Waves, spin and the scales of a Kerr black hole

Read why a spinning black hole can make waves stronger first if the idea is new. Algebra, angular frequency and the energy levels of hydrogen are enough for this chapter.

Measure everything in units of the hole's mass

A black hole of mass $M$ sets its own length and time. Set Newton's constant and the speed of light to one, $G=c=1$. A length is then counted in units of $GM/c^2$ and a time in units of $GM/c^3$. Writing "$r=2M$" means a radius of $2GM/c^2$.

Both columns grow in proportion to the mass. Any dimensionless result, such as a frequency times $M$, holds for every mass at once; only the conversion to kilometres or seconds changes.

The Kerr horizon and how fast it turns

A rotating, uncharged black hole is described by the Kerr solution. Besides $M$ it has one more number, the spin $a=J/M$, the angular momentum per unit mass. In these units $0\le a\le M$. The horizon sits at the larger root of $r^2-2Mr+a^2=0$,

The rotating hole drags space around it. A small object hovering just outside the horizon is forced to circle with the horizon's angular velocity

In the equatorial plane the region from $r_+$ out to $r=2M$ is the ergoregion: inside it nothing can stay at rest relative to distant stars.

Worked example 1. Take $a=0.99M$. Then $\sqrt{1-0.99^2}=0.141$, so $r_+=1.141M$ and $M\Omega_H=0.99/(2\times1.141)=0.434$. For a hole of ten solar masses, $\Omega_H=0.434/(49.3\,\mu\mathrm s)=8.81\times10^3\,\mathrm{rad\,s^{-1}}$. The horizon turns about $1.40\times10^3$ times per second.

Brito, Cardoso and Pani, Superradiance, §4.7

A wave carries angular momentum with its energy

Write a wave that circles the axis as $e^{-i\omega t+im\varphi}$, where $\varphi$ is the azimuthal angle and $m$ an integer. A crest stays at fixed phase $m\varphi-\omega t$, so the pattern turns at angular velocity $\omega/m$. When the wave delivers energy $\mathrm dE$ to an absorber, it also delivers angular momentum

The ratio $m/\omega$ is fixed because the wave keeps a single frequency and a single $m$ while it scatters off an axisymmetric body. Brito, Cardoso and Pani, Superradiance, §3.5.1

When the horizon must give energy away

First consider an ordinary body turning at rate $\Omega$. In the frame that turns with it, the energy it receives from the wave is $\mathrm dE^\prime=\mathrm dE-\Omega\,\mathrm dJ$. Using the ratio above,

An absorber turns ordered wave energy into heat, so $\mathrm dE^\prime\ge0$. When $\omega<m\Omega$ the bracket is negative, and $\mathrm dE$ must be negative: the body loses energy to the wave. That is Zel'dovich's rotational superradiance.

A black hole obeys the same bookkeeping, with the horizon area $A$ in the role of entropy. Its first law is $\mathrm dM=\Omega_H\,\mathrm dJ+(\kappa/8\pi)\,\mathrm dA$, where $\kappa$ is the surface gravity, and the area theorem says $\mathrm dA\ge0$. Substituting $\mathrm dJ=(m/\omega)\,\mathrm dM$,

For $0<\omega<m\Omega_H$ the hole's mass must go down. A wave with $m\le0$ can never satisfy the condition, and neither can any wave when $a=0$.

A trapped wave: the gravitational atom

Now give the field a mass $m_b$. Its wave cannot escape if its energy is below its rest energy, so gravity holds it in bound states, much as a proton holds an electron. The strength of the binding is set by the dimensionless coupling

the ratio of the hole's size $GM/c^2$ to the particle's reduced Compton wavelength $\hbar/(m_bc)$. In units $G=c=\hbar=M=1$ the field mass and the coupling are the same number, and the coupling plays the role of the fine-structure constant.

For small $\alpha$ the states look like hydrogen. The most important one has one unit of angular momentum, $l=m=1$, and no radial node. Its frequency is close to the rest energy,

and its radial profile is $r^\prime e^{-r^\prime/2}$ with $r^\prime=r\alpha^2/M$. The probability of finding the field at radius $r$ peaks at $r=4M/\alpha^2$, so a small coupling gives a large, dilute cloud. Brito, Cardoso and Pani, Superradiance, §5.6.2

The cloud brushes the horizon on every orbit. It grows when $\omega_R<m\Omega_H$ and is slowly swallowed otherwise. Dolan, Instability of the massive Klein-Gordon field on the Kerr spacetime, §II

Worked example 2. Keep $a=0.99M$, so $M\Omega_H=0.434$. At $\alpha=0.3$, $M\omega_R\approx0.3\times(1-0.01125)=0.297$, below $0.434$: the cloud grows. Its peak sits near $r=4M/0.09=44M$, about $660\,\mathrm{km}$ for ten solar masses. At $\alpha=0.45$, $M\omega_R\approx0.439$, just above the horizon rate: the cloud decays. A full numerical solution gives $M\omega_R=0.436$ there, also above, and a decaying cloud.

Lab tasks

1. Predict the threshold frequency for $a=0.7M$ from Table 1.2. In the Lab, set the spin to 0.7 and drag the wavelength marker until the reflected-wave gain changes sign. Compare the crossing with your prediction.
2. Keep the frequency fixed and drag the spin dial down to zero. Explain from the first-law argument why the gain never returns above zero.
3. With spin 0.99, find the coupling where the cloud growth curve drops to zero. Compare it with the estimate $\alpha\approx M\Omega_H$ that follows from $\omega_R\approx m_b$.

Exercises with explained answers

1. Find $r_+$ and $M\Omega_H$ for $a=0.5M$.
2. A wave with $m=2$ and $M\omega=0.6$ meets a hole with $a=0.99M$. Is it amplified?
3. For a hole of ten solar masses, where does the $l=m=1$ cloud peak when $\alpha=0.1$?
4. Why can a wave with $m=-1$ never be amplified by a hole with $a>0$?

Next, prove the gain condition from the wave equation itself and turn these numbers into seconds and electronvolts in conserved flux, physical units and a first growth estimate.