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At fixed temperature and pressure, a closed reacting system reaches thermochemical equilibrium by minimizing Gibbs free energy over nonnegative species amounts subject to conserved elemental abundances. For an allowed stoichiometric vector $\nu$, an interior minimum has ${\sum }_i{\nu }_i{\mu }_i=0$. More generally the chemical potential vector lies in the row span of the elemental conservation matrix at positive concentrations, with the appropriate one-sided inequalities for absent species. The first law of thermodynamics gives the thermodynamic differential, while the Second law of thermodynamics supplies the minimization direction.

For an extensive equilibrium system, the thermodynamic Euler relation gives $E=TS-PV+\mu N$. At zero chemical potential, divide by $V$ to obtain $s=(\rho +P)/T$. Equivalently, $dP=(\rho +P)dT/T$ makesThe first law of thermodynamics identifies this as $dS$; the thermodynamic Euler relation fixes the entropy normalization. For adiabatic expansion satisfying the cosmological perfect-fluid continuity equation, the numerator on the right vanishes, proving cosmological entropy conservation.

For a single-component equilibrium system with extensive internal energy $E(S,V,N)$, scaling its entropy, volume and particle number together gives $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. The Euler theorem for homogeneous functions therefore gives $E=S{\partial }_{S}E+V{\partial }_{V}E+N{\partial }_{N}E$. The first law of thermodynamics identifies these derivatives as $T,-P,\mu$, respectively, proving $E=TS-PV+\mu N$. Dividing by volume yields the entropy density formula $s=(\rho +P-\mu n)/T$. This step fixes the entropy normalization rather than merely identifying its differential. Differentiating the Euler relation and subtracting $dE=TdS-PdV+\mu dN$ gives the Gibbs-Duhem equation $SdT-VdP+Nd\mu =0$. The argument requires extensivity; it is not an assumption about arbitrary nonadditive systems.

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.

Take a fixed comoving volume, whose physical volume is $V=V_{\ast }{a}^3$. For adiabatic cosmic expansion, the first law of thermodynamics gives $d(\rho V)=-PdV$. ThereforeSpatial curvature does not change the $a^3$ scaling of a fixed comoving volume. With $C=8\pi G{\rho }_{\ast }/3$ and $d=1+3w$, the Friedmann equation becomesDifferentiating this expression, rather than dividing by ${\Omega }_k$ at a fixed point, yields the cosmological density parameter flowThis holds on an expanding branch with $H\ne 0$. The spatially flat solution ${\Omega }_k=0$ is a fixed point. For $|{\Omega }_k|\ll 1$, ${\Omega }_k\propto a^{1+3w}$. Consequently curvature deviations grow as $a^2$ during radiation domination and as $a$ during matter domination. A small present curvature therefore requires a much smaller initial deviation: this is the Flatness problem of a decelerating Big Bang. The issue applies to either sign of curvature, not just an open universe. Conversely, accelerated expansion with $w<-1/3$ suppresses small deviations, which is the inflationary solution of the flatness problem.

The stellar adiabatic exponents are fixed-composition, constant-specific entropy derivatives:The chain rule gives ${\Gamma }_1({\Gamma }_2-1)/{\Gamma }_2={\Gamma }_3-1$. Define the pressure derivatives ${\chi }_{\rho }=(\partial \log P/\partial \log \rho {)}_T$ and ${\chi }_T=(\partial \log P/\partial \log T{)}_{\rho }$. The first law of thermodynamics, with $du=Tds+Pd\rho /{\rho }^2$, givesCombining this with $d\log P={\chi }_{\rho }d\log \rho +{\chi }_{T}d\log T$ on an adiabat givesThus the specific-heat ratio is not generally equal to the three stellar adiabatic exponents.

For the mixture, ${\chi }_{\rho }=\beta$ and ${\chi }_T=4-3\beta$. The specific heat capacity at constant volume, obtained by differentiating $u$ at fixed mass density, isTherefore the adiabatic exponents of a monatomic gas-radiation mixture and its specific-heat ratio areIf a relation involving only the exponents is wanted, eliminate $\beta$ from ${\Gamma }_1=\beta +(4-3\beta )({\Gamma }_3-1)$:For a pure monatomic perfect gas, $\beta =1$ and $\gamma ={\Gamma }_1={\Gamma }_2={\Gamma }_3=5/3$. For any calorically perfect gas with constant heat capacities, the same equality holds with its own ${\gamma }_g$. In a genuine gas-radiation mixture, $0<\beta <1$, $\gamma ={\Gamma }_1/\beta$ while the exponents are given separately above.

In the radiation limit, ${\Gamma }_1={\Gamma }_2={\Gamma }_3=4/3$ and the adiabatic temperature gradient is $1/4$. This follows independently from photon entropy: a comoving volume $V$ has $S\propto V{T}^3$, so an adiabat obeys $T\propto V^{-1/3}$ and $P\propto V^{-4/3}$. However, $c_P/c_V$ is singular in the pure-radiation limit, not $4/3$. This is the pure-radiation constant-pressure heat-capacity singularity. The equation $P=a{T}^4/3$ fixes temperature whenever pressure is fixed, so an ordinary constant-pressure temperature derivative is not available; along the mixture limit $\gamma \to \infty$ and $\beta \gamma \to 4/3$. For photons alone, mass-specific quantities additionally require a material mass label. The often quoted radiation index $4/3$ is its pressure-density adiabatic exponent, not a finite constant-pressure/constant-volume heat-capacity ratio.

For a fixed amount of gas, the first law of thermodynamics gives $dU=TdS-pdV$. Define the Gibbs free energy $G=U+pV-TS$, so $dG=-SdT+Vdp$. Equality of its mixed partial derivatives gives the Maxwell relation

The pipe is adiabatic, so the gas receives no heat. The work done on it at the inlet is $p_1{V}_1$, and the work it does at the outlet is $p_2{V}_2$. Neglecting kinetic and gravitational energy changes, the first law of thermodynamics gives $U_2-U_1=p_1{V}_1-p_2{V}_2$. Hence the throttling process is isenthalpic:The pressure drop is irreversible; constant enthalpy does not imply constant entropy.

Consider a species in thermal equilibrium, with energy density and pressure depending on temperature and with vanishing chemical potential, as in the supplied thermodynamic relation. Put $U=\rho (T)V$. The first law of thermodynamics givesSince $dP/dT=(\rho +P)/T$,The differential is therefore exact:After fixing the irrelevant additive entropy zero, the entropy density at zero chemical potential isMore precisely the first law leaves an additive constant in $S$; extensivity fixes that constant for the usual entropy-density normalization. A nonzero chemical potential would instead require $Ts=\rho +P-\mu n$, so the stated formula is not a universal identity for a decoupled massive species with conserved particle number.

For a reversible thermodynamic process that is an adiabatic process in a comoving volume, $\dot{V}=3HV$ and the cosmological perfect-fluid continuity equation gives $\dot{\rho }=-3H(\rho +P)$. Differentiating the extensive expression explicitly, or using the exact differential above, givesThis is cosmological entropy conservation for the closed equilibrium gas with no entropy-producing energy injection.

For radiation in cosmology, $P=\rho /3$, so $s=4\rho /(3T)$. Comparing $s=c_1{g}_{\star }{T}^3$ and $\rho =c_2{g}_{\star }{T}^4$ yields the coefficient ratioWhen all relativistic species share the same temperature and count, the familiar normalizations are $c_1=2{\pi }^2/45$ and $c_2={\pi }^2/30$. More generally the energy and entropy effective counts can differ; the common $g_{\star }$ here uses the assumptions stated in the question.

The three basic modes are radiation, convection and thermal conduction. Radiative transfer dominates transparent atmospheric layers; convection carries heat through sufficiently unstable fluid regions; thermal conduction matters especially in solids or highly conducting dense material.

Consider a chemically homogeneous ideal gas atmosphere and a small, dry parcel displacement, with rapid pressure equilibration and negligible heat exchange during the displacement. The parcel's first law of thermodynamics per unit mass isFor an adiabatic process, $ds=0$. Using hydrostatic equilibrium, $dP/dz=-\rho g$, the parcel followsAfter an upward displacement $\delta z>0$, its temperature relative to its new surroundings isHere $C_p$ is the specific heat capacity at constant pressure, and $N$ is the buoyancy frequency. At the same pressure, a warmer parcel has lower density and accelerates upward. Thus strict instability requires $dT/dz<-g/C_p$. Equivalently, the parcel equation is $\ddot{\delta z}=-N^2\delta z$, withNegative $N^2$ gives growing displacements. The printed non-strict inequality includes the onset boundary: equality is neutrally stable, not a strictly growing instability. This dry parcel buoyancy in a homogeneous atmosphere is the dry Schwarzschild criterion; condensation and composition gradients require modified criteria.

Let $S$ denote specific entropy, reserving $s$ from the first question for envelope thickness. The stellar adiabatic exponents are defined at fixed composition byIt is useful to introduce ${\Gamma }_3-1=(\partial \log T/\partial \log \rho {)}_S$, so $({\Gamma }_2-1)/{\Gamma }_2=({\Gamma }_3-1)/{\Gamma }_1$.

Take the stellar gas to be a fully ionized, nonrelativistic monatomic ideal gas, with fixed specific gas constant $\mathcal{R}$. Its mixture with thermal radiation hasHere $u$ is specific internal energy, and the stellar gas-pressure fraction is $\beta =\mathcal{R}\rho T/P$. The assumption of monatomic gas fixes its heat capacity; the perfect-gas pressure law alone would not determine it.

At fixed $T$ and fixed $\rho$, respectively, the pressure derivatives areFor an isentropic process, the first law of thermodynamics gives $du=Pd\rho /{\rho }^2$. Differentiate the internal energy rather than artificially holding $\beta$ fixed during the perturbation:The parenthesis is $(P/\rho T)[(3/2)\beta +12(1-\beta )]$, givingSince $d\log P={\chi }_{\rho }d\log \rho +{\chi }_{T}d\log T$,Using the relation between the exponents gives the adiabatic exponents of a monatomic gas-radiation mixtureThe requested values, including the adiabatic temperature gradient, areThe pure-radiation and pure-gas rows are understood as limits of the mixture.

For uniform composition, the Schwarzschild criterion for stellar convective instability isTo test whether a radiative configuration becomes unstable, use its required stellar radiative temperature gradient ${\nabla }_{\mathrm{rad}}$ for $\nabla$. Since $dP/dr<0$, the equivalent radial condition isAn outward-displaced parcel then cools less than its new surroundings, remains less dense at the same pressure, and is further accelerated outward.

The requested Eddington-model convective-core mass fraction follows from the usual global constant-$\beta$ closure. At the outer radiative surface, put $m=M$, $L_r=L$, and $P_{\mathrm{rad}}=(1-\beta )P$. Dividing the radiation-pressure gradient by the stellar hydrostatic equation givesThere is no energy generation outside the convective core, so use $L_r=L$ there. The local radiative-gradient formula isThe same representative $\beta$ has been used in the global luminosity closure and at the boundary. Setting this gradient equal to ${\nabla }_{\mathrm{ad}}$ at $m=M_c$ producesThis is $1$, $77/80$, and $5/8$ for $\beta =0,1/2,1$, respectively, with $\beta =0$ a marginal radiation-dominated limit.

There is a consistency qualification to this last model estimate. It cannot be an exact stellar solution if all the printed assumptions are enforced pointwise. An exactly uniform nonzero radiation fraction would give $d{P}_{\mathrm{rad}}/dP=1-\beta$ at every radiative point. The exact equations would then requireIf $L_r$ is constant throughout an envelope with positive mass density, $m$ increases with radius, so this equality cannot hold there. Equivalently uniform $\beta$ fixes the actual gradient to $1/4$, whereas ${\nabla }_{\mathrm{ad}}>1/4$ for $\beta >0$. The boxed core fraction is the intended Eddington closure and boundary estimate, which relaxes exact constancy of $\beta$ in the detailed envelope. The constant-beta radiative-envelope obstruction identifies the missing approximation; it is not legitimate to silently assert exact compatibility.

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.

Take a homogeneous comoving cell of physical volume $V_c=a^3{V}_0$ and total entropy $S=s{V}_c$. At vanishing chemical potential, the reversible first law of thermodynamics readsFor an extensive equilibrium system, the thermodynamic Euler relation is $E=TS-P{V}_c+\mu N$. At $\mu =0$ this givesThe supplied pressure identity verifies consistency directly: $dP=(\rho +P)dT/T$ impliesThe thermodynamic Euler relation fixes the additive normalization that a differential identity alone would leave undetermined.

For adiabatic cosmological expansion, the cosmological perfect-fluid continuity equation gives $\dot{\rho }+3H(\rho +P)=0$, while ${\dot{V}}_c=3H{V}_c$. HenceThis proves cosmological entropy conservation in the stated perfect-fluid approximation. Kinetic equilibrium alone does not guarantee adiabatic evolution: entropy injection, dissipative stresses or out-of-equilibrium reactions would require additional source terms. Those effects are implicitly excluded in this calculation.

Write $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative and $q=\mathbf{u}\cdot \mathbf{B}$ for the cross-helicity density, so $H_c={\int }_{V}qdV$. The ideal magnetohydrodynamic induction equation of ideal magnetohydrodynamics, with $\nabla \cdot \mathbf{B}=0$, givesThe dot product of the Euler momentum equation with the magnetic field has no Lorentz force contribution, because $\mathbf{B}\cdot [(\nabla \times \mathbf{B})\times \mathbf{B}]=0$. Consequently,Here $h=e+p/\rho$ is the specific enthalpy, $T$ the temperature, and $s$ the specific entropy; $e$ is the internal energy per unit mass. The first law of thermodynamics in this convention is $de=Tds-pd(1/\rho )$, so $dh=Tds+dp/\rho$. Substitution, followed by $\nabla \cdot \mathbf{B}=0$, yields the cross-helicity conservation lawThe vector triple product identity $\mathbf{u}\times (\mathbf{u}\times \mathbf{B})=\mathbf{u}q-u^2\mathbf{B}$ puts this in the equivalent form

For the fixed volume, the divergence theorem gives the exact cross-helicity balanceThus cross-helicity is constant when the net source equals the outward flux. Simple sufficient hypotheses are $\mathbf{B}\cdot \nabla s=0$ throughout the volume and $\mathbf{u}\cdot \mathbf{n}=\mathbf{B}\cdot \mathbf{n}=0$ on its boundary. The first condition includes a homentropic flow; the second prevents both relevant boundary fluxes. Periodic boundary conditions, or sufficiently rapid decay at infinity, provide alternatives. These statements concern smooth ideal magnetohydrodynamics: material specific entropy conservation alone, $Ds/Dt=0$, does not eliminate spatial specific entropy gradients along the magnetic field lines.