I am in a space-craft moving at .99c observing the Michelson-Morley experiment. I am perpendicular to one of the interferometer 'arms'. Due to the apparent 'reality' of Lorentz length contraction I would now see a different expected deviation of the interference fringes perhaps large enough so that it would now seem to confirm the existence of the 'aether'.
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$\begingroup$ If the interferometer is with you, you won't actually see any length contraction in your own reference frame. To you, everything would appear to take place at its own usual pace (including the passage of light). It's the distant observer who would disagree on the length and time measurements between your and their reference frames. $\endgroup$– Dhruv SaxenaCommented Apr 3, 2017 at 19:19
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2 Answers
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Think of the MM experiment as comprising a longitudinal light clock, a transverse light clock, and a sensitive way to compare their rates.

Once think of it that way, you can see that the MM null result Is true in any frame of observation. Both arms change rate in the same way, so the comparison always gets the same answer.
answered Aug 16, 2019 at 5:21
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Lorentz length contraction is the consequence of the invariance of light speed in all inertial frames. When you say that you would use this fact to prove that aether exists and light is a wave in this media, the length contraction doesn't work anymore.
You are mixing one theory based on non-existence[*] of the Aether to prove Aether exists, that is simply not possible.
[*] To be exact: Not on its non-existence, but on its non-importance for the propagation of light.
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$\begingroup$ I am just suggesting if length contraction was a real effect then the relativistic observer may see a different fringe pattern. If so then the MA experiment results could be interpreted differently. I am not trying to prove the aether exists. $\endgroup$ Commented Apr 3, 2017 at 1:30
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$\begingroup$ @G.W.Kentwell I'm not sure what your question is about anymore. You wrote "...it would now seem to confirm the existence of the 'aether'." But it is not, even if the effect shows. There is two experiments then. One observer has L and L, you have L and contracted L, that implies something, but you still have two observers and the only thing you can prove at all is the relativity effect. Results can't be interpreted differently in this case. This is the same result Michelson and Morley has received. $\endgroup$ Commented Apr 7, 2017 at 19:49