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Thermal energy versus electrostatic repulsion. The mean translational kinetic energy of one nonrelativistic thermal proton isUsing the center-to-center touching separation $2r_p$, the Coulomb barrier estimate isThus $E_C/⟨E⟩\simeq 274$. Using $r_p$ rather than $2r_p$ as the rough interaction distance doubles the estimate but leaves the conclusion unchanged. Typical solar thermal energies are far below the classical barrier. The two essential ideas are the high-energy tail of the thermal distribution and quantum tunnelling through the Coulomb barrier: neither a mean-energy estimate nor a sharp classical cutoff describes the reacting population.

For two equal-mass protons, let $v$ be their relative speed. In the center-of-mass frame each has speed $v/2$, so their total collision energy isHere ${\mu }_r$ is the reduced mass, not the stellar mean molecular weight.

In the low-energy, nonresonant approximation, the geometrical quantum-wave factor in a collision cross-section is proportional to the squared de Broglie wavelength, hence to $1/E$. The slowly varying short-range reaction probability is absorbed into $S_0$; for proton–proton fusion it also includes the small weak-reaction probability. The rapidly varying exponential is the quantum tunnelling probability. For a Coulomb potential $V(r)=e^2/r$, its leading semiclassical exponent isThe integral follows from $r=(e^2/E){\sin }^2\theta$; its dimensionless integral is $\pi /2$. A finite nuclear radius modifies the short-distance factor absorbed into $S_0$. In this convention $E_B={\pi }^2{\mu }_r{e}^4/(2{\hslash }^2)$: it is a penetration-energy scale, distinct from the classical contact energy $E_C$.

Thermal averaging and pair counting. Each proton samples collision partners with number density $N$ and collision flux $v$, so its collision rate is $N⟨\sigma v⟩$. Multiplying by the number of protons counts each identical pair twice. The thermonuclear reaction rate is thereforeThe relative speed of two independent thermal protons has a Maxwell-Boltzmann velocity distribution with reduced mass $m_p/2$. Changing variables with $E=m_p{v}^2/4$ gives the normalized energy densitySubstituting $v=2\sqrt{E/m_p}$ and the collision cross-section causes the energy powers to cancel:

The Gamow peak and its Gaussian width. Set $\Phi (E)=E/(kT)+2\sqrt{E_B/E}$. It diverges at either endpoint and has its unique minimum whereThus the energy called $E_G$ in this question is the Gamow peak energy,In the stellar tunnelling regime $E_B\gg kT$,This establishes the requested hierarchy in its physical regime; it is not an inequality valid at arbitrary temperature.

At the minimum, $\Phi (E_G)=3E_G/(kT)$ and ${\Phi }^{\prime \prime }(E_G)=3/(2kT{E}_G)$. HenceThe Gaussian standard deviation is $\sqrt{2kT{E}_G/3}$, small compared with $E_G$ when $kT\ll E_G$. Extending the lower endpoint to $-\infty$ relative to the peak then givesCombining this with the thermal prefactor yields the Gaussian approximation to a nonresonant thermonuclear rate,At fixed reactant density and fixed $S_0,E_B$, comparison with the requested temperature form givesThe constant $\beta$ here has dimensions of temperature; it is not inverse temperature. Resonances, plasma screening or variation of $S(E)$ would require corrections to this particular approximation.

The first equation is vertical hydrostatic equilibrium: the upward pressure gradient balances the central mass's vertical gravitational force. For $|z|\ll r$, the point-mass acceleration is $-GMz/(r^2+z^2{)}^{3/2}\simeq -{\Omega }_K^{2}z$, where Keplerian rotation gives ${\Omega }_K^2=GM/r^3$. The disc's self-gravity and vertical inertia are neglected.

The second equation is local thermal balance. The divergence of the vertical radiative flux equals viscous dissipation. In the convention of this alpha disk, $\nu =\alpha p/(\rho \Omega )$, and the dominant Keplerian shear has $rd\Omega /dr=-3\Omega /2$. ThereforeThe radiative flux increases from zero at the symmetric midplane to a positive outgoing value at the upper surface. Radial heat transport, thermal advection and rapid storage of thermal energy are excluded from this local balance.

The third equation is radiative diffusion, equivalentlyIt applies to optically thick material, with $\kappa$ the opacity per unit mass and $\sigma$ the Stefan-Boltzmann constant. Here constant electron-scattering opacity represents Thomson scattering; the formula uses a diffusive, nearly thermal radiation field. Positive upward flux requires temperature to decrease upwards.

The final equation is the ideal gas pressure law, with mean molecular weight ${\mu }_m$ in proton-mass units. Writing $\mathcal{R}=k/({\mu }_m{m}_p)$ gives $p=\mathcal{R}\rho T$. The pressure is gas pressure because radiation pressure is assumed negligible. The four equations express vertical support, local heating balance, diffusive cooling and the gas equation of state.

Consider a small outward displacement $\delta r$ of a fluid element. It adjusts to the ambient pressure while retaining its specific entropy and stellar composition, so an adiabatic process gives $d\log {\rho }_{\mathrm{parcel}}=d\log P/\gamma$. Its excess mass density over the new surroundings is, to first order,Restoring buoyancy requires this bracket to be positive. With $\gamma =5/3$, the Schwarzschild criterion is thereforeEquality is neutral stability. For uniform mean molecular weight, the ideal gas law gives $d\log \rho =d\log P-d\log T$. Since hydrostatic equilibrium pressure decreases outward, the inequality is equivalent to $\nabla =d\log T/d\log P<{\nabla }_{\mathrm{ad}}=2/5$. Dividing the radiative diffusion in a star and hydrostatic equilibrium stellar structure equations givesThis is the stellar radiative temperature gradient test for stellar convective stability.

In the thin upper stellar atmosphere take $m\simeq M$, $L_r\simeq L$ and use gas pressure dominance. The prescribed opacity becomes $\kappa ={\kappa }_0\mu P{T}^{12}/\mathcal{R}$. Dividing the two structure equations then yieldsAt zero optical depth, the atmospheric boundary has $T_s^4=T_e^4/2$ and negligible gas pressure. Integrating from that boundary, with $T_s^{-8}=4T_e^{-8}$, givesTo locate the onset, put $y=T_e^8/T^8$. Logarithmic derivative gives $d\log P/d\log T=4y/(4-y)$, henceThis starts at zero at the outer boundary and increases inward. It first reaches $2/5$ atThe corresponding optical depth is ${\tau }_b=\frac{4}{3}(\sqrt{13/20}-\frac{1}{2})>0$, so the onset lies inside the stellar atmosphere. This is grey-atmosphere convection onset with thirteenth-power opacity.

Let the interior of the fully convective star have $P=K_T{T}^{5/2}$, with $K_T$ spatially constant. At its boundary $T_b/T_e$ is fixed, so the atmospheric pressure formula givesThe Stefan–Boltzmann law, $L=4\pi R^2\sigma T_e^4$ with $\sigma =ac/4$, consequently impliesThe ideal gas equation transforms the interior relation intoFor completeness, the mass-radius scaling follows by taking $\rho ={\rho }_c{\theta }^{3/2}$ and $r=\alpha \xi$ with ${\alpha }^2=5K_{\rho }{\rho }_c^{-1/3}/(8\pi G)$. Hydrostatic equilibrium and mass conservation reduce to the Lane-Emden equation $({\xi }^2{\theta }^{\prime }{)}^{\prime }=-{\xi }^2{\theta }^{3/2}$, with $\theta (0)=1$, ${\theta }^{\prime }(0)=0$. Its fixed dimensionless profile gives $R\propto \alpha$ and $M\propto {\alpha }^3{\rho }_c$. Eliminating ${\rho }_c$ gives $R\propto K_{\rho }{G}^{-1}{M}^{-1/3}$, or $K_{\rho }\propto G{M}^{1/3}R$. Thus $K_T^2\propto M^{-1}{R}^{-3}$ at fixed stellar composition. Equating this with the atmospheric expression gives $L^{17/4}\propto M^2{R}^{19/2}$, and henceThis Hayashi relation with thirteenth-power opacity concerns the interior of a fully convective star beneath a thin stellar atmosphere. The coefficient $K_T$ may vary between stars; spatially constant specific entropy does not mean equal specific entropy across the whole sequence.

Use radiative homology with proton-proton burning and inverse-fourth-power opacity, retaining the composition factors throughout. To distinguish the stellar radius from the stellar gas constant, write them as $R$ and $\mathcal{R}$. With gas pressure alone, introduce the scale factorsThe ideal gas equation gives $\rho ={\rho }_{\ast }b$, $P=P_{\ast }p$ and $T=T_{\ast }p/b$. Substitution into hydrostatic equilibrium and mass conservation givesso $p^{\prime }=-bq/x^2$ and $q^{\prime }=x^{2}b$.

Set $\zeta =Z(1+X)$. The radiative diffusion in a star equation, with the stipulated opacity, becomesHere ${\rho }^2{T}^{-7}={\rho }_{\ast }^2{T}_{\ast }^{-7}{b}^9{p}^{-7}$. ThusThe luminosity equation follows from the specific heating rate of the Proton–proton chain:ConsequentlyThese give all four dimensionless stellar structure equations with their numerical coefficients.

The stellar homology assumption means the dimensionless profiles are shared across the models, not merely that an individual model can be written in dimensionless coordinates. The displayed equations then require $D$ and $E$ to be fixed. Holding the universal constants and ${\kappa }_0,{ϵ}_0$ fixed givesEquating the two cancels $M^6$ and yields $R^6\propto X^2\zeta {\mu }^{-4}$. Therefore the radius and mass-luminosity relation areThe mass-independent radius is a consequence of the particular heating and opacity exponents in this model; it is not a general radius law for main-sequence stars.

Initially $X=X_0$ is the same for all stars. The adopted mean molecular weight $\mu =4/(5X+3)$ also has the same initial value, neglecting its small direct dependence on $Z$. Thus $R\propto Z^{1/6}$. At fixed luminosity, the Stefan–Boltzmann law $L=4\pi R^2\sigma T_e^4$ givesThe stars compared here can have different masses; the luminosity law adjusts the mass to keep $L$ fixed.

For the evolution, take the stellar mass as constant and maintain homogeneous composition. The available energy in the remaining hydrogen is $M{E}_{H}X$, so conservation of energy in the nuclear-powered model givesThis includes the entire mixed hydrogen reservoir, not just the mass in a central burning region.

To determine the turnoff mass of a homogeneously mixed proton-proton-burning population, write the luminosity law as $L=C{M}^6{Z}^{-7/6}H(X)$, whereSeparating variables, the time to evolve from $X_0$ to $f{X}_0$ isFor a common $X_0$ and $0\le f<1$, this integral is a positive finite constant independent of $M,Z,t$. It follows that the turning-off stars satisfyThe integration retains the evolving luminosity and mean molecular weight; replacing $L$ by its initial value is unnecessary.

For the Schwarzschild criterion, displace a small gas parcel upwards. Assume its composition is unchanged, it remains in pressure balance with its surroundings, and it moves quickly enough that heat exchange is negligible. The parcel obeys the adiabatic equation of state $P{\rho }^{-\gamma }=\mathrm{constant}$. Combining this with the ideal gas equation gives $T\propto P^{(\gamma -1)/\gamma }$, so its adiabatic temperature gradient is ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma =2/5$.

Let the surrounding temperature gradient be $\nabla =d\log T/d\log P$. For an upward displacement $\delta \log P<0$, the parcel's temperature relative to its new surroundings isIf $\nabla <{\nabla }_{\mathrm{ad}}$, this is negative: the parcel is cooler and, at equal pressure and composition, denser than its surroundings. It therefore sinks back; a downward displacement likewise gives a restoring buoyancy force. If the inequality is reversed, the displaced parcel continues to rise. The stability condition is thusEquality is the neutral boundary. There is no composition-gradient term because the ideal-gas parcel and the background have the same mean molecular weight.

In the thin upper stellar atmosphere, take a plane-parallel approximation with constant $g$, optical depth increasing inward, and zero overlying pressure $P(0)=0$. Hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. The ideal gas equation converts the stipulated opacity intoDefine $u=1+3\tau /2$, so the given grey atmosphere relation is $T^4=T_e^{4}u/2$. Integrating from the surface givesThis proves the required pressure law. In particular $P^2\propto g{T}_e^{-4\beta }$ times a function only of $u$ and $\beta$.

For the grey-atmosphere convection onset with density-linear opacity, differentiate the two profiles to obtain the radiative temperature gradient:It increases monotonically from zero, since $\beta >1$, and first reaches the adiabatic temperature gradient at $u_b^{\beta -1}=(4\beta +1)/5$. ThereforeBelow this point the radiative stratification is unstable and convection must transport some of the outward energy flux.

At the outer convection boundary, $u_b$ depends only on $\beta$, so $T_b=T_e(u_b/2{)}^{1/4}\propto T_e$ and $P_b\propto g^{1/2}{T}_e^{-2\beta }$ for fixed $\mu ,{\kappa }_0,\beta$. The matching constant in $P=K{T}^{5/2}$ is $K=P_b/T_b^{5/2}$. ConsequentlyThis temperature form of the polytropic relation follows from an ideal monatomic adiabat; its $K$ is not the density-polytrope coefficient in Question 3.

For the deep radiative boundary beneath a power-law-opacity convective envelope, negligible envelope mass and absent local energy generation give $m\simeq M$ and $L_r\simeq L$. Divide the radiative temperature equation by the pressure equation to get the stellar radiative temperature gradientThis is the gradient that would be needed if radiation alone carried $L$, and it is the quantity to compare with $2/5$ even while the actual region is convective. The deeper Kramers' opacity law gives $\kappa ={\kappa }_1\mu P{T}^{-9/2}/\mathcal{R}$. Substituting $P=K{T}^{5/2}$ yieldsAs $T$ rises inward this decreases, so the transition back to a stable radiative interior is atThus $T_{\mathrm{in}}^{7/2}\propto K^{2}L/M$. Using the outer matching and the Stefan–Boltzmann law,The mass and radius cancel. Hence, at fixed opacity coefficients and composition,This uses the same adiabat through the convection zone and distinguishes its inner transition temperature from its outer onset temperature.

Introduce the characteristic scalesThen $\rho ={\rho }_{\ast }b$, $P=P_{\ast }p$ and $T=T_{\ast }t$. The gas-only equation of state gives the important dimensionless identity $p=bt$. The PDF's CNO cycle approximation has temperature power 13; the converted TeX's power 1.3 is an OCR error.

Since $d/dr=R^{-1}d/dx$, the mass and hydrostatic pressure support equation immediately becomeThus $dq/dx=x^{2}b$ and $dp/dx=-bq/x^2$. The stellar structure equations for nuclear heating givesoFor radiative diffusion in a star, $\kappa \rho ={\kappa }_0{\rho }^2{T}^{-3}$, and thereforeThis gives the radiative gradient coefficientIn an efficiently convective region, the monatomic adiabatic temperature gradient is $2/5$. Hence $dt/dx=(2/5)(t/p)(dp/dx)=-2q/(5x^2)$. If the radiative gradient is shallower, radiation suffices; if it is steeper than the adiabatic gradient, the Schwarzschild criterion selects convection and the shallower adiabatic profile. Thus

The regular central conditions are $q(0)=l(0)=0$ with finite positive $p(0),t(0)$. In particular $q\sim b(0)x^3/3$ and $l\sim Eb(0{)}^{2}t(0{)}^{13}{x}^3/3$, so the pressure and temperature derivatives tend to zero at the centre. The normalized surface has $q(1)=l(1)=1$. In the idealized zero-pressure, zero-temperature surface used for stellar homology, $p(1)=t(1)=0$; a physical photosphere supplies small nonzero surface values that are neglected in these bulk homology relations.

The common dimensionless profiles and boundary conditions keep $D,E$ constant. With fixed ${\kappa }_0,{ϵ}_0$, $D$ constant yields $L{\mu }^{-7}{M}^{-5}=\mathrm{constant}$. Substitution into the nuclear coefficient then gives $R^{16}\propto {\mu }^6{M}^{10}$. Consequently the CNO homology with density-dependent inverse-cubic opacity isThe Stefan–Boltzmann law gives $T_e^4\propto L/R^2\propto {\mu }^{25/4}{M}^{15/4}$. Eliminating mass, rather than dropping composition dependence, gives the Hertzsprung-Russell diagram relationThe additive constant depends on the fixed physical and structural coefficients; logarithms refer to fixed choices of units.

For physical evolution time $s$, distinct from the dimensionless temperature $t(x)$, hold stellar mass fixed and put $u=3+5X$, $u_0=3+5X_0$. The mean molecular weight approximation and the homology relation give $L=L_0(u_0/u{)}^7$. Converting hydrogen mass to helium releases energy at rate $L$, so fuel conservation is $M{E}_{H}dX/ds=-L$. ThereforeIntegration with the initial composition proves the homogeneous fuel-depletion luminosity feedback lawIt is valid while hydrogen remains and the homology assumptions hold. Fuel exhaustion occurs when $u=3$, before the formal denominator vanishes; the apparent infinite luminosity is not a physical endpoint of this stellar model.

Use stellar homology: dimensionless density, temperature and luminosity profiles are taken to have the same shape, so the dimensional scale relations follow from the structure equations. Mass conservation and hydrostatic support giveThe ideal gas equation then gives $T_c\propto \mu M/R$, with the dimensional constants held fixed. The specific stellar energy-generation rate has the prescribed hydrogen factor and temperature exponent, soIndependently, radiative diffusion in a star gives $T_c/R\propto \kappa {\rho }_{c}L/(R^2{T}_c^3)$, henceEquating the nuclear supply and radiative transport establishes composition-dependent CNO homology:At one fixed composition, these reduce to $R\propto M^{3/4}$ and $L\propto M^3$ as required.

The Stefan–Boltzmann law gives $T_e^4\propto L/R^2\propto M^{3/2}$, so $T_e\propto M^{3/8}$. Thus the logarithmic Hertzsprung-Russell diagram slope along this zero-age main sequence isIncreasing mass moves upward and leftward on the conventional diagram, whose temperature axis is reversed.

For composition changes, neglect the small heavy-element contribution to the mean molecular weight and set $Y\simeq 1-X$. Then ${\mu }^{-1}=(3+5X)/4$. The fixed trace catalyst abundance is absorbed into ${ϵ}_0$; this is not an assertion that an exactly metal-free star can undergo CNO burning. Substitution into the homology laws givesFor a retained, nonnegligible fixed metallicity $Z$, the corresponding mean-molecular-weight factor is $3+5X-Z$ instead. The displayed requested formula uses the trace-metal approximation.

At fixed mass, differentiating with respect to $X$ at the formal reference $X=1$ givesConsequentlyand the initial fully mixed hydrogen-burning evolutionary track has slopeHydrogen depletion has $dX<0$, so both luminosity and effective temperature initially increase. The star therefore moves upward and leftward, on a shallower logarithmic track than the mass sequence. The figure uses dimensionless luminosity and temperature relative to its initial reference star; the track is computed for declining $X$ at constant mass.

Ordinary cluster main-sequence turn-offs instead evolve toward brighter, cooler, more extended configurations and eventually the giant region. Normal stars are therefore not mixed throughout their entire mass during hydrogen burning: the core composition changes while the envelope retains hydrogen, and composition and entropy gradients develop. Core exhaustion and subsequent shell burning then produce very different structural evolution. Complete mixing is a special evolutionary scenario, not the standard explanation of ordinary giant formation.

These hydrogen-burning scalings must not be extrapolated literally to $X=0$. Their vanishing nuclear coefficient would make the inferred radius tend to zero and the effective temperature singular; in reality hydrogen exhaustion requires a different structural and energy-generation model. A homogeneous remnant may be helium-rich, but its later structure is not described by the same CNO law. Also, the cover-sheet radiation-pressure expression has a temperature-power typo: the physical term is $a{T}^4/3$. Radiation pressure was neglected in this derivation, so that typo does not enter the calculation.

In the radiative layer, take the enclosed mass to be $M_1$, the transported luminosity to be $L$, and the mean molecular weight to be constant. The envelope's own gravity is neglected. Divide the hydrostatic and radiative temperature equations, and then eliminate density using the gas equation of state:Put $p=n+1$ and $b=4+m+n$. For $p,b\ne 0$, integration givesDefining $C^p=pB/b$ and $K=K_P/C^p$ yieldsIn the relevant positive-exponent case, writing $K=T_0^b$ produces the stated form whenever this integration constant is nonnegative. The exceptional values excluded above lead to logarithms: $p=0$ gives $\log P=B{T}^b/b+\mathrm{const}$, $b=0$ gives $P^p=pB\log T+\mathrm{const}$, and both zero give $\log P=B\log T+\mathrm{const}$.

For radiative-envelope convection matching, compute the logarithmic gradient of the pressure relation:At the marginal boundary, the Schwarzschild criterion equates this with the ideal-gas adiabatic gradient ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma$. HenceThe positive real $T_0$ notation requires a nonnegative bracket; a general signed $K$ is the safer integration constant otherwise. For the later case $n=1,m=3$ and a monatomic gas, this bracket is $3/5$, so $T_0$ is of the same order as $T_b$.

Deep below the convective boundary, with the temperature-power term dominating $K$, the pressure law becomes $P\simeq C{T}^{b/p}$. Differentiating this approximation and using hydrostatic balance givesThe general temperature is a term proportional to $1/r$ plus a boundary-dependent constant. Well inside the boundary radius, where that term dominates, it reduces toThe inverse-radius profile is an inner approximation, not an assertion that the boundary term vanishes everywhere.

Now take $n=1,m=3$, so $p=2$, $b=8$. Then $P\simeq C{T}^4$, $\rho \simeq (\mu C/\mathcal{R})T^3$, and define $A_T=\mu G{M}_1/(4\mathcal{R})$ so that $T\simeq A_T/r$. The luminosity supplied by the thin burning layer isBecause the integral is strongly concentrated near $R_1$ and $r_b\gg R_1$, the outer term is negligible. ThusThe factor $\rho ϵ$, rather than merely $ϵ$, is essential since the specified energy-generation rate is per unit mass. This integration uses the deep radiative profile as a leading thin-shell approximation.

Crucially, $C$ is not independent of luminosity. Its defining relation givesSubstitution into the shell integral therefore yields $L^2\propto M_1^{17}{R}_1^{-13}$. The degenerate-core relation gives $R_1\propto M_1^{-1/3}$, soThis degenerate-core shell-burning mass-luminosity relation is specific to the adopted opacity, burning exponent and structural approximations.

Removing some envelope mass while retaining the hydrogen-burning shell and the deep-envelope regime leaves the leading luminosity approximately unchanged at fixed core mass. The envelope can instead change the radius and effective temperature. This independence is not valid for arbitrarily complete stripping: removing the fuel or the pressure-confined shell structure quenches or modifies shell burning, and the assumed giant model then fails.