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?

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    $\begingroup$ The Lorentz transformation must preserve the line element $ds^2 = -c^2dt^2 + dx^2$. The transformation you suggest does not do that. $\endgroup$ Commented Nov 6, 2018 at 16:19
  • $\begingroup$ I'm not sure what is your question. Please help me understand this. $\endgroup$
    – Galilean
    Commented Nov 6, 2018 at 16:36
  • $\begingroup$ Can you give a link to the derivation that you have seen? $\endgroup$
    – md2perpe
    Commented Nov 6, 2018 at 16:43
  • $\begingroup$ The transformation you are talking about is not a Lorentz transformation. That is the transformation in the name of my SE-identity ;), i.e. the Galilean transform. $\endgroup$
    – Galilean
    Commented Nov 6, 2018 at 16:55
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    $\begingroup$ You need to slow down and take things step by step because small, tacit assumptions can quickly lead you astray when working out relativity problems. It is true that S' frame's origin of x'=0 does correspond to x=vt, which is the same result as for a Galilean transform and constituent with one's "intuition". However, that doesn't mean that x'=x-vt for non-zero values of x'. For example, it follows from this last equation that x'=x at t=0, but that's not true. Review your understanding of relativity and Lorentz transforms to see why that is. $\endgroup$
    – user93237
    Commented Nov 6, 2018 at 20:46

1 Answer 1


Indeed, the intuitive formula for the addition of distances is not valid within special relativity theory.

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 \begin{equation} \left( \array{x^\prime \\ t^\prime} \right) \; = \; \left( \array{1 & -v \\ 0 & 1} \right) \left( \array{x \\ t} \right). \end{equation} 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 \begin{equation} \left( \array{x^\prime \\ t^\prime} \right) \; = \; \gamma \left( \array{1 & -v \\ -v/c^2 & 1} \right) \left( \array{x \\ t} \right). \end{equation} 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.


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