And since we've already done explicit application of the Lorentz transforms and a shortcut with time dilation let's do this a third way.So the answer to both your questions is \(t'_f=2\gamma nL/c\) as derived in #62 which is just \(\gamma\) times the answer I gave in #61

Light travelling at speed \(\pm c/n\) in the x direction in the rest frame of the interferometer moves at speed \((\pm c/n-v)/(1\mp v/(nc))\) in the frame of the car and in the +x direction it moves a distance \(x'_{r1}-x'_0=\gamma L(1-nv/c)\) and in the -x direction it moves a distance \(x'_f-x'_{r1}=-2\gamma Lnv/c-\gamma L(1-nv/c)=-\gamma L(1+nv/c)\) where the \(x'\)s are as defined in #62 with \(\theta=0\) so that the interferometer arms are aligned with the +x and +y axes. The total elapsed time is given by distance over speed in the outward direction plus distance over speed in the inward direction which is \[

\frac{\gamma L(1-nv/c)(1-v/nc)}{c/n-v}-\frac{\gamma L(1+nv/c)(1+v/nc)}{-c/n-v}

\]which simplifies to \(2\gamma L n/c\) as expected.

Now let's look at the light in the other arm of the interferometer. Light travelling at speed \(\pm c/n\) in the y direction in the rest frame of the interferometer has x-velocity \(-v\) and y-velocity \(\pm c/\gamma n\) which makes a speed of \(\sqrt{v^2+(c^2-v^2)/n^2}\) and it moves an x distance of \(x'_{r2}-x'_0=-\gamma Lnv/c\) and a y distance of \(y'_{r2}-y'_0=L\) which means a total distance of \(\gamma L\sqrt{1+(n^2-1)v^2/c^2}\) in the upwards direction and the same in the downward direction from symmetry. Again elapsed time is distance over speed for the two legs so it is \[

\frac{\gamma L\sqrt{1+(n^2-1)v^2/c^2}}{\sqrt{v^2+(c^2-v^2)/n^2}}+\frac{\gamma L\sqrt{1+(n^2-1)v^2/c^2}}{\sqrt{v^2+(c^2-v^2)/n^2}}

\]which again simplifies to \(2\gamma Ln/c\) as expected.

TonyYuan take note that this is the right way to do the calculation you messed up in #56 and #58. Comparing the first calculation to yours you had the velocities (your v1 and v2) correct until you added Earth's orbital velocity for no reason but you had the distances travelled incorrect because you failed to allow for the fact that the interferometer is moving in a frame where it is not at rest. In the second calculation you had the velocity (your v3) wrong because you failed to do the transform properly and you had the distances wrong because you again failed to allow for the fact that the interferometer is moving when it's not at rest. So your belief that there ought to be phase differences comes from your failure to do the maths correctly.

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