# MM ether experiment/length contraction

1. Jan 31, 2006

### Pengwuino

Ok I have this equations Michelson and Morley used:

$$Delta t = t_2 - t_1 = \frac{2}{c}(\frac{{l_2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} - \frac{{l_1 }}{{1 - \frac{{v^2 }}{{c^2 }}}})$$

I need to show that if the length is contracted along the direction of motion, the result comes out to be 0. My question is, which direction is the direction of motion? I have a feeling its L1....

2. Feb 1, 2006

### cepheid

Staff Emeritus
Yeah, take a look at how this formula is derived. The factor of 1 - v^2/c^2 comes from the fact that the earth moves at v wrt to the ether (take all this with a grain of salt), and the beam moves at c relative to the ether, so in the Earth frame, the beam is chasing after the target in the direction of motion with a relative speed of c - v on the way there, and rushing back towards the beamsplitter with a relative speed of c + v on the way back.. So if you work it out, the time it takes to make a round trip along the arm in the direction of motion is given the the expression involving l1, so obviously l1 is the arm in the direction of motion.

The expression involving l2 is more complicated because this is motion perpendicular to the earth's motion, which amounts to motion diagonally through the ether.

3. Feb 1, 2006

### Pengwuino

Alright thanks. I needed to use the length contraction to show how delta t would = 0 if applied and it worked! woohoo.