Codex @codex  0

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 $dt=(1-e\cos E)dE/n$.

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.

Work in the planet's rest frame, neglecting its recoil and tidal forces during the close encounter. A hyperbolic Kepler orbit has positive specific orbital energy $\epsilon =v_{\infty }^2/2$ and specific angular momentum $h=b{v}_{\infty }$, where $b$ is the impact parameter. Its orbital eccentricity isAt the pericentre distance, $r_{\min }=h^2/[\mu (1+e_h)]$. Substituting $h^2={\mu }^2(e_h^2-1)/v_{\infty }^2$ yieldsThe two asymptotes have true anomalies $\pm f_{\infty }$, where $\cos f_{\infty }=-1/e_h$. Their directions of motion give the gravitational scattering angle $\theta =2f_{\infty }-\pi$, so $\sin (\theta /2)=-\cos f_{\infty }=1/e_h$. ThereforeAs a useful independent form, $b^2=r_{\min }^2+2\mu r_{\min }/v_{\infty }^2$, displaying gravitational focusing. Here $\mu =G{M}_p$ to the stated small-comet accuracy; the exact relative-coordinate formula replaces it by $G(M_p+M)$.

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.

Put ${\mu }_{\star }=G{M}_{\star }$. Conservation of specific orbital energy and specific angular momentum givesThe last comet expression is the vis-viva equation at the planet's orbit. The planet has only tangential velocity, while the comet's tangential component is $v_t=h/a_p$. Consequently its prograde relative velocity is exactly, before planetary deflection,In the near-parabolic encounter limit $Q\gg a_p$, this reduces toThe displayed approximation also requires $Q\gg a_p$. The assumptions $Q\gg q$ and $q\ll a_p$ alone do not guarantee it: an orbit with $Q=a_p$ encounters the planet near apoapsis and has $v_{\mathrm{rel}}\simeq v_p$, not $\sqrt{3}{v}_p$.

If all coplanar near-parabolic planet-crossing comets are admitted, so that $0\le q\le a_p$, the tangential component ranges from $-\sqrt{2}{v}_p$ to $+\sqrt{2}{v}_p$ when retrograde orbits are included. ThusFor prograde orbits alone the upper limit is $\sqrt{3}{v}_p$; for the stated deep-crossing limit $q\ll a_p$, both prograde and retrograde encounters approach $\sqrt{3}{v}_p$. These are the speeds outside the planet's gravitational well. The surface impact speed is $v_{\mathrm{imp}}=\sqrt{v_{\mathrm{rel}}^2+v_{\mathrm{esc},\mathrm{p}}^2}$, where $v_{\mathrm{esc},\mathrm{p}}$ is the planetary escape velocity; under negligible gravitational focusing the two speeds agree.

Assume negligible fragment release speed relative to a circular parent planetesimal. Its initial speed is $v_b=\sqrt{G{M}_{\star }/r_b}$, but radiation pressure reduces the grain's gravitational parameter to $(1-\beta )G{M}_{\star }$. Its specific orbital energy and specific angular momentum areFor $0\le \beta <1/2$, the resulting dust orbit released from a circular parent ring hasRelease is at periapsis: the inherited tangential speed exceeds the new circular speed. Low-$\beta$ grains remain near the dust birth ring; higher bound $\beta$ gives increasingly eccentric orbits extending well outside it. Uniform release longitude produces an approximately axisymmetric dust halo, with long residence times near apoapsis. At $\beta =1/2$ the conservative release orbit is parabolic; above this radiation-pressure blowout threshold it is initially unbound. For $\beta \ge 1$, zero or repulsive effective gravity also prevents a bound ellipse.

The left panel shows one release longitude; rotating it around the ring fills the halo. At longer times Poynting–Robertson drag also carries bound grains inward, so an initially outward-spanning release distribution does not imply an empty inner region in a continuously supplied, collision-free population.

The specific angular momentum is $h=\sqrt{G{M}_{\star }a(1-e^2)}$, so the periapsis angular speed isAt the exterior mean-motion resonance, neglecting radiation pressure, $n/n_p=p/(p+q)$. Thus at the limiting orbital eccentricity $e=\sqrt{2q/(5p-q)}$,The condition for overtaking the planet in angle isFor the formal second-order equilibrium, it holds for every physical bound value $0<q<5p/3$. To see this, rewrite $q/p=5e^2/(2+e^2)$, so $p/(p+q)=(2+e^2)/(2+6e^2)$. For $0<e<1$, $\sqrt{1+e}/(1-e{)}^{3/2}>1+2e$, whileThe instantaneous angular motion is therefore prograde relative to the planet near periapsis, although its mean relative motion is retrograde. The range $q\ll p$ is the regime where the eccentricity expansion is most reliable.

Put $r_S=2R_g$. On the midplane, circular orbit balance gives $r{\Omega }^2={\partial }_r\Phi =GM/(r-r_S{)}^2$. The specific angular momentum, specific orbital energy and orbital shear parameter of a Paczyński-Wiita circular orbit are thereforeFor a small radial displacement at fixed specific angular momentum, the effective potential stability criterion gives the radial epicyclic frequencyHere ${\kappa }_r$ is the question's epicyclic $\kappa$. Its square is negative for $2R_g<r<6R_g$, so a radial displacement grows instead of undergoing epicyclic motion. The orbit at $r_{\mathrm{in}}=6R_g$ is marginal, and the outer orbits are stable to small radial displacements. At this inner edge,This efficiency belongs to the Paczyński-Wiita potential; it is an approximation to the relativistic black-hole result.

The Euler-Lagrange equations of the particle Lagrangian in a shearing sheet areThe terms coupling $x$ and $y$ are the Coriolis acceleration. The cyclic coordinate $y$ has conserved canonical momentumFor a Newtonian potential of a point mass, $q_0=3/2$ and ${\Omega }_z={\kappa }_r={\Omega }_0$. Set $p_y={\Omega }_0{x}_0/2$ and substitute $\dot{y}=p_y-2{\Omega }_{0}x$ in the radial equation. It becomes $\ddot{x}+{\Omega }_0^2(x-x_0)=0$, a harmonic oscillator equation about the epicyclic guiding center $x_0$. Integration givesThe four real constants in $A,y_0,x_0$ and the two in $B$ account for the six initial position and velocity data.

Expanding the inertial specific angular momentum gives $h=h_0+r_0{p}_y+O(2)$. Thus $p_y$ measures the angular-momentum offset from the reference circular orbit, and $x_0$ specifies the radius of its associated epicyclic guiding center.

The conserved horizontal energy in the rotating frame isIt is the horizontal part of the local Jacobi energy in a shearing sheet, rather than the inertial specific orbital energy. Its positive term is the epicyclic energy; its negative term is the energy of the background shear at guiding-center position $x_0$. Independently, the vertical harmonic oscillator has conserved energy

Let $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$ and $\mathbf{v}=\dot{\mathbf{r}}$. Relative to the centre of mass, the two positions are ${\mathbf{r}}_1=-(M_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=(M_1/M)\mathbf{r}$. Their velocities have the same mass factors. Adding their kinetic energies and angular momenta gives, with reduced mass $\mu =M_1{M}_2/M$,These are the internal orbital quantities, excluding any uniform centre-of-mass motion. Newton's law of universal gravitation givesFor constant masses, differentiating $E$ makes the gravitational work cancel the derivative of the potential energy, so $\dot{E}=0$. Also $\dot{\mathbf{h}}=\mathbf{r}\times \ddot{\mathbf{r}}=0$, since the force is central; hence conservation of angular momentum gives $\dot{\mathbf{J}}=0$.

The eccentricity vector, the normalized attractive Laplace-Runge-Lenz vector, isUsing $\dot{\hat{\mathbf{r}}}=\mathbf{v}/r-\mathbf{r}(\mathbf{r}\cdot \mathbf{v})/r^3$ andshows that its derivative also vanishes. Adding a relative perturbing acceleration $\mathbf{f}$ changes these cancellations only by the perturbing terms. The perturbed Kepler-orbit conservation laws are therefore

For the distant Newtonian gravitational potential, expand each source term about the binary's centre of mass:The dipole vanishes because ${\sum }_i{M}_i{\mathbf{r}}_i=0$. With $\mathbf{a}=\mathbf{r}$, the quadratic mass moment is ${\sum }_i{M}_i{\mathbf{r}}_i{\mathbf{r}}_i=\mu \mathbf{aa}$, so the quadrupole potential of a circular binary isOn the inner binary's fast timescale, the distant companion's position is approximately fixed. Averaging over the inner circular orbit in the common plane gives $⟨{\cos }^2\theta ⟩=1/2$ and henceThis averaged potential function is independent of both time and azimuth. Consequently the companion's energy and axial angular momentum are conserved in the averaged problem. The exact motion has small fast variations; the statement is a secular approximation for a hierarchical system away from an orbital resonance.

For a circular outer orbit, force balance givesso the circumbinary orbital-period correction isConservative binary mass transfer leaves $M$ unchanged, so it does not change the leading monopole force. It changes $\mu$ and $a$, and therefore the quadrupole correction. If the inner orbital angular momentum is conserved, $J_{\mathrm{in}}=\mu \sqrt{GMa}$ gives $a\propto {\mu }^{-2}$ and $Q=\mu a^2\propto {\mu }^{-3}$. The correction is thus largest as the binary becomes strongly unequal in mass and widens, within the hierarchy $a\ll x$.

One must also allow the outer radius to respond. For slow axisymmetric evolution its specific angular momentum $h_3$ remains constant, while its energy need not: the averaged potential now depends on time. A nearly circular outer orbit has $h_3^2=GMx+3GQ/(4x)$ and $P_3=2\pi x^2/h_3$. At fixed $M,h_3$, differentiation givesThus mass exchange inside the binary can produce a small measurable outer-period change even without total mass loss. Nonconservative mass loss additionally changes the leading monopole term.

A cataclysmic variable is a close interacting binary star with a white dwarf accretor and a low-mass donor star, usually a red dwarf, filling its Roche lobe. Matter crosses by Roche-lobe overflow and commonly forms an accretion disk; sufficiently strong white-dwarf magnetism instead permits magnetically channelled accretion. Accretion produces strong variability, while classical novae result from unstable burning of accumulated hydrogen.

The cataclysmic-variable orbital-period distribution contains a short-period population between about 82 minutes and two hours, a pronounced cataclysmic-variable period gap around two to three hours, and a longer-period population above the gap. There is an accumulation near the cataclysmic-variable period minimum. These describe ordinary hydrogen-rich systems; evolved donors and helium-rich systems can have shorter periods. Selection by luminosity and outburst activity affects the observed counts, so $N(P)$ is not simply the intrinsic distribution.

For the contact red dwarf, $R_2=R_L\propto a(M_2/M{)}^{1/3}$. Using its supplied thermal-equilibrium mass-radius relation, $R_2\propto M_2$, gives $a\propto M_2^{2/3}{M}^{1/3}$. Kepler's third law then givesindependent of $M_1$. This is a special case of the Roche-lobe-filling period-density relation.

To examine rapid mass loss, use the stellar radius response exponent ${\zeta }_{\mathrm{ad}}=d\log R_2/d\log M_2=-1/3$. During conservative binary mass transfer, $M$ and $J=M_1{M}_2\sqrt{Ga/M}$ are fixed, and $d{M}_1=-d{M}_2$. ThereforeThe change in overfill is $d\log (R_2/R_L)=({\zeta }_{\mathrm{ad}}-{\zeta }_L)d\log M_2$. Since $d\log M_2<0$, overfill grows, and mass loss runs away, when ${\zeta }_{\mathrm{ad}}<{\zeta }_L$. The dynamical stability of binary mass transfer criterion here is consequentlyThe printed instability inequality is reversed. With the defined $q=M_2/M_1$, the specified radius laws imply the inequality above; $q<q_{\mathrm{crit}}$ is the stable side. Equality is marginal, and the supplied Roche-radius approximation restricts this calculation to its stated mass-ratio range.

Long-term contact is maintained by angular momentum transport out of the orbit. Gravitational-wave emission from a binary system provides one loss mechanism. Magnetic braking of a binary star uses the donor's stellar wind: its magnetic field enforces approximate corotation out to the Alfvén radius, so the wind carries much more specific angular momentum than a nonmagnetic surface outflow. At that radius, roughlyThe geometry contributes order-one factors to the torque. Even when ${\dot{M}}_w\ll |{\dot{M}}_2|$, the large lever arm can remove substantial spin angular momentum. Tidal locking couples donor spin to orbital motion, so the wind torque ultimately brakes the orbit. In practice classical novae can also eject matter; the conservative calculations here isolate the stated idealization.

For the prescribed $\dot{J}/J=-\alpha$, now use the supplied equilibrium law $R_2\propto M_2$ for the slowly evolving donor. Contact implies $\dot{a}/a=(2/3){\dot{M}}_2/M_2$, assuming negligible net wind mass loss. Taking the logarithmic derivative of $J$ givesSince $P\propto M_2$ along this equilibrium sequence,For the dynamically stable mass ratios, this is negative. The result uses thermal equilibrium as well as slow hydrostatic evolution; evolution slow compared only with the dynamical time does not by itself guarantee the radius law $R_2\propto M_2$.

In the disrupted magnetic braking model, the donor above the gap loses mass fast enough to be inflated relative to thermal equilibrium. When the donor becomes a fully convective star, the model assumes a substantial reduction in magnetic braking. The red dwarf then contracts toward equilibrium on its Kelvin-Helmholtz cooling time, becoming smaller than its Roche lobe; accretion largely stops. Gravitational-wave emission from a binary system continues to reduce the separation, carrying the detached binary through the gap until contact is restored near its lower edge.

This explanation requires strong braking immediately above the gap: the donor must already be out of equilibrium and inflated, so there is room to contract and detach when braking drops. With nearly fixed component masses during detachment, $R_L\propto a\propto P^{2/3}$, giving the illustrative inflation factorA weak braking torque that kept the donor in thermal equilibrium just above the gap would not produce this detachment. The model requires a decrease in the torque, not disappearance of the donor's magnetic field.

Interpret $\dot{m}$ as the rest-mass supply rate through a coherent disc; the hole's gravitating mass grows more slowly because radiation carries away binding energy. Initially the hole is a Schwarzschild black hole, whose innermost stable circular orbit is at $R_{\ast }=6G{M}_0/c^2$. The relativistic orbital constants there areThey follow by extremizing the timelike geodesic effective potential for a circular orbit and imposing marginal stability. Assume negligible torque inside the inner edge, so the gas carries this specific angular momentum into the hole. Its initial spin-up torque isfor $M_0=10^8{M}_{\odot }$ and $\dot{m}=0.1M_{\odot }\mathrm{y}{\mathrm{r}}^{-1}$. In SI units this is $9.7\times 10^{41}\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-2}$. A purely Newtonian estimate $\ell \sim \sqrt{G{M}_0{R}_{\ast }}$ gives a similar order of magnitude, but the relativistic value is appropriate at the innermost stable circular orbit.

The maximal Kerr black hole angular momentum at the initial mass is $J_{\max ,0}=G{M}_0^2/c\simeq 8.8\times 10^{64}\mathrm{g}\mathrm{c}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. Holding both that mass and the initial torque fixed would giveInitially ${\dot{M}}_{\mathrm{BH}}=E_{\mathrm{ISCO}}\dot{m}\simeq 0.094M_{\odot }\mathrm{y}{\mathrm{r}}^{-1}$, so the corresponding frozen-coefficient estimate of gravitating mass gain is about $2.7\times 10^7{M}_{\odot }$. These are useful initial spin-growth scales, not a self-consistent time and mass increase for reaching extremality: the target angular momentum itself grows as $M^2$, and the inner orbit changes as the spin increases.

A consistent endpoint estimate is obtained from black-hole spin-up by coherent accretion. Let $a_{\ast }=cJ/(G{M}^2)$ and $r=R_{\mathrm{ISCO}}{c}^2/(GM)$, where $M$ is the current mass. On the prograde Kerr black hole inner-orbit branch, $1\le r\le 6$, the orbital constants may be writtenIgnoring photon capture, an accreted rest mass $d{m}_0$ changes the hole by $dM=Ed{m}_0$ and $dJ=\ell d{m}_0$. Differentiating the spin definition givesSubstitution of the orbital constants gives $dr/d\log M=-2r$, henceAt the formal extremal limit $r=1$, the result isFor constant rest-mass supply, use $r=6(M_0/M{)}^2$ to integrate the elapsed time:Thus the idealized coherent-disc endpoint isIf the quoted $0.1M_{\odot }\mathrm{y}{\mathrm{r}}^{-1}$ were instead the actual gravitating mass-growth rate, the corresponding endpoint time would be $\Delta M_{\mathrm{BH}}/{\dot{M}}_{\mathrm{BH}}\simeq 1.45\times 10^9\mathrm{yr}$. The two rate conventions cannot be interchanged.