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Special Relativity- Photon/Mirror Collision

  1. Jun 22, 2016 #1
    1. The problem statement, all variables and given/known data
    A photon of frequency ν is reflected without change of frequency from a mirror, with an angle of incidence θ. Calculate the momentum transferred to the mirror.

    2. Relevant equations
    E= hν
    Conservation of four-momentum

    3. The attempt at a solution
    If the mirror is in the x-y plane, the incident photon will have initial 4-momentum
    Pγ = (hν , hνsinθ , 0 , -hνcosθ) as the sum of the squares of the spatial momenta must be equal to the square of the time coordinate momentum.
    The initial 4-momentum of the mirror:
    Pm = (m , 0 , 0 , 0) as it is initially at rest in the chosen frame of reference and m is the mass of the mirror.

    After the collision:
    Pγ' = ( hν , hνsinθ, 0 , hνcosθ)
    Pm' = (E , p)

    Applying conservation law:
    Pm + Pγ = Pγ' + Pm'
    (Pm')^2 = (Pm)^2 + (Pγ)^2 + (Pγ')^2 + 2Pm (Pγ-Pγ') - 2Pγ⋅Pγ'
    -E^2 + p^2 = -m^2 + 0 + 0 +2⋅0 -2{ -(hν)^2 + (hνsinθ)^2 - (hνcosθ)^2} (p^2 is the square of the norm of the spatial momenta components)
    -E^2 + p^2 = -m^2 + 4(hνcosθ)^2

    This is where I don't get what to do. I know that E^2 = m^2 + p^2 but if I substitute that in, the momentum term disappears and I'm left with 4(hνcosθ)^2 = 0.

    I also know I get the right answer if E=m, then p= 2hνcosθ , but I don't see how I am able to say that. Is it because the mirror is not a particle in its own right, but rather a system? Or does I have to change to a different reference frame where the mirror will be stationary relative to it?
     
  2. jcsd
  3. Jun 23, 2016 #2

    Orodruin

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    You do not need to know the mirror's mass or original 4-momentum. You also do not need to square any 4-momentum. All you need to do is to apply conservation of momentum.
     
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