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Doppler Shift Special Relativity Question

  1. Mar 28, 2013 #1
    1. The problem statement, all variables and given/known data
    light signal of frequency 33 MHz is reflected off a (radio)mirror that is moving
    toward you at half the speed of light. What is the reflected frequency in Hz?


    2. Relevant equations

    f_f = f*sqrt((1-v/c)/(1+v/c)


    3. The attempt at a solution

    The mirror and light wave are approaching you, so you use the receding/red shift equation (shown above).

    but when you plug in .5c for v and f - 33MHz you get 19.1MHz as the answer which is wrong.

    According to the solution, we need to get rid of the square root because "Doppler Shift from Mirror is the same". What does that mean???
     
    Last edited: Mar 28, 2013
  2. jcsd
  3. Mar 28, 2013 #2

    mfb

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    You don't want the frequency the mirror sees - the light is just reflected there, and relative to the frequency the mirror sees you have the doppler shift a second time.
    Alternatively, you can solve this in a non-relativistic way, just by geometry, in our lab frame.

    As the mirror comes closer, v should be -.5c.
     
  4. Mar 28, 2013 #3
    Using v = -.5, doesn't work out either. The correct answer should be 11MHz.



    I'm trying to think that the frequency of the incoming light is 33MHz. So it travels and it hits the mirror. It reflects and is redshifted to 11MHz... but somehow the math to get there involves only squaring the (c-v)/(c+v) BUT NOT the frequency... so confused




    Here is the solution according to my professor. It will be the very last problem in this pdf: http://panda.unm.edu/Courses/Thomas/Phys262fa10/P262X4S.pdf
     
  5. Mar 28, 2013 #4

    vela

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    It seems to me you should detect a frequency higher than the original 33 MHz. The mirror will see the light blue-shifted as it's moving toward the source, and then the reflected light will again be blue-shifted because the mirror is moving toward you.
     
  6. Mar 29, 2013 #5

    mfb

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    Doppler shift multiplies the frequency by some factor, and you have to do this multiplication twice. Therefore, the result is ##f'=f \sqrt{\frac{c-v}{c+v}} \sqrt{\frac{c-v}{c+v}} = f \frac{c-v}{c+v}##.

    If 11 MHz is the correct answer, the mirror is moving away from us.
     
  7. Mar 29, 2013 #6
    I get what you mean that if the answer is 11MHz the mirror must be moving away. This makes sense because the light will be compressed (thus increasing frequency) for a mirror moving toward us.

    So, from both of your explanations, I'm understanding the following. Please correct me if my reasoning is wrong.

    The mirror is approaching the light and the light is approaching the mirror, so according to the mirror, the light is compressed before it is even reflected. The already compressed light is then reflected and compressed further. Then we see it.

    And we denote this 'double' doppler shift as mfb stated in his previous post.
     
  8. Mar 29, 2013 #7
    So we get

    f_f = f * sqrt((1+v/c)/(1-v/c))*sqrt((1+v/c)/(1-(v/c))
    plug in the values f = 33MHz and v = .5c
    f_f = 99MHz

    We switch the equation to f_f = f*sqrt((1+v/c)/(1-v/c)) because the light and mirror are approaching each other. Correct? or semi-correct?
     
  9. Mar 29, 2013 #8

    mfb

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    The sign of v is just a convention, and you can use the same formula both for approaching/receding mirrors if you use positive/negative v.
     
  10. Mar 30, 2013 #9
    Thank you, mfb and vela!
     
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