I need help to understand a Michelson Morley experiment in a moving inertial Frame I need some help to understand a Michelson Morley experiment in a moving inertial Frame
If there was a Michelson Morley experiment taking place in an inertial frame (S) that was moving relativistically to my inertial frame (S'), then would I measure the distances travelled by the two half beams to be different in my inertial frame (S')?
letting arm A  in be the arm moving in the x direction (moving inertial frame moving also in the x direction) and arm B be the arm perpendicular to the movement. Then assuming that the scientist doing the experiment in the moving inertial frame (S) has set it up so that in his Inertial frame (S) the two arms A and B are the same length. (Length L)
I would expect  the distance travelled by the half beams not to be the same because 
i) from my inertial frame (S')  I would seem the length of arm A contract, (so that the length L would measure (L')  would be less than that as measured in the moving Inertial frame (S) (ie L > L') and the half beam would travel a total distance of less than 2 L.  ie if Lx is the total legth travelled by the half beam in arm  A in the X direction then 2L > Lx
ii) from my inertial frame I would see the half beam in arm B travel in a sort of V shape whereby the length travelled is twice the hypotonuse of triangle of  one side of length L. Therefore  the length travelled by the second beam will be more than 2L ie if Ly is the total legth travelled by the half beam in the arm in the y direction  then 2l < Ly
Therefore Ly > Lx.
Is this correct?
Now how about the time taken for the two half beams to travel the two arms A and B?
The time taken to travel the two arms ( to the mirror and back) is identical in the moving inertail Frame (S) . As measured in my inertial Frame (S') are the times the same?  I would expect them to be the same from the following argument.
Two events in different inertial frames are not simultaneous if they are seperated in space.
But in the Moving Inertial Frame (S) the two half beams combine and leave the eyepiece of the apparatus at the same time (there is no interference) If the two half beams leaves the eyepiece of the apparatus (ie from the same point in space) at the same time in the moving apparatus, and I see this eyepiece from my inertial frame (S') then I will also see a full beam arriving in my inertail frame (S') at the same time. Afterall all if both half beams leave the apparatus at the same time they must both travel the same path in the same time until they reach my inertial frame (S').
My difficulty here is then is that as  the speed of light is constant for all observers in all inertial frames (S and S') then if the length of the distance travelled by the half beams is different from my inertial frame (S') then there must be a diffent time that I would measure for the light to transverse the different distances. 
As far as I can see there are only 3 possibilites with any stationary observer observing a moving michelson morely  experiment.
1) the time it takes for each half beam to traverse each arm is identical, meaning that as the speed of light is constant, the distance travelled is identical.    (which I think is incorrect)
or
2)  The speed of light is different depending on what direction you measure it. (which is surely incorrect)
or 
3)  the length travelled by the beams are different, which means that the measured times are different, but as the whole beam leaves the eyppiece at the same time, then time is three dimensional.
(But I dont seem to be able to find any literatiure about this? 
Could anyone please help here?
However as the two half beams merge and I see a full beam throught the eyepiece, this would mean that  to the observer in inertia frame (S') time dilates or contracts differently depending on direction, which would suggest that time is three dimensional ie different times are measured in a moving inertial frame (S) by an observer in another inertial Frame ( S') depending on which direction the light is travelling in the moving inertial frame (s) 
I am sure that someone can help? 
 A: Yeah, so, it helps to get very clear about what the split light beam is doing in the frame where the apparatus is moving. Let's say it starts out at $(x, y) = (0, 0)$.
We know that the $x$-track is contracted to $L' = L\sqrt{1-\beta^2}$ (where $\beta =v/c$) whereas the $y$-track is not. However, the light does not travel to the point $(L', 0)$ -- it needs to meet up with the mirror at a point $(L' + v~t, 0),$ as some time $t$ is going to elapse before the light hits the mirror, and the mirror is moving. The standard approach for this in classical mechanics works perfectly well: we divide $L'$ by the relative velocity between the end of the $x$-track and the light pulse in our coordinates, which is going to be $c-v = c~(1-\beta).$ Then on reflection it needs to traverse the same distance $L'$ again, but its relative velocity is now going to be $c + v = c~(1+\beta).$ (Technically it is going from $(L' + v~t_1, 0)$ to $(v~(t_1 + t_2), 0)$ over this time $t_2$, if that helps.)
This means that the total time taken for the light to go in the $x$-direction and return is actually, $$T_x = \frac{L'}{c~(1 - \beta)} + \frac{L'}{c~(1 + \beta)}.$$ Now if we just add these fractions like we were all taught, the common denominator will be $c (1 - \beta)(1 + \beta) = c (1 - \beta^2)$ and the numerators will be just $(1 + \beta) + (1 - \beta) = 2$, leaving just: $$T_x = \frac{2 L'}{c ~(1 - \beta^2)} = \frac {2L}{c\sqrt{1 - \beta^2}}.$$Similarly as you've noticed the time to travel the $y$-path needs to be lengthened because instead of making contact with the mirror at $(x, y) = (0, L)$ the mirror has moved forward a distance, so we must make contact at some point $(v~t, L)$ and therefore the distance the light has traveled must be $(c~t)^2 = (v~t)^2 + L^2$ by the Pythagorean theorem.
Solving for $t$ we actually find directly that $t = L/(c\sqrt{1 - \beta^2})$ and it's the same distance back, so the total time elapsed is $T_y = 2t = T_x.$ The time for light to travel both distances must be exactly the same, and we also predict an interference pattern in the eyepiece even though we're not in the rest frame of the eyepiece.
In fact all of these steps of logic are reversible, and when one runs the argument backwards, it derives that there must be this length contraction effect must happen with effect $L' = L \sqrt{1 - \beta^2}.$ That is, this is a standard "Einstein's trains" argument proving that length contraction must exist; you have simply run it backwards. Where I think you were stuck is this fine point about the relative velocities; I think you wanted to write $T_x = 2L'/c,$ which does not make sense in the reference frame where the apparatus is moving; $L'/c$ is neither the time for the light to reach the mirror or for the light to return to the eyepiece.
Anyway, once we have length contraction right, your option (1) was correct, the distance that you think the two light pulses have gone is absolutely 100% the same, which is why you see both pulses move at the speed of light and come back to the eyepiece at the same time. I don't know what "time is three-dimensional" means, but we don't need it in standard relativity theory.
A: You can actually use the Michelson Morley interferometer in a moving frame to derive the equations for time dilation and Lorentz contraction. Comparing the travel time of the branch perpendicular to direction of movement will yield time dilation. Next, by stating that light from both branches must arrive back at the splitter simultaneously (with equal branch lengths) for both the static and the moving frame, you can derive the equation for Lorentz contraction.
(Note that events that are simultaneous in a moving frame at the same position, i.e. two beam arriving back at splitter, are also simultaneous in the static frame)
