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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 constant-density cylindrical rotation $r\Omega (r){\mathbf{e}}_{\varphi }$ and a purely toroidal magnetic field, an axisymmetric radial displacement gives $b_{\varphi }=(B_{\varphi }/r-B_{\varphi }^{\prime }){\xi }_r$. Combining its curvature force with conservation of angular momentum gives the restoring coefficient $\mathcal{D}$ above. A local incompressible radial-vertical mode has ${\omega }^2=(k_z^2/k^2)\mathcal{D}$. Positive $\mathcal{D}$ gives stability in this ideal interchange setting; negative $\mathcal{D}$ permits growth. This is distinct from weak-field magnetorotational instability: a steeply increasing toroidal field can release current energy even without differential rotation.

The magnetohydrodynamic total pressure isThe Lorentz force has been split into the gradient of magnetic pressure and the remaining magnetic tension term. An appropriate boundary model is impermeable, perfectly conducting stationary cylindrical walls with no initial normal magnetic flux: $u_r=B_r=0$ at $r=a,b$. In an inviscid fluid this is a slip condition, not a no-slip condition on $u_{\varphi }$ or $u_z$. It also makes the tangential ideal electric field vanish at the walls. Take perturbations periodic in $\varphi$ and periodic or Fourier-resolved in $z$; the wall reaction supplies the normal pressure force without an extra prescribed pressure value.

Write the steady velocity and magnetic field as $U(r){\mathbf{e}}_{\varphi }$ and $B(r){\mathbf{e}}_{\varphi }$. They satisfy both divergence constraints. Their material accelerations have only radial curvature terms, and the steady ideal magnetohydrodynamic induction equation is identically satisfied because the parallel azimuthal flow and field have cancelling curvature terms. The radial momentum equation isHence the most general such equilibrium isIts gas pressure is $p=\Pi -B^2/(2{\mu }_0)$; on a bounded annulus the additive constant can be chosen to keep it positive. Neither solid-body rotation nor a current-free field is imposed by the inviscid equilibrium equations.

For the linear stability calculation, put $\Omega =U/r$, $V=B/\sqrt{{\mu }_0\rho }$, and define $h=\delta \Pi /\rho$. Use axisymmetric normal modes proportional to $e^{ikz-i\omega t}$. Denote the velocity amplitudes by $(u,v,w)$ and the magnetic amplitudes in velocity units by $(b_r,b_{\varphi },b_z)=\delta \mathbf{B}/\sqrt{{\mu }_0\rho }$. Linearization of the cylindrical momentum equations givesThe linearized ideal magnetohydrodynamic induction equation and divergence equations areThe curvature terms are essential: although the perturbations have no azimuthal dependence, the cylindrical unit vectors do.

For $\omega \ne 0$, $b_r=b_z=0$. Introduce the radial Lagrangian fluid displacement by $u=-i\omega \xi$. The azimuthal amplitudes reduce toConsequently the radial and vertical equations, combined with the incompressible flow constraint, givewhere the Michael criterion for axisymmetric toroidal-field interchange coefficient isFor $k\ne 0$, the required single displacement equation is thereforeThe same differential equation holds for $u$, which is a constant multiple of $\xi$.

Set $y=r\xi$. Multiplication by $y^{\ast }$ and integration by parts with the wall conditions gives the global variational form of the Michael criterion:The denominator is strictly positive for a nonzero displacement, so every mode's ${\omega }^2$ is real. If $\mathcal{D}\ge 0$ everywhere, no exponentially growing mode exists. Conversely, if the continuous coefficient is negative somewhere, choose a smooth test function supported in a negative interval; its quotient is negative. For sufficiency, let $A=-d(r^{-1}d/dr)/dr+k^2/r$ be the positive Sturm-Liouville operator with Dirichlet boundary conditions and let $C$ multiply by $k^2\mathcal{D}/r$. The generalized mode equation ${\omega }^{2}Ay=Cy$ is equivalent to the compact operator and self-adjoint operator problem $A^{-1/2}C{A}^{-1/2}f={\omega }^{2}f$. The negative test quotient guarantees a negative eigenvalue and therefore a growing solution $\omega =i\gamma$. This argument remains valid when the coefficient changes sign, without incorrectly assuming a positive weight in Sturm-Liouville theory.

Thus, for any fixed nonzero vertical wavenumber,Equality is marginal. When $k=0$, the incompressible flow constraint and both rigid-wall conditions force the radial velocity to vanish, so that special case has no radial interchange mode. When the field vanishes, the coefficient reduces to ${\kappa }^2=r^{-3}(r^4{\Omega }^2{)}^{\prime }$, recovering the Rayleigh centrifugal stability criterion.

For a Keplerian disk, ${\kappa }^2={\Omega }^2$. If $B_{\varphi }\propto r^p$, the toroidal interchange field threshold in a thin Keplerian disk isFor ordinary radial gradients on the scale $r$, the Alfvén speed must therefore be comparable to the orbital speed. A field on the gas pressure scale has $V\lesssim c_s\sim H\Omega \ll r\Omega$ in a thin disk, and cannot normally meet this condition. Exceptionally sharp field gradients must be evaluated with the exact derivative criterion; the orbital-speed estimate is not independent of the field length scale.

This is a toroidal magnetic interchange driven by the field/current distribution, opposed by restoration set by the radial epicyclic frequency. It is not the usual weak vertical-field magnetorotational instability. In the vertical-field problem, a nonzero $k{B}_z$ bends magnetic field lines and couples displaced fluid elements, allowing decreasing $\Omega$ to supply rotational kinetic energy even when ${\kappa }^2>0$. Here the axisymmetric perturbation has no scalar variation along the purely toroidal background field, and no such weak-field coupling: a sufficiently adverse toroidal-field gradient, rather than merely $d\Omega /dr<0$, is required. Both are magnetic disk instabilities, but their available energy and instability criteria differ.

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.

Set $\xi =\tanh (z/H)$, so $h=1-{\xi }^2$ and $d/dz=(1-{\xi }^2)H^{-1}d/d\xi$. The vertical equation becomesFor $|\xi |<1$, division by the common factor gives the Legendre differential equation with eigenvalue $k^2{H}^2$. Requiring boundedness at both surfaces $\xi =\pm 1$ selectsThe second independent solution is unbounded at an endpoint. These are the bounded vertical modes of a sech-squared magnetized disk. The division by $H$ is outside the square root, as in the original PDF; the converted TeX places it incorrectly. The magnetic perturbation involves $F_n^{\prime }$, which tends to zero at either surface.

The constant $n=0$ function has $k=0$. The specified magnetic ansatz contains $1/(ik)$ and must not be applied to it. A uniform horizontal velocity with no magnetic perturbation is an epicyclic motion, not a growing magnetic mode; the zero-wavenumber formal degeneracy does not provide growing MRI at arbitrary field strength. Nontrivial magnetically coupled vertical modes have $n\ge 1$.

For $A>0$, the magnetorotational instability grows when $A<3{\Omega }^2$. Indeed, the quadratic in $s^2$ then has a negative constant term, and one positive root. For $A\ge 3{\Omega }^2$, the two roots are nonpositive, since their sum is negative, their product nonnegative, and their discriminant is ${\Omega }^4+16{\Omega }^{2}A>0$. The smallest nonzero vertical eigenvalue is $k_1^2=2/H^2$, so all vertical modes are stable ifThe strict inequality requested gives stability, and equality is marginal for $n=1$. A sufficiently strong field increases magnetic tension. The unstable MRI needs a sufficiently long vertical wavelength to exchange angular momentum without excessive restoring tension; bounded vertical structure imposes a smallest nonzero effective wavenumber. Beyond the finite-thickness magnetorotational instability criterion, no allowed magnetic mode is long enough to remain unstable.

Using $K^2=2q{\mu }_0\rho {\Omega }^2/B_z^2$, the finite-thickness magnetorotational instability criterion is preciselyFor the midplane-symmetric equilibrium with $B_{+}\ne 0$, the absolute radial magnetic field isIts first maximum occurs at $z_{\ast }=\pi /(2K)$. The instability condition puts $z_{\ast }$ strictly inside the disc: $z_{\ast }<H$. The magnitude rises from zero to this maximum, then decreases on a nonempty interval before the surface. Thus the unstable equilibrium exhibits nonmonotonic magnetic bending, as illustrated below.

The link between bending and instability concerns a nonzero imposed surface inclination. If $B_{+}=0$, the unbent equilibrium $B_x=0$ can still have magnetorotational instability; the literal nonmonotonic-bending conclusion does not apply to that degenerate case. Resonances with $\sin (KH)=0$ also require the separate equilibrium analysis in part (b).

The radiative efficiency of black-hole accretion isFor a steady thin disk with negligible stress at the innermost stable circular orbit, matter radiates the binding energy lost before plunging, so $\eta =1-E_{\mathrm{ISCO}}$. Black-hole spin changes both the ISCO radius and its specific orbital energy. A prograde disk around a rapidly rotating Kerr black hole reaches deeper into the potential and is more efficient than a retrograde disk; representative ideal values run from $\eta \simeq 0.057$ for a Schwarzschild hole toward $0.42$ for an extremal prograde Kerr hole, reduced to about $0.3$ when photon capture limits astrophysical spin-up.

In adiabatic Bondi accretion, spherical compression raises the gas's internal energy reversibly, but much of that energy is advected through the horizon. There is no sustained shear stress that converts orbital binding energy into heat at a sequence of radii. A Shakura--Sunyaev thin disk, by contrast, must transport angular momentum outward. Its differential rotation stores free energy, local stress dissipates that energy as heat, and the short cooling time lets an optically thick disk radiate it before accretion. High efficiency therefore requires irreversible heating beyond adiabatic compression.

The likely source is MRI-driven magnetohydrodynamic turbulence. A weak magnetic field couples neighboring annuli; when angular velocity decreases outward, magnetic tension transfers angular momentum outward and amplifies the displacement. The alpha disk prescription replaces the unresolved turbulent stress by $T_{r\varphi }=\alpha P$, or equivalently $\nu \sim \alpha c_{s}H$. It captures the correct dimensional scale because subsonic turbulent motions are bounded by $c_s$ and their largest local eddies by $H$.

Its limitations include the following.

Uniform ${\rho }_0,p_0,$ and ${\mathbf{B}}_0=B_0\hat{\mathbf{z}}$ obey the unperturbed ideal magnetohydrodynamics equations whenIndeed $({\mathbf{v}}_0\cdot \nabla ){\mathbf{v}}_0=0$, while the Coriolis and tidal terms cancel:

For perturbations proportional to $e^{i(\omega t-kz)}$, the horizontal velocity and magnetic perturbations decouple from the compressive variables. Their linear equations areEliminating $B_x,B_y,v_x,v_y$ and writing the Alfvén speed as $v_A^2=B_0^2/(4\pi {\rho }_0)$ givesSince $d{\Omega }_0^2/d\log r=-2q{\Omega }_0^2$, a root has ${\omega }^2<0$ precisely when the constant term is negative:This is the magnetorotational instability criterion.

For a circular Kepler orbit, $q=3/2$. With $K=k^2{v}_A^2$, the unstable branch isMinimizing it givesand henceThe e-folding time is of order the dynamical time, so the growth is rapid: several e-foldings occur in one orbit.

Instability requires $k^2{v}_A^2<3{\Omega }_0^2$, making the critical wavelength of the magnetorotational instabilityFor a thin isothermal disk, $H=c_{s,0}/{\Omega }_0$. Setting ${\lambda }_{\mathrm{crit}}=2H$ gives $v_A^2=3c_{s,0}^2/{\pi }^2$, and thereforeAbove this field strength the shortest unstable vertical MRI wavelength exceeds the full disk thickness, so no such vertical mode fits inside the disk. Magnetic tension then stabilizes this local mode; the result corresponds to a magnetic-to-gas pressure ratio $B^2/(8\pi p)=3/(2{\pi }^2)$.

For $B_{\varphi }\propto r^p$, constant density and Keplerian ${\kappa }^2={\Omega }^2$, the Michael criterion for axisymmetric toroidal-field interchange becomes ${\Omega }^2-2(p-1)v_A^2/r^2$. Instability therefore requires $p>1$ and $v_A^2>r^2{\Omega }^2/[2(p-1)]$. For a field varying on the radial scale $r$, this is an orbital-speed field, much stronger than a gas-pressure-scale field with $v_A\lesssim c_s\sim H\Omega$ in a thin disk. Steep radial field gradients require the exact derivative criterion rather than this scale estimate. The instability is powered by the toroidal field distribution and is distinct from weak vertical-field magnetorotational instability.

// Target: fluid-mechanics.bigb