How LIGO Measures a Thousandth of a Proton Radius Without Violating Quantum Mechanics — Epoche C1
Research notebook — instrumentation reading group, week 9. Notes made while re-reading the discovery paper on GW150914, the first observed binary black-hole merger (Abbott et al. 2016). Purpose of this entry: to reconstruct, step by step, why a displacement of $4\times10^{-18}\,\mathrm{m}$ was measurable at all. The paper quotes a peak strain of $h = 1.0\times10^{-21}$, strain meaning the fractional change in length, $h = \Delta L/L$. Across an arm of length $L = 4\,\mathrm{km}$ that is $\Delta L = hL = 4\times10^{-18}\,\mathrm{m}$, about one two-hundredth of the proton's charge radius of $8.4\times10^{-16}\,\mathrm{m}$. The signal itself swept upward from about $35$ to $250\,\mathrm{Hz}$ in roughly $0.2\,\mathrm{s}$, which fixes the timescales everything below has to work on. The objection I keep meeting, sometimes from physicists, is that quantum uncertainty must forbid this — that nothing can be located to a precision far below the size of a single particle. The objection sounds reasonable and is wrong, and I want the exact place where it fails on record. Step 1 — do not measure a length; measure a difference The first design decision is that no single length is ever determined, and the reason lies in the geometry of the wave. In the transverse–traceless gauge — the coordinate choice in which a passing gravitational wave shows up purely as a distortion of distances in the plane perpendicular to its propagation, with no trace, so that the two transverse directions are affected oppositely — a wave of the "plus" polarisation arriving along $z$ produces fractional length changes $+h/2$ along $x$ and $-h/2$ along $y$. The trace-free condition is what forces the two signs to be opposite; it is not an accident of the example. LIGO is therefore built as a Michelson interferometer: one laser beam is split down two perpendicular arms, and the returning beams are recombined so that their relative phase — the relative timing of the light waves — sets the brightness at the output port. The instrument reads only $\Delta L = L_x - L_y$. This does two jobs at once. Since one arm lengthens by $hL/2$ while the other shortens by the same amount, the difference is $hL$, twice what either arm alone would give; the factor of two is bought by the geometry, not by any effort. And any disturbance common to both arms — a wander in the laser frequency, a slow thermal drift of the whole building — subtracts away, because it enters $L_x$ and $L_y$ identically. Everything that follows assumes this: from here on only differential disturbances count, which is why the noise budget below is a list of things that fail to be common. Step 2 — the photon budget, or where $10^{-18}\,\mathrm{m}$ comes from The output phase is estimated from photon counts, and the precision of that estimate is what sets the floor. Photons from a laser arrive independently at random, following Poisson statistics, so a count of mean $N$ fluctuates by $\sqrt{N}$ and the fractional fluctuation is $1/\sqrt{N}$. An interferometer phase inferred from such a count therefore carries an uncertainty $\delta\phi \approx 1/\sqrt{N}$ — shot noise; the exact numerical factor depends on the readout scheme but is of order one. On the signal side, a differential mirror displacement $\delta L$ shifts the phase by $4\pi\,\delta L/\lambda$: one factor of $2\pi/\lambda$ converts distance to phase, and the extra factor of two is because the light traverses the change twice, out and back. Setting the signal equal to the noise gives the smallest readable displacement, with the photon number simply the collected energy divided by the energy of one photon: $$\delta L = \frac{\lambda}{4\pi\sqrt{N}}, \qquad N = \frac{P\,\tau}{\hbar\omega}.$$ Here $P$ is the light power circulating on the mirrors, $\tau$ the measurement time, $\lambda = 1064\,\mathrm{nm}$ the laser wavelength and $\hbar\omega = 1.9\times10^{-19}\,\mathrm{J}$ the energy of one such photon. The operating numbers: $P = 100\,\mathrm{kW}$ and $\tau = 10\,\mathrm{ms}$, the latter chosen short enough to resolve structure in a signal near $100\,\mathrm{Hz}$. That is $10^{3}\,\mathrm{J}$ collected, hence $N = 10^{3}/(1.9\times10^{-19}) = 5\times10^{21}$ photons, $\delta\phi = 1/\sqrt{N} = 1.4\times10^{-11}\,\mathrm{rad}$, and $$\delta L = \frac{1.064\times10^{-6}}{4\pi}\times 1.4\times10^{-11} \approx 1.2\times10^{-18}\,\mathrm{m},$$ roughly a thousandth of a proton radius, from counting statistics alone. Two remarks on the inputs. The $100\,\mathrm{kW}$ is not laser output; the input laser supplies of order $10^{2}\,\mathrm{W}$, and the arms are Fabry–Pérot cavities — pairs of mirrors between which the light bounces many times before leaking out — which build up the circulating power and, more importantly, multiply the accumulated phase. The number of effective round trips is $2\mathcal{F}/\pi$, where the finesse $\mathcal{F}$ measures how many bounces a photon survives; at the aLIGO arm finesse of a few hundred this is of order $300$ (Aasi et al. 2015). That surplus is the margin the noise budget below will spend. The storage that provides it also costs something: light stored for many round trips cannot follow a displacement that changes faster than the storage time, so the response rolls off above a cavity pole frequency of order tens of hertz, and the high-frequency end of the band is fighting that roll-off as well as the shot noise. Step 3 — why quantum mechanics permits it Where exactly does the "impossible" argument fail? In two independent places, and it is worth separating them because they are different mistakes. First, the uncertainty principle constrains the state of each photon, not the mean of $5\times10^{21}$ of them. Independent single-quantum uncertainties average down as $1/\sqrt{N}$, and the shot-noise formula above is precisely the statement of that averaging. Nothing here beats the principle; the measurement simply repeats a very imprecise determination an enormous number of times. Second, t