Lorentz transformation under symmetric configuration

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Assume there are two observers O1 and O2, each using their own Cartesian coordinate system to measure space and time intervals. O1 uses $(t 1, x 1, y 1, z 1)$ and O2 uses $(t 2, x 2, y 2, z 2)$ . Assume further that the coordinate systems are oriented so that the x1-axis and the x2 -axis overlap but in opposite directions. The y1-axis is parallel to the y2 -axis but in opposite directions. The z1-axis is parallel to the the z2 -axis and in the same direction. The relative velocity between the two observers is v along the x1 or x2 axis. v is defined as a positive number when O1 sees O2 sliding in the direction of x1. 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 O2 appears to O1 in the identical way that frame O1 appears to O2. However, in the standard configuration, if O2 sees O1 going forward then O1 sees O2 going backward.

The Lorentz transformation for frames in symmetric configuration is:

$t1 = \gamma (t2 - \frac{v x2}{c^{2}})\ ,$

$x1 = \gamma (v t2 - x2)\ ,$

$y1 = - y2 \ ,$

$z1 = z2 \ ,$

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

The inverse transformation is:

$t2 = \gamma (t1 - \frac{v x1}{c^{2}})\ ,$

$x2 = \gamma (v t1 - x1)\ ,$

$y2 = - y1 \ ,$

$z2 = z1 \ .$

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

In matrix form the forward symmetric transformation is:

$\begin{bmatrix} c t1 \\ x1 \\ y1 \\ z1 \end{bmatrix} = \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ \beta \gamma&-\gamma&0&0\\ 0&0&-1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c t2 \\ x2 \\ y2 \\ z2 \end{bmatrix}\ .$

where $\beta := \frac{v}{c}$ .

The inverse symmetric transformation is:

$\begin{bmatrix} c t2 \\ x2 \\ y2 \\ z2 \end{bmatrix} = \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ \beta \gamma&-\gamma&0&0\\ 0&0&-1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c t1 \\ x1 \\ y1 \\ z1 \end{bmatrix}\ .$

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

As expected:

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

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Categories: Equations | Minkowski spacetime | Special relativity

Lorentz transformation under symmetric configuration

From Wikipedia, the free encyclopedia

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