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On a steady phase-mixed orbit, line density on a Kepler orbit is proportional to $1/v$, so cross-sectional area is uniform in time. For an optically thin wire of blackbodies in radiative equilibrium,The proof uses specific angular momentum: $dt/r^2=df/h$, and $Ph=2\pi a^2\sqrt{1-e^2}$. If one instead imposes a line density proportional to $v$, area is weighted by $v^{2}dt$, yielding $f={\sigma }_{\mathrm{tot}}(1+e^2)/[4\pi a^2(1-e^2{)}^{3/2}]$. The two distributions agree only for a circular Kepler orbit.

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.

For a small, uniform surface at temperature $T$, enclosed by much larger black walls all at temperature $T_s$, the outward net heat transfer by thermal radiation isAssume an opaque grey body that is a diffuse thermal emitter, with the same emissivity $ϵ$ and radiometric absorptance at the surface's current temperature. Its emitting area $A$ sees only the enclosure: the view factor to the walls is one, with no self-viewing surface, opening or intervening obstruction. The intervening medium must neither absorb nor emit appreciable thermal radiation; a vacuum supplies a useful idealization. The enclosure is maintained at uniform $T_s$. The small-body assumption makes that reservoir description physically reasonable; genuinely black isothermal walls supply the incoming irradiance even without an infinite area ratio. Positive ${\dot{Q}}_{\mathrm{rad}}$ means net loss by the surface; negative means net gain. This expression counts thermal radiation alone, not all heat transfer.^[ref]

The reasoning is a two-way energy balance. Black walls supply irradiance $E=\sigma T_s^4$, so the surface absorbs $P_{\mathrm{abs}}=ϵAE$ and emits $P_{\mathrm{emit}}=ϵA\sigma T^4$. Subtract these to obtain ${\dot{Q}}_{\mathrm{rad}}$. For an opaque surface the remaining fraction $1-ϵ$ of incoming radiant flux is reflected; counting emitted plus reflected outgoing radiant flux minus total incoming radiant flux gives the same result. Kirchhoff's law of thermal radiation justifies equal spectral absorption and emission factors; the grey body assumption lets one factor be taken outside both spectral integrals.

Choose $ϵ=0.8$, $A=0.01{\mathrm{m}}^2$, $T=500\mathrm{K}$ and $T_s=300\mathrm{K}$. These are stipulated model inputs, not measured material properties. Using $\sigma \approx 5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$ givesThe incoming irradiance is $459.300\mathrm{W}{\mathrm{m}}^{-2}$, corresponding to $4.5930\mathrm{W}$ incident on the surface; only $80\%$ is absorbed. At $T=T_s=300\mathrm{K}$, emitted and absorbed powers are both $3.6744\mathrm{W}$, so radiative equilibrium has zero net exchange despite continuing emission. Changing only emissivity to $0.2$ reduces both one-way powers and the net loss by a factor of four, giving $6.1694\mathrm{W}$ net. A low-emissivity coating therefore suppresses both radiative cooling to colder walls and radiative heating from hotter walls.

Near equal temperatures, write $T=T_s+\Delta T$. Taylor's theorem givesFor constant emissivity, the radiative heat transfer coefficient $h_{\mathrm{rad}}$ is approximately $4.8992\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ at $T_s=300\mathrm{K}$ and $ϵ=0.8$. This local linear approximation is useful for small temperature differences; the $200\mathrm{K}$ difference in the worked example needs the full fourth-power expression.

A cooling experiment requires a wider first law of thermodynamics balance. In a uniform-temperature model without a heater,where $C(T)$ is the sample's heat capacity and the last two terms are signed outward convection and thermal conduction losses. A vacuum suppresses convection but does not eliminate thermal conduction through supports or sensor leads. A cooling slope therefore cannot establish emissivity unless these paths and heat capacity are independently constrained. Multiple wall temperatures, incomplete view factors, wavelength-selective materials or appreciable intervening absorption require a more complete radiative transfer calculation.

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 $\Psi$ be the gravitational potential, with ${\nabla }^2\Psi =4\pi G\rho$, and let ${\varpi }^2=x^2+y^2$ measure distance from the rotation axis. Uniform solid-body rotation contributes the centrifugal potential $-{\Omega }^2{\varpi }^2/2$. In the corotating frame, static force balance therefore gives the uniformly rotating stellar hydrostatic equilibrium equationsThe final coefficient follows because ${\nabla }^2(x^2+y^2)=4$.

A tangent displacement along a regular connected equipotential surface has zero dot product with $\nabla \varphi$, and hence with $\nabla P$. Thus $P=P(\varphi )$ locally. Taking the curl of the hydrostatic equilibrium equation gives $\nabla \rho \times \nabla \varphi =0$, so $\rho =\rho (\varphi )$ on such surfaces as well. The Poisson equation then shows that ${\nabla }^2\varphi$ is constant on each equipotential surface.

The magnitude $g=|\nabla \varphi |$ need not be constant. For instance $\varphi =A(x^2+y^2)+B{z}^2$, with positive $A\ne B$, has constant Laplacian but different pole and equator gradient magnitudes on an ellipsoidal equipotential. In a rotating star, flattening and centrifugal acceleration likewise make the effective gravity latitude dependent.

With uniform composition, an equation of state determines $T$ from $P$ and $\rho$; for the supplied ideal gas plus radiation pressure law its right side increases monotonically with $T$. Hence $T=T(\varphi )$, and the radiative conductivity $\chi (\rho ,T)$ is also constant on an equipotential surface. Put $f(\varphi )=-\chi dT/d\varphi$. Then $\mathbf{F}=f(\varphi )\nabla \varphi$. On the stellar surface, below mass shedding, the outward normal is $\nabla \varphi /g$, soThis is the radiative gravity-darkening law. The effective temperature measures emergent flux; it is not the same quantity as the interior material temperature that was constant on an equipotential.

A rapidly rotating star is oblate, with reduced effective gravity and cooler effective temperature around the equator. The poles are hottest. The original sketch uses a Roche model of a uniformly rotating star only to illustrate the shape and surface-flux variation:

The obstruction to purely local radiative equilibrium is visible by taking the divergence:On an equipotential surface, $\rho ϵ$, $f$, $f^{\prime }$ and the second term are constant, but $g^2$ generally varies. Unless $f^{\prime }=0$ or another special circumstance removes the mismatch, $\nabla \cdot \mathbf{F}=\rho ϵ$ cannot hold at every latitude. This is the radiative-equilibrium obstruction in a rotating barotropic star; heat advection by stellar meridional circulation resolves the missing energy transport in the leading hydrostatic treatment.

For the steady circulation, mass conservation gives $\nabla \cdot (\rho \mathbf{v})=0$. The specific enthalpy identity and hydrostatic equilibrium giveConsequentlyBoth $h$ and $\varphi$ are constant on the bounding equipotential surface. By the divergence theorem and steady continuity equation,Integrating the given thermal energy equation therefore gives equipotential luminosity conservation with stellar circulation:Individual circulation streams can cross the surface; only their net steady mass flux has to vanish.

Since $f$ is constant on that surface, integrating the Poisson equation givesThus $dT/d\varphi =-L/[\chi (4\pi Gm-2{\Omega }^{2}V)]$, while $dP/d\varphi =-\rho$. Dividing these derivatives and inserting $\chi =4ac{T}^3/(3\kappa \rho )$ yields the radiative gradient in a uniformly rotating star:The bracket must be positive for the regular outward-gravity configuration being used.

For an interior equipotential, both mass factors must be the enclosed mass $m$. The printed final bracket uses $M$, but its derivation gives $m$; at the outer surface $m=M$ and the printed bracket is recovered. For a direct check, a uniform-density model has $m=\rho V$, so the true correction ${\Omega }^{2}V/(2\pi Gm)={\Omega }^2/(2\pi G\rho )$ is independent of the chosen enclosed volume. Replacing $m$ by a fixed total mass would incorrectly make it volume dependent.

The standard thin-disk dissipation flux, summed over both faces, follows by substituting the Keplerian accretion disk profile into the viscous heating rate:The Stefan–Boltzmann law gives $\mathcal{C}=2\sigma T^4$ because there are two emitting faces. Local radiative equilibrium consequently gives the effective-temperature profile of a zero-torque diskHere $\sigma$ is the Stefan-Boltzmann constant. The scale $T_{\mathrm{in}}$ is not the temperature exactly at the inner boundary: the zero-torque inner boundary condition makes that formal temperature zero. Maximizing $T^4$ shows that the maximum effective temperature of a zero-torque disk occurs at $r=49r_{\ast }/36$, with $T_{\max }\simeq 0.488T_{\mathrm{in}}$.

At equal central mass and accretion rate, characteristic effective temperatures scale as $r_{\ast }^{-3/4}$. Thus, comparing corresponding values of $r/r_{\ast }$,The neutron-star disk is about 180 times hotter. The Planck law and Wien displacement law move its characteristic emission to about 180 times higher frequency, or 180 times shorter wavelength. A white dwarf disk commonly emits in optical and ultraviolet bands, while the hotter neutron star disk can emit in X-rays. Absolute bands require an actual accretion rate; the relative shift follows directly from the stated scaling.

The printed line-density statement is incorrect for a phase-mixed orbit. A steady line density on a Kepler orbit is inversely proportional to speed: for a cross-sectional-area current $J_{\sigma }$, the area per unit arc length is $d\sigma /ds=J_{\sigma }/v$. Equivalently, phase mixing gives $d\sigma =({\sigma }_{\mathrm{tot}}/P)dt$, uniform in mean anomaly, where $P=2\pi \sqrt{a^3/\mu }$ is the orbital period.

For an optically thin population of blackbodies in radiative equilibrium, a fragment absorbs $\sigma L_{\star }/(4\pi r^2)$ and reradiates the same luminosity. Thus the fractional luminosity of a phase-mixed eccentric wire isThe specific angular momentum gives $dt/r^2=d\theta /h$, so the integral is $2\pi /h$. Since $Ph=2\pi a^2\sqrt{1-e^2}$,This derives the intended result after explicitly correcting the density to $1/v$.

The error is consequential. If one instead imposes the literal density $d\sigma /ds\propto v$, its time weighting is $v^{2}dt$. Using $⟨v^2⟩=\mu /a$, $⟨r^{-2}⟩=[a^2\sqrt{1-e^2}{]}^{-1}$ and $⟨r^{-3}⟩=[a^3(1-e^2{)}^{3/2}{]}^{-1}$ givesAt $e=1/2$ this is $5/3$ times the printed result. The inverse-radius averages can also be obtained directly with eccentric anomaly $E$, $r=a(1-e\cos E)$ and $dt=(1-e\cos E)dE/n$.

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.

For a plane-parallel atmosphere with optical depth increasing downward, the radiative transfer equation isIntegrating over solid angle gives the first radiation-field momentIn radiative equilibrium, matter has no net local radiative heating, so the opacity-weighted frequency integral of $J_{\nu }-S_{\nu }$ vanishes. After converting each optical-depth derivative to physical depth and integrating over frequency,Thus the bolometric internal flux is constant with depth, as stated by constant flux in a plane-parallel radiative-equilibrium atmosphere.

In local thermal equilibrium $S_{\nu }=B_{\nu }(T)$. Define the planet's internal effective temperature of a planet by the constant outward flux. The standard Planck law integral givesThe corresponding intrinsic luminosity is $L_{\mathrm{int}}=4\pi R_p^2\sigma T_{\mathrm{eff}}^4$.

For the plane-parallel radiative transfer equationangular integration gives the zeroth radiation-field moment equationThe net radiative heating per unit volume is thereforeRadiative equilibrium requires it to vanish, equivalently that the frequency-integrated radiative flux be independent of depth:In local thermodynamic equilibrium with coherent isotropic scattering,where ${\omega }_{\nu }$ is the single-scattering albedo. Since ${\alpha }_{\nu }(1-{\omega }_{\nu })={\alpha }_{\nu ,\mathrm{abs}}$, the condition becomesConservative scattering redistributes directions but contributes no net material heating.

In an infrared-opaque single-layer greenhouse model, the atmospheric layer obeys $2\sigma T_a^4=\sigma T_s^4$. The outgoing planetary flux is $\sigma T_a^4$, while the globally averaged absorbed stellar flux is $(1-A_B)L/(16\pi a^2)$. Radiative equilibrium therefore givesHence the orbit at which the prescribed surface temperature can be maintained isThis assumes uniform redistribution, constant Bond albedo, unit longwave emissivity, a transparent atmosphere to starlight, and no internal heat. Without the greenhouse layer, replace $8\pi$ by $16\pi$.

WriteThe quasi-steady model is the coupled pairFor a thin shell, $T_s$ is close to $T_m$. A first-order asymptotic expansion givesand thereforeTo leading order $F_c\sim F_{\ast }$, so an initially freezing shell grows linearly:

For a thick shell the conductive correction in the surface balance is small. The radiative equilibrium temperature and its first correction areThus, with $A=a\log (T_m/T_e)$,When $F_w$ is negligible over an intermediate range, this ordinary differential equation gives the square-root growth lawRetaining $F_w$, its implicit solution isConsequently a sketch of $h(t)$ starts approximately linearly, crosses to square-root growth, and approaches $h_{\infty }$ with exponential decay of $h_{\infty }-h$. A shell placed above $h_{\infty }$ instead thins because the basal oceanic heat flux exceeds the conductive loss.

On a regular equipotential surface, $f,f^{\prime },\rho$ and ${\nabla }^2\varphi$ are constant, but $|\nabla \varphi {|}^2$ generally varies. Consequently a flux $\mathbf{F}=f(\varphi )\nabla \varphi$ cannot ordinarily satisfy local radiative equilibrium $\nabla \cdot \mathbf{F}=\rho ϵ$ at all latitudes. Special cases such as $f^{\prime }=0$ or spherical symmetry evade this obstruction. Stellar meridional circulation supplies heat advection in the leading hydrostatic treatment.

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.

Consider an idealized planet that is a sphere of radius $R$, illuminated by parallel sunlight of irradiance $S$, measured on a plane perpendicular to the rays. Its intercepting projected area is $\pi R^2$, whereas its total surface area is $4\pi R^2$. Therefore the incoming power averaged per unit total surface area iswhere $\alpha$ is the broadband Bond albedo. The factor four is a sphere geometry result, not an atmospheric transmission factor. Local sunlight still varies over the illuminated hemisphere and vanishes on the dark side.

Now stipulate a single spatially uniform temperature $T$, no atmosphere, no internal heating, and emissivity one for outgoing thermal radiation. This is a thermal blackbody model using the Stefan–Boltzmann law; the prescribed solar-band reflection is treated separately, so the body is not a perfect absorber at all wavelengths. Let $C>0$ be the effective heat capacity per unit surface area, in $\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$. Conservation of energy gives the ordinary differential equationHere $F$ is net heating per unit surface area, in $\mathrm{W}{\mathrm{m}}^{-2}$, and $\sigma$ is the Stefan-Boltzmann constant. Use $\sigma =5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, rounded from the exact constant reported by NIST. ^[ref] The uniform temperature is a model assumption: generally the surface average of $T^4$ differs from the fourth power of the surface-average temperature.

For fixed Bond albedo $\alpha$, radiative equilibrium requires $F(T_{\ast })=0$, givingWith the illustrative solar constant $S=1360\mathrm{W}{\mathrm{m}}^{-2}$, the global incoming average is $340\mathrm{W}{\mathrm{m}}^{-2}$:

Doubling albedo here lowers the equilibrium temperature by $33.231\mathrm{K}$, rather than halving it: the absorbed fraction changes from $0.7$ to $0.4$, and thermal radiation scales as $T^4$. These stipulated Bond albedos are whole-planet examples, not measurements of bare water and ice. The model predicts neither Earth's modern surface temperature nor its response to actual climate change; it omits the greenhouse effect, spatial heat transfer and seasons.

Each fixed-albedo equilibrium is a stable equilibrium: for $T_{\ast }>0$, the derivative is $F^{\prime }(T_{\ast })=-4\sigma T_{\ast }^3<0$. A small temperature increase produces net cooling, and a small decrease produces net heating. This is the simplest expression of the stabilizing Planck feedback. Allowing albedo to fall with temperature adds a competing term, as in ice-ocean albedo feedback.