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Length Contraction Ether wind

  1. Sep 21, 2012 #1
    Length Contraction "Ether wind"

    1. The problem statement, all variables and given/known data

    Due to the ether wind, anything that moves along it is "contracted".

    3. The attempt at a solution

    Taking L1 to be contracted length,

    (L1)2
    = (L1x)2 + (L1Y)2
    = (L10,X/γ)2 + (L10,Y)2


    But what they wrote is the opposite..
     

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    Last edited: Sep 21, 2012
  2. jcsd
  3. Sep 21, 2012 #2

    TSny

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    Re: Length Contraction "Ether wind"

    The equation (3-3) in the book is an expression for the "normal" (not contracted) length in terms of the components of the contracted length. Your expression is for the contracted length in terms of the components of the normal length.
     
  4. Sep 21, 2012 #3
    Re: Length Contraction "Ether wind"

    Are both equivalent??


    I tried to prove but the square roots got me..
     
  5. Sep 22, 2012 #4
    Michelson Morley Experiment: Rotating setup

    1. The problem statement, all variables and given/known data

    Apparently rotating the setup at an angle doesn't change the time difference between t1 and t2..

    3. The attempt at a solution

    n1xx68.jpg

    But the square roots below are clearly different..
     
  6. Sep 22, 2012 #5

    Simon Bridge

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    Re: Michelson Morley Experiment: Rotating setup

    square roots below what?
     
  7. Sep 22, 2012 #6
    Re: Michelson Morley Experiment: Rotating setup

    I tried changing everything in the square roots to only 'sin' or 'cos' but can't seem to remove the square roots at all..
     
  8. Sep 22, 2012 #7

    Simon Bridge

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    Re: Michelson Morley Experiment: Rotating setup

    I'll have to see your working - is there an attachment missing?
    OH I see ... there is a tinypic.com link that is not rendering for me.
     
  9. Sep 22, 2012 #8
    Re: Michelson Morley Experiment: Rotating setup

    Oh sorry about that, can you view this attachment?
     

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  10. Sep 23, 2012 #9
    Re: Length Contraction "Ether wind"

    bumpp
     
  11. Sep 23, 2012 #10

    Simon Bridge

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    Re: Length Contraction "Ether wind"

    Oh there you are - couldn't access the thread for a while there.
    I'll have a proper look when I have a bit of time.
    [edit]Is this the reworking of the MM experiment using SR?
     
    Last edited: Sep 23, 2012
  12. Sep 23, 2012 #11

    TSny

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    Re: Length Contraction "Ether wind"

    unscientific,

    Note that the time, [itex]t_1[/itex], to travel out and back along [itex]l_1[/itex] is not [itex]2l_1/c[/itex] because ##c## is not the speed of light relative to the apparatus. You're going to have to figure out the speed of the light relative to the apparatus for the light going out along arm [itex]l_1[/itex] and the (different) speed for the light coming back. Likewise for the other arm.
     
    Last edited: Sep 23, 2012
  13. Sep 26, 2012 #12
    Re: Length Contraction "Ether wind"

    I assume:
    1) only the component parallel to ether wind is affected by it
    2) The light is aimed at a smaller angle to compensate for this

    so the speed towards is:
    √[ (c sinθ)2 + (c cosθ - v)2 ]

    and the speed back is:

    √[ (c sinθ)2 + (c cosθ + v)2 ]

    Not sure if that's right..
     
  14. Sep 26, 2012 #13

    TSny

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    Re: Length Contraction "Ether wind"

    That would be ok if θ here is the angle that the light is "aimed at". But in that case, it's not the angle of inclination of the arm as shown in the diagram.
     
  15. Sep 27, 2012 #14
    Re: Length Contraction "Ether wind"

    I'm not sure what you mean..If c is indeed affected by ether wind won't θ be off-balanced?
     
  16. Sep 27, 2012 #15

    TSny

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    Re: Length Contraction "Ether wind"

    Indeed. You'll have to fire the light beam in some direction [itex]\phi[/itex] so that the wind will blow it into the direction θ (along the arm). I was interpreting the direction "aimed at" as the direction fired ([itex]\phi[/itex]). Your expression for the light speed relative to the apparatus is ok if your θ is replaced by [itex]\phi[/itex].

    You might try drawing a velocity addition triangle and using the law of cosines to express the light speed relative to the apparatus in terms of θ without worrying about the value of [itex]\phi[/itex].
     
  17. Sep 27, 2012 #16
    Re: Length Contraction "Ether wind"

    If I correctly understand what you say, then you are copying an error of Michelson! See:

    1. http://en.wikisource.org/wiki/The_Relative_Motion_of_the_Earth_and_the_Luminiferous_Ether :
    " If, however, the light had traveled in a direction at right angles to the earth's motion it would be entirely unaffected."

    I think that this is also what you are saying now.

    2. http://en.wikisource.org/wiki/Influence_of_Motion_of_the_Medium_on_the_Velocity_of_Light :

    "In deducing the formula for the quantity to be measured, the effect of the motion of the earth through the ether on the path of the ray at right angles to this motion was overlooked." [etc]

    Does that help?
     
  18. Sep 27, 2012 #17
    Re: Length Contraction "Ether wind"

    Can't seem to get it to be independent of ∅... it turns out as (θ-∅)
    2yyy3nq.jpg
     
  19. Sep 27, 2012 #18
    Re: Length Contraction "Ether wind"

    I think this question works on the assumption that speed of light 'c' is unaffected by aether. Rather, they are trying to do length contraction here...

    This brings me to my second post:

     
  20. Sep 27, 2012 #19

    TSny

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    Re: Length Contraction "Ether wind"

    Harrylin, I haven't yet had time to look carefully at your links. They look interesting!

    But all I'm saying is that if you let c' represent the speed of light along the arm ##l_1##, then you can express c' in terms of c and θ using relative velocity arguments. See the attached diagram.
     

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    Last edited: Sep 27, 2012
  21. Sep 27, 2012 #20
    Re: Length Contraction "Ether wind"

    In this part they assume speed of light 'c' is the same. They are assuming only the distance contracts..

    This is from A.P. French's Relativity text...chapter 3. I'm finding this text very very difficult, especially chapter 2 and 3 (I've skipped the entire chapter 2) I've already completed the homework problems from Halliday, Resnick and Walker and I find them to be much easier. This is demoralizing..
     
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