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Let $L$ be a core-scale length, $B_{\ast }$ a representative strength of the prescribed poloidal magnetic field, and $v_A=B_{\ast }/\sqrt{{\mu }_0\rho }$ the Alfvén speed. A consistent weakly damped, small-amplitude ordering isThe first ratio is the Lehnert number. It makes the wave slow relative to rotation and allows almost geostrophic flow, with the azimuthal velocity approximately constant along a rotation-axis cylinder. The small wave amplitude gives $|B_{\varphi }|\sim \sqrt{{\mu }_0\rho }|U|\ll B_{\ast }$. Poloidal advection then changes neither the background field nor the wave substantially over one wave period. If there are no-slip boundary layers, also require their damping rate $\sqrt{\nu \Omega }/L\ll \sigma$, or otherwise retain their torque; a small bulk viscosity alone does not remove that boundary torque.

In the rotating magnetohydrodynamic momentum equation, axisymmetry removes the azimuthal pressure derivative. The neglected nonlinear azimuthal advection has size $|{\mathbf{u}}_p|U/L$, and the neglected viscous term has size $\nu U/L^2$. The retained azimuthal Coriolis force is $2\Omega u_r$. A typical compatible local balance has $|u_r|\sim \sigma U/\Omega$, small compared with $U$ when the Lehnert number is small. The azimuthal Lorentz force density can be written using $\nabla \cdot {\mathbf{B}}_p=0$:The physical prefactor is $1/({\mu }_0\rho )$: the extra $i$ printed inside the first reduced momentum equation is a typographical error, also contradicted by the subsequent real-valued system.

In the resistive induction equation, neglecting poloidal advection of the perturbation and magnetic diffusion leaves the winding of the prescribed poloidal magnetic field by differential rotation. ThusThis is the small-amplitude torsional Alfvén wave approximation, with a time-independent background and no external azimuthal forcing.

We now average over the full height of a rotation-axis cylinder. The sphere is impenetrable. Integrating incompressible flow over the volume inside any such cylinder gives zero flux through its cylindrical side, since there is no flux through its spherical caps. ThereforeThis is why the Coriolis force disappears from the cylinder average; it need not vanish locally. An axisymmetric field in the insulating exterior is a potential field with $B_{\varphi }=0$, so continuity gives $B_{\varphi }=0$ on the sphere. Consequently both the end flux terms and the moving-endpoint terms in the magnetic-stress integral vanish. Expanding the divergence givesSince $U$ is independent of $z$, its integrated time derivative is $2z_b{U}_t$. Defining $T={\int }_{-z_b}^{z_b}{B}_r{B}_{\varphi }dz$ givesThe stationary background and the fact that $U/r$ depends only on $r,t$ also givesoTogether these form the cylinder-averaged torsional Alfvén wave system.

Assume $H>0$ on the cylinders carrying the wave, with regular endpoint limits. If $H$ vanishes on an interior cylinder, then $B_r$ and $T$ vanish there and that cylinder has no radial magnetic coupling; the displayed $H^{-1}$ energy term must be interpreted on the coupled region rather than literally dividing by zero. Since $H$ is time independent, differentiation of the proposed conserved energy of a cylinder-averaged torsional wave givesRegularity at the rotation axis makes the lower endpoint vanish, while vanishing magnetic torque at the collapsed outer cylinder makes the upper endpoint vanish. This provesThe magnetic term has a direct interpretation. Decompose $B_{\varphi }=(T/H)B_r+B_{\varphi }^{\perp }$, with $\int B_r{B}_{\varphi }^{\perp }dz=0$. Then $\int B_{\varphi }^{2}dz=T^2/H+\int (B_{\varphi }^{\perp }{)}^{2}dz$, and the induction equation gives $(B_{\varphi }^{\perp }{)}_t=0$. Thus, apart from a common dimensional factor and a stationary orthogonal magnetic component, the conserved quantity is the physical kinetic plus magnetic energy.

For the frequency estimate, put $V=U/r$ and eliminate $T$:Locally this is a wave equation with speedIt is the cylinder-averaged radial field component that matters, not simply the dipole measured at the Earth's surface. Taking a cylinder height of order $6\times 10^6\mathrm{m}$, $\rho \sim 10^4\mathrm{kg}{\mathrm{m}}^{-3}$, and an assumed internal cylindrical radial field $B_{r,\mathrm{rms}}\sim 0.2$--$2\mathrm{mT}$ givesFor a standing wave of radial length $L_c\sim 2$--$3.5\times 10^6\mathrm{m}$, the cyclic frequency is of order $c_T/(2L_c)$ and the angular frequency is of order $\pi c_T/L_c$. HenceGeometry and the assumed internal field can shift the estimate substantially. As a concrete calibration, the 2010 torsional-wave study inferred a cylindrical radial rms field near $2\mathrm{mT}$ and a roughly four-year crossing time, with an observed six-year recurrence. A much weaker assumed interior field instead gives the older multi-decadal estimate. The paper supplies no numerical interior field strength, so a unique numerical frequency cannot be deduced without such an assumption.

Write the horizontal velocity amplitudes as $(u,v)$ and magnetic amplitudes as $(b_x,b_y)$. The perturbation is horizontally uniform, divergence-free, and has no vertical velocity, so the unperturbed density and pressure are consistent at linear order. In the Keplerian shearing sheet, the horizontal components of the linearized ideal magnetohydrodynamic equations areUse the undivided equations at a zero of $F$. If $F^{\prime \prime }=-k^{2}hF$, the magnetic-force coefficient becomes $ik{v}_A^2$; the stratification cancels. The ideal magnetohydrodynamic induction equation similarly reduces toThe azimuthal induction term is field stretching by the background differential rotation. The four amplitudes obey the homogeneous systemIts determinant must vanish for a nonzero normal mode. Put $A=v_A^2{k}^2$. For nonzero $s$, elimination first gives $(s^2+A)u-2\Omega sv=0$ and $s(s^2+A)v+\Omega (s^2-3A)u/2=0$, whose solvability condition is $(s^2+A{)}^2+{\Omega }^2(s^2-3A)=0$. The original determinant extends the same polynomial to marginal $s=0$. ThusThis is the ideal magnetorotational dispersion relation with the midplane Alfvén speed $v_A=B_0/\sqrt{4\pi {\rho }_0}$. The vertical structure enters through the admissible eigenvalues $k$, rather than through a different horizontal dispersion polynomial. No division by a vanishing growth rate is required in the determinant derivation.

For a steady flow, mass conservation gives $\rho w=\text{constant}$ and ${\rho }^{\prime }/\rho =-w^{\prime }/w$. Define the Alfvén velocity ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$ and $C=w^2-v_{az}^2$. Combining the horizontal momentum and MHD induction equations without dividing by $C$ givesDot these identities with $B_x/({\mu }_0\rho )$ and $B_y/({\mu }_0\rho )$, respectively:The vertical momentum equation and the isothermal equation of state giveMultiply by $C$ and eliminate the magnetic derivative to obtainNo division by $w^2-v_{az}^2$ was needed in deriving this necessary relation.

The magnetosonic critical speeds in the $z$ direction areThe plus sign gives the fast magnetosonic wave speed and the minus sign the slow magnetosonic wave speed. Thus the differential coefficient is $(w^2-c_{\mathrm{f}}^2)(w^2-c_{\mathrm{s}}^2)$. A smooth outflow proceeding from below both speeds to above both must normally pass through both magnetosonic critical speeds. At each crossing, the right-hand side must also vanish; this is the regularity at a magnetosonic point condition. The derivative coefficient changes sign at each nondegenerate crossing. For $w>0$ and $w^{\prime }>0$, the driving term must be positive below the slow point, negative between the points and positive above the fast point.

The Alfvén speed component satisfies $c_{\mathrm{s}}^2\le v_{az}^2\le c_{\mathrm{f}}^2$. With downward gravity $g>0$, the gravitational term is consequently nonnegative at the slow point and nonpositive at the fast point. The magnetohydrodynamic shear work contribution must balance it at each point, and can provide the upward driving needed to pass the fast point. For $a=0$ and strictly positive $g$, generic separated slow and fast points cannot satisfy the required zero numerator; special vanishing-gravity or coincident-speed cases need separate treatment.

There is also Alfvén-point compatibility in a plane-parallel sheared flow. At $w^2=v_{az}^2$, the original transverse equations requireThese restrictions are not generally visible as a zero of the scalar differential coefficient, which there equals $-v_{az}^2(v_{ax}^2+v_{ay}^2)$. In particular, a strictly accelerating regular solution must have $B_x=0$ at that point. The scalar relation is therefore a necessary wind equation, not a substitute for regularity of all the original ideal magnetohydrodynamic equations. Degenerate cases such as a purely longitudinal magnetic field can merge characteristic speeds and reduce the number of distinct critical conditions.

Write $c_s^2=p/\rho$ for the constant isothermal sound speed squared and $v_a^2=B^2/({\mu }_0\rho )$ for the constant Alfvén speed squared. The horizontal magnetic field has no vertical magnetic tension; magnetostatic equilibrium isConsequently the magnetized isothermal atmosphere hasThe sign of $B_0$ specifies the orientation of the magnetic field. The adiabatic sound speed is $v_s^2=\gamma p/\rho =\gamma c_s^2$, also constant.

For the proposed normal mode, let $A=\exp (i\mathbf{k}\cdot \mathbf{x}+z/(2H)-i\omega t)$, $a=1/(2H)$, and temporarily write $\mathbf{\xi }$ for its constant complex amplitude. The divergence amplitude is $\Delta =i{k}_x{\xi }_x+i{k}_y{\xi }_y+(i{k}_z+a){\xi }_z$. Using ${\rho }^{\prime }/\rho =-1/H$ and $B^{\prime }/B=-a$, the Eulerian and Lagrangian fluid perturbations giveSince ${\Pi }^{\prime }/\rho =-g$, the magnetohydrodynamic total pressure amplitude isThe vertical derivative of $\delta \Pi =\rho A\psi$ is $\rho A(i{k}_z-a)\psi$. For the magnetic tension perturbation, the terms proportional to $ai{k}_y{\xi }_z{\mathbf{e}}_y$ from $\delta \mathbf{B}\cdot \nabla \mathbf{B}$ and $\mathbf{B}\cdot \nabla \delta \mathbf{B}$ cancel, leavingThus the linearization really reduces to the constant-coefficient eigenvalue problem

For a static ideal magnetohydrodynamics equilibrium with conservative gravitational acceleration, the magnetohydrodynamic energy principle gives a Hermitian operator for the restoring force in the mass density weighted inner product, provided the surface terms vanish. The factor $e^{z/(2H)}$ removes the exponential mass density weight. This can be checked directly, without assuming stability: set $s=v_s^2$, $b=v_a^2$, and $q=s+b$. Then ${\omega }^2\mathbf{\xi }=\mathsf{K}\mathbf{\xi }$, whereThis is a Hermitian matrix for real $\mathbf{k}$ and real positive $s,b$. Therefore every squared frequency ${\omega }^2$ is real; negative values give exponential growth, rather than an oscillation with a complex squared frequency. In an infinite atmosphere these are generalized plane waves, not finite-total-energy global modes.

At ${\omega }^2=0$ and $k_y\ne 0$, the $y$ equation, after cancelling its magnetic terms, gives $s\Delta =g{\xi }_z$. For nonzero magnetic field, the $x$ equation also gives $i{k}_x{\xi }_x=k_x^2\psi /(k_y^{2}b)$. Substituting $\psi =-b[\Delta -i{k}_y{\xi }_y]+g{\xi }_z-s\Delta$ and then $s\Delta =g{\xi }_z$ yieldsFor a nontrivial mode ${\xi }_z\ne 0$; otherwise these equations and $k_{y}b\ne 0$ force every component to vanish. Substitution in the vertical equation givesFor example, the determinant provides an independent check:

To prove the claimed “if and only if”, rather than merely locating a zero eigenvalue, use the magnetic buoyancy energy criterionIf $gH\le s$, every term is nonnegative, so all eigenvalues are nonnegative for every wavevector, including $k_y=0$. If $gH>s$, choose $k_z=0$, any fixed $k_x\ne 0$, and ${\xi }_z=1$, ${\xi }_x=-a/(i{k}_x)$, ${\xi }_y=g/(si{k}_y)$. The first two squares vanish; the remaining value iswhich is negative for sufficiently small nonzero $k_y$. The Rayleigh quotient of this displacement is therefore negative, proving a negative eigenvalue and an unstable normal mode. This uses precisely the assumption that boundary conditions do not exclude the chosen wavelengths. Equality is a neutral threshold, not exponential growth.

Finally the plasma beta is $\beta =p/[B^2/(2{\mu }_0)]=2c_s^2/v_a^2$. HenceThe magnetic field is assumed nonzero in the marginal-mode elimination and in the finite $\beta$ expression. The positive-sound-speed, nonmagnetic limit remains covered by the energy identity; for the usual $\gamma >1$ it is stable. Parker instability occurs because magnetic pressure supports part of the atmosphere's weight while the gas can drain along gently bent magnetic field lines. The resulting buoyant displacement can release Newtonian gravitational potential energy; choosing long wavelengths along the magnetic field makes the stabilizing magnetic tension small. The completed-square calculation quantifies this competition, including the adiabatic process restoring force.

Along the open line, $B_p\sim F/r^2$ and the mass-loading relation with $u_p\to u_{\infty }$ givesThe solutions of part b consequently haveThe azimuthal Alfvén speed therefore approaches the nonzero constantFor an unconfined outflow it is natural to take $\Phi \to 0$ and $h\to 0$ at infinity; also $u_{\varphi }\to 0$. The magnetic term in the magnetohydrodynamic Bernoulli invariant tends toHence the asymptotic energy of a radial magnetohydrodynamic wind is

The post-shock density is $\chi {\rho }_0$, while part c givesUsing the Alfvén speed $v_A=B/\sqrt{4\pi \rho }$ in Gaussian units,The ratio is largest in the equatorial plane, where the field is tangential to the shock:

The wing and its forcing are time independent in the co-moving frame, so after transients have propagated away the perturbation is stationary there. Write the background velocity and field asand take every perturbation to be independent of $y$. The cold-plasma linearized ideal magnetohydrodynamic equations areTheir relevant components areDifferentiate the momentum equation with respect to $x$ and use the two induction relations. With the Alfvén speedthe result is

The adiabatic sound speed and Alfvén speed areThe entropy is held fixed in the sound-speed derivative.

DefineSolving the matching condition gives the magnetohydrodynamic interface wave speedThe weights ${\Sigma }_{\pm }$ are positive, so $v^2$ is a weighted average of the two squared Alfvén speeds. It therefore lies between them, and so does the positive phase speed $v$.

If $B_{y2}=0$ but $B_{y1}\ne 0$, tangential momentum continuity in the formforces $A_1^2=1$. Since the normal components are positive,so the upstream velocity equals the upstream Alfvén speed. This is the limiting switch-off shock condition.

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

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

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

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

In ideal reduced magnetohydrodynamics with constant background sound and Alfvén speeds, perpendicular total-pressure balance gives $\delta p+{\rho }_0{v}_A^2\delta B_{\parallel }/B_0=0$. Put $s=\delta p/p_0-\gamma \delta \rho /{\rho }_0$, $r=\delta \rho /{\rho }_0+s/\gamma$, $c_T=v_A{c}_s/(v_A^2+c_s^2{)}^{1/2}$ and $h=c_s^{2}r/c_T$. Then the parallel momentum, continuity and induction equations yieldThus $f^{\pm }=u_{\parallel }\pm h$ satisfy $D{f}^{\pm }=\mp c_T{\nabla }_{\parallel }{f}^{\pm }$. The two slow-wave populations propagate along bent magnetic field lines and are mixed by the perpendicular Alfvénic turbulence, while the entropy fluctuation is simply advected. They do not force the perpendicular Alfvénic system at this leading order.