# Why shift in fringe is expected in Michelson Morley Experiment?

When I (beginner) learnt Michelson Morley experiment, it was expected that there should be a shift in the fringe pattern when the setup was rotated by 90 deg if the presence of ether is true. But why should there be a shift in fringe after the setup was rotated, assuming the presence of ether?

One may calculate the time that light (electromagnetic waves) need to go through the arms of the interferometer. If the speed of light relatively to the aether is $c$, then $c$ is also the approximate speed through the arm that is approximately orthogonal to the velocity $\vec v$ of the Earth (or interferometer) relatively to the aether. Well, the speed is a bit smaller, $c/\gamma$, due to the transverse Doppler shift.

On the other hand, in the arm that is parallel to $\vec v$, the speed of the aether, the speeds with which the waves propagate are $c+v$ and $c-v$, respectively, depending on the direction (back and forth). We need to calculate the inverse speeds because we want to know the time it takes to get through the arm (back and forth). The inverse speeds are $1/(c\pm v)$ and their average is $$\frac 12 \left[ \frac{1}{c+v} + \frac{1}{c-v} \right] = \frac{c}{c^2-v^2}\approx \frac 1c (1+v^2/c^2)$$ The harmonic average speed is therefore the inverse of this or $$c (1-v^2/c^2)$$ up to subleading corrections. That's smaller than the transverse average speed that was $$c (1-v^2/2c^2)$$ So the difference between the "effective average speeds" through the arms is $$\Delta c = \frac{v^2}{2c}$$ where $v$ is the speed of the Earth relatively to the aether. If you divide the overall distance traveled by a photon (through one of the arms) by the speed, you get the time, and $$\Delta t = \frac{s}{c} - \frac{s}{c-\Delta c} \approx \frac{s\cdot \Delta c}{c^2}\approx \frac{s v^2}{2 c^3}$$ So the two photons need a slightly different time to get through the arms (the total traveled distance is $s$ for a fixed arm). This $\Delta t$ will be translated into a phase shift $2\pi \cdot \Delta t / t_{\rm period}$ is the angle that will decide about the phase shift that may be seen through the shifted interference pattern. ($2\pi$ corresponds to the first nonzero shift that is undetectable.)

Even though the prediction for the phase shift only appears in the second-order, it was already observable by the 19th century interferometers, and the result of the experiment was that this phase shift doesn't exist. So the hypothesis that the Earth is moving relatively to the "environment whose oscillations are perceived as light" was experimentally ruled out.

• I get that rotating the apparatus by 90° would cause the time difference to be opposite of that which was before. What I don't get is why does that matter to the phase difference (and thus the pattern)? Suppose first light ray 'A' was the faster one ,and the slower one was 'B'. Now it's the opposite . But there is no difference between A and B(as far as I know ), so why should the pattern be any different? Commented Jan 29, 2020 at 15:47
• The phase difference is just equal to the time difference multiplied by the (completely constant) angular frequency omega. It doesn't matter whether you talk about time differences or phase differences. Commented Jan 30, 2020 at 16:08
• Yes indeed. But all rotation by 90 does is result in the negative of the original time difference. Wouldn't the phase difference physically be the same both before and after ? Commented Jan 30, 2020 at 16:17
• In reality, there is no shift because relativity works. Before relativity, and that's what the question asks about, people expected the shift. Changing speed or orientation mattered - those weren't symmetries - because the arms of the gadget were moving by different speeds relatively to the aether, the only right frame where the calculations are "easy". Commented Feb 1, 2020 at 5:34
• Apologies for the late reply. I have always been told that we would have one fringe pattern first and then after we have rotated by 90 (and stopped) the pattern is different and this is what we should have detected. Just to be clear, you're saying that the we're instead looking for a shift while we are rotating, yes? This makes more sense to me. Commented Feb 8, 2020 at 13:47

Consider paths A and B. Orbital speed of Earth is v (30km/sec approx.), D is the length of both paths, P is the optical path length.
Supposing that the ether wind is along A, thus $$P_a - P_b = \frac{D v^2}{c^2}$$. Got some fringe pattern! After rotating the apparatus by $$90^{\circ}$$, B comes along the ether wind. So, $$P_b - P_a = \frac{D v^2}{c^2}$$. Got some fringe pattern! Just draw the diagram, rotate it as it is(if you didn't get it) and think about this symmetrical result. This means that a displacement in the fringe pattern equal to $$\frac{2 D v^2}{\lambda c^2}$$ should occur if Ether exists. The displacement as the original paper said was much less than the theoretical one which indicated some other complications like the movement of the solar system altogether which could have lowered v (the measure of relative movement of earth through ether) to display such small displacement in the fringes. So, the experiment was repeated at intervals of 3 months to avoid all ambiguity. The original paper is available online if you want to delve deeper.