Gravitational waves and the chirp mass

Builds on: Sections 10.2, 8.2, 3.2 and D5
Reading time: 55 min
Exercises: 3
Olympiad problems: 2

On 14 September 2015 the two detectors of LIGO, at Hanford in Washington and Livingston in Louisiana, recorded the same faint signal 7 ms apart. It lasted about 0.2 s, and in that time its frequency rose from 35 Hz to a few hundred hertz before it stopped. Played as sound, it rises like a bird's chirp, and such signals are called chirps. It came from two black holes of 36 and 29 solar masses, 1.3 billion light-years away, spiralling together and merging into one of 62 solar masses. The missing 3 solar masses had been carried away as gravitational waves. In this section we see what these waves are and what a chirp tells us.

10.7.1Waves in space

Einstein's general relativity describes gravity as the curving of space and time by mass. When masses move, the curving changes, and the changes spread out at the speed of light as gravitational waves. A wave passing through a ring of floating particles stretches it in one direction and squeezes it at right angles, then the other way round. The fractional change of length is called the strain $h$ .

The waves are extremely weak. Those of GW150914 had a strain of about $10^{- 21}$ at Earth, so they changed the difference between the lengths of LIGO's two 4 km arms by about $4 \times 10^{- 18}$ m, a few thousandths of the width of a proton. LIGO measures this with laser light bounced many times along two arms at right angles, and compares the arms.

Before you readMake a guess first

Two neutron stars orbit each other 50 times a second. At what frequency do they send out gravitational waves?

A binary sends out gravitational waves because its shape keeps turning. The stretching depends on the direction of the line through the two bodies, and not on which end of the line is which, so it repeats every half turn, twice in each orbit.

ResultThe frequency of the waves from a binary(10.17)

f_{GW} = 2 f_{orb} = \frac{1}{π} \sqrt{\frac{G M}{a^{3}}}

$f_{GW}$	the frequency of the gravitational waves
$f_{orb}$	the number of orbits in each second
$M, a$	the total mass of the binary and the separation of the two bodies

The second form uses Kepler's third law (8.1) for a circular orbit.

10.7.2The chirp

The waves carry energy away from the orbit. The energy of a circular orbit of two bodies of any masses is $- G m_{1} m_{2} / (2 a)$ , with $a$ their separation ((3.4), Section 3.7), so as it falls the separation shrinks, the bodies move faster and the waves rise in frequency and strength. The faster the frequency rises, the more massive the pair. Einstein's theory shows that, while the bodies are well apart, the rate of rise depends on the masses only through one combination of them, called the chirp mass.

ResultThe chirp mass and the rise of the frequency(10.18)

\begin{aligned} M & = \frac{(m_{1} m_{2})^{3 / 5}}{(m_{1} + m_{2})^{1 / 5}} \\ \frac{d f_{GW}}{d t} & = \frac{96}{5} π^{8 / 3} {(\frac{G M}{c^{3}})}^{5 / 3} f_{GW}^{11 / 3} \end{aligned}

$M$	the chirp mass
$m_{1}, m_{2}$	the masses of the two bodies
$f_{GW}$	the frequency of the waves

This holds for circular orbits while the separation is many times the Schwarzschild radii, before the last few orbits. For two equal masses

m

the chirp mass is

2^{- 1 / 5} m = 0.871 m

, a little less than either mass.

Show the derivation

A binary on a circular orbit radiates the power $P = \frac{32}{5} G^{4} m_{1}^{2} m_{2}^{2} M / (c^{5} a^{5})$ , with $M = m_{1} + m_{2}$ , which we take from general relativity. Setting $P$ equal to the rate of loss of the orbital energy $- G m_{1} m_{2} / (2 a)$ gives

\frac{d a}{d t} = - \frac{64}{5} \frac{G^{3} m_{1} m_{2} M}{c^{5} a^{3}} .

By (10.17), $f_{GW}$ goes as $a^{- 3 / 2}$ , so the rates of change $\dot{f}$ and $\dot{a}$ obey $\dot{f} / f = - \frac{3}{2} \dot{a} / a$ . Writing $a$ in terms of $f_{GW}$ with $a^{3} = G M / (π^{2} f_{GW}^{2})$ leaves the masses only in the form $m_{1} m_{2} M^{- 1 / 3} = M^{5 / 3}$ , and gives the result.

Integrating (10.18) from a frequency $f$ up to an infinite frequency gives the time left before the merger. A detector that measures $f$ and $d f / d t$ finds the chirp mass directly, and with it the time left.

ResultThe time left before a merger(10.19)

τ = \frac{5}{256} {(\frac{G M}{c^{3}})}^{- 5 / 3} (π f_{GW})^{- 8 / 3}

$τ$	the time left before the merger, when the waves have the frequency $f_{GW}$
$M$	the chirp mass

Here

G M / c^{3}

is a time, 4.92 µs for each solar mass. The inspiral ends a little earlier, at the innermost stable orbit (Section 10.2), where the waves have the frequency

c^{3} / (6^{3 / 2} π G M)

, about 4.4 kHz divided by the total mass in solar masses. The bodies then plunge together and merge within a few more cycles, which these formulas do not describe.

Quick checkThe chirp of GW150914

The black holes of GW150914 had 36 and 29 solar masses. Find the chirp mass and the time left before the merger when the waves had the frequency 35 Hz.

Example 10.13GW150914 at the innermost orbit

Find the frequency of the waves of GW150914, whose black holes had 36 and 29 solar masses, when the inspiral reached the innermost stable orbit, the separation of the two black holes then, and their speed relative to each other.

Solution

The total mass is 65 solar masses, so by the note under (10.19) the frequency there is $4400 / 65 = 68$ Hz, and the orbit turned 34 times a second. The innermost stable orbit is at $6 G M / c^{2} = 3 \times 2.95 \times 65$ km, which is 575 km. On a circular orbit the relative speed is $\sqrt{G M / a} = c / \sqrt{6}$ , which is $1.22 \times 10^{5}$ km/s.

AnswerThe waves reached 68 Hz when the two black holes were 575 km apart and moving at 0.41 of the speed of light relative to each other. The signal kept rising to about 250 Hz during the plunge and the merger that followed, which only computer simulations of general relativity can follow.

Figure 10.7The chirp of two merging bodies

Mass of body 1Mass of body 2

Watch

Chirp mass
From 30 Hz to the innermost orbit
Wave frequency at the innermost orbit

The figure shows, on the left, two bodies 1 and 2 on their shrinking orbits round their centre of mass. On the right, the heavy line is the frequency of the waves against the time left before the merger, both on logarithmic scales, up to the innermost stable orbit, and the shaded band is the range of frequencies that ground-based detectors can hear. The dot marks the moment shown on the left. Below are the last ten cycles of the wave, growing in strength and frequency, with the dashed line at the moment shown on the left. Running the inspiral moves through these ten cycles slowly. The figure can be set to , , or .

A pair of heavy black holes sweeps through the detectors' band in a fraction of a second and about 10 cycles. A pair of neutron stars, with a chirp mass of 1.22 solar masses for two of 1.4, takes 54 s to rise from 30 Hz and makes about 2600 cycles before it merges, which is why the merger of neutron stars in 2017, GW170817, gave such a clean measurement of its chirp mass. Section D5 works through that measurement from the data.

Example 10.14A chirp mass from a measured chirpfinish it

When the waves of GW150914 had the frequency 100 Hz, their frequency was rising at 3240 Hz each second. Find the chirp mass.

The start

Solving (10.18) for the chirp mass gives

\frac{G M}{c^{3}} = {(\frac{5}{96} π^{- 8 / 3} f_{GW}^{- 11 / 3} \frac{d f_{GW}}{d t})}^{3 / 5} .

Your stepNow put in the numbers and divide by 4.92 µs for each solar mass.

solar masses

Common errorsSection 10.7

The frequency of the waves is twice the orbital frequency, not equal to it.
The chirp mass is not the total mass. For two equal masses $m$ it is $0.871 m$ , less than either of them.
The energy carried away in a merger is the difference between the masses before and after, times $c^{2}$ , and not the total mass times $c^{2}$ .
The formulas of this section hold only while the bodies are well apart. Near the innermost stable orbit they no longer apply, and the last moments need computer simulations.

ExercisesSection 10.73 questions

10.7.1
What is the chirp mass of two black holes of 10 solar masses each?
solar masses
10.7.2
At the same frequency, how many times longer does a pair of neutron stars with a chirp mass of 1.22 solar masses have left before merging than a pair of black holes with a chirp mass of 28.1 solar masses?
10.7.3
From the frequency of the waves of a binary and its rate of rise, which quantity can be found directly?

Olympiad problemsPast IOAA problems, restated2 problems

2025Two black holes spiral together15 points

Adapted from IOAA 2025 (Mumbai), theory problem T03. We have restated it in our own words and kept the original data. The solution was last checked on 23 October 2026.

Two black holes in our Galaxy, of masses $m_{1} = 36$ and $m_{2} = 29$ solar masses, move on circular orbits with angular frequency $ω$ round their centre of mass.

(a) With Newton's gravity, find the angular frequency $ω_{ini}$ of the orbit at the time $t_{ini}$ when the separation was 4.0 times the sum of their Schwarzschild radii, in terms of $m_{1}$ , $m_{2}$ and physical constants only, and find its value. (5 points)

(b) The waves have the frequency $f_{GW}$ , with $2 π f_{GW} = 2 ω$ , and it rises as

\frac{d f_{GW}}{d t} = \frac{96 π^{8 / 3}}{5} G^{5 / 3} c^{β} M^{α / 3} f_{GW}^{δ / 3},

where $M = (m_{1} m_{2})^{3 / 5} / (m_{1} + m_{2})^{1 / 5}$ is the chirp mass. Find $α$ , $β$ and $δ$ . (4 points)

(c) The waves were first detected at $t_{ini} = 0$ . Find the time $t_{merge}$ at which $f_{GW}$ becomes very large, in terms of $ω_{ini}$ , $M$ and physical constants, and find its value. (6 points)

I solved it without the solution

Answer(a) $ω_{ini} = c^{3} / (16 \sqrt{2} G (m_{1} + m_{2})) = 1.4 \times 10^{2}$ rad/s. (b) $α = 5$ , $β = - 5$ , $δ = 11$ . (c) $t_{merge} = 5 c^{5} / (256 ω_{ini}^{8 / 3} (G M)^{5 / 3}) = 0.10$ s.

2021The energy of GW1509145 points

Adapted from IOAA 2021 (Bogotá, remote), theory problem Q1. We have restated it in our own words and kept the original data. The problem uses 35 and 30 solar masses, later estimates for the same black holes. The first published values, used in the text, were 36 and 29, and both give 3 solar masses radiated. The solution was last checked on 23 October 2026.

In GW150914 black holes of 35 and 30 solar masses merged into one of 62 solar masses. Ignore their spin and take the energy released, $E_{GW}$ , to have gone entirely into gravitational waves. A supernova releases $E_{SN} = 2 \times 10^{44}$ J. Estimate $E_{SN} / E_{GW}$ . (5 points)

I solved it without the solution

Answer $E_{SN} / E_{GW} = 3.7 \times 10^{- 4}$ , so the merger gave out about 2700 times as much energy as a supernova, almost all of it in a fraction of a second.