 Solution

A fully convective, monatomic ideal gas is approximately isentropic through its deep interior. Its adiabatic index is $5/3$, so $P=K{\rho }^{5/3}$ and $T\propto K{\rho }^{2/3}$. The $n=3/2$ stellar polytrope scaling, or direct substitution of the central homology scalings into $K=P_c/{\rho }_c^{5/3}$, gives $K\propto R{M}^{1/3}$.

Near the photosphere, hydrostatic balance expressed in optical depth gives $P_{\mathrm{ph}}\sim g/{\kappa }_{\mathrm{ph}}$, with $g\propto M/R^2$. Treat $T_{\mathrm{ph}}$ as proportional to $T_{\mathrm{eff}}$ and match to the convective adiabat, neglecting a varying surface entropy jump in this homology approximation. Since $\rho \propto P/T$ and ${\kappa }_{\mathrm{ph}}\propto \rho T^{10}\propto P{T}^9$,The adiabat gives $P\propto K^{-3/2}{T}^{5/2}$, so the surface condition yieldsThe Stefan–Boltzmann law consequently gives $L_{\mathrm{surf}}\propto M^{4/7}{R}^{16/7}$. The same specific nuclear law as before still gives $L_{\mathrm{nuc}}\propto M^7{R}^{-8}$. Equating these scalings produces $R^{72/7}\propto M^{45/7}$. Thus the fully convective homology with tenth-power surface opacity isThen $T_{\mathrm{eff}}\propto M^{3/16}$, and the theoretical HR slope isIts steepness expresses the relatively weak change of effective temperature with mass. The graph using $-\log T_{\mathrm{eff}}$ has the negative of this slope.