 Oblique-field plane-wave magnetoconvection

In thermal-diffusion units, let an imposed field point along $\mathbf{c}=(\sin \alpha ,0,\cos \alpha )$ and consider a Fourier mode of wavevector $(k,0,m)$. Put $q=k^2+m^2$ and $d=k\sin \alpha +m\cos \alpha$. The growth exponent $s$ of its buoyant branch satisfieswhere $\sigma$ is the Prandtl number, $\zeta$ the magnetic-to-thermal diffusivity ratio, $Q$ the Chandrasekhar number, and $R$ the Rayleigh number. The resistive induction equation gives $\hat{\mathbf{b}}=id\hat{\mathbf{u}}/(s+\zeta q)$, and thermal transport gives $\hat{\theta }=\hat{w}/(s+q)$. Projecting momentum perpendicular to the wavevector removes pressure; the vertical component of the projected buoyancy is $R(k^2/q)\hat{\theta }$. Substitution gives the displayed polynomial. The stationary neutral curve is $R=q^3/k^2+Qq{d}^2/k^2$. No vertical boundary conditions are imposed in this plane-wave calculation.