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Write $M=M_1+M_2$, $a=a_1+a_2$, and $\mu =M_1{M}_2/M$, the reduced mass. In the center of mass frame, $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Both components of the circular orbit have the same angular velocity $\Omega$. Thus their total orbital angular momentum isUsing Kepler's third law, ${\Omega }^2=GM/a^3$, and the orbital period $P=2\pi /\Omega$, gives the circular-binary orbital angular momentum

Take ${\dot{M}}_1<0$, so ${\dot{M}}_2=-f{\dot{M}}_1$ and $\dot{M}=(1-f){\dot{M}}_1$. A parcel in an isotropic stellar wind has, on average, the donor star's orbital velocity. Its wind velocity relative to the donor star averages to zero, so its mean specific orbital angular momentum about the center of mass is $j_1=a_1^2\Omega$. The escaping mass per unit time is $-(1-f){\dot{M}}_1$. With negligible stellar spin, no additional wind torque, and internal redistribution of the angular momentum of retained matter, donor-wind angular-momentum loss therefore givesIsotropy is in the donor star's frame: it does not make the escaping orbital angular momentum vanish. This is also different from isotropic re-emission from a binary star, where matter escapes from the accretor.

Logarithmically differentiate the expression for $J$ along a slowly evolving sequence of circular orbits:Substitution of the donor-wind angular-momentum loss and mass rates givesFor constant $f$, the last expression is $-3fd\log M_1/dt-3d\log M_2/dt-2d\log M/dt$. Hence the period invariant for constant-fraction donor-wind mass loss isFor a time-dependent $f$, the differential equation still holds, but this integrated power law does not. At $f=1$ it reduces to the period-product invariant for conservative mass transfer; at $f=0$, $M_2$ is constant and $P{M}^2$ is constant, as in Jeans-mode mass loss.

Set the binary mass ratio $q=M_1/M_2$. Since $a^3=GM{P}^2/(4{\pi }^2)$, the Kepler third law and the preceding orbital period derivative implyThe specified approximation to the Roche lobe gives $\log R_L=\log (0.46)+\log a+\frac{1}{3}(\log M_1-\log M)$. Consequently the donor-wind Roche-lobe response isThis is a local Roche-lobe radius response exponent, so it remains valid instantaneously even if $f$ varies.

The stellar radius response exponent of $R_1\propto M_1^{-n}$ is ${\zeta }_{\ast }=-n$. Maintaining Roche-lobe overflow in exact contact requires ${\dot{R}}_1/R_1={\dot{R}}_L/R_L$, yieldingFor a precise feasibility of donor-wind binary contact test, put ${\zeta }_0=\frac{1}{3}-\frac{4q}{3(1+q)}$ and ${\zeta }_1=2q-\frac{5}{3}$. The Roche-lobe radius response exponent is ${\zeta }_L=(1-f){\zeta }_0+f{\zeta }_1$. Thus a permitted contact fraction exists exactly whenUnless $A={\zeta }_1-{\zeta }_0=2q+4q/[3(1+q)]-2$ vanishes, the required fraction is $f=(-n-{\zeta }_0)/A$. At $q=(\sqrt{10}-1)/3$, both limiting Roche-lobe radius response exponents coincide: contact is possible for any $f$ only if $n=5/3-2q$, and for no $f$ otherwise.

The sign of the overfilling change resolves the failure of contact:If ${\zeta }_{\ast }>\max ({\zeta }_0,{\zeta }_1)$, mass loss makes the donor star underfill its Roche lobe: the system detaches and contact-driven transfer stops. If ${\zeta }_{\ast }<\min ({\zeta }_0,{\zeta }_1)$, mass loss increases the overfilling: transfer is destabilized, and rapid transfer or a common envelope may result. Calling this a failure of dynamical stability of binary mass transfer specifically requires $-n$ to be the adiabatic stellar radius response exponent; a thermal or equilibrium response concerns a different timescale. Additional angular momentum losses or intrinsic stellar expansion can change these outcomes by changing the contact equation.

A cataclysmic variable is a close semidetached binary in which a white dwarf accretes from a cool, usually low-mass donor star filling its Roche lobe. In an ordinary hydrogen-rich system the donor star is often near the lower main sequence. In a nonmagnetic cataclysmic variable (a disk-fed CV), the white dwarf's magnetic field is too weak to control the flow over the disk. This leaves a characteristic disk-fed geometry, summarized in the original schematic below. NASA's introduction to cataclysmic variables describes the basic components.

Roche-lobe overflow passes through the inner Lagrange point $L_1$. A nearly ballistic stream bends in the rotating frame and strikes the outer accretion disk, producing an accretion-disk stream-impact hot spot. Its retained angular momentum prevents direct radial infall. Viscous evolution of an accretion disk transports angular momentum outward while gas moves inward through a nearly Keplerian accretion disk. Close to the white dwarf, gas slows from orbital rotation toward the stellar rotation in an accretion-disk boundary layer. The disk, accretion-disk stream-impact hot spot, accretion-disk boundary layer, white dwarf and donor star all contribute to the spectrum and, where the orbital inclination permits, to the eclipses of an eclipsing binary. In particular, “nonmagnetic” describes the accretor's control of the flow; it does not imply that the donor star cannot sustain a magnetic field.

The gravitational power available from accretion is approximately $L_{\mathrm{acc}}=G{M}_{\mathrm{WD}}{\dot{M}}_{\mathrm{acc}}/R_{\mathrm{WD}}$. For a slowly rotating white dwarf and a thin steady Keplerian accretion disk, the specific energy changes from approximately zero far out to $-G{M}_{\mathrm{WD}}/(2R_{\mathrm{WD}})$ at the inner disk. Roughly half the available power is radiated by the accretion disk; the rest is released in the accretion-disk boundary layer as the orbital kinetic energy is dissipated. Stellar rotation and departures from a steady thin disk change this partition.

A classical nova has a different energy source. Transferred hydrogen accumulates on the white dwarf; compression heats the base of its envelope until temperature-sensitive hydrogen burning accelerates. Electron degeneracy pressure initially weakens the expansion response to heating, helping a thermonuclear runaway develop. The envelope subsequently expands and ejects material, producing a large optical outburst followed by a decline as the ejecta expand and residual burning ends. The white dwarf usually survives, so continued accretion can build another fuel layer. A rough recurrence scale is the ignition-envelope mass divided by the mean accretion rate; both this mass and the rate vary strongly among systems. The event is an envelope eruption, and the retained fraction is not automatically unity. Starrfield, Iliadis and Hix's nova calculations explains this nuclear mechanism. A classical nova is powered by unstable nuclear burning on the white dwarf.

A dwarf nova undergoes recurrent, shorter brightenings powered principally by enhanced gravitational accretion. The hydrogen-ionization disk instability creates cold, mostly neutral and hot, ionized branches of the disk's accretion-disk thermal S-curve, separated by unstable equilibria. In quiescence the cool disk stores matter because inward transport is slow. Once a critical surface density is reached, a heating transition puts the disk into a hotter, more state with higher effective viscosity: the inward accretion rate and luminosity rise and the disk drains. A cooling transition returns it to quiescence, completing the cycle. A persistent increase in the donor star's transfer rate is not required. Sufficiently high transfer rates can keep the disk on its hot stable branch, giving a nova-like variable rather than ordinary disk cycles. Lasota's disk-instability analysis and Hameury's disk-instability review develop this picture. A dwarf-nova outburst is a disk instability, not a white-dwarf thermonuclear explosion. The names classify mechanisms and need not identify permanently distinct binaries: a nova-producing binary can also possess an unstable disk between nuclear eruptions.

The cataclysmic-variable orbital-period distribution is not smooth. For ordinary hydrogen-rich cataclysmic variables, prominent features are a cataclysmic-variable period gap around two to three hours, a cataclysmic-variable period minimum near eighty minutes, and an accumulation near that minimum. These are population features rather than absolute exclusions. Selection effects matter: luminous high-accretion rate systems are easier to find than faint evolved systems. Helium-transferring binaries have a different period range and are not described by the hydrogen-rich minimum. Gänsicke and collaborators' period-minimum study documents the observed accumulation.

The Roche-lobe-filling period-density relation makes the orbital period a measure of donor structure. Combining $R_d\simeq 0.46a(M_d/M{)}^{1/3}$ with Kepler's third law givesAs the donor star loses mass, its stellar radius response exponent $\zeta =d\log R_d/d\log M_d$ impliesA donor with $\zeta >1/3$ evolves toward shorter orbital periods. When its effective response falls below $1/3$, continued mass loss instead lengthens the orbital period: this is the cataclysmic-variable period bounce. A very low-mass donor may be substellar and increasingly affected by electron degeneracy pressure; the ideal degenerate scaling $R_d\propto M_d^{-1/3}$ illustrates the reversal. The precise cataclysmic-variable period minimum depends on thermal disequilibrium and the strength of orbital angular momentum loss. Near a turning point $|\dot{P}|$ is small; for an approximately steady evolutionary flow of systems, the number per period interval scales as $N(P)\propto 1/|\dot{P}|$, explaining the accumulation. Knigge, Baraffe and Patterson's donor-based evolutionary study relates the donor star's response to these period features.

Long-term transfer is driven mainly by losses of orbital angular momentum, rather than by disk outbursts. Magnetic braking of a binary star removes the cool donor star's spin through a magnetized stellar wind. Tidal synchronization makes the orbit replenish that spin, so the wind extracts orbital angular momentum. This is usually the dominant standard driving mechanism above the cataclysmic-variable period gap. Gravitational-wave emission from a binary system supplies a baseline loss, especially important below the gap. For a weak-field, slowly moving circular orbit, the circular gravitational-wave inspiral givesBecause ordinary transfer from the lighter donor star tends to expand its Roche lobe if orbital angular momentum is conserved, an external loss is needed to sustain contact. In the conservative contact approximation, the binary mass-transfer contact equation readswhere the stellar radius response exponent must match the evolutionary timescale and additional donor expansion has been neglected. The positive denominator on the stable branch makes ${\dot{J}}_{\mathrm{ext}}<0$ drive ${\dot{M}}_d<0$. Nova ejecta can introduce additional nonconservative binary mass transfer. Knigge's evolutionary discussion describes the standard loss mechanisms and their limitations.

In the disrupted magnetic braking model, relatively rapid mass loss above the cataclysmic-variable period gap keeps the donor star inflated relative to stellar thermal equilibrium. When the donor approaches the fully convective star transition, the model postulates a substantial reduction in magnetic braking of a binary star. The donor star can contract within its Roche lobe, suppressing Roche-lobe overflow near the upper edge of the gap. Gravitational-wave emission from a binary system continues to shrink the detached binary; near the lower edge the Roche lobe again reaches the donor radius and transfer resumes. No mass transfer is needed during the detached crossing. The density relation predicts a radius ratio $(3/2{)}^{2/3}\simeq 1.31$ between contact at three and two hours if the masses remain nearly fixed, illustrating the required inflation before detachment. The torque reduction is a model ingredient, not a claim that all fully convective stars lose their magnetic fields. Zorotovic and collaborators' detached-binary study tests the predicted detached population in the gap.

The standard formation channel starts with an initially wider binary star containing two main sequence stars. The initially more massive component evolves first and becomes a giant. Unstable Roche-lobe overflow can engulf the companion in a common envelope. Drag causes inward orbital motion, releasing two-body orbital energy and transferring angular momentum to the envelope. If the envelope is expelled before merger, a close detached binary survives, containing the exposed core, which becomes a white dwarf, and the lower-mass companion. This explains how a binary becomes much tighter than the giant progenitor's radius would have allowed. The outcome depends on envelope binding and the efficiency of energy deposition, summarized approximately by the common-envelope energy formalism; ejection is not guaranteed. Ivanova and collaborators' common-envelope analysis discusses the relevant physics and uncertainties.

Subsequent magnetic braking of a binary star and gravitational-wave emission from a binary system shrink the detached binary until its companion fills its Roche lobe. Stable Roche-lobe overflow, for a suitable binary mass ratio and stellar radius response exponent, then creates a cataclysmic variable. Its later nuclear eruptions, disk cycles and secular orbital period evolution occur on different timescales. The formation sequence is a wide binary, envelope ejection, a close detached white-dwarf binary, and angular-momentum-driven contact.

For constant component masses, the center of mass frame has ${\mathbf{r}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=-(M_1/M)\mathbf{r}$. Newton's law of universal gravitation gives the relative equation $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. Substitution in the component angular momenta yieldsHere $\mu =M_1{M}_2/M$ is the reduced mass, while $\mathbf{h}$ is specific angular momentum.

The printed total-energy expression lacks a factor of $\mu$ in its kinetic term. With the stated physical separation coordinate, the dimensionally consistent total two-body orbital energy isThe second expression is specific orbital energy; these two conventions must not be mixed. Its conservation follows directly:For the eccentricity vector, differentiate $GM\mathbf{e}=\dot{\mathbf{r}}\times \mathbf{h}-GM\hat{\mathbf{r}}$. Since $\mathbf{h}$ is constant,The cancellation is the vector triple product identity. Thus all three corrected Kepler integrals are conserved. Dotting the eccentricity relation with $\mathbf{r}$ also gives the useful orbit identity

For ballistic streamline focusing by a moving star, work in the star's rest frame and take the incoming gas to have velocity $v{\mathbf{e}}_x$, with downstream behind the star at $x>0$. A streamline with impact parameter $d>0$ approaches from $(-\infty ,d,0)$. Its upstream specific angular momentum and eccentricity vector areAt its downstream axis crossing, $\mathbf{r}=b{\mathbf{e}}_x$. The orbit identity gives $b=d^2{v}^2/(GM)-b$, hence the collision distance isAt that point, $h_z=b{u}_y=-dv$, while the $y$ component of $GM\mathbf{e}=\mathbf{u}\times \mathbf{h}-GM\hat{\mathbf{r}}$ gives $dv{u}_x=d{v}^2$. ThereforeA symmetric streamline with opposite impact parameter has the opposite transverse velocity. Their shock wave removes the opposing transverse motion while preserving the common downstream component. Thus the remaining velocity is $v{\mathbf{e}}_x$ in the star frame. In the frame where the undisturbed medium is stationary, adding the star's velocity gives zero velocity immediately after this idealized collision.

After transverse kinetic energy is dissipated, the remaining specific orbital energy is ${\epsilon }_{\mathrm{after}}=v^2/2-GM/b$. It is negative if $b<2GM/v^2$, or equivalently if the upstream impact parameter satisfies $d<d_{\mathrm{cap}}=2GM/v^2$. This is the post-shock capture criterion in ballistic accretion; bound axial streams can return to the star. The captured incident mass flux through the corresponding disk gives the ballistic Bondi--Hoyle--Lyttleton accretion rate:This is the pressureless, dissipative capture model, with the shock's transverse energy unavailable to unbind the wake.

For wind mass transfer in a binary star, steady isotropic donor mass loss gives the local wind mass density at the companion:A fast wind has $v_w\gg 2\pi a/P$ and a capture scale small compared with $a$, so its relative incident speed is $v_w$ to leading order. Apply the same Bondi--Hoyle--Lyttleton accretion rate with accretor mass $M_2$ to obtainFinally, Kepler's third law eliminates the unquoted separation, $a=[G(M_1+M_2)P^2/(4{\pi }^2){]}^{1/3}$, giving the rate in terms of the stated orbital period:Equivalently, the fast-wind accretion fraction in a circular binary is $[M_2/(M_1+M_2){]}^2(v_{\mathrm{orb}}/v_w{)}^4$, with $v_{\mathrm{orb}}=2\pi a/P$ the relative circular orbital speed. Its smallness is consistent with using an almost undisturbed isotropic stellar wind.

The printed line-density statement is incorrect for a phase-mixed orbit. A steady line density on a Kepler orbit is inversely proportional to speed: for a cross-sectional-area current $J_{\sigma }$, the area per unit arc length is $d\sigma /ds=J_{\sigma }/v$. Equivalently, phase mixing gives $d\sigma =({\sigma }_{\mathrm{tot}}/P)dt$, uniform in mean anomaly, where $P=2\pi \sqrt{a^3/\mu }$ is the orbital period.

For an optically thin population of blackbodies in radiative equilibrium, a fragment absorbs $\sigma L_{\star }/(4\pi r^2)$ and reradiates the same luminosity. Thus the fractional luminosity of a phase-mixed eccentric wire isThe specific angular momentum gives $dt/r^2=d\theta /h$, so the integral is $2\pi /h$. Since $Ph=2\pi a^2\sqrt{1-e^2}$,This derives the intended result after explicitly correcting the density to $1/v$.

The error is consequential. If one instead imposes the literal density $d\sigma /ds\propto v$, its time weighting is $v^{2}dt$. Using $⟨v^2⟩=\mu /a$, $⟨r^{-2}⟩=[a^2\sqrt{1-e^2}{]}^{-1}$ and $⟨r^{-3}⟩=[a^3(1-e^2{)}^{3/2}{]}^{-1}$ givesAt $e=1/2$ this is $5/3$ times the printed result. The inverse-radius averages can also be obtained directly with eccentric anomaly $E$, $r=a(1-e\cos E)$ and $dt=(1-e\cos E)dE/n$.

On the corrected steady phase-mixed orbit, each fragment crosses a fixed orbital longitude once per orbital period. Hence the cross-sectional-area current through that point isFrom the fractional luminosity of a phase-mixed eccentric wire, ${\sigma }_{\mathrm{tot}}=4\pi a^2\sqrt{1-e^2}f=2Phf$. Therefore the area rate at the foreground crossing of the line of sight isOnly the foreground segment blocks the star; the far-side intersection does not add another occultation current. The orbital area current is constant because the line density varies as $1/v$, even though the local orbital speed varies.

Put $A=\beta \mu /c$. The Poynting–Robertson drag acceleration has $\bar{R}=-2A\dot{r}/r^2$ and $\bar{T}=-A{v}_{\theta }/r^2$. Its work is strictly dissipative:For an unperturbed elliptic orbit, use $dt=r^{2}df/h$, $\dot{r}=(\mu /h)e\sin f$ and $v_{\theta }=(\mu /h)(1+e\cos f)$. Averaging over its orbital period $P$ givesSince $\dot{a}=2a^2\dot{\epsilon }/\mu$, $P=2\pi a^{3/2}/\sqrt{\mu }$ and $h^3={\mu }^{3/2}{a}^{3/2}(1-e^2{)}^{3/2}$, the Poynting–Robertson decay of a circumstellar orbit isFor a circular Kepler orbit this becomes $-2A/a$. This is a secular first-order average, not an exact instantaneous decay law. If static radiation pressure is appreciable, use a bound conservative orbit with central parameter ${\mu }_{\mathrm{eff}}=(1-\beta )\mu$, requiring $0\le \beta <1$, and define its orbital elements accordingly; repeating the work calculation cancels ${\mu }_{\mathrm{eff}}$ and leaves the same coefficient $A=\beta G{M}_{\star }/c$.

Treat $a,e$ as instantaneous osculating orbital elements of the fixed-central-parameter Kepler orbit. The perturbing work changes the specific orbital energy according toThe specific angular momentum and conic equation giveSubstituting proves the Gauss planetary equation for the semi-major axis:Here $f$ denotes true anomaly, not the fractional luminosity used in Question 1. The equation follows from work alone; a normal perturbing acceleration has no instantaneous contribution because the velocity lies in the orbital plane. For slowly averaged evolution, the perturbation must be weak over one orbital period.

Let ${\mu }_p=G{M}_p$, $A=\beta G{M}_{\star }/c$ and $\kappa =A/a_p^2$. A tightly bound circumplanetary orbit lies well inside the Hill sphere, so $a/a_p\ll 1$ and its orbital period is short compared with the planet's year. Treat the stellar flux and stellar direction $\hat{\mathbf{R}}$ as constant during one dust orbit. With stellar-frame velocity ${\mathbf{V}}_p+\mathbf{u}$, the velocity-dependent acceleration isThe terms independent of $\mathbf{u}$, including the leading static radiation pressure, do no net work on an unperturbed closed circular orbit. HenceFor a coplanar circular orbit, $u^2={\mu }_p/a$ and the component along the stellar direction has mean square $u^2/2$. Thus $\dot{a}=2a^2{\dot{\epsilon }}_p/{\mu }_p$ yields the Poynting–Robertson decay of a circumplanetary orbitThe coefficient three assumes coplanarity, which the PDF does not state. For orbital normal $\hat{\mathbf{k}}$, the general circular covariance is $⟨u_i{u}_j⟩=(u^2/2)({\delta }_{ij}-k_i{k}_j)$, givingAn orbital plane normal to the stellar radial direction has coefficient two during that orbit, a counterexample to an orientation-independent coefficient three. Averaging also over the planet's circular stellar orbit, with fixed dust-plane orbital inclination $I$ to it, gives coefficient $3-\frac{1}{2}{\sin }^{2}I$. The conservative forces must be weak enough for the assumed approximately circular, planet-bound orbit to persist; the small planet-to-star mass ratio alone does not ensure this for arbitrary grain $\beta$.

Write $x=1/a$ for the comet's inverse semi-major axis. Its specific orbital energy is $-G{M}_{\star }x/2$, so escape corresponds to $x=0$. The printed quantityhas dimensions of inverse length. It cannot, literally, be a standard variance-per-time diffusion coefficient. Interpret it as the characteristic root mean square step in $x$ per periapsis passage, as in a discrete comet energy diffusion model.

If successive kicks are unbiased and uncorrelated, after $N$ passages the root mean square displacement is $s_x\sqrt{N}$. Starting at $a_0$ gives $x_0=1/a_0$, hence $N_{\mathrm{ej}}\sim (x_0/s_x{)}^2$. Taking the characteristic interval to be the initial orbital period $P(a_0)$ givesHere $P_p=2\pi \sqrt{a_p^3/(G{M}_{\star })}$. In particular, for $a_0\sim a_p$ this is $P_p(M_{\star }/M_p{)}^2/100$. Without specifying the initial semi-major axis and the time per statistically independent encounter, the PDF cannot determine a unique time.

With the convention $⟨(\Delta x{)}^2⟩=2D_{x}t$, the actual diffusion coefficient would be $D_x=s_x^2/[2P(a_0)]$. The estimate requires $N_{\mathrm{ej}}\gg 1$, orbit crossing, and sufficient phase decorrelation. Near escape the orbital period grows, and resonant or secular correlations invalidate the simple constant-step clock. This is a characteristic diffusion scale, not an exact mean first-passage time: even an unbiased Brownian motion on an unbounded half-line has an infinite mean time to reach its absorbing endpoint.

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.

Set $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $M=M_1+M_2$ and $\mu =M_1{M}_2/M$, the reduced mass. Newton's law of universal gravitation gives ${\ddot{\mathbf{x}}}_1=-G{M}_2\mathbf{r}/r^3$ and ${\ddot{\mathbf{x}}}_2=G{M}_1\mathbf{r}/r^3$. Subtraction establishesFor fixed masses, the orbital energy and angular momentum areTaking derivatives gives $\dot{E}=\mu \mathbf{v}\cdot (\ddot{\mathbf{r}}+GM\mathbf{r}/r^3)=0$ and $\dot{\mathbf{J}}=\mu \mathbf{r}\times \ddot{\mathbf{r}}=0$. These are conservation of energy and conservation of angular momentum in the unperturbed Kepler orbit. To see directly that the orbital eccentricity is fixed, put $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ and $\mathbf{n}=\mathbf{r}/r$. The eccentricity vector isSince $\dot{\mathbf{h}}=0$, the cross product identity $\mathbf{r}\times (\mathbf{r}\times \mathbf{v})=\mathbf{r}(\mathbf{r}\cdot \mathbf{v})-r^2\mathbf{v}$ gives $\ddot{\mathbf{r}}\times \mathbf{h}=GM\dot{\mathbf{n}}$, so $\dot{\mathbf{e}}=0$. Its magnitude $e$ is the orbital eccentricity. Equivalently,Thus the semi-major axis and orbital eccentricity of a bound noncollision Kepler orbit are constant, including $e=0$.

Take the centre of mass as origin. The two positions are ${\mathbf{x}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{x}}_2=-(M_1/M)\mathbf{r}$. For $x\gg r$, a Taylor expansion givesThe mass dipole vanishes because ${\sum }_A{M}_A{\mathbf{x}}_A=0$, and the second mass moment tensor is ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$. The gravitational quadrupole potential of a point-mass binary is consequentlyBoth printed tensors are trace-free. Contracting them with the Kronecker delta summed over three spatial dimensions givesThis proves the requested expression through second order. In particular the printed convention is $q_{ij}=\frac{3}{2}{Q}_{ij}$, where $Q_{ij}=\mu (r_i{r}_j-r^2{\delta }_{ij}/3)$ is the standard mass quadrupole moment; the radiation coefficient must be adjusted accordingly. No mass dipole term may be retained in centre of mass coordinates.

To evaluate the quadrupole formula without assuming a circular orbit, put $k=GM$, $u=\dot{r}=\mathbf{n}\cdot \mathbf{v}$ and ${\mathbf{v}}_{\perp }=\mathbf{v}-u\mathbf{n}$. Differentiating the gravitational acceleration gives the kinematic jerkFor $S_{ij}=r_i{r}_j$, the product rule then givesInserting these into the printed mass quadrupole moment convention yieldsThe two tensors in brackets are orthogonal in their tensor contraction, because $\mathbf{n}\cdot {\mathbf{v}}_{\perp }=0$. Their squared norms are $6$ and $2v_{\perp }^2$, respectively. HenceThe instantaneous quadrupole luminosity of a Kepler binary is thereforeThe radial derivative $\dot{r}$ is not the vector speed. The bracket equals $v_{\perp }^2+u^2/12$, so the radiated luminosity is nonnegative for every instantaneous velocity.

For a slowly evolving circular orbit, there is no preferred orbital phase at which to excite a persistent eccentricity vector. More precisely, the rotating mass quadrupole moment emits at twice the orbital frequency, with angular harmonic number two; its gravitational-wave energy and angular-momentum balance is $\dot{E}=\Omega \dot{J}$. The circular Kepler orbit sequence has $J=\mu \sqrt{GMa}$ and $dE/dJ=\Omega$, so this loss is tangent to the sequence. At leading adiabatic order it preserves zero secular orbital eccentricity. This is a circular gravitational-wave inspiral, with a small radial drift rather than an exactly fixed-radius Newtonian circle. The energy flux alone would not establish circularity without this symmetry and angular momentum balance.

To leading radiation order use $r=a$, $u=0$ and ${\mathbf{v}}^2=GM/a$ in the loss formula. Combining it with $E=-GM\mu /(2a)$ givesKeeping the fixed masses and leading quadrupole formula, integration givesThis formal coalescence time displays the strong $a_0^4$ dependence: gravitational-wave emission from a binary system matters far more for close binary stars than for wide ones. The orbital period decreases as $a^{3/2}$ and the gravitational-wave frequency increases, producing a chirp. A detached binary can be driven into Roche-lobe overflow; further evolution then depends on mass-transfer stability. A sufficiently close compact pair may merge. The weak-field, slow-motion, point-mass approximation ceases to apply before literal $a=0$; finite stellar radii or strong relativistic effects determine the final interaction.