ISEGORIA / MATH ENCYCLOPEDIA
Gravitational waves: ripples in the geometry of space
A gravitational wave is a travelling change in distances: it stretches space along one direction and squeezes it along the perpendicular one, then swaps them, a hundred times a second or more. Two black holes spiralling together radiate such waves, sweeping up in frequency and strength in a characteristic chirp whose rate depends on a single combination of their masses. An L-shaped laser interferometer measures the difference between its two arm lengths to a ten-thousandth of a proton’s width, and how much it sees depends on where in the sky the wave comes from.
Before you begin: General relativity, oscillations and Kepler orbits
Predict, manipulate, then test yourself: every experiment ends with practice questions that are generated fresh at three levels. Graphs illustrate the mathematics; they do not replace a proof.
1. A wave that stretches space
Free masses floating in a ring feel nothing push them, yet the distances between them change as a gravitational wave goes through. The wave is transverse: it acts only in the plane across its direction of travel, and it is traceless: what it adds along one axis it takes away along the perpendicular one. There are two independent patterns, plus and cross, the same shape turned through 45°. Five rings are strung along the direction of travel so you can see the deformation move. Mix the polarisations, make them circular, and compare what the two arms of an L-shaped detector record.
Worked example. A wave with \(h_+=10^{-21}\) passing straight down through a detector with 4 km arms lengthens one arm by \(\tfrac12 hL=2\times10^{-18}\) m and shortens the other by the same amount, so the difference is \(4\times10^{-18}\) m, about a two-hundredth of the radius of a proton. A pure cross wave along the same axes changes neither arm to first order.
Watch out. The strain is exaggerated by about \(10^{20}\) so the ring visibly changes shape; the formula above is the first-order (linear) description, valid because real strains are so small. Distances here are proper distances between free masses in the transverse-traceless gauge; the masses do not move in those coordinates, the metric between them does.
Why can a single L-shaped detector miss a wave completely?
Because it measures only the difference of its two arm lengths. A cross-polarised wave aligned with the arms stretches and squeezes along the diagonals, which changes both arms equally to first order, so the difference is zero.
2. The chirp of a binary
Two compact bodies in orbit radiate waves at twice their orbital frequency, lose energy, draw closer, orbit faster and radiate more: a runaway that ends in a plunge. At leading order the whole frequency sweep depends on one combination of the masses, the chirp mass, so measuring how fast the pitch rises weighs the system. The sheet is the wave in the orbital plane, a two-armed spiral moving outward at the speed of light. Choose the masses and the distance and play the last orbits.
Worked example. For 36 and 29 solar masses, roughly the first detected pair, the chirp mass is 28.1 solar masses and the last stable orbit is reached at a wave frequency of 67.6 Hz. From 20 Hz the signal takes 0.81 s and about 24 cycles to get there; at 410 Mpc the strain at that point is \(1.2\times10^{-21}\). Two neutron stars of 1.4 solar masses each have a chirp mass of 1.22 and spend 158 s and about 5000 cycles above 20 Hz.
Watch out. This is the leading-order (Newtonian quadrupole) inspiral: accurate early on, increasingly off as the bodies approach, and stopped at the innermost stable circular orbit of the total mass. The plunge, merger and ringdown that follow need numerical relativity, so the field inside the last outgoing wavefront is left flat rather than invented. In the 3D view distances are in units of \(GM/c^2\) and the bodies are drawn much larger than their horizons.
Why does the frequency rise faster and faster?
The power radiated grows like the sixth power of the orbital frequency times a factor in the masses, and a tighter orbit both radiates more and has less energy to lose before the next step in. So each orbit shrinks the separation by more than the last, and the frequency runs away as \(f\propto\tau^{-3/8}\).
3. Where a detector is deaf
An L-shaped interferometer is not equally sensitive in all directions. A wave arriving from straight overhead lies across both arms and gives the largest signal; one arriving in the plane of the arms, along the line midway between them, changes both arms equally and gives none. The response is the antenna pattern, drawn here in 3D around the detector. Drag the source across the sky, change the wave’s polarisation angle, and switch between the plus response, the cross response and their combination.
Worked example. From \(\theta=40^\circ\), \(\phi=20^\circ\) with \(\psi=0\): \(F_+=\tfrac12(1+0.587)\times0.766=0.608\) and \(F_\times=\cos40^\circ\sin40^\circ=0.492\), so the polarisation-averaged response is \(\sqrt{0.608^2+0.492^2}=0.782\). Averaged over the whole sky and all polarisations, \(F_+^2\) is exactly \(1/5\).
Watch out. This is the long-wavelength limit: the arms are taken to be much shorter than the gravitational wavelength, which holds for ground-based detectors at their working frequencies. \(\theta\) is measured from the zenith of the detector, \(\phi\) from the x arm, and \(\psi\) is the polarisation angle in the plane of the wave.
Why does the network need detectors in different places and orientations?
Because one detector’s response hides the direction: many directions give the same amplitude, and blind spots give none. Detectors far apart see the wave at different times (up to about 10 ms across the Earth), and differently oriented ones see different mixtures of \(F_+\) and \(F_\times\); together these pin down the sky position and both polarisations.