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MMX calculation problem

  1. Jun 3, 2015 #1
    Hi All,

    MMX has been discussed a lot but I could not find solution to my problem. So, I had to start a new thread on this subject! Sorry it is lengthy, you can just read the last paragraph (below red text) for short form.

    I was going through details of the calculations and have problem at one step on the vertical calculation. Before going in to that step, I will detail the general approach of my understanding so far. I am trying to use same symbolism as in Special Relativity (with primes), please bear with me.

    For any diagram, I am using what is in Wikipedia:
    http://en.wikipedia.org/wiki/Michelson–Morley_experiment#Light_path_analysis_and_consequences

    Situation A when there is 100% aether drag: t'/t = 1 for both horizontal and vertical.
    Method 1: velocity relative to local aether is zero.
    Method 2: If distant static aether is used as reference, speed of light will be affected by same extent and compensate the L changes which will result in t'/t = 1.​
    So, anyway we will expect a null result if there were full aether drag. So far I have no problem.

    Situation B when there is 0% aether drag:
    Horizontal L'/L = t'/t = gamma^2 (not yet including length contraction) (still no problem).

    Vertical: Only here I have problem understanding how it can be L'/L = t'/t = gamma.
    We know that the mirrors will have the increased distance of L' = L*gamma (always). But, my thinking is that the light will not be following the mirrors. It will just stay with static aether and move straight up/down (in earlier Situation A it went to the mirror only because it was dragged there by the aether). So, the light path will not be same as the hardware distances.
    This will give only L'/L = t'/t = 1 (or same calculation arrived by Michelson first as Tt = 2L/c). In such case, the beam will hit the top mirror vt distance behind and hit the inclined mirror at 2vt distance behind (compared to static earth expected positions). Basically the confusion is how the light can follow same path whether aether drag is there or not? It will be great if where I am mistaking is pointed out.
    Thanks all and sorry for the lengthy description.
     
  2. jcsd
  3. Jun 3, 2015 #2

    PeterDonis

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    I don't understand; what do you mean by "full aether drag"? Which diagram and/or calculation on the Wikipedia page do you think represents "full aether drag"?

    Once again, which diagram and/or calculation on the Wikipedia page do you think represents "0% aether drag"?

    Remember that the light source is also moving, at the same speed as the mirrors are. The light's motion has a horizontal component because the light source does, not because the mirrors do; the horizontal components of the light's and the mirrors' motions are the same because the mirrors are moving at the same speed as the light source.
     
  4. Jun 3, 2015 #3
    Hi Peter,
    I am just trying to calculate both scenarios so that I can know what to expect if either was true.

    It is not specified in Wikipedia which situation they are calculating but it looks to me as full aether drag since vertical light follows the mirror. So, basically it has only one figure used for one scenario and I am just using the general setup for calculating the other scenario. But null result is expected for full aether drag from what I understand but it is not accepted because it is not consistent with other observations. There is no problem in that scenario. So, to the next scenario (no aether drag).

    There is no difference in the horizontal path length in both scenarios. So, no problem in that part.

    Only the vertical path I have problem. I cannot grasp that the path can be same no matter if there is aether drag or not. If there is no aether drag, how the vertical beam will move/bend toward the top moving mirror from the point of reflection at inclined mirror? That is the problem I cannot understand. My thinking is: It is supposed to be reflected vertically always. If there is aether drag the light will be dragged and if no drag it will stay vertical.

    I hope this clarifies a little better.
    Thanks.
     
  5. Jun 3, 2015 #4

    PeterDonis

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    Are you referring to Figure 4 on the Wikipedia page? Look at the lower part of that figure, with the two animations showing the predicted result of the MMX as analyzed in two different frames. One (lower left) has the mirrors and light source at rest; the other (lower right) has them moving to the right.

    There is an obvious difference in the predicted result for the two cases, but whether this is attributed to "ether drag" is a matter of interpretation. The key point, as I said before, is that in the second case (mirrors and light source in motion), the vertical light beam has the same horizontal component of motion as the light source. That's why it goes up to the mirror and back.

    Given the above, I'm still not sure what you mean by "ether drag". Which of the two animations in Figure 4 (lower left or lower right) do you think corresponds to the case of "full ether drag"? Which one do you think corresponds to the case of "no ether drag"? And how does "ether drag" change if the mirrors and light source are moving vs. at rest?
     
  6. Jun 4, 2015 #5
    Hi Peter,
    Yes, it is the Figure 4.
    I am only using the path length calculation diagram above the animations which is for one scenario (animations do not have details for calculations).

    These calculations were using the aether theory (and trying to test aether flow). So, the movements are relative to aether I would guess. So, if the aether is dragged, it is same as in rest and if it is not dragged then the paths will be different. I understand that Special Relativity explanation will be different. But I am trying to understand how Lorentz' modified theory was able to explain it.

    But today when I read Wikipedia page more carefully, I found some differences.
    The subtitle reads: "Observer resting in the aether" but the text below says "The reflecting mirror is ... moving with velocity [PLAIN]http://upload.wikimedia.org/math/9/e/3/9e3669d19b675bd57058fd4664205d2a.png." [Broken]
    If the observer was outside the experimental frame, they do not contribute to the measurements (that adds to my confusion). Still not clear whether they are calculating for aether drag scenario or no aether drag scenario. It is vital because according to aether theory all speeds (including light) are relative to aether (that's what I understand).


    If you are comparing with "light clock" model that will apply here if outside observer was moving and I understand completely. But if the experiment was moving, it will be different (due to the absolute frame in aether theory). Though it deals too much with old theory, I thought it is good to understand the problems as a learning exercise.
     
    Last edited by a moderator: May 7, 2017
  7. Jun 4, 2015 #6

    PeterDonis

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    Yes. At least, that is the assumption of the aether theory that Michelson and Morley were using to predict a non-null result for the experiment.

    The aether theory with aether drag is not the theory Michelson and Morley were using. They were simply comparing an experiment done at rest relative to the aether, with an experiment done in motion relative to the aether. Aether drag was a separate hypothesis (the idea that the aether around the Earth might be dragged by the Earth, making Earth-bound experiments unable to detect motion relative to the aether); it was introduced by others to try to explain the null result while still preserving aether theory. However, that separate hypothesis of aether drag proved to be unworkable. (It should be easy to see why: the whole idea of the aether was to introduce something physical as a reference for "absolute rest". But if some parts of the aether can be in motion relative to other parts, as the aether drag hypothesis requires, then the aether no longer provides a reference for "absolute rest", so there's no point in hypothesizing it at all.)

    I agree the "observer" language is confusing and doesn't really add anything. The key is whether or not the mirrors and the light source are moving relative to the aether; Figure 4 is simply comparing a scenario where they are all moving (lower right animation) with a scenario where they are all at rest (lower left animation) relative to the aether.

    They're calculating for the no aether drag scenario. See above.
     
  8. Jun 4, 2015 #7
    Hi Peter,
    That was huge help. Reduced lot of problem. I can ignore the aether drag hypothesis also. So, overall I have to think about only half now! Thanks for all the help.

    I tried to calculate the same and I am seeing opposite path. Path length is same but lagging behind and not reaching same point as horizontal beam.
    I have made a figure and can you please check see what is wrong. Thanks.

    The top figure is how the experimenter in the rotating/moving earth frame will perceive. Relative to the experimenter, the mirrors will be static but the light will be dragged behind (due to aether wind or static aether). So, it will come back to the inclined mirror at 2*v*t distance behind and so it may not overlap with the horizontal beam. Notice that the path bends backwards (towards source side) not forward! This is a highly exaggerated view for clarity.

    The bottom figure is how an outside observer (in static aether frame) will perceive. This observer will see that the light goes straight up (because of being in same aether frame) and come back at same path (if it did not miss the moving top mirror) buy by this time the inclined mirror would have moved 2*v*t distance forward. The aqua color mirrors are old positions of the mirrors. Time dilation most likely will compensate for the difference in path lengths. It is not necessary to calculate both but using it as a cross-check for consistency.

    But now I have the separation of 2*v*t between the beams by either view, in where they hit the inclined mirror on return. Length contraction rectifies the horizontal-vertical path length difference. Does the length contraction also makes this separation go away? If that will be the case, then that would have further reduced the vertical path length and the horizontal-vertical path length difference will remain again! A last bit of confusion remains.

    MMX1.jpg
     
  9. Jun 4, 2015 #8

    PeterDonis

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    Ok.

    But to show this, you have to show how the horizontal beam moves, under the assumption of an aether wind. The beam is going against the wind one way, and with the wind the other, and that turns out to mean a different round-trip travel time (again, under the assumption of an aether wind and Newtonian mechanics) than the vertical beam going "crosswind" both ways. That's the whole point of the Wikipedia page calculation.

    I thought we weren't talking about how SR resolves this scenario. Length contraction and time dilation are SR phenomena.
     
  10. Jun 4, 2015 #9

    Nugatory

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    Not necessarily - it used to be called "Fitzgerald contraction" because Fitzgerald suggested it many years before Einstein gave us SR.

    All we're really doing in this thread is reaffirming that the Michelson-Morley experiment did not prove SR (of course!), it falsified all the straightforward ether hypotheses. The non-straightforward ones, with ad hoc assumptions about Fitzgerald contraction, ether drag, and the like survive the MMX negative result but collapse under their own weight.
     
  11. Jun 4, 2015 #10

    Dale

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    Also, aether drag is incompatible with other experiments (ring interferometers).
     
  12. Jun 4, 2015 #11

    Nugatory

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    Good point... Michelson-Morley gets all the press, but it left frame dragging to collapse agonizingly under its own weight like a beached whale. Fortunately it's not the only experiment out there.
     
  13. Jun 5, 2015 #12
    Hi Peter,
    I am getting the horizontal calculations correctly. That's why not focusing on that part. It goes along the same line all the time and the path length are easy to calculate. Also where it will hit the inclined mirror will never change.

    As others have mentioned, I used the Lorentz-Fitzgerald contraction but unnecessarily dragged the Time Dilation without even analyzing much. That was bad. Sorry.

    Only on the vertical side I am not able to get the same path as shown in Wikipedia. So, trying to find where I am missing. Still that shifted return position is the only confusion left!

    Thanks Nugatory and DaleSpam for your comments. I am having trouble with aether theory explanation even with length contraction as my path length expectation in the figure shown in post #7.
     
    Last edited: Jun 5, 2015
  14. Jun 5, 2015 #13
    The shift of 2*v*t I am calculating in my path position calculation comes to about 2 mm (for 10 m arm length). Still I am not understanding why I am getting the path differently but just for the fun of it calculated how much it will be!

    C = 300,000 km/s; v = 30 km/s; L = 10 m
    2*v*t = 2*v*L/C = 600/300,000 = 2/1000 = 2 mm (per 10 m arm length)

    2 mm may be small or big depending on the beam size. But it may all be unnecessary when my point of error is found and corrected! Still it was fun to get a number to it. Hope somebody will find out where I am going wrong in the figure in post 7.
     
  15. Jun 5, 2015 #14

    PeterDonis

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    Yes, agreed. This is actually what I meant to say. Honest. o_O
     
  16. Jun 5, 2015 #15

    PeterDonis

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    But when it hits the inclined mirror changes (meaning, it changes depending on whether the apparatus is at rest relative to the aether, or moving relative to the aether), and it's the difference in the times that the horizontal and vertical beams take to return that is crucial, not the difference in where exactly they hit the inclined mirror.
     
  17. Jun 6, 2015 #16
    The two beams have to overlap back to create Fringe effect I am assuming. So, that's why taking this difference seriously.
    If it was 1 inch beam, 2mm will not make any difference!
    But if it is 1 mm beam, they can stay separately without any Fringe effect. But if this was true, we will see varying effect while rotating the setup. I think no such systemic changes were observed. So, something still wrong in my calculation. Cannot figure out what that can be.
     
  18. Jun 6, 2015 #17

    PeterDonis

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    I haven't been able to find any information online about the beam widths, but I would be surprised if they were as narrow as 1 mm.

    We would only see varying effects when rotating the setup if the original "Newtonian" aether theory were correct. It isn't. Remember that the actual observed results are consistent with SR, not the "Newtonian" aether theory you have been assuming.
     
  19. Jun 7, 2015 #18
    There has been discussion in the early literature about how mirrors may have to be adjusted such that the images interfere well at the same point in the middle of the mirrors; and for that it is necessary for the light to follow the theoretical path as sketched in Wikipedia. However that practical consideration is now quite irrelevant.
    It looks to me as if you forgot two things:

    1. Light reflection from a moving mirror. You wrote: "my thinking is that the light will not be following the mirrors. It will just stay with static aether and move straight up/down". That is wrong: even if it "just stays with static aether", it bounces off a moving mirror. That it must deflect under an angle you can find by means of a Huygens construction.

    2. Length contraction. For a perfect compensation of motion, you must adjust the mirror orientation by accounting for length contraction. That changes the mirror's angle. Next when you repeat the Huygens construction, you will find that the light beam will follow exactly the same path as a beam in rest.

    Many years ago I worked that out for myself, but I recall that I have seen an article about it (perhaps in the AJP). If you like, I can try to find it back.

    PS I found it back: see my old post https://www.physicsforums.com/threa...nt-for-true-length.469311/page-9#post-3140506
     
    Last edited: Jun 7, 2015
  20. Jun 7, 2015 #19
    Hi Peter,
    I am also not able to see the beam width info.
    I understand that this was to "test" the old theory. But I just wanted to see if I can derive and see how their explanation was working even with the "ad hoc" length contraction which is not "classical". I thought the setup and math was easy enough, but not looking so!
     
  21. Jun 7, 2015 #20
    Thanks Harold,
    I have not read about Huygens construction. I have to read about it (the link to the paper in that post no longer works). May be that will help.
    But length contraction and changing angle are a bit confusing. Won't the length contraction be zero and the angle be same as claimed (45 or 90 degrees) for the experimenter on the same frame?
    If we use static aether frame as reference, then the bottom of the figure (in post 7 above) will be the case. Then length contraction will be like one leg in water and one leg in land!
     
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