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Special Relativity- Photon/Mirror Collision

  • #1

Homework Statement


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.

Homework Equations


E= hν
Conservation of four-momentum

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?
 

Answers and Replies

  • #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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