Codex @codex  0

For any indexed cubic reflection,Use ${\lambda }_{\mathrm{Mo}}\simeq 0.07107\mathrm{nm}$ for the unresolved molybdenum $K_{\alpha }$ doublet; the tabulated components are approximately $0.0709300$ and $0.0713590\mathrm{nm}$. These are external spectral constants, not an assertion about the unavailable historical Data Book.^[ref]

Reading broad ring centers in the supplied image more closely gives representative radii $5.96,6.99,10.63,13.23\mathrm{mm}$. Their corresponding estimates are about $0.3471,0.3473,0.3478,0.3480\mathrm{nm}$. Their small scatter understates possible common scale and camera-distance errors. In particular,for a first-order signed perturbation; independent uncertainties are combined in quadrature instead. A reading uncertainty of order $0.2\mathrm{mm}$, plus the printed-scale and setup uncertainty, justifies a deliberately modest precision:Nickel is a plausible identification, since it is FCC with room-temperature edge about $0.352\mathrm{nm}$.^[ref] The source photograph alone does not chemically identify the specimen; nearby elemental lattice constants and the scale uncertainty preclude a rigorous unique assignment. The estimate is not reported with artificial precision inherited from pixel fitting.

For a spherical homogeneous nucleus, the capillarity approximation givesThe first term penalizes creating an interface; below the melting point the bulk free-energy change per volume $\Delta g_V$ is negative and favors growth. Near $T_m$, neglect the temperature dependence of the latent heat. Since the free-energy difference vanishes at $T_m$,Set the radius derivative to zero:The nonzero stationary point is a maximum, since $\Delta G^{\prime \prime }(r^{\ast })=-8\pi \gamma <0$. Subcritical nuclei lower free energy by shrinking; supercritical nuclei lower it by growing. The positive radius and its inverse-supercooling dependence follow from the negative freezing enthalpy. This derivation assumes an isotropic, size-independent interfacial energy and a bulk-like spherical nucleus; it is not a molecular-scale kinetic theory.

The actual ionic radii need not equal the limiting ratio, and electronic charge distributions are not rigid sphere boundaries. In the same unlike-contact hard-sphere model, put $q=r_a/r_c$ and use $a=4(r_a+r_c)/\sqrt{3}$ to getA cation somewhat larger relative to the anion than the limiting size opens gaps between the anions. Starting from $q=1.366$ and reducing $q$ modestly reduces this packing fraction; thus a lower value is expected for that departure from the ideal contact construction. The formula is not monotonic for all imaginable ratios, so no universal inequality should be inferred without specifying radii. Coordination-dependent ionic radii, polarizability, nonionic contributions to bonding, thermal expansion and point defects also limit the model. No numerical real-crystal packing fraction is determined by the supplied data alone.

Eliminate $y$ using $y=1-x^{\prime }$ to get $x^{\prime \prime }+x=-1$. The initial data give $x(0)=1$ and $x^{\prime }(0)=0$, so the simple harmonic motion about the equilibrium $x=-1$ isIndeed, $x^{\prime }+y=1$ and $y^{\prime }-x=1$ identically, and both initial coordinates equal one. This also describes a circle of radius two around $(-1,1)$ in the phase plane.

The complex exponential identity $\sin \theta =(e^{i\theta }-e^{-i\theta })/(2i)$ givesThus the locus is a circle of radius $1/2$ centered at $i/2$, traversed counterclockwise twice as $\theta$ runs through the specified interval. Its period as a parametrized curve is $\pi$. Near $\theta =0$, the Taylor expansion gives $\Re z=\theta +O({\theta }^3)$ and $\Im z={\theta }^2+O({\theta }^4)$, so locally $\Im z=(\Re z{)}^2+O((\Re z{)}^4)$. There is a smooth horizontal tangent, not a cusp: $dz/d\theta =e^{2i\theta }$ is nonzero at each passage through the origin.

The arc length of the helix follows from its parametrization:Thus ${\ell }_h=15\pi \sqrt{5}\mathrm{m}$. Its endpoint is $(0,20,15\pi )$, at height $h=15\pi \mathrm{m}$.

Under the usual intended interpretation that the descending quarter of a circle drops through one radius, its radius is $R=h$. Its arc length is ${\ell }_c=\pi R/2=15{\pi }^2/2\mathrm{m}$, givingThe printed smoothness condition is inconsistent. The terminal tangent vector of the helix is ${\mathbf{r}}^{\prime }(3\pi )=(-10,0,5)$. Every tangent vector to a curve confined to the $yz$ plane has zero $x$ component, so no such regular curve can join this helix with a common tangent line. Moreover, the helix is still rising at that endpoint. Thus no smooth ride satisfying all the literal conditions exists.

Even after allowing a corner, a quarter of a circle reaching the ground does not alone determine its radius: the orientation of its endpoint radii must also be specified. For example, in the $yz$ plane, the circular arctraversed toward decreasing $\theta$, descends monotonically for $0\le {\theta }_0\le \pi /2$ and reaches $z=0$ when $R=h/(\sin {\theta }_0+\cos {\theta }_0)$. Choosing ${\theta }_0=\pi /4$ gives $R=h/\sqrt{2}$, whereas ${\theta }_0=\pi /2$ gives $R=h$. The numerical answer above is conditional on the intended $R=h$ geometry; it does not repair the impossible smooth join.

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

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

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

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

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

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

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

Use $\mathcal{R}$ for the stellar gas constant, reserving $R$ for the stellar radius. The specific internal energy of a monatomic gas obeying the ideal gas law is $e=P/[(\gamma -1)\rho ]=3P/(2\rho )$. At fixed enclosed mass, the first law of thermodynamics gives the heat available to the outgoing luminosity as $ϵ=-{\partial }_{t}e-P{\partial }_t(1/\rho )$ when there is no stellar nuclear fusion heating. ThereforeThe second coefficient includes both the mass density dependence of $e$ and pressure work. This is gravitational energy generation in a homologously contracting ideal-gas star.

Write the characteristic scales asThen $\rho ={\rho }_{0}b$, $P=P_{0}p$, $T=T_{0}p/b$, $m=Mq$ and $L_r=Ll$. Since $d/dr=R^{-1}d/dx$, direct substitution in the stellar hydrostatic equation and mass conservation givesFor radiative diffusion in a star, dividing the temperature equation by $T_0/R$ givesIn particular the fourth power here is of $\mathcal{R}$, not of the radius.

Because $q(x)$ is fixed and $M$ is constant, a fixed mass label has fixed $x$. Thus stellar homology gives ${\partial }_t\rho =-3\rho \dot{R}/R$ and ${\partial }_{t}P=-4P\dot{R}/R$. The heating rate reduces to $ϵ=-3P\dot{R}/(2\rho R)$, positive during Kelvin-Helmholtz contraction. Substitution in $d{L}_r/dr=4\pi r^2\rho ϵ$ now yieldsThe dimensionless profiles fix $D$ and $E$. In a common stellar homology family with fixed stellar composition and opacity normalization, the expression for $D$ impliesIt is independent of radius and constant in time for a particular fixed-mass star. The formula for $E$ gives $\dot{R}=-2EL{R}^2/(3G{M}^2)$, which integrates toWhen the initial-radius term is negligible,This is the Kelvin-Helmholtz contraction time law in the prescribed stellar homology model. Infinite initial radius is a limiting idealization; a finite initial radius retains the first term.

At main sequence arrival, CNO cycle heating with fixed stellar composition has luminosityThe integral is dimensionless and fixed within the stellar homology family. The constant-opacity radiative scaling still gives $L\propto M^3$. Equating the stellar nuclear fusion and transported luminosities givesFinally the Kelvin-Helmholtz contraction law evaluated at $R_{\mathrm{MS}}$ gives $t_{\mathrm{MS}}\propto M^2/(L{R}_{\mathrm{MS}})\propto M^{-34/19}$. In a coeval star cluster, the mass just arriving at the main sequence consequently satisfiesThese results comprise constant-opacity homologous contraction to CNO ignition; their mass exponents compare fixed-composition models with common dimensionless profiles.

Outside the emitting layer, take $m\simeq M_c$, $L_r\simeq L$ and $P=\mathcal{R}\rho T/\mu$. The Kramers' opacity law then reads $\kappa =({\kappa }_0\mu /\mathcal{R})P{T}^{-9/2}$. Division of the radiative diffusion in a star equation by the stellar hydrostatic equation givesWith the outer pressure constant of integration neglected, integration yieldsSubstitute this into hydrostatic equilibrium:Define $B=4\mu G{M}_c/(17\mathcal{R})$. The general antiderivative is $T=B/r+T_0$. The Kramers radiative-zero envelope around a stellar core sets $T_0=0$, givingThis is exact for the idealized boundary $T\to 0$ as $r\to \infty$. With a finite outer radius $R_s$ and temperature $T_s$, instead $T=T_s+B(1/r-1/R_s)$; the displayed profile is the deep-envelope approximation when $R_s\gg R_c$ and $|T_s-B/R_s|\ll B/R_c$. Small outer pressure alone does not eliminate this temperature constant of integration.

The specific hydrogen burning rate becomesConsequentlyIts local radial e-folding length is $R_c/20$, so burning is concentrated within a small fraction of the stellar core radius. The volume heating is even steeper: $\rho ϵ\propto C^2{T}^{93/4}$. Using the exterior envelope profile to estimate the thin shell's luminosity givesA finite but very extended upper limit changes the factor by $1-(R_c/R_s{)}^{81/4}$. The large exponent makes this correction small and further verifies the hydrogen burning thin-shell approximation. Since $C^2\propto M_c/L$ and $B\propto M_c$, the integral implies $L^2\propto M_c^{97/4}{R}_c^{-81/4}$. HenceThis is the fixed-radius-core shell-burning luminosity relation. The proportionality coefficient is a thin-shell estimate: luminosity rises from its value beneath the shell to $L$ through the emitting layer, so treating it as constant there uses the exterior profile rather than resolving the burning region. The scaling holds while its dimensionless shell structure and stellar composition remain fixed.

For consistency, $M_c/R_c$ must provide a hot hydrogen burning shell while the stellar core remains inert to core helium burning; the radiative envelope mass must be much less than $M_c$, and $R_c$ must be well inside an extended radiative envelope. At the shell base the gas must remain a nondegenerate gas and dominated by gas pressure, in particular $a{T}_b^4/3\ll C{T}_b^{17/4}$ with $T_b=B/R_c$. The assumed stellar core sequence must genuinely have nearly fixed $R_c$ over the mass interval considered. A cold helium stellar core supported by a nonrelativistic degenerate electron gas instead has the mass-radius law from part (a), so substituting that law would describe a different model and change the luminosity exponent.

Let $M=M_1+M_2$. The centre of mass condition gives $a_1=a{M}_2/M$, $a_2=a{M}_1/M$. For circular motion the component angular momenta are $J_i=M_i{a}_i^2\Omega$, soUsing Kepler's third law, $a^3{\Omega }^2=GM$, and $\Omega =2\pi /P_{\mathrm{orb}}$ givesThis is the circular-binary orbital angular momentum. Mass received by the companion redistributes angular momentum within the binary star; only the escaping stellar wind removes it under the prescribed model.

The stellar wind from the donor star has specific angular momentumHere ${\dot{M}}_1<0$, ${\dot{M}}_2=-f{\dot{M}}_1$ and $\dot{M}=(1-f){\dot{M}}_1$. Thus donor-wind angular-momentum loss givesTaking the logarithmic derivative of the period expression for $J$ givesUse $M_2/(M_{1}M)=1/M_1-1/M$ to simplify this toFor constant $f$, integration therefore givesThis is the donor-wind period invariant with fixed retention fraction. If $f$ changes during the evolution, the differential equation remains valid but the integrated invariant is instead $P_{\mathrm{orb}}{M}_2^3{M}^2\exp (3\int fd\log M_1)=\mathrm{constant}$. The fixed-power expression cannot be used with a time-dependent exponent without this modification.

For the Roche lobe, write $q=M_1/M_2$. The equivalent separation expression $J=M_1{M}_2\sqrt{Ga/M}$ givesTaking the logarithmic derivative of $R_L=0.46a(M_1/M{)}^{1/3}$ then adds ${\dot{M}}_1/(3M_1)-\dot{M}/(3M)$, yieldingThis donor-wind Roche-lobe response uses only the instantaneous value of $f$ and thus still holds if it varies.

Define $A(q)=2q+4q/[3(1+q)]-2$ and $B(q)=1/3-4q/[3(1+q)]$. The Roche-lobe radius response exponent is ${\zeta }_L=B+Af$, whereas the donor star's stellar radius response exponent is ${\zeta }_{\ast }=-n$. Keeping $R_1=R_L$ requires the two logarithmic derivatives of radius to agree, soFor $A\ne 0$, this determinesA physical mixture of Roche-lobe overflow and stellar wind from the donor star needs $0<f<1$. Equivalently, $-n$ must lie strictly between the pure-wind response $B$ and the conservative response $B+A=2q-5/3$. The values $f=0$ and $1$ are valid limiting prescriptions, respectively pure stellar wind and conservative mass transfer, rather than mixtures. If $f_{\mathrm{req}}$ lies outside $[0,1]$, no allowed division of the lost mass can maintain contact under this radius law and angular-momentum-loss prescription. This is feasibility of donor-wind binary contact.

The likely direction away from contact follows without assigning a nonphysical fraction. For any actual $f$, let $\Delta =\log (R_1/R_L)$. ThenSince ${\dot{M}}_1<0$, a positive value of $A(f_{\mathrm{req}}-f)$ makes the donor star move inside its lobe and detach; a negative value increases overfill and promotes stronger transfer. Thus if $A>0$, $f_{\mathrm{req}}<0$ gives increasing overfill and $f_{\mathrm{req}}>1$ gives detachment; if $A<0$, these outcomes are reversed. A runaway or a new equilibrium would require the donor star's response or the angular-momentum-loss model to change, possibly with additional driving. The contact calculation alone does not prove a particular nonlinear endpoint.

There is also a genuine exceptional denominator: $A=0$ at $q=(\sqrt{10}-1)/3$. At this ratio both limiting lobe responses coincide. Contact is possible only if $-n=B$, namely $n=(\sqrt{10}-2)/(\sqrt{10}+2)$; then every $f$ gives the same instantaneous radius response. Otherwise no fraction works, and the sign of $(-n-B){\dot{M}}_1/M_1$ determines detachment or increasing overfill.

Put $S=u^2+v^2$. At a nonzero equilibrium point the two equations form the homogeneous linear systemIts determinant must vanish, giving $S^2/2-(\mu +\sigma )S+{\mu }^2+{\sigma }^2-1=0$. ThusEach positive real polynomial root gives an antipodal equilibrium point pair; a negative polynomial root does not describe a real equilibrium point. A direction is specified by $\sin 2\theta =S/2-\mu$, $\cos 2\theta =S/2-\sigma$, where $u=\sqrt{S}\cos \theta$, $v=\sqrt{S}\sin \theta$. This is the polar form of the cubic confinement model. The origin is always an equilibrium point.

The origin's eigenvalues are ${\lambda }_{\pm }=\mu \pm \sqrt{1-{\sigma }^2}$. A simple zero eigenvalue occurs on ${\mu }^2+{\sigma }^2=1$, except at $\mu =0$. The symmetry of an odd function $(u,v)\mapsto (-u,-v)$ makes the generic steady bifurcation a pitchfork bifurcation. At a point $(\sigma ,{\mu }_0)$ on this circle with ${\mu }_0\ne 0$, vary $\mu ={\mu }_0+\delta$. Expanding the radius equation gives $S\sim 2{\mu }_0\delta /({\mu }_0+\sigma )$. In a signed, unit-eigenvector centre manifold coordinate, the cubic coefficient therefore has sign $-({\mu }_0+\sigma )/(2{\mu }_0)$.

On the lower semicircle ${\mu }_0=-\sqrt{1-{\sigma }^2}$ the transverse eigenvalue is negative: the pitchfork bifurcation is supercritical for $\sigma <1/\sqrt{2}$ and subcritical for $\sigma >1/\sqrt{2}$. On the upper semicircle it is supercritical for $\sigma >-1/\sqrt{2}$ and subcritical for $\sigma <-1/\sqrt{2}$ in the centre manifold direction, but its transverse eigenvalue is positive, so the bifurcating equilibrium points are not fully attracting. At $(\sigma ,{\mu }_0)=(\pm 1/\sqrt{2},\mp 1/\sqrt{2})$ the cubic vanishes. The radius equation instead gives $\delta =-S^2/(4{\mu }_0)+o(S^2)$, a nonzero quintic degeneracy with amplitude proportional to $|\delta {|}^{1/4}$: these are degenerate pitchforks in the cubic confinement model.

A Hopf bifurcation occurs at $\mu =0$, $|\sigma |>1$, with frequency $\sqrt{{\sigma }^2-1}$. To determine its direction, first note that the divergence is $2\mu -2S$. The Bendixson-Dulac criterion excludes every nonconstant periodic orbit for $\mu \le 0$, since this divergence is negative except possibly at the origin. We can also calculate a nonzero saturation coefficient. Set $a=|\sigma +1|$, $b=|\sigma -1|$ and $V=a{u}^2+b{v}^2$. At $\mu =0$ the linear orbits are ellipses of constant $V$. Direct differentiation givesOn a linear ellipse, write $u=\sqrt{V/a}\cos \varphi$, $v=\sqrt{V/b}\sin \varphi$. The angular speed is constant; the averages of $u^{3}v$ and $u{v}^3$ vanish, while $⟨S⟩=V|\sigma |/({\sigma }^2-1)$. Averaging the cubic terms, or removing their oscillatory parts by a periodic change of coordinates in its normal form, gives for $\varrho =\sqrt{V}$The cubic coefficient is strictly negative. Hence a small attracting limit cycle exists for $\mu >0$, andFinally $(\sigma ,\mu )=(\pm 1,0)$ have a nonzero nilpotent matrix as their linearization with two zero eigenvalues. They are reflection-symmetric double-zero bifurcations where the Hopf bifurcation and steady thresholds meet; an ordinary nonzero-frequency Hopf bifurcation calculation does not apply there. These, the two degenerate pitchfork bifurcation points, and the generic pitchfork bifurcation/Hopf bifurcation loci above exhaust the origin's local loss-of-hyperbolicity possibilities.

The trace and determinant of the origin's linearization are $2\mu$ and ${\mu }^2+{\sigma }^2-1$. Thus its open region of linear asymptotic stability isFor $|\sigma |<1$ this means below the lower semicircle; for $|\sigma |\ge 1$ all negative $\mu$ lie in the open stable region. The origin is a saddle equilibrium inside the unit circle and a source outside it with positive $\mu$.

For nonzero equilibrium points, put $\Delta =2-(\mu -\sigma {)}^2$. Existence requires $\Delta \ge 0$ and $\mu +\sigma +\sqrt{\Delta }>0$. An equivalent region isInside the circle exactly one radius polynomial root is positive, giving two equilibrium points. Outside it, with $\mu +\sigma >0$ and $\Delta >0$, both polynomial roots are positive, giving four. On $\Delta =0$ with $\mu +\sigma >0$, the two radii coalesce in a saddle-node bifurcation at each antipodal point. On the circle a zero radius is the origin and is excluded from the nontrivial count. These distinctions are shown in the parameter sketch.

At $(\sigma ,\mu )=(1/\sqrt{2},-1/\sqrt{2})$ the linearization has one simple zero and one negative eigenvalue, while the cubic centre manifold coefficient vanishes. The quintic coefficient is negative. It is a codimension-two bifurcation with a reflection-symmetric degenerate pitchfork bifurcation: a wedge of coexistence between an attracting origin and an attracting antipodal pair opens between a subcritical pitchfork bifurcation and the nearby saddle-node bifurcation curve. The intervening smaller-radius pair consists of saddles. Indeed at a nonzero equilibrium point, direct differentiation of the polar coordinates equations givesThe smaller-radius branch has negative determinant, while in the coexistence wedge the larger-radius branch has positive determinant and negative trace.

At critical boundaries the linear stability analysis test alone is inconclusive. The origin is nonlinearly attracting at a supercritical lower pitchfork bifurcation, at its stabilizing quintic endpoint, and at the supercritical Hopf bifurcation threshold, although decay is no longer exponential. It is unstable at the subcritical lower pitchfork bifurcation. The double-zero point $(\sigma ,\mu )=(-1,0)$ is also attracting: the positive function $W=v^2+u^4/8$ hasnear the origin away from zero. Here $|u{v}^3|\le (u^2{v}^2+v^4)/2$ and $|u^{3}v|\le (u^4+u^2{v}^2)/2$ prove the bound. The other double-zero point $(1,0)$ has an unstable quartic potential, as the blow-up below shows. The shading records the open linear stability analysis regions; these boundary remarks distinguish nonlinear from exponential stability.

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.