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The specific intensity of a blackbody is direction-independent over its outward hemisphere. Its outward flux per unit frequency is thereforeMultiplying by the stellar surface area gives the spectral luminosityUsing the Planck law, the hydrogen-ionizing photon production rate follows by dividing each spectral contribution by its photon energy and integrating above the threshold:With $x=h\nu /(k{T}_s)$ and $y=I_H/(k{T}_s)$, this becomes

For the initial hydrogen-only photon balance, assume an ionization-bounded nebula at steady photoionization equilibrium, negligible dust absorption and ionizing-photon leakage, negligible collisional ionization, and local reabsorption of photons from direct ground-state recombinations. A ground-state recombination then causes another photoionization and produces no net removal of a proton. Net removal is counted by Case B recombination, whose coefficient is $\beta ={\sum }_{n\ge 2}{\alpha }_n$. Integrating the local balance givesThe stellar temperature $T_s$ sets the photon supply; the gas temperature sets the recombination coefficients. They are not generally equal. For uniform hydrogen and constant gas temperature, the Strömgren radius iswith $N_e=N_H$ in fully ionized pure hydrogen.

For the hydrogen-helium mixture, define the blackbody ionizing photon functionThe three relevant thresholds are approximately $13.6$, $24.6$ and $54.4\mathrm{eV}$, for ${\mathrm{H}}^0$, ${\mathrm{He}}^0$ and ${\mathrm{He}}^{+}$ respectively. Higher stellar temperature increases the relative supply of the harder photons. These cumulative photon rates cannot be assigned independently to the three species: a photon above a helium threshold can instead ionize hydrogen, and helium recombination radiation can also ionize hydrogen. The absorption cross-sections and the diffuse radiation determine the effective allocation.

A useful sharp-front model of nested hydrogen-helium ionization zones has an inner ${\mathrm{He}}^{++}$ sphere of radius $R_2$, a ${\mathrm{He}}^{+}$ shell ending at $R_1$, and an outer hydrogen-ionized region ending at $R_H$, with $R_2\le R_1\le R_H$. Let $n_H,n_{He}$ be uniform nuclear number densities. The electron number densities in these three regions are $n_H+2n_{He}$, $n_H+n_{He}$ and $n_H$. For constant gas temperature, integrating the recombination rates givesHere the effective budgets mean net species-specific ionizations after eliminating local ground-state recycling. Their calculation must include photon competition and any interspecies diffuse contribution; the helium coefficients label the recombining ion's emitted spectrum. Once these budgets are specified, solve the third equation for $R_2^3$, the second for $R_1^3-R_2^3$, and the first for $R_H^3$.

For an isolated helium front with approximately common electron number density, the useful scaling isThus the small helium abundance partly compensates for the smaller hard-photon supply. A cooler central star produces little ${\mathrm{He}}^{++}$ and can have a substantially smaller helium-ionized zone. A sufficiently hot star can ionize helium throughout most of the hydrogen region. If independent radius estimates place a helium front beyond the hydrogen front, the independent photon allocations are inconsistent: the fronts and transfer must be treated together. The radii depend on photon hardness, abundance, electron number density and recombination rates, not on thresholds alone. The sharp-front equations are an approximation; they do not assert three exact independently conserved photon budgets for a real mixed nebula.

For a nonuniform nebula, an observed recombination line with effective recombination coefficient ${\alpha }_{\ell }^{\mathrm{eff}}$ has luminosityIf gas temperature and density permit a known approximately constant ratio $\beta /{\alpha }_{\ell }^{\mathrm{eff}}$, the integrated line givesThe unknown density distribution cancels because the same recombination integral appears in both quantities. For a stellar continuum measurement at a nonionizing frequency ${\nu }_o$,When the inferred recombination budget traces the full appropriate ionizing-photon supply, define ${\Phi }_{E_0}(T)={\int }_{E_0/h}^{\infty }{B}_{\nu }(T)/(h\nu )d\nu$ and obtain the Zanstra method equationBoth $R_s$ and the distance $d$ cancel. Hydrogen and helium recombination lines probe different hardness thresholds, so their ratios and the stellar continuum constrain $T_s$ and test the assumed spectrum and trapping. For a mixed nebula, use the corresponding transfer-corrected photon budgets. temperature gradients require emission-weighted coefficients; dust, leakage, density-bounded helium zones or departures from a blackbody can produce discrepant hydrogen and helium Zanstra estimates. Line fluxes alone without an appropriate budget or continuum normalization need not uniquely determine $T_s$.

Finally, locate the derived $T_s$ and luminosity on the Hertzsprung-Russell diagram. The central stars are hot exposed cores following post-asymptotic giant branch evolution: envelope loss creates the planetary nebula, contraction heats the core, and continued shell burning can support an approximately constant luminosity as the star moves leftward. Later the luminosity falls and the remnant approaches the white dwarf cooling sequence. Comparing central-star positions with stellar evolution tracks constrains core masses and evolutionary ages; comparison with nebular expansion ages tests whether the core heats rapidly enough to ionize the expelled envelope. The central-star locus traces the transition from an envelope-losing giant to a compact stellar remnant.

The first argument is the energy budget. The gravitational energy reservoir is of order $G{M}^2/R$, so the Kelvin-Helmholtz cooling time is $t_{\mathrm{KH}}\sim G{M}^2/(RL)$. For the Sun, $M\simeq 2\times 10^{30}\mathrm{kg}$, $R\simeq 7\times 10^8\mathrm{m}$ and $L\simeq 3.8\times 10^{26}\mathrm{W}$ give $t_{\mathrm{KH}}\sim 3\times 10^7$ years. This is much shorter than the approximately $4.6\times 10^9$-year solar-system age. Chemical binding energies are only of order electronvolts per atom and provide an even smaller reservoir. Thus neither ordinary chemical energy nor continuing gravitational contraction can supply the long-term solar luminosity.

Stellar nuclear fusion provides an adequate reservoir. Conversion of four hydrogen atoms into one helium-4 atom releases about $26.7\mathrm{MeV}$, corresponding to a mass defect of about $0.007$ of the initial mass; some energy leaves as neutrinos. If a fraction $f$ of stellar mass with hydrogen fraction $X$ is available for core burning, the energy scale is $0.007fXM{c}^2$. Taking $f\sim 0.1$ and $X\sim 0.7$ supplies roughly $9\times 10^{43}\mathrm{J}$, enough for several billion years at the solar luminosity, consistent with a stellar nuclear timescale of order $10^{10}$ years. The core temperature and density inferred from stellar structure permit the proton–proton chain; hotter stellar cores favor the CNO cycle.

The most direct evidence is the detection of solar neutrinos. They are created in the weak reactions of the proton–proton chain and can escape from the core, unlike the photons whose transport through the interior is slow. Measurements accounting for neutrino oscillation recover a total neutrino flux consistent with nuclear-powered solar models. Their spectra and reaction-dependent fluxes test the core reactions rather than merely the surface effective temperature.

There is also independent evidence from stellar evolution: the main sequence is a long-lived core-hydrogen-burning phase, more massive luminous stars consume fuel more quickly, and the main-sequence turnoff of a star cluster records its age. Subsequent evolution and the production of heavier elements fit further nuclear reactions. Agreement among these observations, nuclear energy release and stellar-model lifetimes is much stronger than the energy-budget argument alone. The combination of a sufficient nuclear reservoir, measured core neutrinos and the observed evolutionary sequence identifies nuclear reactions as the main long-term power source of ordinary stars.

After complete ionization, each hydrogen nucleus supplies one ion and one Electron, while each helium-4 nucleus supplies one ion and two Electrons. Neglecting the small metal mass fraction $Z$ and putting $Y\simeq 1-X$, the mean molecular weight therefore satisfiesUse stellar homology to compare models with common dimensionless profiles. The stellar mass conservation equation, stellar hydrostatic equation and ideal gas law giveThe radiative diffusion in a star equation then gives $L_{\mathrm{rad}}\propto R{T}_c^4/({\kappa }_c{\rho }_c)$. Inserting the specified opacity yields $L_{\mathrm{rad}}\propto R{T}_c^7/(Z{\rho }_c^2)\propto {\mu }^7{M}^5/Z$. Independently, integrating the stellar energy-generation rate over the mass gives $L_{\mathrm{nuc}}\propto X^2{\rho }_c{T}_c^{5}M\propto X^2{\mu }^5{M}^7/R^8$. Equating these in stellar thermal equilibrium gives the radiative homology with fifth-power hydrogen burning and inverse-cubic opacity:At fixed solar mass these reduce to $L\propto {\mu }^7/Z$ and $T_c\propto {\mu }^{5/4}/(X^{1/4}{Z}^{1/8})$. The stellar homology approximation, rather than merely the local power laws, is what permits these relations between whole stars.

Equal luminosities imply ${\mu }_2/{\mu }_1=(Z_2/Z_1{)}^{1/7}=2^{-1/7}$. Thusso $X_2=0.8$ to one significant figure. Retaining the unrounded hydrogen mass fraction when comparing central temperatures givesThe first model has the higher central temperature.

For homogeneous hydrogen burning at fixed $M,Z$, fuel conservation gives $M{E}_0\dot{X}=-L$. Since $d\mu /dX=-5{\mu }^2/4$ and $L=L_0(\mu /{\mu }_0{)}^7$, the homogeneous fuel-depletion luminosity feedback obeysIntegration with the initial mean molecular weight ${\mu }_0$ givesThis idealized stellar evolution law stops when $X=0$; it does not predict a physically attainable infinite luminosity. Indeed $\mu =4/3$ at fuel exhaustion, giving $t_H=M{E}_0[1-(3{\mu }_0/4{)}^8]/(10{\mu }_0{L}_0)$, strictly before the formal singularity.

A $5M_{\odot }$ star of roughly solar stellar composition begins on the zero-age main sequence with a hot convective core, a mostly radiative envelope, and hydrogen burning dominated by the CNO cycle. The strong temperature dependence of the CNO cycle concentrates the stellar energy-generation rate near the centre, helping produce the convective core. Core mixing replenishes hydrogen within that region; it does not homogenize the entire star. The nuclear fuel supply gives a stellar nuclear timescale of order $10^8$ years. For a numerical comparison, Table 2 of Ekström et al.'s solar-metallicity models gives core hydrogen burning lifetimes of about $88$ and $109$ million years for nonrotating and rotating $5M_{\odot }$ models, respectively.

After central hydrogen exhaustion, an inert helium core contracts and heats through Kelvin-Helmholtz contraction, while a hydrogen-burning shell supplies the growing luminosity. The envelope expands and cools: the track crosses the Hertzsprung gap toward the red giant region. Its deepening convective envelope causes first dredge-up, bringing material altered by hydrogen burning to the surface. The rapid structural crossing has a thermal timescale, roughly $10^5$–$10^6$ years for a star of this mass, rather than the full stellar nuclear timescale; an estimate follows from the Kelvin-Helmholtz cooling time $G{M}^2/(RL)$ with the evolving radius and luminosity.

The helium core reaches ignition before becoming strongly supported by electron degeneracy pressure. Thus a normal $5M_{\odot }$ star starts core helium burning without the helium flash characteristic of lower-mass stars. The Triple-alpha process produces carbon-12, and carbon-12 alpha capture produces oxygen-16; these reactions build a carbon-oxygen core. There is a convective core and continued hydrogen-burning shell activity. Core helium burning lasts order $10^7$ years: the same model table gives approximately $19$ and $18$ million years. The changing core and envelope structure can produce a blue loop in the Hertzsprung-Russell diagram; its extent depends on stellar composition, mixing and stellar rotation.

After central helium exhaustion, the carbon-oxygen core contracts and becomes increasingly supported by electron degeneracy pressure. A helium-burning shell forms above it, beneath the hydrogen-burning shell. The star returns to a cool, luminous Asymptotic giant branch configuration. On this early branch, helium-burning shell activity is an important energy source; the overlying hydrogen-burning shell can weaken during readjustment and resume later. At this mass, a deepening convective envelope can cause second dredge-up. This shell-supported evolution to the first AGB thermal pulse takes order $10^6$ years, with model-dependent duration. As the helium-burning shell becomes sufficiently thin, the thin-shell instability ends steady shell burning and marks the onset of AGB thermal pulses.

The original sketch shows the direction of stellar evolution, with higher effective temperature to the left. It illustrates a possible blue loop; its coordinates and loop shape are schematic, rather than a computed stellar model. The stages before the first AGB thermal pulse are the requested track; the later remnant evolution is described below.

In the usual final evolution, stellar winds and an AGB superwind remove the envelope before the carbon-oxygen core can ignite substantial carbon burning. Once shell fuel is exhausted, the exposed core contracts and heats, potentially ionizing the expelled gas as a planetary nebula, and finally becomes a carbon-oxygen white dwarf that loses its stored thermal energy by cooling. The relevant competition is between the AGB envelope-loss and core-growth timescales, not just the total initial mass.

Much higher late-stage mass loss strips the envelope sooner, shortens the Asymptotic giant branch phase, and allows less core growth, usually producing a smaller carbon-oxygen white dwarf. Much lower mass loss prolongs shell burning and permits greater core growth. In an extreme hypothetical case, retention of the envelope could let a degenerate carbon-oxygen core approach the Chandrasekhar mass and undergo explosive carbon burning: the proposed type 1.5 supernova channel. This is not an inevitable fate for a $5M_{\odot }$ star. Doherty et al.'s evolutionary calculations emphasize the uncertainty of that channel and find that their adopted mass loss removes carbon-oxygen envelopes before such ignition. If carbon burning instead proceeds without disruption, the possible further evolution depends on the resulting core and continued envelope retention; lowering mass loss alone does not establish an electron-capture supernova outcome.

Treat the stars as point masses for the orbital motion, neglecting their spin angular momenta. Let $M=M_1+M_2$. Their distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. The sum of their orbital angular momenta isThus the circular-binary orbital angular momentum is

For conservative binary mass transfer, both $M$ and $J$ are fixed. In terms of the binary mass ratio $q=M_1/M_2$, $M_1=Mq/(1+q)$ and $M_2=M/(1+q)$, henceMass conservation also gives $d{M}_2=-d{M}_1$, soDifferentiating $J\propto M_1{M}_2{a}^{1/2}$ at fixed $M,J$ gives $d\log a/d\log M_1=2(q-1)$. Therefore the stipulated Roche lobe law has the Roche-lobe radius response exponent

Define the overfill by $\Delta =\log (R_1/R_L)$. On a time scale short compared with the slow change of $K$, the donor's response is $d\log R_1=\beta d\log M_1$. ThusDuring mass loss $d\log M_1<0$. If ${\zeta }_L>\beta$, mass loss increases the overfill and therefore drives more mass loss: this is positive feedback, giving more rapid Roche-lobe overflow. If ${\zeta }_L<\beta$, mass loss decreases the overfill and can balance the slow expansion due to stellar evolution. More explicitly, with $k_{\mathrm{ev}}=d\log K/dt>0$, contact requiresA slow negative transfer rate can satisfy this only on the stable side $\beta >{\zeta }_L$. The rapid-transfer condition is thereforeIts time scale depends on the physical donor response represented by $\beta$; it need not be a dynamical time scale if $\beta$ is a thermal-equilibrium mass-radius exponent.

For Roche-lobe overflow feedback with a constant wind torque factor, retain mass conservation to the stated approximation but allow the prescribed angular-momentum loss. Since $d\log J=fd\log M_1$, differentiation givesThe new Roche-lobe radius response exponent isThe same overfill calculation givesA loss of angular momentum has $f>0$ because ${\dot{M}}_1<0$, and it lowers the critical ratio by making the orbit contract more strongly.

For the specified solar-mass donor exponent $\beta =1/2$, the answer isWithout the wind torque this becomes $q>22/23\simeq 0.957$, so donors slightly less massive than their companions can already lie on the rapid-transfer side of this model. For $0<f<11/10$ the lower threshold is $(22-20f)/23$, and for $f\ge 11/10$ every positive $q$ satisfies the strict inequality. There is no finite upper bound on $q$ from this criterion. Since the paper does not specify $f$, the wind case has a conditional range rather than a unique numerical cutoff.

Write $n_{56}$ and $n_{54}$ for the number densities of nickel-56 and iron-54. Both nuclei contain 28 neutrons, while their proton numbers differ by two. Dividing the positive-exponential nuclear statistical equilibrium formulas printed in the PDF therefore gives the nickel–iron nuclear equilibrium ratioThe neutron-density factors cancel. $K$ contains the isotope-dependent statistical and translational mass factors suppressed in the supplied proportionality; it is constant in this simplified calculation. The TeX conversion incorrectly reverses the exponential sign, so the PDF sign is essential here.

In the specified restricted composition, equality of total proton and neutron numbers meansConsequentlyDifferentiation, with the suppressed prefactors held constant, givesUsing the paper's supplied signed energy difference producesWith that negative value, $F$ tends to zero at both temperature endpoints, increases below this temperature and decreases above it. At the stationary point $d^2\log F/d{T}^2=-3/T^2<0$, so it is indeed a maximum for the printed toy data.

There is a genuine physical-data qualification: the stated negative energy difference is inconsistent with interpreting $Q$ as the conventional positive total nuclear binding energy of these actual isotopes. Using atomic masses, whose electron masses cancel in this reaction,The masses used are $1.00782503223\mathrm{u}$, $53.93960899\mathrm{u}$ and $55.94212855\mathrm{u}$, from NIST hydrogen masses, NIST iron masses and NIST nickel masses. Small electronic binding corrections do not change the sign. With a positive $\Delta B$, the simplified $F(T)$ above is strictly decreasing for $T>0$ and has no finite maximum. Thus the numerical maximum follows from the supplied negative parameter, not from consistent physical binding energies; changing the TeX sign to recover a maximum would not repair the PDF's data.

Nuclear statistical equilibrium is appropriate when the relevant forward and inverse strong or electromagnetic reactions are fast compared with changes of temperature, density and composition. It is useful in sufficiently hot late silicon-burning material and in the hottest explosive burning regions of massive stars. Some burning zones reach only quasi-equilibrium, and cooling or expansion eventually freezes the abundances out; an equilibrium expression cannot by itself predict the full ashes when these timescale conditions fail. Temperature, density, conserved baryon number and the electron fraction in stellar matter must all be specified. Negligible free-particle contributions in bulk bookkeeping are not the same as exactly zero free densities in the equilibrium formula.

Strong and electromagnetic reactions preserve total charge and baryon number, so they conserve the proton-to-neutron ratio. Weak reactions need not: electron capture and beta-plus decay convert protons into neutrons and lower it. Such conversions occur in hydrogen-burning networks and in the nitrogen-to-neon chain during helium burning. At high density in advanced stellar cores and core collapse, electron capture becomes particularly important; neutrino emission also affects energy and lepton loss. Beta-minus decay changes the ratio in the opposite direction and can compete during later relaxation. Nuclear equilibrium for strong reactions does not automatically imply weak equilibrium or a fixed proton-to-neutron ratio throughout all stellar evolution.

Let $T=10\mathrm{Gyr}$ and let $t$ denote present stellar age, rather than time measured from the galaxy's birth. With a fixed initial mass function, constant star formation rate also gives a constant number of births per unit time, as in the birth-age distribution under constant star formation. Counting the supplied stellar evolution model's remnants as part of the population, ages have a uniform distribution on $[0,T]$. Integrating the constant birth rate therefore givesOutside this interval the cumulative fraction is zero or one as appropriate.

Write $m=M/M_{\odot }$ for the birth-mass parameter and $M_{\min }=0.2M_{\odot }$. The normalization cancels when integrating the initial mass function:Thus the mass-tail fraction isIt is one at and below the lower cutoff. The normalization constant has no effect on any number fraction.

For evolutionary-state selection by stellar lifetimes, a red giant has completed its main sequence lifetime but not its giant lifetime. Its stellar lifetime regions in a mass-age diagram are bounded byThe lower boundary meets $t=T$ at $m=\sqrt{0.8}$ and the upper boundary meets it at $m=1$. A white dwarf lies above the upper lifetime curve, within the same age interval.

The probability integral transform, applied to the decreasing mass-tail coordinate, makes $X$ uniform on $(0,1)$: $\mathrm{Pr}(X<x)=x$. The independent birth age gives $Y=t/T$ uniform on $(0,1)$, so number fractions are areas in a unit square. Since $m^{-2}=25X$, the red giant region becomes $20X<Y<25X$, a triangle with vertices $(0,0)$, $(1/25,1)$ and $(1/20,1)$. Its area is $\frac{1}{2}(1/20-1/25)=1/200$. The white dwarf region $Y>25X$ is a triangle of area $\frac{1}{2}(1/25)=1/50$. Hence the present individual-star fractions areThe rest are on the main sequence under the given toy lifetime model.

For the binary stars, draw two independent mass-tail coordinates $X_1,X_2$, but only one shared age coordinate $Y$: components born together are coeval. The coeval binary population therefore occupies a uniform unit cube $(X_1,X_2,Y)$. At a fixed age $Y=y$, the state probabilities for either component areTheir evolutionary states have conditional independence given $y$. They generally do not have unconditional independence, because sharing an age correlates their states. The binary fractions below are slice areas averaged over $0<y<1$.

For this stellar population, let $m=M/M_{\odot }$ denote initial mass in units of the solar mass, and let $\tau =t/\mathrm{Gyr}$ denote present stellar age. Constant formation of equal numbers of stars per unit time makes the stellar age a uniform distribution on $[0,10]$ Gyr, so $Y=\tau /10$ is uniform on $[0,1]$. The normalized initial mass function has probability density function $f(m)=0.1m^{-2}$ for $m>0.1$, since ${\int }_{0.1}^{\infty }{m}^{-2}dm=10$. ConsequentlyThese formulae have the stated age and mass domains; outside them the relevant cumulative fractions saturate at zero or one. The probability integral transform makes $X$ uniform on $[0,1]$: $m=0.1/X$ gives $f(m)|dm/dX|=1$. The time-independent initial mass function and constant number formation rate give independence of $X$ and $Y$. All fractions here count objects, including white dwarfs, using the stipulated stellar evolution law.

A red giant hasThus the mass boundaries are $m=9/\tau$ and $m=10/\tau$ for positive $\tau$. Equivalently, for $0.9<m<1$ the red giant region runs from $9/m$ to the age cap $10$, and for $m\ge 1$ it runs from $9/m$ to $10/m$. There are no red giants with $m\le 0.9$. Boundaries have zero probability and their endpoint convention does not affect the fractions. In the uniform $(X,Y)$ square the red giant region is $9X<Y<10X$, with triangle vertices $(0,0)$, $(1/10,1)$ and $(1/9,1)$. The white dwarf region is the triangle $0<X<Y/10$.

At fixed $Y=y$, define the conditional probabilities of a red giant, white dwarf and main sequence star by $g(y)$, $w(y)$ and $s(y)$. Their interval widths areIntegration over the uniform distribution of age gives the individual-star fractionsThe systems form a coeval binary population: the two binary star components have the same age. Their masses are independent, so $(X_1,X_2,Y)$ has uniform probability density function on $[0,1{]}^3$ and their states have conditional independence given $Y$. Unconditional independence of their states would be incorrect: older systems make both evolved states more likely. The law of total probability now givesThe subtraction removes the double counting of systems containing two red giants.

Assume an isolated star of initial mass five solar masses of approximately solar stellar metallicity. Rotation, convective overshooting and mass loss change numerical ages and the extent of a blue loop, so the ages below are estimates. Take zero age at the zero-age main sequence; the pre-main-sequence star phase adds a much shorter contraction time.

The original schematic Hertzsprung-Russell diagram marks the stages discussed below. The strip denotes the approximate Cepheid instability strip; the track is illustrative, rather than a computed stellar model.

The main sequence lasts roughly $8\times 10^7$–$10^8$ years. CNO cycle hydrogen burning is concentrated in a mixed convective core. Its declining hydrogen mass fraction and rising helium mass fraction are nearly uniform there, while the retreating core leaves a composition gradient. The unprocessed envelope remains hydrogen rich. At the terminal-age main sequence, core hydrogen is exhausted and a hydrogen-burning shell takes over.

The inert helium core grows through shell burning. The Schönberg-Chandrasekhar limit is approximatelyBeyond this limit an isothermal nondegenerate core cannot remain in thermal equilibrium with its envelope. Core contraction and envelope expansion carry the star across the Hertzsprung gap. An initial slower shell-burning interval can precede the rapid crossing; the crossing itself is governed by Kelvin-Helmholtz contraction, with a representative $G{M}^2/(RL)\sim 10^5$–$10^6$ years. First dredge-up then mixes hydrogen-processed material into the convective envelope, lowering its hydrogen mass fraction and increasing helium and nitrogen, while leaving a composition discontinuity where the envelope later retreats.

At an age still of order $10^8$ years, core helium burning begins quietly: the core is nondegenerate, so there is no helium flash. The Triple-alpha process and subsequent alpha capture build a carbon-oxygen core. Core helium burning lasts roughly $1.6\times 10^7$–$2.2\times 10^7$ years in representative solar-composition models. Pols's stellar-evolution notes illustrate the substantial dependence of these lifetimes on convective overshooting.

During a blue loop the envelope contracts and effective temperature increases, before the star returns redward. The loop depends on core and envelope structure and the hydrogen discontinuity left by first dredge-up. A loop reaching the Cepheid instability strip produces two further crossings, blueward and redward, in addition to the earlier rapid crossing of the Hertzsprung gap. Cepheid variables pulsate through the opacity mechanism, involving helium ionization. Neither loop extent nor all three strip crossings are guaranteed for every composition or mixing prescription. The hydrogen-burning shell continues moving outward in enclosed mass while central helium is depleted. Lattanzio's five-solar-mass tutorial shows these composition changes.

After helium exhaustion, at an age roughly $1.0\times 10^8$–$1.2\times 10^8$ years, the star ascends the Asymptotic giant branch. An inert carbon-oxygen core is surrounded by helium-rich material and a hydrogen-rich envelope. Second dredge-up mixes helium and hydrogen burning products into the envelope and reduces the hydrogen-exhausted core mass. Subsequently a helium-burning shell and hydrogen-burning shell alternate in importance. A thermal pulse of an asymptotic-giant-branch star causes intershell convection, expansion and temporary quenching of hydrogen burning; third dredge-up can carry carbon and slow neutron-capture process products outward. Between pulses hydrogen burning rebuilds the helium layer. These mechanisms are illustrated in Lattanzio's AGB tutorial.

The following original stellar composition profile sketches distinguish hydrogen exhaustion from helium exhaustion; their mass boundaries are illustrative.

The final giant phases add at most a few million years at this level of accuracy. Strong stellar winds remove the envelope; the hot remnant can illuminate a planetary nebula and then cool as a carbon-oxygen white dwarf. A typical remnant is of order one solar mass. The ordinary isolated $5M_{\odot }$ case does not reach sustained carbon burning and core collapse. Thus the usual endpoint is a carbon-oxygen white dwarf, at a total age of order $10^8$ years, with its subsequent cooling age added separately.

A stellar population is an ensemble of stars described by its formation times, initial masses, composition and present stellar evolution states. Number fractions must be distinguished from mass-weighted fractions. Shared formation times can correlate the states of components of a binary star even when their initial masses have independence.