We suppose that we have a Michelson-Morley interferometer in free fall, will there be no interference: the round trip time in both arms of the interferometer is the same?
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$\begingroup$ Where is the observer? Is the observer also in free fall with the system? $\endgroup$– YejusCommented Dec 19, 2021 at 12:34
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$\begingroup$ Of course, thank you for specifying it. $\endgroup$– The TilerCommented Dec 19, 2021 at 14:12
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1 Answer
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Yes, a Michelson Morley interferometer will have no interference in free fall.
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1$\begingroup$ Thanks, but how did you know? (a reference please) $\endgroup$ Commented Dec 19, 2021 at 14:18
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2$\begingroup$ The LISA Pathfinder mission achieved sub-picometer accuracy in free fall. arxiv.org/abs/1903.08924 $\endgroup$– DaleCommented Dec 19, 2021 at 15:02
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$\begingroup$ Thank you very much even if it is not my goal but it is a very interesting experience and good to know $\endgroup$ Commented Dec 19, 2021 at 16:05
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$\begingroup$ A long time ago I tried to change (accomodate) the Lorentz transformations for acceleration with: v'=γ(v-at), with: c²-a²t²=0, I got: t'(1)=T/[1-(at/c)] and t'(2)=T/[1-(at/c)] and t'(orthogonal)=T/[1-(a²t²/c²)]^(1/2),.... :-) but this is custom-made DIY:-) (DL traduction, My level in English is average, I registered here to improve it, Arabic after French and now English and they say why we are behind (third world) !!!) $\endgroup$ Commented Dec 19, 2021 at 16:29
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$\begingroup$ @TheTiler to get relativistic transformations between non-inertial frames it is best to use tensors. The Lorentz transformations for acceleration won't be meaningful $\endgroup$– DaleCommented Dec 19, 2021 at 20:06