SR: Lorentz Transformation

Date:

30 Sep 2019

Derivation

Step 1

\[\begin{split}\begin{align*} & S'\text{ frame moves at }\vec{v}=(v,0,0) \text{ viewed from } S \text{ frame, then}\\ & {\partial x'^{(a)}\over \partial x^{(b)}}=\Lambda^a_b= \begin{pmatrix} \gamma & -\gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} = \begin{pmatrix} \cosh \Theta & -\sinh\Theta & 0 & 0 \\ -\sinh\Theta & \cosh \Theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\\ & \text{Where }\gamma ={1\over \sqrt{1-v^2} }, \Theta=\tanh^{-1} v \\ & {\partial x^{(a)}\over \partial x'^{(b)}}=(\Lambda^{-1})^a_b= \begin{pmatrix} \gamma & \gamma v & 0 & 0 \\ \gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \end{align*}\end{split}\]