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Angle independence of Michelson-Morley type experiments

  1. Jul 7, 2007 #1
    In all descriptions of the Michelson Morley experiment I know of, the two arms of the interferometer are at right angles to each other.

    Does anyone know of an experiment which, just for completeness, tried other angles? If not, does anyone know of a reference with clear argumentation why it can be ruled out that other angles need to be tested?

    Not that I expect a different result for non-right angles, but (mathematical) proof is all the while better than guesstimation.

    Harald.
     
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  3. Jul 7, 2007 #2

    Meir Achuz

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    Right angles are used so that when the apparatus is turned through 90 degrees, the two directions are interchanged. To interchange the directions for any other angle than right angle, would require flipping the apparatus. This would be a tricky procedure for such a sensitive apparatus.
     
  4. Jul 7, 2007 #3

    StatusX

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    The assumption is that the aether provides an absolute rest frame, so that if the earth is moving through the aether, its velocity determines a priveleged direction in our reference frame. Then if the speed of light was only isotropic in the aether frame, it would have a velocity depending on the angle [itex]\theta[/itex] between the direction the light is travelling in our frame and the priveleged direction mentioned above. What Michelson and Morely attempted to do was measure [itex]v(\theta)[/itex].

    Since we don't initially know the special direction, we actually need to measure the velocity in any direction [itex]\hat n[/itex], and if the aether theory is correct, there should be a direction [itex]\hat m[/itex] such that [itex] v(\hat n) = f( \hat m \cdot \hat n) [/itex].

    So let the two directions of the arms of the appartus be directed along [itex]\hat n_1[/itex] and [itex] \hat n_2[/itex]. They then observed that the time it takes a light beam to travel the length of the arm and back is the same for both arms. Assuming the arms are the same length, this amounts to the equation:

    [tex] \frac{1}{v(\hat n_1)} + \frac{1}{v(-\hat n_1)} = \frac{1}{v(\hat n_2)} + \frac{1}{v(-\hat n_2)}[/tex]

    In theory, this could be verified for any directions [itex]\hat n_1,\hat n_2[/itex], in which case we would arrive at:

    [tex]\frac{1}{v(\hat n)} + \frac{1}{v(-\hat n)} = C[/tex] (*)

    To compare the velocity of light along antiparellel directions you need to agree on a timing convention at distant points. Usually the convention is taken by assuming the two velocities are the same. Another way is to assume that if a clock is moved sufficiently slowly to a new point, it still measures time accurately, and it can be shown this convention agrees with the first. Thus

    [tex]\frac{1}{v(\hat n)} + \frac{1}{v(-\hat n)} = \frac{2}{v(\hat n)} = C[/tex]

    And the speed of light is independent of direction.

    You point out that in practice, only directions [itex]\hat n_1,\hat n_2[/itex] which are perpendicular are compared. In 2 dimensions, it would be impossible to make the step (*) above, since if a is not perpendicular to b, then there is no vector c that is perpendicular to both, which would need to be used to compare them.

    However in 3 dimensions, there is such a vector c for any a and b, namely their cross product. I'm not sure if Michelson and morley actually used this technique, as they weren't so much concerned with measuring the isotropy of light as verifying the aether theory, which could be proved/disproved more simply (eg, by comparing the results at two different points along earths orbit, so that the priveleged direction would necessarily change).
     
    Last edited: Jul 7, 2007
  5. Jul 8, 2007 #4
    Actually, what you want to do is interchange the directions of the arms as measured from the direction of motion. You can always do that, no matter what the angles. Say the arms are at angles A and B measured clockwise with respect to the direction of motion. Now rotate through -(A+B), keeping track of +/- clockwise rotations. You can see that A --> -B and B --> -A. So the angles of the arms with respect to the direction of motion are interchanged. The directions orthogonal to the motion have changed sign, but that happens with right angles too.
     
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