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Use the secular, circular-orbit approximation: the wind escapes fast enough to avoid appreciable interaction with the companion, while the stellar masses change slowly enough that orbital elements describe the evolving binary star. Neglect spin angular momentum. Let $\omega =(GM/a^3{)}^{1/2}$ be the orbital angular speed. The donor's distance from the centre of mass is $a_1=a{M}_2/M$, and the circular-binary orbital angular momentum isA spherically symmetric wind in the donor frame has zero mean additional angular momentum about the donor. Averaging its angular momentum about the binary centre of mass therefore gives the donor's orbital specific angular momentum, $j_1=a_1^2\omega$. Since $\dot{M}<0$ denotes mass removed from the system, ${\dot{J}}_{\mathrm{orb}}=j_1\dot{M}$, andThis is donor-wind angular-momentum loss in the Jeans-mode mass loss approximation.

First allow only wind loss, so ${\dot{M}}_1=\dot{M}$ and ${\dot{M}}_2=0$. The logarithmic derivative of $J_{\mathrm{orb}}$ givesSolving yields $\dot{a}/a=-\dot{M}/M$. By Kepler's third law, $P^2=4{\pi }^2{a}^3/(GM)$, so $\dot{P}/P=-2\dot{M}/M$. Integrating givesWind loss thus expands the separation and lengthens the orbital period. These are secular relations; an impulsive loss on an orbital time could instead generate eccentricity and would need a different analysis.

To test contact, put $D=d\log (R_1/R_L)/dt$. The given short-time stellar radius response exponent gives ${\dot{R}}_1/R_1=-n{\dot{M}}_1/M_1$. Differentiating the Roche lobe approximation givesFor wind alone this becomes ${\dot{R}}_L/R_L=\dot{M}/(3M_1)-4\dot{M}/(3M)$. Hence, with the binary mass ratio $q=M_1/M_2$,Because $\dot{M}$ is negative, increasing overfill, $D_{\mathrm{w}}>0$, occurs precisely whenThis is wind-driven Roche-lobe overflow. For the reverse strict inequality, $D_{\mathrm{w}}<0$ and the donor star retreats inside its Roche lobe, producing a detached binary. At equality the wind is neutral to contact at this order.

Now permit an additional conservative binary mass transfer rate $x={\dot{M}}_2$ to the mass-gaining star. The total escaping wind remains $\dot{M}$, while ${\dot{M}}_1=\dot{M}-x$. Transfer retains orbital angular momentum within the binary under the assumed neglect of spins, so the same wind-loss formula for ${\dot{J}}_{\mathrm{orb}}$ applies. Logarithmic differentiation now givesSubstituting this and ${\dot{M}}_1=\dot{M}-x$ into the two radius derivatives givesSteady contact requires $D=0$. If the wind is driving overflow and $6q<5-3n$, the transfer contribution reduces overfill and a positive contact rate exists:The numerator and denominator are positive in this regime, making ${\dot{M}}_2>0$. This explicitly establishes both the contact transfer rate with Jeans-mode wind loss and its stabilizing sign.

If the wind drives overflow but $6q>5-3n$, positive transfer instead increases $R_1/R_L$: the mass loss required to relieve overfill makes the overfill worse. The formal contact formula would require ${\dot{M}}_2<0$, which is not donor-to-companion transfer. There is no stabilizing positive contact rate; overflow runs away. For a red giant this can lead to a common envelope, followed by envelope ejection into a tighter binary or a merger. The criterion concerns the supplied short-time radius response; the actual runaway timescale and outcome require the donor's dynamical and thermal response. At $6q=5-3n$ transfer is neutral to contact at this order, so it cannot balance a nonzero wind driver. If the wind already causes detachment, the transfer instability condition alone does not initiate overflow.

The event rate counts unordered pairs of reacting nuclei. For unlike species there are $n_i{n}_j$ pairs per unit-volume normalization; for identical species there are approximately $n_i^2/2$ pairs, since counting ordered pairs counts each collision twice. The Kronecker delta therefore supplies exactly the required divisor $1+{\delta }_{ij}$. It is the event rate, not the destruction rate of identical nuclei: one identical-pair reaction destroys two nuclei.

The temperature dependence combines the high-energy tail of the Maxwell-Boltzmann distribution with quantum tunnelling through the Coulomb barrier. The principal exponential factor in the energy integral is $\exp [-E/(k_{B}T)-b/\sqrt{E}]$. Its stationary point, the Gamow peak, has $E_0\propto T^{2/3}$ and exponent $\eta =3E_0/(k_{B}T)\propto T^{-1/3}$. Integrating around this peak gives a prefactor proportional to $T^{-2/3}$, or ${\eta }^2$, when the nuclear factor varies slowly. Thus the supplied thermonuclear reaction rate is locally proportional to ${\eta }^2{e}^{-\eta }$. The slow weak interaction in the proton-proton reaction also makes its overall coefficient very small; its local thermal slope follows the supplied formula.

At fixed number densities, define the local temperature exponent of a thermonuclear reaction by a logarithmic derivative. Since $d\eta /d\log T=-\eta /3$,For proton-proton reactions, $A=1/2$ and $Z_1=1$. At $T_6=15$, ${\eta }_{11}=13.67$ and $\alpha =3.89$. For two helium-3 nuclei, $A=3/2$ and $Z_3^2{Z}_3^2=16$, giving ${\eta }_{33}=49.68$ and $\beta =15.89$. For helium-3 with helium-4, $A=12/7$ and the same charge product gives ${\eta }_{34}=51.95$ and $\gamma =16.65$. HenceThese are local powers near the specified temperature, not constant exponents over an arbitrarily large range. The greater reduced nuclear mass in the $34$ reaction increases its tunnelling exponent and its thermal sensitivity.

For the effective proton-proton reaction network, denote the event rates byThe short-lived beryllium intermediate can be eliminated, so production through $R_{34}$ feeds $n_7$, representing predominantly lithium-7. This rapid conversion is electron capture. Count the nuclei consumed and produced in each reaction: $11$ consumes two protons and makes one deuterium; $21$ consumes a proton and a deuterium and makes one helium-3; $33$ consumes two helium-3 and makes one helium-4 and two protons; $34$ consumes one of each helium isotope; $17$ consumes a proton and lithium-7 and makes two helium-4. It follows thatFor closure, ${\dot{n}}_7={\lambda }_{34}{n}_3{n}_4-{\lambda }_{17}{n}_1{n}_7$. These nuclear terms conserve $n_1+2n_2+3n_3+4n_4+7n_7$, providing an independent stoichiometric check. They describe a fixed local volume; compression or mixing would add transport terms.

The roughly one-second destruction time of deuterium is negligible compared with the proton-burning and helium-3 evolution times. Its abundance rapidly adjusts to the slowly evolving production rate. Set ${\dot{n}}_2\simeq 0$, giving the deuterium quasi-equilibrium in proton-proton burningSubstitution, without setting ${\dot{n}}_1=0$, givesThe helium-3 relaxation time of about $6\times 10^5$ years is likewise short compared with the central proton-burning time and the age of the Sun. On this relaxation time the proton and helium-4 abundances, and the background temperature, can be treated as fixed. Put $A={\lambda }_{33}$, $B={\lambda }_{34}{n}_4$, $C={\lambda }_{11}{n}_1^2/2$. Then ${\dot{n}}_3=C-A{n}_3^2-B{n}_3$, and its nonnegative equilibrium isThe other root is negative and unphysical. Write $n_3=n_{3e}+x$ and subtract the equilibrium equation. Exactly, for a fixed background,Dropping the quadratic displacement givesThus small displacements decay exponentially. A slowly moving equilibrium introduces an additional $-{\dot{n}}_{3e}$ term, negligible to leading order when $\tau$ is short compared with background evolution.

To estimate the temperature for helium-3 freeze-out, use the cooler pp-I-dominated region and neglect $B$ compared with $2A{n}_{3e}$. Then $n_{3e}\simeq n_1\sqrt{{\lambda }_{11}/(2{\lambda }_{33})}$ and ${\tau }^{-1}\simeq n_1\sqrt{2{\lambda }_{11}{\lambda }_{33}}$. With the local powers just obtained and fixed proton number density, $n_{3e}\propto T^{-6}$ and $\tau \propto T^{-10}$. Adopting $t_{\odot }\simeq 4.6\times 10^9\mathrm{yr}$ givesThis is an order-of-magnitude extrapolation of local exponents. Keeping the supplied exponential factors gives the helium-3 freeze-out estimate with Gamow factorswhich yields about $6.8\times 10^6\mathrm{K}$ with the same normalization and fixed-density approximation. Thus the robust estimate is several million kelvin. The supplied data do not fix the local density, composition or pp-II fraction away from the centre, so they cannot determine an exact transition temperature or radius. In particular, using $\tau \propto T^{-16}$ alone would omit the temperature dependence of the equilibrium helium-3 abundance.

The solar hydrogen and helium-3 abundance profiles follow from burning, finite equilibration time and convection. Central hydrogen has been consumed, so $X_1$ is lowest in the centre and increases toward its much less processed outer value. In the hot centre, rapid destruction keeps helium-3 small. Moving outward initially raises its equilibrium abundance because ${\lambda }_{11}/{\lambda }_{33}$ rises strongly as temperature falls. Further out, equilibration is no longer reached within the solar age, and still farther out even production is slow. Hence $X_3$ has an off-centre maximum and then declines toward its small outer abundance. The outer convective envelope mixes each abundance toward a constant value.

The sketch shows these qualitative shapes; the helium-3 ordinate is scaled to its own maximum. The illustrated amplitudes, peak radius and envelope boundary are schematic, not a numerical solar-model solution.

There is a genuine inconsistency in this part. For any nonnegative metallicity distribution function, its maximum is at least its mean. A population of positive mean solar metallicity cannot have a maximum equal to half that mean. For example, a uniform abundance distribution on $[0,2Z_{\odot }]$ has mean $Z_{\odot }$ and maximum $2Z_{\odot }$. Thus the printed maximum-to-mean ratio cannot be proved. The following calculation explains the supplied function, gives the consistent distribution, and also treats the literal birth-metallicity interpretation separately.

Let $f(\mu )$ denote the supplied abundance function. Algebraically it isWith $\delta =1-\mu$, a Taylor series of $(1-\delta {)}^{Y+1}$ givesConsequentlyThis is an early-consumption expansion: its fractional correction is $(1-Y)\delta /3+O({\delta }^2)$. Small yield alone does not make the expression linear throughout most of the gas-consumption history. At fixed $0<\mu <1$, the small-$Y$ expansion instead is

In a well-mixed closed-box model of galactic chemical evolution, the gas fraction of a galaxy also labels the permanently locked stellar mass: $d{M}_s=-M_{g,\mathrm{init}}d\mu$. Newly formed stars inherit the current gas-phase metallicity $Z_g(\mu )$. Therefore the mass-weighted mean stellar metallicity satisfies the cumulative and birth stellar metallicities relationApplying the derivative to $f$ givesThus the supplied function has the natural interpretation of a cumulative mean, not an individual star's birth abundance. Direct integration of $1-u^Y$ verifies that it gives exactly $f$. This is the saturating closed-box enrichment law: it corresponds to effective stellar yield $Y(1-Z_g)$, with $Z_g$ measured as a fraction. The usual constant-yield trace-metal model instead gives $Z_g=-Y\log \mu$. Both agree to first order when $Y|\log \mu |\ll 1$; their exact formulae should not be identified outside that regime.

Let the final gas fraction be ${\mu }_f>0$ and ${\delta }_f=1-{\mu }_f$. In the consistent mean-metallicity interpretation, ${\overline{Z}}_{\ast }({\mu }_f)=Z_{\odot }$ while the most recently formed stars have the maximum $z_f=1-{\mu }_f^Y$. At early consumption,so the corrected result isThe assumption here includes ${\delta }_f\ll 1$, equivalently $2Z_{\odot }/Y\ll 1$ in this approximation. Exactly, $z_f=1-{\mu }_f^Y$ and $Z_{\odot }=f({\mu }_f)$ determine the maximum implicitly; it remains larger than the mean, but its ratio to the mean need not equal two after substantial consumption.

For the mass-weighted metallicity distribution function, stars formed while the gas fraction falls by $-d\mu$ have probability $dP=-d\mu /{\delta }_f$. Writing their individual abundance as $z=1-{\mu }^Y$ gives $\mu =(1-z{)}^{1/Y}$ and the change of variablesThe density is zero outside this interval. Its cumulative probability is $[1-(1-z{)}^{1/Y}]/(1-{\mu }_f)$, so it integrates to one; integrating $zF(z)$ reproduces $f({\mu }_f)$. A star-count distribution has the same form if the initial mass function and survival selection give a fixed number of observed stars per unit locked mass. Luminosity-weighted samples require different weights.

For trace abundances, $(1-z{)}^{1/Y-1}\simeq e^{-z/Y}$ to leading order at fixed $z/Y$. The corresponding ordinary closed-box metallicity distribution isIn the early-consumption regime $z_f\ll Y$, this becomes approximately uniform, $F\simeq 1/(Y{\delta }_f)$ on $[0,Y{\delta }_f]$. Its mean is half its upper endpoint, again producing the corrected maximum $2Z_{\odot }$.

If the supplied $f(\mu )$ is instead retained literally as the individual birth abundance, set $q=f(\mu )$ rather than $z=1-{\mu }^Y$. The observation-time population mean is then ${\delta }_f^{-1}{\int }_{{\mu }_f}^{1}f(u)du$, not $f({\mu }_f)$. Its exact distribution can be specified parametrically usingPositivity follows because $f$ is the average of the increasing function $1-(1-\delta {)}^Y$. Its support is $0<q<f({\mu }_f)$, and normalization follows directly from $dP=d\delta /{\delta }_f$. At early consumption it is uniform with $F_{\mathrm{literal}}\simeq 2/(Y{\delta }_f)$, maximum $Y{\delta }_f/2$ and mean $Y{\delta }_f/4$. Even this literal alternative gives maximum twice the population mean, never half. The distinction between birth abundance and cumulative mean resolves the calculation; the printed inverse maximum-to-mean ratio is false under either interpretation.

For an isolated smooth mass distribution with spherical symmetry, assemble each shell $dM$ at radius $r$ against the Newtonian gravitational potential of its interior mass. The spherical shell theorem givesAssume finite total mass, finite binding energy and no central point-mass self-energy, so $M(r{)}^2/r$ vanishes at both endpoints. Integration by parts then givesFor example, $M(r)\propto r^3$ near a regular center guarantees the central boundary condition, while finite total mass guarantees the one at infinity. A divergent central binding energy would fall outside this formula's finite-energy hypotheses.

For the discrete system use the centre of mass frame, let $N$ be the particle number, and write $M_{\mathrm{tot}}=Nm$. The printed $M$ in the particle-count factor and the subsequent $n$ must be understood consistently as $N$. DefineBy Newton's second law, differentiation givesEach pair contributesTherefore the exact dynamical virial theorem is $\ddot{I}/2=2K+W$. In a stationary statistical state, or on a bounded-motion time average for which the mean $\ddot{I}$ vanishes, it becomesThere is no external pressure, tidal field or boundary work in this isolated point-particle derivation. For a time-averaged system, the following $W$ and gravitational radius refer to that same averaged binding energy.

Using $W=-G(Nm{)}^2/R_g$ and $2K=Nm⟨v_{\mathrm{eq}}^2⟩$, virial equilibrium givesThis is the three-dimensional mean squared speed about the centre of mass. For an isotropic system each one-dimensional velocity dispersion squared is one third of it.

Set the Newtonian gravitational potential to zero at infinity and exclude each particle's own field. The local escape velocity in the instantaneous frozen potential obeys $v_{\mathrm{esc},i}^2=-2{\Phi }_i$, where ${\Phi }_i=-Gm{\sum }_{j\ne i}{r}_{ij}^{-1}$. Since every pair occurs twice, ${\sum }_{i}m{\Phi }_i=2W$. Averaging over the identical particles proves the gravitational rms escape and collapse speeds relationAn actual encounter in a changing potential can exchange energy, so this is the usual local escape-speed definition, not a claim about every many-body trajectory.

Finally, dissipation-free collapse from rest at infinite separation has total energy zero. Conservation of energy at the specified binding energy therefore gives $K_{\mathrm{coll}}=-W$, twice the equilibrium kinetic energy at the same $W$. HenceThis mean squared speed includes coherent infall; a local random velocity dispersion after subtracting the infall velocity need not obey that ratio.

Write ${\sigma }_0^2=⟨v_z^2⟩=v_{\mathrm{rms}}^2/3$ for the one-dimensional thermal velocity dispersion of an isotropic monatomic ideal gas. An isothermal process keeps ${\sigma }_0$ fixed, and the vertical pressure is $P=\rho {\sigma }_0^2$. Hydrostatic equilibrium in the Newtonian gravitational potential $\Phi (z)$ givesPlanar symmetry reduces the Poisson equation for Newtonian gravity to ${\Phi }^{\prime \prime }=4\pi G\rho$. Combining the two equations yieldsThe velocity factor here is a velocity second moment, not the square of a particular particle's instantaneous vertical velocity.

Set $H^2={\sigma }_0^2/(2\pi G{\rho }_0)$ and try the symmetric self-gravitating isothermal slab profile $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$. Its logarithmic derivatives areThe chosen $H$ makes the second expression satisfy the required equation. The central conditions $\rho (0)={\rho }_0$ and ${\rho }^{\prime }(0)=0$ fix the smooth even solution uniquely. In the scale-height convention used here,The undefined constant printed as $a$ in the scale-height denominator must be $G$; this follows both from substitution and dimensions.

Integrating hydrostatic equilibrium, with the additive potential constant chosen so that $\Phi (0)=0$, givesThe orbits in a self-gravitating isothermal slab separate into horizontal and vertical motions. There is no horizontal force, so $x=x_i+v_{x}t$ and $y=y_i+v_{y}t$, with constant horizontal velocities. For collisionless test particles, or between gas collisions, conservation of energy fixes the vertical energyEvery finite $E_z>0$ has two turning points $\pm A$, whereThus the general orbit drifts uniformly in the plane while oscillating periodically across it. A particle with $E_z=0$ stays in the midplane; an orbit with nonzero horizontal velocity is generally not closed.

Near the plane, $\Phi (z)={\sigma }_0^2{z}^2/H^2+O(z^4)$, so the vertical motion is a harmonic oscillator with ${\nu }^2=2{\sigma }_0^2/H^2=4\pi G{\rho }_0$. Far from the plane, $\Phi \sim 2{\sigma }_0^2|z|/H$ and the force tends to a constant magnitude. Indeed the surface density is $\Sigma =\int \rho dz=2{\rho }_{0}H$, and that magnitude is $2\pi G\Sigma$. Large-amplitude vertical motion is therefore approximately uniformly accelerated on each side, rather than sinusoidal. Because this ideal slab extends infinitely in the horizontal directions and its potential grows without bound as $|z|\to \infty$, no finite-energy particle escapes vertically.

For an ideal gas with constant adiabatic index $\gamma >1$, let $c_v$ and $c_p$ be specific heat capacities and $\mathcal{R}=c_p-c_v$. The ideal gas law and the internal energy of an ideal gas give $P=\rho \mathcal{R}T$ and specific internal energy $e=c_{v}T$. Since $\gamma =c_p/c_v$, the internal energy per unit volume is $\rho e=P/(\gamma -1)$. Consequently a star with spherical symmetry hasThe radial integral requires the volume measure $dV=4\pi r^{2}dr$: the first printed expression omits this factor if its differential is read literally as $dr$. The mass-coordinate expression confirms the required volume interpretation.

Define the enclosed mass $M(r)$, total mass $M_{\mathrm{tot}}$ and dimensionless shell label $m=M(r)/M_{\mathrm{tot}}$. Mass conservation gives $M_{\mathrm{tot}}dm=dM=\rho dV$, so the change of variables givesHere $m$ labels a fixed material shell rather than its radial position. In a mass-preserving stellar homology, $r(m)=x{r}_0(m)$ with $x=R/R_0$. Both a shell's radius and thickness gain a factor $x$, so its volume gains $x^3$ and its mass density becomes $\rho (m)=x^{-3}{\rho }_0(m)$. An adiabatic process keeps $P/{\rho }^{\gamma }$ fixed separately in each shell, even if that constant differs between shells. Thus $P=x^{-3\gamma }{P}_0$ and $P/\rho =x^{-3(\gamma -1)}{P}_0/{\rho }_0$. Integration at fixed mass labels provesThe spherical shell theorem gives the Newtonian gravitational potential energy by assembling shells against the mass already inside them:Every radius gains the same factor $x$, while $M$ and $dM$ remain fixed. HenceThis is the homologous adiabatic stellar stability test. The derivative with respect to the physical stellar radius is ${\partial }_R=R_0^{-1}{\partial }_x$. Stationarity at $x=1$ therefore requires $-a{U}_0-W_0=0$, orAfter imposing this equilibrium relation, $E/U_0=x^{-a}-a/x$ andSince $a>0$, a strict energy minimum under this homologous displacement requiresFor $a>1$, contraction sends $E$ to $+\infty$, expansion sends it to $0$ from below, and $x=1$ is the unique minimum, with $E(1)=(1-a)U_0<0$. For $0<a<1$, contraction sends $E$ to $-\infty$, expansion sends it to $0$ from above, and $x=1$ is the unique maximum. When $a=1$, the equilibrium relation makes $E(x)=0$ for every $x$: the scale direction is marginal, not strictly stable. The plotted curves display these three possibilities. This checks homologous radial stability; stability against every stellar displacement needs a fuller perturbation analysis.