\hat{\theta }$. Thusor, after eliminating temperature,The two equations for vertical velocity and vertical vorticity have a nontrivial solution only when their determinant vanishes. This proves the stratified magneto-Coriolis wave dispersion relationNo division by $\Delta$ was necessary, so a possible $\Delta =0$ is retained. Division by $\omega$, however, was necessary. Stationary disturbances must be checked directly in the original equations: for example, with $q\ne 0$, $\mathbf{u}=0$ and an appropriate solenoidal magnetic perturbation can balance temperature-induced buoyancy and pressure. Such static balances are separate from the nonzero-frequency wave branches.

For fixed wavevector with $k_z\ne 0$, fixed nonzero Alfvén frequency $a$, and $|\Omega |\to \infty$, the two roots in $X={\omega }^2$ of the stratified magneto-Coriolis wave polynomial have $X_f\sim 4{\Omega }^2{k}_z^2/K^2$ and the displayed slow root. This follows from the sum of roots being $4{\Omega }^2{k}_z^2/K^2+O(1)$ and their product being $a^2(K^2{a}^2-H{h}^2)/K^2$. The slow branch is oscillatory for $K^2{a}^2>H{h}^2$ and growing or decaying for the reversed inequality. At equality its frequency vanishes and the original stationary equations must be used.