- #1

Kanesan

- 36

- 2

MMX has been discussed a lot but I could not find solution to my problem. So, I had to start a new thread on this subject! Sorry it is lengthy, you can just read the last paragraph (below red text) for short form.

I was going through details of the calculations and have problem at one step on the vertical calculation. Before going into that step, I will detail the general approach of my understanding so far. I am trying to use same symbolism as in Special Relativity (with primes), please bear with me.

For any diagram, I am using what is in Wikipedia:

http://en.wikipedia.org/wiki/Michelson–Morley_experiment#Light_path_analysis_and_consequences

Situation A when there is 100% aether drag: t'/t = 1 for both horizontal and vertical.

Method 1: velocity relative to local aether is zero.

Method 2: If distant static aether is used as reference, speed of light will be affected by same extent and compensate the L changes which will result in t'/t = 1.

So, anyway we will expect a null result if there were full aether drag. So far I have no problem.Method 2: If distant static aether is used as reference, speed of light will be affected by same extent and compensate the L changes which will result in t'/t = 1.

**Situation B**when there is 0% aether drag:

Horizontal L'/L = t'/t = gamma^2 (not yet including length contraction) (still no problem).

**Vertical:**Only here I have problem understanding how it can be L'/L = t'/t = gamma.

We know that the mirrors will have the increased distance of L' = L*gamma (always). But, my thinking is that the light will not be following the mirrors. It will just stay with static aether and move straight up/down (in earlier Situation A it went to the mirror only because it was dragged there by the aether). So, the light path will not be same as the hardware distances.

This will give only L'/L = t'/t = 1 (or same calculation arrived by Michelson first as T

_{t}= 2L/c). In such case, the beam will hit the top mirror vt distance behind and hit the inclined mirror at 2vt distance behind (compared to static Earth expected positions).

**Basically the confusion is how the light can follow same path whether aether drag is there or not?**It will be great if where I am mistaking is pointed out.

Thanks all and sorry for the lengthy description.