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There is a mistake in relativity.

  1. Dec 3, 2009 #1
    I found a verry simple mistake in the Michelson Morley interferometer experiment. If I understand it right, both of interference patterns were the same, right? We have to take into account for two frames of refference though. One for a person observing from Earth, and one for a person observing from a stationary point somewhere in space. The interferometer experiment was made in the Earth's frame of refference which was moving, with the Earth, and there was no experiment for the other frame of refference. Classic physics also proves that this frame of reffernce will not show any difference in the interference patterns. So both theories can NOT be correct. If you now use the theory of relativity for a theoretical frame of refference in space, it will prove that both frames see the same thing (same interference patterns on the table), because relativity says that the speed of light is constant. This is not correct, however, since if we do not "attach" our selves to thinking that light speed is constant ie. fastest, the theoretical stationary frame of refference (from space for say) must show a difference in speeds from the two observation points. Classic physics shows this, if we use a slower speed. I have drawn a simple explanation of this on paper, using a traveling fly. If you want I can post it later, but it is really simple to show this yourself.
  2. jcsd
  3. Dec 3, 2009 #2


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    The Michelson Morley experiment was based on the assumption from classical physics that light was a vibration in a substance filling all of space known as the luminiferous aether, akin to sound waves in air. Sound waves always travel at the same speed relative to the medium they are vibrations in--even if you're moving at a high speed relative to the air, if you make a noise the sound wave will still travel at the same speed relative to the air's rest frame as a sound made by someone at rest relative to the air, which means that relative to yourself sound waves aimed in different directions will travel at different speeds (because they must travel at the same speed relative to the air's rest frame, and you are moving relative to this frame). If light worked the same way, we would expect that if the Earth had a significant velocity relative to the luminiferous aether, observers on Earth would see noticeable differences in the speed of light relative to themselves when the light was traveling in different directions. And the MM experiment was repeated at different points in the year, so even if the Earth happened to be nearly at rest relative to the aether at one point in its orbit, it was assumed it would have a significant velocity relative to the aether at a different point in its orbit.
  4. Dec 3, 2009 #3
    Yes, you are absolutly correct about sound waves, and frames of reference. I did not say however that aether exists. That was an assumption that physisists made a long time ago. Even if aether does not exist it doesn't prove relativity, because nonexistance of one thing does not prove existance of another thing - they have to be related somehow to make that conclusion. To disprove one theory all it takes is one experiment. Simple logic. I'll draw the MM experiment with two frames of refference and you'll see what I am talking about.
  5. Dec 3, 2009 #4


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    Hi WarpSpeed, welcome to PF

    The MM experiment was designed to detect any anisotropy in the speed of light. This anisotropy was predicted by the then-current aether theories of light. The null result was conclusive evidence against the theory being tested and was strong evidence against other aether-based theories.

    It is considered support for SR because one of the postulates of SR includes the isotropy of the speed of light.
  6. Dec 3, 2009 #5


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    But no one argues that the MM proves the whole of relativity, it just disproves the aether theory, which was how classical physicists tried to explain the fact that Maxwell's laws of electromagnetism predict the speed of a light wave is independent of the speed of the emitter (according to the aether theory Maxwell's laws would only work exactly in the rest frame of the aether, where light would have a uniform speed just like sound waves in the rest frame of the air). If you want tons more experiments supporting different aspects of special relativity, look here:


    Do you disagree that the MM punches a hole in the classical aether theory? Because that's the only theory the experiment is taken as disproving (or at least providing strong evidence against)
  7. Dec 3, 2009 #6
    Hmmm, I don't see exactly how it does disprove it - I am not saying that aether exists, I don't know. I only see that this experiment disproves relativity, which it was sorta designed to prove. My reasoning is that there would be a shifted image on the interferometer if relativity was right, yet the interferometer shows the same patter as it would if you shined the same laser directly at the interferometer. Basicly you don't even need to test from a different reference frame. Just had to get my thoughts more clear. I am currently woking on the picture, so it looks good and is very descriptive. Give me a few more hours, almost done.

    Thank you for welcoming me to the PF, I love science!!!
  8. Dec 3, 2009 #7


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    If you think that relativity would predict a shift you're confused. It doesn't matter what frame you analyze it from, all frames make identical predictions about local events like whether two wave crests reach the detector at the same time or different times. And analyzed from the rest frame of the detector, it's obvious that if light travels at the same speed in both directions in this frame, there will be no shift (if you want to analyze the same situation in a different frame, you have to take into account length contraction which will change the length of one or both of the interferometer's 'arms' in any frame where the interferometer is in motion).
  9. Dec 3, 2009 #8
    The experiment took place some while before Einstein proposed his theory. Some even claim that he was unaware of the experiment.

  10. Dec 3, 2009 #9


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    What "other frame of reference"? There is NO "stationary point"! The Michelson-Morley experiment was done a number of different time, at different points in the earth's orbit. The only way it could have come to a null result if it were always oriented in the same way relative to the "aether".

  11. Dec 3, 2009 #10

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    Let me remind everyone that the sticky "IMPORTANT: Read before posting" says:

    This forum is meant as a place to discuss the Theory of Relativity and is for the benefit of those who wish to learn about or expand their understanding of said theory. It is not meant as a soapbox for those who wish to argue Relativity's validity, or advertise their own personal theories. All future posts of this nature shall either be deleted or moved by the discretion of the Mentors.

    If you want to discuss what the MMX says, fine. If you want to argue relativity is wrong, the thread will be closed.
  12. Dec 3, 2009 #11
    HallsoIvy, hi. I think there is always a stationary frame of refference, say one person is standing on the Earth and watching a moving train, and then there is another person on the train - there are the two frames. Everything moves in space so it would be kinda hard to find the exact null speed, you are right. Since the experiment was done on Earth, the frame of refference was basicly "on the train," and always moving, not just around Earth's orbit, but also around the sun etc etc... Here are the pictures I've mayed, the first is for classic physics and the second one is for relativistic. I don't know how much simpler I could make it... I'll write the equations:

    These are the classic physics equations:

    d/c + d/c + d/c + d/c = 4d/c <- t1 for horizontal light.

    d/c + d/c + d/c + d/c = 4d/c <- t2 for vertical light.

    so t1 = t2 for both paths, and of cource there is no shift in patter!!!



    As you can see the values are exactly the same, now there is no point in writing the relativistic equations, because like I said if classic physics are right then the relativity is wrong. You can try and do the relativity eqs, but you'll see that they will not be equal, even if you manage to get both light beams into the interferometer...
  13. Dec 3, 2009 #12
    That is all I am trying to show, and I am just showing proof, and not arguing...

    Thank you
  14. Dec 3, 2009 #13


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    WarpSpeed, did you consider the effects of length contraction on the arms of the interferometer in the frame where the interferometer is moving? If you do you'll find that this frame also predicts there will be no shift in the pattern.
  15. Dec 3, 2009 #14


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    I note that you have the speed of light being c on both arms. That is the core of relativity.

    I think you are confused about which theories say what.
  16. Dec 3, 2009 #15


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    What is "classic physics"? Which light propagation model? You said no aether, so it can only be ballistic theory of light, which is consistent with the MM-result but disproved by other observations:

    May remind you of yout own words: "and I am just showing proof, and not arguing...". So please prove that Relativity is not consistent with the MM-result
  17. Dec 3, 2009 #16
    Hmmm, no, do you mean that objecs contract when they move fast where relativity would play a role?

    BTW I am not trying to argue, it's just that I try to use my knowledge of physics to my best ability and I am always learning. I could be wrong I account for that when I write, and didn't want you guys to think that I act like I know everything. I am just trying to keep this descussion proffesional, with proofs to back up my reasoning.

    So if you can show me a proof that I was wrong I'll accept it as is, I don't argue with things that are right like math and geometry, because these are basicly the axioms and always exist and can't be prooven wrong.

    For now all I can say is that relativity has very percise formulas and definitions, if even one small thing is missing from the proofs it could lead to a very wrong conlusion.

    Just a question though, are you guys willing to discuss this topic further? I have trouble reading science books (but I do anyways, sometimes), and I learn better through communication like this with people who know more about experiments. It's hard to find a teacher who could give you the right pointers, because they have very big time constraints.

    So a little more about my reasoning. I think, if classic physics show no shift, and relativity shows no shift due to mirrors and arm lengths being contracted, then it leads me to another question. I'll try to explain the way I see it. The basis of the theory of relativity is that the speed of light "c" is always constant, and changes speeds slightly in different substances. The basis of classic physics is that light can go faster or slower depending on the frame, from where it originates, and not just being hindered by materials. Am I right? So what I think, if I am correct, is that in the first picture light actually travels at the speed "h" as you can see on the left of with vectors. Meaning that "c" depends on the speed "V" of the frame. And here is my aching question... How can it be that if the two basisis are different but imply the same conclusion? What do you guys think about this statement?

    Thank you for not closing the forum, I just need to get these two theories streightend out, and I think we will come to show that only one type of physics is correct - not saying that relitivity is wrong yet, untill I know more. I even think it would be kinda cool to set up more experiments, that prove either one. I want to be at the sight lol.

    Thank you for reading.

    P.S. Oh nice you guys posted when I was writing. Balistic theory, this is a new word for me, so by what I am guessing my experiment is based on this. A.T. I'll look on wiki you linked and see the theory. But for now I'll just post the above question and hope to see what you all think.
  18. Dec 3, 2009 #17


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    Yes, this is called length contraction. Basically if an object has a length L in the frame where it's at rest, then if you go to another frame where it's moving at speed v, its length (in the direction of motion) is shrunk to [tex]L*\sqrt{1 - v^2/c^2}[/tex].

    So, say the two arms of the interferometer are each 10 light-seconds long in the interferometer's rest frame, at right angles to each other. Then in another frame where the apparatus is moving at 0.6c in the horizontal direction, the vertical arm is still 10 light-seconds long but the horizontal arm is shrunk to 10*sqrt(1 - 0.6^2) = 20*0.8 = 8 light seconds. Now say two wave peaks are released simultaneously at the intersection point, one going vertically and the other going horizontally. The peak traveling horizontally is traveling at 1c while the end of the horizontal arm is moving away at 0.6c, so the closing speed is 1 - 0.6c = 0.4c, and if the arm is 8 light-seconds long in this frame, it takes 8/0.4 = 20 seconds for the peak to catch up with the other end of the horizontal arm. Then the peak is reflected and travels back towards the intersection point, now the intersection point is moving towards the peak at 0.6c so the closing speed is 1.6c, meaning it'll only take 8/1.6 = 5 seconds to travel from the far end back to the intersection point. So, the peak takes a total of 25 seconds to go from the intersection point to the end of the horizontal arm and back.

    Now consider the peak traveling along the vertical arm. In 12.5 seconds, the vertical arm has moved a horizontal distance of 12.5*0.6c=7.5 light seconds, and its vertical height is 10 light seconds, so by the pythagorean theorem if a signal takes 12.5 seconds to travel from the intersection point to the top, it'll have gone a distance of [tex]\sqrt{7.5^2 + 10^2}[/tex] = 12.5 light seconds, showing that the signal was moving at the speed of light. So, this must be the time for the light to get from the intersection point to the top of the vertical arm. In another 12.5 seconds it'll have traveled back down to the intersection point, so the light takes 12.5 + 12.5 = 25 seconds to travel from the intersection point to the top of the vertical arm and back to the intersection point. So you can see that both peaks take 25 seconds to travel from the intersection point to the end of their respective arms and back to the intersection point, meaning there will be no shift, even though we analyzed everything from the perspective of a frame where the interferometer was moving at 0.6c rather than being at rest (and assumed both peaks moved at c in this frame).
  19. Dec 3, 2009 #18


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    It predicts the same result as special relativity for the MM-experiment, so you need other observations to tell which theory is right. Ballistic theory fails on some of them, SR on none, so far.
  20. Dec 3, 2009 #19
    I have to read a bit more about the Doppler Effect. Oh I see there are three different variables involved, but this is for sound waves though right, in a midium? Can you please show me a link where Doppler Effect applyes to light? Because light doesn't travel through a midium, I mean in space... I always though that light had constant frequencies for each color of light no matter if the receiver or the source was moving or not. I think this is only true though for slow speeds. I need more info please. My mind is racing now, I have a feeling that light is not a wave though... don't know, not sure. It has been said that light is a combination of a wave and a particle. I don't know where the wave phenomena comes from. Hope I don't get booted for asking so many questions.

    Thank you for your posts I am learning.
  21. Dec 3, 2009 #20
    Ok. But consider this. When the light at 1c goes verticly it takes 10*1c = 10 light seconds to travel up along the vertical arm. When it goes down again it also takes 10*1c = 10 light seconds. Now, you have to use the pethagorian theorem for the light to catch up to the mirror at the top (because it has moved a distance of 0.4 light seconds), and... you can't.. why, because the light once it's fired never changes direction so it still shoots up verticly as it came from the intersection point - this is by relativity so I am not changing the subject (light speed, more correctly velocity because it's a vector, is not effected by the source, correct?). And more so, the light direction has to change so as to fire right at the top mirror (or where the mirror will end up after), it can't go streight up since the mirror will be long gone in the horizontal direction. So now even if we were just to change the angle of reflection to a bigger angle, aming right at where the top mirror will be after 0.4 light seconds, to make it there the light speed has to be bigger... simply because of another pythagorian theorem.

    Here is the proof: "x" - hypotinuse of the triangle along which the light from the intersection point travels to the moved top mirror (we agree that all of the mirrors have moved horizontally right?); "a" - bottom triangle leg along which the theoretical light travels at 1c for 0.4 light seconds; and "o" - vertical leg of the right triangle where another theoretical light has to travel at 1c for another 10 light seconds when the system is stopped. Say we drop the speed of the table form 0.6c to a still (I don't want to fly in that ship lol) right when the light along the x direction hits the top mirror, but now we go back to the theoretical lights. a = 1c; o = 1c and so x = [tex]\sqrt{1^2c + 1^2c}[/tex] = [tex]\sqrt{2}[/tex] c. And there ya have it's a contradiction. That is what I meant by the second picture I posted, by relativity the light will never hit the interferometer espetialy at speed like these.

    OK. I am ready for boot now lol. Noooo just kidding, I think I can still learn a lot from these forums there are somethings I may not know. But also maybe get a Nobel Prize :)

    Thank you for attention.
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