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Reflection on a moving mirror

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data
    A mirror is moving uniformly in a direction normal to its plane with velocity v=βc. Given the angle of incidence and frequency of an incident photon (θ_e,nu_e in the figure), calculate the reflection angle and the observed new frequency (θ_i,nu_i in the figure).

    Also prove that:
    sin(θ_e)/(1+βcos(θ_e)) = sin(θ_i)/(1-βcos(θ_i))

    2. Relevant equations
    It has been suggested in a couple of threads to use Lorentz transformations, but i seem to have problems dealing with them.


    3. The attempt at a solution
    As said, my attempts to use them have been an utter failure. In the mirror's system the angles should be the same, but i get stuck at this point.
     

    Attached Files:

  2. jcsd
  3. Oct 30, 2012 #2

    tiny-tim

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    welcome to pf!

    hi xayon! welcome to pf! :wink:
    yes, the Lorentz transformation should do it …

    show us how far you've got, and where you're stuck :smile:
     
  4. Oct 30, 2012 #3

    haruspex

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    Does it help to think about the momenta in the X and Y directions?
     
  5. Oct 31, 2012 #4
    Yes, I've thought of the momentum, h*nu. Is the frequency of the incident photon as measured in the mirror frame the same as the one from an external observer?
    The other thing to take into account is that in the mirror frame the angles are the same isn't it? But again, by Doppler effect, the frequency of the photon would be different...
     
  6. Oct 31, 2012 #5

    haruspex

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  7. Nov 2, 2012 #6
    Ok, having Lorentz transformations for the momentum:
    [itex]E'=\gamma(E-\beta cp_{x})[/itex]
    [itex]cp'_{x} =\gamma(cp_{x}-\beta E)[/itex]
    [itex]cp'_y=cp_y[/itex]
    [itex]cp'_z=cp_z[/itex]

    Where the primed frame is the mirror's one.

    As haruspex pointed, I have divided the x and y components of the momentum, getting two expressions for [itex]tan(\theta _{e})[/itex] proving the statement. From that i can get [itex]\theta_{i}[/itex] in terms of [itex]\theta _{e}[/itex] and v.

    Finally getting [itex]\nu[/itex] from the 1st transformation.

    Am I right?
     
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