Extreme mass ratio inspirals, from orbits to forecasts

How orbits precess around a spinning black hole and where they stop being stable

Sandbox Physics · 2026-11-12

This chapter sets up the orbits used in the rest of the course. We work in units with $G=c=1$ and measure lengths and times in the large black hole mass $M$ . We stay in the equatorial plane, where the orbit has two motions: in and out, and around. Their frequencies decide the precession, the last stable orbit and, in later chapters, the gravitational-wave lines.

Coordinates and conventions

We use Boyer–Lindquist radius $r$ and azimuth $\phi$ . The hole has mass $M=1$ and spin $a$ with $|a|<1$ . We let $a$ carry a sign: positive when the orbit turns the same way as the hole, negative when it turns against it. The horizon sits at $r_+=1+\sqrt{1-a^2}$ .

A bound orbit moves between a periapsis $r_p$ and an apoapsis $r_a$ . We describe it with the semi-latus rectum $p$ and eccentricity $e$ , defined by $r_p=p/(1+e)$ and $r_a=p/(1-e)$ . These are labels for the turning points. Only far from the hole do they reduce to the Kepler ellipse parameters.

Reading: Glampedakis & Kennefick (2002) · Abramowicz & Kluźniak (2004)

Circular orbits

For a circular equatorial orbit, the angular frequency seen from far away, the energy per unit mass and the angular momentum per unit mass are

\Omega_\phi=\frac{1}{r^{3/2}+a},\qquad E=\frac{1-2/r+a\,r^{-3/2}}{\sqrt{1-3/r+2a\,r^{-3/2}}},\qquad L=\frac{\sqrt r\,\bigl(1-2a\,r^{-3/2}+a^2/r^2\bigr)}{\sqrt{1-3/r+2a\,r^{-3/2}}}

For

a=0

and large

r

these reduce to Kepler’s

\Omega_\phi=r^{-3/2}

and to

E\approx1-1/(2r)

, rest energy plus the Newtonian binding energy.

Reading: Abramowicz & Kluźniak (2004)

The radial frequency and the last stable orbit

Nudge a circular orbit outward and it oscillates in radius. The angular frequency of that small radial oscillation is the radial epicyclic frequency $\Omega_r$ . Abramowicz and Kluźniak derive it from the second derivative of the effective potential. For the Kerr equatorial plane it takes a compact form.

When $\Omega_r^2$ drops through zero, a nudged orbit no longer returns: the circular orbit is unstable. The radius where that happens is the innermost stable circular orbit, the ISCO. For $a=0$ the bracket vanishes at $r=6$ .

\Omega_r^2=\Omega_\phi^2\left(1-\frac{6}{r}+\frac{8a}{r^{3/2}}-\frac{3a^2}{r^2}\right)

We checked this closed form against a numerical second derivative of the Kerr effective potential; they agree to better than

2\times10^{-6}

over the spins and radii we tested.

Spin $a$	ISCO radius $r_{\rm isco}$	Binding energy $1-E$ at the ISCO	Wave frequency at the ISCO for $M=10^6\,M_\odot$
$0.9$	$2.32$	$15.6\,\%$	$14.6\,\mathrm{mHz}$
$0$	$6$	$5.72\,\%$	$4.40\,\mathrm{mHz}$
$-0.9$	$8.72$	$3.90\,\%$	$2.60\,\mathrm{mHz}$

Reading: Abramowicz & Kluźniak (2004) · Barack & Pound (2019)

How far the periapsis advances

In one radial period $T_r=2\pi/\Omega_r$ the body sweeps an azimuth $\Delta\phi=\Omega_\phi T_r$ . A closed ellipse would sweep exactly $2\pi$ . The excess is the periapsis advance per radial lap.

Worked example. Take $a=0$ and a nearly circular orbit at $r=10$ . Then $\Omega_\phi=10^{-3/2}=0.0316$ and $\Omega_r=\Omega_\phi\sqrt{1-6/10}=0.0200$ . The ratio is $1.581$ , so the periapsis advances by $360^\circ\times0.581=209^\circ$ per lap. Mercury, by comparison, advances by well under a thousandth of a degree.

Far from the hole, expand the square root: $\Omega_\phi/\Omega_r\approx1+3/r$ , which gives $\Delta\phi-2\pi\approx6\pi/r$ . At $r=1000$ this is $1.08^\circ$ ; the exact eccentric-orbit value in the Lab for $p=1000, e=0.4$ is $1.085^\circ$ .

\Delta\phi_{\rm peri}=2\pi\left(\frac{\Omega_\phi}{\Omega_r}-1\right)\;\xrightarrow{\;a=0,\ r\gg1\;}\;\frac{6\pi}{r}

The same formula holds for eccentric orbits once the two frequencies are computed for the whole orbit, as in the third chapter.

Reading: Abramowicz & Kluźniak (2004)

Eccentric orbits, the separatrix and zoom-whirl

An eccentric orbit has the energy, angular momentum and third root of its radial motion fixed by $p$ , $e$ and $a$ . Glampedakis and Kennefick write the angular momentum as $x=L-aE$ and obtain $x^2$ from a quadratic in $x^2$ . Stable bound orbits need $x^2(1+e)(3-e)<p^2$ .

The boundary $p_s(e)$ is called the separatrix. For $a=0$ it is $p_s=6+2e$ , so an orbit with $e=0.6$ must have $p>7.2$ . Just outside the separatrix the radial period grows without bound while the body keeps turning, so $\Delta\phi$ becomes very large. At $a=0, e=0.4$ and $p=7$ , only 0.2 above the separatrix, the advance is $783^\circ$ per lap: more than two extra turns near periapsis.

Spin moves the separatrix. At $e=0.4$ it is 3.90 for $a=0.7$ and 9.82 for $a=-0.9$ . At $p=12, e=0.4$ the advance is $96^\circ$ , $151^\circ$ and $250^\circ$ for $a=0.7$ , $0$ and $-0.7$ .

Reading: Glampedakis & Kennefick (2002)

Try it in the Lab

Measure the advance and find the separatrix

Set the eccentricity to its lowest value and $p=10$ with zero spin. Read the periapsis advance and compare it with the worked example.
Set $e=0.6$ and drag the orange periapsis handle inward. The notice appears at $p=7.3$ , one tenth above the separatrix.
With the orbit fixed, change the spin and watch the advance and the pink ISCO ring move together.

The Lab keeps orbits one tenth outside the separatrix so that the radial period stays finite.

Exercises and explained answers

Show that the bracket in the radial frequency vanishes at

r=6

when

a=0

, and find the circular-orbit energy there.

With $a=0$ the bracket is $1-6/r$ , which is zero at $r=6$ . Then $E=(1-1/3)/\sqrt{1-1/2}=\sqrt{8/9}=0.9428$ . The body has radiated $1-E=5.72\,\%$ of its rest energy by the time it reaches the ISCO.

For

a=0

, at what circular radius does the periapsis advance by exactly one full turn per radial lap?

We need $\Omega_\phi/\Omega_r=2$ , so $1/\sqrt{1-6/r}=2$ and $1-6/r=1/4$ . That gives $r=8$ . Inside $r=8$ the orbit makes more than one extra turn per radial lap.

Why does an orbit running against the spin have a larger ISCO?

With $a<0$ the term $8a/r^{3/2}$ in the bracket is negative, so the bracket reaches zero at larger $r$ . Physically, the frame dragging works against the orbital motion, and the orbit needs more room to stay stable. For $a=-0.9$ the ISCO is at $8.72$ .

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.
Glampedakis & Kennefick (2002), Zoom and whirl: eccentric equatorial orbits around spinning black holes and their evolution under gravitational radiation reaction, Phys. Rev. D 66, 044002
arXiv version 1. Section II, eqs. 1–24: equatorial geodesic constants in terms of p and e, the quartic for L − aE, and the separatrix condition; Appendix A. We did not use the printed integrands for the periods; ours were derived separately and checked numerically.
Abramowicz & Kluźniak (2004), Epicyclic frequencies derived from the effective potential: simple and practical formulae, arXiv:astro-ph/0411709, Astrophys. Space Sci. doi:10.1007/s10509-005-1173-z
arXiv version 1. Eqs. 20–34: circular-orbit angular velocity, energy and angular momentum in a stationary axisymmetric metric, and the radial and vertical epicyclic frequencies from the effective potential.

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