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Match an $n=3/2$ stellar polytrope to the grey-atmosphere convection onset with seventeenth-power opacity. At each instant the convective entropy coefficient obeys $P=K{T}^{5/2}$, $K\propto M^{-1/2}{R}^{-3/2}$ at fixed composition, while the boundary gives $K\propto g^{1/2}{T}_e^{-21/2}$. Thus $T_e^{21/2}\propto M{R}^{1/2}$ and the first displayed luminosity law follows. The stellar virial theorem for a monatomic gas gives total energy $-3G{M}^2/(7R)$, so Kelvin-Helmholtz contraction has $\dot{R}=-7L{R}^2/(3G{M}^2)$. Integrating at fixed mass gives $R^{-67/21}-R_0^{-67/21}\propto t$. The second displayed relation holds after the initial-radius term is negligible. The entropy coefficient is spatially constant but changes during contraction.

Let $\Omega =-{\int }_0^{M}Gmdm/r$ be the gravitational energy. Multiply the stellar hydrostatic equation by $4\pi r^3$ and integrate through the star. Integration by parts and zero surface pressure givewhile its hydrostatic right-hand side is $-{\int }_0^{R}4\pi Gmr\rho dr=\Omega$. Hence the stellar virial theorem is

There is a genuine reference-sheet error to resolve. A monatomic ideal gas with adiabatic exponent $5/3$ has specific internal energy of an ideal gas $u=P/[\rho (5/3-1)]=3P/(2\rho )$, not the printed $3P/\rho$ on PDF page 2. The latter would give total internal energy $-\Omega$ and zero total energy, inconsistent with radiating contraction. The following derivation uses the consistent monatomic value.

A fully convective star is approximately isentropic and has an adiabatic stellar polytrope with index $3/2$. The given polytropic gravitational-energy formula yields $\Omega =-6G{M}^2/(7R)$. The virial relation gives total internal energy $U=-\Omega /2$, so total energy is $U+\Omega =-3G{M}^2/(7R)$. At constant mass with no nuclear heating, Kelvin-Helmholtz contraction obeys $d(U+\Omega )/ds=-L$, whenceHere $s$ denotes physical time.

In the plane-parallel atmosphere, the inward optical depth satisfies $d\tau =-\kappa \rho dr$, so hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. Neglecting radiation pressure, $\rho =\mu P/(\mathcal{R}T)$ and the specified opacity becomes $\kappa =({\kappa }_0\mu /\mathcal{R})P{T}^{16}$. ThereforeThe grey atmosphere relation and the boundary $P(0)=0$ integrate toTo apply the Schwarzschild criterion, put $w=2+3\tau$. Logarithmic differentiation givesThis increases inward, starting from zero. Marginal convection has ${\nabla }_{\mathrm{rad}}=2/5$, so $w_b^3=136/5$. Since $T_b^{12}/T_e^{12}=w_b^3/64$, the grey-atmosphere convection onset with seventeenth-power opacity isThe purely radiative atmosphere is used only up to this boundary.

The convective relation $P=K{T}^{5/2}$ is spatially uniform in its entropy coefficient $K$ at each instant; $K$ can change as the star radiates and contracts. At the convection boundary $w_b$ is a fixed number, so $P_b\propto g^{1/2}{T}_e^{-8}$ at fixed composition and opacity normalization. ThusOn the other hand, the central scalings for the $n=3/2$ stellar polytrope give $P_c\propto G{M}^2/R^4$ and $T_c=\mu P_c/(\mathcal{R}{\rho }_c)\propto \mu GM/(\mathcal{R}R)$. Hence, at fixed composition,Equating the interior and atmosphere expressions yields $T_e^{21/2}\propto M{R}^{1/2}$. Applying the Stefan–Boltzmann law then proves the Hayashi contraction with seventeenth-power opacity relation

At fixed mass write $L=A{M}^{8/21}{R}^{\alpha }$, with constant $A$ and $\alpha =46/21$. The contraction equation is $\dot{R}=-C{R}^{\alpha +2}$, where $C=7A{M}^{8/21}/(3G{M}^2)$. Direct integration givesMultiplying through by $R^{\alpha +1}$ and substituting the current luminosity givesFor the stated very large initial radius, or after that initial-radius term becomes negligible,This is the luminosity-age relation within the contraction model, rather than an assertion that a physical star begins at literally infinite radius.