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Null result of the michelson morley experiment

  1. Sep 6, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that the null effect of the michelson morley experiment can be accounted for if the interferometer arm parallel to motion is shortened by a factor of
    (1-v^2/c^2)^.5

    2. Relevant equations

    V(earth)=29.8 km/s , fringe in MM experiment is .4 ; gamma =sqrt(1-v^2/c^2)

    3. The attempt at a solution

    In the MM original experiment , .4 fringe/(29.8 km/s)^2 = .00045043

    I was supposed to show that the fringe effect from the original experiment is still the same even if I used lorentz transformations right?>

    .4 fringe/(v(earth) * gamma)=.4 fringe/((29.5)*^2*sqrt(1-(29.5)^2/c^2))= .00045043

    My calculator approximates sqrt(1-v^2/c^2 ) to 1
     
    Last edited: Sep 6, 2007
  2. jcsd
  3. Sep 6, 2007 #2
    You can use a binomial approximation to calculate sqrt(1-v^2/c^2)=1-1/2*v^2/c^2+...
     
  4. Sep 6, 2007 #3
    you kinda didn't answer my first question

    Original question:
    I was supposed to show that the fringe effect from the original experiment is still the same even if I used lorentz transformations right?>

    Other than that, Is everything else I solved correct
     
  5. Sep 6, 2007 #4

    learningphysics

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    The total time taken light was expected to take, in the path perpendicular to earth's motion was:

    [tex]T_1 = \frac{2L}{\sqrt{1-v^2/c^2}}[/tex]

    The total time expected parallel to earth's motion was:

    [tex]T_2 = \frac{2Lc}{c^2-v^2}[/tex]

    So if length contraction happens during the parallel path, what happens T2... and hence what does T2 - T1 become?
     
  6. Sep 7, 2007 #5
    if length is contracted during parrallel path, would L = L(0)/gamma ?
     
  7. Sep 7, 2007 #6

    learningphysics

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    Yes, the length along the parallel path would be L/gamma. So substitue L/gamma instead of L in the T2 formula...
     
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