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For a fixed-mass, fully convective monatomic gas protostar, write $P=K{T}^{5/2}$ and $K=K_0{\mu }^{-5/2}{M}^{-1/2}{R}^{-3/2}$. Matching the photosphere to $\kappa P=2g/3$ with $\kappa ={\kappa }_0\rho T^4$ gives $T_e^8=2G\mathcal{R}{\mu }^4{M}^{2}R/(3{\kappa }_0{K}_0^2)$. The Stefan–Boltzmann law gives $L=\pi ac[2G\mathcal{R}/(3{\kappa }_0){]}^{1/2}{K}_0^{-1}{\mu }^{2}M{R}^{5/2}$. The stellar virial theorem gives total energy $-3G{M}^2/(7R)$, so $L=-3G{M}^2\dot{R}/(7R^2)$. Integrating proves $C=(49\pi ac/6)[2\mathcal{R}/(3{\kappa }_{0}G){]}^{1/2}{K}_0^{-1}$. This holds before stellar nuclear fusion, changing opacity or electron degeneracy pressure invalidates the assumed contraction law.

For a monatomic fully convective ideal gas star, uniform specific entropy gives $P=K{T}^{5/2}$ with $K\propto {\mu }^{-5/2}{M}^{-1/2}{R}^{-3/2}$. Combining this with the stellar surface boundary condition $\kappa P=2g/3$ and $\kappa ={\kappa }_0{\rho }^a{T}^b$ givesAll composition comparisons here hold ${\kappa }_0$ fixed. For $a=1,b=4$, $T_e^8\propto {\mu }^4{M}^{2}R$, and the Stefan–Boltzmann law gives $L\propto {\mu }^{2}M{R}^{5/2}$. The stellar virial theorem and Kelvin-Helmholtz contraction then give $R^{-7/2}-R_0^{-7/2}\propto {\mu }^2{M}^{-1}(t-t_0)$, with a positive coefficient. Once the initial-radius term is negligible, $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$.

For a spherical stellar polytrope of finite radius and zero surface pressure, the specific enthalpy is $(n+1)P/\rho$. Hydrostatic equilibrium gives $h+\Phi =-GM/R$, hence $(n+1)\int PdV=-G{M}^2/R-2\Omega$. Eliminating the pressure integral with the stellar virial theorem proves the displayed formula. For a gas adiabatic stellar polytrope, $n=1/(\gamma -1)$ and $U=G{M}^2/[(5\gamma -6)R]$; a general polytropic index need not have this relation to the specific-heat ratio.

This term accounts for thermal storage and compression or expansion in the stellar energy balance equation. At fixed composition the first law of thermodynamics makes it $-TDS/Dt$. Kelvin-Helmholtz contraction can supply positive luminosity, while thermal adjustment may locally absorb energy. It is not simply the full change in gravitational potential energy: the stellar virial theorem also requires changes in internal energy.

Match an $n=3/2$ stellar polytrope to the grey-atmosphere convection onset with seventeenth-power opacity. At each instant the convective entropy coefficient obeys $P=K{T}^{5/2}$, $K\propto M^{-1/2}{R}^{-3/2}$ at fixed composition, while the boundary gives $K\propto g^{1/2}{T}_e^{-21/2}$. Thus $T_e^{21/2}\propto M{R}^{1/2}$ and the first displayed luminosity law follows. The stellar virial theorem for a monatomic gas gives total energy $-3G{M}^2/(7R)$, so Kelvin-Helmholtz contraction has $\dot{R}=-7L{R}^2/(3G{M}^2)$. Integrating at fixed mass gives $R^{-67/21}-R_0^{-67/21}\propto t$. The second displayed relation holds after the initial-radius term is negligible. The entropy coefficient is spatially constant but changes during contraction.

The intrinsic planetary luminosity $L_{\mathrm{int}}$ is the power supplied by interior cooling, contraction, nuclear reactions if present, or internal heating, excluding reflected or passively reradiated stellar light. For fixed mass and no accretion or nuclear source, global conservation of energy gives $L_{\mathrm{int}}=Q_{\mathrm{deep}}-d(U+\Omega )/dt$, where $Q_{\mathrm{deep}}$ is any additional power deposited in the interior, such as tidal heating or Joule heating. For $Q_{\mathrm{deep}}=0$ and a gas-supported quasi-static sphere of fixed specific-heat ratio $\gamma$ and negligible surface pressure, the stellar virial theorem gives $\Omega =-3(\gamma -1)U$, henceFor a dynamically stable gas-supported model take $\gamma >4/3$. For $\gamma =5/3$ this is half the gravitational release. Degenerate giants require a different internal energy reservoir; the half-factor is not universal.

For a homologous stellar polytrope of index $3/2$, take $P=K{\rho }^{5/3}+\mathcal{R}\rho T/\mu$ and zero surface pressure. The stellar virial theorem, with $\Omega =-6G{M}^2/(7R)$, givesThe normalized density profile fixes $A$ independently of mass and radius at fixed composition. Writing ${\overline{T}}_M=\alpha /R-\delta /R^2$ shows that its maximum is ${\alpha }^2/(4\delta )=3\mu G^2{M}^{4/3}/(49\mathcal{R}A)$, at $R=7A{M}^{-1/3}/(3G)$. Below this radius contraction increases the fraction of support provided by electron degeneracy pressure and lowers the thermal temperature. A fixed normalized profile also fixes the ratio of central to mass-averaged temperature, giving a hydrogen-burning minimum mass.

Efficient convection keeps the interior at nearly uniform specific entropy. For the fully ionized monatomic gas, the heat capacity ratio is $5/3$, so the adiabatic process gives $P=K_{\rho }{\rho }^{5/3}$, with $K_{\rho }$ spatially constant. The ideal gas relation $P=\mathcal{R}\rho T/\mu$ then impliesThe constant is spatial, not temporal: loss of specific entropy during contraction changes it. This is a stellar polytrope with polytropic index $n=3/2$. We neglect radiation pressure and take the thin photosphere as an atmospheric matching layer rather than a significant fraction of the stellar mass.

For a fixed dimensionless profile of the Lane-Emden equation, the stellar homology relations have the formwhere $C_{\rho },C_P$ are positive dimensionless structure constants. More explicitly, write $\rho (r)={\rho }_{c}f(x)$ and $m(r)=Mh(x)$ with $x=r/R$ and fixed dimensionless profiles. Mass conservation gives $M=4\pi {\rho }_c{R}^3{\int }_0^{1}f(x)x^{2}dx$. Integrating the hydrostatic pressure support equation, with negligible surface pressure on the interior scale, gives $P_c=(GM{\rho }_c/R){\int }_0^{1}h(x)f(x)x^{-2}dx$. Both integrals are fixed positive numbers, proving the two central scalings. Substituting these relations into $K=P_c/T_c^{5/2}$ givesThus the same polytropic index fixes $K_0$ throughout this homologous sequence, while the dimensional stellar polytrope constant changes.

At the photosphere, whose optical depth is $2/3$, identify the temperature with the effective temperature $T_e$. The ideal gas relation and the interior adiabat give ${\rho }_{\mathrm{ph}}=\mu K{T}_e^{3/2}/\mathcal{R}$. With the prescribed opacity and stellar surface boundary condition,Eliminating $K$ therefore provesUsing the Stefan–Boltzmann law, with $\sigma =ac/4$, gives the luminosity

The gravitational energy of a stellar polytrope is $\Omega =-6G{M}^2/(7R)$ for $n=3/2$. The stellar virial theorem for a monatomic gas gives $2U+\Omega =0$, and hence the total energy is $E=U+\Omega =-3G{M}^2/(7R)$. With no appreciable stellar nuclear fusion or accretion, energy conservation givesCombining the two expressions for $L$ yieldsMultiplication by $-7R^{-9/2}/2$ and integration proves the contraction law with density-linear fourth-power stellar opacity:When the accumulated contraction term dominates the initial-radius term, $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$. Substitution into the central ideal gas relation giveswhich has the stated $t^{2/7}$ behavior when the time origin is negligible. This late-time approximation still requires the assumed fully convective star, fixed composition and opacity law to remain applicable.

A stellar polytrope replaces a detailed equation of state and thermal structure by $P=K{\rho }^{1+1/n}$ for $n>0$, with spatially constant $K$ and polytropic index $n$. Together with spherical hydrostatic equilibrium and mass conservation, it gives a tractable family of stellar density profiles. Introduce the Lane-Emden variables for a stellar polytropeCombining $dP/dr=-Gm\rho /r^2$ with $dm/dr=4\pi r^2\rho$ eliminates $m$ to give $r^{-2}d[r^2{\rho }^{-1}(dP/dr)]/dr=-4\pi G\rho$. Substitution of the pressure law and the dimensionless variables gives the Lane-Emden equationA regular finite-radius model for $0\le n<5$ ends at the first zero ${\xi }_1$. Its radius and mass areThese formulas supply central-to-mean density ratios and stellar homology scalings without solving a full evolutionary model. Eliminating ${\rho }_c$, for $0<n<5$ and $n\ne 3$, gives the polytropic mass-radius relation

An efficient-convection monatomic gas has $n=3/2$. The same index describes a cold, nonrelativistic electron degeneracy pressure law; at fixed composition it gives $R\propto M^{-1/3}$ for a white dwarf. Ultrarelativistic electron degeneracy pressure has $n=3$, for which the mass is independent of ${\rho }_c$ and scales as $(K/G{)}^{3/2}$. This is the origin of the Chandrasekhar limit. A radiation-dominated, approximately constant-entropy interior also approaches $P\propto {\rho }^{4/3}$ and $n=3$. The polytrope of index zero is an incompressible comparison model, useful for illustrating the role of central concentration.

The gravitational energy of a stellar polytrope, $\Omega =-3G{M}^2/[(5-n)R]$, and the stellar virial theorem make polytropes useful for contraction-energy estimates. They also provide structural models for perturbation and stability calculations. A polytrope is a structural approximation, not a complete stellar evolution calculation. One must still specify stellar nuclear fusion, opacity, energy transport, composition and boundary conditions to determine the luminosity and history. Moreover, the structural exponent $1+1/n$ need not equal the adiabatic exponent relevant to a rapid perturbation. Real stars can have different effective indices in their cores and envelopes, as well as ionization zones or varying electron degeneracy pressure.

For a regular, spherical star, hydrostatic equilibrium gives $dP/dr=-Gm(r)\rho (r)/r^2$. Multiply by $4\pi r^3$ and integrate to the surface. Integration by parts givesThe central boundary term vanishes for finite central pressure. Since $dm=4\pi r^2\rho dr$, the right-hand side of the integrated hydrostatic equation isThe last expression is the gravitational energy: adding a thin mass shell $dm$ to the interior mass $m$ contributes $-Gmdm/r$, counting each interacting pair once. HenceWith the specified zero surface pressure this proves the stellar virial theorem in the required form,

To derive the nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture, take fully ionized hydrogen and helium with hydrogen mass fraction $X$, the Electron number density isHere helium nuclei have mass $4m_p$ in the approximation used in the question. Filling the two spin states of each momentum mode up to the Fermi momentum $p_0$ givesThe isotropic momentum flux gives pressure $\frac{1}{3}\int pvdn$. For nonrelativistic Electrons $v=p/m_e$, so the electron degeneracy pressure isSubstitute the composition-dependent $n_e$ and simplify powers of two:Comparing $5/3=1+1/n$ gives the polytropic index $n=3/2$. The result assumes complete Electron degeneracy, $p_0\ll m_{e}c$, and negligible nuclear mass corrections.

Now adopt the stated approximate partially degenerate equation of state, keeping composition and hence $K,\mu$ fixed during contraction. Set $q=m/M$ and define the mass-averaged temperature ${\overline{T}}_M={\int }_0^{1}Tdq$. The stellar virial theorem from part (a) givesFor an index-$3/2$ stellar polytrope, the permitted gravitational-energy formula is $\Omega =-6G{M}^2/(7R)$. Also, using $\bar{\rho }=3M/(4\pi R^3)$,Thus, with $A=3K{\int }_0^1[3\rho /(4\pi \bar{\rho }){]}^{2/3}dq$,The normalized density $\rho /\bar{\rho }$ is a fixed function of $q$ for this homologous polytropic structure. Therefore its integral and $K$ fix $A$, which is independent of $M$ and $R$ for a common composition. The mean here is mass-weighted, rather than a volume average.

For the mass-averaged temperature maximum in a partially degenerate polytrope, put $\alpha =2\mu GM/(7\mathcal{R})$ and $\delta =\mu A{M}^{2/3}/(3\mathcal{R})$, both positive. As a function of $s=1/R$ the temperature is the concave quadraticIt reaches its maximum when $R=2\delta /\alpha =7A{M}^{-1/3}/(3G)$, withAt large radius, contraction increases the temperature as in a thermally supported star. Once Electron degeneracy provides a sufficiently large fraction of the support, further contraction can lower the thermal temperature. The formal zero-temperature radius is $R_{\mathrm{deg}}=\delta /\alpha =R_{\max }/2$, with $R_{\mathrm{deg}}\propto M^{-1/3}$. The continuation to negative temperatures would be unphysical; the adopted sequence only describes its positive-temperature part.

The contraction-limited central temperature has the same mass scaling. Indeed, if the total polytropic pressure is $P=K_{\mathrm{tot}}{\rho }^{5/3}$, the approximate equation of state gives $T=(\mu /\mathcal{R})(K_{\mathrm{tot}}-K){\rho }^{2/3}$. The fixed density profile therefore fixes $T_c/{\overline{T}}_M$. A sufficiently small mass never reaches the central temperature required for sustained hydrogen burning, producing a hydrogen-burning minimum mass. Such an object becomes a cooling, degeneracy-supported brown dwarf instead of settling onto the main sequence. More massive contracting objects can ignite hydrogen before reaching this degeneracy-limited maximum. This conclusion uses the nonrelativistic, fixed-composition model; it does not require extrapolation into the relativistic regime.

Let $\Omega =-{\int }_0^{M}Gmdm/r$ be the gravitational energy. Multiply the stellar hydrostatic equation by $4\pi r^3$ and integrate through the star. Integration by parts and zero surface pressure givewhile its hydrostatic right-hand side is $-{\int }_0^{R}4\pi Gmr\rho dr=\Omega$. Hence the stellar virial theorem is

There is a genuine reference-sheet error to resolve. A monatomic ideal gas with adiabatic exponent $5/3$ has specific internal energy of an ideal gas $u=P/[\rho (5/3-1)]=3P/(2\rho )$, not the printed $3P/\rho$ on PDF page 2. The latter would give total internal energy $-\Omega$ and zero total energy, inconsistent with radiating contraction. The following derivation uses the consistent monatomic value.

A fully convective star is approximately isentropic and has an adiabatic stellar polytrope with index $3/2$. The given polytropic gravitational-energy formula yields $\Omega =-6G{M}^2/(7R)$. The virial relation gives total internal energy $U=-\Omega /2$, so total energy is $U+\Omega =-3G{M}^2/(7R)$. At constant mass with no nuclear heating, Kelvin-Helmholtz contraction obeys $d(U+\Omega )/ds=-L$, whenceHere $s$ denotes physical time.

In the plane-parallel atmosphere, the inward optical depth satisfies $d\tau =-\kappa \rho dr$, so hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. Neglecting radiation pressure, $\rho =\mu P/(\mathcal{R}T)$ and the specified opacity becomes $\kappa =({\kappa }_0\mu /\mathcal{R})P{T}^{16}$. ThereforeThe grey atmosphere relation and the boundary $P(0)=0$ integrate toTo apply the Schwarzschild criterion, put $w=2+3\tau$. Logarithmic differentiation givesThis increases inward, starting from zero. Marginal convection has ${\nabla }_{\mathrm{rad}}=2/5$, so $w_b^3=136/5$. Since $T_b^{12}/T_e^{12}=w_b^3/64$, the grey-atmosphere convection onset with seventeenth-power opacity isThe purely radiative atmosphere is used only up to this boundary.

The convective relation $P=K{T}^{5/2}$ is spatially uniform in its entropy coefficient $K$ at each instant; $K$ can change as the star radiates and contracts. At the convection boundary $w_b$ is a fixed number, so $P_b\propto g^{1/2}{T}_e^{-8}$ at fixed composition and opacity normalization. ThusOn the other hand, the central scalings for the $n=3/2$ stellar polytrope give $P_c\propto G{M}^2/R^4$ and $T_c=\mu P_c/(\mathcal{R}{\rho }_c)\propto \mu GM/(\mathcal{R}R)$. Hence, at fixed composition,Equating the interior and atmosphere expressions yields $T_e^{21/2}\propto M{R}^{1/2}$. Applying the Stefan–Boltzmann law then proves the Hayashi contraction with seventeenth-power opacity relation

At fixed mass write $L=A{M}^{8/21}{R}^{\alpha }$, with constant $A$ and $\alpha =46/21$. The contraction equation is $\dot{R}=-C{R}^{\alpha +2}$, where $C=7A{M}^{8/21}/(3G{M}^2)$. Direct integration givesMultiplying through by $R^{\alpha +1}$ and substituting the current luminosity givesFor the stated very large initial radius, or after that initial-radius term becomes negligible,This is the luminosity-age relation within the contraction model, rather than an assertion that a physical star begins at literally infinite radius.

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 givesThe 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 $\rho (r)={\rho }_{c}f(r/R)$ 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 $P_c=C_{P}G{M}^2/R^4$. HenceThese are homology scalings of central stellar pressure. Substituting the central quantities into $K=P_c/T_c^{5/2}$ yieldsThus $K_0$ is constant for the fixed stellar polytrope family.

At the approximate photosphere, take $T=T_e$ and use the ideal gas relation $\rho =\mu P/(\mathcal{R}T)$. The prescribed opacity givesThe stellar surface boundary condition and surface gravity of a star therefore implyThis is the temperature-four special case of convective contraction with power-law opacity.

Using the Stefan–Boltzmann law, the luminosity isFor $n=3/2$, the supplied gravitational-energy formula is $\Omega =-6G{M}^2/(7R)$. The stellar virial theorem for a monatomic gas gives $2U+\Omega =0$, so the total energy is $E=U+\Omega =-3G{M}^2/(7R)$. During Kelvin-Helmholtz contraction, with constant mass and no significant nuclear source, $L=-dE/dt$. ThusMultiplying by $-(7/2)R^{-9/2}$ and integrating givesThe 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 late-time central-temperature scalingReplacing $t-t_0$ 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.

For a spherical hydrostatic star with negligible surface pressure, the stellar virial theorem isFor a monatomic nonrelativistic perfect gas, $U=\frac{3}{2}\int PdV$, so $2U+\Omega =0$ and the total energy is $E=U+\Omega =\Omega /2$. Half the released binding energy raises the internal energy; only the other half can be radiated. With fixed mass, no nuclear supply and no external work, conservation of energy givesFor the uniform-density stellar model, this becomes the contraction-luminosity evolution equationIt relates luminosity to the contraction rate; it does not by itself prescribe $L(t)$. Given $L(t)$,For constant luminosity, $R(t)=R_0/[1+t/t_0]$ with $t_0=3G{M}^2/(10R_{0}L)$. This describes quasi-static Kelvin-Helmholtz contraction, not dynamical free fall. If the gas has a different constant specific-heat ratio ${\gamma }_g$, the appropriate stellar virial theorem is $3({\gamma }_g-1)U+\Omega =0$, so the radiated fraction of binding release is $(3{\gamma }_g-4)/[3({\gamma }_g-1)]$ rather than universally one half.

For a slowly evolving spherical star, use a Lagrangian enclosed mass coordinate $m$. Let $u$ be specific thermal internal energy and $D/Dt$ follow a mass element. The stellar energy balance equation isThe outward luminosity is energy transported through the mass shell, ${ϵ}_{\mathrm{nuc}}$ is nuclear rest-mass energy released per unit mass and time, and ${ϵ}_{\nu }$ is escaping neutrino power per unit mass. Define the gravothermal stellar energy generation by ${ϵ}_{\mathrm{grav}}=-Du/Dt-PD(1/\rho )/Dt$. At fixed composition it is $-TDS/Dt$ by the first law of thermodynamics. During composition changes the full internal-energy derivative must use the appropriate equation of state, with nuclear rest-mass changes counted only once. If a tabulated nuclear rate already subtracts reaction-neutrino energy, that loss must not be subtracted a second time in ${ϵ}_{\nu }$.

In a nearly steady main sequence star, stellar nuclear fusion supplies most of the luminosity. Before stable hydrogen ignition, Kelvin-Helmholtz contraction releases gravitational energy. In a gas-supported hydrostatic star the stellar virial theorem gives $U\simeq -\Omega /2$, so about half the change in gravitational binding energy heats the star and half is available for radiation. The associated timescale is $G{M}^2/(RL)$. After nuclear burning ends, a white dwarf can shine by losing stored thermal energy; crystallization releases latent heat, and composition separation can add gravitational energy. These are heat and gravitational reservoirs rather than sustained hydrogen fusion.

Stellar nuclear fusion is possible because lighter nuclei can combine into products with larger nuclear binding energy. The energy per reaction is $Q=(m_{\mathrm{initial}}-m_{\mathrm{final}})c^2$, including the relevant particles consistently. Although thermal energies are below the Coulomb barrier, quantum tunnelling permits fusion. The thermonuclear reaction rate averages a cross-section over the thermal relative-speed distribution:For nonresonant charged-particle reactions the energy weighting contains $\exp [-E/(kT)-\sqrt{E_G/E}]$; the compromise between the thermal tail and penetration produces the Gamow peak. A resonance can greatly enhance the rate. Density, abundances and temperature therefore all affect the stellar energy-generation rate, and power laws such as $T^4$ or $T^{16}$ are local approximations.

For the proton–proton chain, first produce deuterium and helium-3:The initial reaction is slow because it converts a proton to a neutron through the weak interaction. It is the bottleneck that allows long hydrogen-burning lifetimes. The alternative pep reaction $p+e^{-}+p\to {}^2\mathrm{H}+{\nu }_e$ also feeds the same chain. There are three principal proton-proton chain branches. In pp I, two helium-3 nuclei terminate the chain:In pp II, helium-3 first captures an existing helium-4 nucleus, then electron capture and proton capture finish the branch:In pp III, the beryllium-7 instead captures a proton:Relative branch weights change with temperature and composition. All branches convert four protons into a net helium-4 nucleus, with two weak conversions and two neutrinos, allowing for the electron consumed in pp II. After including positron annihilation, the atomic-mass energy budget is approximately $26.7\mathrm{MeV}$ per net helium-4 nucleus, but the deposited heat is smaller by the branch-dependent escaping-neutrino energy. Near ordinary low-mass main-sequence conditions, ${ϵ}_{pp}\propto \rho X_H^2{T}^4$ is a useful local approximation.

The CNO cycle provides an alternative catalyzed hydrogen burning route in hotter cores. The CNO-I cycle, also called the CN cycle, consists ofThe carbon seed is regenerated: the net reaction again consumes four protons and produces one helium-4 nucleus, two positrons and two neutrinos. The slow ${}^{14}\mathrm{N}(p,\gamma ){}^{15}\mathrm{O}$ reaction governs the ordinary cycle and causes nitrogen-14 to accumulate. The CNO-II cycle branches through ${}^{15}\mathrm{N}(p,\gamma ){}^{16}\mathrm{O}$, ${}^{16}\mathrm{O}(p,\gamma ){}^{17}\mathrm{F}$, ${}^{17}\mathrm{F}\to {}^{17}\mathrm{O}+e^{+}+{\nu }_e$, and ${}^{17}\mathrm{O}(p,\alpha ){}^{14}\mathrm{N}$, after which the main cycle continues. Thus CNO nuclei are catalytic in the closed cycles, although their relative abundances change. A local rate is ${ϵ}_{\mathrm{CNO}}\propto \rho X_H{X}_{\mathrm{CNO}}{T}^{\eta }$ with a much steeper exponent than the PP chain; the preceding homology problem specifies $\eta =16$. The concentration of this heating near the centre favors a convective core in hotter, more massive hydrogen-burning stars.

Once hydrogen is exhausted in a core, contraction can raise its temperature enough for core helium burning. The Triple-alpha process overcomes the absence of stable mass-5 and mass-8 nuclei by maintaining a tiny transient beryllium-8 population:The Hoyle state, an excited carbon-12 resonance near the three-alpha threshold, makes the second capture efficient enough despite beryllium-8's very short lifetime. The net conversion is $3{}^4\mathrm{He}\to {}^{12}\mathrm{C}$, releasing about $7.27\mathrm{MeV}$. Because three alpha particles are required, the specific rate scales as ${\rho }^2{Y}^3$; a schematic resonant rate isIts local temperature exponent is $-3+44/T_8$, of order forty near $T_8=1$. A degenerate core cannot expand promptly to regulate this increase, giving a helium flash; in a nondegenerate core the stellar thermostat permits stable helium burning. The competing ${}^{12}\mathrm{C}(\alpha ,\gamma ){}^{16}\mathrm{O}$ reaction determines much of the eventual carbon-oxygen mixture.

If the core becomes hot enough, carbon burning follows, for example through ${}^{12}\mathrm{C}+{}^{12}\mathrm{C}\to {}^{20}\mathrm{Ne}+\alpha$ and ${}^{12}\mathrm{C}+{}^{12}\mathrm{C}\to {}^{23}\mathrm{Na}+p$. Neon burning begins with photodisintegration, ${}^{20}\mathrm{Ne}+\gamma \to {}^{16}\mathrm{O}+\alpha$, followed by alpha capture such as ${}^{20}\mathrm{Ne}+\alpha \to {}^{24}\mathrm{Mg}+\gamma$. The first step consumes heat, but the combined rearrangement can release net energy. Oxygen burning includes ${}^{16}\mathrm{O}+{}^{16}\mathrm{O}\to {}^{28}\mathrm{Si}+\alpha$ and channels producing phosphorus and other nearby nuclei. At still higher temperatures, silicon burning proceeds through photodisintegration and particle captures in a reaction network approaching quasi-statistical equilibrium, producing iron-group nuclei. It is not simply direct silicon-plus-silicon fusion; the products depend on the electron fraction and weak-interaction timescale.

The increasing nuclear binding energy per nucleon supplies progressively less energy as products approach the iron group. Further fusion past that region cannot provide sustained net heat to support a core. Massive stars can therefore end with an unstable iron-group core; lower-mass stars do not attain all these burning stages and instead leave remnants such as carbon-oxygen white dwarfs. Heavy-element neutron captures can synthesize nuclei beyond the iron group without being the principal hydrostatic power source.

Finally, stellar neutrino energy loss competes with every heat source. Reaction neutrinos accompany the proton–proton and CNO chains. Hot or dense material also emits neutrino pairs through $e^{-}+e^{+}\to \nu +\bar{\nu }$, plasmon decay, $e^{-}+\gamma \to e^{-}+\nu +\bar{\nu }$, and electron-ion bremsstrahlung. These neutrinos usually escape, unlike the photons whose transport is diffusive; neutrino trapping requires much more extreme collapse conditions. Late burning has smaller fuel-energy reservoirs and strong neutrino cooling, so its duration is much shorter than core hydrogen burning. The surface luminosity is the escaping photon power, $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$, not the sum of photons and unobserved neutrino power. Radiative diffusion in a star, convection and electron conduction redistribute energy, while the energy-balance equation distinguishes true sources, losses and storage. Accretion or mass loss adds boundary energy and mechanical work when present, rather than changing the nuclear reaction budget.

For a quasi-static self-gravitating object of fixed mass, negligible surface pressure, and no accretion, nuclear source or additional deep heating, conservation of energy gives the intrinsic planetary luminosityHere $U$ is internal energy, and $\Omega$ is negative gravitational potential energy. Gravitational contraction makes $\Omega$ more negative and releases power $L_{\mathrm{grav}}\equiv -d\Omega /dt>0$. Some of that release raises the internal energy; it cannot all be radiated while the object maintains hydrostatic equilibrium.

To quantify the split assume thermal gas pressure with constant specific-heat ratio $\gamma >4/3$ and negligible rotation or magnetic support. This excludes the dynamically unstable gas-supported regime. The stellar virial theorem givesIt follows thatFor a monatomic ideal gas, $\gamma =5/3$, so $L_{\mathrm{int}}=L_{\mathrm{grav}}/2$: half the released gravitational energy heats the gas and half escapes. If the density profile remains homologous, write $\Omega =-qG{M}^2/R$, with $q>0$ a fixed structure constant. Thenup to the structure and virial coefficients, where ${\tau }_{\mathrm{KH}}$ is the Kelvin-Helmholtz cooling time.