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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.

The separation of the two bodies is $a$, as is fixed by the supplied Kepler third law ${\omega }_o^2=G(M_1+M_2)/a^3$. The phrase “orbit of radius $a$” cannot denote the first body's distance from the center of mass in that formula.

Expand the companion's Newtonian potential of a point mass about the first body's center. The constant term produces no force; the linear term produces the uniform acceleration of the center and disappears in the translating frame. The leading differential acceleration therefore comes from the quadrupolar tidal forcingwhere $\mathbf{n}(t)$ points toward the companion. It is a degree-two solid spherical harmonic. For a circular nonrotating-frame orbit, its time-dependent components have $m=\pm 2$ and frequency $2{\omega }_o$, because the squared direction cosines repeat twice per orbit. There is also a static $m=0$ component. The real part of one complex harmonic represents the pair of time-dependent components; its normalization and orientation are absorbed into the order-one coefficient $\psi$. This gives $\Psi =\psi (G{M}_2/a^3)r^2{Y}_2^m$ and $\omega =2{\omega }_o$. The expansion is the leading term for $R_1/a$ small; spin is being neglected.

For a forced potential flow with the same temporal frequency, the Euler momentum equation becomesUse $U=A{r}^2{Y}_2^m$. The surface gravity perturbation of a uniform-density sphere is unchanged, as is the vanishing Lagrangian pressure perturbation boundary condition. Integrating the equation and applying that condition givesSince ${\xi }_r(R_1)=2A{R}_1{Y}_2^m$, the real surface displacement isIndeed $ϵ{\omega }_2^2=2G{M}_2/a^3$, so the normalization is consistent. This is a particular forced response away from the resonance; arbitrary free normal modes can be added.

Put $\mu =M_2/M_1$ and $q=(R_1/a{)}^3$. The tidal resonance of a stellar oscillation condition is $4{\omega }_o^2={\omega }_2^2$, henceThe resonance gives $0\le ϵ<1/2$ for finite nonnegative $\mu$, with actual forcing requiring $M_2>0$. At exact resonance the undamped linearization has no bounded response at the forcing frequency: the time-domain particular solution grows proportionally to $t$ times a sinusoid. Thus the frequency-response formula has a pole, rather than describing a finite resonant displacement. Away from that pole, linearization requires the displayed surface displacement to be small compared with $R_1$; the leading quadrupolar tidal forcing approximation separately requires $R_1/a$ small.

The transfer ellipse has semi-major axis and semi-minor axisConservation of angular momentum gives constant areal velocity $h/2$. The satellite sweeps half the ellipse, whose area is $\pi ab/2$, soUsing part (i),This is half the orbital period from Kepler's third law.

The unit binary separation and unit mean motion imply through Kepler third law thatThe centre of mass conditions put $M_1$ at $x_1=-{\mu }_2$ and $M_2$ at $x_2={\mu }_1$. Therefore

Let $a_{\mathrm{ns}}$ be the neutron star's distance from the centre of mass. In a circular orbit,Combining this with Kepler third law,gives the binary mass functionSince $\sin i\le 1$ and $(M_{\mathrm{ns}}+M_{\mathrm{wd}}{)}^2>M_{\mathrm{wd}}^2$,

When a low-mass red giant undergoes stable Roche-lobe overflow, the neutron star receives matter with substantial specific angular momentum, usually through an accretion disk. The accretion torque spins it up and weakens its external magnetic field, producing a recycled pulsar with a millisecond period. Removal of the donor's envelope exposes its helium core as a low-mass white dwarf.

At detachment, the donor mass and core mass both become $M_{\mathrm{wd}}$. Its luminosity is fixed by the red-giant core-mass--luminosity relation, and the supplied radius law therefore makes its final radius $R_L$ a function only of $M_{\mathrm{wd}}$. Cubing the Roche-lobe relation givesEliminating $a$ with Kepler third law yieldsThis is the white-dwarf mass--orbital-period relation: dependence on the neutron-star mass cancels.

For randomly oriented binaries, the isotropic binary inclination distribution isThe observed period gives $M_{\mathrm{wd}}$ from the preceding relation. If $M_{\mathrm{ns}}=1.35M_{\odot }$ is adopted, the measured mass function then givesso the inclination is inferred without astrometry.

Two plausible causes of an apparent shortage of edge-on systems are:

Scatter or bias in the core-mass--period relation can reinforce the second effect.

The component distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$, where $M=M_1+M_2$. Their orbital angular momenta add to

Let $W>0$ be the isotropic wind-loss magnitude, so ${\dot{M}}_1=-W$ and ${\dot{M}}_2=0$. The wind carries star 1's specific orbital angular momentum $j_1=a_1^2\Omega$, and henceUsing Kepler third law to write $J=M_1{M}_2\sqrt{Ga/M}$ and differentiating showsSince $P^2\propto a^3/M$, it follows thatThis is Jeans-mode mass loss.

Now allow transfer to star 2 at rate ${\dot{M}}_2>0$ while the wind continues. ThenThe same conservation of angular momentum calculation givesThe donor response $R\propto M_1^{-n}$ iswhile

Before contact, put ${\dot{M}}_2=0$. The wind drives the star farther into its Roche lobe when $\dot{\log }(R/R_L)>0$. With $q=M_1/M_2$, this condition reduces toIf the inequality is reversed, the Roche lobe expands relative to the donor, so wind loss detaches the star and no wind-driven Roche-lobe transfer is sustained; later nuclear expansion may restore contact.

During stable contact, impose $\dot{R}/R={\dot{R}}_L/R_L$ and solve for the transfer rate. Straightforward algebra givesIn the paper's signed notation $W=-\dot{M}$, this is exactlyIfthe stationary response has the wrong sign: transfer enlarges the overfill rather than removing it. Dynamical stability of binary mass transfer is lost, leading to runaway transfer and usually a common envelope or merger.

Metric compatibility and parallel transport implyUsing $s^t=R^2\Omega s^{\varphi }/f$ givesSubstitution of part d makes this independent of $t$ precisely whenthe relativistic circular-orbit form of Kepler third law. One orbit takes $T=2\pi /\Omega$, during which the spin phase advances by ${\Omega }^{\prime }T=2\pi \sqrt{1-3M/R}$. Relative to the radial direction, the spin therefore lags by the geodetic precession angle

In a circular binary star, the distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. Summing the two orbital angular momenta givesEquivalently, $J=\mu \sqrt{GMa}$ by Kepler third law.

First consider a rapid conservative perturbation. Both $M=M_1+M_2$ and $J$ are fixed, while $d{M}_1=-d{M}_2$. With $q=M_2/M_1$,soThe Roche lobe formula then gives its mass-radius exponentAfter mass loss, dynamical stability of binary mass transfer requires the donor to shrink relative to its lobe. Since $d\log M_2<0$, this means ${\zeta }_{\mathrm{ad}}>{\zeta }_L$, or

For stable secular conservative binary mass transfer, put $m={\dot{M}}_2/M_2$. The angular-momentum and contact conditions areElimination of $\dot{a}/a$ givesHere magnetic braking of a binary star removes orbital angular momentum, while the donor's expansion maintains Roche-lobe overflow.

In a cataclysmic variable, hydrogen-rich material accumulates on a degenerate white dwarf. Degeneracy prevents initial expansion from regulating its temperature, so nuclear ignition produces the thin-shell instability and a classical nova. Nuclear burning of hydrogen to helium releases about $0.007m{c}^2$, whereas the binding energy at a white-dwarf surface is only of order $G{M}_{1}m/R_1$, typically a few $10^{-4}m{c}^2$. Even modest coupling can therefore eject all the newly accreted envelope without disrupting the white dwarf.

Finally suppose every transferred mass element is expelled by isotropic re-emission from a binary star. Then ${\dot{M}}_1=0$, $\dot{M}={\dot{M}}_2$, and expelled matter carries the white dwarf's specific angular momentum $j_1=(M_2/M{)}^2{a}^2\Omega$. HenceOn the other hand, logarithmic differentiation of $J=M_1{M}_2{M}^{-1/2}{G}^{1/2}{a}^{1/2}$ and of the Roche-lobe radius givesEliminating the separation produces

Use the total massWith the stated last-orbit estimate,The Kepler third law then givesA binary's dominant quadrupole radiation has two cycles per orbit, so the last-orbit frequency of a compact binary isOnly one significant figure is meaningful under the approximations.

Since the binary separation and mean motion are both one, the Kepler third law givesThe center of mass condition puts the bodies atin the rotating reference frame. Thus the sketch has $M_1$, $O$, and $M_2$ in that order along the $x$ axis, with $P=(x,y)$, andIn the inertial sketch this entire configuration rotates uniformly about $O$.

At the exterior $(p+q):p$ mean-motion resonance,Using the Kepler third law for $M_{\ast }\gg M_1$ givesThe disturbing function is a Fourier series in integer combinations of the orbital angles. The D'Alembert characteristic permits the eccentric termAway from resonance, terms with rapidly circulating angles average away. Here, however,so ${\varphi }_1$ is a slow resonant argument. Successive astronomical conjunctions then act coherently, making this term dominate the long-period resonant dynamics even though a $q$th-order resonance has coefficient proportional to $e^q$ at small eccentricity.

In the synchronously rotating reference frame, stationary fluid obeyswhere $\Psi$ is the sum of the two gravitational potentials and the centrifugal potential. The circular binary's Kepler third law gives ${\Omega }^2=G(M_1+M_2)/a^3=GM/a^3$, hence

Uniform composition and the assumed central condensation make star 1 approximately barotropic, so one may define the specific enthalpyHydrostatic balance becomes $\nabla (h+\Psi )=0$, or $h+\Psi =\text{constant}$ throughout the star. Since $h$ is a monotone function of $P$ and $\rho$, constant-$P$ and constant-$\rho$ surfaces are equipotential surfaces of $\Psi$.

For a circular orbit, normalization and the value of $h$ giveConsequentlyThis is the relativistic form of the circular-orbit Kepler third law.