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

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.

• Welcome! Please see this guidance about screenshots. Is your quoted text from Wikipedia? A link to the article would be helpful.
– rob
Commented Sep 5, 2021 at 2:41
• 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. Commented Sep 5, 2021 at 2:59
• @AndrewDrysdale Do you understand what $F$ and $F'$ are? Commented Sep 5, 2021 at 3:48
• @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. Commented Sep 5, 2021 at 3:52
• 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. Commented Sep 5, 2021 at 4:32

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 $$v$$.
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 $$x = x'/\gamma + vt$$
• 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). Commented Sep 7, 2021 at 7:47