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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 shell feels only radial gravity and the radial force due to the cosmological constant, so its torque vanishes and its specific angular momentum $L=r^2\dot{\theta }$ is conserved. Multiplyingby $\dot{r}$ and integrating gives the conserved specific orbital energy

For a uniform sphere, assembling concentric shells gives its gravitational potential energyThe cosmological-constant potential per unit mass is $-\Lambda r^2/6$. Since $⟨r^2⟩=3r_{\mathrm{ta}}^2/5$ in a uniform sphere,

The scalar virial theorem weights a potential homogeneous of degree $n$ by $-nW$. Gravity has degree $-1$ and the $\Lambda$ potential degree $2$, so the final state obeysAt turnaround $T=0$, while the virial relation gives $E_f=W_{G,f}/2+2W_{\Lambda ,f}$. Conservation of energy, together with $M=4\pi {\rho }_{\mathrm{ta}}{r}_{\mathrm{ta}}^3/3$, then gives, for $x=r_f/r_{\mathrm{ta}}$ and $\eta =\Lambda /(4\pi G{\rho }_{\mathrm{ta}})$,The root connected continuously to the $\Lambda =0$ solution has $x=1/2-\eta /8+O({\eta }^2)$, equivalentlyto first order in $\eta$. With $\Lambda =0$, virialization occurs at half the turnaround radius. At fixed turnaround state, positive $\Lambda$ makes this equilibrium root slightly smaller because its repulsive quadratic potential enters both energy conservation and the virial relation; sufficiently strong repulsion instead prevents a bound virialized state. Negative $\Lambda$ shifts the root in the opposite direction.

Let $l(R)=R^2\Omega =\sqrt{GMR}$ be the Keplerian specific angular momentum. In a source-free steady interval, the sum of advected and viscously transported angular momentum is constant:Inside the injection radius, $\dot{m}={\dot{m}}_0$. The zero-torque inner boundary condition at $R_{\mathrm{ISCO}}$ sets $C={\dot{m}}_0{l}_{\mathrm{ISCO}}$, and therefore

Outside $R_0$ there is no net mass flow in the stated steady distribution, but it must carry outward the angular momentum deposited by matter moving from $R_0$ to the ISCO. Its constant viscous torque in an accretion disk is thereforeThusThe two expressions agree at $R_0$; the jump in mass flux there is exactly the injected rate.

For $r\gg R$, the multipole expansion isThe center of mass condition removes the mass dipole. Axisymmetry makes the mixed second moments vanish and gives $\int \rho x^{\prime 2}=\int \rho y^{\prime 2}$. Undoing trace reversal therefore yieldswhereThus $a=J_z/m$ is the specific angular momentum, and the cross term is the dipole part of the slowly rotating weak-field metric.

This mode is most plausible for a nearly stationary supermassive black hole immersed in the hot, pressure-supported, X-ray-emitting atmosphere of a massive elliptical galaxy. Such gas is comparatively smooth, its random and rotational speeds can be smaller than its sound speed, and radiative cooling can be slow on the inflow scale.

The ideal Bondi accretion assumptions can fail in several independent ways. The gas may carry enough specific angular momentum to circularize into an accretion disk; a galaxy's stars and dark matter can dominate the gravitational potential outside the black hole's sphere of influence; the hole can move relative to the gas; cooling, conduction, or external heating can destroy adiabaticity or isothermality; a multiphase or clumpy medium is not a uniform reservoir; magnetic fields, turbulence, and viscosity add stresses absent from the spherical model; and AGN feedback, winds, or jets can expel or recirculate the inflow. Time dependence and self-gravity provide further failures. Thus a Bondi estimate is best interpreted as an idealized supply rate, not automatically as the rate crossing the event horizon.

Steady mass conservation gives $\dot{M}=-2\pi R\Sigma u_R$. Substituting this into the angular-momentum equation and integrating from the innermost stable circular orbit with zero torque givesThereforeUsing $\nu =\alpha c_{s}H$, $c_s\simeq H\Omega$, $l=R{u}_{\varphi }$, and $Rd\Omega /dR$ of order $-\Omega$ givesup to the order-unity Keplerian factor $3/2$.

At the sonic transition, $|u_R|\simeq c_s\simeq (H/R)u_{\varphi }$. Hencefor a geometrically thin disk with $\alpha \ll 1$. The specific angular momentum therefore differs only fractionally from $l_{\mathrm{ISCO}}$ before the gas enters the plunging region of a black-hole accretion disk. Its much shorter inflow time then prevents appreciable viscous transport, justifying angular-momentum conservation across the ISCO and the zero-torque boundary condition.

The impact gives fragments a spread of orbital energy and specific angular momentum. Their resulting spread of mean motion lets Keplerian shear stretch a compact dust clump into an arc and then a ring, while the spread of orbital frequencies causes phase mixing. Size-dependent radiation-pressure coefficients immediately place small grains on different eccentric or even radiation-pressure blowout orbits; Poynting–Robertson drag and stellar-wind drag then alter their orbits on longer timescales. Further collisions grind or disperse the clump, and planetary perturbations can accelerate mixing.

These processes depend strongly on $D_f$. Small grains have larger radiation-force-to-gravity ratios and generally shorter collisional or drag lifetimes, while larger fragments remain closer to the parent orbit but can preserve a velocity-dispersion-driven clump for longer. The lifetime also depends on collision location, ejection velocities, optical depth, and orbital radius. A universal fixed $\Delta t$ is therefore a useful population-model approximation, not a literal property of every collision; a size- and event-dependent lifetime distribution is more realistic.

If an exterior planetesimal reaches conjunction just before apoapsis, its radius is increasing. The closer pre-conjunction pull from the trailing inner planet removes more specific angular momentum than the more distant post-conjunction pull restores. Its semi-major axis falls, its mean motion rises, and the next conjunction moves later in its orbit toward apoapsis. A conjunction just after apoapsis produces the reverse imbalance: the stronger post-conjunction pull adds angular momentum, raises the semi-major axis, and shifts the next conjunction earlier. The conjunction phase is therefore restored toward apoapsis.

The symmetric configuration has one of the three conjunction branches at apoapsis and the other two symmetrically placed. It hasso the resonant argument librates about $180^{\circ }$.

Define the surface density of a disk, outward mass flux, internal torque, and surface magnetic torque byThe sign convention makes $\mathcal{G}$ positive for outward angular momentum transport in an ordinary Keplerian accretion disk. Vertical integration of mass conservation gives

The specific angular momentum is $h=r{u}_{\varphi }=r^2\Omega$. Multiply the azimuthal equation by $r$, use the continuity equation to put its left-hand side in conservative form, and integrate over $z$. The assumed decay removes the vertical mass and viscous fluxes, whereas the magnetic surface stress remains:Subtracting $h$ times the integrated mass equation yieldsSince $F_M=-(dh/dr{)}^{-1}({\partial }_r\mathcal{G}-r\mathcal{T})$, substitution in mass conservation gives the required one-dimensional advection-diffusion equation

For constant $v_m$, $M_r$ obeys the linear transport equationThe method of characteristics therefore givesThe localized profile translates inward without changing shape: material initially concentrated near $r_0$ reaches the central star after

Each annulus loses specific angular momentum as its characteristic moves to smaller $r$. The negative magnetic torque transmits that angular momentum along the large-scale field to the anchored interstellar medium. Thus the disk's angular momentum decreases even before mass crosses $r=0$; angular momentum is conserved only after the field and its anchoring medium are included.

The centrifugal criterion concerns axisymmetric disturbances, so set $m=0$. The azimuthal equation givesEliminating $w^{\prime }$ with incompressibility and then $p^{\prime }$ with the axial equation yieldswhere the Rayleigh discriminant isImpermeability at the two solid walls gives $u^{\prime }(r_i)=u^{\prime }(r_o)=0$.

Multiply the equation by $r\overline{u^{\prime }}$, integrate between the walls, and use integration by parts. The boundary terms vanish and one obtainsThe denominator is positive. Therefore $\Phi \ge 0$ throughout the annulus excludes positive real ${\sigma }^2$ and gives centrifugal stability. Since $r^4{\Omega }^2=(r^2\Omega {)}^2$ is the square of the specific angular momentum, Rayleigh's circulation criterion isAn outward decrease of squared specific angular momentum permits an axisymmetric centrifugal instability.

Write $k=\mu (1-\beta )$. Taking the cross product of the equation of motion with $\mathbf{r}$ givesSince $\mathbf{r}\cdot \mathbf{h}=0$ at every time, the orbit lies in the fixed plane perpendicular to the conserved specific angular momentum $\mathbf{h}$. Taking the inner product with $\dot{\mathbf{r}}$ gives the conserved specific orbital energyFor $\beta >1$, the effective inverse-square force is repulsive and the potential term is positive.

The force is central force, so its torque about the star vanishes. The specific angular momentum is therefore constant:Taking the scalar product of $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$ with $\dot{\mathbf{r}}$ givesThus conservation of energy gives the second constant

Put $u=1/r$. Since $h=r^2\dot{\theta }$, the radial equation becomes the Binet equationwhere primes denote derivatives with respect to the polar angle. Choosing the angular origin at periapsis givesWriting the semi-latus rectum as $h^2/\mu =a(1-e^2)$ yields the Kepler orbitHence the specific angular momentum and specific orbital energy are

The kick changes the tangential velocity by $\gamma v_K\cos (\varphi -f)$. With $s=\sqrt{1-e^2}$ and $r/a=s^2/(1+e\cos f)$, the new specific angular momentum isEvery Kepler orbit satisfies $h^{\prime 2}=\mu a^{\prime }(1-e^{\prime 2})$. Combining this identity with the value of $a/a^{\prime }$ found in part (d) giveswhere $a/a^{\prime }$ is the explicit function of $e,f,\gamma ,\varphi$ in part (d).

Here $a_p$ is the planet's semi-major axis, while $a,e,I$ are the particle's semi-major axis, orbital eccentricity, and orbital inclination relative to the planet's plane. The formula assumes the circular restricted three-body problem: the planet-to-star mass ratio is small, the particle has negligible mass, and its motion is approximately heliocentric and Keplerian away from brief encounters. The planet's own orbit is circular.

The Jacobi constant is exactly conserved in that ideal rotating problem. Expressing it in heliocentric orbital elements away from the planet gives the approximately conserved Tisserand parameterThe first term measures the particle's normalized binding energy. The second is twice the component of its specific angular momentum normal to the planet's plane, normalized by $\sqrt{G{M}_{\ast }{a}_p}$.

Define the surface density of a disk and outward radial mass flux byVertical integration of mass conservation eliminates the surface term because $\rho u_z\to 0$, and axisymmetry eliminates the azimuthal derivative. Hence

For $u_{\varphi }\simeq r\Omega (r)$, the specific angular momentum is $h=r{u}_{\varphi }=r^2\Omega$. Multiply the azimuthal momentum equation by $r$, integrate vertically and azimuthally, and define the viscous torque in an accretion diskSubtracting $h$ times the mass equation from the integrated angular-momentum equation givesFor an axisymmetric circular flow, ${\Pi }_{r\varphi }=\mu rd\Omega /dr$. Ifthen

Combining the two conservation laws givesThe first term is the local rate of change of angular momentum per radial interval, $\mathcal{F}h$ is outward advective angular-momentum flux, and $\mathcal{G}$ is outward stress-carried angular-momentum flux.

At the inner edge, the imposed torque isThereforeThe angular-momentum conservation law says that $\mathcal{F}h+\mathcal{G}$ is constant. The magnetic process supplies angular momentum $\Theta$ at the inner boundary; viscous stress passes it outward, and the mass leaving at $r_{\mathrm{out}}$ carries the injected angular momentum together with the angular momentum $\dot{M}{h}_{\mathrm{in}}$ that entered with the mass. A larger torque must therefore move the removal radius outward so that each unit mass can carry more specific angular momentum.

The specific angular momentum of an elliptic Kepler orbit is $h=\sqrt{\mu a(1-e^2)}$. At periapsis, $r=a(1-e)$, soThe mean motion understates instantaneous angular speed near periapsis. A particle in an exterior mean-motion resonance can briefly overtake the planet in angle despite completing full orbits more slowly.

If the relative acceleration is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3+\mathbf{f}$, with constant masses, then the specific orbital energy $\epsilon$, specific angular momentum $\mathbf{h}$ and eccentricity vector obeyThe inverse-square-force terms cancel in each derivative. These identities express perturbing work and torque, and track the instantaneous change of the ellipse's shape and orientation.