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Relativistic Doppler

  1. Oct 16, 2007 #1
    1. The problem statement, all variables and given/known data
    I am told that to a stationary observer, the lights infront of a spaceship has the wavelength = 480 nm when it arrives, and the lights in the back have wavelength 640 nm when the ship is moving away. Both lights (front and back) are the same.

    I have to find the wavelength of the two lights.

    3. The attempt at a solution
    I have tried to use T = T_0 * (1+w/c)/sqrt(1-w^2/c^2) and from there use T-T_1 ... so the w's would go out, but they don't.

    When the spaceship is arriving, the speed is v - when it is moving away, it is -v. From there, I tried the above.

    How do I find the wavelengths? (I have to calculate relativistic).
     
    Last edited: Oct 16, 2007
  2. jcsd
  3. Oct 16, 2007 #2
    Is it possible to find it without using the velocity? (Because I am not told what it is)
     
  4. Oct 16, 2007 #3
    Another thing, related to my first post, is: When the spaceship is arriving and the light is being sent forward - is there any change from that between when the spaceship is moving forward and the light is being emitted backwards?
     
  5. Oct 16, 2007 #4

    Doc Al

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    Staff: Mentor

    This is an equation for period, which is 1/f. Express this in terms of wavelength by using [itex]\lambda = c/f = cT[/itex].

    Now write both equations: one for an approaching source and one for a receding source.

    By combining the two equations, you'll be able to eliminate the velocity and solve for the source wavelength.

    Nothing that's not already contained in the formula.
     
  6. Oct 16, 2007 #5
    Hmm, it seems easier to solve for v first, and then for lambda?
     
  7. Oct 16, 2007 #6

    Doc Al

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    If you combine the two equations correctly, you won't need to solve for v.
     
  8. Oct 16, 2007 #7
    I have:

    1) lamdba/480 = (1 + v/c)/sqrt(1-v^2/c^2)

    2) lamdba/640 = (1 - v/c)/sqrt(1-v^2/c^2)

    I isolated lamdbda, and inserted and found v. But I think I am meant to go the other way? Although it seems much harder
     
  9. Oct 16, 2007 #8

    Doc Al

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    Good.

    Not sure what you mean. If you mean to solve for v, then plug it into the other equation to get lamdba... WAY too hard.

    Hint: Think of clever ways of combining these two equations. (Think of the basic math operations.)
     
  10. Oct 16, 2007 #9
    :smile:

    I can add them, so the v/c-part goes out, but the squareroot in the nominator can't? I tried subtracting, but that gave 2v/c, which doesn't seem much better.
     
  11. Oct 16, 2007 #10

    Doc Al

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    Good! So addition and subtraction don't seem to help. What's next? :wink:
     
  12. Oct 16, 2007 #11
    Ahh.. thanks Doc! :smile:
     
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