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Start in an inertial frame with the inviscid magnetohydrodynamic momentum equation,Here $\rho$ is mass density, $P$ is the isotropic fluid pressure, $\mathbf{v}$ is inertial-frame velocity, $\mathbf{J}$ is electric current density, $\mathbf{B}$ is the magnetic field, and $\Phi$ is gravitational potential per unit mass. In the low-frequency approximation to Ampère's law, $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$. The vector calculus identityseparates the Lorentz force density into magnetic tension and a magnetic pressure gradient. Transform the material acceleration to a rigid frame rotating with constant angular velocity ${\mathbf{\Omega }}_0$, using the stated frame transformation. The rotating-frame magnetohydrodynamic momentum equation isThe final terms are the Coriolis force and centrifugal force densities; $\mathbf{u}$ is the velocity relative to the rotating frame, and $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ uses that frame.

This requires a smooth conducting-fluid continuum with nonrelativistic motion, a constant magnetic permeability ${\mu }_0$, the solenoidal magnetic-field constraint, a negligible displacement current and negligible net electric-charge force. Viscous forces and anisotropic fluid stresses are omitted. A time-dependent frame would add an Euler force. Incompressible flow is not needed for this force equation. Finite electrical resistivity does not change its displayed instantaneous magnetic force; the later frozen-field calculation additionally uses ideal magnetohydrodynamics. A differentially rotating fluid does not define a single rigid frame with spatially varying ${\mathbf{\Omega }}_0$: choose a fixed local reference rate and retain the fluid's shear in $\mathbf{u}$.

The equation describes small plane-wave perturbations of a uniform, stationary plasma in ideal magnetohydrodynamics, with propagation parallel to the background magnetic field. The nonzero field defines $\hat{\mathbf{b}}={\mathbf{v}}_A/v_A$; $c_s$ is the sound speed for the assumed thermodynamic closure. Neglect viscosity, electrical resistivity, gravity and microscopic dispersive effects. A uniform background flow would instead introduce its Doppler-shifted frequency.

Decompose $\mathbf{u}={\mathbf{u}}_{\perp }+u_{\parallel }\hat{\mathbf{b}}$. The displayed matrix equation then becomesThus there are two independent transverse Alfvén wave polarizations with ${\omega }^2=k^2{v}_A^2$, restored by magnetic tension, and a longitudinal acoustic mode with ${\omega }^2=k^2{c}_s^2$, restored by pressure. The transverse modes have no first-order density compression; a strictly longitudinal displacement along the field does not bend it. These are the parallel-propagating ideal magnetohydrodynamic wave branches. In the general oblique problem, fast and slow magnetosonic modes mix compression and magnetic response; their parallel limits coincide with the faster and slower of these speeds. If $c_s=v_A$, all three velocity polarizations share the same frequency magnitude. The velocity equation alone does not enumerate nonpropagating entropy perturbations.

An Alfvén wave is a transverse disturbance restored by magnetic tension. In a uniform conducting fluid with background ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$, a small transverse displacement ${\mathbf{\xi }}_{\perp }(z,t)$ bends the magnetic field lines. Magnetic flux freezing gives ${\mathbf{b}}_{\perp }=B_0{\partial }_z{\mathbf{\xi }}_{\perp }$, and the linear transverse Lorentz force density is $(B_0/{\mu }_0){\partial }_z{\mathbf{b}}_{\perp }$. Thus the Linear Alfvén wave equations reduce toFor this shear polarization, $\delta \rho =\delta p=0$ at linear order. The disturbance transports energy along the background magnetic field; a single traveling wave has equal perturbation magnetic energy and kinetic energy, with ${\mathbf{b}}_{\perp }=\mp \sqrt{{\mu }_0\rho }{\mathbf{u}}_{\perp }$ for propagation along $\pm {\mathbf{B}}_0$. These properties distinguish it from compressive magnetosonic waves.

The ideal magnetohydrodynamics approximation requires strong electrical coupling and wavelengths large compared with microscopic scales. Finite electrical resistivity and viscosity dissipate the wave's energy. In stellar winds and other astrophysical plasmas, Alfvén waves carry disturbances and can contribute to heating when dissipated. The Lundquist liquid-metal Alfvén wave experiment provides a laboratory counterpart: a conducting liquid in a strong applied magnetic field transmits a transverse magnetic disturbance, making the field's mechanical restoring action measurable.

A uniform conducting fluid supports three propagating magnetohydrodynamic wave branches. Ideal magnetohydrodynamics couples fluid velocity to the magnetic field through the induction equation; the Lorentz force then supplies magnetic tension and magnetic pressure restoring forces. Let $\theta$ be the angle between wave vector $\mathbf{k}$ and background field ${\mathbf{B}}_0$, and define Alfvén speed $v_A=B_0/\sqrt{{\mu }_0{\rho }_0}$ and adiabatic sound speed $c_s^2=\gamma p_0/{\rho }_0$.

The shear Alfvén wave hasIts velocity perturbation is perpendicular to both $\mathbf{k}$ and ${\mathbf{B}}_0$. It bends field lines without first-order compression, so magnetic tension is the restoring force. Its group velocity follows the background field; the displayed phase speed along $\mathbf{k}$ is not the full group speed.

The fast magnetosonic wave and slow magnetosonic wave involve coupled compression and magnetic perturbations, with displacement in the plane of $\mathbf{k},{\mathbf{B}}_0$:The plus sign is the fast branch. Parallel propagation gives characteristic speeds $c_s,v_A$; perpendicular propagation gives fast speed $\sqrt{c_s^2+v_A^2}$ while the other two phase frequencies vanish. Degenerate directions require care in assigning polarizations, rather than claiming three distinct nonzero frequencies everywhere.

These waves transport energy and momentum through magnetized stellar atmospheres and astrophysical plasmas. Their restoring mechanisms distinguish them from ordinary acoustic waves. Resistivity, viscosity, ion-neutral friction and kinetic effects introduce damping or change the model; an ideal undamped dispersion relation alone does not calculate coronal heating.

A plasma contains mobile charged species whose collective electromagnetic response matters. On scales large compared with the Debye length, positive and negative charge densities nearly cancel: this is quasineutrality, not the absence of electric currents or electric fields. For electrons of temperature $T_e$ and number density $n_e$, the characteristic plasma screening and oscillation scales areA collective plasma description requires many particles within a screening volume; long-wavelength, sufficiently slow behavior can often be described by magnetohydrodynamics with an appropriate fluid closure.

An individual particle feels the Lorentz force $q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. In a uniform magnetic field, its transverse motion gyrates while its parallel motion remains free, giving a helical path when both components are present. The nonrelativistic cyclotron frequency and gyroradius areWhen the system scale is much larger than $r_L$ and the relevant time scale longer than ${\Omega }_c^{-1}$, the field strongly organizes transport across it, provided collisions do not interrupt the gyromotion. A perpendicular electric field also gives the common guiding-centre drift ${\mathbf{v}}_E=\mathbf{E}\times \mathbf{B}/B^2$.

For a conducting fluid, the macroscopic Lorentz force density has the useful decompositionThese terms are magnetic pressure and magnetic tension. Magnetic tension resists bending, while magnetic-pressure gradients resist nonuniform compression. The plasma beta $\beta =2{\mu }_{0}p/B^2$ compares thermal and magnetic stresses.

With uniform magnetic diffusivity ${\eta }_m$, the resistive induction equation isIf the magnetic Reynolds number $R_m=UL/{\eta }_m\gg 1$, ideal magnetohydrodynamics applies approximately and magnetic flux freezing links the field geometry to fluid motion. Finite diffusivity permits field slippage and magnetic reconnection. Field tension supports Alfvén waves with $v_A=B/\sqrt{{\mu }_0\rho }$; coupled magnetic and gas compression also supports magnetosonic waves. These effects allow a field to transmit forces and angular momentum between distant parts of an astrophysical plasma.

This is magnetic flux freezing in ideal magnetohydrodynamics. In addition to the supplied ideal magnetohydrodynamic induction equation, a physical magnetic field obeys Gauss's law for magnetism, $\nabla \cdot \mathbf{B}=0$.

Parameterize the material surface by two particle labels $a,b$, so $d\mathbf{S}=({\partial }_a\mathbf{x}\times {\partial }_b\mathbf{x})dadb$ and $D\mathbf{x}/Dt=\mathbf{u}$. Differentiating these tangents gives the material surface element evolution lawwhere $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. Thus the material magnetic-flux transport theorem isExpand the curl in the ideal magnetohydrodynamic induction equation:Since $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ is the material derivative, the first three terms cancel the corresponding terms in the surface transport formula. The remaining integrand is $\mathbf{u}\nabla \cdot \mathbf{B}=0$. ThereforeNo incompressible flow assumption is needed: the changing surface area cancels the compression term. This proof assumes smooth velocity and magnetic field and a surface carried with the fluid; the ideal magnetohydrodynamic induction equation by itself, for an arbitrary non-solenoidal vector field, would leave the final divergence term.

Assume the stated outward-decreasing pressure, $L_p>0$, and setFor $|k_{\parallel }|<k_c$, the interchange instability hasAn outward-displaced flux tube carries higher pressure than its new surroundings. In the unfavorable magnetic field-line curvature of the Z-pinch, the resulting imbalance drives it farther outward. Parallel variation bends field lines, and magnetic tension supplies the stabilizing $k_{\parallel }^2{v}_A^2$ term. The fastest local interchange is at $k_{\parallel }=0$, with $\gamma ={\omega }_I$; this endpoint is understood as the limit of the stated Alfvénic ordering. There is no leading $k_y$ dependence once $k_y\gg k_{\parallel }$.

For $|k_{\parallel }|>k_c$, the two real frequencies are $\omega =\pm \sqrt{k_{\parallel }^2{v}_A^2-{\omega }_I^2}$: the displacement oscillates as a curvature-modified Alfvén wave. At $|k_{\parallel }|=k_c$, the leading equation is ${\ddot{\xi }}_x=0$, so the marginal solutions are a constant and a term linear in time; neglected corrections determine finer behavior. For a favorable reversed pressure gradient, $L_p<0$, this drive is stabilizing.

The leading dispersion relation is the same in ideal magnetohydrodynamics. Its adiabatic law gives $\delta p=-{\xi }_x{p}_0^{\prime }-\Gamma p_0\nabla \cdot \mathbf{\xi }$; the compressive part is smaller than the steep-gradient advection by $O(L_p/R)$ or $O(\beta )$, giving the same leading $\delta p=-{\xi }_x{p}_0^{\prime }$. Relaxing those orderings introduces compression, field-strength and field-length changes, and pressure anisotropy; the drift-kinetic equation can then introduce parallel kinetic phase mixing and Landau damping. At larger plasma beta, coupling to compressive branches matters; if pressure anisotropy develops, mirror instability and firehose instability may also become relevant. These are possibilities beyond the derived leading equation, not additional roots already established here.

The wind integrals require a steady axisymmetric magnetohydrodynamic wind in ideal magnetohydrodynamics, with no viscous azimuthal torque or imposed toroidal loop voltage. Axisymmetry by itself is insufficient. Assume smooth connected poloidal flux surfaces with nonzero poloidal magnetic field; functions of the flux label below are understood on each such branch. In cylindrical coordinates, the given poloidal magnetic flux function obeysFor a steady ideal wind, the electric field is $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$ and $\nabla \times \mathbf{E}=0$. Its toroidal component satisfies ${\partial }_z{E}_{\varphi }=0$ and ${\partial }_s(s{E}_{\varphi })=0$, so $E_{\varphi }=C/s$. Regularity at the axis, or equivalently the absence of an imposed loop voltage in an annular domain, gives $C=0$. Thus $({\mathbf{u}}_p\times {\mathbf{B}}_p{)}_{\varphi }=0$ and ${\mathbf{u}}_p=f{\mathbf{B}}_p$. The continuity equation and $\nabla \cdot {\mathbf{B}}_p=0$ now yieldConsequently $\kappa$ is the magnetohydrodynamic mass loading. Including the toroidal magnetic field, rewrite the full velocity asSince $\mathbf{u}\times \mathbf{B}=\omega \nabla \chi$, steady ideal induction gives $0=\nabla \omega \times \nabla \chi$. Hence $\omega =\omega (\chi )$, establishingThe invariant $\omega$ is the field-line angular velocity of an axisymmetric wind, not generally the fluid's angular velocity $\Omega$. When an ideal magnetic surface is anchored to a rigidly rotating star, $\omega$ equals the stellar angular velocity. With no poloidal outflow this reduces to Ferraro's law of isorotation.

To calculate the magnetic torque, use the azimuthal momentum equation. Axisymmetric pressure and Newtonian gravity have no azimuthal components; the magnetic pressure has none either. The azimuthal acceleration and magnetic tension contain the curvature terms characteristic of cylindrical coordinates:Multiplication by $s$ gives the Maxwell torque conservation in an axisymmetric wind equationUsing $u_{\varphi }=s\Omega$, $\rho {\mathbf{u}}_p=\kappa {\mathbf{B}}_p$ and ${\mathbf{B}}_p\cdot \nabla \kappa =0$ provesThis is the total angular momentum transport per unit poloidal magnetic flux. The first term is the material contribution and the second is the contribution of the Maxwell stress tensor. The specific magnetohydrodynamic angular-momentum invariant is $L=\ell /\kappa$ when $\kappa \ne 0$; its normalization differs from the paper's $\ell$.

Define the poloidal Alfvén Mach number bySubstituting $\Omega =\omega +\kappa B_{\varphi }/(\rho s)$ into $\ell$ givesAt a smooth crossing of the Alfvén surface, $M_A^2=1$ and the toroidal magnetic field is finite. The Alfvén-surface regularity condition for an axisymmetric wind is thereforeOne can also expose the apparent singularity explicitly:Finite passage through the Alfvén surface requires the numerator to vanish with the denominator. Equality of the numerator zeros is necessary; a globally smooth wind must also satisfy its other dynamical critical conditions. Here $s_A$ is the cylindrical distance from the rotation axis, not automatically the spherical Alfvén radius.

For a tube carrying outward mass flux $d\dot{M}=\kappa {\mathbf{B}}_p\cdot d\mathbf{S}$, the outward angular momentum flux isThus the stellar torque is ${\dot{J}}_{\star }=-\int \omega s_A^{2}d\dot{M}$, with the integral over the escaping wind. Magnetic stresses give each mass element an effective angular-momentum lever arm equal to the cylindrical Alfvén crossing radius. In a strongly sub-Alfvénic region the fluid approximately corotates with the magnetic surface; beyond that region it can carry the accumulated angular momentum away. For an outward field and positive $\omega$, a super-Alfvénic expanding tube with $s>s_A$ has $B_{\varphi }<0$, so the magnetic stress transports positive angular momentum outwards. If $s_A$ greatly exceeds the stellar surface lever arm, a modest mass-loss rate can cause substantial wind-driven magnetic braking of a solar-type star. An unmagnetized outflow would instead carry approximately the material angular momentum fixed at its launch radius.

Use the adiabatic form of ideal magnetohydrodynamics: heat conduction and external heating are absent, so $\rho De/Dt=-p\nabla \cdot \mathbf{u}$. Infinite electrical conductivity and zero viscosity alone would not remove the thermal source terms on the cover page. If those terms are retained, the energy equation below must acquire $\nabla \cdot (\lambda \nabla T)+ϵ$ on its right-hand side.

Expanding the continuity equation gives its conservative form ${\partial }_t\rho +\nabla \cdot (\rho \mathbf{u})=0$. Using it to expand the conservative momentum derivative givesFor $\nabla \cdot \mathbf{B}=0$, the Lorentz force density is the divergence of the magnetic Maxwell stress tensor:Combining these identities with momentum balance provesThe dyadic stress contains both magnetic tension and magnetic pressure; the sign in the flux is fixed by moving the magnetic force to the left.

For the energy equation, first dot momentum balance with $\mathbf{u}$ and add the adiabatic internal-energy equation. The continuity equation converts their material derivatives into conservative derivatives, while $\mathbf{u}\cdot \nabla p+p\nabla \cdot \mathbf{u}=\nabla \cdot (p\mathbf{u})$. The result isThe ideal magnetohydrodynamic induction equation gives the magnetic counterpart. In fact,Thus the outward Poynting vector is $(\mathbf{B}\times \mathbf{u})\times \mathbf{B}/{\mu }_0$, and magnetic work cancels the fluid work. Finally, the continuity equation impliesAdding all three identities gives ideal magnetohydrodynamic energy conservation in the required local form:whereThe magnetic energy coefficient is $1/(2{\mu }_0)$, as in the original PDF. For a self-consistent, time-dependent gravitational field this local balance should not be confused with a global self-gravitating energy integral: the latter uses gravitational field energy, or the familiar one-half factor in $\int \rho \Phi$ after the field is eliminated.

Work in the shock frame of a thin planar discontinuity, with normal speed $u>0$. A smooth gravitational potential has no finite jump through a vanishingly thin front; its advected contribution cancels using the common mass flux. Integrating the conservation laws gives constant normal mass, momentum and energy fluxes, together with the normal-field and tangential-electric-field conditions of the ideal magnetohydrodynamic shock conditions.

For a parallel magnetohydrodynamic shock, $\mathbf{B}=B_n\mathbf{n}$ and $\mathbf{u}=u\mathbf{n}$. The magnetic solenoidal constraint makes $B_n$ the same on both sides. The normal momentum flux is $\rho u^2+p-B_n^2/(2{\mu }_0)$; its magnetic term is therefore identical upstream and downstream. The Poynting vector is zero because $\mathbf{u}\times \mathbf{B}=0$. Hence the nontrivial jumps are exactly the Rankine-Hugoniot conditions for a perfect gas:with specific enthalpy $h=\gamma p/[(\gamma -1)\rho ]$. Physically, motion along the straight field neither bends it nor compresses its transverse flux. The field consequently provides no changed stress or energy flux in this purely parallel geometry. This conclusion does not apply to a shock that develops a transverse magnetic field.

For a perpendicular magnetohydrodynamic shock, the magnetic field is tangential and the velocity normal. Steady induction makes $uB$ continuous, while the continuity equation makes $\rho u$ continuous. Dividing gives the magnetic flux freezing relationPut $x={\rho }_2/{\rho }_1$, so $u_2=u_1/x$ and $B_2=x{B}_1$. Constant normal momentum flux now reads $\rho u^2+p+B^2/(2{\mu }_0)$. Substitution givesThe magnetic correction is negative for compression: part of the incoming ram pressure builds magnetic pressure instead of thermal pressure.

Here the Poynting vector has normal component $u{B}^2/{\mu }_0$. Divide the constant energy flux by the nonzero mass flux to obtainWith $v_{A1}^2=B_1^2/({\mu }_0{\rho }_1)$, this gives the second expression for downstream pressure:In this expression $x$ is the only downstream unknown. Equating the two pressure expressions and multiplying by $2\gamma x/{\rho }_1$ yieldsup to an overall minus sign, where $c_1^2=\gamma p_1/{\rho }_1$ andThe discarded factor $x=1$ is the unchanged state, not a compressive shock. This derives the polynomial for the compression ratio of a perpendicular magnetohydrodynamic shock without omitting its trivial branch.

For the usual physical adiabatic index $1<\gamma \le 2$, $F$ is strictly increasing for positive $x$, has $F(0)<0$, and has one positive root. Sincethat root is greater than one precisely when $u_1^2>c_1^2+v_{A1}^2$. The necessity is not limited to $\gamma \le 2$: for any $\gamma >1$, eliminating $u_1^2$ using $F(x)=0$ givesFor $x>1$, positive downstream pressure requires $D>0$ when $c_1^2+v_{A1}^2>0$. Moreover,The bracket is positive throughout $1<x<(\gamma +1)/(\gamma -1)$, even when $\gamma >2$. The exactly cold unmagnetized limit gives the familiar strong-shock upper compression ratio directly. Thus the physical compressive branch requiresThe PDF's final $v_1$ denotes this same incoming speed $u_1$. The right side is the squared speed of a perpendicular fast magnetosonic wave: the upstream flow must be super-fast-magnetosonic so that this compressive signal cannot run upstream against the incoming flow. At equality the shock amplitude vanishes into a characteristic disturbance.

Write horizontal components with a subscript $h$. The steady ideal magnetohydrodynamic induction equation gives $\nabla \times (\mathbf{u}\times \mathbf{B})=0$. Since ${\partial }_z=0$, its horizontal components implyThus $(\mathbf{u}\times \mathbf{B}{)}_z=C$ is a spatial constant. A vanishing axial electric field, $E_z=-C=0$, is needed to infer horizontal alignment. It is satisfied, for example, if the constant electric field vanishes at a boundary or at infinity. With this additional condition and ${\mathbf{B}}_h\ne 0$, define the magnetohydrodynamic mass loading $k$ by $\rho {\mathbf{u}}_h=k{\mathbf{B}}_h$, and put $U=u_z-k{B}_z/\rho$. ThenThe steady continuity equation and the solenoidal magnetic-field constraint giveAlso $\mathbf{u}\times \mathbf{B}=U{\mathbf{e}}_z\times \mathbf{B}$. Using the vector identity for the curl of a cross product, together with ${\partial }_z\mathbf{B}=0$ and $\nabla \cdot \mathbf{B}=0$, its curl is ${\mathbf{e}}_z\mathbf{B}\cdot \nabla U$. HenceThese are steady planar ideal-MHD field-line invariants: both quantities are constant along a magnetic field line.

The zero-electric-field condition does not follow from the printed stationarity assumptions alone. Constant $\rho ,p,s$, $\Phi =0$, $\mathbf{u}=u_0{\mathbf{e}}_x$ and $\mathbf{B}=B_0{\mathbf{e}}_y$, with $u_0{B}_0\ne 0$, satisfy every displayed ideal magnetohydrodynamics equation but have perpendicular horizontal velocity and magnetic field. Their $C=u_0{B}_0$ is nonzero, so no $k,U$ give the stated decomposition. The field-line decomposition is valid on the $E_z=0$ branch; as an unrestricted implication, the first claim has this counterexample. The subsequent calculations use that branch.

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.

Use a steady axisymmetric flow governed by ideal magnetohydrodynamics and take the Newtonian gravitational potential to be axisymmetric. Work locally on connected regular magnetic flux surfaces with nonzero poloidal flow. As usual for a wind without an imposed toroidal loop voltage, $E_{\varphi }=0$; regularity on an included rotation axis guarantees this condition. It is needed to exclude steady cross-field flow in an annular domain.

Since $\nabla \varphi ={\mathbf{e}}_{\varphi }/R$, the poloidal magnetic flux function givesThe poloidal magnetic field is divergence-free. The zero toroidal electric field, with $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, gives ${\mathbf{u}}_p\parallel {\mathbf{B}}_p$. Define $k$ by $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$. Steady mass conservation then givesThus $k=k(\psi )$ is the magnetohydrodynamic mass loading, the mass flux per unit poloidal magnetic flux. Subtracting $(k/\rho )\mathbf{B}$ from the full velocity leaves only an azimuthal component. The cross product isThe steady ideal magnetohydrodynamic induction equation requires $0=\nabla \times (\omega \nabla \psi )=\nabla \omega \times \nabla \psi$. Consequently the field-line angular velocity is $\omega =\omega (\psi )$ and

Neither gas pressure nor axisymmetric gravity exerts an azimuthal torque. Multiplying the azimuthal momentum equation by $R$ combines the cylindrical curvature terms into angular-momentum transport:Insert $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$ and use ${\mathbf{B}}_p\cdot \nabla k=0$. It follows thatso the magnetohydrodynamic angular-momentum invariant isThe two terms account respectively for matter and Maxwell stress tensor transport of angular momentum.

Finally, subtracting $\gamma$ times the logarithmic mass density equation from the logarithmic pressure equation gives $D\ln (p/{\rho }^{\gamma })/Dt=0$. The specific entropy of a fixed-composition ideal gas is a function of $p/{\rho }^{\gamma }$. Steadiness and axisymmetry reduce its advection to ${\mathbf{u}}_p\cdot \nabla s=0$. HenceThese flux-label functions are local on connected regular surfaces; disconnected components with the same numerical label need not share constants without an additional matching condition.

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.

Write $F=f(R,z+v(R)t)$, $G=g(R,z-v(R)t)$, $U=F+G$ and $b=F-G$, where $b$ is the toroidal Alfvén velocity. The azimuthal velocity is $U$, while the Alfvén velocity vector is $b{\mathbf{e}}_{\varphi }+v{\mathbf{e}}_z$. Both solenoidal constraints hold automatically. Since there is no meridional flow, the azimuthal magnetohydrodynamic momentum equation and ideal magnetohydrodynamic induction equation reduce toDifferentiating the traveling arguments verifies both equations for arbitrary smooth $f$, $g$ and $v$. In particular, no condition $v^{\prime }=0$ is needed.

The remaining force balance is essential. Define the total pressure per unit mass, including the gravitational potential, byThe radial acceleration is $-U^2/R$, and magnetic curvature tension contributes $-b^2/R$. The radial and axial momentum equations are consequentlyThey are compatible if and only ifThis is the pressure compatibility of nonlinear torsional Alfvén profiles. It is also sufficient locally: when $FG$ is independent of $z$, chooseA stationary axisymmetric $\Phi$ satisfying the Poisson equation can be absorbed in this pressure. Gravity need not vanish. If the axis is included, smoothness requires $U,b=O(R)$ and a regular axial field; $f,g=O(R)$ and smooth even $v$ give the usual regular extension. The pressure constant may be chosen to make $p$ positive on a bounded region where the other terms are bounded.

The compatible profiles have the following interpretation. At a cylinder with $v\ne 0$, the variables $a=z+vt$ and $c=z-vt$ are independent as $z,t$ vary. Compatibility reads $f_a(R,a)g(R,c)+f(R,a)g_c(R,c)=0$. If both profiles are nonzero, division gives $f_a/f=-g_c/g=\kappa (R)$, independent of both traveling arguments. Thereforelocally on such a region. Alternatively, one profile vanishes identically and the other is unrestricted. Choosing a point where one profile is nonzero shows that a nontrivial opposite profile obeys the same linear ordinary differential equation everywhere on a connected axial interval, so isolated zeros do not create additional overlapping branches.