How are 'fringe shifts' in the Michelson-Morley experiment calculated?

[DrBanana]

TL;DR Summary

I want to understand why rotating the interferometer results in a fringe shift at all

Ok for the interferometer let the arm lengths be $L$, speed of light relative to the aether be $c$ and the speed
of the setup relative to the aether be $v$. By calculation you can show that the distance between the two 'heads'
of the beams of light is $\frac{v^2L}{c^2}$, and using the convention of the Wikipedia article, we denote this difference as a difference of wavelengths, so $\Delta \lambda = \frac{v^2L}{c^2}$.Now rotating the setup by ninety degrees, it's intuitive that the absolute value of the difference doesn't change. But what I don't understand is why this time we take $\Delta \lambda _2=-\frac{v^2L}{c^2}$. That is, why does the 'order' of the lengths matter here? Both beams have the same wavelength and the same frequency, so the interference pattern should be the same, so why is that not the case?Also I'd like to know how to calculate fringe shifts (given path differences) in the first place. In my textbook's optics chapter there's only talk of Young's double slit experiment, which is for two light sources that are separated by some distance, which is not the case here.

 

[BvU]

Now rotating the setup by ninety degrees, it's intuitive that the absolute value of the difference doesn't change. But what I don't understand is why this time we take $\Delta \lambda _2=-\frac{v^2L}{c^2}$. That is, why does the 'order' of the lengths matter here? Both beams have the same wavelength and the same frequency, so the interference pattern should be the same, so why is that not the case?

Rotating the setup switches the $T$ for the blue and the red light path in the figure

so if there is a difference in one orientation, it reverses sign in the other.

Also I'd like to know how to calculate fringe shifts (given path differences) in the first place. In my textbook's optics chapter there's only talk of Young's double slit experiment, which is for two light sources that are separated by some distance, which is not the case here.

The essence is in the path length difference, whatever the cause

$\$

 

[Ibix]

TL;DR Summary: I want to understand why rotating the interferometer results in a fringe shift at all

Also I'd like to know how to calculate fringe shifts (given path differences) in the first place. In my textbook's optics chapter there's only talk of Young's double slit experiment, which is for two light sources that are separated by some distance, which is not the case here.

If you have the path difference and the speed then the path difference divided by speed gives you the arrival time difference. Multiply by the frequency to get the phase difference in cycles, which is equal to the fringe shift in fringes.

Note that rotating the interferometer by 90° switches the path difference sign, so the path fringe shift between the two cases is twice the fringe shift between the two arms.

 

[DrBanana]

If you have the path difference and the speed then the path difference divided by speed gives you the arrival time difference. Multiply by the frequency to get the phase difference in cycles, which is equal to the fringe shift in fringes.

Note that rotating the interferometer by 90° switches the path difference sign, so the path fringe shift between the two cases is twice the fringe shift between the two arms.

So my question is, why is the path difference a 'signed' quantity? Why are we not only working with the absolute value of the difference?

 

[PeterDonis]

why is the path difference a 'signed' quantity?

What you are calling a "path difference" is actually a phase difference, which is a signed quantity. The Wikipedia article does talk about the beams being out of phase, but it would be clearer if it used the term "phase difference" (which comes from the path difference, but is signed because the phase relationship switches direction if the apparatus is rotated by 90 degrees--it's the same magnitude of difference, but which phase is ahead of the other is reversed).

 

[Ibix]

So my question is, why is the path difference a 'signed' quantity? Why are we not only working with the absolute value of the difference?

Because it's a difference of two quantities, the light path lengths, and a difference can be negative. You're free to define it as $l_1-l_2$ or $l_2-l_1$, but not change half way.

Typically you introduce a bit of tilt in one arm so you get fringes - a series of vertical lines. When you lengthen one arm the fringes move right; when you shorten it they move left. They don't stop and reverse at some point, so you don't want to try to model the fringe shift with an absolute value.

 

[Mister T]

Now rotating the setup by ninety degrees, it's intuitive that the absolute value of the difference doesn't change.

But as you rotate you will notice the pattern changing, and you will see that it doesn't become the same pattern as it was before the $90 ^{\circ}$ rotation.

 

[DrBanana]

What you are calling a "path difference" is actually a phase difference, which is a signed quantity. The Wikipedia article does talk about the beams being out of phase, but it would be clearer if it used the term "phase difference" (which comes from the path difference, but is signed because the phase relationship switches direction if the apparatus is rotated by 90 degrees--it's the same magnitude of difference, but which phase is ahead of the other is reversed).

Thank you, I don't believe I've gotten the full picture but it shouldn't be too difficult from here.

 

[Ibix]

Thank you, I don't believe I've gotten the full picture but it shouldn't be too difficult from here.

One way to do it is to calculate the flight time along one arm when it is oriented at an angle $\theta$ to the ether wind. Then you can immediately write down the phase difference between two arms at 90° to one another and see the expected phase change as you rotate the interferometer. (It's easiest to do this in the ether rest frame.) That should produce a phase difference that varies as the sine of the angle (obviously with a zero when the arms are 45° either side of the ether wind) with a large enough amplitude to be detectable (if it were present!) when you plug in our orbital velocity.

 

[PeterDonis]

Moderator's note: a separate discussion of the negative Michelson-Morley result has been moved to another thread:

https://www.physicsforums.com/threads/implications-of-negative-michelson-morley-result.1061865/