 LIGO displacement readout and control

An ideal single-round-trip Michelson interferometer converts differential length into phase difference $\delta \varphi =4\pi \Delta L/\lambda$. If its dark-port power scale is $P_0$ and its static phase offset is ${\varphi }_0$, thenExactly at a dark fringe, ${\varphi }_0=0$, power is quadratic in small displacement and loses its sign. A deliberate offset supplies an optical reference field: interference with the small signal field produces a linear term. Equivalently, $|E_{\mathrm{ref}}+E_{\mathrm{sig}}{|}^2$ contains $2\mathrm{Re}(E_{\mathrm{ref}}^{\ast }{E}_{\mathrm{sig}})$. The photodetector measures power, not a signed displacement directly.

Advanced LIGO uses this DC homodyne principle with a small differential arm offset and an output mode cleaner. Its feedback control maintains optical resonance and alignment, using actuation distributed over the suspension stages.^[ref] The simple formula above omits resonant storage: arm and recycling cavities give a frequency-dependent optical gain and delay, so no constant pass-count multiplier describes the entire observing band.

To see why a locked detector can still measure a disturbance, use a linear feedback control model. Let $x_{\mathrm{ext}}$ be the external differential displacement, $d_{\mathrm{err}}$ the residual readout, $d_{\mathrm{ctrl}}$ the recorded control command, $C(f)$ the sensing response and $A(f)$ the actuation response. With the convention $x_{\mathrm{res}}=x_{\mathrm{ext}}-A{d}_{\mathrm{ctrl}}$, reconstruction givesHere tildes denote Fourier transforms; $C$ has readout units per metre and $A$ metres per command unit. Feedback suppresses residual motion but records the compensating command. Omitting that command would suppress the reconstructed low-frequency signal too. In the actual GW150914 experiment, a modulated auxiliary laser exerted calibrated radiation pressure on a mirror to measure the response; periodic excitations monitored calibration.^[ref]