University Physics IV · Frontiers of Modern Physics · 15.8
Gravitational Waves & Detection
Two black holes stir spacetime and a kilometres-long vacuum twitches by a few thousandths of a proton radius. This lesson works out where that number comes from, why the sweep of the note matters more than its loudness, and how an instrument is built to hear something that faint.
Build the model
Connect the measurement to the mechanism.
Write the metric as flat spacetime plus a small perturbation, g = η + h with |h| ≪ 1, keep first-order terms only, and Einstein's equations collapse to a wave equation: spacetime carries transverse waves at c with two independent polarisations, plus and cross, set 45° apart. Where can they come from? Not a varying monopole, since total mass-energy is conserved; not a varying mass dipole, since that would mean a changing total momentum.
The first moment free to vary is the quadrupole, so the amplitude goes as the second time derivative of the mass quadrupole moment multiplied by G/c⁴ = 8.3×10⁻⁴⁵ m J⁻¹. That prefactor is the whole cost of the subject: GW150914 radiated three solar masses of energy, briefly outshining every star in the observable universe, and still arrived as a strain of 10⁻²¹ — four attometres across a 4 km arm. The compensation is that an inspiralling binary is a clock as well as a source.
The sweep df/dt fixes the chirp mass with no reference to distance, orientation or amplitude calibration, and once the chirp mass is known the amplitude returns the distance, so one waveform is both a mass measurement and a standard siren. The model's limit is built into it: the expansion is in v/c and in h, so it degrades as the orbit tightens and fails outright at merger, where numerical relativity must supply what linearised theory cannot.
- Simple definition
- A gravitational wave is a propagating strain in spacetime — a fractional change h in the proper distance between free masses, travelling at the speed of light, transverse to its direction of travel and carrying two polarisations.
- Example
- GW150914 reached h ≈ 1.0×10⁻²¹ at Earth: one 4.0 km LIGO arm grew by 2.0×10⁻¹⁸ m while the perpendicular arm shrank by the same, a differential change of 4.0×10⁻¹⁸ m — about one two-hundredth of a proton radius.
h = 1.0×10⁻²¹ across L = 4.0 km is ΔL = 4.0×10⁻¹⁸ m — twice what either arm does alone.
h dimensionless; L the arm length in m; ΔL the differential change in m
Monopole is closed off by mass-energy conservation and dipole by momentum conservation, so a perfectly spherical collapse radiates nothing.
Q the mass quadrupole moment in kg m²; r the distance in m; G/c⁴ = 8.3×10⁻⁴⁵ m J⁻¹
The sweep alone fixes Mc — no distance, no orientation, no amplitude calibration enters.
Mc = (m₁m₂)³⁄⁵/(m₁+m₂)¹⁄⁵; f the GW frequency in Hz; GM_☉/c³ = 4.925 μs
With Mc already fixed by the sweep, the measured h returns D directly — no Cepheids, no rungs.
D the luminosity distance in m; face-on, optimally placed source; leading order, and only below fISCO; GM_☉/c² = 1.477 km
56 s from 30 Hz for Mc = 1.18 M_☉; 10 ms from 100 Hz for Mc = 30 M_☉.
τ in s, counted from the instant the wave passes frequency f
68 Hz for a 65 M_☉ binary, yet GW150914 swept past 250 Hz — that gap is numerical relativity's territory.
M the total mass; a Schwarzschild test-particle estimate, not a merger prediction
Linearise the metric, and exactly two polarisations survive
Put gμν = ημν + hμν with every |hμν| ≪ 1 and drop everything of second order. In the transverse-traceless gauge the ten components of hμν collapse to two independent functions, h₊ and h_×, whose deformation patterns sit 45° apart, both transverse to the direction of travel. What a wave does to matter is easiest to see with a ring of freely falling masses: h₊ squeezes the ring along one axis while stretching it along the perpendicular one, then reverses half a period later. In TT coordinates the masses never move; what changes is the proper distance between them, which is why a detector reads a fractional length change h = ΔL/L rather than a force. Because the pattern is quadrupolar, rotating the source through 180° reproduces it, so a circular binary radiates at twice its orbital frequency. It also dictates the instrument: two perpendicular arms, one stretched while the other shrinks, doubling the signal and rejecting motion common to both.
Monopole and dipole are closed off, so the quadrupole leads
Expand the source in multipoles and ask which one can vary. The monopole is the total mass-energy: conserved, so its second time derivative vanishes and it cannot radiate. The mass dipole Σmᵢ rᵢ has a first derivative equal to the total momentum, itself conserved for an isolated system, so its second derivative vanishes too. Here gravity parts company with electromagnetism: an accelerating charge radiates a dipole field because charge-to-mass ratios differ from one particle to the next, but the gravitational 'charge' is the mass itself, and momentum conservation kills the analogue. The first moment allowed to change is the quadrupole, giving h ≈ (2G/c⁴r) d²Q/dt². The consequences are sharp and testable: a perfectly spherical collapse or explosion radiates nothing at all, and a neutron star spinning about an exact symmetry axis radiates nothing either. It needs a bump — a 'mountain' perhaps centimetres high — before its quadrupole moment varies in time at all.
Why the strain is 10⁻²¹ when the power is 10⁴⁹ W
The prefactor G/c⁴ = 8.3×10⁻⁴⁵ m J⁻¹ is the reason gravitational-wave astronomy needed a century. Try a laboratory source: a 1000 kg steel bar, 2.0 m long, spun about its centre at 100 revolutions per second. Its quadrupole moment has magnitude of order ML²/12 and varies at twice the spin rate, so d²Q/dt² ≈ (ML²/12)(2ω)² = ⅓ML²ω² = ⅓(1000)(4.0)(628.3)² = 5.3×10⁸ kg m² s⁻². At r = 100 m the strain is h ≈ 2(8.3×10⁻⁴⁵)(5.3×10⁸)/100 = 8.8×10⁻³⁸ — sixteen orders of magnitude below what LIGO records. Now run it the other way. GW150914 converted about 3.0 M_☉, or 5.4×10⁴⁷ J, into gravitational waves, peaking near 3.6×10⁴⁹ W, more than the combined light of every star in the observable universe. Even that, at 410 Mpc = 1.3×10²⁵ m, arrives as h ≈ 10⁻²¹. Only stellar masses moving at a fair fraction of c are ever going to be detectable.
Read the chirp: sweep gives the mass, amplitude the distance
As the orbit radiates it tightens, so f rises and the rate of rise obeys df/dt ∝ Mc⁵⁄³ f¹¹⁄³. Invert it and a measured sweep hands you the chirp mass with nothing else supplied: a rate of 0.20 Hz s⁻¹ at 30 Hz gives Mc = 1.18 M_☉ and 56 s to merger, identifying a neutron-star binary from timing alone before a single photon arrives. With Mc fixed, h = (4/D)(GMc/c²)⁵⁄³(πf/c)²⁄³ inverts for D, which is what makes a compact binary a standard siren: the distance comes from general relativity plus detector calibration, with no rung of the astronomical distance ladder underneath it. Two degeneracies bound the claim. The amplitude also carries the inclination of the orbit and the antenna response for that patch of sky, so amplitude alone cannot separate a face-on distant binary from an edge-on nearer one. And the mass the waveform reports is redshifted: the source-frame chirp mass is Mc/(1+z).
The instrument: a Fabry-Perot Michelson held on a dark fringe
A Michelson turns a differential arm change into a phase difference Δφ = 4πΔL/λ. With 1064 nm light and ΔL = 4×10⁻¹⁸ m that is 4.7×10⁻¹¹ rad on a single pass — hopeless. So each arm becomes a Fabry-Perot cavity storing the light for a few hundred round trips, an effective path of order 10³ km, and power recycling builds the circulating power to about 10⁵ W. The output port is held dark, so only differential motion delivers light to the photodiode. From there the noise floor sets the band: below roughly 10 Hz, ground motion and fluctuating local gravity gradients; from about 10 to 100 Hz, thermal noise in the suspensions and mirror coatings; above that, photon shot noise, which falls as 1/√P — more power helps, until radiation-pressure fluctuations pushing the mirrors grow as √P and the two meet at the standard quantum limit. Near 150 Hz the design strain noise is about 4×10⁻²⁴ Hz(−1/2); across the ~100 Hz a stellar-mass merger occupies, that is 4×10⁻²³, so a 10⁻²¹ signal shows up at a signal-to-noise ratio of a few tens.
Where linearised theory quits, and what two events bought
The post-Newtonian series is an expansion in v/c, and by the last orbits v/c approaches 0.5, where it stops being useful. The Schwarzschild test-particle estimate puts the innermost stable circular orbit at f ≈ 4.40 kHz × (M_☉/M), which is 68 Hz for GW150914's 65 M_☉ total — yet the recorded signal kept climbing to about 250 Hz. Those last cycles and the ringdown are precisely what linearised theory cannot supply; numerical relativity, stable only since 2005, produces those waveforms, and the search matched-filters the data against banks of them. What the events bought: GW150914 gave 36 + 29 M_☉ merging to 62 M_☉ with roughly 3 M_☉ radiated, arriving 6.9 ms apart at the two LIGO sites. GW170817 gave a neutron-star binary at 40 Mpc, about 100 s in band counting from roughly 24 Hz, a γ-ray burst 1.7 s after merger and a kilonova days later — pinning the wave speed to within about 4×10⁻¹⁶ of c, tying short γ-ray bursts to neutron-star mergers, and delivering the first standard-siren Hubble constant.
Change one variable at a time
Make the relationship visible.
Push the chirp mass to 45 M☉: fewer cycles fit the same 344 ms window and the ISCO line slides left, since a heavier binary merges at a lower frequency. Then move distance or inclination — amplitude drops while every frequency stays put: the sweep fixes the mass, the amplitude the distance.
LARGEST h DRAWN1.50 ×10⁻²¹
4 km ARM THERE6.01 ×10⁻¹⁸ m
ISCO FREQUENCY64 Hz
CYCLES IN VIEW12.0
Live interpretationLARGEST h DRAWN: 1.50 ×10⁻²¹. 4 km ARM THERE: 6.01 ×10⁻¹⁸ m. ISCO FREQUENCY: 64 Hz. CYCLES IN VIEW: 12.0
Catch the common trap
Explain before calculating.
A neutron star spins at 300 Hz. Under what condition does it radiate gravitational waves, and at what frequency?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyGW150914 reached a peak strain of h = 1.0×10⁻²¹ at LIGO, whose arms are L = 4.0 km long. Find the change in each arm and the differential change between them, then compare that with the proton charge radius, 8.4×10⁻¹⁶ m.
- A gravitational wave acts on lengths in proportion, not by pushing: for an optimally oriented wave on a Michelson, one arm changes by δLₓ = +½hL while the perpendicular arm changes by δLy = −½hL.
- Each arm: ½ × 1.0×10⁻²¹ × 4.0×10³ m = 2.0×10⁻¹⁸ m, one lengthening and the other shortening by that amount.
- The interferometer reads only the difference: ΔL = δLₓ − δLy = hL = 1.0×10⁻²¹ × 4.0×10³ m = 4.0×10⁻¹⁸ m. This is the figure normally quoted as 'the arm-length change', and it is twice what either arm does on its own.
- Compare with the proton: 4.0×10⁻¹⁸ ÷ 8.4×10⁻¹⁶ = 4.8×10⁻³, about one two-hundredth of a proton radius. Nothing is being resolved at that scale — the reading is a phase shift accumulated over hundreds of round trips of a beam carrying about 10⁵ W, which at 1064 nm (photon energy hc/λ = 1.9×10⁻¹⁹ J) is some 5×10²³ photons a second, not a picture of a mirror moving.
AnswerEach arm moves 2.0×10⁻¹⁸ m in opposite senses, so the differential change is ΔL = 4.0×10⁻¹⁸ m — roughly 1/210 of a proton radius.
MediumFor GW150914 the detector-frame chirp mass is Mc ≈ 30 M_☉ and the luminosity distance about 410 Mpc. Estimate the strain as the gravitational-wave frequency passes 150 Hz, using h = (4/D)(GMc/c²)⁵⁄³(πf/c)²⁄³. Take GM_☉/c² = 1477 m and 1 Mpc = 3.086×10²² m. The measured peak was about 1.0×10⁻²¹ — account for the difference.
- Put the mass in length units: GMc/c² = 30 × 1477 m = 4.43×10⁴ m. Raise it to the 5/3 power: (4.43×10⁴)⁵⁄³ = 5.55×10⁷ m⁵⁄³.
- Put the frequency in inverse length: πf/c = π × 150 ÷ (3.00×10⁸ m s⁻¹) = 1.57×10⁻⁶ m⁻¹, and (1.57×10⁻⁶)²⁄³ = 1.35×10⁻⁴ m(−2/3).
- Multiply the two: 5.55×10⁷ × 1.35×10⁻⁴ = 7.50×10³ m, and the factor 4 gives 3.00×10⁴ m.
- Convert the distance: D = 410 × 3.086×10²² m = 1.27×10²⁵ m, so h = 3.00×10⁴ ÷ 1.27×10²⁵ = 2.4×10⁻²¹.
- Two separate things sit between that number and the recorded 1.0×10⁻²¹, and it is worth keeping them apart. First, orientation: the formula assumes a face-on orbit sitting where the detector is most sensitive, and GW150914 was neither, so an inclination factor and an antenna-pattern factor each cut the amplitude. Second, validity: this is the leading-order inspiral amplitude, and 150 Hz is already more than twice this binary's ISCO near 68 Hz, so it is being read outside the regime it was derived for. Together those account for the factor of 2.4 down to what was measured — the estimate is worth an order of magnitude and a factor of two or three, not a decimal place.
Answerh ≈ 2.4×10⁻²¹ read at 150 Hz for an optimally oriented source; the observed peak of 1.0×10⁻²¹ is that estimate reduced by a combined inclination-and-antenna factor of about 0.4, with the leading-order formula itself already extrapolated past this binary's ISCO.
HardA binary is tracked from the moment its gravitational-wave frequency passes 30 Hz, where the sweep rate is df/dt = 0.20 Hz s⁻¹. (a) Find the chirp mass from df/dt = (96/5)π⁸⁄³(GMc/c³)⁵⁄³ f¹¹⁄³, taking GM_☉/c³ = 4.925 μs. (b) Find the time to coalescence from 30 Hz. (c) A γ-ray burst arrives 1.7 s after the merger from a host 40 Mpc away; bound the fractional difference between the wave speed and c.
- Gather the constants: 96/5 = 19.2 and π⁸⁄³ = 21.17, so their product is 406.5. At f = 30 Hz, f¹¹⁄³ = 30(3.667) = 2.61×10⁵.
- Solve for the mass in seconds: (GMc/c³)⁵⁄³ = (df/dt) ÷ (406.5 × 2.61×10⁵) = 0.20 ÷ 1.06×10⁸ = 1.89×10⁻⁹ s⁵⁄³.
- Take the 3/5 power: GMc/c³ = (1.89×10⁻⁹)(0.6) = 5.83×10⁻⁶ s, so Mc = 5.83×10⁻⁶ ÷ 4.925×10⁻⁶ = 1.18 M_☉ — a neutron-star-mass binary, read from timing alone.
- Time to merger: τ = (5/256)(GMc/c³)(−5/3)(πf)(−8/3). Here (GMc/c³)(−5/3) = 1 ÷ 1.89×10⁻⁹ = 5.30×10⁸, and (π × 30)⁸⁄³ = 94.25⁸⁄³ = 1.84×10⁵, so (πf)(−8/3) = 5.44×10⁻⁶. Then τ = 0.01953 × 5.30×10⁸ × 5.44×10⁻⁶ = 56 s.
- Light travel time over 40 Mpc = 1.23×10²⁴ m is 1.23×10²⁴ ÷ 3.00×10⁸ = 4.1×10¹⁵ s. A 1.7 s lead or lag accumulated over that journey bounds |v − c|/c ≲ 1.7 ÷ 4.1×10¹⁵ = 4×10⁻¹⁶.
- State what the chirp mass does not give. It is one combination of m₁ and m₂, so the individual masses need higher post-Newtonian terms, and the value measured is redshifted, the source-frame chirp mass being Mc/(1+z).
AnswerMc ≈ 1.18 M_☉, τ ≈ 56 s from 30 Hz, and |vgw − c|/c ≲ 4×10⁻¹⁶ — the GW170817 numbers, which is how a neutron-star merger was flagged nearly a minute before its γ-ray burst.