# Lorentz tranformation of law of reflection

## Homework Statement

Consider an inertial frame $S$ with coordinates $x^{\mu}=(t,x,y,z)$, and a frame $S'$ with coordinates $x^{\mu'}$ 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 ball hitting the wall (travelling in the $x$-$y$ plane) and the reflected angle? What about the velocity before and after?

## The Attempt at a Solution

From the point of view of $S'$, the incident angle equals the reflected angle.

The frame $S'$ travels parallel to the $y$-axis of the frame $S$. Therefore, $y'=-x' \implies \gamma\ y = - x$.

Therefore, from the point of view of the $S$ frame, the wall has a gentler slope (i.e., slope of smaller magnitude).

Will the trajectories also change their directions in the $S$ frame?

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The fact that the wall has a gentler slope in the $S$ frame is a true fact. However, I think that it is not of much use in solving the problem. It can, however, help in developing a feel for how the wall and the trajectories of the incident and the reflected particles appear from the $S$ frame. In particular, I think that, in the $S$ frame, (due to Lorentz contraction only along the $y$-direction, and no Lorentz contraction along the $x$- direction) the incident angle decreases and the reflected angle increases (with respect to the Lorentz-transformed normal to the wall) with the increase of the velocity of the $S'$ frame. This is illustrated in the figure below. My additional progress with the problem so far: Using the figure above,

$1 = \text{tan}\ 45^{\circ} = \text{tan}\ \phi'_{i}=\frac{Y'_{i}}{X'_{i}}=\frac{\gamma\ Y_{i}}{X_{i}}=\gamma\ \text{tan}\ \phi_{i}$ so that $\text{tan}\ \phi_{i}=\frac{1}{\gamma}$

and

$1 = \text{tan}\ 45^{\circ} = \text{tan}\ \phi'_{f}=\frac{X'_{f}}{Y'_{f}}=\frac{X_{f}}{\gamma\ Y_{f}}=\frac{1}{\gamma}\ \text{tan}\ \phi_{f}$ so that $\text{tan}\ \phi_{f}=\gamma$. Furthermore, using the figure above,

$\text{tan}\ (\phi'_{i}-\theta'_{i})=\frac{y'_{i}}{x'_{i}}=\frac{\gamma\ y_{i}}{x_{i}}=\gamma\ \text{tan}\ (\phi_{i}-\theta_{i})$

and

$\text{tan}\ (\phi'_{f}-\theta'_{f})=\frac{x'_{f}}{y'_{f}}=\frac{x_{f}}{\gamma\ y_{f}}=\frac{1}{\gamma}\ \text{tan}\ (\phi_{f}-\theta_{f})$

so that

$\text{tan}\ (\phi'_{i}-\theta'_{i})=\text{tan}\ (\phi'_{f}-\theta'_{f})$

$\implies \gamma\ \text{tan}\ (\phi_{i}-\theta_{i}) = \frac{1}{\gamma}\ \text{tan}\ (\phi_{f}-\theta_{f})$

$\implies \gamma^{2}\ \frac{\text{tan}\ \phi_{i}-\text{tan}\ \theta_{i}}{1+\text{tan}\ \phi_{i}\text{tan}\ \theta_{i}}=\frac{\text{tan}\ \phi_{f}-\text{tan}\ \theta_{f}}{1+\text{tan}\ \phi_{f}\text{tan}\ \theta_{f}}$

$\implies \gamma^{2}\ \frac{\frac{1}{\gamma} - \text{tan}\ \theta_{i}}{1+\frac{1}{\gamma}\text{tan}\ \theta_{i}}=\frac{\gamma-\text{tan}\ \theta_{f}}{1+\gamma\ \text{tan}\ \theta_{f}}$

$\implies \gamma^{2}\ \frac{1- \gamma\ \text{tan}\ \theta_{i}}{\gamma +\text{tan}\ \theta_{i}}=\frac{\gamma-\text{tan}\ \theta_{f}}{1+\gamma\ \text{tan}\ \theta_{f}}$

Am I doing ok? Am I heavy weather out of a simple problem?