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## Homework Statement

Consider an inertial frame S with coordinates ##x^μ = (t, x, y ,z)##, and a frame S' with coordinates ##x^{μ'}## related to S by a boost with velocity parameter v along the y-axis. Imagine we have a wall at rest in S', lying along the line x' = -y'. From the point of view of S, what is the relationship between the incident angle of a particle hitting the wall (traveling in the x-y plane) and the reflected angle? What about the velocity before and after?

## Homework Equations

Lorentz Transformations

##θ_i = incident~ angle, ~~~θ_f = reflected~ angle##

##γ = Lorentz~ factor##

## The Attempt at a Solution

We know that ##θ_i' = θ_f'##. Since S' is moving along the y-axis, the wall would appear to be contracted only in the y direction. Suppose the wall has length L, then the y component of the wall as viewed in S is

##y = \frac{L~sin(φ')}{γ}##

The x component of the wall is the same as in the S' frame, so the angle ##φ## that the wall makes with respect to the x axis is

##φ = tan^{-1}(\frac{\frac{L~sin(φ')}{γ}}{L~cos(φ')}) = tan^{-1}(\frac{tan(φ')}{γ})##

This angle ##φ## will be smaller compared to ##φ'##, their difference ##φ' - φ## is the increase in angle the observer in S will see for the incident angle (decrease for the reflected angle), so the incident and reflected angles as seen in S are

##θ_i = θ_i' + (φ' - φ). ~~~ θ_f = θ_f' - (φ' - φ)##

The velocity before and after in frame S are just related by Lorentz transformations to the velocity before and after in the frame S'.

I'm not sure if my argument for the incident and reflected angle is correct, can anyone help check my work? Thanks.