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Kelvin's circulation theorem states that a smooth inviscid barotropic fluid subject to a conservative body force preserves circulation around every closed material curve. Write the pressure term as $\nabla w$ with $dw=dp/\rho$, and the body force per mass as $-\nabla \Phi$. Then the Euler equations give $D\mathbf{u}/Dt=-\nabla (w+\Phi )$. For a moving parametrization $\mathbf{x}(s,t)$ with ${\partial }_t\mathbf{x}=\mathbf{u}$,The last integral is the increment of a single-valued scalar around a closed curve. ConsequentlyConstant density is a special barotropic case with $w=p/\rho$. Viscosity, nonconservative forcing or baroclinic pressure gradients can invalidate this conservation law; the exterior-cylinder conclusions use the smooth inviscid hypotheses.

In the dominant central Newtonian gravitational potential, conservation of energy for gas falling from rest at infinity gives $u^2/2=G{M}_c/R$, hence $u=a{R}^{-1/2}$ with $a=(2G{M}_c{)}^{1/2}$. The characteristic stellar speed has the same radial scaling. For random stellar directions, the cross term in $|\mathbf{u}-\mathbf{v}{|}^2$ averages to zero, so the typical relative speed is also $V=b{R}^{-1/2}$. With the stated kinetic-energy estimate, an rms convention gives $V^2\simeq 4G{M}_c/R$; another convention changes only the constant $b$.

In the steady spherical inflow approximation, mass conservation gives $A=4\pi R^2\rho u$. ThusThis leading calculation treats the inward gas flux as essentially undepleted over the region used. Substantial distributed removal by stars would make the local flux a function of $R$, an important limitation discussed below.

Insert these scalings in the Bondi--Hoyle--Lyttleton accretion rate. The factors $R^{3/2}$ from $V^{-3}$ and $R^{-3/2}$ from $\rho$ cancel:This is quadratic accretion in a central Keplerian potential. Write its fixed coefficient as $\lambda =\Lambda$. Integrating $d(1/M)/dt=-\lambda$ givesThe cancellation of radius means that orbital variation of $R$ alone does not change this coefficient, provided the assumed steady gas profile, central mass and relative-velocity scaling remain fixed.

Conserve the number of stars along these deterministic mass trajectories. Inverting the map and taking its Jacobian determinant givesThe mass-spectrum transport under quadratic accretion is thereforeFor a narrow but nonzero initial mass interval, the mapping stretches the high end especially strongly. If its endpoints are $m_1<m_2$, their evolved ratio is $(m_2/m_1)(1-\lambda m_{1}t)/(1-\lambda m_{2}t)$, which can become large as $t$ approaches $1/(\lambda m_2)$ from below. In an intermediate high-mass range with $\lambda Mt\gg 1$, mapped back into a portion of the initial interval where $F_0$ changes little,Here $M_{\ast }$ is near $1/(\lambda t)$ on the populated side of the initial interval. Finite initial support gives a finite upper mass cutoff at every time before the earliest blow-up; this is an approximate broad-range power law, not a literal infinite tail of a finite pre-blow-up population. A strictly monochromatic initial population remains monochromatic under identical accretion coefficients and does not generate a spectrum.

The limitations of this accretion model are substantial. The star cluster must remain centrally dominated, and both the central mass and gas supply must remain effectively fixed. Stellar capture depletes the gas flux and eventually invalidates the $R^{-3/2}$ profile. Gas pressure, angular momentum, shocks, radiative feedback and winds can reduce capture relative to the cold Bondi–Hoyle accretion estimate; a broad relative-speed distribution is not faithfully represented by one typical speed, especially because capture scales as $V^{-3}$. A growing star can perturb the surrounding potential, overlap capture regions, change its orbit through momentum loading, or compete with its neighbors for the same gas. Finally, number conservation excludes mergers, new star formation and stellar losses. The formal finite-time mass divergence cannot persist under a finite available gas mass, so the inferred mass-spectrum slope applies only while these approximations remain adequate.

A stellar polytrope has a spatially constant coefficient $K$ in the relation $P=K{\rho }^{1+1/n}$; $n$ is its polytropic index. This relates pressure to mass density and need not describe an adiabatic gas. Combining hydrostatic equilibrium with mass conservation eliminates the enclosed mass:For $\rho ={\rho }_c{\theta }^n$, the polytropic equation of state gives $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Choose the Lane-Emden variables for a stellar polytrope withSubstitution then gives the dimensionless Lane-Emden equation:A regular stellar centre has $\theta (0)=1$, ${\theta }^{\prime }(0)=0$ and $m(0)=0$. In particular the regular central expansion is $\theta =1-{\xi }^2/6+O({\xi }^4)$. For an isolated finite-radius stellar polytrope with negligible external pressure, the surface is the first zero ${\xi }_1$ of $\theta$, and $R=\alpha {\xi }_1$. A core embedded in an envelope can instead end at a positive matching pressure, before this zero.

The integrated Lane-Emden equation is ${\xi }^2{\theta }^{\prime }=-{\int }_0^{\xi }{s}^2\theta (s{)}^{n}ds$. Thus mass conservation gives the Lane-Emden mass formula directly:For the proposed functional form, differentiation givesChoosing $\gamma =-1/2$ removes the extra polynomial factor; equality to $-{\theta }^5$ then requires $3\beta =1$. The regular polytrope of index five is thereforeIts enclosed mass and enclosed mean mass density areThere is no finite zero: its isolated radius is infinite. Nevertheless its total mass converges, since the outer mass density falls as $r^{-5}$:The zero mean mass density is a limiting volume average, not a statement that the material has zero mass density at finite radius.

To turn this density model into a CNO cycle burning model, use gas-dominated pressure and uniform mean molecular weight. The ideal gas law and the polytropic equation of state then imply $T/T_c=\theta$. Without this thermal assumption a polytropic equation of state for total pressure does not alone fix the temperature profile. Since the stellar energy-generation rate is power per unit mass, its contribution to luminosity is $ϵdm$, giving the luminosity integral of an index-five polytrope:Put $u=\xi /\sqrt{3+{\xi }^2}$. Then $\xi =\sqrt{3}u/\sqrt{1-u^2}$ and the dimensionless integral becomes $3\sqrt{3}{\int }_0^1{u}^2(1-u^2{)}^{8}du$. Evaluation givesThe PDF's power of $\alpha$ in the proposed luminosity scaling is a genuine dimensional error: it must be three, rather than two. The volume element supplies ${\alpha }^3$; replacing it by ${\alpha }^2$ would give power divided by length, and could not be repaired by a dimensionless numerical constant.

The polytrope of index five is unlikely to model a real burning stellar core well. An embedded core has a finite boundary, and a complete isolated star cannot have this infinite radius. More importantly, strongly temperature-sensitive CNO cycle heating concentrates the luminosity centrally and tends to produce a convection zone there. A well-mixed gas-pressure-dominated convective core is closer to an adiabatic stellar polytrope of index $3/2$ than to index five. A finite truncation would also change the integral defining $A$; its value above pertains to the full mathematical model.

In stellar homology, write $r=Rx$, $m=Mf(x)$ and $\rho ={\rho }_{c}h(x)$, with common dimensionless profiles. Integrating hydrostatic equilibrium from the surface, where the pressure is negligible, givesThe profile integral is common to the sequence. Similarly, mass conservation gives $M=4\pi {\rho }_c{R}^3{\int }_0^1{x}^{2}h(x)dx$. These prove the homology scaling of central stellar pressure:The ideal gas law consequently gives $T_c\propto \mu M/R$, with the gravitational and gas constants absorbed into the coefficient.

The radiative diffusion in a star equation has a characteristic gradient $T_c/R$. Hence $L\propto R{T}_c^4/({\kappa }_c{\rho }_c)$. Using Kramers' opacity law, and comparing fixed-composition homologous models, this becomesThe volume integral of the stellar energy-generation rate $ϵ={ϵ}_0\rho T^{\nu }$ instead givesThe dimensionless integrals behind these relations do not depend on $M$ along a homologous sequence. In stellar thermal equilibrium the two luminosities agree, proving Kramers homology with a variable nuclear exponent:For the specified proton–proton chain exponent $\nu =7/2$, the mass factor cancels. ThusThe model captures two important features of the lower main sequence: hydrogen burning primarily uses the proton–proton chain, and gas rather than radiation pressure supplies the principal support. A radiative interior and a Kramers' opacity law approximation are useful near the Sun. It is a simplified local model: the Sun has a convective envelope, the coolest main sequence stars become fully convective, and real opacities and burning exponents vary. In particular a literally constant radius is not a realistic law over the entire lower main sequence.

On this proton–proton chain branch $T_c\propto M$, because $R$ and mean molecular weight are fixed. Normalizing to the Sun gives the upper mass estimateFor the specified CNO cycle exponent $\nu =23/2$, the homology relation instead reads $R^{14}\propto M^8$. Substituting into the radiative luminosity relation givesThis branch retains the deliberately idealized fully radiative assumptions, even though real CNO cycle cores are often convective.

For a Hertzsprung-Russell diagram, use the Stefan–Boltzmann law, $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$. On the proton–proton chain branch $T_{\mathrm{eff}}\propto M^{11/8}$, so $L\propto T_{\mathrm{eff}}^4$. On the CNO cycle branch $T_{\mathrm{eff}}\propto M^{57/56}$, giving $L\propto T_{\mathrm{eff}}^{292/57}$. Thus the upper branch rises slightly more steeply toward the hot, luminous upper left in a logarithmic Hertzsprung-Russell diagram. The sketch below matches the approximate branches continuously at their respective burning crossovers; the true crossover has both heating terms and is smooth.

The proton–proton chain begins with reactions between hydrogen nuclei, so its specific heating scales as $X_H^2\rho T^{7/2}$ in this approximation. The catalytic CNO cycle involves hydrogen and CNO nuclei, and scales as $X_H{Z}_{\mathrm{CNO}}\rho T^{23/2}$. Here $X_H$ is the hydrogen mass fraction and $Z_{\mathrm{CNO}}$ the CNO mass fraction, not the total abundance of every metal. ConsequentlyKeeping $X_H$ fixed while dividing $Z_{\mathrm{CNO}}$ by 256 gives the CNO crossover temperatureThe mass estimate uses the unchanged proton–proton chain branch below crossover, together with the stipulated unchanged opacity and mean molecular weight.

The low-CNO proton–proton chain sequence therefore lies on the same locus, but extends to twice the transition mass. Beyond its later crossover the low-CNO CNO cycle sequence has the same logarithmic slope $292/57$ but a different normalization. At a fixed mass where both sequences are CNO-dominated, lowering the nuclear coefficient by 256 givesThus those low-CNO models are smaller, hotter and modestly more luminous at the same mass. At the same effective temperature, their CNO branch is lower in luminosity than the solar-composition CNO branch; these comparisons hold different quantities fixed.

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.

Let $g>0$ be the magnitude of the downward gravity and write ${\alpha }_T$ for the thermal expansion coefficient, to distinguish it from the dynamo coefficient of the previous question. The conductive state has $\mathbf{u}=0$, $\mathbf{B}=B_0\hat{\mathbf{z}}$ and $T_0(z)=T_r-\Delta Tz/d$. Its gravitational term is absorbed in a background hydrostatic pressure.

For the dimensional two-dimensional fields, choose the stream function and Cartesian magnetic flux function conventionsThese automatically give zero divergence. Define the Jacobian determinant bracket $J(f,h)=f_x{h}_z-f_z{h}_x$, so $\mathbf{u}\cdot \nabla h=J(\psi ,h)$. The full magnetic flux function, including the imposed field, must be retained in the magnetic force; using the perturbation alone would lose its linear restoring term.

The scalar vorticity is $(\nabla \times \mathbf{u}{)}_y=-{\nabla }^2\psi$. Also $(\nabla \times \mathbf{B}{)}_y=-{\nabla }^{2}A$, and henceThe buoyancy perturbation is $g{\alpha }_T(T-T_0)\hat{\mathbf{z}}$, whose curl in the $y$ direction is $-g{\alpha }_T(T-T_0{)}_x$. Taking the curl of the momentum equation and multiplying by minus one eliminates pressure and therefore givesWe have restored ${\mu }_0$ in this dimensional expression. The printed magnetic force uses units in which ${\mu }_0=1$, or equivalently a correspondingly rescaled field.

Use length scale $d$, time scale $d^2/\kappa$, velocity scale $\kappa /d$, stream-function scale $\kappa$, magnetic flux-function scale $B_{0}d$, and temperature perturbation scale $\Delta T$. Thus the dimensionless total flux function isHere $x,z$ now denote coordinates divided by $d$, and the same $T$ notation in the second identity denotes the physical temperature. The dimensionless parameters for thermal-diffusion scaling of planar magnetoconvection areThese are the Rayleigh number, Chandrasekhar number, Prandtl number and magnetic-to-thermal diffusivity ratio. The nondimensional magnetic-force coefficient is $B_0^2{d}^2/({\mu }_0{\rho }_r{\kappa }^2)=\sigma \zeta Q$, so omitting the factor $\zeta$ would use a different magnetic parameter.

The resistive induction equation becomes $A_t+J(\psi ,A)=\zeta {\nabla }^{2}A$. Since $J(\psi ,x)=-{\psi }_z$, its perturbation equation has source $+{\psi }_z$. Similarly, advection of the conductive temperature gradient gives $J(\psi ,-z)=-{\psi }_x$, producing the source $+{\psi }_x$ in the perturbation heat equation. The resulting nonlinear equations are thereforeThis derives both forcing signs from the background fields, rather than selecting them by a stream-function convention after the fact.

For the specified roll mode put $K=\pi /\lambda$ and $p_L=K^2+{\pi }^2$, the positive eigenvalue of minus the Laplacian on each spatial factor. Writeand take $\psi =(p_L/K)a{f}_s$, $\Theta =b{f}_T$, $\chi =\lambda c{f}_A$, with amplitudes depending on $\tau =p_{L}t$. These factors satisfy the printed boundary conditions: sine factors make the normal velocity and $B_x=-{\chi }_z$ vanish at the required faces; the second derivatives of the velocity factors give the stress-free conditions; $\Theta$ vanishes at the top and bottom and ${\Theta }_x$ vanishes at the sides.

Discard quadratic perturbation terms. Since ${\nabla }^{2}A={\nabla }^2\chi$, the magnetic force linearizes to $J(x,{\nabla }^2\chi )=({\nabla }^2\chi {)}_z$. The needed derivatives areThe heat and induction equations give $b^{\prime }=a-b$ and $c^{\prime }=a-\zeta c$. In the vorticity equation, its left side is $-(p_L^3/K)a^{\prime }{f}_s$ and its three right-side contributions are $\sigma [(p_L^3/K)a-RKb+\zeta Q{p}_L\pi \lambda c]f_s$. Consequently, withthe three-amplitude vertical-field magnetoconvection system isPrimes now mean $d/d\tau$. They are genuine time derivatives, as the original PDF shows.

For its linear stability, seek amplitudes proportional to $e^{s\tau }$. Eliminating $b,c$ or taking the determinant giveswhereAssume the physical range $\sigma >0$, $\zeta >0$, $q\ge 0$. The Routh-Hurwitz stability criterion gives strict decay exactly when $A_2>0$, $A_3>0$ and $A_1{A}_2>A_3$. At $r=0$ these hold. As $r$ increases, the first loss of stability can occur through a zero eigenvalue or a conjugate imaginary pair.

The stationary threshold isAt this value, $P(s)=s(s^2+A_{1}s+C)$ withIf $C>0$, the other two roots have negative real parts and the zero root is simple. Its derivative with respect to $r$ is $ds/dr=\sigma \zeta /C>0$. This is stationary onset; the reflection symmetry of the roll permits a generic pitchfork bifurcation. The linear calculation does not determine whether the nonlinear branch is supercritical or subcritical.

For oscillatory onset set $s=i\omega$, $\omega \ne 0$. Separating real and imaginary parts gives ${\omega }^2=A_2$ and $A_1{A}_2=A_3$. Solving the latter for $r$ yieldsA genuine Hopf bifurcation onset requires ${\omega }_H^2>0$, soAt this threshold $P(s)=(s+A_1)(s^2+{\omega }_H^2)$, giving the third root $-A_1<0$. The pair crosses transversely: differentiating $P(s,r)=0$ at $i{\omega }_H$ gives

The first instability of the conductive state is therefore stationary at $r_s$ if $\zeta \ge 1$, or if $0<\zeta <1$ and $q<q_{\ast }$. It is oscillatory at $r_H<r_s$ if $0<\zeta <1$ and $q>q_{\ast }$. At $q=q_{\ast }$, the two thresholds meet and $P(s)=s^2(s+A_1)$: the frequency is zero and neither a simple pitchfork nor a nonzero-frequency Hopf test applies. The oscillatory marginality in vertical-field magnetoconvection formula must not be used where its frequency squared is negative.

For conversion back to physical parameters, $R_{s,H}={\lambda }^2{p}_L^3{r}_{s,H}/{\pi }^2$, while the dimensional Hopf frequency is $(\kappa p_L/d^2){\omega }_H$. These are onset conditions for the specified spatial mode. Identifying the globally first mode would additionally require comparison over the allowed spatial modes. Nonlinear saturation and the stability of the bifurcating finite-amplitude states are not determined by this linear three-amplitude system alone.

Magnetic flux concentration and flux expulsion describe redistribution of an existing field, not necessarily creation of net flux. For a weak field in a prescribed conducting flow, the resistive induction equation isIts ideal part combines transport, stretching and compression:Thus a converging flow can intensify a component, and stretching can increase magnetic energy even if the signed net flux is zero. At large magnetic Reynolds number $\mathrm{Rm}=UL/\eta$, diffusion is initially slow on the eddy scale, so magnetic flux freezing carries field lines with the flow. These statements do not imply that an ideal closed eddy can simply lose its initially enclosed flux.

The missing step is the production of small scales by differential circulation. In a two-dimensional incompressible eddy, the Cartesian magnetic flux function obeysDifferent streamlines have different circulation rates. An initially smooth field is wound into fine folds; diffusion and reconnection then smooth the variation of $A$ inside the eddy. As $A$ becomes nearly uniform there, its gradient and hence the interior magnetic field become small. The surviving flux lies mainly near the circulation boundary and adjacent separatrices, where stronger gradients remain. This is flux expulsion. With exactly zero diffusivity, field topology and the flux of material loops remain frozen, so a relaxed flux-free interior is not obtained by pure advection alone. Rigid rotation also fails to generate the differential winding responsible for this process.

The phase mixing in magnetic flux expulsion estimate makes the enhanced role of small diffusion explicit. If the angular-velocity gradient is of order $U/L^2$, a wound radial wavenumber grows as $k_r\sim Ut/L^2$. The accumulated damping exponent is of orderIt becomes order one atThis is much shorter than the large-scale diffusion time $(L/U)\mathrm{Rm}$ at large $\mathrm{Rm}$, but longer than one turnover time. It is a kinematic differential-shear estimate, not a universal time for every turbulent or rapidly evolving convection cell.

A related concentration mechanism occurs at converging flow boundaries. For a locally vertical bundle, conservation of flux gives $B\mathcal{A}\simeq \mathrm{constant}$, so reducing its cross-section increases $B$. In a simple converging boundary region with strain rate $U/L$, balancing strain and transverse diffusion givesThis is the thickness estimate for that local strain-diffusion model; it need not be the thickness or energy scaling of every closed-eddy flux-expulsion problem. Stronger field eventually exerts Lorentz force feedback and changes the flow. The equipartition magnetic field $B_{\mathrm{eq}}=\sqrt{{\mu }_0\rho }U$ is a useful indication that the kinematic assumption can fail, not a universal upper bound on concentrated fields.

At the Sun's photosphere, hot upflows form solar granules, and their spreading horizontal motion transports weak vertical field toward cooler intergranular lanes and downflows. A thin-layer illustration, neglecting vertical transport terms, is ${\partial }_t{B}_z+{\nabla }_h\cdot ({\mathbf{u}}_h{B}_z)=0$. It gives $D_h{B}_z/Dt=-B_z{\nabla }_h\cdot {\mathbf{u}}_h$: horizontal convergence intensifies the vertical flux density. This surface transport is related to the redistribution by convective eddies, but is not identical to the closed-streamline mechanism in the preceding planar example. Larger-scale supergranulation also gathers magnetic elements toward its boundaries, helping organize the solar magnetic network.

Concentration can be further intensified by compressible thermal dynamics. In the thin-tube picture of convective collapse, an unstable downward flow drains gas, reduces internal pressure and allows contraction at approximately conserved flux. Lateral balance $p_i+B^2/(2{\mu }_0)\simeq p_e$ then permits field stronger than the local kinetic equipartition estimate. This evacuation mechanism is excluded by a strict Boussinesq approximation. It is not a description of every evolving surface concentration: gas pressure, dynamic stresses and magnetic curvature can all matter when thin-tube equilibrium fails. The resulting strong, intermittent magnetic elements are associated with intergranular structure; their radiative appearance also depends on thermal structure and opacity. The organized fields of sunspots require a broader account of flux supply and magnetic feedback, rather than attribution to weak-field expulsion alone.

Thus convection redistributes and concentrates flux, finite diffusion enables classical expulsion, and nonlinear magnetic and thermal feedback determines the strong-field outcome. An expulsion argument alone neither supplies the original solar field nor establishes a self-sustaining solar dynamo.

The vertical hydrostatic equilibrium equation balances the gradient of the total pressure against gravity. In a thin disk with Keplerian rotation, expanding the central gravitational field about the midplane gives vertical acceleration $-{\Omega }^{2}z$, so $p_z=-\rho {\Omega }^{2}z$. Disk self-gravity is neglected in this approximation.

The flux-gradient equation is local thermal balance: the upward radiative flux increases with height because differential rotation dissipates energy. For a Newtonian effective dynamic viscosity $\mu$, the volumetric heating rate is $\mu (r{\Omega }^{\prime }{)}^2$, which becomes $9\mu {\Omega }^2/4$ for Keplerian rotation. The symbol $\mu$ here is dynamic viscosity, not kinematic viscosity; the latter is $\nu =\mu /\rho$.

The radiative diffusion equation applies to an optically thick medium in local thermodynamic equilibrium. Its diffusion coefficient is $c/(3\kappa \rho )$, and the equilibrium radiation energy density is $u_{\mathrm{rad}}=4\sigma T^4/c$. HenceThe minus sign means energy flows down the temperature gradient. Here $\sigma$ is the Stefan-Boltzmann constant, $\kappa$ the opacity per unit mass, and $c$ the speed of light. Constant Thomson scattering opacity is the adopted transport approximation; local thermodynamic equilibrium is an additional assumption behind the radiation equation of state.

Finally the equation of state adds ideal gas gas pressure $p_g=\rho kT/({\mu }_m{m}_H)$ and isotropic radiation pressure $p_r=u_{\mathrm{rad}}/3=4\sigma T^4/(3c)$. The Boltzmann constant is $k$, and ${\mu }_m{m}_H$ is the mean mass per gas particle. Both contributions supply the total pressure entering vertical hydrostatic equilibrium.

Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.