#+TITLE:Lorentz Transformation #+SETUPFILE: ../math_options.org #+LATEX_HEADER: \usepackage{bm} #+LATEX_HEADER: \usepackage{mathtools} #+LATEX_HEADER: \newcommand{\dcoff}[1]{\frac{\text{d}}{\text{d}x} #1} #+LATEX_HEADER: \newcommand{\sdcoff}[1]{\frac{\text{d}#1}{\text{d}x}} #+LATEX_HEADER: \newcommand{\deriv}[2]{\frac{\text{d}}{\text{d}x} #1 &= #2} As a direct consequence of the second postulate, it follows, that for 2 events in spacetime describing the propagation of a beam of light it must hold \begin{equation*} c^2(t_2 - t_1)^2 - (x_2 - x_1)^2 - (y_2 - y_1)^2 - (z_2 - z_1)^2 = 0 \end{equation*} To abbreviate we introduce the following expressions \begin{align*} t &= t_2 - t_1\\ x &= x_2 - x_1\\ y &= y_2 - y_1\\ z &= z_2 - z_1 \end{align*} Furthermore, because this must also hold in any other reference frame, for example $\Sigma'$, we have: \begin{equation*} c^2t^2 - x^2 - y^2 - z^2 = c^2{t'}^2 - {x'}^2 - {y'}^2 - {z'}^2 = 0 \begin{end*} We introducing the Minkowski metric $\eta$ and rewrite this using matrices \begin{equation*} \sum_{\mu\nu} \eta_{\mu\nu} {x'}_\mu {x'}_\nu = \sum_{\alpha\beta} \eta_{\alpha\beta} x_\alpha x_\beta \end{equation*} Let us now consider some inertial system $\Sigma'$ that is moving away in respect to $\Sigma$ with some constant speed $v$ in the x direction. We are interested in the transformation that will allow us to convert the coordinates between this 2 systems. Further development of our last equation yields: \begin{align*} \sum_{\alpha\beta} \eta_{\alpha\beta} x_\alpha x_\beta &= \sum_{\mu\nu} \eta_{\mu\nu} \left(\sum_\alpha \Lambda_{\mu\alpha} x_\alpha \right) \left(\sum_\beta \Lambda_{\nu\beta} x_\beta \right)\\ &= \sum_{\mu\nu\alpha\beta} \eta_{\mu\nu} \Lambda_{\mu\alpha} \Lambda_{\nu\beta} x_\alpha x_\beta \end{align*} From this we notice \begin{align*} \eta_{\alpha\beta} &= \sum_{\mu\nu} \eta_{\mu\nu} \Lambda_{\mu\alpha} \Lambda_{\nu\beta}\\ &=\sum_{\mu\nu} (\Lambda^\text{T})_{\alpha\mu} \eta_{\mu\nu} \Lambda_{\nu\beta}\\ &=\sum_{\nu} \left(\sum_\mu (\Lambda^\text{T})_{\alpha\mu} \eta_{\mu\nu}\right) \Lambda_{\nu\beta} \end{align*} And thus \begin{equation*} \eta = \Lambda^\text{T}\eta\Lambda \end{equation*}