linearity of the Lorentz transformation

linearlorentz.pdf

linearlorentz.tex

\documentclass{article}

\usepackage{wcsetup}
\usepackage{parskip}

\begin{document}

Consider an inertial reference frame~$s$,
and another inertial reference frame~$s'$
moving at a velocity~$v$ relative to~$s$.
Define the transformation~$L : \R^2 \to \R^2$
by
\begin{equation*}
  L((x, ct)) = (\gamma (x - \beta c t), \gamma (c t - \beta x)),
\end{equation*}
where $\beta = v / c$ and~$\gamma = (1 - \beta^2)^{-1/2}$.
This is the \noun{Lorentz transformation}.

\bigskip

\begin{claim*}
  The Lorentz~transformation is a linear transformation.
\end{claim*}
\begin{proof}
  Let~$(x_1, c t_1)$ and~$(x_2, c t_2)$ be events
  in the reference frame~$s$.
  Then
  \begin{align*}
    &\mathrel{\phantom=} L((x_1, c t_1) + (x_2, c t_2)) \\
    &= L((x_1 + x_2, ct_1 + ct_2)) \\
    &= L((x_1 + x_2, c(t_1 + ct_2))) \\
    &=
    (\gamma ((x_1 + x_2) - \beta c (t_1 + t_2)),
    \gamma (c (t_1 + t_2) - \beta (x_1 + x_2))) \\
    &=
    ((\gamma (x_1 - \beta c t_1)) + (\gamma (x_2 - \beta c t_2)),
    (\gamma (c t_1 - \beta x_1)) + (\gamma (c t_2 - \beta x_2))) \\
    &=
    (\gamma (x_1 - \beta c t_1), \gamma (c t_1 - \beta x_1)) +
    (\gamma (x_2 - \beta c t_2), \gamma (c t_2 - \beta x_2)) \\
    &= L((x_1, c t_1)) + L((x_2, c t_2)),
  \end{align*}
  so $L$~is additive.

  Now let~$(x, c t)$ be an event in the reference frame~$s$,
  and let~$\lambda \in \R$ be a scalar.
  Then
  \begin{align*}
    &\mathrel{\phantom=} L(\lambda (x, c t)) \\
    &= L((\lambda x, \lambda c t)) \\
    &= L((\lambda x, c (\lambda t))) \\
    &=
    (\gamma (\lambda x - \beta c (\lambda t)),
    \gamma (c (\lambda t) - \beta (\lambda x))) \\
    &= (\lambda \gamma (x - \beta c t),
    \lambda \gamma (c t - \beta x)) \\
    &= \lambda (\gamma (x - \beta c t), \gamma (c t - \beta x)) \\
    &= \lambda \, L((x, c t)),
  \end{align*}
  so $L$~is $1$-homogeneous.

  Therefore, $L$~is a linear operator.
\end{proof}

\end{document}