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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.

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.

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.

Take the gas pressure $P$ to vanish at the exterior boundary and the atmosphere to be thin enough that surface gravity of a star is constant. With constant electron-scattering opacity, negligible radiative acceleration, and hydrostatic equilibrium in optical depth,If the radiative force is retained, $dP/dr=-\rho (g-\kappa F/c)$ for gas pressure, and the same result holds with the constant positive effective gravity $g_{\mathrm{eff}}=g-\kappa F/c$. A nonzero exterior gas pressure instead gives $d\log P/d\tau =1/(\tau +\kappa P(0)/g_{\mathrm{eff}})$. Total gas-plus-radiation pressure has the nonzero radiative surface term from part (ii), so the advertised exact $1/\tau$ formula presumes gas pressure with zero exterior gas pressure.

Because $H\ll R$ and $M_{\mathrm{env}}\ll M$, the envelope has nearly constant area $4\pi R^2$ and nearly constant surface gravity of a star $g=GM/R^2$. Plane-parallel hydrostatic equilibrium isSince $d\Sigma /dz=-\rho$, one has $dP/d\Sigma =g$. Taking the pressure to vanish with $\Sigma$ at the surface gives

The spherical stellar energy equation is $d{L}_r/dr=4\pi r^2\rho ϵ$. With $r\simeq R$ and $F=L_r/(4\pi R^2)$,Division by $d\Sigma /dr=-\rho$ givesLikewise, radiative diffusion in a star givesand therefore

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.