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Thread: Michelson & Morley X stream vs. Up/Down Stream

  1. #1 Michelson & Morley X stream vs. Up/Down Stream 
    Time Lord
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    Mar 2007
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    I don't really know what, if anything, to conclude from this, but something I noticed about the effect that was predicted to be observed by Michelson and Morley's interferometer (assuming there had been an aether), is that, if you measure the apparent speed of light as a ratio, instead of an absolute value, then the cross stream speed would always be the square root of the upstream/downstream speed.

    For example, suppose the apparatus were moving at .2 C in the forward direction. The beam of light crosses the distance upstream at an apparent speed of .8 C, and then returns at an apparent speed of 1.2 C.

    Time of travel = D/.8C + D/1.2C = 1.2D + .8D)/.96C = 2D/.96C Which implies an average apparent speed of .96 C

    For the apparent cross stream speed, we can use a right triangle constructed from the vectors where the hypotenuse is the light's actual path, one leg is the path of the mirror the light is traveling to, and third leg is the path the light appears to be following (since it appears to be going perfectly sideways to an observer located in the apparatus.)

    So, using C as the speed for the hypotenuse, and .2C as the speed for the mirror, C^2(1 - .04) = .96 C^2 and then we have to take the square root. 0.980 So that's my point, the ratio is the same until we take the square root.

    Using other values, like .1 or .3, it still always comes out the same.

    Upstream downstream for .1 is D/.9C + D/1.1C = D(1.1 + .9)/.99C = 2D/.99
    Cross Stream is C^2(1-.01) = .99C^2 (and then you have to take the square root 0.995)

    Upstream downstream for .3 is D/.7C + D/1.3C = D(1.3 + .7)/ .91C = 2D/.91C
    Cross Stream is C^2(1-.09) = .91C^2 (and then you have to take the square root 0.954)


    And.... that second thing is identical to the Lorentz factor anyway. For a speed of .3C, 1 - v^2/c^2 could be written as 1 - .3^2/1 = .91 (Remembering that you have to take the square root, and get 0.954 I just can't find a good square root symbol on my keyboard or I'd include it.) I think the reason why they're identical is because if you divide any number by its own square root, you .... get it's square root.

    That's what happens, of course, when we apply the Lorentz factor. If the apparent upstream/down stream speed (d/t) appeared to be .96C because the apparatus is moving at .2C, and we then shorten our own perception of distance (which slows the apparent speed of the light), by dividing that velocity by .980, we get .980 as its apparent speed, and now the cross stream speed is identical to the upstream/downstream speed. (Which is why the Lorentz factor exists.)

    But, we could have achieved the same effect by simply taking the square root of the apparent upstream/downstream speed.


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  3. #2  
    Time Lord
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    In short, my point is that the cross stream speed factor is always related to the upstream/downstream speed factor by way of being it's square root. (Just in case nobody really wants to read that big long description above.)


    I know that's probably not a big deal. It just makes me wonder what alternative ways the data could be interpreted, being as how the Lorentz contraction factor isn't the only way to make the speeds match..... or kind isn't.


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