Codex @codex  0

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

A stellar polytrope has a spatially constant coefficient $K$ in the relation $P=K{\rho }^{1+1/n}$; $n$ is its polytropic index. This relates pressure to mass density and need not describe an adiabatic gas. Combining hydrostatic equilibrium with mass conservation eliminates the enclosed mass:For $\rho ={\rho }_c{\theta }^n$, the polytropic equation of state gives $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Choose the Lane-Emden variables for a stellar polytrope withSubstitution then gives the dimensionless Lane-Emden equation:A regular stellar centre has $\theta (0)=1$, ${\theta }^{\prime }(0)=0$ and $m(0)=0$. In particular the regular central expansion is $\theta =1-{\xi }^2/6+O({\xi }^4)$. For an isolated finite-radius stellar polytrope with negligible external pressure, the surface is the first zero ${\xi }_1$ of $\theta$, and $R=\alpha {\xi }_1$. A core embedded in an envelope can instead end at a positive matching pressure, before this zero.

The integrated Lane-Emden equation is ${\xi }^2{\theta }^{\prime }=-{\int }_0^{\xi }{s}^2\theta (s{)}^{n}ds$. Thus mass conservation gives the Lane-Emden mass formula directly:For the proposed functional form, differentiation givesChoosing $\gamma =-1/2$ removes the extra polynomial factor; equality to $-{\theta }^5$ then requires $3\beta =1$. The regular polytrope of index five is thereforeIts enclosed mass and enclosed mean mass density areThere is no finite zero: its isolated radius is infinite. Nevertheless its total mass converges, since the outer mass density falls as $r^{-5}$:The zero mean mass density is a limiting volume average, not a statement that the material has zero mass density at finite radius.

To turn this density model into a CNO cycle burning model, use gas-dominated pressure and uniform mean molecular weight. The ideal gas law and the polytropic equation of state then imply $T/T_c=\theta$. Without this thermal assumption a polytropic equation of state for total pressure does not alone fix the temperature profile. Since the stellar energy-generation rate is power per unit mass, its contribution to luminosity is $ϵdm$, giving the luminosity integral of an index-five polytrope:Put $u=\xi /\sqrt{3+{\xi }^2}$. Then $\xi =\sqrt{3}u/\sqrt{1-u^2}$ and the dimensionless integral becomes $3\sqrt{3}{\int }_0^1{u}^2(1-u^2{)}^{8}du$. Evaluation givesThe PDF's power of $\alpha$ in the proposed luminosity scaling is a genuine dimensional error: it must be three, rather than two. The volume element supplies ${\alpha }^3$; replacing it by ${\alpha }^2$ would give power divided by length, and could not be repaired by a dimensionless numerical constant.

The polytrope of index five is unlikely to model a real burning stellar core well. An embedded core has a finite boundary, and a complete isolated star cannot have this infinite radius. More importantly, strongly temperature-sensitive CNO cycle heating concentrates the luminosity centrally and tends to produce a convection zone there. A well-mixed gas-pressure-dominated convective core is closer to an adiabatic stellar polytrope of index $3/2$ than to index five. A finite truncation would also change the integral defining $A$; its value above pertains to the full mathematical model.

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

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

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

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

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

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

Efficient convection makes the interior follow a nearly adiabatic process and maintains uniform specific entropy. For the fully ionized monatomic ideal gas, the specific-heat ratio is $5/3$, so the adiabatic temperature gradient gives $d\log T/d\log P=2/5$. Integrating at a fixed time givesThe coefficient is spatially constant because the composition and specific entropy are uniform. It can change as the star loses energy. Equivalently the gas is an adiabatic stellar polytrope with $P=K_{\rho }{\rho }^{5/3}$ and polytropic index $n=3/2$; $K_{\rho }$ and the pressure-temperature coefficient $K$ are different quantities.

This fixes the dimensionless Lane-Emden equation profile throughout a stellar homology sequence. Writing $\rho (r)={\rho }_{c}f(r/R)$ in the mass integral gives ${\rho }_c=C_{\rho }M/R^3$, with a fixed positive shape constant $C_{\rho }$. Integrating the stellar hydrostatic equation through the same profile, with negligible surface pressure compared with central pressure, gives $P_c=C_{P}G{M}^2/R^4$. HenceThese are homology scalings of central stellar pressure. Substituting the central quantities into $K=P_c/T_c^{5/2}$ yieldsThus $K_0$ is constant for the fixed stellar polytrope family.

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

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

When the accumulated contraction term dominates the initial-radius term, this yields $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$. Combining it with the central ideal gas temperature gives the late-time central-temperature scalingReplacing $t-t_0$ by $t$ gives the stated late-time form. The asymptotic step also requires a time long enough to forget the initial radius, not merely a choice of zero for the age.

Let $f=0.46$. At Roche-lobe overflow contact, the supplied equilibrium radius relation givesKepler's third law then impliesThus the Roche-lobe period-mass relation for a linear donor radius law isHere $R_{\odot }$ and $M_{\odot }$ denote the solar radius and solar mass. Cancellation of the total mass is the special Roche-lobe-filling period-density relation combined with $R_2\propto M_2$.

Between eruptions the conservative binary mass transfer has $M$ constant and ${\dot{M}}_1=-{\dot{M}}_2$. Neglecting spin, writeIts logarithmic derivative givesDifferentiating the Roche lobe relation consequently givesFor $\dot{J}=0$, the Roche-lobe radius response exponent is ${\zeta }_L=2q-5/3$. If the donor follows its specified equilibrium radius sequence, ${\dot{R}}_2/R_2={\dot{M}}_2/M_2$. Its logarithmic overflow therefore changes bySince ${\dot{M}}_2<0$, $q<4/3$ gives negative feedback in this equilibrium-response model, whereas $q>4/3$ gives increasing overflow. For $q>4/3$, loss of donor mass shrinks its Roche lobe faster than its modeled stellar radius, increasing transfer and hence amplifying the original disturbance. The $q>4/3$ comparison is a formal extension; the initial system is specified to have $q<1$.

The word dynamically needs a qualification. Thermal equilibrium specifies ${\zeta }_{\mathrm{eq}}=1$, but actual dynamical stability of binary mass transfer depends on the adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}$, which need not equal one. Stability after a rapid small mass loss requires ${\zeta }_{\mathrm{ad}}>{\zeta }_L$, because $d\log M_2<0$. The equilibrium relation alone does not supply this adiabatic response. As a formal counterexample to deriving a universal dynamical threshold from it, a radius model with a dimensionless entropy label $S$, $R(M_2,S)=C{M}_2^2{e}^S$, can have equilibrium $S_{\mathrm{eq}}=S_0-\log (M_2/M_{\ast })$, giving $R_{\mathrm{eq}}\propto M_2$ but ${\zeta }_{\mathrm{ad}}=2$ at fixed $S$. At $q=3/2$, ${\zeta }_L=4/3<2$, so that model is dynamically stable despite the extrapolated $q>4/3$ condition. This shows the missing response hypothesis, not a claim that this toy radius law describes the given low-mass star.

A more relevant illustration is an ideal fully convective adiabatic stellar polytrope of index $3/2$. At fixed entropy its polytropic mass-radius relation is $R\propto M_2^{-1/3}$, giving ${\zeta }_{\mathrm{ad}}=-1/3$ and a dynamical critical ratio $q=2/3$ under the same Roche lobe approximation. For example, $q=0.9<4/3$ can be unstable dynamically even though the equilibrium comparison has negative feedback. The requested $4/3$ threshold is the intended result with instantaneous response $R_2\propto M_2$; it is not a universal dynamical criterion. The subsequent cycle calculation uses the supplied equilibrium sequence and presumes a physically stable, sufficiently slow transfer regime.

Two important angular-momentum-loss mechanisms are gravitational-wave emission from a binary system and magnetic braking of a binary star. Gravitational waves directly remove orbital energy and angular momentum. A magnetized stellar wind removes donor spin angular momentum with a large magnetic lever arm; tidal synchronization replenishes that spin from the orbit, so the net effect is orbital angular-momentum loss. Treating the wind mass loss as small compared with the Roche-transfer flow is consistent with the conservative approximation used here; otherwise the mass and angular-momentum equations must both be modified. Neglecting the stored spin does not eliminate the torque transmitted through it.

Such losses shrink the Roche lobe and offset the donor's tendency to move inside it. Maintaining contact on the equilibrium sequence requires ${\dot{R}}_L/R_L={\dot{M}}_2/M_2$, hence the binary mass-transfer contact equation becomesFor $q<4/3$ and $\dot{J}<0$, this drives ${\dot{M}}_2<0$ and maintains the modeled semi-detached binary.

Now take $\delta m>0$ for the mass expelled from the accretor in one classical nova. The white dwarf loses $\delta M_1=-\delta m$, while the donor has $\delta M_2=0$ and the total mass change is $\delta M=-\delta m$. Isotropic ejection in the accretor rest frame supplies no mean kick and carries the accretor's specific orbital angular momentum. Its orbital radius is $a_1=a{M}_2/M$, soThis is the angular-momentum budget for isotropic re-emission from a binary star, without an additional spin or frictional torque on the ejecta. The general logarithmic change of $J=M_1{M}_2\sqrt{Ga/M}$ isUsing $M/M_1=1+q$ gives the first-order nova changesThe donor mass and its assumed equilibrium radius are unchanged during the eruption. It is therefore left inside the enlarged Roche lobe, and Roche-lobe overflow ceases in this sharp-contact model. An extended atmosphere, irradiation-induced donor expansion or additional ejecta torques would require a different treatment.

The stipulated circular orbit is an important idealization. Isotropic loss that is slow compared with the orbital period can preserve circularity. If it is truly impulsive compared with that period, an initially circular point-mass orbit instead acquires eccentricity. With unchanged velocity and position, its new specific orbital energy is ${\epsilon }^{\prime }=GM/(2a)-G(M-\delta m)/a=-G(M-2\delta m)/(2a)$ and its squared specific angular momentum stays $h^2=GMa$. The Kepler orbit energy and eccentricity relations then give $a^{\prime }=a(M-\delta m)/(M-2\delta m)$ and $e^{\prime }=\delta m/(M-\delta m)$. The latter still has the same first-order separation increase. The question assumes circularity; its first-order circular budget is consistent with adiabatic orbital mass loss or with subsequent circularization that changes the separation only at second order. Work symbolically with $\delta m$, since the printed numerical layer mass has no explicit unit.

Finally put ${\mathcal{L}}_J=-\dot{J}>0$, constant over both intervals as specified, and let $J,M,q$ denote their leading-order values over one small cycle. While detached, masses and donor radius are fixed, so ${\dot{R}}_L/R_L=2\dot{J}/J=-2{\mathcal{L}}_J/J$. The fractional gap $4\delta m/(3M)$ closes afterOnce contact resumes, the contact equation gives $-{\dot{M}}_2=3M_2{\mathcal{L}}_J/[J(4-3q)]$. Accumulation of the next layer takesDividing yields the detached-to-semidetached time ratioBoth times are first-order in $\delta m/M$. Corrections to either time are second-order in that small mass fraction, so the displayed ratio has relative corrections of order $\delta m/M$. The positive denominator, negligible eruption duration relative to the cycle, fixed donor radius during detachment and common external loss rate are all essential to this idealized nova-induced binary detachment cycle.

For Roche-lobe overflow in the stipulated stellar thermal equilibrium, $R_2=R_L$. Using the mass-radius relation and Kepler's third law,The total mass cancels, givingHere $R_{\odot }$ is the solar radius. This is the Roche-lobe-filling period-density relation specialized to $R_2\propto M_2$.

Neglecting spin, the angular momentum of a circular orbit is $J=M_1{M}_2\sqrt{Ga/M}$. Between classical novae the transfer conserves total mass, so $\dot{M}=0$ and ${\dot{M}}_1=-{\dot{M}}_2$. Its logarithmic derivative givesIn particular, for conservation of angular momentum,The source PDF has $\dot{J}=0$ here; the TeX's $J=0$ is a transcription defect. For $q>4/3$, the Roche-lobe radius response exponent exceeds the equilibrium stellar radius response exponent $1$. Since ${\dot{M}}_2<0$, the Roche lobe then shrinks faster than the donor star; the overfill increases and drives more transfer, giving positive feedback. Equality at $q=4/3$ is marginal in this linear response test.

Two qualifications matter. First, $q>4/3$ is a formal extrapolation outside the stated $q<1$ range, and the supplied Roche lobe approximation need not remain accurate there. Second, true dynamical stability of binary mass transfer compares ${\zeta }_L$ with the adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}$, not ${\zeta }_{\mathrm{eq}}=1$. The stellar thermal equilibrium argument supplies the displayed feedback threshold under the imposed radius law, rather than a necessary and sufficient physical dynamical threshold. For example, the polytropic mass-radius relation for a fully convective adiabatic stellar polytrope with index $3/2$ gives ${\zeta }_{\mathrm{ad}}=-1/3$; within this same Roche lobe approximation its dynamical threshold is $q=2/3$. Thus $q<4/3$ by itself does not guarantee dynamical stability.

Gravitational-wave emission from a binary system removes orbital energy and angular momentum. Magnetic braking of a binary star provides another important loss: a magnetized stellar wind carries away donor spin, while tidal locking couples that spin to the orbit. Tides alone redistribute angular momentum and are not an external sink. Matter expelled from the system can also carry orbital angular momentum, although appreciable continuous mass loss would require nonconservative modifications of the preceding equations. Setting ${\dot{R}}_L/R_L={\dot{R}}_2/R_2$ gives the binary mass-transfer contact equationFor $q<4/3$, a negative external $\dot{J}$ therefore sustains negative ${\dot{M}}_2$ along the stipulated stellar thermal equilibrium contact sequence, provided the system is otherwise stable and can remain thermally relaxed.

During the classical nova, the ejecta carry the white dwarf's specific angular momentum, not zero angular momentum. With ${\Omega }^2=GM/a^3$ and the accretor's distance from the centre of mass $a_1=a{M}_2/M$, isotropic escape givesThe slow ejection relative to the orbital period permits the adiabatic circular orbit approximation; the donor star does not transfer appreciable additional mass during it. Take $\delta M_1=-\delta m$, $\delta M_2=0$ and $\delta M=-\delta m$. Differentiating $J=M_1{M}_2\sqrt{Ga/M}$ givesSince $M/M_1=1+q$, this yieldsThese are first-order formulae with errors of order $(\delta m/M{)}^2$. The orbital widening agrees with Jeans-mode mass loss. The donor star's radius is unchanged to this order by the assumed ejection, so its Roche lobe expands away from it, causing nova-induced binary detachment. This idealization neglects changes in the donor from irradiation or interaction with ejecta; those are not supplied in the model.

Let $\Gamma =-\dot{J}>0$ be the fixed external loss rate. During the detached binary phase both component masses are fixed, and ${\dot{R}}_L/R_L=2\dot{J}/J=-2\Gamma /J$. Closing the fractional gap $4\delta m/(3M)$ takesIn the subsequent semidetached binary, the binary mass-transfer contact equation gives $-{\dot{M}}_2=3M_2\Gamma /[J(4-3q)]$. Accumulating a fresh layer of mass $\delta m$ therefore takesThroughout these first-order estimates $J,M,q$ can be evaluated at the start of the cycle: their fractional changes during it are of order $\delta m/M$. The limit $\Gamma =0$ is excluded, since no finite reconnection time follows without an external shrinkage mechanism.

Consider a spherical stellar polytrope with $P=K{\rho }^{1+1/n}$, constant $K$, $0<n<5$, zero surface pressure and finite radius $R$. Its specific enthalpy, measured from zero density, is $h={\int }_0^{P}d{P}^{\prime }/\rho =(n+1)P/\rho$. Hydrostatic equilibrium gives $h+\Phi =\Phi (R)=-GM/R$. Integrating over mass therefore yieldsThe static stellar virial theorem gives $3\int PdV=-\Omega$, and eliminating the pressure integral proves the gravitational energy of a stellar polytrope:For an ideal gas with constant specific-heat ratio, the accompanying internal energy isTo express both energies using $\gamma$ alone one additionally identifies the model as an adiabatic stellar polytrope, so $n=1/(\gamma -1)$. ThenFor $\gamma =5/3$ these become $\Omega =-6G{M}^2/(7R)$ and $U=3G{M}^2/(7R)$. A general polytropic index need not be determined by the thermodynamic specific-heat ratio; without the adiabatic identification both $n$ and $\gamma$ remain.

A fully convective monatomic ideal gas is an adiabatic stellar polytrope with index $n=3/2$. Its polytropic mass-radius relation givesAt the base of a thin grey atmosphere, the stellar surface boundary condition and electron-scattering opacity giveThe ideal-gas equation evaluated on the convective adiabat givesCombining these relations,The Stefan–Boltzmann law then gives

A fully convective monatomic perfect-gas star follows an $n=3/2$ adiabatic stellar polytrope. At the photosphere, $\tau =2/3$ and hydrostatic equilibrium in optical depth givesFor an ideal gas on an adiabat,Evaluating this at the photosphere givesThe $n=3/2$ polytropic mass-radius relation is $R\propto K{M}^{-1/3}$, soFinally the Stefan–Boltzmann law givesThis is the specified Hayashi track.