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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.

For a homogeneous ideal gas of constant specific-heat ratio $\gamma >1$, the Schwarzschild criterion applied to the square-root exponential atmospheric profile givesFor $T_0>0$ and $\alpha >0$, put $s=\sqrt{T-T_0}$ and $A={\nabla }_{\mathrm{ad}}$. The quadratic inequality $A\alpha s^2-2s+A\alpha T_0<0$ has solutions precisely when $A\alpha \sqrt{T_0}<1$, between $s_{\pm }=[1\pm \sqrt{1-(A\alpha {)}^2{T}_0}]/(A\alpha )$. This is a local instability of the imposed temperature gradient; actual convection adjusts that gradient and the prescribed profile need not survive.

For a homogeneous dry ideal gas in a hydrostatic adiabat, $dT/dz=-g/C_p$, where $z$ increases upwards and $C_p$ is the specific heat capacity at constant pressure per unit mass. It follows from $dT/dP=1/(\rho C_p)$ and hydrostatic equilibrium. The altitude version of the Schwarzschild criterion is $dT/dz<-g/C_p$ for instability; equality is marginal. Moisture, composition gradients and nonideal equations of state modify the criterion.

For a dry, homogeneous ideal gas in hydrostatic equilibrium, a displaced parcel undergoing an adiabatic process follows $dT/dz=-g/C_p$, where $C_p$ is specific heat capacity at constant pressure. At a common pressure its buoyancy depends on its temperature difference from the environment, giving $N^2=(g/T)(dT/dz+g/C_p)$. Negative $N^2$ means growing displacements; zero $N^2$ means neutral stability. This is the homogeneous dry limit of the Schwarzschild criterion.

Assume an ideal-gas grey atmosphere with $P(0)=0$, constant surface gravity of a star $g$, $T^4=T_e^4(1+3\tau /2)/2$ and $\kappa ={\kappa }_0\rho T^{4\beta +1}$ for $\beta >1$. Using $\rho =\mu P/(\mathcal{R}T)$ in $dP/d\tau =g/\kappa$ and integrating givesDifferentiation yields $d\log T/d\log P=[(1+3\tau /2{)}^{\beta -1}-1]/[2(\beta -1)]$. Applying the Schwarzschild criterion with monatomic adiabatic gradient $2/5$ gives ${\tau }_b=\frac{2}{3}[((4\beta +1)/5{)}^{1/(\beta -1)}-1]$. Since ${\tau }_b$ depends only on $\beta$, matching to $P=K{T}^{5/2}$ fixes $K\propto g^{1/2}{T}_e^{-2\beta -5/2}$.

Take an ideal gas grey atmosphere with $T^4=T_e^4(3\tau +2)/4$, constant surface gravity of a star $g$, vanishing external pressure and $\kappa ={\kappa }_0{P}^{1/2}{T}^8$. Integrating hydrostatic equilibrium in optical depth gives $P^{3/2}=8g[1/2-(3\tau +2{)}^{-1}]/({\kappa }_0{T}_e^8)$. Its logarithmic temperature-pressure gradient is $9\tau /16$, so the Schwarzschild criterion for a monatomic gas is crossed at ${\tau }_c=32/45$. The radiative continuation beyond that depth is unstable and must be replaced by convective transport.

In a thin ideal-gas grey atmosphere with $T^4=T_e^4(1/2+3\tau /4)$, constant enclosed mass $M$ and luminosity $L$, and $\kappa ={\kappa }_0\rho T^{13}$, divide the stellar hydrostatic equation by radiative diffusion in a star. This gives $PdP/dT=16\pi acGM\mathcal{R}{T}^{-9}/(3{\kappa }_{0}L\mu )$. Integrating from $P=0$, $T^4=T_e^4/2$, yieldsLogarithmic derivative gives $\nabla =T^8/T_e^8-1/4$. For a monatomic gas gas, the Schwarzschild criterion reaches equality at $\nabla =2/5$, hence $T/T_e=(13/20{)}^{1/8}$. The purely radiative profile cannot be continued past this point as a convectively stable model.

The original PDF has absolute-value bars around both gradients; they are missing from the transcription. The stability argument must respect the sign of the upward coordinate.

Displace a small parcel upward by $\xi$. Assume it remains in pressure equilibrium with its surroundings but exchanges negligible heat, so it follows an adiabatic process. For a perfect gas with adiabatic index $\gamma$, $T\propto p^{(\gamma -1)/\gamma }$ along that parcel's adiabat. Its temperature gradient is thereforeThe last step uses hydrostatic equilibrium, and $c_p$ is the specific heat capacity at constant pressure. At its new height, the temperature difference between the parcel and the environment is $(T_{\mathrm{ad}}^{\prime }-T^{\prime })\xi$. At equal pressure, its relative density excess is the negative of its relative temperature excess. The upward buoyancy acceleration is thusThis is the squared Brunt–Väisälä frequency. Convective stability requires $N^2>0$, or, in signed-gradient form,For a temperature decreasing upward, this is exactly the PDF's Schwarzschild criterionThe absolute-value condition is also sufficient more generally, although a sufficiently steep temperature inversion remains stable even when that stronger condition fails. Reversing the signed inequality, as the bar-free transcription does, would instead describe an unstable temperature decrease.

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.

Consider a small outward displacement $\delta r$ of a fluid element. It adjusts to the ambient pressure while retaining its specific entropy and stellar composition, so an adiabatic process gives $d\log {\rho }_{\mathrm{parcel}}=d\log P/\gamma$. Its excess mass density over the new surroundings is, to first order,Restoring buoyancy requires this bracket to be positive. With $\gamma =5/3$, the Schwarzschild criterion is thereforeEquality is neutral stability. For uniform mean molecular weight, the ideal gas law gives $d\log \rho =d\log P-d\log T$. Since hydrostatic equilibrium pressure decreases outward, the inequality is equivalent to $\nabla =d\log T/d\log P<{\nabla }_{\mathrm{ad}}=2/5$. Dividing the radiative diffusion in a star and hydrostatic equilibrium stellar structure equations givesThis is the stellar radiative temperature gradient test for stellar convective stability.

In the thin upper stellar atmosphere take $m\simeq M$, $L_r\simeq L$ and use gas pressure dominance. The prescribed opacity becomes $\kappa ={\kappa }_0\mu P{T}^{12}/\mathcal{R}$. Dividing the two structure equations then yieldsAt zero optical depth, the atmospheric boundary has $T_s^4=T_e^4/2$ and negligible gas pressure. Integrating from that boundary, with $T_s^{-8}=4T_e^{-8}$, givesTo locate the onset, put $y=T_e^8/T^8$. Logarithmic derivative gives $d\log P/d\log T=4y/(4-y)$, henceThis starts at zero at the outer boundary and increases inward. It first reaches $2/5$ atThe corresponding optical depth is ${\tau }_b=\frac{4}{3}(\sqrt{13/20}-\frac{1}{2})>0$, so the onset lies inside the stellar atmosphere. This is grey-atmosphere convection onset with thirteenth-power opacity.

Let the interior of the fully convective star have $P=K_T{T}^{5/2}$, with $K_T$ spatially constant. At its boundary $T_b/T_e$ is fixed, so the atmospheric pressure formula givesThe Stefan–Boltzmann law, $L=4\pi R^2\sigma T_e^4$ with $\sigma =ac/4$, consequently impliesThe ideal gas equation transforms the interior relation intoFor completeness, the mass-radius scaling follows by taking $\rho ={\rho }_c{\theta }^{3/2}$ and $r=\alpha \xi$ with ${\alpha }^2=5K_{\rho }{\rho }_c^{-1/3}/(8\pi G)$. Hydrostatic equilibrium and mass conservation reduce to the Lane-Emden equation $({\xi }^2{\theta }^{\prime }{)}^{\prime }=-{\xi }^2{\theta }^{3/2}$, with $\theta (0)=1$, ${\theta }^{\prime }(0)=0$. Its fixed dimensionless profile gives $R\propto \alpha$ and $M\propto {\alpha }^3{\rho }_c$. Eliminating ${\rho }_c$ gives $R\propto K_{\rho }{G}^{-1}{M}^{-1/3}$, or $K_{\rho }\propto G{M}^{1/3}R$. Thus $K_T^2\propto M^{-1}{R}^{-3}$ at fixed stellar composition. Equating this with the atmospheric expression gives $L^{17/4}\propto M^2{R}^{19/2}$, and henceThis Hayashi relation with thirteenth-power opacity concerns the interior of a fully convective star beneath a thin stellar atmosphere. The coefficient $K_T$ may vary between stars; spatially constant specific entropy does not mean equal specific entropy across the whole sequence.

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.

Let $\Omega =-{\int }_0^{M}Gmdm/r$ be the gravitational energy. Multiply the stellar hydrostatic equation by $4\pi r^3$ and integrate through the star. Integration by parts and zero surface pressure givewhile its hydrostatic right-hand side is $-{\int }_0^{R}4\pi Gmr\rho dr=\Omega$. Hence the stellar virial theorem is

There is a genuine reference-sheet error to resolve. A monatomic ideal gas with adiabatic exponent $5/3$ has specific internal energy of an ideal gas $u=P/[\rho (5/3-1)]=3P/(2\rho )$, not the printed $3P/\rho$ on PDF page 2. The latter would give total internal energy $-\Omega$ and zero total energy, inconsistent with radiating contraction. The following derivation uses the consistent monatomic value.

A fully convective star is approximately isentropic and has an adiabatic stellar polytrope with index $3/2$. The given polytropic gravitational-energy formula yields $\Omega =-6G{M}^2/(7R)$. The virial relation gives total internal energy $U=-\Omega /2$, so total energy is $U+\Omega =-3G{M}^2/(7R)$. At constant mass with no nuclear heating, Kelvin-Helmholtz contraction obeys $d(U+\Omega )/ds=-L$, whenceHere $s$ denotes physical time.

In the plane-parallel atmosphere, the inward optical depth satisfies $d\tau =-\kappa \rho dr$, so hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. Neglecting radiation pressure, $\rho =\mu P/(\mathcal{R}T)$ and the specified opacity becomes $\kappa =({\kappa }_0\mu /\mathcal{R})P{T}^{16}$. ThereforeThe grey atmosphere relation and the boundary $P(0)=0$ integrate toTo apply the Schwarzschild criterion, put $w=2+3\tau$. Logarithmic differentiation givesThis increases inward, starting from zero. Marginal convection has ${\nabla }_{\mathrm{rad}}=2/5$, so $w_b^3=136/5$. Since $T_b^{12}/T_e^{12}=w_b^3/64$, the grey-atmosphere convection onset with seventeenth-power opacity isThe purely radiative atmosphere is used only up to this boundary.

The convective relation $P=K{T}^{5/2}$ is spatially uniform in its entropy coefficient $K$ at each instant; $K$ can change as the star radiates and contracts. At the convection boundary $w_b$ is a fixed number, so $P_b\propto g^{1/2}{T}_e^{-8}$ at fixed composition and opacity normalization. ThusOn the other hand, the central scalings for the $n=3/2$ stellar polytrope give $P_c\propto G{M}^2/R^4$ and $T_c=\mu P_c/(\mathcal{R}{\rho }_c)\propto \mu GM/(\mathcal{R}R)$. Hence, at fixed composition,Equating the interior and atmosphere expressions yields $T_e^{21/2}\propto M{R}^{1/2}$. Applying the Stefan–Boltzmann law then proves the Hayashi contraction with seventeenth-power opacity relation

At fixed mass write $L=A{M}^{8/21}{R}^{\alpha }$, with constant $A$ and $\alpha =46/21$. The contraction equation is $\dot{R}=-C{R}^{\alpha +2}$, where $C=7A{M}^{8/21}/(3G{M}^2)$. Direct integration givesMultiplying through by $R^{\alpha +1}$ and substituting the current luminosity givesFor the stated very large initial radius, or after that initial-radius term becomes negligible,This is the luminosity-age relation within the contraction model, rather than an assertion that a physical star begins at literally infinite radius.

In the radiative layer, take the enclosed mass to be $M_1$, the transported luminosity to be $L$, and the mean molecular weight to be constant. The envelope's own gravity is neglected. Divide the hydrostatic and radiative temperature equations, and then eliminate density using the gas equation of state:Put $p=n+1$ and $b=4+m+n$. For $p,b\ne 0$, integration givesDefining $C^p=pB/b$ and $K=K_P/C^p$ yieldsIn the relevant positive-exponent case, writing $K=T_0^b$ produces the stated form whenever this integration constant is nonnegative. The exceptional values excluded above lead to logarithms: $p=0$ gives $\log P=B{T}^b/b+\mathrm{const}$, $b=0$ gives $P^p=pB\log T+\mathrm{const}$, and both zero give $\log P=B\log T+\mathrm{const}$.

For radiative-envelope convection matching, compute the logarithmic gradient of the pressure relation:At the marginal boundary, the Schwarzschild criterion equates this with the ideal-gas adiabatic gradient ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma$. HenceThe positive real $T_0$ notation requires a nonnegative bracket; a general signed $K$ is the safer integration constant otherwise. For the later case $n=1,m=3$ and a monatomic gas, this bracket is $3/5$, so $T_0$ is of the same order as $T_b$.

Deep below the convective boundary, with the temperature-power term dominating $K$, the pressure law becomes $P\simeq C{T}^{b/p}$. Differentiating this approximation and using hydrostatic balance givesThe general temperature is a term proportional to $1/r$ plus a boundary-dependent constant. Well inside the boundary radius, where that term dominates, it reduces toThe inverse-radius profile is an inner approximation, not an assertion that the boundary term vanishes everywhere.

Now take $n=1,m=3$, so $p=2$, $b=8$. Then $P\simeq C{T}^4$, $\rho \simeq (\mu C/\mathcal{R})T^3$, and define $A_T=\mu G{M}_1/(4\mathcal{R})$ so that $T\simeq A_T/r$. The luminosity supplied by the thin burning layer isBecause the integral is strongly concentrated near $R_1$ and $r_b\gg R_1$, the outer term is negligible. ThusThe factor $\rho ϵ$, rather than merely $ϵ$, is essential since the specified energy-generation rate is per unit mass. This integration uses the deep radiative profile as a leading thin-shell approximation.

Crucially, $C$ is not independent of luminosity. Its defining relation givesSubstitution into the shell integral therefore yields $L^2\propto M_1^{17}{R}_1^{-13}$. The degenerate-core relation gives $R_1\propto M_1^{-1/3}$, soThis degenerate-core shell-burning mass-luminosity relation is specific to the adopted opacity, burning exponent and structural approximations.