Lorentz transformation under symmetric configuration

In physics, the Lorentz transformation converts between two different observers' measurements of space and time, where one observer is in constant motion with respect to the other.

Assume there are two observers $O_1$ and $O_2$ , each using their own Cartesian coordinate system to measure space and time intervals. $O_1$ uses $(t_1, x_1, y_1, z_1)$ and $O_2$ uses $(t_2, x_2, y_2, z_2)$ . Assume further that the coordinate systems are oriented so that the $x_1$ -axis and the $x_2$ -axis overlap but in opposite directions. The $y_1$ -axis is parallel to the $y_2$ -axis but in opposite directions. The $z_1$ -axis is parallel to the $z_2$ -axis and in the same direction. The relative velocity between the two observers is $v$ along the $x_1$ or $x_2$ axis. $v$ is defined as a positive number when $O_1$ sees $O_2$ sliding in the direction of $x_1$ . Also assume that the origins of both coordinate systems are the same. If all this holds, then the coordinate systems are said to be in symmetric configuration.

In this configuration, frame $O_2$ appears to $O_1$ in the identical way that frame $O_1$ appears to $O_2$ . However, in the standard configuration, if $O_2$ sees $O_1$ going forward then $O_1$ sees $O_2$ going backward. This symmetric configuration is equivalent to the [Lorentz transform#Lorentz transformation for frames in standard configuration|standard configuration]] followed by a mirror reflection of the x and y-axes. For the stationary $v=0$ case, this reduces to only the reflections, whereas the standard form reduces to the identity transformation.

The Lorentz transformation for frames in symmetric configuration is:

$t_1 = \gamma (t_2 - \frac{v x_2}{c^{2}})\ ,$

$x_1 = \gamma (v t_2 - x_2)\ ,$

$y_1 = - y_2 \ ,$

$z_1 = z_2 \ ,$

where $\gamma�:= 1/\sqrt{1 - v^2/c^2}$ is the Lorentz factor.

The inverse transformation is:

$t_2 = \gamma (t_1 - \frac{v x_1}{c^{2}})\ ,$

$x_2 = \gamma (v t_1 - x_1)\ ,$

$y_2 = - y_1 \ ,$

$z_2 = z_1 \ .$

The above forward and inverse transformations are identical. This offers mathematical simplicity.

In matrix form the forward symmetric transformation is:

$\begin{bmatrix} c t_1 \\ x_1 \\ y_1 \\ z_1 \end{bmatrix} = \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ \beta \gamma&-\gamma&0&0\\ 0&0&-1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c t_2 \\ x_2 \\ y_2 \\ z_2 \end{bmatrix}\ .$

where $\beta�:= \frac{v}{c}$ .

The inverse symmetric transformation is:

$\begin{bmatrix} c t_2 \\ x_2 \\ y_2 \\ z_2 \end{bmatrix} = \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ \beta \gamma&-\gamma&0&0\\ 0&0&-1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c t_1 \\ x_1 \\ y_1 \\ z_1 \end{bmatrix}\ .$

A single transformation matrix is used for both the forward and the inverse operation.

As expected:

$\begin{bmatrix} \gamma&-\beta \gamma&0&0\\ \beta \gamma&-\gamma&0&0\\ 0&0&-1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ \beta \gamma&-\gamma&0&0\\ 0&0&-1&0\\ 0&0&0&1\\ \end{bmatrix} = \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} .$