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After complete ionization, each hydrogen nucleus supplies one ion and one Electron, while each helium-4 nucleus supplies one ion and two Electrons. Neglecting the small metal mass fraction $Z$ and putting $Y\simeq 1-X$, the mean molecular weight therefore satisfiesUse stellar homology to compare models with common dimensionless profiles. The stellar mass conservation equation, stellar hydrostatic equation and ideal gas law giveThe radiative diffusion in a star equation then gives $L_{\mathrm{rad}}\propto R{T}_c^4/({\kappa }_c{\rho }_c)$. Inserting the specified opacity yields $L_{\mathrm{rad}}\propto R{T}_c^7/(Z{\rho }_c^2)\propto {\mu }^7{M}^5/Z$. Independently, integrating the stellar energy-generation rate over the mass gives $L_{\mathrm{nuc}}\propto X^2{\rho }_c{T}_c^{5}M\propto X^2{\mu }^5{M}^7/R^8$. Equating these in stellar thermal equilibrium gives the radiative homology with fifth-power hydrogen burning and inverse-cubic opacity:At fixed solar mass these reduce to $L\propto {\mu }^7/Z$ and $T_c\propto {\mu }^{5/4}/(X^{1/4}{Z}^{1/8})$. The stellar homology approximation, rather than merely the local power laws, is what permits these relations between whole stars.

Equal luminosities imply ${\mu }_2/{\mu }_1=(Z_2/Z_1{)}^{1/7}=2^{-1/7}$. Thusso $X_2=0.8$ to one significant figure. Retaining the unrounded hydrogen mass fraction when comparing central temperatures givesThe first model has the higher central temperature.

For homogeneous hydrogen burning at fixed $M,Z$, fuel conservation gives $M{E}_0\dot{X}=-L$. Since $d\mu /dX=-5{\mu }^2/4$ and $L=L_0(\mu /{\mu }_0{)}^7$, the homogeneous fuel-depletion luminosity feedback obeysIntegration with the initial mean molecular weight ${\mu }_0$ givesThis idealized stellar evolution law stops when $X=0$; it does not predict a physically attainable infinite luminosity. Indeed $\mu =4/3$ at fuel exhaustion, giving $t_H=M{E}_0[1-(3{\mu }_0/4{)}^8]/(10{\mu }_0{L}_0)$, strictly before the formal singularity.

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.

Mechanical balance and equipotentials. Let $\Phi$ be the Newtonian gravitational potential and $\varpi$ the cylindrical distance from the rotation axis. For rigid angular velocity $\Omega$, a fluid element at rest in the rotating frame has no Coriolis acceleration. The centrifugal acceleration is ${\Omega }^2\varpi \hat{\varpi }$, so hydrostatic equilibrium becomesUsing the Poisson equation for gravity and ${\nabla }^2(x^2+y^2)=4$ gives

On any regular connected equipotential surface, every tangent displacement satisfies $d\varphi =0$ and therefore $dP=0$. Locally $P=P(\varphi )$, with $dP/d\varphi =-\rho$. This makes $\rho$ a function of $\varphi$ as well. Equivalently, taking the curl of the hydrostatic equation gives $\nabla \rho \times \nabla \varphi =0$. Thus $P,\rho$ and ${\nabla }^2\varphi$ are constant on each surface. The last conclusion follows from the modified Poisson equation; it does not constrain the magnitude of the first derivative $|\nabla \varphi |$ to be constant there.

For uniform composition, an equation of state $P=P(\rho ,T)$ determines $T$ uniquely at fixed $P,\rho$ in the usual stellar regime. For the gas-plus-radiation equation, the derivative with respect to $T$ is positive. Hence $T=T(\varphi )$, and the opacity and radiative conductivity are also functions of $\varphi$.

Surface flux and the requested sketch. Define $f(\varphi )=-\chi dT/d\varphi$, positive for the outward radiative flux. ThenOn an outer equipotential surface, $f$ is constant and the outward normal is parallel to $\nabla \varphi$. Its effective gravity is $g=|\nabla \varphi |$, so $F_n=fg$. The Stefan–Boltzmann law $F_n={\sigma }_{\mathrm{SB}}{T}_e^4$ gives the radiative gravity-darkening law,A rapidly rotating star is oblate. Centrifugal support reduces the effective gravity near the equator, which is therefore cooler and dimmer; the poles are hotter. The original sketch below uses a subcritical Roche model of a uniformly rotating star to show this geometry and the corresponding temperature variation. The material temperature on an interior equipotential is not the same quantity as the emergent effective temperature defined by surface flux.

Why purely local radiative balance fails. Differentiating the flux givesOn one equipotential, $f,f^{\prime },{\nabla }^2\varphi$ and the proposed source $\rho ϵ(\rho ,T)$ are all constant, but $|\nabla \varphi {|}^2$ generally varies with latitude. Unless $f^{\prime }=0$ or there is another special cancellation, no single value of $\rho ϵ$ can match this flux divergence at all points on the surface. This is the radiative-equilibrium obstruction in a rotating barotropic star. Slow stellar meridional circulation can transport heat to compensate; it is distinct from abandoning hydrostatic balance at leading order.

Integrated energy balance with circulation. The steady continuity equation is $\nabla \cdot (\rho \mathbf{v})=0$. Combining the entropy–enthalpy relation with hydrostatic balance givesIntegrate the energy equation over $V$ and apply the divergence theorem:Uniform composition makes $h=h(\rho ,T)$ constant on the enclosing equipotential, and $\varphi$ is constant there by definition. Pull $(h+\varphi )$ out of the last integral; the remaining net mass flux is zero by steady continuity. Thus the equipotential luminosity conservation with stellar circulation isThis does not require $\mathbf{v}\cdot \hat{n}=0$ pointwise on $S$: inward and outward circulation may cancel in the net mass flux.

Finally, $f$ is constant on $S$ and the divergence theorem applied to the effective potential givesTherefore $-\chi dT/d\varphi =L/(4\pi Gm-2{\Omega }^{2}V)$. Since $dP/d\varphi =-\rho$,The mass $m$ is the mass enclosed by this surface, not necessarily the total stellar mass. The denominator is positive for the regular hydrostatic surfaces used here; near loss of centrifugal confinement the assumed stellar model must be reconsidered. The derivation uses slow circulation, retaining hydrostatic/equipotential thermodynamics to leading order while allowing a nonzero advective heat flux.

Use $\mathcal{R}$ for the stellar gas constant, reserving $R$ for the stellar radius. The specific internal energy of a monatomic gas obeying the ideal gas law is $e=P/[(\gamma -1)\rho ]=3P/(2\rho )$. At fixed enclosed mass, the first law of thermodynamics gives the heat available to the outgoing luminosity as $ϵ=-{\partial }_{t}e-P{\partial }_t(1/\rho )$ when there is no stellar nuclear fusion heating. ThereforeThe second coefficient includes both the mass density dependence of $e$ and pressure work. This is gravitational energy generation in a homologously contracting ideal-gas star.

Write the characteristic scales asThen $\rho ={\rho }_{0}b$, $P=P_{0}p$, $T=T_{0}p/b$, $m=Mq$ and $L_r=Ll$. Since $d/dr=R^{-1}d/dx$, direct substitution in the stellar hydrostatic equation and mass conservation givesFor radiative diffusion in a star, dividing the temperature equation by $T_0/R$ givesIn particular the fourth power here is of $\mathcal{R}$, not of the radius.

Because $q(x)$ is fixed and $M$ is constant, a fixed mass label has fixed $x$. Thus stellar homology gives ${\partial }_t\rho =-3\rho \dot{R}/R$ and ${\partial }_{t}P=-4P\dot{R}/R$. The heating rate reduces to $ϵ=-3P\dot{R}/(2\rho R)$, positive during Kelvin-Helmholtz contraction. Substitution in $d{L}_r/dr=4\pi r^2\rho ϵ$ now yieldsThe dimensionless profiles fix $D$ and $E$. In a common stellar homology family with fixed stellar composition and opacity normalization, the expression for $D$ impliesIt is independent of radius and constant in time for a particular fixed-mass star. The formula for $E$ gives $\dot{R}=-2EL{R}^2/(3G{M}^2)$, which integrates toWhen the initial-radius term is negligible,This is the Kelvin-Helmholtz contraction time law in the prescribed stellar homology model. Infinite initial radius is a limiting idealization; a finite initial radius retains the first term.

At main sequence arrival, CNO cycle heating with fixed stellar composition has luminosityThe integral is dimensionless and fixed within the stellar homology family. The constant-opacity radiative scaling still gives $L\propto M^3$. Equating the stellar nuclear fusion and transported luminosities givesFinally the Kelvin-Helmholtz contraction law evaluated at $R_{\mathrm{MS}}$ gives $t_{\mathrm{MS}}\propto M^2/(L{R}_{\mathrm{MS}})\propto M^{-34/19}$. In a coeval star cluster, the mass just arriving at the main sequence consequently satisfiesThese results comprise constant-opacity homologous contraction to CNO ignition; their mass exponents compare fixed-composition models with common dimensionless profiles.

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.

Use radiative homology with proton-proton burning and inverse-fourth-power opacity, retaining the composition factors throughout. To distinguish the stellar radius from the stellar gas constant, write them as $R$ and $\mathcal{R}$. With gas pressure alone, introduce the scale factorsThe ideal gas equation gives $\rho ={\rho }_{\ast }b$, $P=P_{\ast }p$ and $T=T_{\ast }p/b$. Substitution into hydrostatic equilibrium and mass conservation givesso $p^{\prime }=-bq/x^2$ and $q^{\prime }=x^{2}b$.

Set $\zeta =Z(1+X)$. The radiative diffusion in a star equation, with the stipulated opacity, becomesHere ${\rho }^2{T}^{-7}={\rho }_{\ast }^2{T}_{\ast }^{-7}{b}^9{p}^{-7}$. ThusThe luminosity equation follows from the specific heating rate of the Proton–proton chain:ConsequentlyThese give all four dimensionless stellar structure equations with their numerical coefficients.

The stellar homology assumption means the dimensionless profiles are shared across the models, not merely that an individual model can be written in dimensionless coordinates. The displayed equations then require $D$ and $E$ to be fixed. Holding the universal constants and ${\kappa }_0,{ϵ}_0$ fixed givesEquating the two cancels $M^6$ and yields $R^6\propto X^2\zeta {\mu }^{-4}$. Therefore the radius and mass-luminosity relation areThe mass-independent radius is a consequence of the particular heating and opacity exponents in this model; it is not a general radius law for main-sequence stars.

Initially $X=X_0$ is the same for all stars. The adopted mean molecular weight $\mu =4/(5X+3)$ also has the same initial value, neglecting its small direct dependence on $Z$. Thus $R\propto Z^{1/6}$. At fixed luminosity, the Stefan–Boltzmann law $L=4\pi R^2\sigma T_e^4$ givesThe stars compared here can have different masses; the luminosity law adjusts the mass to keep $L$ fixed.

For the evolution, take the stellar mass as constant and maintain homogeneous composition. The available energy in the remaining hydrogen is $M{E}_{H}X$, so conservation of energy in the nuclear-powered model givesThis includes the entire mixed hydrogen reservoir, not just the mass in a central burning region.

To determine the turnoff mass of a homogeneously mixed proton-proton-burning population, write the luminosity law as $L=C{M}^6{Z}^{-7/6}H(X)$, whereSeparating variables, the time to evolve from $X_0$ to $f{X}_0$ isFor a common $X_0$ and $0\le f<1$, this integral is a positive finite constant independent of $M,Z,t$. It follows that the turning-off stars satisfyThe integration retains the evolving luminosity and mean molecular weight; replacing $L$ by its initial value is unnecessary.

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.

Use a characteristic midplane temperature $T$, density $\rho$ and disk scale height $H$. For an ideal gas of fixed composition, dominant gas pressure gives $c_s^2\propto T$, where $c_s$ is the sound speed. Vertical hydrostatic equilibrium and the alpha disk prescription giveNumerical factors in the vertical averaging do not affect the exponents. The stipulated opacity becomesFor an optically thick, radiatively efficient disk, radiative diffusion gives a cooling rate per unit area proportional to $T^4/(\kappa \Sigma )$. The viscous heating rate is proportional to $\nu \Sigma {\Omega }^2$. ThusEquating heating and cooling gives $T\propto \alpha {\Sigma }^{1/2}{\Omega }^{-1/2}$, soAt fixed $\alpha$ and central mass, $\Omega \propto R^{-3/2}$. Therefore the gas-pressure alpha disk with inverse-density opacity satisfies

For thermal stability of an accretion disk, perturb $T$ at fixed $R$ and $\Sigma$, since the thermal adjustment is faster than radial mass redistribution. The logarithmic heating and cooling slopes are respectively one and two. At equilibrium, a fractional temperature increase therefore makes the excess heating $Q^{+}-Q^{-}$ decrease, and the disk cools back. Thus the local thermal equilibrium is stable within the assumed gas-pressure, radiative-diffusion closure.

For viscous stability of an accretion disk, use the thermally equilibrated relation between viscosity and surface density. It givesLocally freezing the background radius in the Keplerian viscous diffusion equation, a short radial Fourier mode has growth rate $-3k_R^2{\partial }_{\Sigma }(\nu \Sigma )<0$. Density enhancements diffuse rather than grow, so the disk is locally viscously stable as well.

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.