(a) Write out the matrix that describes a Galilean transformation (Eq. 12.12).

(b) Write out the matrix describing a Lorentz transformation along the yaxis.

(c) Find the matrix describing a Lorentz transformation with velocity v along the x axis followed by a Lorentz transformation with velocity $\overline{\mathbf{v}}$along they axis. Does it matter in what order the transformations are carried out?

Write the values of$\overline{t},\overline{x},\overline{y}$ and $\overline{z}$using Galilean transformation.

$$\begin{gathered} {\mathit{x}}^{\mathbf{0}}\mathbf{=}\mathit{c}\mathit{t} \\ \overline{\mathbf{x}}\mathbf{=}\mathit{x}\mathbf{-}\mathit{v}\mathit{t} \\ \overline{\mathbf{y}}\mathbf{=}\mathit{y} \\ \overline{\mathbf{z}}\mathbf{=}\mathit{z} \end{gathered}$$

Also,

$\overline{\mathbf{t}\mathbf{}}\mathbf{=}\mathit{t}$

(a)

Write a matrix that describes a Galilean transformation.

$\left[\begin{array}{c}ct\\ \overline{x}\\ \overline{y}\\ \overline{z}\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ -\beta & 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}ct\\ x\\ y\\ z\end{array}\right]$

Therefore, the matrix that describes a Galilean transformation is

$\left[\begin{array}{c}ct\\ \overline{x}\\ \overline{y}\\ \overline{z}\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ -\beta & 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}ct\\ x\\ y\\ z\end{array}\right]$

(b)

Write the values of $\overline{t},\overline{x},\overline{y}$and $\overline{z}$using Lorentz transformation along the y-axis.

$$\begin{gathered} \overline{\mathbf{t}\mathbf{}}\mathbf{=}\mathit{y}\left(1-\beta t\right) \\ \overline{\mathbf{x}}\mathbf{=}\mathit{x} \\ \overline{\mathbf{y}}\mathbf{=}\mathit{y}\left(y-\beta t\right) \\ \overline{\mathbf{z}}\mathbf{=}\mathit{z} \end{gathered}$$

Write a matrix that describes a Lorentz transformation along the y-axis.

$\left[\begin{array}{c}t\\ \overline{x}\\ \overline{y}\\ \overline{z}\end{array}\right]=\left[\begin{array}{cccc}\gamma & 0& -\gamma \beta & 0\\ 0& 1& 0& 0\\ -\gamma \beta & 0& \gamma & 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}t\\ x\\ y\\ z\end{array}\right]$

Therefore, the matrix that describes a Lorentz transformation along the y-axis is.

$\left[\begin{array}{c}t\\ \overline{x}\\ \overline{y}\\ \overline{z}\end{array}\right]=\left[\begin{array}{cccc}\gamma & 0& -\gamma \beta & 0\\ 0& 1& 0& 0\\ -\gamma \beta & 0& \gamma & 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}t\\ x\\ y\\ z\end{array}\right]$

(c)

Write the matrix for Lorentz transformation with velocity v along the x-axis.

$A=\left[\begin{array}{cccc}\gamma & -\gamma \beta & 0& 0\\ -\gamma \beta & \gamma & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

Write the matrix for Lorentz transformation with velocity along the y-axis.

$B=\left[\begin{array}{cccc}\overline{\gamma }& 0& -\overline{\gamma }\overline{B}& 0\\ 0& 1& 0& 0\\ -\overline{\gamma }\overline{B}& 0& \overline{\gamma }& 0\\ 0& 0& 0& 1\end{array}\right]$

Take the product of matrices A and B.

$$\begin{gathered} A\times B=\left[\begin{array}{cccc}\gamma & -\gamma \beta & 0& 0\\ -\gamma \beta & \gamma & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{cccc}\overline{\gamma }& 0& -\overline{\gamma }\overline{B}& 0\\ 0& 1& 0& 0\\ -\overline{\gamma }\overline{B}& 0& \overline{\gamma }& 0\\ 0& 0& 0& 1\end{array}\right] \\ A\times B=\left[\begin{array}{cccc}\gamma \overline{\gamma }& -\gamma \overline{\gamma }\beta & -\overline{\gamma }\overline{\beta }& 0\\ -\gamma \beta & \gamma & 0& 0\\ -\overline{\gamma }\gamma \overline{\beta }& \gamma \overline{\gamma }\beta \overline{\beta }& \overline{\gamma }& 0\\ 0& 0& 0& 1\end{array}\right] \end{gathered}$$

Yes, the order does matter, in the other order, bar and no-bars would be switched, and this forms a different matrix.

Therefore, the matrix describing a Lorentz transformation along the x and y-axis are $A=\left[\begin{array}{cccc}\gamma & -\gamma \beta & 0& 0\\ -\gamma \beta & \gamma & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$and $B=\left[\begin{array}{cccc}\overline{\gamma }& 0& -\overline{\gamma }\overline{B}& 0\\ 0& 1& 0& 0\\ -\overline{\gamma }\overline{B}& 0& \overline{\gamma }& 0\\ 0& 0& 0& 1\end{array}\right]$respectively. Yes, the order does matter, in the other order, bar and no-bars would be switched, and this forms a different matrix.