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A stationary mass density disturbance can be balanced by a changed azimuthal velocity in a shearing sheet. The continuity and azimuthal equations give zero radial velocity, while the radial equation balances Coriolis force against pressure and gravity. This mode is separate from the density-wave pair in the softened Toomre dispersion relation and is lost if one divides every equation by frequency.

A shearing-wave ansatz in a uniform inviscid isothermal shearing sheet removes background advection when the radial wavevector grows as $k_x(t)=k_x(0)+S{k}_{y}t$. The razor-thin disk Poisson kernel reduces the pressure-plus-gravity amplitude to $[c_s^2-2\pi G{\Sigma }_0/k]{\tilde{\Sigma }}^{\prime }/{\Sigma }_0$. Continuity couples it to $i{k}_x{v}_x+i{k}_y{v}_y$, while the two momentum equations have rotational couplings $-2\Omega v_y$ and $(2\Omega -S)v_x$. The linearized vortensity amplitude remains constant, because the barotropic pressure and gravitational forces have zero curl and the background vortensity is uniform.

In a radially stratified shearing sheet, one can normalize the scalar buoyancy perturbation by writing $b_x=-N^2\theta$ and $D_t\theta =u_x+\xi {\nabla }^2\theta$. With this convention $\theta$ has dimensions of length, rather than temperature; $\xi$ is thermal diffusivity.

Let $c_s^2=(dP/d\Sigma {)}_0$ be the barotropic closure of a razor-thin disk sound-speed squared and ${\kappa }_r$ the radial epicyclic frequency, with ${\kappa }_r^2=2(2-q_r){\Omega }^2$ for orbital shear parameter $q_r$. For a radial axisymmetric normal mode in a uniform barotropic shearing sheet, let $U_x,U_y$ be velocity amplitudes and $\sigma$ the surface-density amplitude. Linear momentum gives $-i\omega U_x=2\Omega U_y-ik{c}_s^2\sigma /{\Sigma }_0$ and $-i\omega U_y=-{\kappa }_r^2{U}_x/(2\Omega )$; mass conservation gives $\omega \sigma =k{\Sigma }_0{U}_x$. Eliminating the amplitudes for $\omega \ne 0$ proves the displayed dispersion relation. The rotational term is epicyclic restoration; the pressure term is acoustic restoration. Without a perturbed gravitational potential, no self-gravity term belongs in the formula.

In an incompressible shearing sheet, a single real-wavevector velocity and magnetic Fourier mode transverse to that wavevector has identically zero quadratic self-advection and cross-advection. Advecting its phase by background shear and stretching its spatially uniform field give the displayed evolution laws. Their opposite contributions keep the Alfvén frequency constant. The resulting viscous and resistive wave-amplitude equations are exact finite-amplitude solutions; physical pressure adjusts to the quadratic magnetic-pressure terms.

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.

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.

Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.

Choose a reference circular orbit at $r_0$, and use local coordinates $x=r-r_0$, $y=r_0[\varphi -{\Omega }_{0}t]$, with the frame rotating at $\Omega ={\Omega }_0=\Omega (r_0)$. Linearizing differential rotation in the shearing sheet gives the background flowThus the velocity perturbation is measured relative to this background. The local orbital shear parameter is $q=2A/\Omega =-d\ln \Omega /d\ln r$; Keplerian rotation has $q=3/2$ and $A=3\Omega /4$. Rotation supplies the Coriolis force, while the linear shear supplies the terms proportional to $A$.

For constant density, $\psi$ is the total-pressure potential after background force balance has been subtracted:up to a function of time and any prescribed perturbation potential that is absorbed into it. The magnetic-pressure gradient has been combined with the gas-pressure gradient, leaving magnetic tension as $\mathbf{B}\cdot \nabla \mathbf{B}/({\mu }_0\rho )$. The pressure also enforces incompressibility. In the unstratified shearing box here, the tidal/background terms have already been removed in deriving the perturbation equations; no independent vertical stratification is being evolved.

The additional conditions are $\boxed{\nabla \cdot \mathbf{u}=0,\nabla \cdot \mathbf{B}=0}$: incompressible flow and the solenoidal magnetic field constraint. The scalar $\Psi$ combines pressure per unit density with magnetic pressure per unit density and any conservative gravitational and centrifugal potentials. In particular, the Lorentz force identity separates magnetic tension from $-\nabla (B^2/2{\mu }_0\rho )$. Adding this latter term to the rotating-flow pressure potential gives the pressure appearing in the stated equation; the Coriolis force remains explicit.

Work in magnetic velocity units $\mathbf{B}/\sqrt{{\mu }_0\rho }$, and write $\mathbf{U}=-2Ax{\mathbf{e}}_y$ for the background shearing sheet flow. Its advective acceleration is zero, and $2\Omega {\mathbf{e}}_z\times \mathbf{U}=4\Omega Ax{\mathbf{e}}_x=-\nabla (-2\Omega A{x}^2)$, so the stated quadratic background pressure potential gives an exact balance. A spatially uniform mean magnetic field must evolve asThis is stretching of radial field into azimuthal field by differential rotation.

For the shearing wave phase $\Theta =\mathbf{k}(t)\cdot \mathbf{x}$, advection by the background givesIt vanishes identically ifThe changing wavevector describes the shear of initially planar wavefronts. The mean-field stretching and wavevector evolution cancel in the Alfvén frequency:Hence $\boxed{{\omega }_A=\mathbf{k}\cdot \bar{\mathbf{b}}\text{is constant}}$.

These solutions are exact finite-amplitude disturbances, not merely a linearization. To see this, let ${\mathbf{u}}^{\prime }=\mathrm{Re}(\mathbf{v}{e}^{i\Theta })$ and ${\mathbf{B}}^{\prime }=\mathrm{Re}(\mathbf{b}{e}^{i\Theta })$ in magnetic velocity units. The divergence constraints give $\mathbf{k}\cdot \mathbf{v}=\mathbf{k}\cdot \mathbf{b}=0$. Thus both real disturbances are pointwise perpendicular to k, whereas all their spatial derivatives are along k. Every quadratic term ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}^{\prime }$, ${\mathbf{B}}^{\prime }\cdot \nabla {\mathbf{B}}^{\prime }$, ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{B}}^{\prime }$ and ${\mathbf{B}}^{\prime }\cdot \nabla {\mathbf{u}}^{\prime }$ vanishes, including terms involving conjugate amplitudes. This extends the exact incompressible shearing wave to an exact magnetic shearing wave. The gas pressure can compensate quadratic magnetic-pressure terms even though the combined $\Psi$ contains only its stated background and wave harmonic.

The remaining wave terms are therefore preciselyThe shear terms arise from perturbations acting on $\nabla \mathbf{U}$, the magnetic terms from the uniform field acting on the wave, and the last terms from diffusion. Pressure enforces the moving-wavevector velocity constraint: differentiating it givesUsing this pressure leaves $\mathbf{k}\cdot \mathbf{v}=0$ invariant. Differentiating $\mathbf{k}\cdot \mathbf{b}$ in the induction equation similarly leaves it zero. Thus the evolution is consistent with both constraints for initially transverse amplitudes.

Now take positive $\Omega$, so the stipulated condition implies $0<A<2\Omega$, and take magnetic Prandtl number $\nu /\eta =1$, with $\nu =\eta >0$. For a vertical wavevector $\mathbf{k}=k{\mathbf{e}}_z$, use the equal-diffusivity magnetorotational channel formThe two horizontal momentum equations, after adding them, give $\dot{V}=(A-\nu k^2)V$. Either component then gives $i{\omega }_{A}B=(A-2\Omega )V$. The first induction equation gives $({\partial }_t+\nu k^2)B=i{\omega }_{A}V$. These equations are simultaneously solved byThe second induction component gives the same relation, since $B_y=-B_x$. Exponential growth occurs when $\boxed{\nu k^2<A}$. The quarter-cycle phase between magnetic and velocity perturbations is required by the factor i in the Fourier equations.

For Keplerian rotation, expanding the local shear gives $2A=-(r{\Omega }^{\prime })=3\Omega /2$, hence $A=3\Omega /4$. Therefore ${\omega }_A^2=15{\Omega }^2/16$. With $v_{Az}={\bar{B}}_z/\sqrt{{\mu }_0\rho }$ and ${\omega }_A=k{v}_{Az}$, the growth condition for this prescribed polarization becomesA weaker field would require a shorter wavelength to retain this mode's selected Alfvén frequency, making diffusive damping too strong. The required wavelength must also fit any actual vertical domain; no vertical boundary conditions are specified here.

The field bound is for the requested equal-component mode, rather than a universal necessary condition for all magnetorotational instability. This distinction matters if all vertical wavelengths are admitted. Indeed, let $s=\gamma +\nu k^2$ for a general horizontal polarization. Eliminating its two magnetic amplitudes gives $(s+{\omega }_A^2/s)v_x-2\Omega v_y=0$ and $(s+{\omega }_A^2/s)v_y+[2(\Omega -A)-2A{\omega }_A^2/s^2]v_x=0$. Their determinant givesFor a Keplerian disk, ${\kappa }^2={\Omega }^2$. As a concrete counterexample to a universal reading of the bound, in consistent dimensionless units take $\Omega =1$, $A=3/4$, $\nu =\eta =1$, $v_{Az}=0.1$, and $k=0.01$. The positive root gives $s\simeq 0.00173205$ and $\gamma =s-0.0001>0$, even though $v_{Az}^2=0.01<1.25$. This mode has a different polarization and a much longer wavelength. Real finite-height discs restrict such wavelengths; the PDF supplies no such restriction.

Finally, the horizontally averaged radial flux of azimuthal momentum is the sum of Reynolds stress and magnetic tension stress:Multiplication by the local radius gives an outward angular-momentum flux. Equal radial and azimuthal velocities give the largest positive product at fixed horizontal kinetic energy, while opposite magnetic components give the largest positive $-B_x{B}_y$ at fixed horizontal magnetic energy. Both follow from $2|ab|\le |a{|}^2+|b{|}^2$. This stress-maximizing magnetorotational channel polarization therefore makes both contributions reinforce outward angular momentum transport. It extracts shear energy at rate $-T_{xy}{\partial }_x{U}_y=2A{T}_{xy}>0$, explaining why these correlations are especially effective in driving accretion.

Consider a local shearing sheet centered at a circular orbit. Take uniform mass density, background velocity ${\mathbf{u}}_0=-2Ax{\mathbf{e}}_y$, rotation $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$ and a weak uniform vertical magnetic field ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$. HereUse ideal magnetohydrodynamics, neglect viscosity, magnetic diffusion and vertical stratification, and take incompressible axisymmetric disturbances proportional to $e^{ikz-i\omega t}$. They have horizontal velocity and field perturbations. Let $\mathbf{b}=\delta \mathbf{B}/\sqrt{{\mu }_0\rho }$ be the magnetic perturbation in velocity units, $v_A=B_0/\sqrt{{\mu }_0\rho }$ the Alfvén speed and ${\omega }_A=k{v}_A$ the Alfvén frequency. Magnetic pressure can be included in the perturbation's total pressure; the horizontal Lorentz force for this mode is the magnetic tension term $i{\omega }_A\mathbf{b}$.

Linearizing momentum and induction givesThe term $-2A{b}_x$ winds a radial magnetic perturbation into an azimuthal one. Omitting it would remove the shear's magnetic energy source and give the wrong stability criterion.

A useful elimination that also covers neutral modes introduces the fluid displacement $\mathbf{\xi }$. Frozen-in flux gives $\mathbf{b}=i{\omega }_A\mathbf{\xi }$, while the Eulerian perturbation velocity is $u_x={\dot{\xi }}_x$, $u_y={\dot{\xi }}_y+2A{\xi }_x$, because the displacement samples the gradient of the basic flow. The equations become the spring model of magnetorotational instability:For the stated time dependence, their determinant isThus the ideal magnetorotational dispersion relation isIt follows without dividing by $\omega$, so the neutral stability boundary is included.

To analyze it, put $a={\omega }_A^2$ and $Y={\omega }^2$. ThenFor a hydrodynamically stable rotation law, ${\kappa }_r^2>0$, the sum of the two roots is positive. Their product is negative precisely when $0<a<4\Omega A$. One squared frequency is then negative, giving an exponentially growing mode $\omega =i\gamma$ and its decaying partner. Therefore the magnetorotational instability, or MRI, has criterionAngular velocity must decrease outwards, although specific angular momentum may increase outwards and satisfy Rayleigh's circulation criterion. This is why a Keplerian disc can be stable to the axisymmetric hydrodynamic disturbances of Question 2 yet unstable once magnetic coupling is introduced. At zero magnetic frequency the unstable root becomes neutral, while sufficiently large magnetic frequency restores stability. For increasing angular velocity, $A<0$ with $\Omega >0$, this ideal vertical-field instability is absent.

The growth rate isFor $0<A<\Omega$, differentiation with respect to $a$ sets $\sqrt{{\kappa }_r^4+16{\Omega }^{2}a}=4{\Omega }^2$, yielding the maximum growth rate of ideal magnetorotational instability:For Keplerian rotation, $A=3\Omega /4$, soThus the fastest growth is on an orbital timescale. The field strength selects its wavelength, $k_{\max }\propto 1/v_A$, rather than the maximum ideal growth rate. At fixed $k$, growth does approach zero as $B_0\to 0$; retaining the finite maximum in that limit requires shorter wavelengths, which magnetic diffusion can eventually exclude. Unstable wavelengths satisfy $\lambda >2\pi v_A/\sqrt{4\Omega A}$ in this local model. A finite disc must accommodate such a wavelength, so a sufficiently strong field can suppress the available vertical modes through excessive tension.

Physically, a bent field line acts like a spring joining neighboring fluid elements. The inner element rotates faster and pulls the outer element forward, transferring angular momentum outward. The inner element loses angular momentum and moves inward; the outer element gains it and moves outward. In a rotation law with outward-decreasing angular velocity, these motions increase their angular-velocity difference and the magnetic coupling produces further separation. A weak restoring spring therefore creates a feedback instability by enabling angular-momentum exchange. Without that exchange, each displaced element keeps its angular momentum and a positive epicyclic frequency restores its orbit. Strong magnetic tension instead prevents the separation. The free energy is the differential rotation, not the initial magnetic energy.

The instability provides an important route to turbulence and angular momentum transport in sufficiently conducting accretion discs. The radial flux of azimuthal momentum includes the stressThe first term is a Reynolds stress and the second is the magnetic contribution from the Maxwell stress tensor, with the sign appropriate to momentum flux. For the growing branch, take $u_x$ real and positive. The linear equations give $u_y/u_x=({\gamma }^2+a)/(2\Omega \gamma )>0$ and $b_y/b_x=({\gamma }^2+a-4\Omega A)/(2\Omega \gamma )<0$. The last sign follows from $({\gamma }^2+a)({\gamma }^2+a-4\Omega A)+4{\Omega }^2{\gamma }^2=0$. Thus both the velocity and magnetic contributions carry angular momentum outward. The associated energy extraction is $-T_{xy}rd\Omega /dr=2A{T}_{xy}>0$. Inward accretion can then accompany outward angular-momentum transport, and dissipation of the resulting fluctuations heats the disc. This gives a physical mechanism behind an effective turbulent-viscosity description such as an alpha disk, whereas microscopic viscosity alone is generally too weak for rapid accretion.

MRI is an exponentially growing magnetic instability of outward-decreasing angular velocity; hydrodynamic shearing-wave amplification is transient and does not by itself establish sustained turbulence. Nonlinear saturation, field regeneration and transport efficiency depend on field geometry, boundaries and thermodynamics. Poor conductivity or important nonideal magnetic effects can suppress or modify the ideal instability, so the local dispersion relation is a mechanism and growth criterion rather than a universal prescription for the saturated stress.

The magnetorotational instability extracts energy from differential rotation through magnetic tension. It can destabilize a disk whose specific angular momentum increases outward, even though such a disk satisfies Rayleigh's circulation criterion. The distinction is that an unmagnetized displaced parcel approximately conserves its own angular momentum, whereas a field line transfers angular momentum between parcels. If angular velocity decreases outward, an outward-displaced parcel lags its inner partner. Magnetic tension transfers angular momentum to the outer parcel and removes it from the inner one, making their radial separation increase. A sufficiently stiff magnetic connection suppresses the separation instead. The instability therefore requires a finite range of field-line-bending strengths.

Make this mechanism quantitative in the local shearing sheet of a cylindrical rotation profile $\Omega (r)$. Put $q=-d\log \Omega /d\log r$, and take $0<q<2$ so the unmagnetized flow is stable. Its radial epicyclic frequency is ${\kappa }^2=2(2-q){\Omega }^2$. Let the uniform field be vertical, $B_z{\mathbf{e}}_z$, with constant mass density, and use an axisymmetric horizontal disturbance proportional to $e^{st+ikz}$. Write $\mathbf{h}=\mathbf{b}/\sqrt{{\mu }_0\rho }$ and ${\omega }_A=k{B}_z/\sqrt{{\mu }_0\rho }$. The linearized ideal magnetohydrodynamics equations areThe first pair contains Coriolis acceleration and the epicyclic response; the last equation is the shear-winding of a radial field perturbation into an azimuthal one. Omitting that winding term would remove the driving mechanism.

Eliminating $h_r,h_{\varphi }$ gives $(s^2+{\omega }_A^2)u_r=2\Omega s{u}_{\varphi }$ and $s(s^2+{\omega }_A^2)u_{\varphi }+\Omega [(2-q)s^2-q{\omega }_A^2]u_r=0$. Their determinant is the ideal magnetorotational dispersion relationIts discriminant as a quadratic in $s^2$ is ${\kappa }^4+16{\Omega }^2{\omega }_A^2>0$. Since $2{\omega }_A^2+{\kappa }^2>0$, a positive $s^2$ exists precisely when its constant term is negative:Thus angular velocity, rather than specific angular momentum, must decrease outward. A Keplerian disk has $q=3/2$, ${\kappa }^2={\Omega }^2$ and is hydrodynamically stable, yet magnetic modes with $0<k^2{v}_A^2<3{\Omega }^2$ grow.

The growing root is $s^2=-{\omega }_A^2-{\kappa }^2/2+\frac{1}{2}\sqrt{{\kappa }^4+16{\Omega }^2{\omega }_A^2}$. Differentiation with respect to ${\omega }_A^2$ gives the maximum growth rate of ideal magnetorotational instability:For Keplerian shear these are $3\Omega /4$ and $15{\Omega }^2/16$. The ideal maximum is set by shear; reducing field strength reduces the fastest wavelength instead of reducing this maximum growth rate. This is why the exactly zero-field problem differs from a sequence of weak-field problems with progressively larger wave number. At fixed wave number, growth does tend to zero as the field vanishes. The wavelength must fit the actual disk, and diffusion eventually matters at sufficiently short scales.

The instability has the required angular-momentum-transfer sign. In a growing mode, $u_{\varphi }/u_r=(s^2+{\omega }_A^2)/(2\Omega s)>0$. At the fastest mode this ratio is one, while induction gives $h_{\varphi }=-h_r$ after allowing for their common Fourier phase. Thus the averaged Reynolds stress minus Maxwell stress tensor, $\rho ⟨u_r{u}_{\varphi }-h_r{h}_{\varphi }⟩$, is positive: angular momentum moves outward. Its product with $q\Omega$ is the energy supplied by the background shear. These local mechanisms and coefficient conventions are also checked in the primary MRI stability calculation.

More general fields require separating field-line bending from field gradients and curvature. At leading local order, for a stationary compatible field in an incompressible flow with axisymmetric wavevector $(k_r,0,k_z)$, the bending frequency isPressure elimination projects the radial force by $f=k_z^2/k^2$. The radial momentum equation becomes $s{u}_r-2f\Omega u_{\varphi }=i{\omega }_A{h}_r$; the azimuthal and induction equations retain the form above. Eliminating the amplitudes gives the oblique axisymmetric magnetorotational dispersion relationHere $k_z\ne 0$. The unstable range is $0<{\omega }_A^2<-fd{\Omega }^2/d\log r$. The poloidal magnetic field therefore determines whether an axisymmetric perturbation bends a field line. A locally uniform toroidal magnetic field does not enter $\mathbf{k}\cdot \mathbf{B}$ for axisymmetry, and can contribute magnetic pressure that is projected out of this incompressible leading-order problem. Thus a toroidal component does not suppress the weak-field MRI merely by being present. Conversely a purely toroidal field has no ordinary axisymmetric local bending MRI in this approximation. This does not imply that every globally curved toroidal equilibrium is stable.

A steady general poloidal magnetic field must also be compatible with the equilibrium rotation. The azimuthal ideal magnetohydrodynamic induction equation contains ${\partial }_t{B}_{\varphi }=r{\mathbf{B}}_p\cdot \nabla \Omega$. A stationary axisymmetric equilibrium with no meridional flow consequently requires ${\mathbf{B}}_p\cdot \nabla \Omega =0$, Ferraro's law of isorotation. A radial field crossing cylindrical differential-rotation surfaces would otherwise wind secularly; it cannot simply be substituted into a stationary normal-mode calculation. Boundary conditions and the along-field eigenvalue problem replace a freely chosen local $k$ in nonuniform or bounded equilibria. Strong enough field-line tension can then stabilize all admissible wavelengths.

An explicit axisymmetric effect of nonuniform field strength and curvature is obtained with a purely toroidal field $B_{\varphi }(r){\mathbf{e}}_{\varphi }$, constant mass density and cylindrical rotation. A radial fluid displacement changes the field byThis follows directly from $\mathbf{b}=\nabla \times (\mathbf{\xi }\times \mathbf{B})$ and incompressibility. Its radial curvature force, after absorbing the magnetic-pressure perturbation, is $-2B_{\varphi }{b}_{\varphi }/({\mu }_0\rho r)$. The axisymmetric azimuthal momentum equation is $s{u}_{\varphi }+(2\Omega +r{\Omega }^{\prime })u_r=0$. Combining it with the radial force and then projecting out pressure givesThis is the local Michael criterion for axisymmetric toroidal-field interchange. The new term can either stabilize or destabilize: for $B_{\varphi }\propto r^p$, $\mathcal{D}={\kappa }^2+2(1-p)v_{A\varphi }^2/r^2$. A current-free $B_{\varphi }\propto r^{-1}$ is stabilizing; $B_{\varphi }\propto r$ leaves this axisymmetric interchange coefficient unchanged; a sufficiently steep increase with $p>1$ can destabilize even a centrifugally stable rotation law. That last instability can persist without differential rotation, so its energy source is the magnetic-current equilibrium rather than the ordinary MRI. Nonaxisymmetric current-driven modes are outside the requested essay scope.

For mixed poloidal and toroidal fields, the full induction perturbation contains $(\mathbf{B}\cdot \nabla )\mathbf{\xi }-(\mathbf{\xi }\cdot \nabla )\mathbf{B}-\mathbf{B}\nabla \cdot \mathbf{\xi }$, and the linear Lorentz force contains both $(\nabla \times \mathbf{b})\times \mathbf{B}$ and $(\nabla \times \mathbf{B})\times \mathbf{b}$. These terms explain why gradients and curvature cannot be replaced everywhere by a scalar $k^2{v}_A^2$: they can draw on equilibrium-current energy and couple the radial and vertical motions. Compressibility introduces additional magnetic pressure and magnetosonic responses; a strong toroidal field can alter the dispersion even though it disappeared from the leading incompressible weak-field result.

Finally, a vertically or radially stratified general field also supports part of the equilibrium weight through magnetic pressure. A displaced flux tube carries its frozen-in mass-to-flux relation, so its mass density at the surrounding total pressure can differ from its new environment. This introduces magnetic buoyancy instability and competition with ordinary buoyancy. Axisymmetric interchange displacements can access this energy without an azimuthal wave number, while poloidal tension resists variation along the field. Growth that survives when $d\Omega /dr=0$ must not be attributed to shear-driven MRI. The general-field stability problem therefore combines shear, field-line tension, current/curvature forces, compressibility, buoyancy and boundary constraints; the uniform-field result isolates the basic angular-momentum feedback mechanism.

The steady incompressible shearing sheet equation isThe final term follows from the local tidal potential $-\Omega S{x}^2$. For the given linear velocity, the advective acceleration is $-\alpha \beta (x{\mathbf{e}}_x+y{\mathbf{e}}_y)$, while the Coriolis acceleration is $2\Omega \beta x{\mathbf{e}}_x+2\Omega \alpha y{\mathbf{e}}_y$. Equating components givesThe cross derivatives vanish, so the pressure Hessian of an elliptical shearing-sheet vortex integrates toThe additive constant is fixed by exterior matching or the pressure reference. No zero-pressure free-boundary condition is imposed on an incompressible vortex patch: it is embedded in a surrounding fluid.

Let $k=|k_x|$, $c_s^2=(dP/d\Sigma {)}_0$ and ${\kappa }^2=2\Omega (2\Omega -S)$. Linearize the barotropic closure of a razor-thin disk and shearing sheet about the given state. Write the radial and azimuthal velocity perturbations as $u$ and $v$. Axisymmetry removes advection by the background shear, givingThe softened potential is ${\Phi }_1=-2\pi G{\Sigma }_1{e}^{-kϵ}/k$. Eliminating the velocity components for the compressive branch gives the softened Toomre dispersion relationThe Coriolis/shear combination supplies the radial epicyclic restoring term, pressure supplies the short-wave term, and self-gravity lowers the squared frequency. The full three-variable determinant also has a stationary axisymmetric geostrophic mode, with $u=0$ and azimuthal flow balancing the pressure-plus-gravity gradient. The displayed relation describes the density-wave pair; eliminating by division by $\omega$ must not silently deny that stationary mode.

Write the perturbation velocity as $(u,v)$, the density amplitude as ${\sigma }_1$, and ${\gamma }_d=ϵ\Omega$. Axisymmetry removes advection by the background azimuthal flow, but the radial perturbation advects the Keplerian shear: $u{\partial }_x{U}_{g,y}=-3\Omega u/2$. Combining this with the Coriolis force gives the linearized shearing sheet equationsThe razor-thin disk Poisson kernel supplies the self-gravity term. The $\Omega /2$ coefficient, rather than $2\Omega$, is essential: it includes the perturbed advection of the background velocity.

Put $A=c^2{k}^2-2\pi G{\sigma }_0|k|$ and ${\omega }^2={\Omega }^2+A$. The determinant of the three amplitude equations isExpanding it gives the dust gravitational dispersion relation with gas dragUsing a determinant avoids division by $s$ and retains the neutral/secular branch. The radial epicyclic frequency of this Keplerian shearing sheet is $\Omega$, so the three terms in ${\omega }^2$ represent rotational support, self-gravity and dust pressure.

The secular gravitational instability needs $|k|<k_{\mathrm{crit}}=2\pi G{\sigma }_0/c^2$. A finite radial extent $L$ removes arbitrarily small wavenumbers; the smallest available one is $k_{\min }\sim \alpha /L$, where $\alpha$ is a boundary-dependent constant of order unity. With $H=c/\Omega$ and the dust Toomre parameter,Thus the finite-size secular gravitational criterion is $\boxed{Q\lesssim L/H}$ at the rough accuracy requested. For example, a periodic radial interval has $k_{\min }=2\pi /L$ and gives $Q<L/(\pi H)$ in this idealized model. The numerical prefactor is not universal, but the scaling is. Finite size therefore restores a practical threshold for the otherwise long-wavelength instability. Growth must additionally occur within the disk lifetime, and a global wavelength comparable to disk radius lies beyond the strictly local shearing sheet approximation.