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One and two-way transformations

The transformation of Galileo is a one-way transformation, i.e., it uses only the one-way speed of light, which for simplicity is assumed to be instantaneous. The transformation of Lorentz is a the two-way transformation, which uses the universal two-way speed of light. The following approach defines two different one-way transformations, which combine to equal the two-way Lorentz transformation. Note that β = v/c; 1/γ² = 1 − β²; and γ = 1/γ + β²γ.

Galilean transformation:  {x}' \mapsto x-vt;\; \; {t}' \mapsto t.

Dual Galilean transformation:  {x}' \mapsto x;\; \; {t}' \mapsto t-wx.

These could be combined with a selection factor κ of zero or one:

{x}' \mapsto x - \epsilon vt;\; \; {t}' \mapsto t-(1-\epsilon )wx.

Lorentz transformation (boost): {x}' \mapsto \gamma (x-vt);\; \; {t}' \mapsto \gamma (t-vx/c^{2}).

General Lorentz boost (see here): {x}' \mapsto \gamma (x-vt);\; \; {t}' \mapsto \gamma(t-k^{2}vx)

with \gamma =\left (1-\frac{v^{2}}{c^{2}} \right )^{-1}  and k = 1/c for the Lorentz boost.

General dual Lorentz boost:  {x}' \mapsto \gamma_{2} (x-kwt);\; \; {t}' \mapsto \gamma_{2} (t-wx)

with \gamma_{2} =\left(1-\frac{w^{2}}{k^{2}} \right)^{-1}and k = 1/c.

The following approach defines two different one-way transformations, which combine to equal the one-way Lorentz transformation. Note that β = v/c and 1 − β² = 1/γ². Also 1/γ + β²γ = γ.

The one-way Galilean transformation (G):

\begin{pmatrix} {x}'\\ {t}' \end{pmatrix} = \begin{pmatrix} 1 & -v \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ t \end{pmatrix} =\begin{pmatrix} x-vt\\ t \end{pmatrix}

\begin{pmatrix} {x}'\\ {ct}' \end{pmatrix} = \begin{pmatrix} 1 & -\beta \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ ct \end{pmatrix} = \begin{pmatrix} x-\beta ct \\ ct \end{pmatrix}

The inverse Galilean transformation (G-1):

\begin{pmatrix} {x}'\\ {t}' \end{pmatrix} = \begin{pmatrix} 1 & v \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ t \end{pmatrix} = \begin{pmatrix} x+vt \\ t \end{pmatrix}

\begin{pmatrix} {x}'\\ {ct}' \end{pmatrix} = \begin{pmatrix} 1 & \beta \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ ct \end{pmatrix} = \begin{pmatrix} x+\beta ct \\ ct \end{pmatrix}

The two-way Lorentz transformation (L):

\begin{pmatrix} {x}'\\ {t}' \end{pmatrix} = \begin{pmatrix} \gamma & -\gamma v \\ -\gamma v/c^{2} & \gamma\end{pmatrix} \begin{pmatrix} x \\ t \end{pmatrix} = \begin{pmatrix}\gamma (x-vt)\\ \gamma (t-vx/c^{2}) \end{pmatrix}

\begin{pmatrix} {x}'\\ {ct}' \end{pmatrix} = \begin{pmatrix} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{pmatrix} \begin{pmatrix} x \\ ct \end{pmatrix} = \gamma \begin{pmatrix}x-\beta ct\\ ct-\beta x \end{pmatrix}

Note the metric (M) is compatible with the Lorentz transformation (L−1ML = M):

\begin{pmatrix} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma\end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma\end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

The (left) para-Galilean transformation (F = LG−1) such that FG = L:

\begin{pmatrix} \gamma & 0 \\ -\gamma v/c^{2} & 1/\gamma \end{pmatrix} \begin{pmatrix} 1 & -v \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \gamma & -\gamma v \\ -\gamma v/c^{2} & \gamma \end{pmatrix}

\begin{pmatrix} \gamma & 0 \\ -\beta \gamma & 1/\gamma \end{pmatrix} \begin{pmatrix} 1 & -\beta \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{pmatrix}

This can be better expressed with the transpose (GT) such that GTJG = L:

\begin{pmatrix} 1 & 0 \\ -v/c^{2} & 1 \end{pmatrix} \begin{pmatrix} \gamma & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1/\gamma \end{pmatrix} \begin{pmatrix} 1 & -v \\ 0 & 1 \end{pmatrix} =\begin{pmatrix}\gamma & 0 \\ -\gamma v/c^{2} & 1 \end{pmatrix} \begin{pmatrix} 1 & -v \\ 0 & 1/\gamma \end{pmatrix}

\begin{pmatrix} 1 & 0 \\ -\beta & 1 \end{pmatrix} \begin{pmatrix} \gamma & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1/\gamma \end{pmatrix} \begin{pmatrix} 1 & -\beta \\ 0 & 1 \end{pmatrix} =\begin{pmatrix}\gamma & 0 \\ -\beta \gamma & 1 \end{pmatrix} \begin{pmatrix} 1 & -\beta \\ 0 & 1/\gamma \end{pmatrix} = \begin{pmatrix} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{pmatrix}

Here is the right para-Galilean transformation (F) such that GF = L:

\begin{pmatrix} 1 & -v \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1/\gamma & 0 \\ -\gamma v/c^{2} & \gamma \end{pmatrix} = \begin{pmatrix} \gamma & -\gamma v \\ -\gamma v/c^{2} & \gamma \end{pmatrix}

\begin{pmatrix} 1 & -\beta \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1/\gamma & 0 \\ -\beta \gamma & \gamma \end{pmatrix} = \begin{pmatrix} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{pmatrix}

This is expressed with the transpose (GT) such that GJ−1GT = L:

\begin{pmatrix} 1 & -v \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1/\gamma & 0 \\ 0 & \gamma \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -v/c^{2} & 1 \end{pmatrix} = \begin{pmatrix} \gamma & -\gamma v \\ -\gamma v/c^{2} & \gamma \end{pmatrix}

\begin{pmatrix} 1 & -\beta \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1/\gamma & 0 \\ 0 & \gamma \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -\beta & 1 \end{pmatrix} = \begin{pmatrix} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{pmatrix}

Either way, the one-way Galilean transformation is compatible with the two-way Lorentz transformation. The matrices J and L are congruent.

From GTJG = L we see that the Galilean transformation (G) represents one-way light as instantaneous, with a turn-around transformation (J). The round trip with turn-around comprises the Lorentz transformation (L).

Last updated April 23, 2020.

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