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B Question about Michelson-Morley experiment

  1. Jun 4, 2016 #1
    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 here.

    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?
     
  2. jcsd
  3. Jun 4, 2016 #2
    That seems correct, but I can't get the linked simulation to work. What is required?
     
  4. Jun 4, 2016 #3
    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°.
     
  5. Jun 4, 2016 #4
    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.
     
  6. Jun 4, 2016 #5

    Ibix

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    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.
     
    Last edited: Jun 4, 2016
  7. Jun 4, 2016 #6
    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

    [tex]t_B = \frac{L}{c^2-v^2} \left(c|\cos \theta|-v \cos \theta\right)[/tex]

    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.
     
    Last edited: Jun 4, 2016
  8. Jun 4, 2016 #7

    Ibix

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

    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.
     
  9. Jun 4, 2016 #8
    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).
     
  10. Jun 4, 2016 #9

    Ibix

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    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.
     
  11. Jun 4, 2016 #10
    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
     
  12. Jun 5, 2016 #11
    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

    [tex]c^2 t^2 = L^2 \sin^2 \theta + (L \cos \theta -v t)^2[/tex]

    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

    [tex]c^2 t^2 = L^2 \sin^2 \theta + (L \cos \theta +v t)^2[/tex]

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

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

    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

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

    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.
     
  13. Jun 6, 2016 #12
    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...
     
    Last edited: Jun 6, 2016
  14. Jun 6, 2016 #13

    Ibix

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    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 travelling 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.
     
  15. Jun 6, 2016 #14
    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.
     
  16. Jun 6, 2016 #15

    Ibix

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    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
    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,
     
  17. Jun 6, 2016 #16
    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.
     
  18. Jun 6, 2016 #17
    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.
     
  19. Jun 7, 2016 #18
    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.
     
  20. Jun 7, 2016 #19
    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.
     
  21. Jun 7, 2016 #20

    Ibix

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    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.
     
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