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For zero mean velocity and magnetic field in a shearing box, the perturbation energy obeys $\dot{E}\le 2(|A|-4{\pi }^2\eta /L^2)E$ when kinematic viscosity and magnetic diffusivity agree. Bound the work of Reynolds stress and Maxwell stress tensor by $2|A|E$, and use the spectral gap of a shearing-periodic box to bound dissipation below. Thus the displayed bound on magnetic Reynolds number guarantees exponential decay and excludes sustained turbulence.

In a steady axisymmetric star, a torque due to Reynolds stress, Maxwell stress tensor or viscosity must balance advection of specific angular momentum by stellar meridional circulation. Writing the nonadvective angular momentum transport flux as ${\mathbf{F}}_J$ gives the displayed balance. Thus stresses that maintain differential rotation also drive a meridional response. The stellar thermal-wind balance supplies a separate leading force constraint, not a replacement for this torque equation.

In the constant-stress overlap layer, the law of the wall gives $U_y=u_{\ast }/(\kappa y)$ and the Reynolds stress gives $-\overline{u^{\prime }{v}^{\prime }}\simeq u_{\ast }^2$. Thus ${\nu }_T=u_{\ast }\kappa y$ and local balance gives $P\simeq ϵ=u_{\ast }^3/(\kappa y)$. Combining with the K-epsilon model viscosity formula gives the displayed constant $k$. The energy diffusion term then vanishes. The dissipation diffusion term does not: it is $u_{\ast }^4/({\sigma }_{ϵ}{y}^2)$. Balancing it against the dissipation-equation source terms yields the displayed Von Kármán constant relation. Neglecting dissipation diffusion would incorrectly require equal production and destruction constants.

The work of the Reynolds stress against the mean velocity gradient transfers kinetic energy from the mean flow into fluctuations. In a plane shear flow $U(y)$ this is $P=-\overline{u^{\prime }{v}^{\prime }}{U}_y$. The eddy viscosity closure gives $P={\nu }_T{U}_y^2\ge 0$. Buoyancy production and transport are separate terms in the turbulent kinetic energy budget.

A mixing length is the characteristic transverse displacement over which a turbulent parcel retains its streamwise momentum before mixing with its surroundings. Taylor expansion of a mean velocity profile gives a fluctuation of magnitude $l|U_y|$. Combining comparable transverse fluctuations with down-gradient Reynolds stress transport gives the closure $\overline{u^{\prime }{v}^{\prime }}=-l^2|U_y|U_y$. This is a phenomenological approximation, not a consequence of the Navier-Stokes equation. In a free jet, a prescribed local mixing length must also be compatible with centreline regularity and vanishing outer momentum flux.

A mixing-length closure uses a prescribed mixing length $\ell$ to approximate turbulent transport. For mean shear ${\partial }_{z}U$, a standard model gives eddy viscosity ${\nu }_t={\ell }^2|{\partial }_{z}U|$ and down-gradient Reynolds stress. Scalar-diffusion versions replace momentum transport by an appropriate scalar flux. The choice of length and correlations is a modelling assumption rather than an exact closure of the Navier-Stokes equation.

Rotating anisotropic convection can produce an azimuthal-meridional velocity correlation even without a gradient of the mean rotation rate. Its Reynolds stress transports angular momentum and can drive differential rotation, whereas a purely viscous closure only erases shear. Energy comes from the thermally driven convection; the Coriolis force changes the correlations but does no work on a parcel.

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

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

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

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

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

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

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

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

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

For stationary horizontally homogeneous means, incompressibility gives ${\bar{w}}_z=0$ and the impermeable boundary makes $\bar{w}=0$ throughout. With no imposed horizontal mean pressure gradient, the balances reduce toThe last equation includes the vertical Reynolds stress; omitting it requires an additional small-normal-stress approximation. The total vertical buoyancy flux is $B_0$ and the horizontal momentum flux is a signed constant $J_M$ with $|J_M|=U_{\ast }^2$. These are fluxes per unit reference density.

The local down-gradient closure defines eddy viscosity and eddy diffusivity byA parcel moving upwards typically retains the smaller momentum or scalar value from its previous height in an increasing mean profile; its fluctuation then correlates negatively with upward motion. This motivates positive coefficients, but is a closure assumption, not a universal theorem about turbulent or convective fluxes. The mean gradients obeyFor positive applied traction on a fluid above a bottom boundary, $J_M=+U_{\ast }^2$ and the mean velocity decreases away from the forcing boundary. The usual atmospheric convention specifies the stress absorbed by the boundary and has the opposite sign of $J_M$; scalar and Richardson-number results are unchanged. At large Reynolds number and Péclet number, outside the molecular wall sublayers, $K_M\gg \nu$ and $K_B\gg \kappa$, so molecular transport can be neglected in these flux-gradient relations. Large bulk numbers do not justify doing this exactly at a smooth solid wall.

Define the perturbation energy per unit mass by $E=⟨v^2/2+B^2/(2{\mu }_0\rho )⟩$. Dot the momentum equation with $\mathbf{v}$ and the induction equation with $\mathbf{B}/({\mu }_0\rho )$, then take volume averages. The Coriolis force does no work. Pressure work and nonlinear advection are divergences and integrate to zero, as does the background $x{\partial }_y$ transport. The two magnetic exchange terms add toThe velocity shear term supplies $2A⟨v_x{v}_y⟩$, while magnetic shear supplies $-2A⟨B_x{B}_y⟩/({\mu }_0\rho )$. Integration by parts gives the viscous and resistive losses as $-\nu ⟨|\nabla \mathbf{v}{|}^2⟩$ and $-\eta ⟨|\nabla \mathbf{B}{|}^2⟩/({\mu }_0\rho )$. For a solenoidal field $\mathbf{F}$, another integration by parts showsAll boundary terms cancel by the shearing-periodic identities. The resulting balance isThe source is work extracted from maintained differential rotation by Reynolds stress and the magnetic part of the Maxwell stress tensor; internal magnetic/kinetic exchange has canceled.

For the decay bound, the elementary inequality $2|ab|\le a^2+b^2$ givesA uniform spectral gap of a shearing-periodic box is needed; ordinary radial periodicity cannot simply be assumed. At each time the compatible Fourier modes areThey are orthogonal over the box. Their wavevectors areFor every nonconstant mode, $|\mathbf{k}|\ge 2\pi /L$, where $L=\max (L_x,L_y,L_z)$: if $m$ or $p$ is nonzero, its corresponding unsheared component supplies the bound; otherwise $n\ne 0$ and the radial component does. Zero mean excludes precisely the constant mode. The Parseval identity therefore proves the time-uniform Poincaré inequalityApply it componentwise to $\mathbf{v}$ and $\mathbf{B}$. With $\nu =\eta$, the energy decay criterion for a viscous-resistive shearing box becomesHenceThis excludes sustained finite-amplitude turbulence as well as infinitesimal growth. It is a sufficient stability condition, not a claim that every box above the threshold must become turbulent. The zero-mean restrictions matter: an imposed uniform magnetic field or bulk epicycle is not controlled by this spatial spectral-gap inequality.

A Rossby wave is a displacement of a materially conserved potential vorticity distribution whose induced balanced velocity propagates that displacement. In a quasi-geostrophic approximation interior, take a locally uniform zonal current $U$ and meridional background PV gradient $B={\overline{Q}}_y$. A small meridional parcel displacement $\eta$ produces $q^{\prime }=-B\eta$. The linear potential-vorticity equation isFor a shallow-water vertical mode, potential-vorticity inversion is $q^{\prime }=({\nabla }_H^2-L_D^{-2}){\psi }^{\prime }$ and $v^{\prime }={\psi }_x^{\prime }$. Substitution of $e^{i(kx+ly-\omega t)}$ givesThe same structure holds for other vertical modes, with their own deformation lengths, and for barotropic waves with $L_D^{-1}=0$. A topographic or shear-induced PV gradient can replace the planetary one: locally $B=\beta -U^{\prime \prime }$ in a barotropic zonal current, with the corresponding bottom contribution when present.

The restoring process can be seen directly by choosing $\eta ={\eta }_0\cos (kx-\omega t)$ in one horizontal dimension with no mean flow. Then $q^{\prime }=-B\eta$ and ${\psi }^{\prime }=B\eta /(k^2+L_D^{-2})$. The resulting $v^{\prime }={\psi }_x^{\prime }$ is a quarter wavelength out of phase with the displacement and requires $\omega =-Bk/(k^2+L_D^{-2})$. A PV anomaly is therefore not merely carried passively: its nonlocal balanced velocity moves the displaced contours.

For $B>0$, the intrinsic zonal phase velocity isChanging the sign of $k$ gives the conjugate representation of the same real pattern, not an eastward intrinsic phase branch. More generally the intrinsic phase-velocity vector has nonpositive zonal component $-B{k}^2/[|\mathbf{k}{|}^2(k^2+l^2+L_D^{-2})]$. This one-way phase property contrasts with acoustic and internal gravity waves, whose frequency equations have two independent propagation branches. The direction reverses when the relevant PV gradient reverses; a sufficiently strong current can make the laboratory phase move eastward. Thus one-way refers to propagation relative to the mean flow and a specified gradient, not an absolute geographic direction in every problem.

Scale dependence is supplied by potential-vorticity inversion. At short horizontal scales, a fixed PV anomaly creates ${\psi }^{\prime }\sim q^{\prime }/|\mathbf{k}{|}^2$, so the intrinsic phase speed is small, of order $B/|\mathbf{k}{|}^2$. At long scales with finite $L_D$, it approaches $-B{L}_D^2$. The group velocity need not have the phase direction:In particular, short nearly zonal waves can carry a packet eastward relative to the current while their crests move westward. A group-velocity statement requires a slowly varying, propagating packet and a smooth dispersion branch; turning regions and critical levels require further analysis.

Interior PV gradients are not the only support. In a stratified quasi-geostrophic approximation fluid the inversion operator includes ${\partial }_z[(f^2/N^2){\psi }_z^{\prime }]$. The boundary buoyancy $b^{\prime }=f{\psi }_z^{\prime }$ also supplies inversion data. A meridional boundary-buoyancy gradient is equivalent, in the inversion problem, to a boundary PV sheet. Consequently Boundary Rossby waves can exist even if the interior PV gradient vanishes. Their perturbations decay away from the boundary over a horizontal-wavenumber-dependent Rossby height; opposite boundaries can support oppositely directed intrinsic waves because their effective sheet gradients have opposite signs.

The Eady model makes this concrete. For $U=\Lambda z$, constant $N$ and horizontal wave-vector magnitude $k_h$, put $K=N{k}_h/|f|$. Zero interior PV gives ${\psi }_{zz}^{\prime }-K^2{\psi }^{\prime }=0$, and each rigid boundary has the buoyancy conditionA lower isolated boundary has ${\psi }^{\prime }\propto e^{-Kz}$, so $c_{\mathrm{lower}}=U(0)+\Lambda /K$. An upper isolated boundary has ${\psi }^{\prime }\propto e^{K(z-H)}$, so $c_{\mathrm{upper}}=U(H)-\Lambda /K$. For positive shear, the lower wave is intrinsically eastward and the upper wave intrinsically westward, entirely consistently with their opposite effective PV gradients.

In the finite-depth Eady dispersion relation, $\gamma =KH/2$. For $\gamma \gg 1$, $\tanh \gamma$ and $\coth \gamma$ both tend to one. The square root therefore tends to $\gamma -1$, givingThese are the two uncoupled Eady edge waves, not a pair of opposite branches on a single boundary. Their overlap is exponentially small in $KH$. For smaller separation, the interaction can phase-lock the counterpropagating boundary waves; the square root becomes imaginary for $0<KH<2.399\dots$, yielding Eady instability. The short-wave limit is stable and illustrates why interior-PV-only reasoning misses boundary wave propagation and baroclinic instability.

The governing balance requires a small Rossby number, weak fractional height or density perturbations, stable stratification where relevant, and frequencies slow relative to $|f|$. The local gradient must change little across a wavelength; on a planetary beta-plane $\beta \ell /|f|\ll 1$ is the associated small parameter. A hydrostatic stratified example also requires a small vertical-to-horizontal aspect ratio and $N$ large compared with its slow frequency. For reference, midlatitude $|f|\sim 10^{-4}{\mathrm{s}}^{-1}$ and $\beta \sim 2\times 10^{-11}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$. Atmospheric deformation lengths of order $10^6\mathrm{m}$ give long-wave speeds of order $\beta L_D^2\sim 20{\mathrm{m}\mathrm{s}}^{-1}$ and synoptic-to-planetary times of days to weeks. An oceanic baroclinic deformation length of order $3\times 10^4\mathrm{m}$ instead gives about $0.02{\mathrm{m}\mathrm{s}}^{-1}$ and much longer basin-crossing times. Structural constants and background advection matter; these are scale estimates, not universal measured speeds.

Finite-amplitude Rossby waves can overturn or strongly filament PV contours, especially near a critical layer where their phase speed approaches the local current. Smooth inviscid motion still conserves each parcel's PV. Apparent mixing comes from the resulting fine structure under coarse-graining or from weak diffusion; it can flatten a coarse-grained background PV gradient and sharpen neighboring jet-edge gradients. A simplest description of Rossby-wave PV mixing and zonal momentum is a downgradient fluxFor barotropic zonal averages with zero mean meridional transport, $u^{\prime }=-{\psi }_y^{\prime }$, $v^{\prime }={\psi }_x^{\prime }$ and $q^{\prime }={\nabla }_H^2{\psi }^{\prime }$. Zonal integration by parts gives the Taylor identity for quasi-geostrophic fluxThus mixing down a positive PV gradient produces westward acceleration in the mixing region: wave-generated Reynolds stress redistributes zonal momentum. The inversion relation ${\overline{Q}}_y=\beta -{\overline{U}}^{\prime \prime }$ then relates PV homogenization to a reshaped current. Total zonal momentum cannot simply disappear in an isolated domain; integrated stress convergence is a boundary flux, so compensating momentum changes or boundary forcing must accompany localized drag.

In small conservative waves with $B>0$, the quadratic wave activity is $\mathcal{A}=\overline{q^{\prime 2}}/(2B)$. Multiplying the linear PV equation by $q^{\prime }/B$ gives ${\mathcal{A}}_t=-\overline{v^{\prime }{q}^{\prime }}$ for a zonal average on a locally uniform current. Together with the momentum equation this yields ${\partial }_t(\overline{U}+\mathcal{A})=0$ during conservative growth or decay. In a commonly used momentum convention the Rossby-wave pseudomomentum is $-\mathcal{A}$; signs must be stated rather than inferred from the word activity. Breaking and absorption invalidate the reversible small-wave description and deposit the wave's momentum through its flux convergence. A steady undamped conservative wave cannot provide sustained local drag without a flux divergence or boundary exchange, as expressed by the non-acceleration theorem for quasi-geostrophic waves.

Rossby propagation is the slow, direction-selective motion produced by PV advection coupled to balanced inversion. Interior gradients and boundary sheets share that mechanism; inversion determines the scale dependence, while wave packets, boundary-wave interaction and breaking govern how energy and momentum reach other regions.

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.

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.

Write $\nu =\mu /\rho$ and average the horizontal Navier-Stokes equation, with zero-mean fluctuations and incompressible flow. The mean flow is stationary, has no vertical velocity and has no horizontal pressure gradient. The mean advection of $U(z)$ is therefore zero. Incompressibility rewrites the averaged nonlinear fluctuation term as a divergence:These correlations transport mean momentum and produce Reynolds stress. Horizontal statistical homogeneity makes the horizontal derivatives of the averaged correlations vanish. It does not require the instantaneous turbulent derivatives themselves to vanish. Only the vertical divergence remains, givingMultiplying by density and integrating once gives constant-stress Reynolds-averaged wall flow:Here ${\tau }_d$ is physical shear stress. Positive mean shear commonly has $\overline{u^{\prime }{w}^{\prime }}<0$, representing downward transport of positive horizontal momentum. The Reynolds term then contributes positively to the total stress.

The printed target formula has an extra vertical derivative in the Reynolds contribution. That term has dimensions of stress per length, while $\mu U_z$ has dimensions of stress, so the formula as printed is inconsistent. The derivative belongs in the stress-divergence equation above; the boxed integrated stress is the corrected result. Part b also uses a kinematic rather than physical stress, so the density factor must be tracked explicitly.

Define the friction velocity $u_{\ast }=\sqrt{{\tau }_d/\rho }$ for positive wall shear. In the near-wall region turbulence is suppressed and molecular viscosity carries the stress. With no slip, $U\simeq u_{\ast }^{2}z/\nu$. The viscous wall coordinate $z^{+}=z{u}_{\ast }/\nu$ becomes order one at the characteristic viscous sublayer lengthThe numerical extent of viscosity dominance depends on the crossover convention and wall condition; the scaling is the required viscous length.

In the Boussinesq approximation, replace density by a constant reference density ${\rho }_0$ in inertia and incompressible flow, retaining the small density departure only in the gravitational force. Write $\rho =\bar{\rho }(z)+{\rho }^{\prime }$ and subtract the resting hydrostatic pressure. The upward buoyancy perturbation acceleration and squared buoyancy frequency areThus stable density stratification has $N^2>0$. Let $p$ henceforth denote the pressure anomaly divided by ${\rho }_0$. The ideal, nonrotating Boussinesq equations areThe buoyancy equation follows by materially conserving the total buoyancy $b_0(z)+\sigma$, where $b_0^{\prime }=N^2$; it does not assume constant $N$.

For an $x$-periodic flow of period $P$, define $\bar{a}=P^{-1}{\int }_0^{P}adx$. An infinite-domain average with vanishing end contributions works as well. The necessary pressure condition is no mean zonal pressure gradient, $\overline{p_x}=0$: periodic pressure satisfies it, whereas a pressure containing a term proportional to $x$ does not. Use incompressible flow to write the zonal momentum equation in conservative form,Averaging its zonal derivative terms yields the exact Reynolds stress relation

For two-dimensional small disturbances about rest, the Linearized Boussinesq equations are $u_t=-p_x$, $w_t=-p_z+\sigma$, ${\sigma }_t=-N^{2}w$, and $u_x+w_z=0$. Differentiate the two momentum equations to eliminate pressure: $(u_z-w_x{)}_t=-{\sigma }_x$. Differentiating in $x$ and using $u_{xz}=-w_{zz}$ gives ${\nabla }^2{w}_t={\sigma }_{xx}$. One further time derivative givesThe equation remains valid for $N=N(z)$. No vertical derivative of $N$ was taken: ${\partial }_x^2(N^{2}w)=N^2{w}_{xx}$.