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For a grey atmosphere with $T^4=\frac{3}{4}{T}_{\mathrm{eff}}^4(\tau +2/3)$ and $P^2=P_0^2\log (1+3\tau /2)$, logarithmic differentiation givesA monatomic perfect gas of uniform composition has adiabatic temperature gradient $2/5$. The Schwarzschild criterion therefore predicts the first marginally unstable layer at ${\tau }_c=\frac{2}{3}(e^{4/5}-1)$, and convection for larger optical depth. Beyond that onset, the assumed purely radiative profile is no longer a self-consistent transport solution.

A constant radiative-diffusion coefficient and constant gravity give $T^4=T_0^4+4AP/g$. Thus ${\nabla }_{\mathrm{rad}}=AP/(g{T}^4)<1/4$ for $T_0>0$. A diatomic ideal gas has adiabatic temperature gradient $2/7$, so this idealized profile never becomes convective. A physical radiative-convective boundary requires additional opacity, flux, or thermodynamic variation.

The global Eddington closure $L=(1-\beta )4\pi cGM/\kappa$, combined with a representative stellar gas-pressure fraction, makes the exterior stellar radiative temperature gradient approximately $M/(4m)$ when the luminosity is constant outside the core. Matching it to the adiabatic temperature gradient estimates the displayed boundary mass. This is a closure estimate: imposing exactly uniform $\beta$ throughout a finite-mass radiative envelope also conflicts with constant luminosity, as described by the constant-beta radiative-envelope obstruction.

For a grey atmosphere with $T^4=T_e^4(2+3\tau )/4$, an ideal gas and $\kappa ={\kappa }_0\rho T^{17}$, hydrostatic equilibrium in optical depth integrates to $P^2=512\mathcal{R}g[1/8-(2+3\tau {)}^{-3}]/(9\mu {\kappa }_0{T}_e^{16})$, taking $P(0)=0$. The logarithmic radiative gradient is $[(2+3\tau {)}^3-8]/48$. Equating it to the monatomic adiabatic temperature gradient $2/5$ gives $(2+3{\tau }_b{)}^3=136/5$ and the displayed onset temperature. Beyond this point the radiative profile must be replaced by convective transport.

Buoyancy and the convective criterion. Consider a small fluid parcel displaced outward by $\delta r$. Assume that sound waves keep its pressure equal to the ambient pressure, while heat exchange is slow enough that its displacement is an adiabatic process. Composition is uniform, so the parcel and its surroundings have the same mean molecular weight. For an ideal gas, $P\propto \rho T$, and an adiabatic parcel obeys $P\propto {\rho }^{\gamma }$. If $\delta \ln P<0$ is the pressure change on moving outward, thenThe density excess of the parcel at its new location is thereforeIf $\nabla >(\gamma -1)/\gamma$, this excess is negative: the outward-displaced parcel is lighter and accelerates farther outward. A downward displacement similarly continues downward. Conversely, a smaller gradient gives a restoring buoyancy force. The Schwarzschild criterion is thus stability for $\nabla <{\nabla }_{\mathrm{ad}}$ and instability for $\nabla >{\nabla }_{\mathrm{ad}}$, where the adiabatic temperature gradient is ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma$. For a monatomic gas,A composition gradient would require the additional density effect in the Ledoux criterion; it is absent in this calculation.

Opacity and pressure. The supplied opacity is characteristic of a cool hydrogen-rich stellar atmosphere dominated by negative hydrogen ion opacity. Mostly neutral hydrogen can attach electrons supplied by partially ionized metals; bound-free and free-free absorption involving the resulting $H^{-}$ ions produces a steep positive temperature dependence. The weak density dependence and strong temperature sensitivity can be represented by local power-law fits. At fixed composition the given pressure form is equivalent to $\kappa \propto {\rho }^{1/2}{T}^{17/2}$, because $P\propto \rho T$. This is an approximate fit over the relevant cool-atmosphere regime, not an exact opacity law at all temperatures; molecular opacity or hydrogen ionization eventually changes the regime.

For a geometrically thin atmosphere take the surface gravity of a star to be constant, $g_s=GM/R^2$. Combining $dP/dr=-\rho g_s$ with $d\tau /dr=-\kappa \rho$ gives hydrostatic equilibrium in optical depth,The grey atmosphere relation gives $T^8=T_e^8(3\tau +2{)}^2/16$. With negligible external pressure, $P(0)=0$, integration yieldsAn imposed nonzero external pressure would add $P(0{)}^{3/2}$; the boundary assumption is what fixes the integration constant here.

The factor in parentheses is $3\tau /[2(3\tau +2)]$. HenceThe grey-atmosphere convection onset with eighth-power temperature opacity occurs when this reaches $2/5$, soThe radiative continuation below this boundary would be unstable. Efficient convection instead keeps the deeper star approximately adiabatic.

Matching to the convective interior. Write $P=K{\rho }^{5/3}$. Eliminating density using the ideal gas law givesat fixed composition. Since ${\tau }_c$ is a fixed number, the matching temperature $T_b$ is proportional to $T_e$, while the matching pressure satisfies $P_b^{3/2}\propto g_s{T}_e^{-8}$. ConsequentlyThe polytropic mass-radius relation for index $3/2$ gives $K\propto R{M}^{1/3}$. Raising the matching relation to the ninth power and using $g_s\propto M/R^2$ therefore givesFinally the Stefan–Boltzmann law $L=4\pi R^2{\sigma }_{\mathrm{SB}}{T}_e^4$ yields the Hayashi relation with eighth-power temperature opacity,Composition and the opacity normalization are fixed in this comparison. The entropy coefficient $K$ is spatially constant in each convective model, but need not be the same for stars with different masses or radii.

For the Schwarzschild criterion, displace a small gas parcel upwards. Assume its composition is unchanged, it remains in pressure balance with its surroundings, and it moves quickly enough that heat exchange is negligible. The parcel obeys the adiabatic equation of state $P{\rho }^{-\gamma }=\mathrm{constant}$. Combining this with the ideal gas equation gives $T\propto P^{(\gamma -1)/\gamma }$, so its adiabatic temperature gradient is ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma =2/5$.

Let the surrounding temperature gradient be $\nabla =d\log T/d\log P$. For an upward displacement $\delta \log P<0$, the parcel's temperature relative to its new surroundings isIf $\nabla <{\nabla }_{\mathrm{ad}}$, this is negative: the parcel is cooler and, at equal pressure and composition, denser than its surroundings. It therefore sinks back; a downward displacement likewise gives a restoring buoyancy force. If the inequality is reversed, the displaced parcel continues to rise. The stability condition is thusEquality is the neutral boundary. There is no composition-gradient term because the ideal-gas parcel and the background have the same mean molecular weight.

In the thin upper stellar atmosphere, take a plane-parallel approximation with constant $g$, optical depth increasing inward, and zero overlying pressure $P(0)=0$. Hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. The ideal gas equation converts the stipulated opacity intoDefine $u=1+3\tau /2$, so the given grey atmosphere relation is $T^4=T_e^{4}u/2$. Integrating from the surface givesThis proves the required pressure law. In particular $P^2\propto g{T}_e^{-4\beta }$ times a function only of $u$ and $\beta$.

For the grey-atmosphere convection onset with density-linear opacity, differentiate the two profiles to obtain the radiative temperature gradient:It increases monotonically from zero, since $\beta >1$, and first reaches the adiabatic temperature gradient at $u_b^{\beta -1}=(4\beta +1)/5$. ThereforeBelow this point the radiative stratification is unstable and convection must transport some of the outward energy flux.

At the outer convection boundary, $u_b$ depends only on $\beta$, so $T_b=T_e(u_b/2{)}^{1/4}\propto T_e$ and $P_b\propto g^{1/2}{T}_e^{-2\beta }$ for fixed $\mu ,{\kappa }_0,\beta$. The matching constant in $P=K{T}^{5/2}$ is $K=P_b/T_b^{5/2}$. ConsequentlyThis temperature form of the polytropic relation follows from an ideal monatomic adiabat; its $K$ is not the density-polytrope coefficient in Question 3.

For the deep radiative boundary beneath a power-law-opacity convective envelope, negligible envelope mass and absent local energy generation give $m\simeq M$ and $L_r\simeq L$. Divide the radiative temperature equation by the pressure equation to get the stellar radiative temperature gradientThis is the gradient that would be needed if radiation alone carried $L$, and it is the quantity to compare with $2/5$ even while the actual region is convective. The deeper Kramers' opacity law gives $\kappa ={\kappa }_1\mu P{T}^{-9/2}/\mathcal{R}$. Substituting $P=K{T}^{5/2}$ yieldsAs $T$ rises inward this decreases, so the transition back to a stable radiative interior is atThus $T_{\mathrm{in}}^{7/2}\propto K^{2}L/M$. Using the outer matching and the Stefan–Boltzmann law,The mass and radius cancel. Hence, at fixed opacity coefficients and composition,This uses the same adiabat through the convection zone and distinguishes its inner transition temperature from its outer onset temperature.

Introduce the characteristic scalesThen $\rho ={\rho }_{\ast }b$, $P=P_{\ast }p$ and $T=T_{\ast }t$. The gas-only equation of state gives the important dimensionless identity $p=bt$. The PDF's CNO cycle approximation has temperature power 13; the converted TeX's power 1.3 is an OCR error.

Since $d/dr=R^{-1}d/dx$, the mass and hydrostatic pressure support equation immediately becomeThus $dq/dx=x^{2}b$ and $dp/dx=-bq/x^2$. The stellar structure equations for nuclear heating givesoFor radiative diffusion in a star, $\kappa \rho ={\kappa }_0{\rho }^2{T}^{-3}$, and thereforeThis gives the radiative gradient coefficientIn an efficiently convective region, the monatomic adiabatic temperature gradient is $2/5$. Hence $dt/dx=(2/5)(t/p)(dp/dx)=-2q/(5x^2)$. If the radiative gradient is shallower, radiation suffices; if it is steeper than the adiabatic gradient, the Schwarzschild criterion selects convection and the shallower adiabatic profile. Thus

The regular central conditions are $q(0)=l(0)=0$ with finite positive $p(0),t(0)$. In particular $q\sim b(0)x^3/3$ and $l\sim Eb(0{)}^{2}t(0{)}^{13}{x}^3/3$, so the pressure and temperature derivatives tend to zero at the centre. The normalized surface has $q(1)=l(1)=1$. In the idealized zero-pressure, zero-temperature surface used for stellar homology, $p(1)=t(1)=0$; a physical photosphere supplies small nonzero surface values that are neglected in these bulk homology relations.

The common dimensionless profiles and boundary conditions keep $D,E$ constant. With fixed ${\kappa }_0,{ϵ}_0$, $D$ constant yields $L{\mu }^{-7}{M}^{-5}=\mathrm{constant}$. Substitution into the nuclear coefficient then gives $R^{16}\propto {\mu }^6{M}^{10}$. Consequently the CNO homology with density-dependent inverse-cubic opacity isThe Stefan–Boltzmann law gives $T_e^4\propto L/R^2\propto {\mu }^{25/4}{M}^{15/4}$. Eliminating mass, rather than dropping composition dependence, gives the Hertzsprung-Russell diagram relationThe additive constant depends on the fixed physical and structural coefficients; logarithms refer to fixed choices of units.

For physical evolution time $s$, distinct from the dimensionless temperature $t(x)$, hold stellar mass fixed and put $u=3+5X$, $u_0=3+5X_0$. The mean molecular weight approximation and the homology relation give $L=L_0(u_0/u{)}^7$. Converting hydrogen mass to helium releases energy at rate $L$, so fuel conservation is $M{E}_{H}dX/ds=-L$. ThereforeIntegration with the initial composition proves the homogeneous fuel-depletion luminosity feedback lawIt is valid while hydrogen remains and the homology assumptions hold. Fuel exhaustion occurs when $u=3$, before the formal denominator vanishes; the apparent infinite luminosity is not a physical endpoint of this stellar model.

Efficient convection makes the interior follow a nearly adiabatic process and maintains uniform specific entropy. For the fully ionized monatomic ideal gas, the specific-heat ratio is $5/3$, so the adiabatic temperature gradient gives $d\log T/d\log P=2/5$. Integrating at a fixed time givesThe coefficient is spatially constant because the composition and specific entropy are uniform. It can change as the star loses energy. Equivalently the gas is an adiabatic stellar polytrope with $P=K_{\rho }{\rho }^{5/3}$ and polytropic index $n=3/2$; $K_{\rho }$ and the pressure-temperature coefficient $K$ are different quantities.

This fixes the dimensionless Lane-Emden equation profile throughout a stellar homology sequence. Writing $\rho (r)={\rho }_{c}f(r/R)$ in the mass integral gives ${\rho }_c=C_{\rho }M/R^3$, with a fixed positive shape constant $C_{\rho }$. Integrating the stellar hydrostatic equation through the same profile, with negligible surface pressure compared with central pressure, gives $P_c=C_{P}G{M}^2/R^4$. HenceThese are homology scalings of central stellar pressure. Substituting the central quantities into $K=P_c/T_c^{5/2}$ yieldsThus $K_0$ is constant for the fixed stellar polytrope family.

At the approximate photosphere, take $T=T_e$ and use the ideal gas relation $\rho =\mu P/(\mathcal{R}T)$. The prescribed opacity givesThe stellar surface boundary condition and surface gravity of a star therefore implyThis is the temperature-four special case of convective contraction with power-law opacity.

Using the Stefan–Boltzmann law, the luminosity isFor $n=3/2$, the supplied gravitational-energy formula is $\Omega =-6G{M}^2/(7R)$. The stellar virial theorem for a monatomic gas gives $2U+\Omega =0$, so the total energy is $E=U+\Omega =-3G{M}^2/(7R)$. During Kelvin-Helmholtz contraction, with constant mass and no significant nuclear source, $L=-dE/dt$. ThusMultiplying by $-(7/2)R^{-9/2}$ and integrating givesThe sign makes the radius decrease as time increases.

When the accumulated contraction term dominates the initial-radius term, this yields $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$. Combining it with the central ideal gas temperature gives the late-time central-temperature scalingReplacing $t-t_0$ by $t$ gives the stated late-time form. The asymptotic step also requires a time long enough to forget the initial radius, not merely a choice of zero for the age.

The stellar adiabatic exponents are fixed-composition, constant-specific entropy derivatives:The chain rule gives ${\Gamma }_1({\Gamma }_2-1)/{\Gamma }_2={\Gamma }_3-1$. Define the pressure derivatives ${\chi }_{\rho }=(\partial \log P/\partial \log \rho {)}_T$ and ${\chi }_T=(\partial \log P/\partial \log T{)}_{\rho }$. The first law of thermodynamics, with $du=Tds+Pd\rho /{\rho }^2$, givesCombining this with $d\log P={\chi }_{\rho }d\log \rho +{\chi }_{T}d\log T$ on an adiabat givesThus the specific-heat ratio is not generally equal to the three stellar adiabatic exponents.

For the mixture, ${\chi }_{\rho }=\beta$ and ${\chi }_T=4-3\beta$. The specific heat capacity at constant volume, obtained by differentiating $u$ at fixed mass density, isTherefore the adiabatic exponents of a monatomic gas-radiation mixture and its specific-heat ratio areIf a relation involving only the exponents is wanted, eliminate $\beta$ from ${\Gamma }_1=\beta +(4-3\beta )({\Gamma }_3-1)$:For a pure monatomic perfect gas, $\beta =1$ and $\gamma ={\Gamma }_1={\Gamma }_2={\Gamma }_3=5/3$. For any calorically perfect gas with constant heat capacities, the same equality holds with its own ${\gamma }_g$. In a genuine gas-radiation mixture, $0<\beta <1$, $\gamma ={\Gamma }_1/\beta$ while the exponents are given separately above.

In the radiation limit, ${\Gamma }_1={\Gamma }_2={\Gamma }_3=4/3$ and the adiabatic temperature gradient is $1/4$. This follows independently from photon entropy: a comoving volume $V$ has $S\propto V{T}^3$, so an adiabat obeys $T\propto V^{-1/3}$ and $P\propto V^{-4/3}$. However, $c_P/c_V$ is singular in the pure-radiation limit, not $4/3$. This is the pure-radiation constant-pressure heat-capacity singularity. The equation $P=a{T}^4/3$ fixes temperature whenever pressure is fixed, so an ordinary constant-pressure temperature derivative is not available; along the mixture limit $\gamma \to \infty$ and $\beta \gamma \to 4/3$. For photons alone, mass-specific quantities additionally require a material mass label. The often quoted radiation index $4/3$ is its pressure-density adiabatic exponent, not a finite constant-pressure/constant-volume heat-capacity ratio.

Assume $R_p/R_{\star }=0.10$, $T_{\star }=5800\mathrm{K}$, uniform disk-averaged thermal emission, a nearly blackbody stellar continuum, negligible reflected light at $20\mu \mathrm{m}$, and a flat response over the narrow band. If the eclipse depth is normalized to the out-of-eclipse star-plus-planet flux, the planet/star flux ratio is $r=0.005/(1-0.005)=0.005025$; using $r=0.005$ changes the estimate by less than one percent. The exoplanet secondary eclipse then givesUsing the Planck law at $\lambda =20\mu \mathrm{m}$ gives $T_1\simeq 3.1\times 10^3\mathrm{K}$, or about $3000\mathrm{K}$ at the precision justified by the radius assumption. The Rayleigh-Jeans law instead gives $T_1\simeq 0.5T_{\star }\simeq 2900\mathrm{K}$, a useful rough check. This is a brightness temperature, not the bolometric planetary equilibrium temperature. With zero Bond albedo and global redistribution, the latter would be only about $1250\mathrm{K}$ at $0.05\mathrm{AU}$; a thermal window can sample hotter, deeper gas.

A continuum photon normally emerges from an optical depth of order unity. In hydrostatic equilibrium,The emission-pressure degeneracy of a brightness temperature means that pressure cannot be uniquely recovered from the measured brightness temperature without $g$ and an opacity model. For an explicit nominal estimate, take $g=20\mathrm{m}{\mathrm{s}}^{-2}$ and an effective continuum mass absorption coefficient ${\kappa }_{20}=2\times 10^{-4}{\mathrm{m}}^2\mathrm{k}{\mathrm{g}}^{-1}$. Then $P_{\tau \sim 1}\sim g/{\kappa }_{20}=10^5\mathrm{Pa}$, or approximately one bar. In a hydrogen-rich atmosphere, collision-induced absorption by transient ${\mathrm{H}}_2$ pairs supplies continuum opacity even without ordinary molecular lines; its density dependence makes the constant-opacity calculation an illustrative estimate. With a normal deep atmospheric pressure-temperature profile, temperature increases at larger pressures, eventually following an interior adiabatic temperature gradient.

Two mechanisms for $T_1\ne T_2$ are wavelength-dependent sampling of a vertically varying temperature, and wavelength-dependent weighting of a horizontally nonuniform dayside. In the first, carbon monoxide-rich opacity near $4.5\mu \mathrm{m}$ can sample cooler gas around $0.1$ bar while the $20\mu \mathrm{m}$ continuum samples hotter gas around one bar. In the second, hot and cool areas contribute different proportions of the total flux in different bands because the Planck function depends nonlinearly on temperature. The vertical mechanism alone produces the illustrated pair of temperatures; a single isothermal, horizontally uniform blackbody would have the same brightness temperature in both bands.

If the J-band observation is thermal emission, its low-opacity window normally samples at least as deeply as the continuum at $20\mu \mathrm{m}$. On the illustrated non-inverted profile, $T_3\gtrsim T_1>T_2$. The ordering is conditional: molecular-line absence alone does not specify the relative continuum opacities, and an inversion or reflected-light contamination can change the inference. The exoplanet emission spectrum measures weighted contribution regions rather than three exact thermometers at known pressures.

Let $S$ denote specific entropy, reserving $s$ from the first question for envelope thickness. The stellar adiabatic exponents are defined at fixed composition byIt is useful to introduce ${\Gamma }_3-1=(\partial \log T/\partial \log \rho {)}_S$, so $({\Gamma }_2-1)/{\Gamma }_2=({\Gamma }_3-1)/{\Gamma }_1$.

Take the stellar gas to be a fully ionized, nonrelativistic monatomic ideal gas, with fixed specific gas constant $\mathcal{R}$. Its mixture with thermal radiation hasHere $u$ is specific internal energy, and the stellar gas-pressure fraction is $\beta =\mathcal{R}\rho T/P$. The assumption of monatomic gas fixes its heat capacity; the perfect-gas pressure law alone would not determine it.

At fixed $T$ and fixed $\rho$, respectively, the pressure derivatives areFor an isentropic process, the first law of thermodynamics gives $du=Pd\rho /{\rho }^2$. Differentiate the internal energy rather than artificially holding $\beta$ fixed during the perturbation:The parenthesis is $(P/\rho T)[(3/2)\beta +12(1-\beta )]$, givingSince $d\log P={\chi }_{\rho }d\log \rho +{\chi }_{T}d\log T$,Using the relation between the exponents gives the adiabatic exponents of a monatomic gas-radiation mixtureThe requested values, including the adiabatic temperature gradient, areThe pure-radiation and pure-gas rows are understood as limits of the mixture.

For uniform composition, the Schwarzschild criterion for stellar convective instability isTo test whether a radiative configuration becomes unstable, use its required stellar radiative temperature gradient ${\nabla }_{\mathrm{rad}}$ for $\nabla$. Since $dP/dr<0$, the equivalent radial condition isAn outward-displaced parcel then cools less than its new surroundings, remains less dense at the same pressure, and is further accelerated outward.

The requested Eddington-model convective-core mass fraction follows from the usual global constant-$\beta$ closure. At the outer radiative surface, put $m=M$, $L_r=L$, and $P_{\mathrm{rad}}=(1-\beta )P$. Dividing the radiation-pressure gradient by the stellar hydrostatic equation givesThere is no energy generation outside the convective core, so use $L_r=L$ there. The local radiative-gradient formula isThe same representative $\beta$ has been used in the global luminosity closure and at the boundary. Setting this gradient equal to ${\nabla }_{\mathrm{ad}}$ at $m=M_c$ producesThis is $1$, $77/80$, and $5/8$ for $\beta =0,1/2,1$, respectively, with $\beta =0$ a marginal radiation-dominated limit.

There is a consistency qualification to this last model estimate. It cannot be an exact stellar solution if all the printed assumptions are enforced pointwise. An exactly uniform nonzero radiation fraction would give $d{P}_{\mathrm{rad}}/dP=1-\beta$ at every radiative point. The exact equations would then requireIf $L_r$ is constant throughout an envelope with positive mass density, $m$ increases with radius, so this equality cannot hold there. Equivalently uniform $\beta$ fixes the actual gradient to $1/4$, whereas ${\nabla }_{\mathrm{ad}}>1/4$ for $\beta >0$. The boxed core fraction is the intended Eddington closure and boundary estimate, which relaxes exact constancy of $\beta$ in the detailed envelope. The constant-beta radiative-envelope obstruction identifies the missing approximation; it is not legitimate to silently assert exact compatibility.

Assume a homogeneous ideal gas, constant specific-heat ratio $\gamma >1$, and an upward displacement that is an adiabatic process and maintains pressure balance with its surroundings. If the ambient temperature falls faster with decreasing pressure than the parcel's adiabatic temperature gradient, the parcel becomes hotter and less dense than its surroundings; buoyancy amplifies the displacement. This is the Schwarzschild criterion:The square-root exponential atmospheric profile hasAn atmospheric thermal inversion with $\alpha <0$ has $\nabla <0$ and is stable to these ordinary adiabatic displacements. For $\alpha >0$, put $s=\sqrt{T-T_0}$. The condition becomes the quadratic inequalityintersected with $s\ge 0$ and $T_0+s^2>0$. In the usual case $T_0>0$, the maximum gradient occurs at $s=\sqrt{T_0}$, namely $T=2T_0$, and is $1/(\alpha \sqrt{T_0})$. Hence a convectively unstable interval exists precisely whenEquality in the existence condition gives a single neutrally stable point. A molecular-hydrogen-dominated hot Jupiter with rotational modes active but vibrational excitation and dissociation negligible has approximately $\gamma =7/5$, so $A=2/7$. The existence condition is then $\alpha \sqrt{T_0}<7/2$. For $T_0\le 0$ the general inequality above remains valid, but its roots must be intersected with the positive-temperature domain; the positive-$T_0$ maximum formula must not be reused. Composition gradients require the Ledoux criterion, and dissociation or variable heat capacities change $A$. Sustained convection normally adjusts a superadiabatic profile toward an adiabatic temperature gradient.

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.