Ideal magnetorotational dispersion relation

For a vertical magnetic field in a homogeneous shearing sheet, a horizontal normal mode with growth rate $s$ and vertical Alfvén frequency ${\omega }_a=k{v}_{Az}$ satisfiesThe two roots for $s^2$ are real. For $q>0$, a growing magnetic mode exists precisely when $0<{\omega }_a^2<2q{\Omega }^2$. Magnetic tension is strong enough to couple radially displaced elements, yet weak enough to allow their separation to increase. Ogilvie's lecture on magnetorotational instability derives this relation and its finite-thickness boundary conditions.

For local cylindrical shear $q=-d\ln \Omega /d\ln r$ with $0<q<2$, maximizing the positive squared-growth root of the ideal magnetorotational dispersion relation gives the displayed formulas. For a Keplerian disk these are $s_{\max }=3\Omega /4$ and ${\omega }_A^2=15{\Omega }^2/16$. Field strength selects the fastest wavelength, while the maximum ideal rate is set by shear. Finite domains and magnetic diffusion can exclude that wavelength.

In the local incompressible weak-field limit, take an axisymmetric wavevector $(k_r,0,k_z)$ and ${\omega }_A=\mathbf{k}\cdot \mathbf{B}/\sqrt{{\mu }_0\rho }$. Eliminating the pressure projects the radial force by $k_z^2/k^2$. This yields the displayed dispersion relation, where $\kappa$ is the radial epicyclic frequency. For a hydrodynamically stable rotation profile with outward-decreasing angular velocity, instability requires $0<{\omega }_A^2<-(k_z^2/k^2)d{\Omega }^2/d\ln r$. A purely toroidal field has no axisymmetric local field-line-bending frequency and therefore no ordinary weak-field MRI in this approximation.

For an axisymmetric vertical mode of a shearing sheet with a vertical magnetic field, radial buoyancy frequency $N_r$, and $A=k^2{v}_A^2>0$, the nonstationary amplitudes obey ${\sigma }^4+(2A+N_r^2+{\Omega }^2){\sigma }^2+A(A+N_r^2-3{\Omega }^2)=0$. The two roots for ${\sigma }^2$ are real. A growing mode exists exactly when $A+N_r^2<3{\Omega }^2$. If the advected buoyancy displacement variable is retained as an independent amplitude, the full characteristic polynomial also has a stationary factor $\sigma$.

For $-5{\Omega }^2<N_r^2<3{\Omega }^2$ and a continuum of vertical wavenumbers, the growing branch of the radially stratified magnetorotational dispersion relation is maximized at $k^2{v}_A^2={\Omega }^2-(N_r^2+{\Omega }^2{)}^2/(16{\Omega }^2)$, with ${\sigma }_{\max }=(3{\Omega }^2-N_r^2)/(4\Omega )$. For $N_r^2\le -5{\Omega }^2$ the optimum moves to the $k\to 0$ boundary and the limiting rate is $\sqrt{-N_r^2-{\Omega }^2}$. For $N_r^2\ge 3{\Omega }^2$ there is no exponential instability. Boundaries that discretize the allowed wavenumbers can alter the actual maximum.

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 $s=\Omega \sqrt{(\sqrt{17}-3)/2}$.