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One approach uses exoplanet transmission spectra. Three aerosol signatures are:

A second approach uses the dayside spectrum and brightness obtained with exoplanet secondary eclipses, optionally supported by orbital phase variations. Its three useful aerosol signatures are:

Transmission measures aerosol extinction along the limb; eclipse/reflection measurements test dayside scattering and emission. No single flattened feature proves clouds: small atmospheric scale height, low molecular abundances, stellar heterogeneity and thermal-gradient degeneracies must be considered. Joint wavelength coverage and geometries make the inference stronger.

At the stellar surface boundary condition, match the interior to a stellar atmosphere, rather than setting the temperature and pressure to zero at an arbitrary radius. At a photosphere of radius $R$, the enclosed mass is $M$, the luminosity is $L$, and the outward radiative flux defines the effective temperature through $F=L/(4\pi R^2)={\sigma }_{\mathrm{SB}}{T}_{\mathrm{eff}}^4$. With no external illumination the incoming specific intensity satisfies $I_{\nu }(0,\mu )=0$ for $\mu <0$. A deeper boundary is matched to the nearly isotropic radiative diffusion field. A grey atmosphere commonly matches near $\tau =2/3$; under the Eddington surface boundary condition this gives $T=T_{\mathrm{eff}}$. For roughly constant gravity and opacity, the total pressure rises by $g\tau /\kappa$ relative to its outer boundary value; if radiation supplies appreciable support, the gas-pressure gradient uses the effective gravity $g-\kappa F/c$ instead.

Take $z$ increasing outward and $\mu$ the outward direction cosine. If $j_{\nu }$ is emission per unit path length and solid angle, the radiative transfer equation isDefine inward-increasing optical depth by $d{\tau }_{\nu }=-\rho {\kappa }_{\nu }dz$ and the source function by $S_{\nu }=j_{\nu }/(\rho {\kappa }_{\nu })$. Then $\mu {\partial }_{{\tau }_{\nu }}{I}_{\nu }=I_{\nu }-S_{\nu }$. This states the emission-coefficient convention; if $j_{\nu }$ is defined per unit mass instead, the emission term is $\rho j_{\nu }$.

For frequency-integrated intensity, introduce the radiation-field momentsThe first angular approximation writes $I\simeq J+3\mu H$. Its odd part does not contribute to $K$, soThis is the Eddington closure approximation, not an exact description of the escaping angular distribution. An approximate surface condition takes the outgoing hemisphere to have direction-independent intensity $I_0$ and the incoming hemisphere to be dark. Its moments give $F=\pi I_0$ and $J(0)=I_0/2=F/(2\pi )$. Integrating the moment equations with the closure then gives $J=3F(\tau +2/3)/(4\pi )$; in thermal equilibrium, $J=a_{\mathrm{rad}}c{T}^4/(4\pi )$ gives $T^4=(3/4)T_{\mathrm{eff}}^4(\tau +2/3)$. This derives the usual approximate photospheric matching condition rather than imposing a zero surface temperature. For conservative coherent isotropic scattering, $S=J$. Integrating the transfer equation gives $dH/d\tau =0$ and $dK/d\tau =H$, the moment conditions for radiative equilibrium.

The exponential ansatz in the PDF has a genuine consistency problem. Compute its moments before making any approximation. For a nonsingular real intensity on all $-1\le \mu \le 1$, take $|ϵ|<1$ and defineThe continuous zero limits are $A_0=1$, $A_1=0$, $A_2=1/3$. Direct integration yieldsFor a decaying nontrivial correction, $\delta >0$ and $\gamma \ne 0$. Exact constant flux then requires $A_1=0$. But $\mathrm{atanh}ϵ/ϵ>1$ for every nonzero real $|ϵ|<1$, so $A_1=0$ forces $ϵ=0$. Thus exact radiative equilibrium cannot imply a small nonzero $ϵ$ for this ansatz.

Substitution into the full transfer equation makes the obstruction sharper. The exponential coefficient must satisfy, for every $\mu$,Comparing the constant and linear coefficients gives $A_0=1$ and $\delta =A_0ϵ=ϵ$. Hence regular real coefficients require $\delta =ϵ=0$, eliminating any decaying correction. This is the obstruction to a single smooth exponential mode in conservative grey transfer. There is no exact nontrivial real solution of the stipulated form with a small nonzero decay parameter. Choosing, for example, $ϵ=\delta =0.1$ and $\gamma \ne 0$ supplies an explicit counterexample to the claimed exact flux constancy, because $A_1\ne 0$.

Several requested conclusions remain useful as formal approximations, provided they are identified as such. If $\delta >0$, the large-depth exponential vanishes, giving $F\to 4\pi \alpha /3$, hence $\alpha =3F/(4\pi )$ for the asymptotic physical flux. It also gives $c{P}_{\mathrm{rad}}\to (4\pi /3)J$. Since $A_0=1+{ϵ}^2/3+O({ϵ}^4)$, neglecting second-order angular errors yields the formal relation $\delta \simeq ϵ$. However $A_1=-ϵ/3+O({ϵ}^3)$, so the finite-depth flux still varies at first order. The formal relation is not an exact transfer or constant-flux solution.

There is also a flux-normalization mismatch in the PDF's last displayed relation. With the same physical $F$ used above, the Hopf function of a grey atmosphere is defined byThe PDF's coefficient $3F/4$ would require redefining its $F$ as physical flux divided by $\pi$; it cannot simultaneously use the earlier $\alpha =3F/(4\pi )$ for the same physical flux. With the consistent physical-flux normalization, formal moment matching givesIn the intended small-$ϵ$ approximation this isThe approximate outgoing surface ansatz then gives the limb darkening ratioFor finite $ϵ$, $\gamma =-0.14/A_0$ gives a parameter-dependent ratio, not a unique exact value. The two Hopf endpoint values alone do not determine the exact emergent intensity: the formal solution of the radiative transfer equation uses the entire source profile, $I(0,\mu )={\int }_0^{\infty }J(t)e^{-t/\mu }dt/\mu$ for $\mu >0$. Nor does the simple linear-plus-exponential ansatz satisfy the exact no-incoming-radiation condition for every negative $\mu$. Thus $2.75$ is a clearly labelled approximate disk-centre-to-limb result, not an exact consequence of inconsistent premises.

For a quasistatic spherical star with fixed total mass, the stellar structure equations in radius areHere $s$ is the specific entropy of a moving mass shell, and the entropy term accounts for release or storage of thermal and gravitational energy. In thermal equilibrium it vanishes. An equation of state, opacity, energy-generation law and composition evolution close the equations. The absence of mass loss fixes the outer mass coordinate; it does not require the star to be in thermal equilibrium.

The radiative temperature gradient needed to carry the full luminosity isHere the actual temperature gradient is $\nabla =d\log T/d\log P$. For uniform composition the Schwarzschild criterion says that a radiative layer is stable if ${\nabla }_{\mathrm{rad}}<{\nabla }_{\mathrm{ad}}$, with marginal stability at equality, where the adiabatic temperature gradient is $(\partial \log T/\partial \log P{)}_s$. Use $\nabla ={\nabla }_{\mathrm{rad}}$ there. If ${\nabla }_{\mathrm{rad}}>{\nabla }_{\mathrm{ad}}$, convection carries some of the flux. Efficient convection gives $\nabla \simeq {\nabla }_{\mathrm{ad}}$; inefficient surface convection needs a transport prescription and can be superadiabatic. For composition gradients, use the Ledoux criterion, with threshold ${\nabla }_{\mathrm{ad}}+\varphi {\nabla }_{{\mu }_{\mathrm{mol}}}/{\delta }_{\mathrm{th}}$ instead. Here ${\nabla }_{{\mu }_{\mathrm{mol}}}=d\log {\mu }_{\mathrm{mol}}/d\log P$, $\varphi =(\partial \log \rho /\partial \log {\mu }_{\mathrm{mol}}{)}_{P,T}$ and ${\delta }_{\mathrm{th}}=-(\partial \log \rho /\partial \log T{)}_{P,{\mu }_{\mathrm{mol}}}$, using the mean molecular weight ${\mu }_{\mathrm{mol}}$.

Above the thin burning shell, take constant luminosity $L$, constant mean molecular weight ${\mu }_{\mathrm{mol}}$, ideal-gas pressure $P=\mathcal{R}T\rho$ with $\mathcal{R}=k_B/({\mu }_{\mathrm{mol}}{m}_u)$, and the Kramers' opacity law $\kappa ={\kappa }_0\rho T^{-7/2}$. Then the temperature equation isTo implement the printed assumption that $P,\rho$ and $M_r$ are all power laws while retaining mass conservation, write $M_r\propto r^m$, $\rho \propto r^b$, $P\propto r^p$ and $T\propto r^d$. The mass equation gives $m=b+3$; the ideal gas law gives $d=p-b$; hydrostatic equilibrium gives $d=m-1$. The diffusion equation gives $d-1=2b-2-13d/2$. Solving this linear system gives the self-gravitating Kramers power-law envelope:The coefficients can also be matched consistently. If $r_c$ is the core boundary and $M_c$ its enclosed mass,Radiative transfer fixes the constant luminosity throughwhose right-hand side is independent of radius for these exponents. Thus this is a solution of all four envelope equations, not just a dimensional estimate.

Matching the boundary temperature to the isothermal core and extrapolating the idealized power law to the specified photosphere givesFor reference $M_R/M_c=(R/r_c{)}^{1/11}=2500^{1/10}\simeq 2.19$, so the envelope mass is not negligible in this particular solution. The given $M_c\simeq M_{\odot }$ fixes the core-radius normalization $r_c=(11/42)G{M}_c/(\mathcal{R}10^7\mathrm{K})$; a numerical absolute radius needs the unspecified molecular weight. The exponent ratio also gives $\nabla =d/p=5/21<2/5$, so the fully ionized monatomic version of this idealization is convectively stable.

A different common approximation neglects the envelope's self-gravity and sets $M_r\simeq M_c$ in the force equation. It cannot obey the exact mass equation with a nonzero density and exactly constant $M_r$. Under that additional approximation, the same calculation instead givesThis is the constant-core-mass power-law opacity radiative envelope limit, not the full power-law solution with changing enclosed mass. Stating which approximation is used resolves the otherwise different numerical radius ratios.

A compact nearly isothermal core, a thin luminosity-producing shell and a much larger cool envelope describe a shell-burning red giant. The full power-law profile is a deliberately simple model. The Kramers' opacity law and an ionized ideal gas are not reliable at a $4000\mathrm{K}$ photosphere; partial ionization, other opacity sources and a convective envelope usually modify real cool giants. The quoted radius ratios are extrapolations within the stipulated model.

Let $f(\mathbf{p})d^{3}p$ be particle number per unit volume in a momentum element, including the spin-state count in $f$. This is the local momentum part of a phase-space distribution function, normalized by the number density $n=\int f(\mathbf{p})d^{3}p$. For an isotropic distribution with particle energy $E(p)=\sqrt{m^2{c}^4+p^2{c}^2}$, the kinetic pressure of an isotropic gas is the average momentum flux:The kinetic energy density, explicitly excluding rest mass, isFor $p\ll mc$, $E-m{c}^2=p^2/(2m)$ and $\mathbf{v}=\mathbf{p}/m$, giving $P=2u_{\mathrm{kin}}/3$. For $p\gg mc$, $E\simeq pc$ and $v\simeq c$, giving $P\simeq u_{\mathrm{kin}}/3$. ThusNeither relation assumes a Maxwellian distribution. At intermediate momenta neither constant ratio is exact, and anisotropic distributions require a pressure tensor rather than this scalar pressure.

For a classical Maxwell-Boltzmann distribution, $f=n(2\pi m{k}_{B}T{)}^{-3/2}\exp [-p^2/(2m{k}_{B}T)]$. Three Gaussian component integrals give $⟨p^2⟩=3m{k}_{B}T$, so the ideal gas hasIn a fully ionized mixture, sum over independent species to obtain $P_{\mathrm{gas}}=\rho k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$, where the mean molecular weight counts ions and free Electrons. This excludes partial-ionization and interaction corrections. A dilute classical ultrarelativistic gas still has $P=n{k}_{B}T$ but $u_{\mathrm{kin}}\simeq 3n{k}_{B}T$; its momentum distribution is proportional to $e^{-pc/(k_{B}T)}$ instead of the nonrelativistic Gaussian.

For a thermal photon gas, the two polarization states and zero photon chemical potential give the Planck photon distributionUse $x=pc/(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation constant and photon equation of state are thereforeLikewise $n_{\gamma }=2\zeta (3)(k_{B}T/\hslash c{)}^3/{\pi }^2$. Unlike a gas with a fixed particle number, photons do not have pressure proportional to baryonic mass density.

For fully degenerate Electrons, the Fermi-Dirac distribution becomes a filled momentum sphere with two spin states. Counting them gives the Fermi momentumHere ${\mu }_e$ is the mean molecular weight per electron. The equation of state of a cold electron gas follows by integrating momentum flux up to $p_F$:Its two limits areThe corresponding kinetic energy densities are $3n_e[p_F^2/(2m_e)]/5$ and $3n_e{p}_{F}c/4$, respectively. Thus in the high-density relativistic limit the Electron pressure scales as ${\rho }^{4/3}$ and is nearly independent of temperature. This is the ideal noninteracting, fixed-composition Electron result, not a universal equation of state at nuclear densities where captures, interactions and the composition change.

The boundaries on a stellar equation-of-state regime diagram concern the Electron component. Define the electron relativistic density threshold and thermal electron relativistic threshold byA degenerate Electron gas is nonrelativistic well below ${\rho }_{\ast }$ and ultrarelativistic well above it: this is an approximately vertical division on a log-density plot. A nondegenerate Electron gas instead becomes thermally relativistic near $T_{\ast }$: this is an approximately horizontal division. The regimes have broad crossovers rather than a discontinuity at either line.

The exact electron Fermi temperature, subtracting Electron rest energy, isFor Electrons with negligible thermal pairs, strong degeneracy requires $T\ll T_F$, whereas $T\gg T_F$ gives a nondegenerate gas. In the pair-rich regime the actual Electron and Positron distributions must instead be treated with their chemical potentials, as discussed below. The two limiting degeneracy boundaries areThese slopes $2/3$ and $1/3$ explain the bent degeneracy boundary on the logarithmic graph. A thermal-wavelength test $n_e[2\pi {\hslash }^2/(m_e{k}_{B}T){]}^{3/2}\sim 2$ gives the same nonrelativistic density/temperature scaling, with an order-one definition of the crossover.

The radiation-to-gas pressure boundary in the nondegenerate fully ionized regime isRadiation dominates above this line. Once Electrons are strongly degenerate, compare radiation pressure with $P_e(\rho )$ instead of continuing the ideal-gas comparison into that regime. The radiation-to-degeneracy pressure boundary is $T=[3P_e(\rho )/a_{\mathrm{rad}}{]}^{1/4}$, with logarithmic slopes $5/12$ and $1/3$ in the nonrelativistic and ultrarelativistic limits. Degeneracy of the Electrons and dominance of their pressure are distinct criteria.

For the pair curve, distinguish baryonic net Electrons from thermally created Electrons and Positrons. Chemical equilibrium with photons requires opposite Electron/Positron chemical potentials when the one-particle energy includes rest energy, as in the distributions below. At low temperature and low degeneracy the zero-chemical-potential density per charge species isCharge neutrality gives $n_{-}-n_{+}=n_b=\rho /({\mu }_e{m}_u)$, while the nondegenerate equilibrium product is $n_{-}{n}_{+}=n_0^2$. HencePairs become important when $n_0$ is comparable with or exceeds $n_b$, approximately the electron-positron thermal pair abundance curve ${\rho }_{\mathrm{pair}}(T)={\mu }_e{m}_u{n}_0(T)$. Below its density at a fixed temperature, pairs dominate over the net charge Electrons. The exponential makes the low-temperature portion steep in a log-log plot. For ${\mu }_e=2$, the full zero-potential integral gives approximate pair-marker densities $1.80\times 10^3\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $10^9\mathrm{K}$ and $7.52\times 10^5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $3\times 10^9\mathrm{K}$. At relativistic temperature use the full zero-chemical-potential Fermi-Dirac distribution instead; it gives $n_0\to [3\zeta (3)/(2{\pi }^2)](k_{B}T/\hslash c{)}^3$, so the high-temperature pair curve approaches slope three in log-density versus log-temperature. Strong net-electron degeneracy suppresses Positrons and requires the full chemical-potential-dependent distribution.

The original diagram uses fully ionized helium, ${\mu }_e=2$ and ${\mu }_{\mathrm{mol}}=4/3$, only to set numerical locations. It plots the full $T_F$ curve and the zero-chemical-potential pair integral, rather than extending their asymptotes into the crossover. The pair boundary is deliberately an approximate abundance marker, not a phase transition; at high temperature the nonrelativistic ion approximation and fixed-composition model also have limits. For $T\gg T_{\ast }$ with abundant nondegenerate pairs, the combined Electron/Positron energy density is $7a_{\mathrm{rad}}{T}^4/4$, so photons plus pairs have $u=11a_{\mathrm{rad}}{T}^4/4$ and $P=u/3$. Real stellar matter adds Coulomb effects, partial ionization, nuclear reactions and, at sufficiently high density, nuclear-matter physics beyond the ideal regime map.

On a length scale $\ell$, pressure communicates support over a sound-crossing time $\ell /v_s$, whereas self-gravity acts over a time of order $(G{\rho }_0{)}^{-1/2}$. When the former is longer, gravity can amplify an overdensity before pressure smooths it. Thus Jeans instability occurs at scales of order $\ell \gtrsim v_s/\sqrt{G{\rho }_0}$; the precise numerical factor depends on whether scale means wavelength or inverse wavenumber.

Take a uniform static background, small perturbations, an inviscid barotropic fluid, and $\delta p=v_s^2\delta \rho$ with $v_s^2=(dp/d\rho {)}_0>0$. An infinite Newtonian homogeneous density cannot itself have zero background gravitational acceleration while satisfying the unmodified Poisson equation for Newtonian gravity. Specify the Jeans swindle: subtract the homogeneous field, or regard an external prescription as balancing it, and apply Poisson's equation only to perturbations. The linearized continuity equation and Euler equations for an inviscid fluid areTaking a time derivative of the first equation and substituting the second yields ${\partial }_t^2\delta \rho =v_s^2{\nabla }^2\delta \rho +4\pi G{\rho }_0\delta \rho$. A plane wave $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ therefore givesFor $0<k<k_J$ there is exponential growth at rate $\sqrt{4\pi G{\rho }_0-v_s^2{k}^2}$. At $k_J$ the density mode is neutral to this linear order, and above it there are acoustic oscillations. The critical wavelength is $2\pi /k_J=\sqrt{\pi }{v}_s/\sqrt{G{\rho }_0}$. Transverse velocity perturbations decouple as zero-frequency vortical modes in this idealization.

For the attractive Yukawa potential, assume a real screening parameter $\alpha \ge 0$. Its Green function obeys $({\nabla }^2-{\alpha }^2)(-Gm{e}^{-\alpha r}/r)=4\pi Gm{\delta }^{(3)}(\mathbf{r})$, including the delta source. Thus $\delta {\varphi }_{\mathbf{k}}=-4\pi G\delta {\rho }_{\mathbf{k}}/(k^2+{\alpha }^2)$. The same fluid elimination gives Jeans instability with Yukawa gravity:There exists a growing nonzero-wavenumber mode precisely whenEquality gives no exponentially growing nonzero mode. If screening exceeds this threshold, the finite-range gravitational response is too weak to overcome pressure at any wavelength. Otherwise all sufficiently long but nonconstant waves remain unstable, with a longer minimum unstable wavelength than in Newtonian gravity. For $\alpha >0$ the long-wave gravitational contribution also scales as $k^2$, so the growth rate tends to zero as $k\to 0$; screening does not produce a second positive lower cutoff in wavenumber. A homogeneous screened background can have the constant gravitational potential $-4\pi G{\rho }_0/{\alpha }^2$, with zero force, so no Newtonian background subtraction is needed for nonzero screening. The limits $k\to 0$ and $\alpha \to 0$ are nonuniform. An exact $k=0$ density change is not one of the growing finite-wavelength modes. In the pressureless edge case $v_s=0$, every finite $k>0$ is unstable for positive density and finite screening.

Let $K={\rho }_0|\mathbf{u}{|}^2/2$ be the kinetic energy per unit volume. Dot the momentum equation with ${\rho }_0\mathbf{u}$. The Coriolis acceleration does no work because $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$. Incompressibility turns the advection and pressure terms into divergences, yieldingThus $\mathbf{F}=(K+P)\mathbf{u}$ is the energy flux and $S={\rho }_0{\Omega }^2(3x-2qz)u_x$ is the source in the requested kinetic-energy form.

To decide about conservation of energy, absorb the ordinary radial shearing-sheet tidal potential ${\Phi }_t=-3{\Omega }^2{x}^2/2$ into the density. Its material derivative cancels the $3x$ work term:For $q=0$ this is the usual conservative Jacobi energy in a shearing sheet, subject to vanishing boundary energy flux. For $q\ne 0$, the extra body force has nonzero curl, $\nabla \times [-2{\Omega }^{2}qz{\mathbf{e}}_x]=-2{\Omega }^{2}q{\mathbf{e}}_y$, so no scalar potential can absorb it. A potential $2{\Omega }^{2}qxz$ would necessarily introduce a vertical force $-2{\Omega }^{2}qx{\mathbf{e}}_z$, absent from the model. Therefore this vertical-shear model does not generically conserve the standard total mechanical energy: its maintained shear/irradiation background can supply or remove energy. This is the energy balance of the incompressible vertical-shear model.

Write the proposed background as $\mathbf{U}=U(x,z){\mathbf{e}}_y$, $U=-3\Omega x/2+q\Omega z$. It satisfies incompressibility, and $\mathbf{U}\cdot \nabla \mathbf{U}=U{\partial }_y\mathbf{U}=0$. A constant pressure has zero gradient. The Coriolis acceleration is $-2\Omega {\mathbf{e}}_z\times \mathbf{U}=2\Omega U{\mathbf{e}}_x=-(3x-2qz){\Omega }^2{\mathbf{e}}_x$, which exactly cancels the prescribed body force. Thus it is a steady state.

The background is ordinary Keplerian radial shear together with vertical shear: ${\partial }_{x}U=-3\Omega /2$ and ${\partial }_{z}U=q\Omega$. At a given radius, different heights have different azimuthal velocities. The small $q$ in this question measures vertical shear; it is distinct from the orbital shear parameter, whose radial value here remains $3/2$. The vertical shear instability draws on this height-dependent differential rotation.

Set $\mathbf{k}=(k_x,0,k_z)\ne 0$. Incompressibility gives $(\mathbf{k}\cdot \tilde{\mathbf{u}})f^{\prime }(\xi )e^{st}=0$. Since $f$ is nonconstant and differentiable, it follows that $\mathbf{k}\cdot \tilde{\mathbf{u}}=0$; thereforeThe background has no advection of a $y$-independent perturbation. Its only remaining cross-advection term is $({\mathbf{u}}^{\prime }\cdot \nabla )\mathbf{U}=(-3\Omega u_x^{\prime }/2+q\Omega u_z^{\prime }){\mathbf{e}}_y$. The velocity/time terms have spatial factor $f$, while $\nabla P^{\prime }=\tilde{P}\mathbf{k}{g}^{\prime }(\xi )e^{st}$. For nonzero pressure amplitude, $g^{\prime }$ must therefore be a constant multiple of $f$; absorbing that multiple into $\tilde{P}$ gives $g^{\prime }=f$. An additive spatial constant in $g$ is a pressure gauge. If $\tilde{P}=0$, there is no such restriction on the unused pressure profile.

Matching coefficients givesBecause the quadratic perturbation term vanishes identically, these amplitudes give an exact single-phase incompressible perturbation of a shear flow, not merely a first-order truncation. Any sufficiently differentiable profile with $g^{\prime }=f$ works locally; boundary conditions may restrict the profiles in a particular domain.

Eliminate pressure by multiplying the radial equation by $k_z$ and subtracting $k_x$ times the vertical equation. For $k_z\ne 0$, incompressibility gives ${\tilde{u}}_z=-(k_x/k_z){\tilde{u}}_x$, and the pressure-free relation is $s(k_x^2+k_z^2){\tilde{u}}_x/k_z=2\Omega k_z{\tilde{u}}_y$. Substitution in the azimuthal equation yieldsThus the vertical shear instability occurs when $2q{k}_x{k}_z>k_z^2$, or $2q(k_x/k_z)>1$ for $k_z\ne 0$. Otherwise the nonzero modes oscillate; equality is marginal in the exponential-growth sense. With $q>0$, growing disturbances have $k_x/k_z>1/(2q)$ and are strongly inclined in wavevector space.

Put $R=k_x/k_z$. Maximizing $(2qR-1)/(1+R^2)$ gives $q{R}^2-R-q=0$. For $q>0$, the unstable maximizing root isMore precisely its bracket is $1-q^2/2+O(q^4)$, and $R_{\max }\sim q^{-1}$. This is the maximum growth rate of the vertical shear instability. For negative $q$, the maximum uses the corresponding negative root and is asymptotically $\Omega |q|$. At $q=0$ there is no exponentially growing mode. The undivided formula handles $k_z=0$: $s^2=0$, not an exponential instability; an allowed vertical velocity can instead force a secular azimuthal change. The excluded $\mathbf{k}=0$ is a spatially uniform disturbance and does not belong to the stipulated nonconstant one-phase family.

For a normal mode proportional to $e^{ikx-i\omega t}$, the linear equations giveFor $\omega \ne 0$, eliminate $U_y=-i\Omega U_x/(2\omega )$ and $\hat{\sigma }=k{\Sigma }_0{U}_x/\omega$. The density-wave dispersion relation in a non-self-gravitating disk isThe ${\Omega }^2$ term is the radial epicyclic frequency squared for Keplerian rotation: Coriolis force and background shear restore an orbital displacement. The $c_s^2{k}^2$ term is acoustic restoration by a pressure gradient. Long wavelengths are mainly epicyclic motion; short wavelengths are mainly acoustic waves. The given model contains no disk self-gravity term.

For completeness, the full three-variable determinant is proportional to $\omega ({\omega }^2-{\Omega }^2-c_s^2{k}^2)$. Its additional $\omega =0$ branch has $U_x=0$ and $2\Omega U_y=ik{c}_s^2\hat{\sigma }/{\Sigma }_0$. This is a stationary pressure/Coriolis balance, the zero-frequency balanced mode of an axisymmetric disk, rather than a propagating density wave. At $k=0$ it reduces to a uniform surface-density change. Dividing by $\omega$ is legitimate only for the wave branches just derived.