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For no incident radiation and outward intensity approximated as direction independent, the surface angular moments are $F=\pi I_0$ and $P_{\mathrm{rad}}(0)=2\pi I_0/(3c)$. Integrating the radiation momentum equation with the Eddington closure approximation yields the grey atmosphere law $T^4=3T_{\mathrm{eff}}^4(\tau +2/3)/4$. This is an approximate angular boundary condition, not isotropy across both hemispheres.

The Hopf function encodes the departure of a plane-parallel grey atmosphere from a purely linear source profile. With physical bolometric radiative flux $F$ and standard mean intensity $J$, it is defined by $J=3F[\tau +q(\tau )]/(4\pi )$. In thermal radiative equilibrium, $T^4=3T_{\mathrm{eff}}^4[\tau +q(\tau )]/4$. The Eddington closure approximation with its approximate surface condition gives constant $q=2/3$; exact angular transfer generally gives a depth-dependent function.

Write $u=\cos \theta$ for the direction cosine and $B=j/\kappa =\sigma T^4/\pi$ for the radiative transfer source function. Integrating the radiative transfer equation $u{\partial }_{\tau }I=I-B$ over solid angle givesRadiative equilibrium makes the radiative flux $F$ constant, so $J=B=j/\kappa$.

For $I=A+Cu$, the radiation-field moments areConsequently the Eddington closure approximation holds. Substituting into the radiative transfer equation, with $B=J=A$, gives $u{A}^{\prime }+u^2{C}^{\prime }=Cu$. Equality for every $u$ requires $A^{\prime }=C$ and $C^{\prime }=0$, with $C=3F/(4\pi )$. The stipulated inward hemispheric radiative flux at the surface isPutting this equal to zero fixes $A_0=2C/3$, hence $A=C(\tau +2/3)$. Using $A=\sigma T^4/\pi$ and the effective temperature definition $F=\sigma T_e^4$ gives the grey atmosphere lawThe positivity limitation of a linear Eddington intensity matters here: the formal surface intensity $I(0,u)=C(2/3+u)$ is negative for $u<-2/3$. Thus this angular ansatz gives the requested Eddington surface boundary condition for approximate moments; vanishing integrated inward flux is not an exact, nonnegative, pointwise no-incoming radiation field.

Neglecting radiation pressure, the stellar hydrostatic equation and the definition of optical depth give hydrostatic equilibrium in optical depth, $dP/d\tau =g/\kappa$. Since $T^4=T_0^4(1+3\tau /2)$, putting $v=T^4$ givesTaking $P(0)=0$ and $\alpha >0$, the grey-atmosphere pressure with power-law opacity is thereforeThe pressure formula printed in the PDF is missing the factor $1/\beta$. Differentiating its right-hand side would otherwise give $dP/d\tau =\beta g/\kappa$. For $\beta =0$ the correct limit is $P^{\alpha }=2\alpha g\log (T^4/T_0^4)/(3{\kappa }_0{T}_0^4)$.

At the specified matching location, the luminosity relation makes $F=L_r/(4\pi r^2)=\sigma T^4$, hence $T=T_e$, $T^4=2T_0^4$ and $\tau =2/3$. Multiplying the corrected pressure relation by ${\kappa }_0{T}^{4-4\beta }/g$ givesThis agrees with the PDF's final surface boundary condition for a power-law-opacity grey atmosphere. It also extends to negative $\beta$ wherever the model applies; at $\beta =0$, its continuous limit is $P\kappa /g=4\alpha \log 2/3$.

Write $u=\cos \theta$, $S=j/\kappa$ and use axial symmetry. The radiative flux is $F=2\pi {\int }_{-1}^{1}uIdu$. Integrating the radiative transfer equation over directions givesThere are no energy sources in the atmosphere, so radiative equilibrium makes $F$ constant. Consequently

For the linear angular approximation $I=A+Cu$, the radiation-field moments areThus it satisfies the Eddington closure approximation, andSince $F$ is constant, $C^{\prime }=0$. Substitution in the radiative transfer equation, using $S=A$, gives $u{A}^{\prime }=Cu$, hence $\boxed{A^{\prime }=C}$.

In this moment approximation, the inward hemispheric flux at the surface isSetting this boundary moment to zero gives $A_0=2C/3$. ThereforeCombining it with the thermal source function and $F=\sigma T_e^4$ gives the grey atmosphere temperature law:

The surface condition is an approximate angular-moment condition. If the affine intensity were interpreted as an exact nonnegative boundary radiation field, no incident radiation would require $I(0,u)=0$ for every $u<0$; a linear polynomial vanishing on that interval would have $A_0=C=0$ and no outward flux either. The derived formal profile indeed has $I(0,-1)=-F/(4\pi )<0$. This positivity limitation of a linear Eddington intensity is why one uses the Eddington surface boundary condition as a moment closure, not the affine ansatz as an exact emergent angular intensity. An outward direction-independent intensity with no inward intensity gives the same boundary moments.

For the gas-pressure-dominated atmosphere, combine $dP/dz=-\rho g$ with $d\tau /dz=-\kappa \rho$. Thus the hydrostatic equilibrium in optical depth is $dP/d\tau =g/\kappa$. Let $w=T^4$. The grey atmosphere law implies $dw/d\tau =3T_0^4/2$. With the specified power-law opacity,At the outer vacuum limit take $P=0$ and $T=T_0$. Since $\alpha ,\beta >0$, integration givesThis is the grey-atmosphere pressure with power-law opacity relation.

At the matching radius the local radiative flux is $L_r/(4\pi r^2)=\sigma T^4$. In the thin plane-parallel atmosphere approximation this equals $F=\sigma T_e^4$, so $T=T_e$ and $T^4=2T_0^4$. Multiplying the pressure relation by ${\kappa }_0{T}^{4-4\beta }/g$ yields the required surface pressure-opacity relationHere $P\kappa$ is a product, not a pressure subscript.

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.

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.

This is a semi-grey irradiated atmosphere in radiative equilibrium. The quantities have these meanings:

The model assumes a static plane-parallel atmosphere, separate grey stellar and thermal bands with constant opacities, local thermodynamic equilibrium for thermal emission, negligible scattering, and transport by radiative transfer rather than convection. The Eddington closure approximation sets the thermal angular-moment ratio to $1/3$, and the upper boundary supplies the usual $2/3$ term. Treating the incoming radiation with representative direction cosine $1/\sqrt{3}$ gives its attenuation $e^{-\sqrt{3}\gamma \tau }$. Intrinsic flux enters from below, while the imposed external stellar flux is absorbed from above.

With constant gravity and hydrostatic equilibrium, $\tau \simeq {\kappa }_{\mathrm{IR}}P/g$, allowing conversion to an atmospheric pressure-temperature profile. This last relation additionally assumes constant thermal opacity. The grey treatment describes the energy balance approximately; it does not resolve individual molecular absorption lines.

Measure optical depth inward, so $d\tau /dr=-\kappa \rho$ and $\tau =0$ at the exterior boundary. The printed integral is shorthand with inconsistent integration labels; this differential definition fixes its meaning. In a plane-parallel atmosphere in radiative equilibrium, $F=\sigma T_{\mathrm{eff}}^4$ is constant and $d{P}_{\mathrm{rad}}/d\tau =F/c$.

For the Eddington surface boundary condition, approximate the outgoing frequency-integrated intensity by a constant $I_0$ over the outward hemisphere, with no incoming radiation. The angular moments at the surface areIntegrating the moment equation then gives $P_{\mathrm{rad}}=F(\tau +2/3)/c$. The Eddington closure approximation, with local thermodynamic equilibrium and radiative equilibrium, identifies the radiation energy density with $a_{\mathrm{rad}}{T}^4$ and sets $P_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$. Using $a_{\mathrm{rad}}c=4\sigma$ givesThe photosphere is represented by $\tau =2/3$, where $T=T_{\mathrm{eff}}$; the exterior boundary instead has $T(0)=2^{-1/4}{T}_{\mathrm{eff}}$. The closure and angular boundary condition are approximations, rather than exact isotropy of radiation at the surface.

The Eddington closure approximation with the standard surface boundary condition gives the grey atmosphere profileTogether with the specified pressure law, logarithmic differentiation givesConsequently the radiative temperature gradient isFor a monatomic perfect gas, $\gamma =5/3$ and the adiabatic temperature gradient is $(\gamma -1)/\gamma =2/5$. Uniform composition makes the Schwarzschild criterion applicable. The first marginal layer therefore satisfies $\log (1+3{\tau }_c/2)=4/5$, giving the convection onset in a logarithmic-pressure grey atmosphereThe assumed profile becomes convectively unstable for $\tau >{\tau }_c$; further inward, a self-consistent model must account for convective transport.

Put $\mu =\cos \theta$. For $I_{\nu }=A_{\nu }+B_{\nu }{\mu }^3$, the mean intensity isbecause the cubic term is odd. The K-integral, or second angular moment, isbecause the ${\mu }^5$ contribution is also odd. HenceThis intensity obeys the Eddington closure approximation even though it is not isotropic.

For $I=A+C\mu$, the angular moments areandThus this angular form satisfies the Eddington closure approximation, $c{P}_r=4\pi J/3$, and

Part (i) gives $S=J=A$. Substitution in $\mu dI/d\tau =I-S$ yieldsMatching powers of $\mu$ gives $A^{\prime }=C$ and $C^{\prime }=0$, the latter being constant flux. Hence $A=C\tau +A_0$.

At the surface there is no incoming intensity. Applying this condition to the outgoing hemisphere in the moment approximation,so $A_0=F/(2\pi )=2C/3$. ConsequentlyNow $A=J=j/\kappa =\sigma T^4/\pi$ and the effective temperature is defined by $F=\sigma T_e^4$. It follows thatAt the top of the grey atmosphere,

For a plane-parallel atmosphere in radiative equilibrium, constant surface gravity of a star $g$ and opacity $\kappa ={\kappa }_0{P}^{\alpha -1}{T}^{4-4\beta }$, match the stellar structure interior where the radiative flux satisfies $F=\sigma T^4$. The grey atmosphere law then puts the surface at optical depth $\tau =2/3$ with $T^4=2T_0^4$. Integrating hydrostatic equilibrium in optical depth from $P(0)=0$ gives $P^{\alpha }=2\alpha g(T^{4\beta }-T_0^{4\beta })/(3\beta {\kappa }_0{T}_0^4)$. Multiplication by ${\kappa }_0{T}^{4-4\beta }/g$ proves the displayed boundary condition for $\alpha >0$, $\beta \ne 0$. At $\beta =0$ the integral is logarithmic and the continuous limit is $P\kappa /g=4\alpha \log 2/3$. The relation assumes negligible radiation pressure and describes the Eddington closure approximation, rather than exact angular transfer.