# Michelson-Morley Confusion

1. Aug 5, 2014

### teodorakis

hi, i am stucked with a confusion about michelson-morley experiment. I am not going into mathemtaical detail, the result of the experiment indicates that vertical and horizontal light waves meet at the same time not producing any shift. In an inertial reference frame that's very obvius but in a moving reference frame( and as far as i understood from the experiment we put ourselves as an observer of a moving reference frame) how do we equate the time needed for light to travel in horizontal and vertical clocks in fact sync clocks in a stationary reference frame isn't synched relative to a moving observer so,
In this experiment do we observe the situation(In a staionary reference frame)?
Then how do we equate the times of vertical and horizontal clocks?( which evidently lead to the length contraction.)
I hope i can define my problem.

Last edited: Aug 5, 2014
2. Aug 5, 2014

### Staff: Mentor

We start with a single light source that we split and send in the two directions with a half-silvered mirror. Because it's one light source we know that the high spots and low spots in the electromagnetic waves are happening at the same time even though we don't know exactly when that time is.

Thus, when we bring them back together, any shift in the interference pattern has to be due to either one path being longer than the other, or the light traveling faster along one path than the other, or a combination of the two. So we do the experiment, then we rotate the whole apparatus so that the directions are different but the distances are the same, and check the fringes again. If the directions relative to the earth's motion matter, the fringes should change - and they don't.

3. Aug 5, 2014

### teodorakis

Thank you for the answer, i will try to be more precise, when they first saw the results of mm experiment for an express solution length contraction theory is rised. With the proper amount of length contraction for a relative speed v there is no interference pattern as the experiment results but i didn't get that if we are in the same inertial reference frame(earth) we shouldn't realize the contraction right?
Then another question comes to my mind: are we observers of the experiment? If so the light clocks shouldn't be synchronized and there should be an interference pattern?

4. Aug 5, 2014

### teodorakis

i mean think about an observer with a relative speed v to the mm experiment device and does he/she see the same results as the stationary observer see? I think he shouldn't because of relativity of simultaneity. He should see both clocks dilate the same amount but the light beam arrive to the starting point at different times right? But when we try to explain the mm experiment the moving observer doesn't conclude such a thing. and due to length contraction, we say the light arrives the starting point at the same time ahhhhh finally i think i can made my point. Please help with this problem it really affects me bad:)

5. Aug 5, 2014

### Staff: Mentor

When the experiment was first done, the hypothesis being tested was that even in an inertial frame the speed of light would be different in different directions because of that frame's movement with respect to the hypothetocal ether. Thus, the experiment was a single-frame experiment: given that we're in a single inertial frame, can we detect that frame's motion relative to the ether? One of the early (and now discredited) explanations for the observed null result was that distances contracted along the direction of motion relative to the ether.

I don't understand what you mean by asking if we are "observers of the experiment", but I think you're still confused about how the time synchronization. There is only clock in the entire experiment, and that's the initial light source. We can use the high and low points of that electromagnetic wave as time references throughout the entire experiment.

6. Aug 5, 2014

### Staff: Mentor

Both observers see exactly the same pattern of light and dark bands on the interferometer screen. Relativity of simultaneity doesn't come into it because they both only need to look at the pattern once and it's unchanging.

7. Aug 6, 2014

### vanhees71

I think the confusion is due to the fact that usually the Michelson-Morley experiment is used to introduce special relativity, using Galileo space-time first and then coming to the conclusion that it is not consistent with what's observed concerning light or more generally electromagnetism; the long story's conclusion is that Maxwell's equations are not Galileo invariant and one has to use the Minkowski space-time to describe Nature more correctly.

Ironically usually the textbooks don't discuss the MM experiment in terms of Special relativity to explicitly show that the null result is well explained by it. A detailed analysis can be found in

Reinhard A. Schumacher, Special relativity and the Michelson–Morley interferometer, Am. J. Phys. 62, 609 (1994)
http://scitation.aip.org/content/aapt/journal/ajp/62/7/10.1119/1.17535

8. Aug 6, 2014

### Chronos

I am curious, do you think M&M misinterpreted, or should have questioned their own experimental results?

9. Aug 6, 2014

### teodorakis

Yes i think my confusion is about this. But unfortunately full text is not available with the link you sent. So to explain the null result with SR briefly do we just accept the fact that we are in a an inertial reference frame with the mm device and we don't "observe" any sr effects so the null result comes naturally. But if we are in a reference frame that have a relative velocity v to the mm device than we have to observe an interference pattern because of relativity of simultaneity(just like the popular lightening strike of the train), that's the new confusion for me.
Thank you so much for the time you allocated.

10. Aug 6, 2014

### ghwellsjr

I don't think MMX had anything to do with relativity of simultaneity or the lightening strike of the train. The null result of MMX was manifested as a static image of light and dark patterns visible to the experimenter as the experiment (and the experimenter) was rotated. I'm not sure an observer stationary on the ground and watching the rotating apparatus would have been able to see what the experimenter could see (looking through an eyepiece) but even if the image was projected on a screen, all observers, moving or not, with respect to the apparatus would see the same static image. Why do you think it could possibly be any different?

11. Aug 6, 2014

### teodorakis

I probably miss something about synchorinization of clocks, think of the mm device as two light clocks they are obviously in synch in resting frame but are they in synch in moving reference frame?
How do we , (or do we?) automatically assume that for an observer in a moving frame see the both clocks synchorinized?
Here I come up with a solution like the light leaves the source in one particular x location (for both observers)
that's event 1 and comes back to the same x location(that's event 2) that's why events must be simultenous for a moving observer as the observer in rest. Moving observer just see the time dilation effect, is this right?

12. Aug 6, 2014

### Staff: Mentor

Synchronization is not an issue for MMX. What MMX tests for is anisotropy in the 2 way speed of light wrt the detectors. If the speed of light wrt the device is anisotropic then the MMX shows interference fringes, otherwise it does not.

13. Aug 6, 2014

### ghwellsjr

I thought MMX would always show interference fringes, no matter what, the only question was whether or not they would shift around as the experiment was rotated. Did I get that wrong?

14. Aug 6, 2014

### Staff: Mentor

Hmm, that is possible. I will need to look that up to refresh my mind.

15. Aug 6, 2014

### Staff: Mentor

No, you're right. I suppose that if the two paths had lengths that differed by exactly one-half wavelength, then there would be no interference fringe, but the experiment would still work - any anisotropy and the fringe would only disappear in one particular orientation, so that's just a special case of "would shift around"

16. Aug 7, 2014

### ghwellsjr

That's probably close enough of an analogy to understand why all observers will see the same null result of MMX. However, in the case of light clocks, what they are seeing is not the Time Dilation effect. Rather, they are seeing the Relativistic Doppler effect.