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+#+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*}