- #1

crick

- 43

- 4

**reference frame where the interferometer is moving.**Consider the picture ##2.##

Setting ##ab_1=ac=L## and ##\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}## (##v## is the velocity of the interferometer), the time taken for path ##aba_1## should be

$$t_1=\frac{2L}{c}\gamma$$

For the other ray the path length should be ##aca## (but it's contracted), therefore

$$t_2=\frac{2L}{c}\frac{1}{\gamma}$$

How can possibly be ##t_1=t_2##?