• Some Mathematica code

Fix two observers, one with frame $S$ with coordinates $(x,y,t)$ and another with frame $S'$ with coordinates $(\xi ,\eta , \tau)$, such that origin of $S'$ moves with velocity $v$ along the $x$-axis of $S$, with the origins coinciding when $t=\tau=0$.

Postulate 1: The speed of light is constant for all inertial observers.

Lemma 2: For all events $A$. $\eta_A = y_A$.

Lemma 3: The transformations between $S$- and $S'$-coordinates are linear.

Derivation of Lorenz transformation Consider a vertical bar with endpoints at $(vt,0,t)$ and $(vt,L,t)$ at time $t$. At time $t=0$ a pulse of light is emitted from the bottom of the bar to the top of the bar. If this takes $T$ seconds to take place, the bar will have moved a transverse distance of $vT$ along the $x$-axis, and hence travel a total distance $R=\sqrt{L^2+v^2T^2}$. On the other hand, the light pulse travels at the speed of light $c$ to $R=cT$. Equating these we can solve for $T$ to find that

$$ T=\frac{L}{\sqrt{c^2-v^2}} $$

Letting $\gamma = (1-v^2/c^2)^{-1}$ we can write this as $T=\gamma L/c$ and we can conveniently express the coordinates of the event $E_1$ of the light pulse hitting the top of the bar as

$$ [E_1]_S = \left( \gamma L v/c, L, \gamma L / c \right) $$

On the other hand, from the perspective of Observer 2, the bar is at rest, and hence the light pulse appears to travel vertically from $(\xi ,\eta)=(0,0)$ up to $(\xi,\eta)=(0,L)$. By the constancy of the speed of light, Observer 2 views this as only taking $T'=L/c$, and to summarize,

$$ [E_1]_{S'}=(0,L,L/c) $$

Consider now a vertical bar, also length $L$ at rest in Observer 1’s reference frame, and a pulse of light fired from the bottom to the top. Let $E_2$ denote the event of this pulse of light reaching the top of the bar. Clearly

$$ [E_2]_S=(0,L,L/c) $$

On the other hand, from the perspective of Observer 2, the bar is moving with velocity $v$ along the negative $\xi$-axis. By analogous reasoning to the above we find

$$ [E_2]_{S'}=(-\gamma L v/c, L, \gamma L/c) $$

Ignoring the $y$- and $\eta$-coordinates and focusing on transforming between $(x,t)$ and $(\xi,\tau)$, we have by linearity,

$$ \begin{align*} \xi &= A(x-vt) + Bt\\ \tau &= E(x-vt) + Ft\end{align*} $$

for some coefficients $A,B,E,F$. Using first the reference event $E_1$ whose coordinates we found in both reference frames, we have

$$ \begin{align*} 0 &= A(0) + B(L/c) &&\Longrightarrow B=0 \\ L/c &= E(0) + F(\gamma L/c) &&\Longrightarrow F=\gamma^{-1}\\ \end{align*} $$

And then using event $E_2$ and the values of $B$ and $F$ we already found,