Need serious help with deriving the Lorentz Transformation from Time Dilation and Length Contraction

Question

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.

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 $x = x'/\gamma + vt$