= Oblique-field plane-wave magnetoconvection
{title2=$R_{mathrm{stat}}=(k^2+m^2)^3/k^2+Q(k^2+m^2)(k\sin\alpha+m\cos\alpha)^2/k^2$}
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 satisfies
$$
q(s+q)\left[(s/\sigma+q)(s+\zeta q)+Q\zeta d^2\right]=Rk^2(s+\zeta q),
$$
where $\sigma$ 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 $\widehat{\mathbf b}=id\widehat{\mathbf u}/(s+\zeta q)$, and thermal transport gives $\widehat\theta=\widehat w/(s+q)$. Projecting momentum perpendicular to the wavevector removes pressure; the vertical component of the projected buoyancy is $R(k^2/q)\widehat\theta$. Substitution gives the displayed polynomial. The stationary neutral curve is $R=q^3/k^2+Qqd^2/k^2$. No vertical boundary conditions are imposed in this plane-wave calculation.
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