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?
[Physics] How does LIGO work
gravitational-wave-detectorsgravitational-wavesinterferometryligooptics
Related Solutions
It is a misconception that LIGO is a very accurate instrument, it has an uncertainty in calibration which is on the order of 10%. This means that the measured strain amplitude of GW150914 of $1.0 \cdot 10^{-21}$ could easily have been $1.1 \cdot 10^{-21}$. Note that this is just a scaling error.
LIGO is however extremely sensitive, it can measure relative length variations on the order of $10^{-22}$, but only in a bandwidth between 10 and 2000 Hz. At lower frequencies, the measurement fluctuates by several orders of magnitude more. You need to do band-pass filtering to reveal the actual signal.
As already mentioned in Chris's answer, a Michelson interferometer can only measure incremental changes in the path length difference. It does not say anything about the absolute length of the arms, and not even about the absolute difference in arm lengths. For a perfect Michelson interferometer, the resulting power on the photodiode is$$P = \frac{P_0}{2} \left(1 + \sin\left(4 \pi \frac{L_1 - L_2}{\lambda}\right)\right)\,,$$which will only tell you how much the difference in the arm lengths changes over time.
Still, there are reasons why you want to have the long arms as equal as possible. For a simple bench-top interferometer with a crappy laser diode, the path lengths need to be reasonably similar, otherwise you run into problems with the coherence length. This is not an issue for LIGO, they use Nd:YAG lasers which already have a coherence length measured in kilometers when running alone. These lasers are further pre-stabilized by locking them on ~16 meter suspended cavities, and finally the laser frequency is locked on the average length of the two 4 km long arms. The resulting line-width of the laser is on the order of 10 mHz, so a coherence length larger than $10^{10}$ meters ...
You still want to make the length of the 4 km arms pretty equal, since any imbalance would couple residual noise of the laser frequency to the differential length measurement. With standard GPS-based surveying methods, the mirrors are positioned with an accuracy on the order of millimeters. There is no need to do this much more accurate, since there are other sources of asymmetry that can couple frequency noise to the differential measurement, such as the differences in absorption and reflection of the mirrors used in the two arms.
One minor detail, which is extremely important in this context, which you perhaps missed is that the beam-splitter is partially silvered at the lower surface, which implies that the appropriate location of the point O is at the lower surface of the beam-splitter (at the glass-air interface).
If we take this detail into account, the explanation is very simple:
- There are $\pi$ phase changes at steps B1-2 and B2-2, but these are common for both beams, and hence do not contribute to any net relative phase difference. (These reflections are depicted as such in the edited diagram)
The only relative phase difference arises due to complete reflection at B1-3 (reflection at the outer, lower surface of the beamsplitter.) This occurs for only one of the two beams, not for both. Hence, net relative phase difference is $\pi$ radians. (This is represented in the diagram for the red colored beam. Note that this is not true for the other beam, shown in blue.)
(IMPORTANT) There is no phase change of $\pi$ in the reflected beam, arising at the original division of amplitude point O, originally. This is because the division took place at the lower interface, where the interface was glass-air and not air-glass. If we invoke the Stokes' relation, you have a phase change of $\pi$ on a reflection at a rare-dense interface. This doesn't fit the description. (Hence, no phase shift of $\pi$ for the blue beam in the figure.)
Hence, the total relative phase difference between the two coherent beams, when they recombine, is only $\pi$ radians.
(Original image edited on suggestion by Floris).
Best Answer
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.
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