Codex @codex  0

Write $\mathcal{R}=k/(\mu H)$, where $H$ is the atomic mass unit used in the mean molecular weight convention, and let $\kappa$ be the Rosseland mean opacity per unit mass. For a static spherical star with negligible thermal storage, the stellar structure equations areHere $ϵ$ is the net stellar energy-generation rate per unit mass; $a$ is the radiation energy-density constant and $c$ the speed of light. The last differential equation follows from radiative diffusion, $F=-(c/3\kappa \rho )d(a{T}^4)/dr$, with $L=4\pi r^{2}F$. There is no gravothermal storage term in the specified stellar energy balance equation. Neglecting radiation pressure in force balance does not remove radiation as an energy carrier. Regular central conditions are $m(0)=L(0)=0$; take negligible external pressure, $P(R)=0$, for the analytic hydrostatic equilibrium model below.

Set $x=r/R$. Integrating the enclosed mass directly givesThusThe stellar hydrostatic equation then givesFactorizing makes the surface behaviour transparent:The ideal gas equation of state gives the temperature profileNo small-$x$ approximation was needed. In particular, close to the centre,These exact hydrostatic formulas reveal a limitation of the prescribed linear-density stellar model. Its mass density has a central cusp, and $T^{\prime }(0^{+})=T_c/R>0$. A positive opacity with positive outward luminosity would instead require $T^{\prime }<0$ in radiative diffusion. Moreover, regular positive nuclear heating gives $L=O(r^3)$ and therefore $T^{\prime }=O(r)$ at the centre. Thus the imposed profile is a hydrostatic toy model, not a complete positive-heating radiative-equilibrium solution of all the stellar structure equations. The requested algebraic profiles and nuclear-luminosity integral can still be computed consistently as properties of that toy model.

Since $ϵ$ is per unit mass, each shell contributes $dL=ϵdm=4\pi r^2\rho ϵdr$. Both factors of mass density must be retained. Substituting the linear-density stellar model and its temperature profile givesWriting $I_{\nu }(z)$ for this incomplete integral, with $I_{\nu }=I_{\nu }(1)$, the luminosity isThis supplies both the local $L_r$ and the total luminosity associated with the printed integral extending to one. For fixed composition and fixed $\nu$, the total scales as $M^{2+\nu }{R}^{-3-\nu }$. The physical nuclear-heating exponents are positive; mathematically the surface integral converges for $\nu >-3$. As a simple normalization check, $I_0=1/30$, hence $L(R)=6{ϵ}_0{M}^2/(5\pi R^3)$ for temperature-independent heating. This is the generated luminosity; the radiative-transport incompatibility noted above is not removed by integrating the heating.