# How does LIGO work?

LIGO is described as working as an interferometer, like a Michelson-Morley interferometer but with many reflections along the arms to increase the sensitivity. In MMs work it was assumed that the mirrors were held in a rigid relationship and so differences in light speed along the orthogonal arms would show up as a phase shift. But in LIGO the mirrors at the ends of the arm are assumed to be test particles that will move with the gravitational waves distortion of space. Wouldn't this distortion also affect the proper distance the photons travel so that there would be no phase shift? How can the mirrors and the photons react differently to the change in spacetime metric?

## 2 Answers

This is an excellent question! The LIGO arm cavities are about 4 km, which takes a photon roughly $$10^{-5} s$$ to traverse. On top of that, as you mentioned a typical photon will bounce around a few hundred times (maybe 500 times) inside the cavity. So a typical photon may spend, say, a few milliseconds in the cavity. Another way to say this, is that we expect the photons in the cavity to be replenished with a frequency of about 100 Hz. This is known as the pole frequency of the LIGO interferometer.

Now a typical gravitational-wave signal from a compact binary coalescence will chirp (increase in frequency) from when it enters the detector's sensitive band (around 20 Hz or so) to when the binary finally merges. The scaling is $$f \sim (t_c-t)^{-3/8}$$, where $$t$$ is the observation time and $$t_c$$ is the time of coalescence, and this is valid to leading order in $$v^2/c^2$$, where $$v$$ is the orbital velocit). Typical merger frequencies are of order a few hundred Hz for black holes, and 1000-2000 Hz for binary neutron star. This scaling means that a binary will tend spend much more time at low frequencies, below 100 Hz, than above it.

In this regime, the photons enter and leave the cavity before the gravitational wave makes a full oscillation. So what is being measured? Well, what is measured is the phase difference of two photons at the output of the interferometer, who made a round trip along two different arms. Two photons that entered the interferometer at the same time will have the same phase when they leave, but they will leave at different times because the round trip times will be different. Therefore by comparing photons that exit the interferometer at the same time, LIGO is essentially measuring the difference in "entry times" to the interferometer for a photon that went on one path vs another. The bottom line is that we have to think of these photons as traveling waves, and what LIGO measures is the change in the behavior of traveling waves as a GW passes through the instrument.

For high frequency gravitational waves, there is some loss above the pole frequency, essentially because of the effect that the OP is saying. The standard picture that "the arm lengths are oscillating differentially so the flight time of a photon is different in different arms" ends up mapping to a calculation that is leading order in the spacetime curvature; higher order terms can be included (in fact the calculation can be done exactly to linear order in the metric perturbation) and the effect of these terms is indeed to reduce the ability of the detector to detect GWs.

[Optional, more advanced technical note] Actually a very subtle point is that because LIGO is a Fabry-Perot cavity, with perfect clocks they could measure GWs with only one arm. In principle LIGO could put the cavity on resonance, and then any GW passing by would change the arm length and knock the cavity away from resonance. However the issue is that there are no perfect clocks; there is frequency noise in the laser which can't be reduced as low as is needed to detect GWs. So in fact what is done in practice is to actively control the interferometer degrees of freedom to hold the interferometer on resonance (or "locked"). The gravitational wave channel, is actually derived from what displacements need to be added to the "natural" motion of the mirrors to keep the interferometer locked; this is called the error term. To maintain resonance of both arms, the light in one cavity can be used as a reference clock for the other cavity. In the end, two arms are still needed, but not because of the analogy with the Michelson interferometer usually given, but because the frequency noise cancels out of the differential arm channel.

Some references:

• So one of the arms is that light clock that’s so often used in explanations of special relativity? – Roman Odaisky Mar 2 at 21:55
• @RomanOdaisky I don't think it's quite the same thing. First, to draw an analogy with this thought experiment, we would need to work in special relativity, so we would need to remove all gravitational waves and any other gravitational effects. In this situation, what the LIGO setup does is to cancel any frequency noise in the output of the interferometer -- essentially both arms have the same uncertainty in the frequency of the light, so this can be cancelled, but neither observer is sure about their own clock. I think Einstein synchronization assumes each observer has a perfect clock. – Andrew Mar 2 at 23:48
• So you actively control the resonance, using the error term as the output (that then has to be compared with the other arm)? That's essentially the same operation as a common FM radio, just with different hardware. – AI0867 Mar 4 at 10:19
• @AI0867 Of course there is more to LIGO, but yes there are a lot of radio frequency engineering that goes into the control system and into LIGO's readout scheme. If you are interested there is a lot more detail and references to further details in the Advanced LIGO instrument paper: arxiv.org/abs/1411.4547 – Andrew Mar 4 at 11:40

It is my understanding that LIGO measures tidal forces --- i.e the difference in the direction and strength of gravity from the local vertical --- and these forces can be significant even when the geometry of space is little affected. The change in the separation of the mirrors is due to their $${\bf f}=m{\bf a} =m (\delta {\bf g})$$ motion and not because $$\int \sqrt{ds^2}$$ changes. Thus the interferometer part of LIGO works just like any other optical intererferance --- except that its precision is in the order of $$10^{-15}$$cm $$\approx 10^{-10} \lambda$$.

• I agree that LIGO measures tidal forces. But I would say "force on the mirror" picture and "proper time elapsed in transit" are just two different ways of ways of talking about and calculating what is measured. The "force on the mirror" picture lets you analyze the interferometer like any other interferometer (as you said), while phrasing things in terms of coordinate invariant quantities like the proper time is more natural from a GR theorist point of view and also is valid when the wavelength of the gravitational wave is smaller than the interferometer. – Andrew Mar 2 at 0:54