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Let $u=u_r<0$ for accretion. The steady spherical continuity equation and radial Euler equations for an inviscid fluid giveFor a barotropic fluid, $dp/dr=c_s^{2}d\rho /dr$. Dividing continuity equation by $r^2\rho u$ gives ${\rho }^{\prime }/\rho =-2/r-u^{\prime }/u$. Eliminating this derivative yieldsAt a smooth transonic critical point, the coefficient of $du/dr$ vanishes. A finite derivative then requires the right side to vanish simultaneously:These are the sonic and regularity conditions. Otherwise the derivative is singular and the proposed smooth transonic passage fails. The local slope must also be a real root compatible with the desired branch. For example, defining $C=\rho d(c_s^2)/d\rho$ at the point and differentiating both sides givesThis explains why simultaneous vanishing is a necessary regularity condition, rather than an automatic proof that every potential admits a critical crossing.

Define the barotropic enthalpy function by $dh=dp/\rho$. The momentum equation integrates to the Bernoulli equationChanging the reference mass density in $h$ changes only the constant $B$. For an isothermal closure this logarithmic function is a barotropic barotropic pressure potential; it should not be confused with the constant thermodynamic specific enthalpy of an ideal gas held at fixed temperature.

For isothermal transonic accretion in the Paczyński-Wiita potential, assume $M>0$ and $c_s>0$, and define $r_B=GM/(2c_s^2)$ and $\beta =16c_s^2/c^2$. The critical equation iswhose two roots are $r_G+\frac{1}{2}{r}_B(1\pm \sqrt{1+\beta })$. Only the plus root lies outside $r_G$. ThereforeFor this isothermal case $C=0$, and the slope relation reduces toThe positive slope is the inward-accretion solution connecting a small inward speed at large radius to the supersonic inward branch. Thus the exterior critical point is a genuine nondegenerate transonic point.

Choose the reference mass density as ${\rho }_0$, so $h=c_s^2\log (\rho /{\rho }_0)$ and $B=0$ from the conditions at infinity. At $u_c=-c_s$,Substitution into $\dot{M}=4\pi r_c^2{\rho }_c{c}_s$ givesThis is the rate selected by the smooth transonic accretion solution. Arbitrary static or subsonic formal solutions are not assigned this rate merely by specifying conditions at infinity. As $r_G/r_B\to 0$, the result tends to the isothermal Bondi accretion rate $\pi e^{3/2}{\rho }_0(GM{)}^2/c_s^3$, a useful normalization check.

The flux representation gives $B_R=-{\psi }_z/R$ and $B_z={\psi }_R/R$. Therefore $\nabla \cdot \mathbf{B}=0$ and $\mathbf{B}\cdot \nabla \psi =0$. With magnetohydrodynamic mass loading $k=k(\psi )$,Thus the steady continuity equation is satisfied.

The cross product in the ideal magnetohydrodynamic induction equation simplifies toHence steady induction requires $\mathbf{B}\cdot \nabla (v/R)=0$. On a regular connected poloidal flux surface, this integrates locally toThis is the field-line angular velocity of an axisymmetric wind. It need not equal the gas angular velocity when $B_{\varphi }\ne 0$, because $u_{\varphi }=R\omega +k{B}_{\varphi }/\rho$. The regular-surface qualification matters: if $\psi$ is constant and the field is purely toroidal, induction alone cannot turn an arbitrary $v(R,z)/R$ into a function of $\psi$.

Put $\mathbf{F}=(\nabla \times \mathbf{B})\times \mathbf{B}$ and $E=|\mathbf{u}{|}^2/2+\Phi$. Dot the pressure-free momentum equation with $\mathbf{u}$. Since $\mathbf{B}\cdot \mathbf{F}=0$, axisymmetry givesThe azimuthal momentum equation in cylindrical coordinates similarly givesSubtracting $\omega$ times the second relation from the first, and using $\mathbf{B}\cdot \nabla \omega =0$, proves that $E-\omega R{u}_{\varphi }$ is constant along a flowing regular poloidal flux surface. For nonzero mass loading its corotating energy invariant of an axisymmetric magnetic wind isThis is the invariant labeled $ϵ$ in the question. It is a corotating mechanical energy, distinct from the total magnetohydrodynamic Bernoulli invariant, which includes electromagnetic energy flux. In standard notation it equals total specific energy minus $\omega$ times the total specific angular-momentum invariant.

Near the footpoint, negligible $k{B}_{\varphi }/\rho$ gives $u_{\varphi }\simeq R\omega$, while the poloidal velocity is parallel to the poloidal magnetic field. The gas approximately corotates and slides along a rotating field line as a bead on a wire, with effective potentialA decreasing effective potential converts corotating potential energy into poloidal kinetic energy. This is the mechanism of magnetocentrifugal acceleration; local launching does not alone guarantee global escape.

For an attractive point mass, take $\Phi =-GM/(R^2+z^2{)}^{1/2}$ and ${\omega }^2=GM/R_0^3$. Parameterize the outward straight line by $R=R_0+s\sin \alpha$, $z=s\cos \alpha$. The first derivative of ${\Phi }_{\mathrm{eff}}$ vanishes at $s=0$ because the footpoint is in circular radial balance. The meridional Hessian there is ${\omega }^2\mathrm{diag}(-3,1)$, soThus the thirty-degree magnetocentrifugal launching criterion in its strict negative-curvature form isAbove this angle the equilibrium is unstable to an outward displacement; below it the near-footpoint potential initially rises. The source's proportionality constant must be negative to describe attraction.

The marginal straight-line launch at thirty degrees has zero quadratic curvature. For precisely the straight line and point-mass potential here, let $x=s/R_0$ and $a=\sin \alpha$. Direct expansion givesAt $\alpha =\pi /6$, this is $-7x^3/16+O(x^4)$, so the outward direction is downhill at cubic order. Consequently the strict inequality is the nondegenerate quadratic criterion; it should not be interpreted as excluding every one-sided nonlinear marginal launch at equality. An exactly resting bead remains an equilibrium until displaced.

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.

For this force, $\dot{V}=V$, hence $V(t)=e^{t}v$ and $X(t)=x+(e^t-1)v$. Invert the characteristic flow map: the initial velocity for an endpoint $(x,v)$ is $e^{-t}v$, and the initial position is $x-(1-e^{-t})v$. Constancy along each characteristic curve givesThere is no amplitude factor here: this is scalar advection, not the conservative continuity equation for a compressible density.

Let $h$ be the water-layer thickness and $z=0$ the horizontal bedrock. In the hydrostatic approximation, with constant ice-interface pressure,The prescribed two-wall stress balance givesIn this subglacial current with constant viscous wall layers, the uniform-pressure ice roof is treated as a movable confining boundary; the closure neglects inertia in comparison with pressure and wall stress. The bulk may be turbulent, while the specified wall viscous boundary layer supplies the linear stress relation. A different turbulent drag law would give a different transport coefficient and similarity exponent.

Without melting, the depth-integrated continuity equation gives $h_t+q_x=0$, henceTo write this in the PDF's displayed plus-sign form, its constant must beThe signed value is essential. If the printed $\gamma$ were interpreted as a positive diffusivity, its equation would drive water uphill and be backward parabolic. For example, linearizing around a uniform depth $H$ would give a Fourier perturbation growth rate $+\gamma H^2{k}^2$, whereas the derived physical equation damps it at $-D{H}^2{k}^2$. The plus-sign form is consistent only with the negative $\gamma$ above.

For the motionless uniform layer, the heat equation reduces to $T_{zz}=0$. The lower temperature and upper heat-transfer condition determine the quasistatic temperature of a conductively cooled film:When $\alpha h/\kappa \ll 1$, $T_s=T_0-\Delta T\alpha h/\kappa +O(\Delta T(\alpha h/\kappa {)}^2)$. For a varying film let its vertical velocity scale as $Uh/L$, as required by incompressible flow. Relative to vertical thermal diffusion, horizontal thermal diffusion is smaller by $(h/L{)}^2$, and either horizontal or vertical thermal advection is smaller by $U{h}^2/(\kappa L)$. Time variation on the film's transport timescale $L/U$ has the same small factor. Thus the leading heat equation is again $T_{zz}=0$ at each $x,t$ when both specified dimensionless parameters are small. This argument retains the small $\alpha h/\kappa$ assumption. If thickness were externally varied on a faster timescale $t_h$, one would additionally need $h^2/(\kappa t_h)\ll 1$; the spatial conditions alone do not control arbitrary rapid time dependence.

Define $b={\gamma }^{\prime }\Delta T\alpha /\kappa >0$. The surface tension at the surface is ${\gamma }_s={\gamma }_0+bh$, so the Marangoni stress is ${\gamma }_{s,x}=b{h}_x$. With capillary pressure $p=-{\gamma }_0{h}_{xx}$ to leading order, horizontal Stokes flow in the film solves $\mu u_{zz}=p_x$, $u(0)=0$, $\mu u_z(h)={\gamma }_{s,x}$. HenceThe continuity equation gives the thermocapillary thin-film equationTo justify using constant surface tension in the capillary pressure, compare the two volume fluxes on horizontal scale $L$: their magnitudes are ${\gamma }_0{h}^4/(3\mu L^3)$ and $b{h}^3/(2\mu L)$. When both mechanisms are retained at leading order, their balance gives $bh/{\gamma }_0=O((h/L{)}^2)\ll 1$. Thus replacing ${\gamma }_s$ by ${\gamma }_0$ in the curvature term incurs only a relative small correction. Equivalently, small $\alpha h/\kappa$ ensures this if ${\gamma }^{\prime }\Delta T/{\gamma }_0$ is bounded; if it were parametrically large, the extra small surface-tension-variation condition would have to be imposed separately.

The capillary pressure is higher beneath a crest than beneath a trough, so pressure-driven volume flux drains the crest and smooths thickness. The Marangoni effect acts oppositely: a thicker region has a colder surface and therefore larger surface tension. Surface velocity is pulled towards that region, supplying more liquid and amplifying the thickness variation. The arrows below represent these two mechanisms separately.

Put $a_c={\gamma }_0/(3\mu )$ and $a_m=b/(2\mu )$. Substitution of $h=\hat{h}H$, $x=\hat{h}X/ϵ$, $t=\hat{t}\tau$ gives coefficients $a_m{ϵ}^2\hat{t}$ and $a_c{ϵ}^4\hat{t}/\hat{h}$. Making both equal to one yieldsIn these scales the fractional surface tension change is $b\hat{h}/{\gamma }_0=2{ϵ}^2/3$, making the previous approximation explicit.

The linear stability analysis about $H=1$ gives ${\eta }_{\tau }+{\eta }_{XX}+{\eta }_{XXXX}=0$. For a normal mode $e^{s\tau +ikX}$,Long waves with $0<|k|<1$ grow, $|k|>1$ decay, and $k=0,\pm 1$ are neutral. The neutral $k=0$ mode changes the mean film thickness. The dispersion relation is even in $k$, with its two maxima at $\pm 1/\sqrt{2}$.

For a steady positive film with zero volume flux, divide $H^2{H}_X+H^3{H}_{XXX}=0$ by $H^3$ and integrate to obtain $H_{XX}+\ln H=C$. Choosing $\hat{h}$ as specified fixes $C=0$. Multiplication by $H_X$ and another integration then give the zero-flux thermocapillary film profiles first integralFor $H>0$, $V^{\prime }=\ln H$ and $V^{\prime \prime }=1/H>0$: the unique minimum is $V(1)=-1$, while $V(0^{+})=0$ and $V(e)=0$. This potential energy interpretation classifies the profiles without assuming a sinusoidal shape at finite amplitude.

For $E=-1$, the only profile is the uniform film $H=1$. For $-1<E<0$, there are two positive turning heights $H_{-}<1<H_{+}<e$ and a smooth periodic film oscillates between them. Its period isFor $E=0$, the maximum is $e$ and the lower endpoint is zero. A drop reaches zero thickness in finite distance, with zero limiting slope but unbounded curvature since $H_{XX}=-\ln H$. Its half-width isThus the formal limiting profile consists of drops of peak height $e$ and footprint width $2\sqrt{\pi e}$, with zero limiting contact angle. Identical drops can touch, or a zero-thickness region can separate them in the degenerate zero-flux model. These are limiting wet-region solutions, not everywhere positive twice differentiable films; the singular contact region is not resolved by lubrication theory. A useful parametric drop profile isFinally let $E=-1+\delta E$ with small positive $\delta E$. Writing $H=1+\eta$ gives $V=-1+{\eta }^2/2+O({\eta }^3)$ and $H_{XX}=-\eta +O({\eta }^2)$. The small-amplitude steady profile is $H=1+\sqrt{2\delta E}\cos (X-X_c)+O(\delta E)$, with period tending to $2\pi$. Its wavenumber tends to the neutral boundary $|k|=1$ of the unstable band, not to the fastest-growing wavenumber.

Take $\Delta \rho >0$ and $\tau >0$, with $x$ pointing along the imposed shear stress, and put $G=\Delta \rho g$. Relative to the upper fluid's hydrostatic reference, the hydrostatic pressure in the current is $p=G(h-z)$. Thus the horizontal pressure gradient is $G\nabla h$. The lubrication theory momentum balance, lower no-slip boundary condition, and imposed upper shear stress giveIts depth-integrated volume flux isDefine $D=G/(3\mu )$ and $A=\tau /(2\mu )$. The continuity equation gives the shear-driven viscous gravity current equationThe two terms have opposite roles: imposed shear stress carries fluid downstream, whereas the hydrostatic pressure gradient spreads it from thick regions towards thin ones.

For the approximation, a representative horizontal velocity is $U\sim \tau H/\mu +G{H}^3/(\mu L)$, with vertical velocity $UH/L$. Small slopes require $H/L\ll 1$. Horizontal fluid inertia relative to vertical viscous resistance is ${\rho }_{c}U{H}^2/(\mu L)$, where ${\rho }_c=\rho +\Delta \rho$. Hence sufficient small parameters, expressed without an unknown velocity, areFor the hydrostatic pressure and normal-stress approximation also require $\mu U/(GHL)\ll 1$. Its gravity-driven part is already $(H/L{)}^2$; its shear-driven part addsThis condition controls viscous normal stress, vertical viscous corrections and the normal projection of shear at a slightly tilted interface relative to $GH$. With it and the previous inertia bounds, vertical inertia is small as well. Time variations here are on the transport timescale $L/U$; separately imposed rapid forcing would need its own unsteady inertia bound. A small ordinary Reynolds number is a stronger sufficient restriction, but the reduced inertia ratios above are what the thin-layer momentum balance directly requires.

In the steady far-downstream regime, cross-stream derivatives dominate gravity-driven spreading. Dropping the smaller downstream gravity flux leavesWrite $Y=y_N(x)$ and let $h_c$ be a typical central thickness. Balance gives $Y^2\sim (D/A)h_c^{2}x$, while conservation of the source volume flux gives $Q\sim A{h}_c^{2}Y$. Therefore the downstream similarity of a shear-driven gravity current hasThe omitted downstream gravity flux relative to the imposed shear flux is $D{h}_c^2/(Ax)\sim (Y/x{)}^2\ll 1$, so this approximation is self-consistent in the stated regime.

For the full profile put $w=h^2$. Then $h^3{h}_y=(w^2{)}_y/4$ andThis is a porous medium equation with downstream distance as its evolution coordinate. Conservation of $\int wdy=Q/A$ and a similarity solution $w=x^{-1/3}f(\eta )$, $\eta =y/x^{1/3}$, givewhere symmetry gives zero integration constant at $\eta =0$. Inside the positive region, $f^{\prime }=-2A\eta /(3D)$, so $f=(A/(3D))({\eta }_N^2-{\eta }^2)$. Requiring the total downstream volume flux to be $Q$ fixes the coefficient:Consequentlyfor $|y|<y_N$, with $h=0$ outside. The squared height is a parabolic Barenblatt solution; the height cross-section is a semicircular profile after rescaling its axes. The cross-stream volume flux vanishes at the edges even though the height slope becomes singular there. The profile describes the outer lubrication theory region, not a resolved microscopic front.

Near the point source the two horizontal dimensions are comparable, say $\ell$. Upstream spreading arrests where outward gravity-driven volume flux balances downstream shear-driven volume flux:Eliminating $h_{\ast }$ givesThis is a scaling estimate, not a determination of a numerical prefactor. The associated depth is $h_{\ast }\sim (\mu Q/(\Delta \rho g){)}^{1/4}$. The same horizontal scale is obtained by setting $y_N\sim x$ in the downstream similarity solution, confirming where that solution fails.

For the line source the steady volume flux per unit width is $q=A{h}^2-D{h}^3{h}_x$. On the downstream constant-height branch, $q=Q_{2d}$ gives $h_d^2=Q_{2d}/A$. Upstream there is no net flux through the finite nose, so $q=0$. Thus $h{h}_x=A/D$, and continuity of height at the source givesThe line-source upstream reach under imposed shear is thereforeIn the upstream part, shear-driven and gravity-driven volume fluxes cancel. The volume flux per unit width jumps by exactly $Q_{2d}$ at the source. As in the point-source profile, the square-root nose is a formal outer solution with an unresolved steep front; its finite upstream reach follows from flux balance, not from assuming a small slope all the way to the nose.

Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ and $\mathbf{L}=(\nabla \times \mathbf{B})\times \mathbf{B}/{\mu }_0$. The continuity equation converts a material specific-energy balance into a conservative energy density balance. Dot the ideal magnetohydrodynamic momentum equation with $\mathbf{u}$, and use the time independence of the Newtonian gravitational potential:For the energy density $U=p/(\gamma -1)$ associated with internal energy, the adiabatic pressure equation givesThe ideal magnetohydrodynamic induction equation and the cross-product divergence identity give the magnetic energy balanceIndeed $(\mathbf{u}\times \mathbf{B})\cdot (\nabla \times \mathbf{B})=-{\mu }_0\mathbf{u}\cdot \mathbf{L}$. The magnetic work cancels the kinetic magnetic work. The remaining pressure terms are $-\nabla \cdot (p\mathbf{u})$. Adding all three balances proves ideal magnetohydrodynamic energy conservation:whereThe last term is the Poynting vector with the ideal electric field $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$. A time-dependent imposed potential would instead supply the source $\rho {\partial }_t\Phi$.

Apply the continuity equation and integrate by parts, with the stated vanishing boundary terms and finite moments. The second mass moment tensor satisfiesand differentiation again givesInsert the stress-divergence equation from part (a). The two force integrals becomeThe first term is $4K_{ij}$. Hence the magnetized-fluid tensor virial theorem isHere $\mathcal{T}$ is the volume-integrated stress, whose sign differs from some gravitational potential-energy tensor conventions. The derivation also requires the advective mass-moment surface terms to vanish; this is automatic for an isolated sufficiently decaying configuration.

The linearized azimuthal Euler momentum equation isFor the time dependence $e^{-i\omega t}$ and $u_R^{\prime }=-i\omega {\xi }_R$, it givesThis expresses conservation of the displaced element's specific angular momentum. It applies directly to nonzero-frequency modes, with the zero-frequency limit taken in the displacement formulation.

The radial advective acceleration supplies $-2\Omega v_{\varphi }$, while the pressure force perturbation is $\delta \rho \nabla p/{\rho }^2-\nabla \delta p/\rho$. Eliminate $v_{\varphi }$ and retain the vertical equation. Under the Cowling approximation, $\delta \Phi =0$, soThe continuity equation gives $\delta \rho =-\nabla \cdot (\rho \mathbf{\xi })$. The Lagrangian adiabatic relation $\Delta p/p=\gamma \Delta \rho /\rho$, with $\Delta \rho =-\rho \nabla \cdot \mathbf{\xi }$, givesHere $\kappa$ is the radial epicyclic frequency. These formulas define the axisymmetric adiabatic displacement operator.

Turbulence usually makes the onset of dust gravitational instability of an astrophysical disk harder in three complementary ways. It increases the dust random velocity dispersion, raising $c$ and the Toomre parameter; it stirs dust vertically, reducing the self-gravity enhancement of a razor-thin disk; and it mixes density enhancements through turbulent diffusion. The last effect is particularly important for a slowly growing secular gravitational instability, which can be erased before it amplifies appreciably.

A simple diffusion model adds $D{\nabla }^2\sigma$ to the dust continuity equation, giving a damping scale of order $D{k}^2$. Growth then has to compete with mixing as well as pressure and rotational support. In the long-wavelength weak-drag regime where ${\omega }^2\simeq {\Omega }^2$, the indicative competition iswith further pressure and finite-thickness corrections. This is a model-dependent long-wave estimate, not a substitution into the original cubic without changing its continuity equation. In an infinite domain, diffusion proportional to $k^2$ need not remove growth proportional to $|k|$ at every arbitrarily long wavelength; in a finite disk, the remaining wavelengths and their growth times may be inadequate. The conclusion is therefore a higher practical collapse threshold and slower growth, rather than guaranteed stability for all turbulent flows.

Turbulence can also concentrate dust through coherent structures or pressure maxima, increasing local surface density and encouraging collapse. Its net quantitative effect requires a model for stirring, diffusion, thickness and concentration. The deterministic fixed gas flow used in the preceding parts does not specify those statistics. This competition is turbulent mixing of a secularly unstable dust layer.

Use a local, homogeneous razor-thin disk approximation in a frame rotating with constant $\Omega$. The unperturbed planar velocity is zero in this frame, and the large-scale gravitational and centrifugal forces balance. Neglect viscosity, magnetic fields, thickness and background gradients across a wavelength. Take small planar disturbances, with wavelength short compared with the galaxy's background scale but long enough for a fluid description. The barotropic closure of a razor-thin disk is $P=K{\Sigma }^{\gamma }$, with $K$ fixed and a positive derivativeSolid-body rotation has no shear and has radial epicyclic frequency $\kappa =2|\Omega |$. By rotational symmetry of the local model choose a wavevector along $x$, and write perturbations proportional to $e^{i(kx-\omega t)}$.

Let ${\Sigma }_1,u,v,{\Phi }_1$ be the surface density, $x$-velocity, $y$-velocity and Newtonian gravitational potential amplitudes. The linearized continuity equation and Euler equations with the Coriolis force areThe perturbing Newtonian gravitational potential solves the Poisson equation for Newtonian gravityIts decaying solution is proportional to $e^{-|k||z|}$. The jump in its derivative is $-2|k|{\Phi }_1(0)=4\pi G{\Sigma }_1$, giving the razor-thin disk Poisson kernelEliminating $v$ from the two momentum equations, and using ${\Sigma }_1=k{\Sigma }_{0}u/\omega$, gives the density-wave branchThe complete linear system also has a zero-frequency balanced mode; it is not the growing density-wave branch, and division by $\omega$ in this elimination excludes it. The result is the uniformly rotating gas-sheet dispersion relation: pressure opposes compression at large wavenumber, self-gravity promotes compression, and rotation provides epicyclic support.