How energy loss sets the cycles left before the last stable orbit

Two clocks that differ by the mass ratio

Write the small mass as $\mu=qM$ . At $r=10$ around a non-spinning hole one orbit takes $2\pi\,r^{3/2}\approx199\,M$ . For $M=10^6\,M_\odot$ , where $M$ corresponds to 4.93 seconds, that is about sixteen minutes. The fall from there to the ISCO takes about 1.4 years when $q=10^{-5}$ .

The ratio of the two times scales as $q$ . Over any single orbit the energy changes by a fraction of order $q$ , so the orbit is a geodesic to that accuracy. This is what “adiabatic” means here.

Reading: Barack & Pound (2019)

Energy balance

Barack and Pound state the balance laws used at this order: averaged over an orbit, the energy the orbit loses equals the energy carried to infinity plus the energy that falls into the horizon. For a circular orbit we only need the energy law. Per unit of the small mass,

Here $E(r)$ is the circular-orbit energy and $\mathcal F$ the energy flux per unit $\mu$ . The wave of a circular orbit oscillates at $2\Omega_\phi$ , so it completes $\Omega_\phi/\pi$ cycles per unit time. Dividing by the radial speed turns both into integrals over radius.

\frac{dE}{dt}=-\mathcal F_\infty-\mathcal F_H,\qquad t(r_0\to r_1)=\int_{r_1}^{r_0}\frac{dE/dr}{\mathcal F}\,dr,\qquad N=\int_{r_1}^{r_0}\frac{\Omega_\phi}{\pi}\,\frac{dE/dr}{\mathcal F}\,dr

Both integrals grow as

1/q

, because the flux is proportional to

q

when written per unit

\mu

Reading: Barack & Pound (2019)

A worked example in the Newtonian limit

Take the Newtonian orbit: $E=1-1/(2r)$ , $\Omega_\phi=r^{-3/2}$ , and the quadrupole flux of a circular binary, $\mathcal F=\tfrac{32}{5}\,q\,\Omega_\phi^{10/3}=\tfrac{32}{5}\,q\,r^{-5}$ . Then $dE/dr=1/(2r^2)$ and the integrals can be done by hand.

From $r=10$ to $r=6$ with $q=10^{-5}$ this gives $t=1.70\times10^{7}\,M$ and $N=226\,842$ cycles.

t=\frac{5}{256}\,\frac{r_0^4-r_1^4}{q},\qquad N=\frac{r_0^{5/2}-r_1^{5/2}}{32\pi\,q}

These are the classic Newtonian chirp formulas, here with the small body’s mass much less than the large one.

The relativistic energy halves the count

Now keep the quadrupole formula for the flux but use the exact Kerr $E(r)$ and $\Omega_\phi$ . The slope $dE/dr$ is smaller than the Newtonian one: $0.00341$ against $0.005$ at $r=10$ , and it falls to zero at the ISCO, which is where the orbit stops being able to lose energy gradually.

Less energy to lose per unit radius means a faster fall. From $r=10$ to just outside the ISCO the count drops to $N=112\,045$ and the time to $8.96\times10^{6}\,M$ , about 1.40 years for $M=10^6\,M_\odot$ . Because both integrals scale as $1/q$ , the table below follows from one calculation.

Mass ratio $q$	Cycles from $r=10$ to the ISCO	Years for $M=10^6\,M_\odot$
$10^{-4}$	$11\,205$	$0.140$
$10^{-5}$	$112\,045$	$1.40$
$10^{-6}$	$1\,120\,453$	$14.0$

How good is the quadrupole flux?

Fujita and Tagoshi solved the Teukolsky equation for circular equatorial orbits and tabulated the luminosity summed through $\ell=6$ . Their Table VII lets us test our flux directly. Far from the hole the table, the fourth post-Newtonian series of Tagoshi and Sasaki and the quadrupole formula line up as expected: at $r=100$ and zero spin the first two agree to five figures and the quadrupole is 2.6% high.

Close in the quadrupole formula is off by tens of percent, in a direction that depends on spin and radius. The table lists the ratio of our flux to the published value.

For zero spin the errors change sign between $r=6$ and $r=10$ and largely cancel: the tabulated flux gives 112,981 cycles from $r=10$ to the ISCO, only 0.8% more than our 112,045. For spin 0.9 the quadrupole flux is too large everywhere on that range and gives 15% too few cycles between $r=10$ and $r=6$ . The tutorial in this series recomputes these numbers from the table.

Radius	Spin $-0.9$	Spin $0$	Spin $0.9$
$r=6$	$0.506$	$0.879$	$1.194$
$r=10$	$0.889$	$1.041$	$1.154$
$r=20$	$1.026$	$1.069$	$1.101$

Reading: Fujita & Tagoshi (2004) · Tagoshi & Sasaki (1994)

What this order leaves out

The horizon term. A spinning hole can absorb energy or, when the orbit turns faster than the horizon, $\Omega_\phi>\Omega_H=a/(2r_+)$ , give some of its rotational energy back to the orbit. The Fujita–Tagoshi table and our Lab both omit this term.

The next order in the phase. Barack and Pound write the orbital phase as $\varphi=\eta^{-1}\bigl[\varphi^{(0)}+\eta\,\varphi^{(1)}+O(\eta^2)\bigr]$ , with $\eta$ the mass ratio. The adiabatic term $\varphi^{(0)}$ needs only the averaged dissipative part of the first-order self-force, which is what energy balance supplies. The next term is of order one radian and needs the full first-order self-force plus part of the second-order one. Matched filtering over hundreds of thousands of cycles cannot ignore a radian.

\varphi=\frac{1}{\eta}\,\varphi^{(0)}(\tilde t)+\varphi^{(1)}(\tilde t)+O(\eta)

This is Barack and Pound’s eq. (74) multiplied out, with

\tilde t=\eta t

the slow time.

Reading: Barack & Pound (2019) · Brito

Try it in the Lab

Change the mass ratio and count the cycles

Read the wave cycles left for the default orbit.
Move the mass ratio from $10^{-5}$ to $10^{-4}$ . The count drops tenfold and the orbit picture stays the same.
Switch to Final circular orbits and compare the remaining years with the default orbit.

The Lab counts cycles at adiabatic order, along circular orbits, with the quadrupole flux.

Exercises and explained answers

Derive

N=(r_0^{5/2}-r_1^{5/2})/(32\pi q)

from the Newtonian energy and quadrupole flux.

$dE/dr=1/(2r^2)$ and $\mathcal F=\tfrac{32}{5}q\,r^{-5}$ give $dt/dr=5r^3/(64q)$ . Multiply by $\Omega_\phi/\pi=r^{-3/2}/\pi$ to get $dN/dr=5r^{3/2}/(64\pi q)$ , and integrate: $\tfrac{2}{5}\cdot\tfrac{5}{64\pi q}=\tfrac{1}{32\pi q}$ .

Why does the relativistic count stay finite even though

dE/dr

vanishes at the ISCO?

The integrand is proportional to $dE/dr$ , so a vanishing slope makes the integrand small, not large. Physically, near the ISCO a small loss of energy moves the orbit a long way inward, so the body crosses the last stretch quickly. The adiabatic picture itself fails right at the ISCO, which is why the integral is stopped just outside it.

The quadrupole flux is 15% to 19% above the published flux between

r=10

and

r=6

for spin 0.9. Should the cycle count be too high or too low?

Too low. A larger flux drains the orbital energy faster, so the body spends less time at each radius. The recomputation gives 222,719 cycles with the quadrupole flux and 260,673 with the table, a ratio of 0.854.

Primary sources and what we read in each

Barack & Pound (2019), Self-force and radiation reaction in general relativity, Rep. Prog. Phys., doi:10.1088/1361-6633/aae552
arXiv version 2, dated 19 November 2018. Section 5.2 and eqs. (72)–(74): the two-timescale phase expansion and what the adiabatic and post-adiabatic orders need; Section 5.3 on transient resonances; Section 6.1 and eqs. (77)–(78), the orbit-averaged balance laws, including negative horizon fluxes.
Fujita & Tagoshi (2004), New numerical methods to evaluate homogeneous solutions of the Teukolsky equation, Prog. Theor. Phys. 112, 415
arXiv version 1. Table VII, the gravitational-wave luminosity through l = 6 for circular equatorial orbits at seven spins; Table III, the mode-by-mode luminosity at r = 6M. The unit normalisation and the reading as flux to infinity are our inference, checked against the quadrupole and post-Newtonian limits.
Tagoshi & Sasaki (1994), Post-Newtonian expansion of gravitational waves from a particle in circular orbit around a Schwarzschild black hole, Prog. Theor. Phys. 92, 745
arXiv version 1. Eq. (43): the energy flux to infinity relative to the quadrupole value, through fourth post-Newtonian order; used only as an independent weak-field check of the flux table.
Brito, Cardoso & Pani, Superradiance, Lecture Notes in Physics 971 (2020), arXiv:1501.06570 version 8
arXiv version 8, Section 6.7, eq. 6.29 only: the energy flux into the horizon of a spinning hole changes sign when the orbital frequency exceeds the horizon frequency.

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