Conserved flux, physical units and a first growth estimate

This chapter assumes the results of Waves, spin and the scales of a Kerr black hole: the units $G=c=M=1$, the horizon radius $r_+$ and its rotation rate $\Omega_H$.

The radial equation

A scalar field of mass $\mu$ (in units of $1/M$) on a Kerr background separates into a time factor $e^{-i\omega t}$, an angular factor $e^{im\varphi}S(\theta)$ and a radial function $R(r)$. The radial function obeys

with $\Delta=r^2-2r+a^2$, $K=(r^2+a^2)\omega-am$ and $\lambda=\Lambda+a^2\omega^2-2am\omega$. The number $\Lambda$ comes from the angular equation; it equals $l(l+1)$ for a non-rotating hole and shifts slightly with spin. Dolan, Instability of the massive Klein-Gordon field on the Kerr spacetime, §II

$\Delta$ vanishes at both horizons, $r_\pm=1\pm\sqrt{1-a^2}$. Near $r_+$ the term $K^2/\Delta$ dominates, and $K(r_+)=2r_+(\omega-m\Omega_H)$. Every quantity in the amplification condition sits in this one boundary value.

A Wronskian that cannot change

Take a massless wave, $\mu=0$, at real frequency. Then every coefficient of the radial equation is real, so the complex conjugate $R^\ast$ solves the same equation. Multiply the equation for $R$ by $R^\ast$, subtract the conjugate equation multiplied by $R$, and the potential terms cancel:

The bracket is a Wronskian. It is the same number at every radius, so its value next to the horizon must equal its value far away.

It is cleaner to evaluate it for $\psi=\sqrt{r^2+a^2}\,R$ written in the tortoise coordinate $r_\ast$, which runs from $-\infty$ at the horizon to $+\infty$ far away. Physically, nothing may come out of the horizon, so close to it the wave only falls in, $\psi\to T\,e^{-ik_Hr_\ast}$ with $k_H=\omega-m\Omega_H$. Far away it is the sum of an incoming and a reflected wave, $\psi\to I\,e^{-i\omega r_\ast}+R\,e^{i\omega r_\ast}$. Here $R$ now names the reflected amplitude. Equating the conserved quantity at the two ends gives

Brito, Cardoso and Pani, Superradiance, §4.4, Eq. (4.53)

The left side is the energy flux that disappears between the incoming and the reflected wave. For $\omega>0$ the reflected wave is stronger than the incoming one, $|R|>|I|$, exactly when $\omega<m\Omega_H$. The proof needs no approximation and no small parameter. The gain is measured by

Worked example 1. Our solver integrates the radial function outward from the horizon for $a=0.99$ and $M\omega=0.3$. The Wronskian, sampled at six radii from $1.14M$ to $267M$, stays at $0.305359584$ to a relative spread of $3.6\times10^{-10}$. Across all 100 frequencies of the spin $0.99$ slice, the two sides of the flux balance agree to better than $4.2\times10^{-11}$. That is the first test any amplification solver should pass. The full recomputation is described in recomputing a published amplification table.

From geometric units to seconds and electronvolts

To return to ordinary units, multiply a dimensionless time by $GM/c^3$, a length by $GM/c^2$ and an energy by $\hbar c^3/(GM)$:

The last factor is $\hbar$ divided by the time unit, $6.58\times10^{-16}\,\mathrm{eV\,s}$ over $4.93\,\mu\mathrm s$. A cloud's coupling $\alpha=M\mu$ therefore fixes the particle's rest energy,

and its growth rate $M\omega_I$, the imaginary part of $M\omega$, fixes the time in which the field amplitude grows by a factor $e$,

The number of particles in the cloud is proportional to the amplitude squared, so it grows as $e^{2\omega_It}$.

The same dimensionless solution describes both rows. A survey of heavier holes is a survey of lighter particles.

What an observer could connect to these numbers

None of these quantities is observed directly. A spin is inferred from X-ray spectra of accreting binaries or from the waveform of a black-hole merger, each with its own model and error bars. A cloud in which bosons annihilate in pairs would radiate nearly monochromatic gravitational waves at about twice the boson frequency, $f_{\mathrm{GW}}\approx2m_bc^2/h$. For $m_bc^2=2.67\times10^{-12}\,\mathrm{eV}$ that is $1.29\,\mathrm{kHz}$, inside the band of ground-based detectors. Baryakhtar, Galanis, Lasenby and Simon, Black hole superradiance of self-interacting scalar fields, Fig. 13

Connecting a growth rate to such a measurement needs more than this chapter computes. The rate below describes one mode in the linear regime. It tells how fast a small cloud grows. How large the cloud becomes depends on the spin-down history.

A first estimate: the hydrogenic growth rate

For small $\alpha$ the cloud lives far from the hole, and the growth can be found by matching the hydrogen-like wave function to the solution near the horizon. For the $l=m=1$ ground state the result is

Brito, Cardoso and Pani, Superradiance, §5.6.2, Eq. (5.27)

The bracket is the superradiance condition in disguise. With $\omega\approx\mu=\alpha$ and $\Omega_H=a/(2r_+)$, the condition $\omega<m\Omega_H$ reads $2\alpha r_+<a$. The steep factor $\alpha^9$ comes from how little of a large, dilute cloud reaches the horizon.

Worked example 2. Take $a=0.99$, $r_+=1.141$ and $\alpha=0.1$. The bracket is $0.99-0.228=0.762$, so $M\omega_I\approx0.762\times10^{-9}/48=1.59\times10^{-11}$. The exact continued-fraction solution is $7.39\times10^{-12}$, smaller by a factor $2.1$. For ten solar masses the estimate gives an e-fold time of $36$ days and the exact solution $77$ days.

Blank cells are couplings where the exact cloud no longer grows. The estimate gets the real part of the frequency right to better than $0.4$ percent everywhere in the table, but the growth rate only to within a factor of a few. Near each cut-off the two curves cross on their way to zero, and that crossing produces apparent agreement such as $1.07$ at $a=0.9$ and $\alpha=0.3$.

Assumptions behind every number in this chapter

1. The field is a test field: the cloud is light enough not to change the Kerr geometry.
2. The field is free. Self-interactions, such as those of an axion, are left out.
3. Only one mode, $l=m=1$ with no radial node, is followed.
4. The hole is isolated: no accretion and no companion.
5. Growth is linear, so it can never stop by itself. A cloud stops growing only when it has taken enough spin for $\Omega_H$ to fall to the cloud's frequency, and that history is not computed here.

As a scale only, a cloud holding one percent of a ten-solar-mass hole at $\alpha=0.42$ contains about $2\times10^{76}$ bosons. Growing from a single boson takes $\ln(2\times10^{76})\approx176$ e-folds of the particle number, which is $88$ amplitude e-fold times, roughly eight hours. Saturation, spin-down and the cloud's own radiation decide the rest.

Lab tasks

1. In the Lab, open the advanced group and change the black-hole mass from $10$ to $100$ solar masses. Predict how the e-fold time and the boson rest energy change, then check the readouts.
2. Pick the preset Right at the threshold. Read the gain chart near $\omega=m\Omega_H$ and confirm that the curve crosses zero there, as the flux identity requires.
3. Set spin $0.99$ and compare the solid and dashed growth curves at small and at large coupling. Find the coupling where the dashed curve is about twice the solid one.

Exercises with explained answers

1. Show that $K(r_+)=2r_+(\omega-m\Omega_H)$.
2. Why does the Wronskian argument fail for complex $\omega$?
3. For $M=4\times10^6\,\mathrm M_\odot$ and $\alpha=0.2$, what is the boson rest energy?
4. The cloud with $a=0.99$ and $\alpha=0.2$ has $M\omega_I=1.61\times10^{-9}$. What is its e-fold time for ten solar masses?
5. A cloud has $\omega=0.38/M$. Below which spin does it stop growing?

The hydrogenic formula is a guide, not an answer. The next chapter computes the exact rate and shows how to know that it is right: solving for the bound cloud's growth rate reproducibly.