- #1

Parallax-Err01

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## 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

**P**γ = (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:

**P**m = (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:

**P**γ' = ( hν , hνsinθ, 0 , hνcosθ)

**P**m' = (E ,

**p)**

Applying conservation law:

**P**m +

**P**γ =

**P**γ' +

**P**m'

(

**P**m')^2 = (

**P**m)^2 + (

**P**γ)^2 + (

**P**γ')^2 + 2

**P**m

**⋅**(

**P**γ-

**P**γ') - 2

**P**γ

**⋅P**γ'

-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?