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The microwave background is due to the expansion of the Universe where the wavelengths of radiations are stretched by spacetime.

As in the LIGO experiment, in the presence of gravitational wave, will the wavelength of laser fluctuate with spacetime as well?

Machine
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  • The microwave background is due to thermal radiation decoupling from the plasma in the early universe. The modulations in that radiation are primarily from acoustic oscillations in the plasma. The modes caused by suggested primordial gravitational waves have not been observed and may not be observable because of magnetic fields and dust in the modern universe. – CuriousOne Feb 12 '16 at 04:38
  • @CuriousOne Thanks! But I think we are addressing different issues. Let's say the spacetime is a rubber sheet which will expand as the universe inflates, and stretch and squeeze in the presence of a gravitational wave. In the former case, the wavelengths of radiations get longer as the rubber band stretches. But in the latter case, will the wavelength of a laser follow the deformation of the rubber sheet? – Machine Feb 12 '16 at 18:09
  • I know that's not your question, I am simply trying to make sure you and others don't get false ideas what the CMB is and how it is related to gravitational waves. Light follows space, that's why it changes wavelength because of the expansion of the universe and the same is happening in a gravitational wave. – CuriousOne Feb 12 '16 at 19:51
  • @CuriousOne If the wavelength fluctuates in pace with the space, then the optical path will remain constant in unit of wavelength, and so the gravitational wave will not be observed from the interference experiment. – Machine Feb 12 '16 at 21:56
  • We have a detailed answer to your last question here: http://physics.stackexchange.com/questions/153657/can-ligo-measure-anything, and that answer has a link to Kip S. Thornes chapter about it, but in short... that's why one needs two interferometer arms, which are changed by different amounts. A single arm interferometer would not detect the wave, if I understand the theoretical explanation correctly. – CuriousOne Feb 12 '16 at 22:16

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