Let me take you through the logic in my head...

  • In frame S, you have coordinate $x$
  • Transform to frame S' with velocity $v$ so the coordinate is now $x' = \gamma x$
  • Now treat the S' frame as if you started there.
  • Transform to a frame S'' moving at velocity $-v$. The coordinate is now $x'' = \gamma x'$
  • However, S'' and S are the same frame so $x'' = x$
  • So $x = \gamma x' = \gamma ^2 x$

I would like to clarify, I'm using t=0 for everything here

How does this make sense, what am I missing?

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    $\begingroup$ That's not a Lorentz transformation. $\endgroup$
    – my2cts
    Commented May 9, 2021 at 8:52
  • 3
    $\begingroup$ The transformation you are using is just wrong. The Lorentz boost of velocity v along the $x$-axis is given by $x'=\gamma (x-vt)$. Check en.wikipedia.org/wiki/Lorentz_transformation $\endgroup$ Commented May 9, 2021 at 8:54
  • $\begingroup$ @MatteoCampagnoli But t=0. $\endgroup$ Commented May 9, 2021 at 8:58
  • 8
    $\begingroup$ @DescheleSchilder but $t'\neq 0$ $\endgroup$
    – OON
    Commented May 9, 2021 at 9:01
  • $\begingroup$ I've edited the question, why can't you do this with t=0 for the whole thing? $\endgroup$ Commented May 9, 2021 at 9:05

2 Answers 2


The Lorentz transformation always transfoms not only coordinates but also time. In fact you can consider it as a sort of a "rotation" in $(t,x)$ "plane".

Whe you start with an event $(0,x)$ in the new frame it will have $t'=-\gamma \frac{v}{c^2} x$, $x'=\gamma x$. You see that even though all events $(0,x)$ are simultaneous in the initial frame, in the new frame they have different $t'$. This is known as a relativity of simultaneity and in my experience most "paradoxes" in special relativity originate from people forgetting about this fact.

Now if you apply the reverse Lorentz transformation you have $$x''=\gamma x'+\gamma v t'=\gamma^2(1-\frac{v^2}{c^2})x=x$$ Similarly you will get $t''=0$

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented May 11, 2021 at 15:42

The first answer gives you insight into simultaneity. Although $t'\neq 0$ everywhere in $S'$ (except for $x'=x=0$), this is not the root of the problem. Even if you take $t'\neq 0$ the paradox arises. The real problem is that you can't find, in general, the correct transformation between $S$ and $S''$ by reference to an intermediate frame $S'$ (so you can't say that $x''=\gamma (x'-vt')$). You'll get trouble indeed:

In $S'$ (for $t=0$):

$$x'=\gamma x$$ $$t'=-\gamma \frac{vx}{c^2}$$

Now assume for $x''$ ($\gamma$ is the same as in the two transformations above because $S''$ moves relative to $S'$ with speed $v$):

$$x''=\gamma (x'-vt')$$

Filling in the expressions for $x'$ and $t'$ gives:

$$x''=\gamma(\gamma x+\gamma\frac{v^2}{c^2}x),$$



which, if $x''=x$, would give:


which is obviously nonsense (when $v=0$ the equality holds but the problem is that you have set $v=v$). So the assumption that $x''=\gamma (x'-vt')$ (as you suggest) is just wrong. You can set $x''=\gamma x'+\gamma v t'$ though, which in combination with the inverse Lorentz transformation $x=\gamma x'+\gamma v t'$ gives you $x''=x$.

This problem can only be resolved by stating the correct (and direct) transformation (for $x$) between $S''$ and $S$:


where $v'$ is the relative velocity between $S$ and $S''$ (which is not the same as the sum of the relative velocities $S-S'$ and $S'-S''$, in accordance with the addition rule of velocities $v'=\frac{v+v}{1+\frac{v^2}{c^2}}$) and $\gamma'$ the with $v'$ associated Lorentz factor. This indeed gives $x=x''$ if $S''$ doesn't move relative to $S$ and, if you set $x''=x$, you get no nonsensical result but instead, you get the sensical result that $v'=0$). You can't find this transformation in the way described above.

So, to answer your question in short, you can't set $x''=\gamma x'$ (or $x''=\gamma (x'-vt')$) in the first place. When you do so you arrive at the wrong result $x''={\gamma}^2 x$ or $x''={\gamma}^2x(1+\frac{v^2}{c^2})=\frac{1+\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}x$. So it's your fourth "bullet" that deserves attention. Only in the special case that the velocity of $S''$ is $-v$ you can write $x''=\gamma (x'-vt')=\gamma (x'+vt')$ (giving $x''=x$). If the velocity of $S''$ were $v$ you couldn't write $x''=\gamma (x'-vt')$ though. So much ado about a minus sign...(with thanks to @J.Murray).

  • 1
    $\begingroup$ This answer is wrong, and the last paragraph essentially amounts to something like "the Lorentz group is not a group" or something like that. $\endgroup$
    – fqq
    Commented May 10, 2021 at 16:31
  • $\begingroup$ @fqq I don't say that two Lorentz transformations performed after one another don't give a new Lorentz transformation. I just say that when you go from S to S′ and subsequently from S′ to S′′ you can't do that in the way as assumed (by assuming that x′′=γ(x′−vt′), which leads to a nonsensical result, as I've shown). $\endgroup$ Commented May 10, 2021 at 16:46
  • $\begingroup$ You are saying that Lorentz transformations don't have inverses/velocities cannot be added, ascribing your wrong formulas to others in the process. I will not reply further. $\endgroup$
    – fqq
    Commented May 10, 2021 at 16:51
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    $\begingroup$ OP is saying that $S''$ is moving with velocity $\color{red}{-}v$ with respect to $S'$, so the line following "Now assume for $x''$ ..." should read $x'' = \gamma(x'\color{red}{+} vt')$, which resolves the problem. $\endgroup$
    – J. Murray
    Commented May 10, 2021 at 18:58
  • 1
    $\begingroup$ It's true that two boosts with velocities $u$ and $v$ are generally not equivalent to a boost with velocity $u+v$, but in the special case that $v=-u$, the two boosts compose to give the identity transformation. You can see this e.g. from the velocity addition rule. $\endgroup$
    – J. Murray
    Commented May 10, 2021 at 19:06

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