Magnetorotational instability (MRI)

The magnetorotational instability destabilizes a magnetized differentially rotating flow when angular velocity decreases outward. Magnetic tension couples neighboring fluid elements, allowing the inner one to lose angular momentum and fall inward while the outer one gains angular momentum and moves outward.

For $0<A<2\Omega$ and equal kinematic viscosity and magnetic diffusivity, a vertical wavevector admits $\mathbf{v}=(V,V,0)$ and $\mathbf{b}=(i{\omega }_{A}V/A,-i{\omega }_{A}V/A,0)$ in Alfvén-speed units. Substitution into the magnetorotational instability equations gives the displayed frequency and growth rate. In a Keplerian disk, growth of this polarization requires $B_z^2/({\mu }_0\rho )>5\nu \Omega /4$. This is a bound for this selected channel and its wavelength, not a universal threshold for every MRI mode in an unbounded domain.

For complex velocity amplitudes $v_x=v_y=V$ and magnetic amplitudes $b_x=-b_y=B$ in Alfvén-speed units, phase averaging gives the displayed stress. For fixed horizontal velocity energy, equal radial and azimuthal components maximize their positive Reynolds stress. Opposite equal magnetic components maximize the outward tension contribution of the Maxwell stress tensor, since $-B_x{B}_y$ is then positive. These conclusions follow from $2|ab|\le |a{|}^2+|b{|}^2$. The vertical channel transports angular momentum outward and extracts energy from decreasing angular velocity.

For an incompressible vertical-field mode, frozen-in magnetic tension supplies a spring of frequency ${\omega }_A$. Its displacement satisfies ${\ddot{\xi }}_x-2\Omega {\dot{\xi }}_y-4\Omega A{\xi }_x=-{\omega }_A^2{\xi }_x$ and ${\ddot{\xi }}_y+2\Omega {\dot{\xi }}_x=-{\omega }_A^2{\xi }_y$. Their determinant gives the ideal magnetorotational dispersion relation. A weak spring transfers angular momentum from the faster inner element to the slower outer element, causing inward and outward migration that increase their separation when angular velocity decreases outward. A sufficiently strong spring instead supplies a restoring force.

// Target: physics.bigb

For ideal magnetohydrodynamics at constant mass density $\rho$, a shearing sheet with velocity $-q\Omega x{\mathbf{e}}_y+{\mathbf{v}}_h(z,t)$ and magnetic field ${\mathbf{B}}_h(z,t)+B_z{\mathbf{e}}_z$ obeysHorizontal invariance and vanishing vertical velocity eliminate nonlinear advection. These equations retain Coriolis acceleration, orbital shear and magnetic tension. The vertical pressure adjusts to balance gravity and magnetic pressure.

In the steady state of the horizontally invariant magnetized shearing-sheet equations, let $B_x(\pm H)=\pm B_{+}$ and $B_y(\pm H)=0$. For $q>0$ and $B_z\ne 0$, the symmetric equilibrium hasThis follows by combining azimuthal induction with radial magnetic tension balance to obtain $B_x^{\prime \prime }+K^2{B}_x=0$. Nonzero $B_{+}$ requires $\sin (KH)\ne 0$; even homogeneous additions at $\cos (KH)=0$ are excluded by imposing midplane symmetry.

For nonzero imposed surface inclination, magnetic bending in an incompressible disk is nonmonotonic when $KH>\pi /2$: the first maximum of $|B_x|$ occurs at $z=\pi /(2K)<H$, followed by a decrease. This is exactly the finite-thickness magnetorotational instability criterion for the same slab. The zero-inclination equilibrium can still be unstable, so nonmonotonic bending is a property of the forced profile, rather than a condition on every possible unstable equilibrium.

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

For a vertical magnetic field in a Keplerian disk, axisymmetric vertical MRI modes are unstable when $k^2{v}_A^2<3{\Omega }^2$. The shortest unstable wavelength is ${\lambda }_{\mathrm{crit}}=2\pi v_A/(\sqrt{3}\Omega )$.