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Along the pp-chain homologous sequence, $R$ is constant and therefore $T_c\propto M$. Normalizing to the Sun gives the transition mass

For the CNO cycle law $ϵ\propto \rho T^{23/2}$, the nuclear scaling becomesThe opacity law is unchanged, so $L_{\mathrm{rad}}\propto M^{11/2}{R}^{-1/2}$. Equating them givesand then

The effective temperature follows from the Stefan–Boltzmann law $L=4\pi R^2\sigma T_e^4$. Thus the pp branch has $T_e\propto M^{11/8}$ and $L\propto T_e^4$, whereas the CNO branch hasOn a Hertzsprung-Russell diagram, both branches rise toward larger luminosity and, conventionally, leftward toward larger temperature. They join near $4M_{\odot }/3$; the CNO branch has the steeper $L$--$T_e$ logarithmic slope and lies to the cooler side of the extrapolated constant-radius branch at fixed luminosity.

If $X$ is the hydrogen mass fraction and $X_{\mathrm{CNO}}$ the combined carbon-nitrogen-oxygen mass fraction, the reaction-pair and catalyst abundances give approximatelyThe CNO nuclei are catalysts, so their abundance multiplies the CNO-cycle rate rather than being consumed by the completed cycle.

For $I=A+C\mu$, the angular moments areandThus this angular form satisfies the Eddington closure approximation, $c{P}_r=4\pi J/3$, and

Part (i) gives $S=J=A$. Substitution in $\mu dI/d\tau =I-S$ yieldsMatching powers of $\mu$ gives $A^{\prime }=C$ and $C^{\prime }=0$, the latter being constant flux. Hence $A=C\tau +A_0$.

At the surface there is no incoming intensity. Applying this condition to the outgoing hemisphere in the moment approximation,so $A_0=F/(2\pi )=2C/3$. ConsequentlyNow $A=J=j/\kappa =\sigma T^4/\pi$ and the effective temperature is defined by $F=\sigma T_e^4$. It follows thatAt the top of the grey atmosphere,

For a homologous star, hydrostatic equilibrium, mass conservation, and the ideal gas equation give the central scalingswhere $\mu$ is the mean molecular weight. Integrating the nuclear energy-generation law over a fixed homologous profile givesRadiative stellar structure gives independentlyEquating the two luminosities yieldsAt fixed zero-age composition, the stars are therefore homologous with

The effective temperature satisfies $L=4\pi R^2\sigma T_e^4$, so $T_e\propto M^{3/8}$. The zero-age main sequence consequently hasIt is a steep line rising toward high luminosity and high temperature on a Hertzsprung-Russell diagram.

For fully ionized hydrogen and helium with $Y=1-X$,while $\kappa \propto 1+X$. At fixed mass,and the corresponding radius relation is

At the pure-hydrogen zero-age point $X=1$,whereasHenceandAs hydrogen is consumed, $X$ falls, so both $L$ and $T_e$ rise. In the usual diagram with temperature increasing leftward, the evolutionary track initially moves upward and leftward from the zero-age main sequence.