---
title: "Conserved flux, physical units and a first growth estimate"
description: "Use the conserved wave flux to prove the gain condition, convert geometric units to seconds and electronvolts, and estimate a cloud's growth rate."
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language: "en"
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last_modified: "2026-11-12"
---

[Spinning black holes and boson clouds](https://sandboxphysics.com/stories/black-hole-superradiance/)

# Conserved flux, physical units and a first growth estimate

Sandbox Physics · 2026-11-12

The wave equation around a Kerr black hole conserves one combination of the incoming, reflected and transmitted amplitudes, and that combination alone proves that a wave is amplified exactly when $0<\omega<m\Omega_H$. With the hole's mass put back in, a growth rate in units of $1/M$ becomes seconds and a coupling becomes a particle mass in electronvolts. A short hydrogenic formula then gives a first estimate of how fast a cloud grows, and a check against an exact solution shows how far that estimate can be trusted.

[Open the black hole superradiance Lab](https://sandboxphysics.com/labs/black-hole-superradiance/)

This chapter assumes the results of [Waves, spin and the scales of a Kerr black hole](https://sandboxphysics.com/stories/black-hole-superradiance/waves-spin-and-kerr-scales/): 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

$$
\frac{\mathrm d}{\mathrm dr}\left(\Delta\frac{\mathrm dR}{\mathrm dr}\right)+\left[\frac{K^2}{\Delta}-\lambda-\mu^2r^2\right]R=0,
$$

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](https://arxiv.org/abs/0705.2880v2)

$\Delta$ 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:

$$
\frac{\mathrm d}{\mathrm dr}\left[\Delta\left(R^\ast R^\prime-RR^{\ast\prime}\right)\right]=0 .
$$

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

$$
\omega\left(|I|^2-|R|^2\right)=(\omega-m\Omega_H)\,|T|^2 .
$$

[Brito, Cardoso and Pani, Superradiance, §4.4, Eq. (4.53)](https://arxiv.org/abs/1501.06570v8)

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

$$
Z=\frac{|R|^2}{|I|^2}-1 .
$$

**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](https://sandboxphysics.com/stories/black-hole-superradiance/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 $\hbar c^3/(GM)$:

$$
\frac{GM}{c^3}=4.93\,\mu\mathrm s\,\frac{M}{\mathrm M_\odot},\qquad
\frac{GM}{c^2}=1.48\,\mathrm{km}\,\frac{M}{\mathrm M_\odot},\qquad
\frac{\hbar c^3}{GM}=1.34\times10^{-10}\,\mathrm{eV}\,\frac{\mathrm M_\odot}{M}.
$$

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,

$$
m_bc^2=\alpha\times1.34\times10^{-10}\,\mathrm{eV}\,\frac{\mathrm M_\odot}{M},
$$

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 $e$,

$$
\tau=\frac{1}{\omega_I}=\frac{GM/c^3}{M\omega_I}.
$$

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

**Table 2.1** One cloud, two black holes: $a=0.99$, $\alpha=0.421$, $M\omega_I=1.50\times10^{-7}$ from the continued-fraction solution of the next chapter.

| Black-hole mass | Boson rest energy | Amplitude e-fold time |
| --- | --- | --- |
| $10\,\mathrm M_\odot$ | $5.6\times10^{-12}\,\mathrm{eV}$ | $328\,\mathrm s$ |
| $4\times10^6\,\mathrm M_\odot$ | $1.4\times10^{-17}\,\mathrm{eV}$ | $4.2\,\mathrm{yr}$ |

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](https://arxiv.org/abs/2011.11646v2)

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

$$
M\omega_I\approx\frac{1}{48}\left(a-2\alpha r_+\right)\alpha^9 .
$$

[Brito, Cardoso and Pani, Superradiance, §5.6.2, Eq. (5.27)](https://arxiv.org/abs/1501.06570v8)

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.

**Table 2.2** Hydrogenic estimate divided by the exact growth rate.

| $\alpha$ | $a=0.5$ | $a=0.7$ | $a=0.9$ | $a=0.99$ |
| --- | --- | --- | --- | --- |
| $0.05$ | $1.15$ | $1.29$ | $1.44$ | $1.53$ |
| $0.10$ | $1.08$ | $1.41$ | $1.86$ | $2.15$ |
| $0.20$ |  | $0.79$ | $2.23$ | $3.54$ |
| $0.30$ |  |  | $1.07$ | $4.67$ |

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?

Answers

1. At the horizon $\Delta=0$, so $r_+^2+a^2=2r_+$. Then $K(r_+)=2r_+\omega-am=2r_+\left(\omega-m\,a/(2r_+)\right)=2r_+(\omega-m\Omega_H)$.
2. For complex $\omega$ the coefficient $K^2$ is complex, so $R^\ast$ no longer solves the same equation and the potential terms do not cancel. A growing or decaying mode is not in a steady state, and its flux changes with time.
3. $m_bc^2=0.2\times1.34\times10^{-10}\,\mathrm{eV}/(4\times10^6)=6.7\times10^{-18}\,\mathrm{eV}$.
4. $\tau=4.93\times10^{-5}\,\mathrm s/1.61\times10^{-9}=3.06\times10^4\,\mathrm s$, about $8.5$ hours.
5. Growth stops when $\Omega_H=a/(2r_+)$ falls to $0.38$ with $m=1$. Writing $r_+=1+\sqrt{1-a^2}$ and solving gives $a=4\omega/(1+4\omega^2)=1.52/1.578=0.963$. This is the switch-off spin the Lab shows for this frequency.

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](https://sandboxphysics.com/stories/black-hole-superradiance/solving-the-bound-cloud-reproducibly/).

## Try it in the Lab

Open the advanced group and change the black-hole mass; check that the e-fold time and boson energy scale as you predicted. [Open the black hole superradiance Lab](https://sandboxphysics.com/labs/black-hole-superradiance/)

## Sources

1. [Dolan, Instability of the massive Klein-Gordon field on the Kerr spacetime, §II](https://arxiv.org/abs/0705.2880v2)
2. [Brito, Cardoso and Pani, Superradiance, §4.4, Eq. (4.53)](https://arxiv.org/abs/1501.06570v8)
3. [Baryakhtar, Galanis, Lasenby and Simon, Black hole superradiance of self-interacting scalar fields, Fig. 13](https://arxiv.org/abs/2011.11646v2)

## Keep reading

Spinning black holes and boson clouds · Reading path

### From an amplified wave to a growing cloud

1. [Article **Why can a spinning black hole make the waves around it ever stronger?**](https://sandboxphysics.com/stories/black-hole-superradiance/why-a-spinning-black-hole-amplifies-waves/)
2. [Article **Waves, spin and the scales of a Kerr black hole**](https://sandboxphysics.com/stories/black-hole-superradiance/waves-spin-and-kerr-scales/)
3. [You are here **Conserved flux, physical units and a first growth estimate**](https://sandboxphysics.com/stories/black-hole-superradiance/flux-units-and-a-first-growth-estimate/)
4. [Article **Solving for the bound cloud's growth rate reproducibly**](https://sandboxphysics.com/stories/black-hole-superradiance/solving-the-bound-cloud-reproducibly/)
5. [Article **Recomputing a published superradiance amplification table**](https://sandboxphysics.com/stories/black-hole-superradiance/recomputing-a-published-amplification-table/)
6. [Article **How self-interactions change boson-cloud limits from spins and gravitational waves**](https://sandboxphysics.com/stories/black-hole-superradiance/reading-self-interacting-boson-cloud-limits/)