Compound Lorentz transformations

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 18.

The Lorentz transformations can be written in matrix form as

$\displaystyle \Lambda_{x}=\left[\begin{array}{cccc} \gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (1)$

where the 0 (${ct}$) component is the first row and first column, followed by the 1, 2, and 3 directions in order. This matrix is for relative motion along the 1 axis.

The Galilean transformations can be written as a matrix as well, where the first coordinate is just ${t}$ rather than ${ct}$:

$\displaystyle \Gamma=\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ -v & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (2)$

or if we want to use the same symbols as in the Lorentz case, where the top row of ${\Gamma}$ is a ${ct}$ coordinate, we can write

$\displaystyle \Gamma=\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ -\beta & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (3)$

The Lorentz transformation along the 2 (${y}$) axis is obtained by putting the transformation terms in row and column 2:

$\displaystyle \Lambda_{y}=\left[\begin{array}{cccc} \gamma & 0 & -\beta\gamma & 0\\ 0 & 1 & 0 & 0\\ -\beta\gamma & 0 & \gamma & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (4)$

If we apply a Lorentz transformation first in the ${x}$ and then in the ${y}$ direction (with different relative velocities), we get the compound matrix:

 $\displaystyle \Lambda_{y}\Lambda_{x}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} \gamma_{y} & 0 & -\beta_{y}\gamma_{y} & 0\\ 0 & 1 & 0 & 0\\ -\beta_{y}\gamma_{y} & 0 & \gamma_{y} & 0\\ 0 & 0 & 0 & 1 \end{array}\right]\left[\begin{array}{cccc} \gamma_{x} & -\beta_{x}\gamma_{x} & 0 & 0\\ -\beta_{x}\gamma_{x} & \gamma_{x} & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} \gamma_{y}\gamma_{x} & -\gamma_{y}\gamma_{x}\beta_{x} & -\beta_{y}\gamma_{y} & 0\\ -\beta_{x}\gamma_{x} & \gamma_{x} & 0 & 0\\ -\beta_{y}\gamma_{y}\gamma_{x} & \gamma_{y}\gamma_{x}\beta_{x}\beta_{y} & \gamma_{y} & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (6)$

Note that although ${\Lambda_{x}}$ and ${\Lambda_{y}}$ are both symmetric, their product is not. This means that applying the transformations in the opposite order gives a different result.

 $\displaystyle \Lambda_{x}\Lambda_{y}$ $\displaystyle =$ $\displaystyle \Lambda_{x}^{T}\Lambda_{y}^{T}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\Lambda_{y}\Lambda_{x}\right)^{T}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} \gamma_{y}\gamma_{x} & -\beta_{x}\gamma_{x} & -\beta_{y}\gamma_{y}\gamma_{x} & 0\\ -\gamma_{y}\gamma_{x}\beta_{x} & \gamma_{x} & \gamma_{y}\gamma_{x}\beta_{x}\beta_{y} & 0\\ -\beta_{y}\gamma_{y} & 0 & \gamma_{y} & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (9)$