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For constant opacity, the ratio $T^3/\rho =3c\lambda \mathcal{R}/(4\sigma )$ cancels the height dependence in the radiative diffusion coefficient. Since $T_z=-{\Omega }^{2}z/[4(1+\lambda )\mathcal{R}]$, the radiative flux isConsequently $F_z=c\lambda {\Omega }^2/[\kappa (1+\lambda )]$. Equating this to the viscous heating rate givesThus the dynamic viscosity is uniform in height even though pressure and mass density decrease. Its units are mass per length per time, since $c/\kappa$ has those units. A pointwise pressure-proportional alpha prescription would be incompatible with this exact profile, which is why the subsequent alpha relation is vertically averaged.

Take the positive constant ratio $\lambda =p_r/p_g$. The equation of state impliesThis is a polytropic equation of state of index three. Differentiating the pressure and using vertical hydrostatic equilibrium gives $4(1+\lambda )\mathcal{R}\rho T_z=-\rho {\Omega }^{2}z$. HenceDefine the semi-thickness $H$ by $T(H)=0$ and put $s=1-z^2/H^2$. Then the constant gas-radiation ratio vertical disc structure iswithThe surface density is the integral across both faces. The supplied polynomial integral gives ${\int }_0^1(1-x^2{)}^{3}dx=16/35$, so $\Sigma =(32/35){\rho }_{0}H$. Substitution yieldsAll coefficients here are local in radial position; the derivation fixes the vertical profile rather than a global accretion rate. The zero-temperature surface is a formal boundary of this bulk model; optically thick diffusion ultimately fails in a thin photospheric layer.

The first relation is vertical hydrostatic equilibrium. In a thin disc, the central object's gravity near the midplane is $g_z=-{\Omega }^{2}z$; the pressure gradient balances it. The second is an equation of state combining ideal gas pressure with isotropic radiation pressure. Write $\mathcal{R}=k/({\mu }_m{m}_p)$, so $p_g=\mathcal{R}\rho T$ and $p_r=4\sigma T^4/(3c)$; $\sigma$ is the Stefan-Boltzmann constant and ${\mu }_m$ the mean molecular weight in proton-mass units.

The third is local energy balance: the increase of upward radiative flux with height equals viscous dissipation. Here $\mu$ is dynamic viscosity, not mean molecular weight. A Keplerian disc has shear $rd\Omega /dr=-3\Omega /2$, so its heating rate is $\mu (r{\Omega }^{\prime }{)}^2=9\mu {\Omega }^2/4$. The fourth is optically thick radiative diffusion, with opacity per unit mass $\kappa$ and diffusion directed down the temperature gradient. Midplane symmetry gives $F(0)=0$ and oppositely directed fluxes from the two faces. The four relations express vertical force balance, gas–radiation thermodynamics, local heating, and diffusive cooling.

The required pressure integral uses ${\int }_0^1(1-x^2{)}^{4}dx=128/315$. With $\Sigma =(32/35){\rho }_{0}H$ and $p_0/{\rho }_0={\Omega }^2{H}^2/8$,The vertically averaged alpha closure for a mixed-pressure layer therefore givesThe height-integrated dynamic viscosity is $2H\mu =8cH\lambda /[9\kappa (1+\lambda )]$. Equate it to $\bar{\nu }\Sigma$ to obtainThese are integrated equalities, not an imposed pointwise relation $\mu (z)\propto p(z)$.

Now keep $\alpha$, the opacity and molecular constants independent of radius, and use Keplerian rotation $\Omega \propto r^{-3/2}$. In the gas-dominated limit, $\lambda \ll 1$ but positive, the last equality gives $\lambda \propto H\Omega \Sigma$. The column-density relation gives $\lambda \Sigma \propto H^7{\Omega }^6$. Eliminating $\lambda$ yields $H^6\propto {\Sigma }^2{\Omega }^{-5}$, whenceThis is the gas-pressure branch of a mixed-pressure alpha disk. In the radiation-dominated limit, $\lambda /(1+\lambda )\to 1$, so $H\propto (\Omega \Sigma {)}^{-1}$ andFor viscous stability of an accretion disk, linearize the Keplerian viscous diffusion equation at fixed radius: a local mass density perturbation has diffusion coefficient $3{\partial }_{\Sigma }(\bar{\nu }\Sigma )$. The gas branch has $\bar{\nu }\Sigma \propto {\Sigma }^{5/3}$ and a positive derivative, so it smooths perturbations. The radiation branch has $\bar{\nu }\Sigma \propto {\Sigma }^{-1}$ and a negative derivative, giving radiation-pressure viscous instability. The gas-pressure branch is viscously stable; the radiation-pressure branch is viscously unstable in this closure. This concerns radial mass-transport stability on wavelengths where the vertically averaged thin-disc model applies.

The steady incompressible shearing sheet equation isThe final term follows from the local tidal potential $-\Omega S{x}^2$. For the given linear velocity, the advective acceleration is $-\alpha \beta (x{\mathbf{e}}_x+y{\mathbf{e}}_y)$, while the Coriolis acceleration is $2\Omega \beta x{\mathbf{e}}_x+2\Omega \alpha y{\mathbf{e}}_y$. Equating components givesThe cross derivatives vanish, so the pressure Hessian of an elliptical shearing-sheet vortex integrates toThe additive constant is fixed by exterior matching or the pressure reference. No zero-pressure free-boundary condition is imposed on an incompressible vortex patch: it is embedded in a surrounding fluid.

Write $B=2\Omega -S$, $A(t)=c_s^2-2\pi G{\Sigma }_0/k(t)$, and $w=i{k}_x{\tilde{v}}_y-i{k}_y{\tilde{v}}_x$. Differentiation, including ${\dot{k}}_x=S{k}_y$, givesThe two pressure/self-gravity cross terms cancel. Continuity gives ${\dot{\tilde{\Sigma }}}^{\prime }=-i{\Sigma }_0(k_x{\tilde{v}}_x+k_y{\tilde{v}}_y)$, so $\dot{w}=B{\dot{\tilde{\Sigma }}}^{\prime }/{\Sigma }_0$. ConsequentlyThe conserved amplitude is linearized vortensity: the perturbation of the absolute vertical vorticity divided by surface density,A two-dimensional inviscid barotropic fluid materially conserves this potential vorticity. Both pressure enthalpy and gravitational forces are gradients, so they create no baroclinic vorticity source. Since $q_0$ is uniform, its linear perturbation obeys $D_0{q}^{\prime }=0$, explaining why a shearing wave has constant vortensity amplitude.

Apply the shearing-wave ansatz to each disturbance. Acting on its phase givesThus a spatially uniform amplitude system exists whenFor a nonzero horizontal wavenumber $k=|\mathbf{k}|$, the razor-thin disk Poisson kernel follows from $({\partial }_z^2-k^2){\tilde{\Phi }}_d^{\prime }=4\pi G{\tilde{\Sigma }}^{\prime }\delta (z)$. The potential decays above and below the sheet as ${\tilde{\Phi }}_{d,m}^{\prime }{e}^{-k|z|}$. Integrating across the midplane gives the jump $-2k{\tilde{\Phi }}_{d,m}^{\prime }=4\pi G{\tilde{\Sigma }}^{\prime }$, henceSubstitute this potential and the phase evolution into the linearized equations. The barotropic shearing-wave gravitational amplitude system isThe self-gravity sign is attractive and opposes the pressure response. The uniform $k=0$ mass density shift requires its own gravitational reference and is not obtained by dividing by $k$.

Let ${\Sigma }_0$ be the uniform background surface density, ${\mathbf{u}}_0=-Sx{\mathbf{e}}_y$, and write $\Sigma ={\Sigma }_0+{\Sigma }^{\prime }$, $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$, ${\Phi }_d={\Phi }_{d0}+{\Phi }_d^{\prime }$. The background sheet has no horizontal gravitational gradient; its vertical field is irrelevant to the stated planar equations. Put $D_0={\partial }_t-Sx{\partial }_y$. Keeping first-order terms givesThe $-S{v}_x$ contribution comes from $\mathbf{v}\cdot \nabla {\mathbf{u}}_0$, while the Coriolis acceleration supplies the remaining coupling. The prescribed tidal potential has no perturbation, and its background force already balances the background Coriolis term. The isothermal pressure linearizes to $c_s^2\nabla {\Sigma }^{\prime }/{\Sigma }_0$.

For $k_y=0$ and nonzero $k_x$, the wavenumber is constant. Continuity gives ${\dot{\tilde{\Sigma }}}^{\prime }=-i{k}_x{\Sigma }_0{\tilde{v}}_x$, and the conserved vortensity relation givesDifferentiate continuity and substitute radial momentum and this identity. The forced axisymmetric density mode with vortensity obeysThe constant forcing represents the stationary vortical or balanced component, which is lost if every mode is assumed to have nonzero frequency. Let $k_{\ast }=\pi G{\Sigma }_0/c_s^2=\kappa /(Q{c}_s)$. Completing the square givesFor $Q>1$, every nonzero radial wavenumber has positive ${\omega }_k^2$. The solution is a constant balanced offset $-2\Omega {\Sigma }_0^2{\tilde{q}}^{\prime }/{\omega }_k^2$ plus bounded sine and cosine oscillations. Pressure stabilizes short wavelengths, rotation stabilizes long wavelengths, and self-gravity is strongest at intermediate wavelengths.

For $Q<1$, the Toomre stability criterion fails in the bandThere ${\omega }_k^2<0$ and the homogeneous solutions grow or decay exponentially. The fastest rate is $s_{\max }=\kappa \sqrt{Q^{-2}-1}$ at $|k_x|=k_{\ast }$. Outside the band the modes oscillate; at its edges the frequency vanishes.

At $Q=1$, ${\omega }_k^2\ge 0$ and it vanishes only at $|k_x|=\kappa /c_s$. At that critical wavenumber,This Toomre marginal algebraic growth is spectrally marginal, not bounded for all data: a nonzero conserved vortensity gives quadratic secular growth, and even zero vortensity can allow the linear term. Other wavenumbers still oscillate about a balanced offset. $Q>1$ gives bounded axisymmetric modes, $Q<1$ allows exponential instability, and $Q=1$ has a zero-frequency marginal mode with possible algebraic growth. These statements are axisymmetric; nonaxisymmetric shearing disturbances can have transient swing behavior. For $k_x=k_y=0$, continuity instead fixes the uniform mass density shift, while the velocities execute epicyclic motion; no $1/k$ gravitational formula is used.

Write $u=u_x$ and $B=B_z$. In this geometry the Lorentz force density is purely longitudinal and equals $-{\partial }_x(B^2/2{\mu }_0){\mathbf{e}}_x$. The ideal magnetohydrodynamic induction equation becomes $B_t+{\partial }_x(uB)=0$. Together with the continuity equation and adiabatic pressure evolution, the four primitive-variable equations areThey have the required quasilinear form ${\mathbf{U}}_t+A{\mathbf{U}}_x=0$ withHere the notation $u_x$ in derivatives means ${\partial }_{x}u$, whereas $u$ denotes the longitudinal velocity. Spatial dependence is one-dimensional; time dependence is allowed in the wave problem.

Put $c_f^2=\gamma p/\rho +B^2/({\mu }_0\rho )$, and assume $\rho >0$ and $c_f>0$. Expansion of the characteristic polynomial givesTwo independent eigenvectors for ${\lambda }_0$ can be chosen as $(1,0,0,0{)}^T$ and $(0,-B/{\mu }_0,0,1{)}^T$. For the two propagating branches, useful eigenvectors areThe repeated branch advects an entropy disturbance and a disturbance with unchanged total gas-plus-magnetic pressure. The $\pm$ branches are perpendicular fast magnetosonic waves, with speed $c_f$ relative to the fluid.

On a uniform background, a Fourier amplitude $\hat{\mathbf{U}}{e}^{i(kx-\omega t)}$ obeys $(kA-\omega I)\hat{\mathbf{U}}=0$. Thus the eigenvalues are the phase velocities $\omega /k$ for $k\ne 0$, and the four dispersion branches are $\omega =ku$ twice and $\omega =k(u\pm c_f)$. The reduced geometry retains neither an independent transverse velocity nor an additional propagating Alfvén wave.

To construct each finite-amplitude simple wave in magnetohydrodynamics, trace a state-space curve with tangent $r_{\pm }$, using $\rho$ as its parameter. Integrating the pressure and magnetic components givesThe constants $K$ and $b$ are the entropy and transverse magnetic flux freezing invariants along this state curve. HenceSince $A{\mathbf{U}}_{\rho }={\lambda }_{\pm }{\mathbf{U}}_{\rho }$, substituting this curve into the original system reduces it to ${\rho }_t+{\lambda }_{\pm }(\rho ){\rho }_x=0$. By the chain rule,Thus the characteristic speed itself satisfies the Inviscid Burgers equation. For initial speed $L(a)$, its smooth solution is ${\lambda }_{\pm }=L(a)$ along $x=a+tL(a)$, until the characteristic curves intersect. This derivation supplies genuine state curves for simple waves, not merely a linear dispersion relation.

In the inertial frame the steady convective acceleration is $-R{\Omega }^2{\mathbf{e}}_R$. For a barotropic fluid, $\nabla p/\rho =\nabla h$, where $h$ is the barotropic enthalpy function. The equilibrium momentum equation therefore giveson each connected fluid region. Uniform rotation is essential: it makes the centrifugal acceleration derivable from the centrifugal potential $-{\Omega }^2{R}^2/2$.

The Cowling approximation sets the gravitational-potential perturbation to zero while retaining background gravity. The barotropic pressure force linearizes as $\delta (\nabla p/\rho )=\nabla \delta h$, with $\delta h=p^{\prime }/\rho =c_s^2{\rho }^{\prime }/\rho$. Denote its mode amplitude by $W$. The advective time derivative on a scalar mode is ${\partial }_t+\Omega {\partial }_{\varphi }$, so the stated positive-frequency exponential produces $i\sigma$ with $\sigma =\omega +m\Omega$. Linearizing the radial centrifugal term produces $-2\Omega v_{\varphi }$, while linearizing the azimuthal convective term produces $2\Omega v_R$. ThusThese are the pressure-gradient and Coriolis acceleration terms in the corotating-frame form. Independently, linearizing continuity equation givesThe equilibrium mass density is axisymmetric, so no additional azimuthal background-density derivative appears.

For $\sigma \ne 0$ and $D=4{\Omega }^2-{\sigma }^2\ne 0$, invert the horizontal momentum system:Inserting these into the continuity equation, dividing by $i\sigma$ and multiplying by $D$ givesThe two terms proportional to $\rho W_R$ cancel; this cancellation leaves precisely the background-density derivative in the last term. Since $-D/{\sigma }^2=1-4{\Omega }^2/{\sigma }^2$, this is exactlyThe singular frequency cases require the original velocity equations rather than this inversion.

In the low-frequency anelastic approximation for a rotating barotropic star, neglect the left side. Spherical background mass density satisfies ${\rho }_R/R={\rho }_z/z$, understood by its smooth limiting form on the equatorial plane. For $W=z{R}^m$,The remaining equation reduces toWriting $q=2\Omega /\sigma$, the bracket factors as $(m+1-q)(q+1)$. The regular desired branch is thereforeThe other algebraic factor corresponds to $\sigma =-2\Omega$, where the eliminated horizontal system is singular; it is not classified by this pressure-equation inversion. A constant-density background makes the bulk factor identically zero but does not invalidate the displayed solution.

For the regular branch, an explicit velocity check is particularly informative. Up to a common mode normalization,It satisfies $R{v}_R+z{v}_z=0$ and $\nabla \cdot \mathbf{v}=0$. Thus $\nabla \cdot (\rho \mathbf{v})=0$ for every spherical mass density profile, directly verifying the anelastic continuity condition. The motion is tangential to spherical shells and is the sectoral inertial mode of a slowly rotating barotropic star. For single-valued azimuthal modes, $m$ is a positive integer. With the source's exponential convention,For $m\ge 2$ the pattern is retrograde relative to the star but prograde in the inertial frame. This is a leading slow-rotation anelastic mode, not an exact solution of the compressible equations whose left-hand term was discarded.

Consider an unforced incompressible flow with localized vorticity and sufficient smoothness for integration by parts. Define the hydrodynamic impulse per unit density by $\mathbf{I}=\frac{1}{2}\int \mathbf{x}\times \mathbf{\omega }dV$, denoting the vector impulse by $\mathbf{I}$ to distinguish it from the later scalar $L$. The vorticity equation can be writtenThe supplied identity, integrated over all space, gives $\frac{1}{2}\int \mathbf{x}\times \nabla \times \mathbf{F}dV=\int \mathbf{F}dV$, because $\mathbf{F}$ is localized and its surface terms vanish. ThereforeThe second integral is a vanishing surface integral. The vector identity $\mathbf{u}\times \mathbf{\omega }=\nabla (|\mathbf{u}{|}^2/2)-(\mathbf{u}\cdot \nabla )\mathbf{u}$ and incompressibility reduce the first to surface terms proportional to the quadratic velocity. Those vanish for $\mathbf{u}=O(R^{-3})$. Thus

For a large spherical control volume define ${\mathbf{P}}_R={\int }_{V_R}\mathbf{u}dV$. The impulse relation is $\mathbf{I}=(3/2){\lim }_{R\to \infty }{\mathbf{P}}_R$ for this localized field. The factor is important: impulse is not simply the integral of velocity, since an $R^{-3}$ velocity tail has a nonvanishing contribution in the impulse integration by parts involving a factor of $R$.

Using pressure per unit density $\Pi$ and the Navier-Stokes equation givesThe advective flux is $O(R^{-4})$, the pressure force is $O(R^{-1})$, and the viscous term is $O(\nu R^{-2})$ under the regular far-field derivative decay. All vanish as $R\to \infty$. The localized eddy cannot change its total impulse without a net external momentum input. Pressure redistributes momentum internally but its resultant force on an increasingly distant control surface tends to zero. At finite $R$, boundary forces need not vanish exactly; the all-space limit is the invariant statement.

Write $u^2=⟨u_x^2⟩$ for the one-component variance and $Q_{LL}=u^{2}f(r)$. Isotropy gives $Q_{ij}=u^2[(f-g)n_i{n}_j+g{\delta }_{ij}]$, and incompressibility gives $g=f+r{f}^{\prime }/2$. Taking the trace yieldsTherefore, with a regular correlation at the origin,For $L>0$, the longitudinal velocity correlation has the tail $u^{2}f(r)\sim L/(4\pi r^3)$. Its physical source is the velocity induced far from a localized eddy with nonzero hydrodynamic impulse. The hydrodynamic Biot-Savart kernel gives the dipolar far fieldThe absence of a vorticity monopole and the surviving impulse dipole give the $r^{-3}$ power. Random eddy impulses thus produce long-range velocity correlations despite the vorticity itself being concentrated in localized structures. Incompressibility also explains how a distant pressure response can redistribute velocity without transporting local vorticity there.

There is no inconsistency with integrability of the trace correlation. At leading order, $g\sim -f/2$, soThe longitudinal and transverse dipolar tails cancel in the trace. The scalar Saffman integral can remain finite even though individual tensor components have long-range, direction-dependent tails.

Multiply the Kármán-Howarth equation for $C(r,t)$ by $4\pi r^2$ and integrate in $r$. Integration by parts gives the exact boundary-flux expressionAt the origin smooth isotropic correlations give $K=O(r)$ or smaller and $C^{\prime }=O(r)$, so both boundary terms vanish. At infinity the usual regular power-law decay $K=O(r^{-4})$, with the corresponding derivative decay, makes the first term tend to zero. A sufficiently smooth integrable trace correlation makes $r^2{C}^{\prime }\to 0$ as well. Under these physical far-field conditions,The exact required conditions are the limits of these two boundary fluxes. A bare big-$O$ condition on $K$ without control of its derivatives is not by itself a rigorous sufficient condition: the smooth tail $K(r)=r^{-4}\sin (r^2)$ is $O(r^{-4})$, but $r^{-1}(r^{4}K{)}^{\prime }=2\cos (r^2)$ has no zero limit. The intended power-law assumption excludes such increasingly rapid far-field oscillations; this qualification is needed in a literal proof.

Physically, $L$ is the large-volume momentum variance divided by volume. Internal advective and pressure forces exchange momentum among eddies but do not provide a net impulse to the whole unforced field. For a large spherical region, the variance balance is driven by its boundary momentum flux. The stated correlation decay makes that contribution negligible per unit volume as the region grows. Thus the conservation law preserves momentum-fluctuation density, not kinetic energy. kinematic viscosity can remove kinetic energy while leaving this integral invariant.