Radiative envelope

A radiative envelope transports most of its energy by radiative diffusion. Its stratified entropy profile gives a different rapid mass-loss response from a convective envelope.

For a negligible-mass ideal-gas envelope above a core of mass $M_c$, with constant luminosity $L$ and Kramers' opacity law, division of hydrostatic equilibrium by radiative diffusion gives $PdP/dT=16\pi acG{M}_c\mathcal{R}{T}^{15/2}/(3{\kappa }_{0}L\mu )$. Taking the outer pressure constant to be negligible yields $P=C{T}^{17/4}$, where $C^2=64\pi acG{M}_c\mathcal{R}/(51{\kappa }_{0}L\mu )$, and $\rho =(\mu C/\mathcal{R})T^{13/4}$. Hydrostatic equilibrium then gives $dT/dr=-4\mu G{M}_c/(17\mathcal{R}{r}^2)$, so $T=B/r+T_0$ with $B=4\mu G{M}_c/(17\mathcal{R})$. The radiative-zero choice $T_0=0$ is exact for a vanishing-temperature boundary at infinity and a deep-envelope approximation for a very extended finite star. It is not implied by small surface temperature alone at a specified finite radius.

For a constant-luminosity radiative envelope with ideal-gas pressure and Kramers' opacity law, impose power laws on enclosed mass, density, pressure and temperature while retaining $M_r^{\prime }=4\pi r^2\rho$. If their exponents are $m,b,p,d$, respectively, mass conservation gives $m=b+3$, hydrostatic balance gives $d=m-1$, the gas law gives $d=p-b$, and diffusion gives $d-1=2b-2-13d/2$. Solving gives $(m,b,p,d)=(1,-32,-42,-10)/11$. In particular $R/r_c=(T_c/T_R{)}^{11/10}$. Neglecting envelope self-gravity instead sets $M_r$ approximately constant and changes the exponents to $d=-1,b=-13/4,p=-17/4$; that approximation is not an exact solution of mass conservation with nonzero density.

A thin radiative envelope with constant enclosed mass, luminosity and mean molecular weight, ideal gas pressure and opacity $\kappa ={\kappa }_0{\rho }^n{T}^{-s}$ has $a=n+1$, $b=n+s+4$, and $C=16\pi a_{\mathrm{rad}}cGM/[3{\kappa }_0(\mu m_u/k_B{)}^{n}L]$. Combining radiative diffusion in a star with hydrostatic equilibrium gives ${\nabla }_{\mathrm{rad}}=P^a/(C{T}^b)$, so $P^a-P_0^a=(aC/b)(T^b-T_0^b)$ for nonzero exponents. Zero exponents give logarithmic limits. The model must be stopped or modified if the Schwarzschild criterion predicts convection.

For $P=C(T^b+K{)}^{1/p}$ with $p=n+1$ and $b=4+n+m$, the logarithmic radiative temperature gradient is $(p/b)(1+K/T^b)$. At a marginal boundary with an efficient ideal-gas convective envelope, the Schwarzschild criterion equates it to $(\gamma -1)/\gamma$. Hence $K=T_b^b[(b/p)(\gamma -1)/\gamma -1]$. The notation $K=T_0^b$ with real positive $T_0$ requires a nonnegative bracket. Vanishing $p$ or $b$ needs the corresponding logarithmic integration instead of this power form.

In a negligible-mass, non-burning stellar envelope with $P=K{T}^{5/2}$ and Kramers' opacity law $\kappa ={\kappa }_1\rho T^{-7/2}$, the stellar radiative temperature gradient is ${\nabla }_{\mathrm{rad}}=3{\kappa }_1\mu K^{2}L/(16\pi ac\mathcal{R}GM{T}^{7/2})$. It decreases inward as the temperature rises, so the inner edge of efficient convection is where it equals $2/5$. If outer matching gives $K\propto g^{1/2}{T}_e^{-2\beta -5/2}$, then $T_b^{7/2}\propto K^{2}L/M\propto T_e^{-4\beta -1}$, since $g=GM/R^2$ and $L=4\pi R^2\sigma T_e^4$. Hence $T_b\propto T_e^{-(8\beta +2)/7}$.