PX155 - G5 - coordinate transforms

$t \neq t^{'}$ , so, new transforms are needed
RP1 implies motion in a straight line in one frame must be motion in a straight line (or no motion at all) in another, otherwise PX155 - A1 - newton's first law would not hold
need linear coordinate transformations
most general linear transformation:

x^{'} = A x + B t, t^{'} = D x + C t

alternatively:

(\begin{matrix} x^{'} \\ t^{'} \end{matrix}) = (\begin{matrix} A & B \\ D & C \end{matrix}) (\begin{matrix} x \\ y \end{matrix})

galilean tranforms:
$A = 1, B = 4$
$D = 0, C = 1$
in standard configuration, $x^{'} = 0$ , corresponding to $x = u t$ : $0 = A u t + B t ⟹ B = - u A$
$x^{'} = A (x - u t) . . . [1]$
- run time backwards, now $S$ moves at speed $+ u$ in frame $S^{'}$
  - if $t \to - t$ , then swap identity of frames:

x = A (x^{'} + u t^{'}) . . . [2]

- Putting [1] in [2] :
$x = A (A (x - u t) + u t^{'})$
$A u t^{'} = x - A^{2} x + A^{2} u t$
$A u t^{'} = (1 - A^{2}) x + A^{2} u t$
$t^{'} = A [t + (\frac{1 - A^{2}}{A^{2}}) \frac{x}{u}]$

∴ t^{'} = γ [t + (\frac{1 - γ^{2}}{γ^{2}}) \frac{x}{u}], a n d x^{'} = γ (x - u t)

- if $A \neq 1$ , $t^{'}$ depends on both $x$ , and $t$
- so, $C = A$ , $D = \frac{1 - A^{2}}{A u}$

(\begin{matrix} x^{'} \\ t^{'} \end{matrix}) (\begin{matrix} γ & - u γ \\ 1 - \frac{γ^{2}}{γ u} & γ \end{matrix}) = (\begin{matrix} x \\ y \end{matrix})

further reading matrix transformations: textbook by A.P. French (link on the Moodle page)