Question

Write the Lorentz boost along the + axis. Express it in terms of B and y Write the matrix for a boost of the same speed, but in the negative x dırection. Show that the product of these two matrices is the identity matrix. transformation as a 2 × 2 matrix for a

Answer 1

$M_{x}=\begin{pmatrix} \frac {1}{\sqrt {1-\frac {v^2}{c^2}}} &\frac {-\frac {v}{c}}{\sqrt {1-\frac {v^2}{c^2}}} \\ \frac {-\frac {v}{c}}{\sqrt {1-\frac {v^2}{c^2}}} & \frac {1}{\sqrt {1-\frac {v^2}{c^2}}} \end{pmatrix}$

$M_{x}=\begin{pmatrix} \gamma&-\gamma \beta \\ -\gamma \beta & \gamma \end{pmatrix}$ (Boost in +x direction)

$M_{-x}=\begin{pmatrix} \gamma&\gamma \beta \\ \gamma \beta & \gamma \end{pmatrix}$ (Boost in -x-direction)

$M_{-x}M_x=\begin{pmatrix} \gamma&-\gamma \beta \\ -\gamma \beta & \gamma \end{pmatrix}\begin{pmatrix} \gamma&\gamma \beta \\ \gamma \beta & \gamma \end{pmatrix}$

$M_{-x}M_x=\begin{pmatrix} \gamma^2-\gamma^2\beta^2 &\gamma^2 \beta-\gamma^2\beta \\ \gamma^2\beta +\gamma^2\beta & -\gamma^2\beta^2+\gamma^2 \end{pmatrix}$

$M_{-x}M_x=\begin{pmatrix} \gamma^2(1-\beta^2 )&0 \\ 0&\gamma^2(1-\beta^2) \end{pmatrix}$

$1-\beta^2=\frac {1}{\gamma^2}$

$M_{-x}M_x=\begin{pmatrix}1&0 \\ 0&\1 \end{pmatrix}=I$