# How does Lorentz transforming forwards, then backwards, stay consistent?

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?

• That's not a Lorentz transformation. – my2cts May 9 at 8:52
• 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 – Matteo Campagnoli May 9 at 8:54
• @MatteoCampagnoli But t=0. – Deschele Schilder May 9 at 8:58
• @DescheleSchilder but $t'\neq 0$ – OON May 9 at 9:01
• I've edited the question, why can't you do this with t=0 for the whole thing? – C.J. Broughton May 9 at 9:05

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 $$t''=0$$

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind May 11 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:

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

so

$$x''={\gamma}^2x(1+\frac{v^2}{c^2})=\frac{1+\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}x,$$

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

$$x={\gamma}^2x(1+\frac{v^2}{c^2})=\frac{1+\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}x,$$

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 $$S$$:

$$x''=\gamma'(x-v't),$$

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).

• This answer is wrong, and the last paragraph essentially amounts to something like "the Lorentz group is not a group" or something like that. – fqq May 10 at 16:31
• @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). – Deschele Schilder May 10 at 16:46
• 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. – fqq May 10 at 16:51
• 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. – J. Murray May 10 at 18:58
• 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. – J. Murray May 10 at 19:06