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The two axial invariants in steady planar ideal-MHD field-line invariants give the displayed identity. When $V-U\ne 0$, a finite smooth flow cannot reach horizontal Alfvén speed. When $V=U$, an everywhere Alfvénic branch is possible even with $B_z\ne 0$: constant $\rho$ and $\mathbf{u}=\mathbf{B}/\sqrt{{\mu }_0\rho }$ give an example. The nondegenerate exclusion must not be applied to this branch.

The Alfvén Mach number compares a characteristic flow speed with the Alfvén speed $v_A=B/\sqrt{{\mu }_0\rho }$. In an axisymmetric magnetohydrodynamic wind, the relevant poloidal quantity is $M_{Ap}=|{\mathbf{u}}_p|\sqrt{{\mu }_0\rho }/|{\mathbf{B}}_p|$. With magnetohydrodynamic mass loading $\rho {\mathbf{u}}_p=\kappa {\mathbf{B}}_p$, its square is ${\mu }_0{\kappa }^2/\rho$. The Alfvén surface is the locus $M_{Ap}=1$, and a smooth crossing requires the corresponding Alfvén-surface regularity condition for an axisymmetric wind. The total-speed and poloidal-speed definitions should be distinguished when there is substantial rotation or a strong toroidal magnetic field.

At $w^2=v_{az}^2$, the transverse steady equations require $w{B}_x{w}^{\prime }=0$ and $w{B}_y{w}^{\prime }=a(w{B}_x-B_z{v}_x)$. These conditions can be lost by division when eliminating transverse magnetic derivatives. A scalar equation displaying the magnetosonic critical speeds need not display every compatibility condition at an Alfvén speed.

The Alfvén velocity is the signed vector ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$, whose magnitude is the Alfvén speed. It follows the orientation of the magnetic field, unlike the nonnegative scalar speed.

Put $c_1^2=\gamma p_1/{\rho }_1$ and $v_{A1}^2=B_1^2/({\mu }_0{\rho }_1)$. The mass, momentum and energy jump conditions for a perpendicular magnetohydrodynamic shock factor into the unchanged state $x=1$ andFor the compressive physical branch $x>1$, positive downstream pressure implies $x<(\gamma +1)/(\gamma -1)$ when $\gamma >1$ and the upstream sound speed or Alfvén speed is nonzero. Indeed the polynomial impliesIt then gives $u_1^2>c_1^2+v_{A1}^2$: the upstream flow exceeds the perpendicular fast magnetosonic wave speed. The exactly cold unmagnetized strong-shock limit reaches the upper compression bound.

In a local isotropic hydrodynamic inertial-range model with only $ϵ$ and $k$, dimensional analysis fixes the Kolmogorov 1941 theory spectrum. A mean magnetic field supplies the independent Alfvén speed, permitting the arbitrary dimensionless function in the displayed formula even before anisotropy adds parallel scales. Cascade-time assumptions such as weak decorrelated interactions or critical balance are therefore necessary to select an MHD spectrum.

For sound speed $v_s$ much larger than Alfvén speed $v_a$, and angle $\theta$ between the wavevector and magnetic field, the fast magnetosonic wave has $v_f^2=v_s^2+v_a^2{\sin }^2\theta +O(v_a^4/v_s^2)$. The slow magnetosonic wave has $v_{\mathrm{slow}}^2=v_a^2{\cos }^2\theta [1-(v_a^2/v_s^2){\sin }^2\theta +O(v_a^4/v_s^4)]$. The fast mode is predominantly longitudinal sound. The slow mode becomes almost incompressible, polarized in the plane of the wavevector and magnetic field; the Alfvén wave is polarized perpendicular to that plane. The latter two are predominantly magnetic-tension waves with equal leading phase speeds.

For a singly charged ion species in Gaussian units, ${\omega }_{pi}=(4\pi n{e}^2/m_i{)}^{1/2}$ is its plasma frequency and $d_i=c\sqrt{m_i/(4\pi n{e}^2)}$. The ion skin depth marks a characteristic scale below which ion and electron motion can separate in a magnetized plasma. Together with the Alfvén speed, it satisfies $v_A{d}_i=c{B}_0/(4\pi en)$, the coefficient of guide-field propagation in electron magnetohydrodynamics.

The Lehnert number compares an Alfvén speed to a rotation speed on length scale $L$. With $\mathrm{Le}\ll 1$, magnetic-wave frequencies are small compared with the rotation frequency, allowing nearly geostrophic flow. Ideal weakly damped magnetic waves also require $\eta /L^2$ and $\nu /L^2$ to be small compared with their wave frequency; a small Lehnert number alone does not guarantee weak diffusion.

A plane-parallel magnetostatic equilibrium with constant downward gravitational acceleration $g$, horizontal magnetic field, constant isothermal sound speed $c_s$, and constant Alfvén speed $v_a$ hasIndeed the magnetohydrodynamic total pressure is $(c_s^2+v_a^2/2)\rho$, and its vertical derivative is $-g\rho$. The magnetic pressure therefore increases the scale height above that of an unmagnetized atmosphere.

Neighboring magnetic surfaces with different Alfvén speeds carry the same axial waveform at different rates. For a traveling profile $f(R,z+v(R)t)$, differentiation produces the displayed term, so transverse gradients grow even without axial wave steepening. Small magnetic diffusivity or viscosity can eventually act strongly on these gradients. The ideal process redistributes phase rather than directly dissipating wave energy.

Stellar rotation is a problem of coupled force balance, heat transport and angular momentum transport. A star need not rotate as a rigid body: in the Sun, the convective envelope rotates faster near the equator than near the poles, while much of the radiative interior rotates more nearly uniformly. The tachocline is the thin transition near the base of the convection zone. Surface tracking and Doppler effect measurements constrain the exterior flow; solar rotational mode splitting in helioseismology constrains interior stellar rotation through integrals of the form $\delta {\omega }_{n\ell m}\simeq m\int K_{n\ell m}\Omega dV$. Multiple kernels are needed, so surface rotation is not a measurement of every depth.

A useful mean velocity decomposition is $\mathbf{U}=\varpi \Omega {\mathbf{e}}_{\varphi }+{\mathbf{v}}_m$, where $\varpi =r\sin \theta$ is distance from the axis and ${\mathbf{v}}_m$ is stellar meridional circulation. In a frame of constant angular velocity ${\mathbf{\Omega }}_0$, use the relative velocity $\mathbf{V}=\mathbf{U}-{\mathbf{\Omega }}_0\times \mathbf{r}$. Its momentum equation includesThe Coriolis force changes the direction of a parcel's motion but does no work. Thermal driving, release of differential rotational kinetic energy, contraction, or external torques supply the energy of the flows. Solid-body stellar rotation in a suitable hydrostatic equilibrium does not by itself require a meridional flow.

For slowly circulating, approximately axisymmetric material, the leading mechanical balance in an inertial frame is $\nabla p/\rho =-\nabla \Phi +\varpi {\Omega }^2{\mathbf{e}}_{\varpi }$. Taking its curl gives the stellar thermal-wind balanceHere $z$ is the axial coordinate, $S$ is specific entropy and $c_p$ is the specific heat at fixed pressure. The last form uses approximately radial hydrostatic equilibrium and uniform composition. If the material is barotropic, the right-hand side vanishes and angular velocity is constant on cylinders. Small latitudinal entropy differences can instead support the observed noncylindrical differential rotation. Stresses and inertial terms modify this leading balance. The equation relates shear to baroclinicity; it does not give the circulation speed or explain how the shear is maintained.

That second question requires an angular-momentum budget. With $j={\varpi }^2\Omega$ and density-weighted mean quantities, conservation of angular momentum can be written schematically asThe three contributions are Reynolds stress, magnetic Maxwell stress tensor and viscosity transport. Appropriate density weighting replaces the simple correlations in a strongly compressible average. Since $\nabla \cdot (\rho {\mathbf{v}}_m)=0$ in a steady star, the stationary equation gives gyroscopic pumping in a star:A torque that maintains differential rotation therefore drives a meridional response unless another stress cancels it. Conversely the circulation transports angular momentum and feeds back on the rotation. An assumed meridional pattern without its driving torque is not a closed dynamical model.

In the convection zone, outward heat transport drives convection. Rotation deflects the rising and falling parcels and makes their correlations anisotropic. Nondiffusive convective angular momentum transport can drive shear even from an initially uniform rotation rate; treating every Reynolds stress as a positive eddy viscosity would only smooth shear and miss its source. A limiting parcel that conserves $j$ when it moves outwards slows relative to uniform rotation, whereas rapid rotational constraint and correlated horizontal motions change the sign and direction of the net transport. Thus the solar equator-fast pattern requires the actual convective correlations, not a universal parcel argument. The Rossby number $u_c/(2\Omega {\ell }_c)$ compares convective inertia with rotational deflection: deep, slower, larger-scale convection is more rotationally constrained than the fastest photospheric motions. Turnover times ${\ell }_c/u_c$ range from minutes near the visible surface to weeks or longer at greater depth. On these times the Reynolds stresses can redistribute angular momentum and drive circulation, while the star's total spin changes much more slowly.

In a radiative region, rotation distorts equipotential surfaces. Under uniform rotation and uniform composition, leading hydrostatic equilibrium makes pressure, density and temperature constant on each such surface, but effective gravity varies along it. Radiative diffusion then produces a flux $\mathbf{F}=f({\Phi }_{\mathrm{eff}})\nabla {\Phi }_{\mathrm{eff}}$. Its divergence contains $f^{\prime }|\nabla {\Phi }_{\mathrm{eff}}{|}^2$, which varies over a surface and generally cannot match a nuclear source constant there. This is the radiative-equilibrium obstruction in a rotating barotropic star. Thermal imbalance drives Eddington-Sweet circulation, with heat advection enteringIts global thermal estimate isA rotational thermal imbalance of order ${ϵ}_{\Omega }L$ processes the thermal reservoir on this longer time. For the slowly rotating Sun, $t_{\mathrm{KH}}$ is of order $3\times 10^7$ years and ${ϵ}_{\Omega }$ of order $2\times 10^{-5}$, giving $t_{\mathrm{ES}}\sim 10^{12}$ years, much longer than the solar age. Stable composition gradients impede the flow further. Thus classical radiative Eddington-Sweet circulation cannot by itself establish the Sun's interior rotation rapidly. In a rapidly rotating star the same estimate is shorter and rotational mixing can compete with its evolutionary lifetime. Torque-driven convection-zone circulation and radiative thermal circulation have different controlling balances.

Magnetic fields introduce a further coupling. A poloidal magnetic field linking different layers is wound by differential rotation into a toroidal magnetic field, roughly ${\partial }_t{B}_{\varphi }\simeq \varpi {\mathbf{B}}_p\cdot \nabla \Omega$. The resulting Maxwell stress tensor transport angular momentum. Communication along a field occurs on an Alfvén speed crossing time $t_A\sim \ell \sqrt{{\mu }_0\rho }/B$, which can be far shorter than molecular diffusion if a coherent connecting field exists. In steady ideal axisymmetric induction, absence of continuing winding requires ${\mathbf{B}}_p\cdot \nabla \Omega =0$; this is constancy along the field, not automatically solid-body rotation throughout the star. The topology and strength of an interior field matter. Solar dynamo action, especially the interaction of shear with poloidal-field regeneration, couples stellar rotation to the magnetic cycle. Magnetic feedback can change shear and circulation on dynamical or cycle times; an approximately eleven-year activity cycle and twenty-two-year magnetic polarity cycle are distinct from secular spin-down.

A magnetized stellar wind also supplies an external torque. It enforces approximate corotation out to an effective Alfvén radius $R_A$, giving wind-driven magnetic braking of a solar-type star $\dot{J}\sim -{\dot{M}}_w\Omega R_A^2$. Its spin-down time is $t_J\sim I/({\dot{M}}_w{R}_A^2)$, much longer than a convective turnover time. For slowly changing moment of inertia and an unsaturated field scaling that gives $\dot{J}=-K{\Omega }^3$, integration gives ${\Omega }^{-2}={\Omega }_0^{-2}+2Kt/I$, the Skumanich rotation law. A growing or contracting star also changes $I$ and can spin up or down even without torque. Differential spin-down can create shear between layers unless internal transport couples them.

Finally, shear can excite Kelvin-Helmholtz instability, and convection can launch internal gravity waves into the stable radiative interior. Wave damping deposits angular momentum; the net transport depends on excitation and selective absorption and is not fixed by a wave-crossing time alone. Such stresses, magnetic coupling and circulation are candidate agents for the nearly uniform radiative rotation and the narrow tachocline. Molecular kinematic viscosity has a diffusion time ${\ell }^2/\nu$ generally too long to couple the entire solar interior over its age, whereas a turbulent effective kinematic viscosity can act much faster but is not appropriate to every stable layer. A circulation has an advection time $\ell /U_m$, thermal diffusion has time ${\ell }^2/\chi$, and magnetic diffusion has time ${\ell }^2/{\eta }_m$; these must be compared with the local stellar rotation and evolutionary times rather than conflated with them. The solar dynamical time $\sqrt{R^3/(GM)}$ is about half an hour, its rotation period is of order a month, and secular magnetic braking takes an evolutionary time. Rapid force adjustment, convective maintenance of shear, cyclic magnetic feedback and slow global spin evolution are separate processes. Their coupled conservation equations and timescale hierarchy are the basis of a consistent account of solar 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.

Continue on the branch of steady planar ideal-MHD field-line invariants, with $k\ne 0$ and positive $\rho ,T$. Adiabatic entropy transport gives $\mathbf{u}\cdot \nabla s=0$. As all quantities are $z$-independent,The axial component of the ideal magnetohydrodynamic momentum equation has no pressure or gravitational term. Magnetic tension givesSince $k$ is constant along a magnetic field line, this becomesSubtracting the two axial invariants yields the useful compatibility relationThe squared ratio of horizontal flow speed to horizontal Alfvén speed is $M_h^2={\mu }_0{k}^2/\rho$; vector equality with the horizontal Alfvén velocity additionally requires $k>0$. If $V-U\ne 0$, the right side is a nonzero constant along the field line, so a regular finite $B_z$ cannot permit $M_h^2=1$ anywhere. This is the nondegenerate exclusion.

However, $B_z\not\equiv 0$ does not imply $V-U\ne 0$. The Alfvénic degeneracy of steady planar MHD is a genuine exception: take constant $\rho ,p,s$, $\Phi =0$, and a constant magnetic field with nonzero horizontal and axial components, and setEvery equation holds, but the horizontal velocity equals the horizontal Alfvén velocity everywhere despite $B_z\ne 0$. Therefore the printed pointwise exclusion needs a nondegeneracy hypothesis, such as $V\ne U$. The general conclusion is the boxed compatibility relation; when $V=U$, it says $(M_h^2-1)B_z=0$ and allows the Alfvénic branch.

Linearizing the ideal magnetohydrodynamic induction equation about a static magnetic field and using $u^{\prime }={\partial }_t\xi$ gives the fluid Lagrangian displacement-generated magnetic field perturbation $b=\nabla \times (\xi \times B)$. The two solenoidal vector field conditions reduce the curl identity toIts divergence vanishes because it is a curl. Directly, the two products of first derivatives in ${\partial }_i{b}_i$ cancel after exchanging their indices, while the remaining terms differentiate $\nabla \cdot \xi$ or $\nabla \cdot B$, both zero. Similarly the integrated linear continuity equation gives ${\rho }^{\prime }=-\nabla \cdot (\rho \xi )$, so $\boxed{{\rho }^{\prime }=-\xi \cdot \nabla \rho }$.

Let $\Pi =p+B^2/(2{\mu }_0)$ be the magnetohydrodynamic total pressure. Magnetostatic equilibrium and the momentum equation in the linearized ideal magnetohydrodynamic equations areThere is no perturbed gravitational potential here because the field is fixed. Multiply by ${\xi }_i^{\ast }$ and integrate. The pressure term vanishes by the assumed boundary cancellation and $\nabla \cdot {\xi }^{\ast }=0$. Thuswhere $\mathcal{M}=-\int {\xi }_i^{\ast }(B_j{\partial }_j{b}_i+b_j{\partial }_j{B}_i)dV$.

Integrate the first magnetic term along $B$, using $\nabla \cdot B=0$, then insert the expression for $b$. This gives the positive term $\int |(B\cdot \nabla )\xi {|}^{2}dV$ and two mixed derivative terms. For clarity, the troublesome term transforms asThe first term on the right cancels the other mixed term. Combining the remaining derivative with the term containing ${\partial }_j{B}_k{\partial }_k{B}_i$ givesDifferentiating the equilibrium relation now combines the gradient of mass density and magnetostatic equilibrium terms:We obtain the incompressible magnetic displacement energy identityThe two Hessian matrices are real symmetric, making their Hermitian form evaluations real. The last term is a squared norm, and $I>0$ for a nonzero displacement in fluid with positive mass density. Thus $\boxed{{\sigma }^2\text{ is real}}$. It may be negative, in which case one temporal branch has exponential growth; real squared frequency is not itself a proof of stability.

For the atmosphere with a horizontal magnetic field, $\Phi =gz$ has zero Hessian matrix and the equilibrium equation isConsequently the interchange stability of an incompressible magnetized atmosphere is governed by the numeratorFor ${\xi }_z=\sin ky$, the magnetic tension term is zero: these fluid Lagrangian displacements interchange neighboring magnetic field lines without varying along them. A mass density increasing with height gives negative energy and instability; decreasing mass density gives positive energy for these vertical interchanges. If ${\rho }_z\le 0$ everywhere, the displayed form is nonnegative for every admitted incompressible displacement. If the gradient changes sign, suitably localized divergence-free interchange trials detect an unstable inverted layer; For example, take ${\xi }_z=f(z)e^{iky}$ and ${\xi }_y=i{f}^{\prime }(z)e^{iky}/k$, with $f$ supported in that layer. Their divergence is zero, and neither component varies along the magnetic field.

For ${\xi }_z=\sin kx$, magnetic field lines are bent. Averaging over a full horizontal wavelength gives the trial Rayleigh quotientunder the stipulated admissible boundary/integrability conditions. It is a trial quotient, not a claim that this displacement is an exact eigenfunction for every profile. mass density decreasing upward is stable here too. For inverted mass density, magnetic tension can stabilize sufficiently short waves along the magnetic field, but does not remove the transverse interchange instability. Locally the two contributions have the familiar scale $-g{\rho }_z/\rho +k^2{v}_A^2$, with the Alfvén speed $v_A=B/\sqrt{{\mu }_0\rho }$.

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.

Put $a=c/(4\pi en)$ and retain terms linear in $\delta \mathbf{B}$. The electron magnetohydrodynamics induction equation becomesHere the curl of $(\nabla \times \delta \mathbf{B})\times B_0\hat{\mathbf{z}}$ reduces to $B_0{\partial }_z\nabla \times \delta \mathbf{B}$, since the background magnetic field is constant and the divergence of a curl vanishes. For a plane wave $\delta \mathbf{B}=\mathbf{b}{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, the linear equation and the solenoidal constraint areOn the transverse plane, the operator $i\mathbf{k}\times$ has eigenvalues $\pm k$ by the helicity decomposition of a transverse Fourier mode. Equivalently, squaring the equation and using the vector triple product gives ${\omega }^2=a^2{B}_0^2{k}_{\parallel }^2{k}^2$. Since $a{B}_0=v_A{d}_i$, where $v_A$ is the Alfvén speed and $d_i$ the ion skin depth,The two branches have opposite circular polarizations. The specified electron-only induction model describes a whistler wave; the paper calls these branches kinetic Alfvén waves. The following cascade calculation uses the specified model and its dispersion relation, without adding the pressure response needed for the usual kinetic Alfvén interpretation.

Let the equilibrium ideal gas be at rest, with uniform ${\rho }_0,p_0$ and ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$. This local wave calculation neglects gravity, viscosity and magnetic diffusion. Define the adiabatic sound speed $c\equiv v_s=(\gamma p_0/{\rho }_0{)}^{1/2}$, the Alfvén speed $v_a=B_0/({\mu }_0{\rho }_0{)}^{1/2}$, and a wavevector $\mathbf{k}=k(\sin \theta ,0,\cos \theta )$, where $k=|\mathbf{k}|$ and $\theta$ is its angle to the equilibrium magnetic field. Perturbations have dependence $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$ and phase speed $v=\omega /k$.

The linearized ideal magnetohydrodynamic equations for mass conservation, pressure, induction and momentum giveThe relation between pressure and mass density follows from eliminating $\mathbf{k}\cdot {\mathbf{u}}^{\prime }$ between the two material evolution equations; it applies to the nonzero-frequency wave sector. The induction result automatically has $\mathbf{k}\cdot \mathbf{b}=0$.

The $y$ MHD wave polarization is perpendicular to both the field and wavevector. It has $\delta p=\delta \rho =0$ and obeysThus the transverse Alfvén wave hasIn the $x$-$z$ plane the two remaining velocity components instead obey the symmetric matrix eigenproblemTaking its determinant gives $v^4-(c^2+v_a^2)v^2+c^2{v}_a^2{\cos }^2\theta =0$. Solving this quadratic equation in $v^2$ derives both magnetoacoustic wave phase speeds,The plus sign is the fast magnetosonic wave, the minus sign the slow magnetosonic wave. The radicand is nonnegative, since it equals $[(c^2-v_a^2{)}^2+4c^2{v}_a^2{\sin }^2\theta ]/4$. Both squared speeds are nonnegative. Each nonzero mode has forward and backward propagation; an advected entropy disturbance is a separate zero-frequency mode in this static equilibrium.

Set $ϵ=v_a^2/c^2\ll 1$. Expanding the square root in part (a) givesHence, to first absolute order in $ϵ$,The high-pressure limit of magnetosonic waves can also be expressed as nonnegative speeds. Retaining the next squared-speed term gives the first relative correction for the slow mode:The exact Alfvén speed along the propagation direction remains $v_a|\cos \theta |$.

The fast magnetosonic wave is predominantly an acoustic wave, with velocity almost parallel to $\mathbf{k}$ and appreciable mass density and gas pressure variations. Its magnetic pressure correction is weak. The Alfvén wave is exactly incompressible, polarized perpendicular to the $\mathbf{k}$-${\mathbf{B}}_0$ plane and restored by magnetic tension.

The slow magnetosonic wave is polarized inside that plane and almost perpendicular to $\mathbf{k}$. Indeed, the leading sound speed part of the matrix in (a) annihilates a vector perpendicular to $\mathbf{k}$, so only the smaller magnetic restoring force acts on that MHD wave polarization. Its compression is smaller by order $ϵ$, with gas pressure changes almost cancelling magnetic pressure changes. It is often called the pseudo-Alfvén wave MHD wave polarization in this high pressure limit. Thus two distinct almost-incompressible magnetic MHD wave polarizations have the same leading speed, while the third is predominantly sound. For exactly perpendicular propagation the Alfvén wave and slow magnetosonic wave frequencies vanish; for parallel propagation the two transverse magnetic MHD wave polarizations are degenerate.

Use the standard uniform-density incompressible flow version of ideal magnetohydrodynamics, in which $\nabla \cdot \mathbf{u}=0$ and pressure enforces that constraint. The finite-$\gamma$ compressible adiabatic pressure equation is replaced by this pressure constraint. The consequences of retaining that cover-sheet equation literally as an additional condition are given below.