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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*}
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