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A finite-amplitude ideal-MHD Alfvén wave can rotate the transverse magnetic field at constant magnitude, with constant mass density, pressure and longitudinal components. Its transverse velocity obeys ${\mathbf{u}}_{\perp }=\mathbf{C}-s{\mathbf{B}}_{\perp }/\sqrt{{\mu }_0\rho }$ and its speed is constant along the wave. The constant magnetic pressure prevents compressive acceleration.

Parker instability is an undular magnetic buoyancy instability of a gravitationally stratified atmosphere supported partly by magnetic pressure. A bent horizontal magnetic field can permit gas to drain along its magnetic field lines, leaving buoyant rising regions. Magnetic tension resists bending, so wavelengths and boundary conditions affect the instability threshold. For a magnetized isothermal atmosphere, the unrestricted-wavevector threshold is given by the magnetic buoyancy energy criterion.

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.

Write $P(x)$ for the outer pressure extrapolated to the wall. The normal balance in the thin boundary layer is $p_y={\overline{F}}_y$, givingThe spatially varying magnetic pressure deficit drives a tangential pressure gradient. For fixed forcing as $\omega \to \infty$, layer inertia is small compared with viscous dissipation; explicitly the ratio is $U{\delta }^2/(\nu b)$. The leading tangential balance isThe outer gradient $P^{\prime }$ contributes only at higher inner order. Matching the leading outer shear gives $u_y\to 0$, and the physical no-slip boundary condition gives $u(x,0)=0$. Two integrations yield the magnetic skin-layer streaming slip:For the present field amplitude,Using incompressibility and $v(x,0)=0$ also givesHence $v/u=O(\delta /b)$ in the layer, except at symmetry zeros where a componentwise ratio is inappropriate. In the matching region $\delta \ll y\ll b$, $u\sim U$ and $v\sim -y{U}^{\prime }$, up to the smaller displacement term $\delta U^{\prime }/2$.

The effective outer boundary conditions are tangential slip $u_{\mathrm{bulk}}(x,0)=U(x)$ and zero leading normal velocity $v_{\mathrm{bulk}}(x,0)=0$. These describe the bulk flow extrapolated through the unresolved magnetic skin layer; the actual wall remains at rest. For $x>0$ the slip is negative, and for $x<0$ it is positive. Fluid converges along the wall toward the wire, turns upward near $x=0$, and returns outward farther above the wall.

The following original streamline sketch uses the small-Reynolds number bulk Stokes flow for this slip. It illustrates that circulation without claiming that the unspecified bulk Reynolds number fixes a unique complete flow. With $X=x/b$, $Y=y/b$, $Z=X+iY$, a dimensionless stream function isIts factor in brackets is holomorphic in $Y>0$, so $\Psi$ solves the biharmonic equation; its wall derivative is ${\Psi }_Y(X,0)=-X/(1+X^2{)}^3$. Thus it satisfies both effective wall conditions and gives an exact creeping-flow illustration of the derived slip.

The relevant competition is between energy reduction and magnetic flux freezing. Ideal evolution transports the entire connectivity of the magnetic field lines, including the separation and linkage of flux regions. Viscous dissipation can remove kinetic energy but cannot permit magnetic reconnection between such regions. As each region relaxes toward lower magnetic energy, neighboring flux regions can approach with different limiting tangential fields. A smooth transition can become increasingly thin; its field remains bounded while its curl, and therefore its electric current density, becomes large.

The limiting object is a current sheet. If $\mathbf{n}$ points from its minus to its plus side, its surface electric current density isFor a tangential discontinuity, ${\mathbf{B}}_{\pm }\cdot \mathbf{n}=0$. The normal balance in magnetostatic equilibrium requiresThus a finite jump of tangential field can coexist with equilibrium: the jump of magnetic pressure is balanced by the ordinary pressure jump.

This need not happen for every initial field; a field already in smooth magnetostatic equilibrium is an immediate exception. The expectation concerns topology-constrained relaxation for which a smooth limiting arrangement is obstructed. Current-sheet formation is consistent with the relaxation framework developed in Moffatt's primary analysis. Retaining complete frozen connectivity is essential; minimizing energy subject only to the value of total magnetic helicity would generally permit field rearrangements forbidden by this evolution.

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.

A density deficit in a flux tube is not by itself an instability calculation. For example, a straight magnetized tube in otherwise field-free gas at the same temperature has lateral balance $p_i+B_i^2/(2{\mu }_0)=p_e$. The perfect gas law then gives ${\rho }_i<{\rho }_e$, so the tube rises unless some other force balances its weight deficit. This configuration simply lacks magnetostatic equilibrium. A magnetic buoyancy instability, in contrast, starts from a force-balanced atmosphere and asks whether a small allowed displacement grows. A balanced horizontal magnetic layer can be stable or unstable; the presence of magnetic pressure alone does not decide this.

Assume a smooth equilibrium, with downward gravitational acceleration $g>0$, and write $\mathcal{R}$ for the gas constant per unit mass. The ambient total pressure and force balance areWe use ideal magnetic flux freezing on the displacement time scale and rapid lateral pressure equilibration. A bodily displacement perpendicular to a straight horizontal tube does not change its length or bend its field. If its area is $\mathcal{A}$, conservation of magnetic flux and mass per unit length gives $B\mathcal{A}=\mathrm{constant}$ and $\rho \mathcal{A}=\mathrm{constant}$. Thus the tube preservesEfficient heat exchange fixes its temperature to the ambient temperature at its new height, not to its original temperature. This is the thermal assumption that distinguishes the requested criterion from the adiabatic one.

Take a tube initially at $z_0$ and let $f_0=B_0(z_0)/\rho (z_0)$. After a small upward displacement $\xi$, let ${\rho }_i(z)$ be the density it would have in lateral balance at $z=z_0+\xi$. Its preserved $f_0$ and imposed local temperature giveAt the initial position ${\rho }_i=\rho$. Differentiate this relation at $z_0$:For the ambient atmosphere the same differentiation must also differentiate $f(z)=B_0(z)/\rho (z)$. It givesSubtracting isolates the actual displaced density contrast:The pressure-gradient force on the tube balances the weight of the displaced ambient gas, so its leading vertical buoyancy acceleration isThe coefficient multiplying $\xi$ is positive precisely when the tube becomes lighter on upward displacement. Thus the isothermal interchange criterion for magnetic buoyancy isWhere $B_0>0$, this is simply $d(B_0/\rho )/dz<0$, or $d\log B_0/dz<d\log \rho /dz$. The squared form is independent of the field's choice of direction. Equality is neutral at this order, and the reversed inequality is restoring for these displacements. A layer with an interval satisfying the strict inequality admits local growing interchange disturbances under the assumed pressure/thermal equilibration. The calculation does not address field-bending undular modes, finite heat-transfer times or magnetic diffusion; those can change the criterion. In particular, the localized interchange criterion for a magnetized atmosphere with an adiabatic gas response is not interchangeable with this fast-thermal-exchange result.

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.

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.

For $k_y=0$, the minimized energy density from part (c) isIf the bracket is nonnegative throughout, this density and the omitted pressure square are nonnegative for every displacement, so there is no negative-energy trial. Conversely, a strict negative value at a point of a smooth equilibrium persists in a small interval. Choose a nonzero smooth ${\xi }_z$ supported there, set ${\xi }_y=0$, and choose ${\xi }_x$ as in part (c). This makes the pressure square zero and gives a strictly negative integral. Therefore the localized interchange criterion for a magnetized atmosphere isThese disturbances interchange flux tubes without along-field variation; both gas and magnetic pressure contribute to their effective compressional stiffness. Equality alone is marginal and does not give a negative energy integral.

Set $P=\gamma p$, so $T=P+C$, and for $k_y\ne 0$ put $q=i{k}_y{\xi }_y$. This variable can be chosen freely. The minimized transverse-displacement energy from part (c) has densityComplete the square in $q$ to obtainThis identity also applies at a zero of $B$, where $C=0$ and the first square contributes nothing. If $-(\ln \rho {)}^{\prime }\ge \rho g/P$ everywhere, every term is nonnegative, including the pressure and transverse-tension terms removed earlier. Thus no nonzero along-field wavenumber gives an unstable trial.

For the converse, suppose the strict reverse inequality holds at an interior point. On a sufficiently small closed interval around it, $-g{\rho }^{\prime }-{\rho }^2{g}^2/P$ is uniformly negative and $C$ is bounded. Choose a sufficiently small but nonzero $|k_y|$ so that addition of $C{k}_y^2$ keeps it negative there. Choose a smooth supported ${\xi }_z$ in that interval and set $q=\rho g{\xi }_z/P$, making the first square vanish. This determines a finite ${\xi }_y=q/(i{k}_y)$. Finally choose ${\xi }_x$ by part (c) with sufficiently large finite $|k_x|$; its remaining tension energy can be made smaller than the strict negative margin. All displacements vanish near the boundaries. The full energy is therefore negative.

We have proved the long-wavelength undular magnetic buoyancy criterion:Under the assumed absence of interchange instability, the new unstable range is $\rho g/(\gamma p+C)\le -(\ln \rho {)}^{\prime }<\rho g/(\gamma p)$. Along-field motion allows material to drain and removes magnetic pressure from the limiting buoyancy stiffness, while tension at nonzero $k_y$ remains stabilizing. The statement is about existence of an allowed nonzero wavenumber; a fixed prescribed $k_y$ retains the term $C{k}_y^2$, and a finite imposed periodic length could exclude the required long wavelengths.

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.

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.