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Relative velocity

  1. Aug 18, 2008 #1
    1. The problem statement, all variables and given/known data
    I was reading up on an experiment,The Michelson Morley experiment, in preparation for Physics lab. Then there was this problem. The textbook stated that the velocity of the light ray that was traveling perpendicularly to the ether wind was \root{c^2-v^2}. Shouldn't it have been \root{c^2+v^2} since the light ray was supposed to be directed perpendicularly to the ether wind? Also, shouldn't the velocity in this case \root{c^2+v^2} have been subjected to changes since the ether wind was now also blowing against it?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 19, 2008 #2
    is anyone able to help? My professor has come up with an explanation for me but I still don't get it!! I feel hard pressed to ask him again.
     

    Attached Files:

  4. Aug 19, 2008 #3

    Doc Al

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    You might find this analysis helpful: The Michelson-Morley Experiment
     
  5. Aug 19, 2008 #4
    I still have problem understanding this:

    The light was reflected perpendicularly to the ether wind, this is not similar to the swimmer's example because the swimmer was swimming at an angle. I agree that the velocity the swimmer would be actually swimming was \root{v1^2-v2^2}, however if the light ray was aimed perpendicular to the ether wind shouldn't the resultant velocity be \root{v^2+c^2}?
     
  6. Aug 19, 2008 #5

    Doc Al

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    I understand what you're saying. The way I look at it is that the beam was aimed such that it traveled perpendicular to the direction of the ether wind. In order for it to travel in such a direction the light beam must have been angled slightly into the ether wind (just like the swimmer, who swims directly across the river, must angle himself upstream). (What they really do is just align the beam and mirrors so that the reflected beam returns along the same path.)
     
  7. Aug 20, 2008 #6
    hmmmm, i see now doc AI, as long as the beam were to travel perpendicularly to hit the mirror, it was then assumed that they have already directed the beam off at an angle and the ether wind has already shifted the ray to the perpendicular position! (when actually it has not of course)

    I just have one more question:
    For the swimmer example, lets say the swimmer has a swimming rate of 5 feet/s and is now swimming upstream and the water is flowing at a rate of 2 feet/s. The link you gave me showed that the swimmer should be swimming at 3 feet/s. How is this the case? This is only the case if it was assumed that the mass of water crashing against him is the same as his mass. Isn't this so and a very large and incorrect assumption?
     
  8. Aug 20, 2008 #7

    Doc Al

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    Mass has nothing to do with it. It's just a question of understanding relative velocity. The swimmer can swim 5 feet/s with respect to the water. The water is moving 2 feet/s with respect to the land. Thus, if he swims directly against the current, the swimmer's speed is 5 - 2 = 3 feet/s with respect to the land.

    In general (if you can catch on to my notation: a = swimmer; b = water; c = land):

    [tex]\vec{V}_{a/c} = \vec{V}_{a/b} + \vec{V}_{b/c}[/tex]
     
  9. Aug 20, 2008 #8
    oh i see now doc ai, the swimmer swims 5 feet wrt to the water! I understand now, I thought that 5 feet/s was with respect to the land! It didnt make sense of course.

    However, (so sorry), I have one last question:

    the speed of waves is supposed to be emitted independent of the speed of the source. (sound etc.) In the Michelson-Morley experiment, they thought that this was the case and the reason the light waves still caught up with them and the velocity of the earth was because of the ether wind right?

    Then is it true to say that the direction of the light emitted is dependent on the source since there is no such thing as the ether?
     
  10. Aug 21, 2008 #9

    Doc Al

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    I'm not exactly sure what you're asking, so I'll just rephrase what you probably already know. The idea was that light traveled through the ether, and since the Earth was rocketing along in its orbit, we should be able to detect some "ether wind" with respect to us. But we didn't.
    Again, I'm not sure of your reasoning, but it's certainly true that the apparent direction of a "ray" of light does depend on the frame doing the observing. To get more on this, look up "stellar aberration".
     
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