Seriously struggling right now, and I could really use some help.

I am trying to follow this derivation of Lorentz Transformations from time dilation and length contraction below:

Time dilation and length contraction

The transformation equations can be derived from time dilation and length contraction, which in turn can be derived from first principles. With $O$ and $O'$ representing the spatial origins of the frames $F$ and $F'$, and some event $M$, the relation between the position vectors (which here reduce to oriented segments $OM$, $OO'$ and $O'M$) in both frames is given by: $$OM = OO' + O'M$$

Using coordinates $(x,t)$ in $F$ and $(x',t')$ in $F'$ for event $M$, in frame $F$ the segments are $OM = x$, $OO' = vt$ and $O'M = x'/\gamma$ (since $x'$ is $O'M$ as measured in $F'$): $$x=vt+x'/\gamma$$

Likewise, in frame $F'$, the segments are $OM = x/\gamma$ (since $x$ is $OM$ as measured in $F$), $OO' = vt'$ and $O'M = x'$: $$x/\gamma =vt'+x'$$

By rearranging the first equation, we get $$x'=\gamma (x-vt)$$ which is the space part of the Lorentz transformation. The second relation gives $$x=\gamma (x'+vt')$$ which is the inverse of the space part. Eliminating $x'$ between the two space part equations gives $$t'=\gamma (t-vx/c^{2})$$ which is the time part of the transformation, the inverse of which is found by a similar elimination of $x$: $$t=\gamma (t'+vx'/c^{2})$$

I cannot understand why $O'M = x'/\gamma$ or their reasoning of "since $x'$ is $O'M$ as measured in $F'$." I would seriously appreciate any help with this, as it has been driving me insane for more hours than I'd like to admit.

  • 1
    $\begingroup$ Welcome! Please see this guidance about screenshots. Is your quoted text from Wikipedia? A link to the article would be helpful. $\endgroup$
    – rob
    Sep 5, 2021 at 2:41
  • $\begingroup$ I am sorry for violating that rule, but I am running out of time to work on this. I would give anything right now for an answer. $\endgroup$ Sep 5, 2021 at 2:59
  • $\begingroup$ @AndrewDrysdale Do you understand what $F$ and $F'$ are? $\endgroup$ Sep 5, 2021 at 3:48
  • $\begingroup$ @VincentThacker F is the inertial reference frame centered on O, F' is the inertial reference frame center on O'. I am extremely sorry if I am missing something simple, I just started taking Special Relativity several days ago. I am also sorry for making you fix my post. $\endgroup$ Sep 5, 2021 at 3:52
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    $\begingroup$ I don't think that this derivation has any sense. Important is the inverse : to derive time dilation and length contraction from the Lorentz transformation. $\endgroup$
    – Frobenius
    Sep 5, 2021 at 4:32

1 Answer 1


It helps (in this case: it doesn't always) to imagine some observer O sitting at O and making measurements, and some other observer O' sitting at O', with O and O' moving apart with speed $v$.

There is an event M. Say a firecracker goes off, at some sharply defined position and sharply defined time.

In the F' frame, the distance O'M is $x'$. That's the definition of $x'$. It's the co-ordinate along the axis of M as measured by O'. O' measures this by putting down a ruler, which is at rest in his/her frame, and noting the distance. (Any other measurement method is equivalent to doing this.)

In the F frame, these rulers are contracted by the $\gamma$ factor. So O says the ruler of O' is miscalibrated and shorter than its markings declare, and the distance from O' to M (at time $t$) is $x'/\gamma$. The distance from O to M is $x$ - again that's by definition. The distance from O to O' is $vt$. Distances add, so $x = x'/\gamma + vt$

  • $\begingroup$ Wait, I thought that the observer at O in frame F would measure the proper length relative to the F' frame since O did not move relative to the start- and end-points of O'M? $\endgroup$ Sep 6, 2021 at 4:32
  • $\begingroup$ I think I get it now. O'M is the length that is being measured, and F' is the reference frame whose origin remains at rest relative to the length, so it will have the proper length in this situation. O, on the other hand, has been seeing the length move during its motion, and is therefore not at rest relative to the length O'M. But what about time? $\endgroup$ Sep 6, 2021 at 4:44
  • $\begingroup$ Don't think about lengths. They are secondary constructs. Events are the primary construct. They have co-ordinates $x,y,z,t$. A length is built from two measurement events (matching an object with the scale on a ruler). $\endgroup$ Sep 7, 2021 at 7:47

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