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Fix time dependence $e^{-i\omega t}$ and use pressure amplitude for $\psi$. Write the incident acoustic plane wave as ${\psi }_i=A{e}^{i\alpha x-i\beta z}$, where $\alpha =k\cos \theta$, $\beta =k\sin \theta >0$. “Perfectly reflecting” specifies zero absorbed energy but does not uniquely select a boundary condition: a rigid wall has a Neumann boundary condition, whereas a pressure-release surface has a Dirichlet boundary condition. We first give the rigid-wall result, and then the alternative pressure-release result.

For the rigid surface the normal-velocity condition, without an immaterial normalizing factor, isExpand $\psi ={\psi }^{[0]}+{\psi }^{[1]}+{\psi }^{[2]}+\cdots$ in the height amplitude, scaling a fixed differentiable profile. The flat total field is ${\psi }^{[0]}=A{e}^{i\alpha x}(e^{-i\beta z}+e^{i\beta z})$. Taylor expansion givesLet $\hat{h}(q)=\int h(x)e^{-iqx}dx$ and $\beta (q)=\sqrt{k^2-q^2}$ on the outgoing sheet: it is positive real for propagating modes and positive imaginary for evanescent modes. Fourier transforming this boundary derivative gives $2A(k^2-\alpha q)\hat{h}(q-\alpha )$. The outgoing angular spectrum therefore yields the small-height Neumann scattering fieldFor an infinite stationary surface the Fourier quantities may be interpreted as random distributions or through a finite illumination window and its limiting mean. The correction is linear in $h$ and $h^{\prime }$, so zero mean height and stationarity giveOnly the rough correction has zero mean at first order; the flat reflected wave remains.

The first nontrivial coherent correction depends on correlation, not just the r.m.s. height. Define $C_h(s)=⟨h(x+s)h(x)⟩$, its power spectrum of surface height $S_h(q)=\int C_h(s)e^{-iqs}ds$, and ${\sigma }^2=(2\pi {)}^{-1}\int S_h(q)dq$. At second order the flat-field third normal derivative and mixed derivative vanish at zero, leavingFor each first-order outgoing Fourier component, its derivative data are $g_1(q)=2A(k^2-\alpha q)\hat{h}(q-\alpha )$, its second normal derivative is $i\beta (q)g_1(q)$, and its tangential derivative is $q{g}_1(q)/\beta (q)$. Averaging their products with $h,h^{\prime }$ gives the mean derivativeDividing by the specular outgoing derivative $i\beta$ therefore gives oblique coherent reflection from a stationary rigid rough surface:This second-order expression requires the relevant spectral moments to exist and the boundary expansion to remain controlled. Small $|kh|$ alone does not control arbitrarily large slopes or high-frequency roughness. The given stationarity and $\sigma$ determine the first-order mean above, but do not determine this spectrum-weighted correction numerically.

For a pressure-release surface, instead impose $\psi (x,h)=0$. The flat reflected coefficient is $-1$, and Taylor expansion gives ${\psi }^{[1]}(x,0)=2i\beta Ah{e}^{i\alpha x}$. Thus small-height Dirichlet scattering givesThe zero-mean first correction again vanishes on averaging. At second order ${\psi }^{[2]}(x,0)=-h{\partial }_z{\psi }^{[1]}(x,0)$, givingIf the roughness spectrum is concentrated near $q=\alpha$ after the incident tangential shift, the additional long-correlation approximation yields $r_N\simeq 1-2{\beta }^2{\sigma }^2$ and $r_D\simeq -1+2{\beta }^2{\sigma }^2$. A local Gaussian random-height phase model would then give $r_N=e^{-2{\beta }^2{\sigma }^2}$ and $r_D=-e^{-2{\beta }^2{\sigma }^2}$, but Gaussian statistics and that local approximation were not supplied. They must not be inferred from stationarity and r.m.s. height alone.

Translation invariance of the stationary surface statistics implies that a single incident tangential wavenumber retains that wavenumber in its coherent mean. Hence, away from the surface, writeThe unknown effective reflection coefficient $r_b$ is referenced to the mean plane $z=0$. It may be the rigid or pressure-release coefficient just derived, or a more general rough-boundary response. Lossless microscopic reflection does not require $|r_b|=1$: diffuse scattering removes power from the coherent specular field.

For the added planar upper interface, keep the same tangential wavenumber $\alpha$. Let ${\beta }_2=\sqrt{k_2^2-{\alpha }^2}$ on its outgoing sheet. The normal acoustic impedances in the layer and upper medium areContinuity of pressure and normal velocity gives the reflection coefficient for upward incidence from the layer and transmission coefficient for downward incidence from above:If a separate impedance sheet rather than an ordinary fluid interface is intended, its specified effective impedance replaces the upper termination; a sheet impedance has not otherwise been given. These formulas correspond to the stated two-fluid interface.

Let the upper incident pressure amplitude at $z=d$ be $I$. If $D$ and $U$ denote downward and upward layer amplitudes at that plane, their interface and lower-reflection relations areSolving, or summing the geometric series of repeated round trips, gives the coherent acoustic layer above a rough reflector:The expression describes the field at heights where the common homogeneous layer exists; perturbative mean fields near the moving rough boundary need their physical trace evaluated separately. For an upper incident wave normalized as $e^{i\alpha x-i{\beta }_{2}z}$, set $I=e^{-i{\beta }_{2}d}$. With a matched upper medium, $r_{12}=0$ and $t_{21}=1$, the formula reduces to the single lower-reflection field. Zeros of the denominator give ideal layer resonances, interpreted with outgoing continuation or small loss.

This is a coherent specular description: a single effective $r_b$ closes the downward/upward mean amplitudes. Diffuse paths reflected back from the upper interface can modify the mean response; they must either be neglected in this approximation or included in the effective lower reflection coefficient for the actual layered environment. Without a full roughness spectrum and boundary type, a unique numerical mean field cannot be specified beyond this general coefficient form.

Use the flat-surface Boussinesq approximation, with small thermal expansion, constant properties, negligible surface deformation, and constant surface tension. The top is impermeable and shear-free, while the solid walls are no-slip. Temperature-dependent surface tension would introduce a separate Marangoni stress, which is not included in this buoyancy-driven model. The stated time scale requires $\kappa$ to mean thermal diffusivity; if the material property given is a dimensional thermal conductivity $k$, use $\kappa =k/({\rho }_0{c}_p)$.

Take $z$ positive downwards and define the nondimensional streamfunction by $u={\psi }_z$, $w=-{\psi }_x$. Velocity is scaled by $\kappa /L$, streamfunction by $\kappa$, and temperature is measured from the cold-end surface value. Define the Prandtl number and horizontal-length Rayleigh number bywhere ${\alpha }_T$ is the thermal expansion coefficient. Put $J(\psi ,f)={\psi }_z{f}_x-{\psi }_x{f}_z$, and let ${\nabla }^2={\partial }_x^2+{\partial }_z^2$. Taking the curl of the Boussinesq equations eliminates pressure and gives the full nonlinear vorticity-temperature systemThe buoyancy sign follows from the downward coordinate: warmer fluid has body acceleration $-g{\alpha }_T\Delta TT{\mathbf{e}}_z$.

Choose the same streamfunction constant, zero, on the connected boundary. The mechanical and thermal boundary conditions areThe upper stress-free boundary condition reduces to ${\psi }_{zz}=0$, since $\psi =0$ there already gives ${\psi }_{xx}=0$. An arbitrary additive constant in $T$ changes only the reference hydrostatic pressure.

In steady infinite-Prandtl-number convection, momentum inertia disappears but thermal advection remains:For a shallow lake, with fixed $\mathrm{Ra}$ as $ϵ\to 0$, the leading interior temperature is $T_0=x$. This is an outer solution: it does not satisfy the insulated sidewall derivative right at $x=0,1$. The leading global conductive solution, which resolves those end regions, isIt is harmonic, has $T_c(x,0)=x$, and satisfies the insulated sides and bottom. Away from sidewall layers of horizontal width $O(ϵ)$ it reduces to $x$ with exponentially small end corrections. These layers also turn the flow and enforce the sidewall no-slip boundary condition.

In the shallow interior the dominant momentum derivatives are vertical, so ${\psi }_{0,zzzz}=-\mathrm{Ra}$. Integrating with the surface and bottom conditions gives the shallow horizontal convection with an insulated bottom solution, with $y=z/ϵ$:Surface flow is toward the colder end, with a deeper return flow. The depth-integrated volume flux is zero because ${\psi }_0(ϵ)={\psi }_0(0)=0$. The bulk expressions apply away from the turning regions, not unchanged on the vertical walls. A useful small-advection condition in the end layers is $\mathrm{Ra}{ϵ}^4\ll 1$, automatic for fixed Rayleigh number in the stated thin limit.

The heat flux in these units is $\mathbf{q}=(uT-T_x,wT-T_z)$. For the leading bulk temperature, the horizontal transport through a vertical section per unit out-of-plane width isThus leading advective horizontal transport vanishes: both layers have the same temperature at a given $x$, and their opposing volume fluxes cancel. The leading conductive transport is from the hot end toward the cold end. In dimensional form it is $-{\rho }_0{c}_p\kappa \Delta Tϵ=-k\Delta Tϵ$ per unit width. Heat enters through the warmer portion of the maintained upper surface and leaves through its colder portion, with the conductive end adjustments particularly important near the two insulated sidewalls. It cannot enter through the insulated sides or bottom.

For the advection correction in shallow horizontal convection, set $T=x+\Theta (z)$ in the bulk. Since $w_0=0$ and $T_x=1$, the steady heat equation gives ${\Theta }_{zz}=u_0={\psi }_{0,z}$. The conditions $\Theta (0)=0$ and ${\Theta }_z(ϵ)=0$ imply ${\Theta }_z={\psi }_0$ and henceFor example the bottom bulk temperature is $T(x,ϵ)=x-\mathrm{Ra}{ϵ}^5/320$. The correction makes the returning deeper fluid slightly colder. Its horizontal advective heat transport isThus the transport including this bulk correction is $\boxed{{\mathcal{H}}_x=-ϵ-19{\mathrm{Ra}}^2{ϵ}^9/1451520}$, both terms directed toward the colder end.

Vertical transport must distinguish local flux from net transport. In the leading interior $w_0=T_{0,z}=0$, whereas the temperature correction gives a local downward conductive flux $q_z=-{\Theta }_z=-{\psi }_0>0$. The end circulation gives an upward advective transport: with impermeable sidewalls and $T_0\simeq x$, integration by parts gives ${\int }_0^{1}w{T}_{0}dx=-{\int }_0^{1}x{\psi }_{x}dx={\int }_0^1\psi dx\simeq {\psi }_0(z)<0$. It compensates that downward conduction. More generally, steady heat conservation and insulated sides give $d[{\int }_0^1{q}_{z}dx]/dz=0$, and the insulated impermeable bottom sets the constant to zero. ThereforeThere is no net vertical heat transport through the lake, despite its local recirculating heat flux and its nonzero horizontal heat transport.

Take the equilibrium mass density to be constant and absorb any conservative equilibrium force into the pressure. The linearized ideal magnetohydrodynamic induction equation, using the solenoidal vector field conditions, isFor $\mathbf{B}=Jr{\mathbf{e}}_{\varphi }$, differentiation of the cylindrical basis is important. Acting on a vector whose cylindrical components have azimuthal dependence $e^{im\varphi }$ gives $(\mathbf{B}\cdot \nabla )\mathbf{u}=imJ\mathbf{u}+J{\mathbf{e}}_z\times \mathbf{u}$. Also $(\mathbf{u}\cdot \nabla )\mathbf{B}=J{\mathbf{e}}_z\times \mathbf{u}$. These basis contributions cancel in induction, giving $i\omega \mathbf{b}=imJ\mathbf{u}$.

The perturbation of the Lorentz force isBoth of the first two terms now contribute a basis-rotation term. With $\tilde{p}=p_1/\rho +\mathbf{B}\cdot \mathbf{b}/({\mu }_0\rho )$, the linearized ideal magnetohydrodynamic momentum equation is thereforeThe magnetic part of $\tilde{p}$ is the perturbation of magnetic pressure, not an extra restoring force to retain after projecting out pressure.

Put $D={\omega }^2-m^2{\tilde{J}}^2$, where $\tilde{J}=J/\sqrt{{\mu }_0\rho }$. For $mJ\ne 0$, substitute $\mathbf{u}=\omega \mathbf{b}/(mJ)$ into the momentum equation and multiply by $mJ$. Taking a curl removes the modified pressure. Sincewe obtain the uniform-current toroidal-field curl reductionA second curl, using $\nabla \times \nabla \times \mathbf{b}=-{\nabla }^2\mathbf{b}$, givesThe divided expression assumes $D\ne 0$. If $mkJ\ne 0$, the undivided curl equation shows that $D=0$ would force $\mathbf{b}=0$, so this restriction loses no such magnetic mode. The $k=0$ degenerate limit can instead be taken in the undivided equations; it has no growing branch of the kind considered below. Axisymmetric $m=0$ disturbances are likewise excluded by the velocity elimination, and induction gives no nonzero-frequency magnetic disturbance in this particular equilibrium.

For a vector Laplacian eigenmode with ${\alpha }^2+k^2>0$, elimination givesThus both squared frequencies are real. The lower branch satisfies ${\omega }_{-}^2\ge {\tilde{J}}^2(|m{|}^2-2|m|)$, so it is positive for $|m|>2$. For $|m|=2$ it is positive when $\alpha \ne 0$, but the printed strict-positivity statement needs an endpoint qualification: $|m|=2$, $\alpha =0$ permits a neutral mode unless boundary conditions exclude it.

This is a genuine counterexample to the unqualified endpoint, not merely a loose algebraic bound. The neutral quadrupolar perturbation of a uniform-current toroidal fieldis regular at the axis, is solenoidal, and has $\nabla \times \mathbf{b}=k\mathbf{b}$, hence ${\nabla }^2\mathbf{b}=-k^2\mathbf{b}$. Also ${\mathbf{e}}_z\times \mathbf{b}=-i\mathbf{b}$, so the non-gradient part of the linearized Lorentz force vanishes for $m=2$. The gas-pressure perturbation $p_1=-\mathbf{B}\cdot \mathbf{b}/{\mu }_0$ cancels its magnetic pressure gradient, as required by $\tilde{p}=0$. Taking its real part gives a real disturbance. Decay at radial infinity or an appropriate homogeneous radial boundary condition would exclude it, restoring strict positivity for the allowed eigenmodes; no such condition is printed.

For $|m|=1$ the lower branch is unstable precisely whenAt equality it is marginal. With the time convention $e^{i\omega t}$, the growing member has $\omega =-is$ andThe other branch remains oscillatory. Actual admissibility of a chosen $\alpha$ depends on the radial domain and its boundary conditions.

Write the inward radial velocity as $u_R=-v(R)$, with $v>0$. Steady spherical mass conservation and the radial Euler equations for an inviscid fluid giveFor a barotropic fluid, define the sound speed $c_s^2=dp/d\rho >0$ and the barotropic enthalpy function by $dh=dp/\rho$. Choosing $\Phi (\infty )=0$, integration gives the Bernoulli equationThe continuity equation gives ${\rho }^{\prime }/\rho =-2/R-v^{\prime }/v$. Eliminating mass density from the momentum equation produces the barotropic spherical accretion equationThe sonic point is where the inward Mach number $v/c_s$ equals one. A smooth solution cannot cross the vanishing coefficient on the left with a nonzero right-hand side, so a regular sonic radius must obeyConversely a continuous flow that is subsonic at infinity and supersonic inward must cross unit Mach number; smooth passage selects a simultaneous zero of the two sides, hence a regular sonic point. The associated critical accretion branch, rather than an arbitrary subsonic solution with smaller flux, is selected by these outer and inner requirements.

For the polytropic equation of state, set $\Gamma =1+1/m$ to distinguish its exponent from any separately specified perturbation exponent. Assume $K>0$ and first take the usual positive index $m>0$. ThenFor the attractive mixed potential, assume $\lambda ,\mu >0$. The sonic condition and Bernoulli equation giveHence polytropic accretion in a mixed inverse-power potential reduces toFor $0<m\le 1/2$ all nonzero coefficients in the left side are nonnegative for $R_s>0$, with at least one strictly positive: there is no finite positive root. For $m>1/2$ the constant term is negative and the leading coefficient positive, giving exactly one positive root. Define $B=(2m-3)\mu /4$ and $C=(2m-1)\lambda /2$. ThenThis is the finite-radius transonic range for the two positive attractive terms.

The critical point has real distinct crossing slopes. To check this rather than just solve the sonic equalities, put $U=\lambda /R_s^2$, $V=\mu /(2R_s)$ and $x=R_s{v}_s^{\prime }/v_s$. Differentiating the barotropic spherical accretion equation at its simultaneous zero givesIts discriminant is $8m[2(2m-1)U+(2m-3)V]/(U+V)$. The sonic Bernoulli equation says $(2m-1)U+(2m-3)V=2m{c}_{\infty }^2$, so this discriminant is strictly positive for $m>1/2$. The accretion crossing is the slope for which $d\log (v/c_s)/d\log R<0$, namely the smaller root. The two slopes have opposite signs of this Mach-number derivative. This is the regular transonic saddle, not a singular vertical tangent.

The corresponding mass accretion rate follows without an additional arbitrary mass density normalization:One may equivalently replace ${\rho }_s$ by $[c_{s,s}^2/(\Gamma K){]}^m$. With a nonzero $\lambda$ term, inner free fall has $v\sim \sqrt{2\lambda }/R$ and continuity gives $\rho \propto R^{-1}$. Thus $v^2/c_s^2\propto R^{-2+1/m}$ increases without bound precisely in the positive-index range $m>1/2$, consistent with the required inner supersonic branch.

The assumptions on the parameters matter. If $\lambda =0$ and $\mu >0$, the finite sonic radius instead requires $m>3/2$, the usual Bondi accretion range; $m=3/2$ is the zero-radius limiting critical case. If $\mu =0$ and $\lambda >0$, the condition remains $m>1/2$. With arbitrary signs of $\lambda ,\mu$, one must use the quadratic and retain only positive roots with $c_{s,s}^2>0$ and real admissible crossing slopes; attraction was not explicitly given as a sign condition in the source.

If positive polytropic index was not intended as an implicit physical convention, there is a further mathematical range. For $K>0$, positive compressibility also permits $m<-1$, giving $0<\Gamma <1$. Both potential terms attractive then give a positive sonic root as well:The same mass accretion rate formula applies; its crossing-slope discriminant is positive. The range $-1<m<0$ has $c_s^2<0$ and no physical acoustic sonic point, $m=-1$ has constant pressure and zero sound speed, and $m=0$ does not define the printed equation of state. Thus the usual answer $m>1/2$ assumes $m>0$ and positive attractive coefficients; these conventions should not be confused with additional printed restrictions.

Assume a nonrotating, nonmagnetic equilibrium, inviscid adiabatic perturbations, and no dissipation. Let $g={\varphi }^{\prime }(R)>0$, so hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use $\mathbf{\xi }{e}^{i\omega t}$ for the fluid displacement. To avoid a source ambiguity, write Eulerian fluid perturbations as $p_1,{\rho }_1,{\varphi }_1$, and use $\Delta$ for quantities following the displaced material element, as in Eulerian and Lagrangian fluid perturbations. They are related by $\Delta p=p_1+{\xi }_R{p}^{\prime }$ and similarly for mass density and potential.

The printed identity needs two corrections: its compression denominator must be $\gamma p$, not $\gamma \rho$, and the pressure and potential appearing in this form of the identity must be Eulerian fluid perturbations. The printed denominator has the wrong dimensions for an energy density. The claimed Lagrangian interpretation does not produce the displayed buoyancy-compression split; the following derivation specifies the consistent version.

Linearized mass conservation, the adiabatic relation, and the Poisson equation giveConsequently $p_1=-\gamma p\nabla \cdot \mathbf{\xi }-{\xi }_R{p}^{\prime }$. Linearizing the Euler equations for an inviscid fluid about the static star gives the linear adiabatic stellar oscillation equationsRequire regularity at the center, a vanishing Lagrangian surface pressure, and the usual matching to a decaying exterior gravitational potential. Because $p$ and $\rho$ vanish at the surface, the pressure boundary term vanishes for regular displacements; compactly supported trials also suffice below. The perturbed Newtonian gravitational potential solves Laplace's equation outside the star and must be included there when its field energy is computed.

Multiply the momentum equation by $-{\mathbf{\xi }}^{\ast }$ and integrate. The pressure term is $-{\int }_V{p}_1\nabla \cdot {\mathbf{\xi }}^{\ast }dV$. Let $d=\nabla \cdot \mathbf{\xi }$. The adiabatic relation and hydrostatic equilibrium implyThe pressure and equilibrium-gravity terms then combine pointwise aswhere $N^2=g[(1/\gamma )p^{\prime }/p-{\rho }^{\prime }/\rho ]$ is the stellar buoyancy frequency. The potentially troublesome cross terms cancel exactly.

For the perturbed self-gravity term, integration by parts givesThus the corrected stellar displacement energy identity isThe right side is a real quadratic form. With these conservative boundary conditions, the force operator is a self-adjoint operator. Its Rayleigh quotient characterizes squared mode frequencies, and a negative admissible trial quotient proves an unstable spectral component; it does not require the trial displacement itself to be an exact eigenfunction.

For a generic compressive disturbance of scale $\ell$ in a region with slowly varying equilibrium coefficients, estimate ${\rho }_1\sim \rho \xi /\ell$, $p_1\sim \gamma p\xi /\ell$ and ${\varphi }_1\sim 4\pi G{\rho }_1{\ell }^2$. The self-gravity and compression energy densities have orderswith adiabatic sound speed $c_s^2=\gamma p/\rho$. Under the plausible stellar estimate $c_s^2\sim G\bar{\rho }{a}^2$, global modes can have comparable gravity and pressure contributions, while their ratio for small-scale compressive modes is of order $(\rho /\bar{\rho })(\ell /a{)}^2$. Thus perturbations with $\ell \ll c_s/\sqrt{4\pi G\rho }$ admit the Cowling approximation. This estimate compares two nonzero compressive terms; it cannot alone justify neglecting gravity for a trial deliberately chosen to have $p_1=0$.

For an explicit convective trial, choose $c<d<e<b$ and a nonzero smooth bumpLet $Y$ be a normalized spherical harmonic of degree $\ell \ge 1$, let $L=\ell (\ell +1)$, and let ${\nabla }_{\Omega }$ be the unit-sphere gradient. The pressure-free localized stellar convective trial isIt is supported away from the center and surface, so it satisfies every needed mechanical boundary condition. Since ${\Delta }_{\Omega }Y=-LY$,Therefore $p_1=-\gamma p\nabla \cdot \mathbf{\xi }+\rho g{\xi }_R=0$ exactly. Its buoyancy energy is ${\int }_d^e\rho N^2{R}^2|f{|}^{2}dR<0$, whereas its kinetic norm ${\int }_d^e\rho R^2(|f{|}^2+L|h{|}^2)dR$ is positive. In the Cowling approximation this gives a negative Rayleigh quotient, and hence $N^2<0$ in an interior shell implies convective instability. Retaining perturbed self-gravity only adds a nonpositive term, so the same trial proves instability in the full conservative problem too.

There is also a quantitative small-scale check for this very trial. Its mass density amplitude is ${\rho }_1=(\rho N^2/g)fY$, independent of $\ell$. Write ${\rho }_1=s(R)Y$ and ${\varphi }_1=F(R)Y$. The radial Poisson equation givesBy the Cauchy-Schwarz inequality, $A\le 4\pi G(\int R^4|s{|}^{2}dR{)}^{1/2}(A/L{)}^{1/2}$, and so the angular-degree bound on perturbed stellar self-gravity isIt tends to zero with horizontal scale $R/\ell$, while the strictly negative buoyancy integral is fixed. This justifies the small-scale neglect of gravity even though the trial's compression term vanishes.

The magnetorotational instability extracts energy from differential rotation through magnetic tension. It can destabilize a disk whose specific angular momentum increases outward, even though such a disk satisfies Rayleigh's circulation criterion. The distinction is that an unmagnetized displaced parcel approximately conserves its own angular momentum, whereas a field line transfers angular momentum between parcels. If angular velocity decreases outward, an outward-displaced parcel lags its inner partner. Magnetic tension transfers angular momentum to the outer parcel and removes it from the inner one, making their radial separation increase. A sufficiently stiff magnetic connection suppresses the separation instead. The instability therefore requires a finite range of field-line-bending strengths.

Make this mechanism quantitative in the local shearing sheet of a cylindrical rotation profile $\Omega (r)$. Put $q=-d\log \Omega /d\log r$, and take $0<q<2$ so the unmagnetized flow is stable. Its radial epicyclic frequency is ${\kappa }^2=2(2-q){\Omega }^2$. Let the uniform field be vertical, $B_z{\mathbf{e}}_z$, with constant mass density, and use an axisymmetric horizontal disturbance proportional to $e^{st+ikz}$. Write $\mathbf{h}=\mathbf{b}/\sqrt{{\mu }_0\rho }$ and ${\omega }_A=k{B}_z/\sqrt{{\mu }_0\rho }$. The linearized ideal magnetohydrodynamics equations areThe first pair contains Coriolis acceleration and the epicyclic response; the last equation is the shear-winding of a radial field perturbation into an azimuthal one. Omitting that winding term would remove the driving mechanism.

Eliminating $h_r,h_{\varphi }$ gives $(s^2+{\omega }_A^2)u_r=2\Omega s{u}_{\varphi }$ and $s(s^2+{\omega }_A^2)u_{\varphi }+\Omega [(2-q)s^2-q{\omega }_A^2]u_r=0$. Their determinant is the ideal magnetorotational dispersion relationIts discriminant as a quadratic in $s^2$ is ${\kappa }^4+16{\Omega }^2{\omega }_A^2>0$. Since $2{\omega }_A^2+{\kappa }^2>0$, a positive $s^2$ exists precisely when its constant term is negative:Thus angular velocity, rather than specific angular momentum, must decrease outward. A Keplerian disk has $q=3/2$, ${\kappa }^2={\Omega }^2$ and is hydrodynamically stable, yet magnetic modes with $0<k^2{v}_A^2<3{\Omega }^2$ grow.

The growing root is $s^2=-{\omega }_A^2-{\kappa }^2/2+\frac{1}{2}\sqrt{{\kappa }^4+16{\Omega }^2{\omega }_A^2}$. Differentiation with respect to ${\omega }_A^2$ gives the maximum growth rate of ideal magnetorotational instability:For Keplerian shear these are $3\Omega /4$ and $15{\Omega }^2/16$. The ideal maximum is set by shear; reducing field strength reduces the fastest wavelength instead of reducing this maximum growth rate. This is why the exactly zero-field problem differs from a sequence of weak-field problems with progressively larger wave number. At fixed wave number, growth does tend to zero as the field vanishes. The wavelength must fit the actual disk, and diffusion eventually matters at sufficiently short scales.

The instability has the required angular-momentum-transfer sign. In a growing mode, $u_{\varphi }/u_r=(s^2+{\omega }_A^2)/(2\Omega s)>0$. At the fastest mode this ratio is one, while induction gives $h_{\varphi }=-h_r$ after allowing for their common Fourier phase. Thus the averaged Reynolds stress minus Maxwell stress tensor, $\rho ⟨u_r{u}_{\varphi }-h_r{h}_{\varphi }⟩$, is positive: angular momentum moves outward. Its product with $q\Omega$ is the energy supplied by the background shear. These local mechanisms and coefficient conventions are also checked in the primary MRI stability calculation.

More general fields require separating field-line bending from field gradients and curvature. At leading local order, for a stationary compatible field in an incompressible flow with axisymmetric wavevector $(k_r,0,k_z)$, the bending frequency isPressure elimination projects the radial force by $f=k_z^2/k^2$. The radial momentum equation becomes $s{u}_r-2f\Omega u_{\varphi }=i{\omega }_A{h}_r$; the azimuthal and induction equations retain the form above. Eliminating the amplitudes gives the oblique axisymmetric magnetorotational dispersion relationHere $k_z\ne 0$. The unstable range is $0<{\omega }_A^2<-fd{\Omega }^2/d\log r$. The poloidal magnetic field therefore determines whether an axisymmetric perturbation bends a field line. A locally uniform toroidal magnetic field does not enter $\mathbf{k}\cdot \mathbf{B}$ for axisymmetry, and can contribute magnetic pressure that is projected out of this incompressible leading-order problem. Thus a toroidal component does not suppress the weak-field MRI merely by being present. Conversely a purely toroidal field has no ordinary axisymmetric local bending MRI in this approximation. This does not imply that every globally curved toroidal equilibrium is stable.

A steady general poloidal magnetic field must also be compatible with the equilibrium rotation. The azimuthal ideal magnetohydrodynamic induction equation contains ${\partial }_t{B}_{\varphi }=r{\mathbf{B}}_p\cdot \nabla \Omega$. A stationary axisymmetric equilibrium with no meridional flow consequently requires ${\mathbf{B}}_p\cdot \nabla \Omega =0$, Ferraro's law of isorotation. A radial field crossing cylindrical differential-rotation surfaces would otherwise wind secularly; it cannot simply be substituted into a stationary normal-mode calculation. Boundary conditions and the along-field eigenvalue problem replace a freely chosen local $k$ in nonuniform or bounded equilibria. Strong enough field-line tension can then stabilize all admissible wavelengths.