For a family in <stellar homology>, dimensionless radial profiles are fixed. <Mass conservation> and <hydrostatic equilibrium> then give the central scalings
$$
\rho_c\propto\frac{M}{R^3},
\qquad
\frac{P_c}{R}\sim\frac{GM\rho_c}{R^2},
\qquad
\boxed{P_c\propto\frac{GM\rho_c}{R}\propto\frac{GM^2}{R^4}}.
$$
The <ideal gas> equation of state, with fixed <mean molecular weight>, consequently gives
$$
\boxed{T_c\propto\frac{P_c}{\rho_c}\propto\frac{M}{R}},
$$
where fixed dimensional constants such as $G$ are suppressed in homology relations.

The <proton–proton chain> law $\epsilon\propto\rho T^{7/2}$ gives the nuclear luminosity
$$
L_{\rm nuc}\sim M\epsilon_c
\propto M\rho_cT_c^{7/2}
\propto M^{11/2}R^{-13/2}.
$$
On the other hand, <radiative diffusion in a star> gives
$$
L_{\rm rad}\propto\frac{RT_c^4}{\kappa_c\rho_c}.
$$
Using the <Kramers opacity law> $\kappa_c\propto\rho_cT_c^{-7/2}$,
$$
L_{\rm rad}\propto\frac{RT_c^{15/2}}{\rho_c^2}
\propto M^{11/2}R^{-1/2}.
$$
Thermal equilibrium requires $L_{\rm nuc}=L_{\rm rad}$. Their common mass factor cancels, leaving $R^{-13/2}\propto R^{-1/2}$, and hence
$$
\boxed{R=\text{constant}},
\qquad
\boxed{L\propto M^{11/2}}.
$$

This model captures the gas-pressure support, pp-chain burning, and strongly mass-dependent luminosity of the <lower main sequence>, including the Sun approximately. Its fully radiative assumption is an idealization: the Sun has a convective envelope, and sufficiently low-mass <red dwarfs> become largely or fully convective, so real radii are not exactly constant.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 / 1 / i / Solution

Because $H\ll R$ and $M_{\rm env}\ll M$, 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> is
$$
\frac{dP}{dz}=-\rho g.
$$
Since $d\Sigma/dz=-\rho$, one has $dP/d\Sigma=g$. Taking the pressure to vanish with $\Sigma$ at the surface gives
$$
\boxed{P=g\Sigma}.
$$

The spherical stellar energy equation is $dL_r/dr=4\pi r^2\rho\epsilon$. With $r\simeq R$ and $F=L_r/(4\pi R^2)$,
$$
\frac{dF}{dr}=\rho\epsilon.
$$
Division by $d\Sigma/dr=-\rho$ gives
$$
\boxed{\frac{dF}{d\Sigma}=-\epsilon}.
$$
Likewise, <radiative diffusion in a star> gives
$$
\frac{dT}{dr}=-\frac{3\kappa\rho L_r}{16\pi ac r^2T^3}
\simeq-\frac{3\kappa\rho F}{4acT^3},
$$
and therefore
$$
\boxed{\frac{dT}{d\Sigma}=\frac{3\kappa F}{4acT^3}}.
$$

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 / 3 / i / Solution

In an optically thick stellar region, radiative diffusion carries luminosity according to
$$
\frac{dT}{dr}=-\frac{3\kappa\rho L_r}{16\pi ac r^2T^3}.
$$


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 Incoming links: Radiative diffusion in a star