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At fixed initial zero-metal composition, equating parts (i) and (ii) gives $R\propto M^{3/4}$ and $L\propto M^3$. Therefore $T_{\mathrm{eff}}\propto (L/R^2{)}^{1/4}\propto M^{3/8}$, and the fully radiative zero-age line on the Hertzsprung-Russell diagram has slope

For the convective sequence, equating parts (i) and (iii) at fixed composition givesIts slope is consequently

For a fully ionized zero-metal hydrogen-helium mixture with $Y=1-X$, the mean molecular weight obeysFor a radiative star of fixed mass,At the pure-hydrogen point $X=1$,andHenceAs hydrogen burns, $X$ decreases, so both $L$ and $T_{\mathrm{eff}}$ increase. The track moves upward and toward higher temperature from the zero-age main sequence, with initial slope $d\log L/d\log T_{\mathrm{eff}}=256/53\simeq 4.83$.

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 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.

After substantial envelope loss on the Asymptotic giant branch, a low- or intermediate-mass star's exposed core heats while its radius shrinks, often at approximately constant luminosity during continued shell burning. It may ionize the expelled envelope as a planetary nebula. After the remaining nuclear fuel is depleted, its luminosity decreases and it joins a white dwarf cooling track. Evolutionary tracks on the Hertzsprung-Russell diagram connect the central-star temperature and luminosity with core mass and elapsed time.

With proton–proton chain energy generation $ϵ\propto \rho T^4$ and Kramers' opacity law $\kappa \propto \rho T^{-7/2}$, homology relations for radiative stars giveThe Hertzsprung-Russell diagram relation has $d\log L/d\log T_{\mathrm{eff}}=284/69$, with effective temperature increasing toward the left in the usual plotting convention.