# Lorentz Transformation Matrices

Express the vector $\left( \begin{matrix} \begin{matrix} x' \\ y' \end{matrix} \\ \begin{matrix} z' \\ t' \end{matrix} \end{matrix} \right) =\left( \begin{matrix} \begin{matrix} \gamma (x-vt) \\ y \end{matrix} \\ \begin{matrix} z \\ \gamma (t-vx/{ c }^{ 2 }) \end{matrix} \end{matrix} \right)$

as a $$4\times4$$ matrix for scalars $$\gamma, v, c$$. Show that the determinant of this matrix is 1. Find the inverse matrix. Verify that the inverse matrix is precisely the Inverse Lorentz Transformation (with no boost in the y and z axes).

Solution

The Lorentz Transformation: \begin{align*} x' &= \gamma (x-vt) \\ y' &= y \\ z' &= z \\ t' &= \gamma (t - vx / {c}^{2}) \\ \end{align*}

or

$\begin{pmatrix} x' \\ y' \\ z' \\ t' \\ \end{pmatrix} = \begin{pmatrix} \gamma & 0 & 0 & -\gamma v \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma {v}/{c}^{2} & 0 & 0 & \gamma \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ t \\ \end{pmatrix}$

Since the determinant is 1, we can just find the adjugate matrix. The adjugate matrix can be found by transposing the cofactor matrix. Hence,

$\begin{pmatrix} x \\ y \\ z \\ t \\ \end{pmatrix} = \begin{pmatrix} \gamma & 0 & 0 & \gamma v \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \gamma {v}/{c}^{2} & 0 & 0 & \gamma \\ \end{pmatrix} \begin{pmatrix} x' \\ y' \\ z' \\ t' \\ \end{pmatrix}$

We retrieve the Inverse Lorentz Transformation: \begin{align*} x &= \gamma (x'+vt') \\ y &= y' \\ z &= z' \\ t &= \gamma (t' + vx' / {c}^{2}) \\ \end{align*}

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
4 years, 4 months ago

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