Question about Michelson-Morley experiment

[FranzDiCoccio]

Hi,

I think I have a good grasp of the principles underlying the Michelson-Morley experiment, but I am not sure about a detail.
In basic textbooks a simplified situation is presented whereby the ether wind is along one of the two arms of the interferometer. Simple classical physics calculations allow to determine the delay between the two light beams.
This delay determines the feature of the interference fringes on a screen.

The interferometer is then rotated about a vertical axis, and the fringe shift is measured during this rotation.
That would be the change in position of one of the interferenge fringes, right?

I expect that when the rotation angle is 90° the fringe pattern is the same as at angle 0°. Indeed rotating by 90° simply swaps the role of the two beams. But the relative delay should stay the same.
This seems to be in agreement with the original data I have seen here and there (e.g. wikipedia page), in which the reading at 0° and 90° is the same (and at 180° of course).

Could you confirm this?

I wanted to create a simulation of the experiment, but I found a nice one http://www.kcvs.ca/site/projects/specialRelativity.html#The_MichelsonMorley_Experiment.

However it seems to me that it is in contradiction with the above discussion.
The interference patterns at 0° and 90° are different
I have not had the time to check their calculations, but something seems off...
I'd say that both waves (the blue and red one) should move when the interferometer is rotated.

Can anybody comment on this?

 

[pixel]

I expect that when the rotation angle is 90° the fringe pattern is the same as at angle 0°. Indeed rotating by 90° simply swaps the role of the two beams. But the relative delay should stay the same.

That seems correct, but I can't get the linked simulation to work. What is required?

 

[FranzDiCoccio]

Hi pixel,

thanks for your help. It is a swf file, so flash. Actually right now I'm having difficulties to open it myself, but I swear that earlier today it worked all right.
I managed to download the file and open it locally with the Adobe flash player, though.

Let me clarify further my question: would it make sense to compare just the interference patterns at 0° and 90° or is it essential to consider the angles in between those?
Based on my reasoning above that those two patterns would be undistinguishable, I'd go for the latter: consider all the angles (or a subset of them, realistically).

One thing that confuses me is that, in the most basic configuration, the interference pattern would be concentric circles.
The authors of the applet consider the interference of two 1D plane waves. I'm not sure whether they're simplifying the situation or perhaps considering portion of fringes far from the center. I'd say the former.

Anyway, I think a simulation, however simplified, should reflect the fact that the interference pattern is the same when either of the arms of the interferometer parallel or at right angles with the ether velocity (angles 0°, 90°, 180° and 270°). I am talking about a simulation of what Michelson and Morley expected to see based on the ether theory.

In the meanwhile I have found the original paper by M&M.
The original apparatus was a bit more complex than the schematic one in basic textbooks. I think it would give straight fringes, but I'm not sure.
Still, the configuration seems fairly symmetric, so I expect the above reasoning to be accurate even for te original apparatus. And, as I mention, the measurements seem to confirm that, since the expected position of the brightest fringe is the same at 0°, 90°, 180°.

 

[FranzDiCoccio]

Update... I'm more confused now.
It seemed logical to conclude that when the arms of the interferometer are at 45° (w.r.t. the direction of the ether) the delay between the two beams is zero, because they're affected by the "ether wind" in the same exact way. So the (expected) interference pattern would be the same as if the ether was not there at all. In any case, it should be the same as at an angle of 135°.
So, I'm not sure why the expected reading at 135° is different from that at 45°.

I'd expect something looking more like a sin² of the angle, rather than a sin.


Maybe I'm oversimplifying the problem.

 

[Ibix]

What Michelson and Morley thought they were doing was rotating their interferometer in ether that was moving past their apparatus at some speed $v$ in a Newtonian universe. It's easiest to analyse in the rest frame of the ether, in which case the instrument is moving at speed $-v$ and light is always doing $c$ (it isn't expected to be of constant speed in the rest frame of the interferometer). You only actually care about two of the arms of the interferometer - the beam injector and readout arms don't matter because they are shared by both halves of the light. So what is the travel time along one interferometer arm?

If the arm is of length $L$ and arranged at an angle $\theta$ to the $+x$ direction, and the half-silvered mirror passes through the origin at $t=0$ then in general it is at $(x,y)=(-vt,0)$ and the mirror at the end of the arm is at $(L\cos\theta-vt,L\sin\theta)$. Light emitted from the origin at time $t=0$ arrives at the arm end at $t=t_B$, which is given by $$c^2t_B^2=(L\cos\theta-vt_B)^2+L^2\sin^2\theta$$ You can solve that for $t_B$ fairly simply. Then, by a similar calculation, you can derive the time $t=t_R$ at which the pulse returns to the half silvered mirror (which isn't where it started - which observation may help you to develop intuition about the behaviour of the experiment). I make it $$t_R=\frac 2{c^2}Lv\gamma\sqrt{\frac{c^2}{\gamma^2v^2}+\cos^2\theta}$$where $\gamma=\sqrt{1-v^2/c^2}$ is the Lorentz gamma factor. Do check my maths.

This is enough to get you the time difference in the return of the pulses as a function of the angle that the apparatus makes to the ether wind - one arm is at $\theta$ and the other at $(\theta+\pi/2)$. It's easy enough to show that $\cos^2\theta=\cos^2(\theta+\pi/2)$ for $\theta=n\pi/2$ (Edit: make that $n\pi/2+\pi/4$) where $n$ is an integer. So Michelson and Morley expected zeroes every quarter turn, assuming I haven't messed up the maths.

Of course, the point about the experimental data is that it does not show this at all. The result is zilch to the precision of the experiment - hence relativity theory. Any pattern you are seeing in the data is experimental error. It's quite possible that it is a systematic error, which you might expect to have a periodicity of some fraction of $2\pi$ that may or may not match the above calculation. But you can't predict what without going into enough detail about the actual setup to be able to understand their systematic errors.

 

[FranzDiCoccio]

thanks Ibix, that's really interesting!
Here (Europe) is pretty late, and I'm a bit too tired to follow your reasoning right now, but I'll definitely give another try tomorrow.
I did manage to find the solution to the first equation which, if I did not make stupid mistakes, should be

$$t_B = \frac{L}{c^2-v^2} \left(c|\cos \theta|-v \cos \theta\right)$$

It looks about right... for $\theta=0$ and $\theta=180$ it gives the two results I expect.
In order to fully understand the matter I need to work out all of the calculations.

I canceled out what I wrote earlier because I understood that I was making a stupid mistake.

 

[Ibix]

Here (Europe) is pretty late, and I'm a bit too tired to follow your reasoning right now, but I'll definitely give another try tomorrow.

I'm also in Europe, which is why I'm suggesting you check my maths...

I did manage to find the solution to the first equation which, if I did not make stupid mistakes, should be

$$t_B = \frac{L}{c^2-v^2} \left(c|\cos \theta|-v \cos \theta\right)$$

That doesn't look right - it's the solution to a quadratic so I'd expect it to have a square root somewhere. Maybe it simplifies - I didn't bother.

But, at an intuitive (and perhaps naive) level I still fail to see how the delay between the two beams could be different in the 135° and 45° cases.

That is where the four zeroes happen. I messed up the end of the maths above - the solutions for $\cos^2\theta=\cos^2(\theta+\pi/2)$ occur at $n\pi/2+\pi/4$ - in other words, at the four 45° configurations. I've added a correction above.

 

[FranzDiCoccio]

I canceled out part of my previous messages because it dawned on me that I was making a stupid mistake.
The angles 45° and 135° are not equivalent. The angle 45° should be equivalent to 45°+180°=125°.
In both cases the two beams "do the same thing", and the delay between them should be 0.
135° is completely different, and should be the same as -45°.

Sorry about all the fuss... Should not post when it is this late.

About the solution: it does simplify. And, as I say, it checks with what is known for $\theta=0$ and $\theta=\pi$ (radians, of course).

 

[Ibix]

About the solution: it does simplify. And, as I say, it checks with what is known for $\theta=0$ and $\theta=\pi$ (radians, of course).

I think those are the only two points our expressions agree... I'll post tomorrow when I've checked if I've made any other stupid errors in my calculations.

 

[FranzDiCoccio]

Ok, I'm pretty confused now.

If I get it right think you're saying that the zeroes in Michelson and Morley's expected results correspond to the arms of the interferometer being at $n \frac{\pi}{2}$ with its velocity wrt ether.

Better continue tomorrow, I agree.
Thanks again,
bye

 

[FranzDiCoccio]

So, I've been looking better into the calculations. As you spotted, my solution of yesterday was wrong. I forgot one term. Sorry about that, and about claiming it was correct.

Let me recap. The apparatus moves along the x direction, and its arms are at an angle $\theta$ wrt to it.
When the beam moves in the opposite direction as the mirror, the distance it has to cover along the x direction to reach the mirror is shortened by $vt$. The distance it has to cover along the y direction only depends on the angle: $L \sin \theta$.
According to the Pitagorean theorem, the equation to solve is

$$c^2 t^2 = L^2 \sin^2 \theta + (L \cos \theta -v t)^2$$

Then the beam bounces off the mirror and its direction reverses. Since now its velocity along x has the same direction as the apparatus, the distance it has to cover along x is lengthtened by $vt$. Nothing really changes along y. The equation is now

$$c^2 t^2 = L^2 \sin^2 \theta + (L \cos \theta +v t)^2$$

The solution to the first equation is
$$t_-= \frac{L}{c^2-v^2} \left(\sqrt{v^2 \cos^2 \theta+ (c^2-v^2)}-v \cos \theta\right)$$

where I discarded the solution corresponding to negative time. The solution to the second equation is
$$t_+= \frac{L}{c^2-v^2} \left(\sqrt{v^2 \cos^2 \theta+ (c^2-v^2)}+v \cos \theta\right)$$
(simply change $v\to - v$).
The total time is
$$t_1=t_++t_-= \frac{2L}{c^2-v^2}\sqrt{v^2 \cos^2 \theta+ (c^2-v^2)}$$

This applies to one arm. But if the first arm (the one containing the source) is at an angle $\theta$, the other is at an angle $\theta+\pi/2$, as you observe.
Now $\cos^2 (\theta + \frac{\pi}{2}) = \sin^2 \theta$.

Therefore, I'd say that the total time for the second beam is

$$t_2=t_+'+t_-'= \frac{2L}{c^2-v^2}\sqrt{v^2 \sin^2 \theta+ (c^2-v^2)}$$.

If my calculations are correct, I'd say that 1) the periodicity of the interference pattern is not $2\pi$ but $\pi$ (only squares of sines and cosines in the total times).
This should mean that the interference pattern is the same at angles $n \pi /2$ and $(n+1) \pi/4$ (but different from the previous.

My calculations seem to be correct at least for the "basic textbook" cases $0,\, \pi /2$.

PS: sorry, I had to edit a bunch of times because of silly typos in my formulas.

 

[FranzDiCoccio]

I think I might have a clue about my doubt.

Let me re-state the “problem”. I am trying to make sense of this figure in Michelson and Morley experiment.
Based on my calculation above, I'd expect the same "reading" at angles $0,\, \pi/2,\, \pi, 3/2 \pi, 2 \pi$. Likewise, I expect the same "reading" at angles $\pi/4,\, 3/4 \pi,\, 5/4 \pi,\, 7/4 \pi$.
The reason for this would be that the interference pattern depends only on the times taken by the beams to reach the screen, and these times depend on $\cos^2 \theta$, where $\theta$ is the angle between the velocity of the apparatus and one of its arms (e.g. the injector).

I'm actually not sure any more about that. I think that, to the lowest approximation, the interference pattern would be the same as that of two sources located at different distances from the screen, on the same line perpendicular to it.

I think that such a pattern not only depends on the distance between the two sources along the line, but also on their distance from the screen.
The above calculation basically deals with the distance between the sources (the delay between their arrival on the screen).

However, the distance of the sources from the screen should vary as well. This depends on what happens in the injector and readout arm.

These two stretches do not affect the distance between the sources, but they do affect the position of the sources (the overall time each beam takes to go from the source to the screen).

It's like moving the screen closer or farther away from the beam splitter. The interference pattern has to change.

However, if these two stretches do not have the same length, the interference pattern depends on the angle, and I see no reason why it should be the same at angles $0,\, \pi/2,\, \pi, 3/2 \pi, 2 \pi$...

So: would one see any difference between the interference pattern at $0$ and $\pi/2$?

Any insight would be really appreciated.


PS. One wants the two interferometer arms beyond the beamsplitter to be the meaningful ones, because they usually determine the real features of the interference pattern.
If their length is much larger than the injector and readout arm, the contribution of the last two can probably be ignored. This is actually the case with M&M interferometer, thanks to multiple reflections...

 

[Ibix]

I am trying to make sense of this figure in Michelson and Morley experiment.
Based on my calculation above, I'd expect the same "reading" at angles $0,\, \pi/2,\, \pi, 3/2 \pi, 2 \pi$. Likewise, I expect the same "reading" at angles $\pi/4,\, 3/4 \pi,\, 5/4 \pi,\, 7/4 \pi$.
The reason for this would be that the interference pattern depends only on the times taken by the beams to reach the screen, and these times depend on $\cos^2 \theta$, where $\theta$ is the angle between the velocity of the apparatus and one of its arms (e.g. the injector).

Taking the difference between your results $t_1$ and $t_2$, noting that $v^2-v^2\cos2\theta=v^2\sin^2\theta$, Taylor expanding the square root and neglecting terms of $O(v^4/c^4)$ and higher, I get that the difference in arrival times for the pulses traveling along the two arms is proportional to $\cos 2\theta$. The diagram you linked is labelled North/South at one end, East/West in the middle and South/North at the other, so showing a full sine wave for only a half turn. I think it's agreeing with you - unless I'm misunderstanding something.

 

[FranzDiCoccio]

Ok, you mean it has periodicity $\pi$ instead of $2\pi$.
I think one can also use $\cos^2 \theta = \frac{1}{2}(1+\cos 2\theta)$ and $\sin^2 \theta = \frac{1}{2}(1-\cos 2\theta)$


But I'd expect $\pi/4$ (NE-SW) and $3 \pi/4$ (SE-NW) to be the same, anyways. For these angles $\cos^2 \theta = \sin^2 \theta = \frac{1}{2}$, so I'm not understanding why there is a "peak" at $\pi/4$ and a "trough" at $3 \pi/4$...

I'm assuming that NS correspond to angle 0.

Thanks for your help.

 

[Ibix]

Ok, you mean it has periodicity $\pi$ instead of $2\pi$.

Yes. In fact, the Wikipedia page says so just above the diagram you linked: The expectation was that the effect would be graphable as a sine wave with two peaks and two troughs per rotation

I'm assuming that NS correspond to angle 0.

Apparently not, since all the mathematics we've done says that $\theta=0$ does not give zero path length difference (which is intuitively reasonable as well). I can't tell from the Wikipedia page what they use as zero - although note that the reproduction of Figure 5 from their paper shows the interferometer as an X not a +, which would be consistent with their zero of angle being when the beams are at 45° to the motion. Have a read of the paper (Wikisource appears to have it here: https://en.wikisource.org/wiki/On_the_Relative_Motion_of_the_Earth_and_the_Luminiferous_Ether) and see what they say,

 

[FranzDiCoccio]

I'll read that, thanks a lot! Cool resource, didn't notice it.

But I took a look at M&M original paper and, from what I understand, they are not graphing the path difference. They are keeping track of the position of the brightest fringe in the interference pattern. So I think the first point should be trivially 0, irrespective of the relative angle of the arm and velocity, because the fringe has not moved yet.
I think the subsequent zeros mean "the fringe is where it was when we started".

As you say, $\theta=0$ does not give zero path difference, but it gives the same path difference as $\pi/2$ (the "internal" beams are simply swapped) and $\pi$ (the beams go through the same paths as 0, but in the opposite order), so I expect the same position.

But, if they start from 45°, the peak is at 90° and the trough is at 180°, which I do not understand.

Another thing I notice is that the position of the source on the table is marked "1" in the first figure, which makes me think they started from that aligned to the velocity of the Earth. In their paper they do the above calculations in the particular cases $0$ and $\pi$. It would be weird to plot a graph where now 0 means 45°.
Unfortunately their paper is not very clear about what they have actually done.

 

[Mister T]

Hi,
I expect that when the rotation angle is 90° the fringe pattern is the same as at angle 0°. Indeed rotating by 90° simply swaps the role of the two beams. But the relative delay should stay the same.

Not sure I'm understanding your question. The apparatus is adjusted (by changing the length of one arm) so the brightest fringe is visible in the center of the viewing window. If rotating the apparatus by 90° one way causes a change in the position of that fringe, then you've detected the presence of the ether. In principle, any amount of rotation would produce a change in the position, but a 90° rotation would produce the maximum amount. Rotating the apparatus in the other direction should allow one to determine the direction of the ether drift, if I'm thinking about this correctly.

 

[FranzDiCoccio]

Hi Mister T,

I am interested in what M&M should have seen if ether was really there (the expected curve).
I do not understand why you say that a 90° rotation would produce the maximum amount of change in the position of the fringe.
That is the whole point of my question. I'd like to understand why.

It seems to me that the interference pattern at 90° should be the same as at 0°, if the injector and detector arms are ignored.

Let us call the two arms A and B. Let $t_A=t_1$ and $t_B=t_2$ the times the beams take to get to the eyepiece at angle 0°. These are the times that you find in textbooks and in M&M's paper, I think.
Rotating by 90° you simply swap the beams: $t_A=t_2$ and $t_B=t_1$.
Rotating by another 90° you swap the beams back: $t_A=t_1$ and $t_B=t_2$.
Each beam goes back and forth from the beam-splitter to its mirror and back to the beam splitter. It seems to me that the order in which they do that does not affect the time they take.
In any of the above cases you have one beam with time $t_1$ and another with time $t_2$. They should produce the same interference pattern, right? Where does the change come from?

So, it seems to me that the direction matters only when the source and injector beams are considered, but since these arms are much shorter, I expect the effect of the other two beams to dominate.

 

[FranzDiCoccio]

I was wondering whether the "reading of the screw-band on the micrometer" in M&M's paper was actually a differential reading.
That is, they noted the number of turns of the screw to move the mirror into the next position.
That would mean that their graph is basically a derivative, and the derivative of a function of $\cos 2\theta$ would include a $\sin 2\theta$ factor.

This seems a bit far fetched, though.

 

[Ibix]

I think you are worrying about nothing. Give or take a constant starting angle they are predicting exactly what we calculated.

It occurs to me that they can't actually know what angle to call zero ahead of time because they don't know the velocity of the Sun with respect to the ether. So they just pick a convenient direction, call it zero and expect $\sin^2 (\theta+\phi)$ instead.

I think Mister T's analysis is a bit off, but raises an important point. The way to zero an interferometer is to adjust it until the fringes say it's matched - but that doesn't give you equal arm lengths in the case of an ether wind. It gives you equal path lengths in the ether rest frame. There's no way to truly zero the interferometer in this case without knowing the velocity of the interferometer with respect to the ether. You just have to record the micrometer readings and show that they trace out a sinusoid. So, I think Michelson and Morley were expecting their results to have the form $A+B\sin (\theta+\phi)$ where B is the prefactor we calculated and $A$ and $\phi$ are constants that just describe the incidental details of how they set up their experiment.

It so happens that the noise has a vaguely sinusoidal component with maximum slope in the NS direction, so that's where they've fitted their model.

 

[FranzDiCoccio]

Hi Ibix,

I can agree that M&M wouldn't know the direction of the apparatus wrt ether. I'd say that they assumed that for some reason the ether was at rest with the Sun, but I'm not sure about that. If that was the case, I think it would have been possible to establish the direction of the Earth with respect to ether. But let us forget about that.

Let me summarize my reasoning once more, and please point out my mistake.

1) They are looking at an interference pattern. The interference pattern is determined by the delay between the two arrival times, which I called $t_1(\theta)$ and $t_2(\theta)$.

2) The interference pattern is the same when the delay $\Delta t = t_1-t_2$ is the same in absolute value. I have seen calculations considering the difference with sign, but I really don't understand that. Consider an angle $\theta'$ such that $t_1(\theta')=t_2(\theta)$ and $t_2(\theta')=t_1(\theta)$.
In both cases we have two beams interfering with the same delay. It does not matter who "gets there first", i.e. the sign does not matter.

3) If the above expressions are right, I expect that the interference pattern repeats every $\pi/2$. In particular, I expect it to be the same at $\theta=0$ and $\theta=\pi/2$ . The function $|t_1(\theta)-t_2(\theta)|$ certainly does that.

I'm making all of this fuss mainly because I've heard people (and read textbooks) claiming that M&M expected that interference pattern at $\theta=0$ and $\theta=\pi/2$ to be different.
As if, by comparing these two interference patterns (and them alone) they could (in principle) tell if the ether was there or not.

I still can't wrap my head around that. I think that, even in a basic explanation of the experiment, one should be aware of the fact that the angles between $0$ and $\pi/2$ were necessary. I mean, these are necessary even if, for some reason, one has a precise idea of the direction of the velocity.

As a byproduct of my reasoning, I do not understand why there are only 3 nodes between $0$ and $\pi$ in the original figure. Surely $\Delta t=0$ when the angle is an odd multiple of $\pi/4$.

Edit: the number of nodes is ok. What I do not understand is why the expected figure has a period $\pi$ instead of $\pi/2$. The presence of a peak and a trough in $[0, \, \pi]$ means that there are angles that differ by $\pi/2$ that produce different interference patterns. I do not understand why is that.

 

[Mister T]

I think Mister T's analysis is a bit off, but raises an important point. The way to zero an interferometer is to adjust it until the fringes say it's matched - but that doesn't give you equal arm lengths in the case of an ether wind. It gives you equal path lengths in the ether rest frame.

Well, it gives you a difference in path lengths that's either zero or a multiple of a whole wavelength in, as you say, an ether rest frame. But there is no way to be sure of the motion of the apparatus through the ether, so that is not a relevant experimental consideration. What is relevant is that the fringe shift is zero. Rotating the apparatus should change the position of the fringe(s) if there's an ether. So I still don't understand the OP's point. Sorry for being so dense, but I just don't get it.

 

[Mister T]

I do not understand why you say that a 90° rotation would produce the maximum amount of change in the position of the fringe. That is the whole point of my question. I'd like to understand why.

When the apparatus is adjusted so that the brightest fringe is visible in the center of the viewing screen, nothing is really known about the length of each arm. It is (or at least was) not possible to measure those lengths with a precision that approaches the wavelength. Thus, all that is known is that the fringe is aligned. Now, if you assume there is no ether, then you can conclude that the difference in arm lengths is a whole number multiple of wavelengths. But doing so means you accept the famously null result of the experiment before you have demonstrated that it's so!

 

[PeterDonis]

gravity will effect (change) the frequency and wavelength of light, but NOT its speed

This is a coordinate-dependent interpretation. Other coordinates can be chosen in which the frequency or wavelength is constant but the speed varies.

 

[Ibix]

Edit: the number of nodes is ok. What I do not understand is why the expected figure has a period $\pi$ instead of $\pi/2$. The presence of a peak and a trough in $[0, \, \pi]$ means that there are angles that differ by $\pi/2$ that produce different interference patterns. I do not understand why is that.

We showed that the difference in arrival times varied as $\sin^2\theta$ which has period $\pi$. The position of the central fringe is a measure of the difference in arrival times so is predicted to have period $\pi$.

 

[Ibix]

Well, it gives you a difference in path lengths that's either zero or a multiple of a whole wavelength in, as you say, an ether rest frame.

The brightest/darkest (most contrast-y) fringe should always correspond to zero optical path difference, because light sources are not perfectly coherent. The zeroth order fringe is where the beam is interfering with itself. The first order fringe is the beam interfering with a slightly delayed version of itself, and the delay means that some of the light is coherent and some is not - so you get a fainter interference pattern overlaid by a faint version of the incoherent "two torches shining on the same patch of wall".

As you say, the only relevance this has here is that the brightest fringe is easy to keep track of, and it is the track if this fringe that is the experimental output.

Rotating the apparatus should change the position of the fringe(s) if there's an ether. So I still don't understand the OP's point. Sorry for being so dense, but I just don't get it.

I think the problem is that he's not quite understanding the output of the interferometer.

OP: As you rotate the interferometer you expect the fringe position to shift as $\sin^2(\theta+\phi)$ (which may go up or down because you don't know the value of the constant $\phi$ a priori) and you adjust the position of one of the mirrors to make it not move - i.e. you tweak the value of $L$ for one of the arms to cancel out the change from changing $\theta$. You plot the change you made to the arm length, which ought to have the same $\sin^2$ behaviour. That's what's plotted in M&M's figure 7, which you linked to earlier, I think.

 

[FranzDiCoccio]

Thanks to everybody for contributing to this post. I think maybe I've finally got it.

So, I reiterate. My question is about what happens to the interference pattern if there is an ether. The expected result.
In an elementary, very schematic version of the experiment (such as the one I find in several texbooks) you have two equal-length arms.
It is further assumed that initially the injector arm of the interferometer is aligned with the apparatus velocity (angle $\theta=0$).
Then the apparatus is rotated (i.e. $\theta$ is changed) and the interference patterns are compared. This is elementary. One assumes there is a way of discerning the interference patterns by looking at them. You take a snapshot of the interference patterns against a gridded background and measure how much it has moved.
Think of the simulation I originally linked, where the interferometer arms remain untouched, and the change in the interference pattern is demonstrated.

If this was the case, I think one should see what I describe above. The interference pattern would change as a function of $|t_1-t_2|$, and the period would be $\pi/4$ instead of $\pi/2$.

Of course the experiment is a whole different matter. In order to measure the shift of the fringe one has to change the length of one of the arms, as some of you points out (e.g. Mr. T).
This is done by adjusting the position of one of the mirrors, which changes the length of one of the arms, which in turn changes one of the two times (say, $t_1$).

I think this adjustment is the sign I could not account for. When $\theta=0$ I'll have to turn the screw in one direction in order to center the brightest fringe. When $\theta=\pi/2$ I get the same interference pattern, but I'll have to turn the screw in the opposite direction in order to center the brightest fringe, because the beam which used to have the shortest time (virtual source closer to the screen) now has the longest (source farther from the screen).
It is in this sense that the interferometer measures the difference in the arrival time. Sorry for being so dense.
I was misled by the simplistic approach of some books, where one does not change anything in the interferometer and simply looks at the interference pattern.

I think that the two adjustments one has to carry out in the $\theta=0$ and $\theta=\pi$ cases are rigorously different. However they are both extremely small, and I'm pretty sure that at the first perturbative level they coincide.

One final remark: I think that the expected velocity of the apparatus was much less than $c$, otherwise one could measure an anisotropy in the speed of light, e.g. using a technique like Fizeau's. This means that the change is never so large to cause a shift of a whole wavelength, but only a tiny fraction of it.

 

[Ibix]

I think you've got it. You can usually distinguish which way the pattern is moving from some cue or other. In fact, from the look of the vertical fringes, I suspect M&M simply injected plane waves then tilted their arm end mirrors slightly so that the plane waves crossed at a slight angle. The result is straight line fringes at the intersections of the planes, and these fringes shift left or right as the path difference increases or decreases.

Nowadays we'd just stick a CCD in the output arm, record the fringe position as a function of $\theta$ and call it done. M&M, not having access to fast vibration free photography, had to touch their apparatus anyway. So they went for the easy pen-and-paper method of tweaking the arm length and writing down the numbers.

 

[FranzDiCoccio]

I think you've got it. You can usually distinguish which way the pattern is moving from some cue or other. In fact, from the look of the vertical fringes, I suspect M&M simply injected plane waves then tilted their arm end mirrors slightly so that the plane waves crossed at a slight angle. The result is straight line fringes at the intersections of the planes, and these fringes shift left or right as the path difference increases or decreases.

Nowadays we'd just stick a CCD in the output arm, record the fringe position as a function of $\theta$ and call it done. M&M, not having access to fast vibration free photography, had to touch their apparatus anyway. So they went for the easy pen-and-paper method of tweaking the arm length and writing down the numbers.

Yes, exactly. My confusion was ultimately a matter of "what the interference pattern looks like" vs "how do I use an interferometer to measure the difference in two interference patterns".

The second is how M&M worked, and the recorded change has a periodicity $\pi/2$.

But if, as you say, they had a ccd --and the interferometer arms had always the same length $L$ -- wouldn't they have observed a periodicity $\pi/4$ instead of $\pi/2$? The position of the brightest fringe would be the same for angle whose difference is $\pi/2$.

Thanks again. I did struggle but I think I have a clearer picture now.
And I think some of my books are misleading.

 

[Mister T]

The brightest/darkest (most contrast-y) fringe should always correspond to zero optical path difference, because light sources are not perfectly coherent.

Of course. I have no idea now what I was thinking!