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Say I am cycling past a Michelson interferometer set up. Someone at rest with the experiment will see both beams travelling a distance $L_0$ and hence conclude that $T_{AB}=T_{AC}=\frac{2L_0}{c}$.

However, according to me, the beam AB will not be travelling $L_0$, but a contracted distance $\frac{L_0}{\gamma}$, hence $T_{AB}=\frac{2L_0}{\gamma c}$. As AC is perpendicular to me and hence there is no relative motion along that axis, I don't observe AC to have any length contraction. However, I do still observe the whole light clock effect or time dilation. According to the observer at rest with the set up, the light beam travels along a horizontal line. According to me however, it travels diagonally, and the overall path $ACA$ describes a triangle. Hence, I see that $T_{AC}=\frac{2L_0 \gamma}{c}$.

This seems to imply I will observe interference while the observer at rest with the set up will not.

Note: $T_{AB}$ and $T_{AC}$ describe the time taken to travel the overall paths $ABA$ and $ACA$ respectively.

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    $\begingroup$ Your $T_{AB}$ formula has not taken into account that in your (bike) frame the beam splitter moves between the events where light departs and light comes back. And the top mirror moves as well. You have to account for all that, then it comes out right. Oh and by the way you have the beam splitter at the wrong angle. $\endgroup$ – Andrew Steane Mar 16 at 14:58
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    $\begingroup$ This has been treated in detail on the site several times already. I did a detailed calculation under physics.stackexchange.com/q/383461/520. See also physics.stackexchange.com/q/276574/520, physics.stackexchange.com/q/14362/520. And probably others. $\endgroup$ – dmckee Mar 16 at 16:10

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