= Solution
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 gives
$$
\boxed{P=K(t)T^{5/2}.}
$$
The 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 $\rho(r)=\rho_cf(r/R)$ 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 $P_c=C_PGM^2/R^4$. Hence
$$
\boxed{\rho_c\propto M/R^3,\qquad P_c\propto GM^2/R^4,\qquad
T_c=\frac{\mu P_c}{\mathcal R\rho_c}\propto\frac{\mu GM}{\mathcal R R}.}
$$
These are <homology scalings of central stellar pressure>. Substituting the central quantities into $K=P_c/T_c^{5/2}$ yields
$$
\boxed{K=K_0\mu^{-5/2}M^{-1/2}R^{-3/2},\qquad
K_0=C_\rho^{5/2}C_P^{-3/2}\mathcal R^{5/2}G^{-3/2}.}
$$
Thus $K_0$ is constant for the fixed <stellar polytrope> family.
At the approximate <photosphere>, take $T=T_e$ and use the <ideal gas> relation $\rho=\mu P/(\mathcal RT)$. The prescribed <opacity> gives
$$
\kappa P=\frac{\kappa_0\mu}{\mathcal R}P^2T^3
=\frac{\kappa_0\mu}{\mathcal R}K^2T^8.
$$
The <stellar surface boundary condition> and <surface gravity of a star> therefore imply
$$
\boxed{T_e^8=\frac{2G\mathcal R}{3\kappa_0K_0^2}\,\mu^4M^2R.}
$$
This is the temperature-four special case of <convective contraction with power-law opacity>.
Using the <Stefan–Boltzmann law>, the <luminosity> is
$$
L=4\pi\sigma\left(\frac{2G\mathcal R}{3\kappa_0}\right)^{1/2}
K_0^{-1}\mu^2M R^{5/2}.
$$
For $n=3/2$, the supplied gravitational-energy formula is $\Omega=-6GM^2/(7R)$. The <stellar virial theorem> for a monatomic gas gives $2U+\Omega=0$, so the total energy is $E=U+\Omega=-3GM^2/(7R)$. During <Kelvin-Helmholtz contraction>, with constant mass and no significant nuclear source, $L=-dE/dt$. Thus
$$
\dot R=-\frac{28\pi\sigma}{3}
\left(\frac{2\mathcal R}{3\kappa_0G}\right)^{1/2}
K_0^{-1}\mu^2M^{-1}R^{9/2}.
$$
Multiplying by $-(7/2)R^{-9/2}$ and integrating gives
$$
\boxed{R^{-7/2}-R_0^{-7/2}
=\frac{98\pi\sigma}{3}
\left(\frac{2\mathcal R}{3\kappa_0G}\right)^{1/2}
K_0^{-1}\mu^2M^{-1}(t-t_0).}
$$
The 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 \b[late-time central-temperature scaling]
$$
\boxed{T_c\propto\mu^{11/7}M^{5/7}(t-t_0)^{2/7}.}
$$
Replacing $t-t_0$ 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.
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