I'm trying to understand the sensitivity of the LIGO interferometer. I've been reading around lots of discussion of how they manage noise cancellation between the two detectors, achieving a very pure laser signal, the many reflections to change the effective length of the interferometer arms to over 1000km, and other impressive tricks of engineering to achieve this remarkable feat of measurement. What I still can't get my head around is how, even in a perfect world with no noise at all, a phase shift of I guess $10^{-7}$ - $10^{-11}$ cycles or so (depending on the sensitivity claim you are taking) can show up as a signal.

From my fairly basic understanding of how an interferometer works. I am imagining that if both beams are initially in phase then a $10^{-7}$ phase shift would change combined amplitude by much less than this (1$0^{-14}$?) If they were at $\pi / 2$ then I guess the amplitude change would be approximately equivalent to the phase shift?

I appreciate that the computational details of how all this works are going to be a whole lot more complicated, but I would really appreciate some pointers on how best to understand the sensitivity challenge here:

  • Is it a case of measuring tiny amplitude variations in the combined laser signal?
  • Is the size of these amplitude variations (as a proportion of amplitude) equivalent to the size of the length variations (x number of reflections and as a proportion of laser wavelength)?
  • If yes to each of the above, am I right in thinking they are detecting laser amplitude oscillations down to $10^{-10}$ ish times typical amplitude?
  • If no to the first two points, please set me straight in any way you can! Many thanks in advance.
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    $\begingroup$ Commenting because I don't have sources to hand: The beams are held in antiphase, I believe and constantly adjusted to keep them there by some kind of feedback loop. The amount and direction of those adjustments is the actual output. I seem to to recall that the limiting factor is "shot noise" ie the fact that the detectors see a whole number of photons. It looks (from the waveforms they have published) as though they get a readout every 1 or a few milliseconds, so the limit of the signal they can detect is about 1000 photons/second. Roughly $10^{-16} W$. The beam power is 100s of KW. $\endgroup$ Jul 8, 2020 at 8:16

1 Answer 1


The LIGO interferometer uses a homodyne detection technique. Basically, the light travelling in each arm of the interferometer is derived from the same laser source and is combined in the output channel and falls onto a photodiode.

The interferometer is operated so that when there is no gravitational wave (GW) passing through the instrument, the beams combine to produce a dark fringe (i.e. they are set to destructively interfere). There is a small offset from this, but basically the phase difference between the combiniing beams is close to $\pi$.

The phase difference caused by a GW, due to the changing length of one arm with respect to the other, can be derived as $$ \Delta \phi \simeq 2\pi \left(\frac{2hL}{c}\right) \left(\frac{c}{\lambda}\right) = \frac{4\pi}{\lambda} hL \ , $$ where $L$ is the length of the arms, $\lambda$ is the laser wavelength and $h$ is the strain amplitude of the fravitational wave signal. Actually, it is a little bit more complex than this, since the arms act as Fabry-Perot resonators which means the light effectively travels backwards and forwards many times in the arms (about 300 for LIGO, i.e. $L$ is effectively 1200 km).

For a typical dimensionless GW strain of $h \sim 10^{-21}$, $\lambda = 1064$ nm, then $\Delta \phi \sim 10^{-8}$ and is modulated at the frequency of the GW (typically 20-2000 Hz).

The problem then reduces to combining $$ E_{\rm tot} = E_0\sin (\omega_l t) + E_0 \sin (\omega_l t + \alpha + \Delta \phi)\ ,$$ where $E$ is the electric field in each arm, $\omega_l$ is the angular frequency of the laser and $\alpha$ is the offset phase between the arms (close to $\pi$).

Using the identity $\sin a + \sin b = 2 \cos[(a-b)/2] \sin[(a+b)/2$ and squaring the total E-field to get an intensity: $$I = 4E^2 \cos^2[(\alpha + \Delta \phi)/2]\, \sin^2[\omega_l t +(\alpha + \Delta \phi)/2] $$

Since $\omega_l$ is much greater than the GW frequency and much higher than can be sampled by any photo-sensitive detector, then the second term in the product above can be replaced by its time-average of $1/2$. If we now identify the total power $P_{\rm in}=E^2$ as the average input power to each arm of the interferometer and note that $\cos^2 (a/2) = (\cos(a)+1)/2$ and $\Delta \phi \lll 1$ $$ I = P_{\rm in} \left[1 + \cos(\alpha + \Delta \phi ) \right] \simeq P_{\rm in} \left[1 + \cos(\alpha) -\Delta \phi \sin(\alpha)\right] = 2P_{\rm in} \left[ \cos^2 (\alpha/2) - \frac{\Delta \phi}{2}\sin \alpha \right]\ .$$

It is the second term inside the bracket that contains the signal of the GW. That signal is proportional to the power in the interferometer and the phase difference between the arms. Note that although the signal-to-(shot) noise is mathematically maximised when $\alpha=\pi$, this would mean the SNR was 0/0 ! In practice there is always some other noise present so $\alpha$ is shifted a little bit away from $\pi$ - Fricke et al. (2012) suggests that $\alpha \sim \pi+ 6\times 10^{-5}$ is used.

The power input into each arm is about 600 W (the 100s of kW Steve Linton mentions in a comment is after accounting for the Fabry-Perot resonator, which I did above by talking about an "effective $L$"). In the absence of other forms of noise then photon counting (shot noise) becomes the limiting factor and is proportional to the square root of the power.

The output signal is the modulated GW signal discussed above which is recorded by detecting photons with the photodiodes. The response function that translates the photodiode signal into a strain is determined by acting on the test masses/mirrors with precisely calibrated lasers modulated at GW frequencies that can produce monochromatic phase shifts in the arm lengths.

  • $\begingroup$ What they're saying are GW's are so violently microscopic, so absurdly abstract, so amorphous that it could literally be the vibrations from a nearby squirrel scratching it's bum. $\endgroup$ Jul 8, 2020 at 13:24
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    $\begingroup$ @WhitePrime, well it could be, except that there are at least 15 orders of magnitude or so of damping between any squirrel and the test mass; a squirrel doesn't scratch it's bum at 100 Hz, or in such a way that it resembles the chirp of a black hole merger; and you'd have to have a similar squirrel scratching its bum in both Louisiana and Washington at almost the same time in order to simulate a detection. But otherwise, yes, "seismic" background noise means that LIGO can't operate much below 20 Hz. $\endgroup$
    – ProfRob
    Jul 8, 2020 at 13:55
  • $\begingroup$ This is a wonderful answer and will take some time to read, is one reason for the small but definite offset to allow the determination of the sign of the strain? In a conventional interferometer (e.g.) with say $\lambda/100$ resolution they can just use quadrature detection, but that's not an option here. If it sat exactly on a null then the intensity would be proportional to strain squared wouldn't it? $\endgroup$
    – uhoh
    Jul 9, 2020 at 8:50
  • $\begingroup$ @uhoh The strain is an amplitude so I'm not sure what you mean by a sign? If you follow the maths above you'll see that if the offset was exactly $\pi$ you would only get a quadratic term in the power like $P_{\rm in} (\Delta \phi)^2/2$, which is basically zero. Of course the shot noise would also be correspondingly small. In the real world there is real other sources of noise though so you need to boost the signal and the maths above shows that you can do that by offsetting away from $\pi$, at the expense of increasing the shot noise too, but increasing the overal SNR. $\endgroup$
    – ProfRob
    Jul 9, 2020 at 9:06
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    $\begingroup$ @uhoh I get what you mean now. Yes, you are sensitive to $h$ and not $h^2$. $\endgroup$
    – ProfRob
    Jul 9, 2020 at 9:09

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