OK, so let's verify the main desired consequence of the Lorentz transformation: that everyone observes the same speed of light.

Observers will measure the speed of light by calculating how long it takes the light going towards $+x$ cross a rod of length $L=x_{2}−x_{1}$ laid in the x axis at position $X1$.

TODO image.

Each observer will observe two events:

- $(t_{1},x_{1},y_{1},z_{1})$: the light touches the left side of the rod
- $(t_{2},x_{2},y_{2},z_{2})$: the light touches the right side of the rod

Supposing that the standing observer measures the speed of light as $c$ and that light hits the left side of the rod at time $T1$, then he observes the coordinates:

$t_{1}x_{1}t_{2}x_{2} =T1=X1=cL =X1+L $

Now, if we transform for the moving observer:
and so the moving observer measures the speed of light as:

$t_{1}x_{1}t_{2}x_{2} =γ(t_{1}−c_{2}vx_{1} )=γ(x_{1}−vt_{1})=γ(t_{2}−c_{2}vx_{2} )=γ(x_{2}−vt_{2}) $

$c_{′} =t_{2}−t_{1}x_{2}−x_{1} =(t_{2}−c_{2}vx_{2} )−(t_{1}−c_{2}vx_{1} )(x_{2}−vt_{2})−(x_{1}−vt_{1}) =(t_{2}−t_{1})−c_{2}v (x_{2}−x_{1})(x_{2}−x_{1})−v(t_{2}−t_{1}) =1−c_{2}v t_{2}−t_{1}x_{2}−x_{1} t_{2}−t_{1}x_{2}−x_{1} −v =1−c_{2}v cc−v =cc−v c−v =c $