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Write $\mathcal{R}=k/(\mu H)$, where $H$ is the atomic mass unit used in the mean molecular weight convention, and let $\kappa$ be the Rosseland mean opacity per unit mass. For a static spherical star with negligible thermal storage, the stellar structure equations areHere $ϵ$ is the net stellar energy-generation rate per unit mass; $a$ is the radiation energy-density constant and $c$ the speed of light. The last differential equation follows from radiative diffusion, $F=-(c/3\kappa \rho )d(a{T}^4)/dr$, with $L=4\pi r^{2}F$. There is no gravothermal storage term in the specified stellar energy balance equation. Neglecting radiation pressure in force balance does not remove radiation as an energy carrier. Regular central conditions are $m(0)=L(0)=0$; take negligible external pressure, $P(R)=0$, for the analytic hydrostatic equilibrium model below.

Set $x=r/R$. Integrating the enclosed mass directly givesThusThe stellar hydrostatic equation then givesFactorizing makes the surface behaviour transparent:The ideal gas equation of state gives the temperature profileNo small-$x$ approximation was needed. In particular, close to the centre,These exact hydrostatic formulas reveal a limitation of the prescribed linear-density stellar model. Its mass density has a central cusp, and $T^{\prime }(0^{+})=T_c/R>0$. A positive opacity with positive outward luminosity would instead require $T^{\prime }<0$ in radiative diffusion. Moreover, regular positive nuclear heating gives $L=O(r^3)$ and therefore $T^{\prime }=O(r)$ at the centre. Thus the imposed profile is a hydrostatic toy model, not a complete positive-heating radiative-equilibrium solution of all the stellar structure equations. The requested algebraic profiles and nuclear-luminosity integral can still be computed consistently as properties of that toy model.

Let $z$ increase outward and define the inward optical depth by $d\tau /dz=-\bar{\kappa }\rho$. With constant outward internal radiative flux $F=\sigma T_{\mathrm{int}}^4$, radiative diffusion givesThe constant $q$ is a boundary condition; diffusion alone does not determine it. For an unirradiated grey atmosphere with the Eddington closure approximation and Eddington surface boundary condition, $q=2/3$, soThe internal effective temperature of a planet is defined by its intrinsic cooling flux, not by its incident stellar heating. Since $dT/d\tau >0$ and $d\tau /dz<0$, temperature decreases outward. Towards the thin upper layers, $T\to 2^{-1/4}{T}_{\mathrm{int}}\simeq 0.84T_{\mathrm{int}}$; if the density and optical-depth gradient vanish there, $dT/dz\to 0$. This is the upper nearly isothermal atmosphere. The diffusion approximation itself fails at small optical depth: the grey boundary closure supplies the approximate continuation. Small irradiation perturbs this intrinsic-flux solution.

Young, self-luminous gas giants at wide orbital separations can have intrinsic cooling dominate their photospheric budget. They are favorable for exoplanet direct imaging, especially in the infrared, where their own thermal radiation is easier to separate from the host light. A strongly irradiated hot Jupiter instead has a large stable outer radiative region set mainly by stellar heating; it can be nearly isothermal over a broad pressure range or develop an atmospheric thermal inversion if stellar light is absorbed sufficiently high. Inversion is not inevitable for every strongly irradiated planet. Deep convection begins below its radiative-convective boundary.

The intrinsic curve follows the grey formula. The irradiated curves illustrate possible shapes only; they are not solutions for a specified opacity model. Smaller optical depth corresponds to greater altitude.

The four equations describe, respectively, vertical hydrostatic equilibrium in the central object's gravity, the vertical divergence of radiative flux balancing viscous dissipation, optically thick radiative diffusion, and the equation of state combining ideal gas gas pressure with radiation pressure. Here $\mu$ is dynamic viscosity, not the mean molecular weight ${\mu }_m$; $\kappa$ is the opacity, $\sigma$ the Stefan-Boltzmann constant, and $\Omega$ the local orbital frequency.

In the radiation-pressure limit, $p=4\sigma T^4/(3c)$, so radiative diffusion givesCompare this with vertical hydrostatic equilibrium. Wherever $\rho \ne 0$,At fixed radius, constant opacity makes $dF/dz=c{\Omega }^2/\kappa$. The viscous dissipation equation therefore yieldsindependent of height within the optically thick radiation-supported layer. The result assumes the supplied gravity, heating and diffusion equations; optically thin surface corrections are outside that approximation.

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.

Use a gas-pressure-supported, optically thick alpha disk, with $\mathcal{R}=k/({\mu }_m{m}_p)$ and positive parameters. Here $\sigma$ is the Stefan–Boltzmann constant in the Stefan–Boltzmann law, and $F_{+}$ and $F_{-}$ below are heating and cooling flux estimates per disk face. Hydrostatic equilibrium and the ideal gas relation give $\mathcal{R}T\sim {\Omega }^2{H}^2$, while $\rho \sim \Sigma /H$. Integrating viscous heating over one semi-thickness and estimating radiative diffusion giveHere the stipulated negative hydrogen ion opacity exponent is $\kappa ={\kappa }_0{\rho }^{1/3}{T}^{10}$, which is a local approximation distinct from other empirical opacity fits. Substituting $T\sim {\Omega }^2{H}^2/\mathcal{R}$ and equating heating and cooling givesDropping numerical constants of order one is essential here. This is the negative-hydrogen-opacity alpha-disk thickness scaling. Its angular thickness estimate isFor a central mass $M$, Keplerian rotation gives $\Omega =(GM/r^3{)}^{1/2}$, so $H/r\propto r^{53/82}\Sigma (r{)}^{-7/41}$. The isothermal sound speed $c_s^2=\mathcal{R}T$ isAn adiabatic sound speed differs by the constant factor $\sqrt{\gamma }$ if the adiabatic index is fixed. For $\Sigma \propto r^{-p}$, constant $\alpha$, and fixed composition, the disk aspect ratio grows as $r^{(53+14p)/82}$, so there is disk flaring when $p>-53/14$; $c_s\propto r^{(6+7p)/41}$. The problem supplies no numerical parameters or specified $\Sigma (r)$, so no unique numerical $H/r$ or unconditional radial power can be inferred.

If the additional standard closure of a constant inward accretion rate well outside a zero-torque edge is intended, $\bar{\nu }\sim \alpha \Omega H^2$ and $\bar{\nu }\Sigma \simeq \dot{M}/(3\pi )$ imply $\Sigma \propto (\alpha \Omega H^2{)}^{-1}$. Eliminating it gives $H^9\propto {\Omega }^{-38/3}$ at fixed $\dot{M},\alpha$, hence $H/r\propto r^{10/9}$ and $c_s\propto r^{11/18}$. These stronger powers use that extra steady-flow assumption. All these local scalings cease to apply if the thin disk or the stated opacity regime fails.

In the vertical equations, $\mu$ denotes dynamic viscosity, while ${\mu }_m$ is the mean molecular weight in units of the proton mass $m_p$. The symbol $k$ denotes the Boltzmann constant. For the intended hydrostatic equilibrium calculation, put $P=K{\rho }^{\gamma }$ with $K>0$, $\gamma >1$, and set $\mathcal{R}=k/({\mu }_m{m}_p)$. Dividing the hydrostatic equation by $\rho$ and integrating from the symmetric midplane givesThe specific enthalpy vanishes at the free surface. Hence the polytropic vertical structure in stellar gravity isOn $|z|<H$, the ideal gas relation gives $T=T_m(1-z^2/H^2)$ and $P=P_m(1-z^2/H^2{)}^{\gamma /(\gamma -1)}$; the vacuum extension sets $\rho =P=0$ outside. The formula needs $\gamma >1$ for a finite positive semi-thickness and assumes the harmonic vertical gravity remains valid through this height.

There is a genuine inconsistency if all four printed structure equations are imposed literally with $F=0$ and finite positive opacity: radiative diffusion then requires $dT/dz=0$, whereas the polytropic temperature above has a nonzero gradient away from the midplane. The intended result is valid for the hydrostatic polytropic equation of state with the radiative closure replaced or neglected, for example an adiabatic opaque limit. It is not an exact simultaneous solution of the stated finite-opacity radiative system. With that radiative system and a nonvacuum ideal gas, $\gamma =1$ instead gives an isothermal solutionwhich has no finite free surface. For $\gamma \ne 1$, constant $T$ and constant $K$ would force constant density, contradicting hydrostatic balance on a nontrivial vertical interval. This is the zero-radiative-flux obstruction for a polytropic ideal gas.

Take $z$ to increase inward, so the positive coefficient $A$ describes an inward-increasing temperature. Constant gravity and hydrostatic equilibrium give $dP/dz=\rho g$, henceMatching the radiative temperature gradient to the adiabatic temperature gradient gives the formal local boundary relationThe same result follows from radiative diffusion: for constant upward thermal flux $F$ and Rosseland mean opacity $\kappa$, $A=3\kappa F/(16{\sigma }_{\mathrm{SB}})$.

There is an important limitation to treating $A$ as constant over the entire radiative layer. Integration from an irradiated outer boundary givesFor a diatomic ideal gas, ${\nabla }_{\mathrm{ad}}=2/7>1/4$, so this profile cannot actually reach a radiative-convective boundary. Formally, imposing a constant ${\nabla }_{\mathrm{ad}}$ would givewhich is positive only for $0<{\nabla }_{\mathrm{ad}}<1/4$. A finite boundary for a normal molecular atmosphere requires additional opacity, flux, or thermodynamic variation. The local matching formula is usable near a real boundary, but constant $A$ is not a complete global model of it. This is the convective stability of a constant-opacity irradiated atmosphere.

For the intended order-of-magnitude scaling, suppose the local values of $g/A$ and ${\nabla }_{\mathrm{ad}}$ are comparable for Jupiter and a hot Jupiter, and assume $T_b$ scales with planetary equilibrium temperature. Equal absorbed-flux factors around the same stellar luminosity give $T_b\propto a^{-1/2}$. Taking $a_J=5.2\mathrm{AU}$ and a representative close-in orbit $a_h=0.05\mathrm{AU}$ yieldsThis illustrates how irradiation can push a boundary much deeper. It is a conditional estimate calibrated from the supplied $0.1\mathrm{bar}$ reference, not a self-consistent prediction of the globally constant-$A$ model. Different intrinsic cooling flux, opacity, gravity, or atmospheric metallicity of a giant planet can substantially alter it.

Here $\beta =p_{\mathrm{rad}}/p_{\mathrm{gas}}$ is the radiation-to-gas pressure ratio, not plasma beta. Write the opacity as ${\kappa }_{\mathrm{op}}$ to distinguish it from the radial epicyclic frequency. The radiation pressure is $p_{\mathrm{rad}}=4\sigma T^4/(3c)$, so radiative diffusion can be written asSince $\beta$ is independent of height, $p_{\mathrm{rad}}=\beta p/(1+\beta )$. Substitution of hydrostatic equilibrium, $p^{\prime }=-\rho {\Omega }^{2}z$, givesDifferentiating and using the viscous heating equation, together with $r{\Omega }^{\prime }=-q\Omega$, yieldsThus the effective dynamic viscosity $\rho \nu$ is independent of height. If the surfaces are at $z=\pm H$, the vertically integrated kinematic viscosity satisfiesUse the previous result and the specified Eddington accretion rate convention,to obtain the full thicknessThe constant-pressure-ratio vertical disk model therefore has thickness proportional to the inward accretion rate. This is a full surface-to-surface thickness, rather than a density disk scale height. Its use as a thin disk requires $2H\ll r$.

The usual Pressure-opacity zones of a Shakura--Sunyaev thin disk, in order of increasing cylindrical radius, are an inner region dominated by radiation pressure and electron-scattering opacity, a middle region dominated by gas pressure and electron-scattering opacity, and an outer region dominated by gas pressure and the Kramers' opacity law, often approximated by free-free opacity. The boundaries depend on mass, mass accretion rate, and the alpha disk viscosity parameter. At sufficiently large radius, accretion-disk self-gravity or changes in ionization can invalidate this three-zone model.

For the inner region writeHere $F$ is the standard thin-disk dissipation flux through one face, and $R_{\ast }$ is the innermost stable circular orbit treated as a zero-torque inner boundary. The given factor $1/2$ in the dissipation rate is appropriate to one face. For Keplerian rotation, it is $(9/8)\nu \Sigma {\Omega }_K^2$.

Let $H$ denote the disk scale height or vertical half-thickness, not the full thickness. In vertical hydrostatic equilibrium, the leading vertical acceleration is ${\Omega }_K^{2}z$, so a one-zone estimate givesIn the inner zone, $P_c\simeq P_{\mathrm{rad},\mathrm{c}}=a_{\mathrm{rad}}{T}_c^4/3$. For an optically thick medium with constant electron-scattering opacity, radiative diffusion gives the emergent one-face fluxEquivalently, combine $F=-c({\kappa }_{\mathrm{es}}\rho {)}^{-1}d{P}_{\mathrm{rad}}/dz$ with vertical pressure balance. Equating the radiative flux to the dissipated flux yields the radiation-supported inner-disk height:Order-unity vertical-profile factors depend on the precise height convention. The central mass cancels at fixed physical mass accretion rate, because both vertical gravity and local dissipation are proportional to $G{M}_{\mathrm{BH}}/R^3$.

If ${\dot{m}}_{\mathrm{Edd}}=L_{\mathrm{Edd},\mathrm{es}}/({\eta }_0{c}^2)$ and $r_g=G{M}_{\mathrm{BH}}/c^2$, this can also be writtenThe radiation-supported height rises from the formal zero at $R_{\ast }$ toward the approximately constant value $H_0=3{\kappa }_{\mathrm{es}}\dot{m}/(8\pi c)$. That zero is not a reliable description of the physical plunging region: the radiation-dominated approximation and the Newtonian zero-torque formula fail very close to the inner edge.

Beyond the inner zone, the height grows slowly with radius rather than staying on that plateau forever. Far from the inner boundary, the standard gas-pressure scalings give $H\propto R^{21/20}$ in the electron-scattering zone and $H\propto R^{9/8}$ in the Kramers zone. These follow from $H\propto T_c^{1/2}{R}^{3/2}$ with $T_c\propto R^{-9/10}$ and $R^{-3/4}$, respectively. A schematic profile is

vertical half-thickness H
  ^
  |                                               / outer gas/Kramers
  |                                          ____/
  |                                     ____/ middle gas/scattering
  |              ______________________/
  |          ___/ inner radiation/scattering plateau
  |       __/
  |     _/
  +----|----------------------------------------------> radius R
      R_*              zone boundaries are schematic

The height formula also tests the thin disk assumption. For a nonrotating hole with $R_{\ast }=6r_g$, the largest $H/R$ in this inner approximation occurs at $R=(9/4)R_{\ast }$. At $\dot{m}={\dot{m}}_{\mathrm{Edd}}$ and ${\eta }_0=0.1$, it is about $0.37$, so a disc this close to the conventional Eddington accretion rate is only marginally thin. Slim accretion disk effects and advected heat can then matter.

Analyze a local temperature perturbation at fixed radius and fixed surface density of a disk $\Sigma$. The thermal timescale of an accretion disk is shorter than its viscous timescale, so mass redistribution is negligible during the perturbation. Let $Q^{+}$ be viscous heating and $Q^{-}$ radiative cooling, measured consistently through one face or both faces, and define net cooling by $\dot{Q}=Q^{-}-Q^{+}$. A temperature increase is unstable if it decreases net cooling.

With gas pressure dominant, vertical hydrostatic equilibrium and the ideal gas law giveFor the alpha disk prescription with temperature-independent $\alpha$, $\nu =\alpha c_{s}H\propto T_c$, and the Keplerian rotation heating law isOne must not hold the steady-state $\dot{m}$ fixed while perturbing the temperature: the stress and heating respond through $\nu$, even though $\Sigma$ is temporarily fixed.

At an equilibrium temperature $T_0$ where $Q^{+}=Q^{-}=Q_0$, if $Q^{-}\propto T_c^q$ and $Q^{+}\propto T_c$, thenThe cooling exponent therefore decides the sign.

For hot, ionized, optically thin gas, the usual intended cooling model is thermal bremsstrahlung, whose volume emissivity scales as ${\rho }^2{T}_c^{1/2}$. Integrated vertically,Thus $q=0$, andA hotter annulus expands vertically, lowering its density enough to offset the direct increase in bremsstrahlung emission; viscous heating still rises.

For an optically thick medium with the Kramers' opacity law, $\kappa \propto {\rho }_c{T}_c^{-7/2}\propto T_c^{-4}$ at fixed $\Sigma$. Thermalized radiative diffusion givesIts falling opacity lets a hotter annulus radiate much more efficiently.

For an optically thick medium with dominant electron-scattering opacity, $\kappa$ is nearly independent of temperature. Provided enough true absorption is present to thermalize the radiation,This conclusion differs from the thermal instability of a radiation-pressure-dominated alpha disk, where the heating has a different temperature dependence. The question explicitly excludes that pressure regime.

The first conclusion needs its cooling assumption stated. $T>10^4\mathrm{K}$ alone does not guarantee that thermal bremsstrahlung dominates; atomic-line cooling, ionization changes, external illumination, or energy advection can change the answer. For a general optically thin emissivity ${\rho }^2\Lambda (T_c)$, the corresponding exponent isAccordingly, under the same fixed-$\Sigma$ alpha prescription,The unstable/stable/stable classification is the standard free-free/diffusion answer, rather than a universal result specified by temperature and optical depth alone.

Deep in an optically thick grey atmosphere, write $I_{\nu }=B_{\nu }+\delta I_{\nu }$ and retain the first spatial-gradient correction in the transfer equation:Angular and frequency integration then gives the radiative diffusion fluxFor a thin plane-parallel atmosphere, constant Rosseland mean opacity $\kappa$, negligible external irradiation, and radius nearly equal to $R$, radiative equilibrium gives $F=L/(4\pi R^2)$. ThereforeWith hydrostatic balance $dP/dz=-\rho g$, the equivalent pressure form iswhere $g=GM/R^2$.

For a perfect gas, $c_s^2\sim kT/({\mu }_m{m}_p)$. At the margin of the Toomre stability criterion,Integrating viscous heating through the disk and applying radiative diffusion givesEquating these fluxes yieldsThe kinematic viscosity is $\bar{\nu }\sim \alpha c_s^2/\Omega$, soBecause a Keplerian accretion disk has $\Omega \propto r^{-3/2}$,

The first equation is vertical hydrostatic equilibrium: the pressure gradient balances the vertical gravity of the central mass, ${\Omega }^{2}z$, and the disk's own potential ${\Phi }_d$. The second is the plane-parallel Poisson equation for disk self-gravity. The third balances the vertical increase of radiative flux against local viscous heating in a Keplerian alpha disk, and the fourth is the optically thick radiative diffusion law.

With $P/\rho \sim c_s^2$, $\Sigma \sim \rho H$, and all terms in hydrostatic balance comparable,The first comparison gives $H\sim c_s/\Omega$; inserting it into the second gives $\Omega c_s\sim G\Sigma$. Henceup to the order-one constants deliberately omitted by the scaling argument.

The photon-trapping radius is where the outward radiative diffusion time equals the inward inflow time. Radiation generated inside it is predominantly advected toward the accretor.

The inner Shakura--Sunyaev thin disk is supported mainly by radiation pressure and has nearly constant electron-scattering opacity. Equating radiative diffusion with local dissipation and imposing vertical hydrostatic equilibrium gives the characteristic half-thicknessThe radiation-supported height approaches a plateau away from the inner boundary. The formal zero at that boundary is outside the reliable radiation-dominated approximation. At an Eddington accretion rate defined using efficiency $0.1$, this height can make the inner flow only marginally thin.

A radiation-supported vertical disk has vertical hydrostatic equilibrium supplied mainly by radiation pressure. In a constant-opacity, optically thick layer in gravity $-{\Omega }^{2}z$, radiative diffusion gives $F=c{\Omega }^{2}z/\kappa$. This statement applies at a fixed radius and does not resolve the optically thin surface.

A radiative envelope transports most of its energy by radiative diffusion. Its stratified entropy profile gives a different rapid mass-loss response from a convective envelope.

In a reflection-symmetric astrophysical disk, the vertical mass coordinate measures mass per unit area above the midplane. It satisfies $dm=\rho dz$, so ${\partial }_m={\rho }^{-1}{\partial }_z$ inside the material. If the total surface density is $\Sigma$, the upper face is $m=\Sigma /2$; the normalized coordinate $\zeta =2m/\Sigma$ ranges from zero to one. Integrals weighted by mass density become ordinary integrals in $m$, which is especially useful for radiative diffusion and the density-weighted mean kinematic viscosity.

For positive finite opacity and nonvacuum ideal gas satisfying radiative diffusion, zero flux implies $dT/dz=0$. Write $P=\mathcal{R}\rho T$, where $\mathcal{R}=k_B/({\mu }_m{m}_p)$ for constant mean molecular weight ${\mu }_m$ in proton-mass units. If a polytropic equation of state $P=K{\rho }^{\gamma }$ with constant $K>0$ is also imposed, then $T=(K/\mathcal{R}){\rho }^{\gamma -1}$. For $\gamma \ne 1$, constant temperature would force constant density, contradicting vertical hydrostatic equilibrium $P^{\prime }=-\rho {\Omega }^{2}z$ on an open interval away from the midplane. The familiar finite polytropic slab for $\gamma >1$ is therefore a solution of the hydrostatic/adiabatic closure, not of this finite-opacity zero-radiative-flux system. For $\gamma =1$, the consistent isothermal profile is $\rho ={\rho }_m\exp [-{\Omega }^2{z}^2/(2\mathcal{R}{T}_m)]$, with no finite surface.