During the detection of gravitational waves by LIGO machine, why wasn't the laser light used in the detection, also stretched as a result of the stretching of space time fabric?


Yes, it stretches and squeezes the light, i.e., the wavelength, and does tthe opposite (lowers and raises) the frequency, and keeps the speed of light the same. What the interferometer sees is that the light in the longer path (because that path was stretched) had to travel further, a peak took longer to arrive (in the stretch portion, the opposite in the squeeze).

So, say if a second Ray of light which did not see that stretching (keep to stretching, we mean alternatively stretching and sneezing in concert with the gravitational wave), and at some instant say it arrived at a peak, then the first one would not be at a peak, there would be a phase difference. The interferometer is sensitive to that phase difference. If that first one was at a trough, it'd cancel out with the second Ray.

The fact that the light going through the gravitational wave arrives a little later allows us to know that the gravitational wave was there. The wavelength is indeed stretched (and later squeezed), but the speed of light is the same, and so it arrives a little later (for a given phase)

In fact, of course, LIGO uses perpendicular legs, and when one is stretched the other one is squeezed, and it changes, so we see the changing phase difference as an electrical wave on the output.

This is answered and described in LIGOs FAQ at https://www.ligo.caltech.edu/page/faq

  • $\begingroup$ You have correctly given the standard argument as to why a gravitational wave effects the interference pattern. But why did you assume the coordinate speed of light along the two axis was c ? The speed of light is invariant when looking into a frame related to yours by a Lorentz transformation (ie: it leaves the Minkowski metric unchanged), but Schwarzschild (Shapiro delay) and gravitational waves change the Minkowski metric and thus change the coordinate speeds of light. In fact in just the right way so the wave fronts' times of flight are unchanged and the same along both arms. $\endgroup$ – Gary Godfrey Jan 23 '17 at 22:29
  • $\begingroup$ What is your definition of time of flight? A certain peak leaving and arriving, or if not what? $\endgroup$ – Bob Bee Jan 24 '17 at 1:00
  • $\begingroup$ Note that my answer is the same as the FAQ. They also state that c is constant. Btw, this is tricky, there's other previous answers on the site, and discussion. $\endgroup$ – Bob Bee Jan 24 '17 at 2:03
  • $\begingroup$ So as it is said, it does not measure distances, but time differences of arrival of a peak $\endgroup$ – Bob Bee Jan 24 '17 at 2:31
  • $\begingroup$ And for your ref., and others who might not have seen it, here is a presentation of the right heuristic way to think about it. See Saulson at aapt.org/doorway/tgrutalks/Saulson/…. He is/was one of the leaders at LIGO, said it took him years to get it right, though others knew how to do it right. Non-trivial, and everybody gets confused. $\endgroup$ – Bob Bee Jan 24 '17 at 2:50

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