Finite-thickness magnetorotational instability criterion

If the horizontal magnetic field perturbations vanish at $z=\pm H$, their allowed vertical wavenumbers are $k_m=m\pi /(2H)$, $m=1,2,\dots$. The lowest wavenumber is unstable exactly whenIf that magnetic mode is stable, all higher magnetic modes are stable as well. A weak field admits longer unstable wavelengths that fit inside the slab; sufficiently strong magnetic tension stabilizes every allowed wavelength.

For density proportional to ${\mathrm{sech}}^2(z/H)$, the magnetic vertical equation $F^{\prime \prime }+k^2{\mathrm{sech}}^2(z/H)F=0$ becomes the Legendre differential equation under $\xi =\tanh (z/H)$. Bounded solutions at both surfaces are Legendre polynomials, with $k_n=\sqrt{n(n+1)}/H$. The $n=0$ constant function has no magnetically coupled finite wavenumber. The smallest MRI-relevant eigenvalue is $k_1^2=2/H^2$, which makes sufficiently strong magnetic tension stabilize all allowed modes.

The growing root of the ideal magnetorotational dispersion relation is maximized continuously at $a=v_A^2{k}^2/{\Omega }^2=15/16$, with growth $3\Omega /4$. Discrete vertical modes instead have $a_n=2n(n+1)/\beta$. The fastest one is found by comparing admissible integers on either side of the continuous maximum, rather than rounding a vertical wavenumber blindly. At $\beta =24$, $n=3$ gives $s=\Omega \sqrt{(\sqrt{17}-3)/2}$.