Oblique-field plane-wave magnetoconvection ($R_{mathrmstat}=(k^2+m^2{)}^3/k^2+Q(k^2+m^2)(k\sin \alpha +m\cos \alpha {)}^2/k^2$)

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 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.

At inclination $\alpha =\arctan \sqrt{2}$ and vertical wavenumber one, the nonmagnetic minimizing wavenumber $k=-1/\sqrt{2}$ also satisfies $\mathbf{c}\cdot \mathbf{k}=0$. The stationary threshold is the sum of $F(k)\ge 27/4$ and a nonnegative magnetic penalty that vanishes there. Therefore $k_c=-1/\sqrt{2}$ and $R_c=27/4$ for every $Q\ge 0$. The selected disturbance is constant along the imposed field, so the linear resistive induction equation produces no magnetic perturbation and there is no field-line tension penalty.

For fixed vertical wavenumber one, write the stationary threshold as $F(k)+QG(k)$, with $F=(1+k^2{)}^3/k^2$ and $G=(1+k^2)(\cos \alpha +k\sin \alpha {)}^2/k^2$. The nonmagnetic minima are $k=\pm 1/\sqrt{2}$ with $F=27/4$; a small positive field selects the negative one. Expanding the stationarity condition gives $k_c=a-Q{G}^{\prime }(a)/F^{\prime \prime }(a)+O(Q^2)$, $a=-1/\sqrt{2}$, andFor large $Q$, the magnetic penalty forces $k$ toward its unique zero $b=-\cot \alpha$. Since $G^{\prime \prime }(b)>0$, $k_c=b-F^{\prime }(b)/[Q{G}^{\prime \prime }(b)]+O(Q^{-2})$ and $R_c={\csc }^6\alpha /{\cot }^2\alpha -F^{\prime }(b{)}^2/[2Q{G}^{\prime \prime }(b)]+O(Q^{-2})$. Thus strongly magnetized rolls align their wavevector perpendicular to the imposed field, rather than continuing to raise their stationary threshold proportionally to field strength.