Exercise 1.01

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 (traveling in the $x-y$ plane) and the reflected angle? What about the velocity before and after?

To solve this problem, let's consider the change of velocity at the moment of collision.

The impact is characterized by the following transformation of the 4-velocity.

$$\tilde{U}^{\mu'}=\Theta ^{\mu'}{}_{\nu'} U^{\nu'} \qquad \text{with}\qquad  \Theta ^{\mu'}{}_{\nu'}=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 \\  0 & 0 & 0 & 1 \end{pmatrix} $$

The boost goes along the $y$-axis.

$$ \Lambda ^{\mu'}{}_{\mu}=\begin{pmatrix} \gamma(v) & 0 & -v\gamma(v) & 0 \\ 0 & 1 & 0 & 0 \\ -v\gamma(v) & 0 & \gamma(v) & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\qquad\Lambda ^{\mu}{}_{\mu'}=\begin{pmatrix} \gamma(v) & 0 & v\gamma(v) & 0 \\ 0 & 1 & 0 & 0 \\ v\gamma(v) & 0 & \gamma(v) & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

We can put the 4-velocity back into the initial frame by applying the inverse transformation.

$$U^{\mu}=\Lambda ^{\mu}{}_{\mu'}U^{\mu'} =\begin{pmatrix} \gamma(v) & 0 & v\gamma(v) & 0 \\ 0 & 1 & 0 & 0 \\ v\gamma(v) & 0 & \gamma(v) & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} \gamma(u) \\ u_{x} \\ u_{y} \\ 0  \end{pmatrix} =\begin{pmatrix} \gamma(v)\gamma(u)+v\gamma(v)u_{y} \\ u_{x} \\ \gamma(v)\gamma(u)v+\gamma (v)u_{y} \\ 0 \end{pmatrix}$$

$$ \tilde{U}^{\mu}=\Lambda ^{\mu}{}_{\mu'}\underbrace{ \Theta^{\mu'}{}_{\nu'} U^{\nu'} }_{ \tilde{U}^{\mu'} } =\begin{pmatrix} \gamma(v) & 0 & v\gamma(v) & 0 \\ 0 & 1 & 0 & 0 \\ v\gamma(v) & 0 & \gamma(v) & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} \gamma(u) \\ -u_{y} \\ -u_{x} \\ 0 \end{pmatrix} =\begin{pmatrix} \gamma(v)\gamma(u)-v\gamma(v)u_{x} \\ -u_{y} \\ \gamma(v)\gamma(u)v-\gamma (v)u_{x} \\0\end{pmatrix}$$

From $U^{\mu}$ and $\tilde{U}^{\mu}$ we can compute the angles... 

$$\tan\left( \frac{\pi}{4}-\theta_\text{incident} \right)=\frac{U^{1}}{U^{2}}\qquad \iff \qquad \theta_\text{incident}= \frac{\pi}{4}-\arctan\left( \frac{U^{1}}{U^{2}} \right)$$

$$\boxed{ \theta_\text{incident}= \frac{\pi}{4}-\arctan\left(\frac{\gamma(v)\gamma(u)v+\gamma (v)u_{y}}{u_{x}} \right) }$$

$$\tan\left( \frac{\pi}{4}-\theta_\text{reflected} \right)=\frac{\tilde{U}^{1}}{\tilde{U}^{2}}\qquad \iff \qquad \theta_\text{reflected}= \frac{\pi}{4}-\arctan\left( \frac{\tilde{U}^{1}}{\tilde{U}^{2}} \right)$$

$$\boxed{ \theta_\text{reflected}= \frac{\pi}{4}-\arctan\left(\frac{-u_{y}}{\gamma(v)\gamma(u)v-\gamma (v)u_{x}} \right) }$$

... and the velocities.

$$\boxed{ v_\text{incident}=\sqrt{ U_{i}U^{i} }=\sqrt{ u_{x}^{2}+\Big(\gamma(v)\gamma(u)v+\gamma (v)u_{y}\Big)^{2} } }$$

$$\boxed{ v_\text{reflected}=\sqrt{ \tilde{U}_{i}\tilde{U}^{i} }=\sqrt{ u_{y}^{2}+\Big(\gamma(v)\gamma(u)v-\gamma (v)u_{x}\Big)^{2} } }$$