# The derivation of the Lorentz transformation: addition of distances

In the derivation of the Lorentz transformation, one has a reference frame, $$S$$, at rest and another, $$S'$$, moving away at constant speed $$v$$. At time $$t$$ there is an event at a point $$x$$ in $$S$$. The same event has coordinate $$x'$$ in $$S'$$. At this time, the origin of $$S'$$ is at the point $$x = vt$$ in $$S$$ so by adding distances one gets $$vt + x' = x$$ which gives $$x' = x - vt$$. This is not in agreement with the Lorentz transformation but seems like a very simple addition of distances. Can somebody please explain what is wrong?

• The Lorentz transformation must preserve the line element $ds^2 = -c^2dt^2 + dx^2$. The transformation you suggest does not do that. – John Rennie Nov 6 '18 at 16:19
If it were, the light from the tail lights of a receding car would travel towards us at a speed lower than $$c$$, and the light from its headlights would travel away from us faster than $$c$$. Special relativity theory was specifically developed to explain the negative empirical finding that there are no detectable differences in the speed of light regardless of the relative motion of the light source and the observer.
As noted by another user above, your derivation assumes that we can transform coordinates in the "stationary" coordinate system to coordinates in the "moving" system by means of the Galilean transformation $$$$\left( \array{x^\prime \\ t^\prime} \right) \; = \; \left( \array{1 & -v \\ 0 & 1} \right) \left( \array{x \\ t} \right).$$$$ According to special relativity theory, the location of an event in the "moving" coordinate system is in fact related to the coordinates in the "stationary" system by the Lorentz transformation $$$$\left( \array{x^\prime \\ t^\prime} \right) \; = \; \gamma \left( \array{1 & -v \\ -v/c^2 & 1} \right) \left( \array{x \\ t} \right).$$$$ Here, $$\gamma = 1/\sqrt{1 - v^2/c^2}$$ is the Lorentz factor, which is also equal to $$(\det{T})^{-1/2}$$, with $$T$$ being the matrix above.