{wiki}

Suppose that a rod has is length $L$ measured on a rest frame $S$ (or maybe even better: two identical rulers were manufactured, and one is taken on a spaceship, a bit like the <twin paradox>).

Question: what is the length $L'$ than an observer in frame $S'$ moving relative to $S$ as speed $v$ observe the rod to be?

The key idea is that there are two events to consider in each frame, which we call 1 and 2:
* the left end of the rod is an observation event at a given position at a given time: $x_1$ and $t_1$ for $S$ or $x'_1$ and $t'_1$ for $S'$
* the right end of the rod is an observation event at a given position at a given time : $x_2$ and $t_2$ for $S$ or $x'_2$ and $t'_2$ for $S'$
Note that what you visually observe on a photograph is a different measurement to the more precise/easy to calculate two event measurement. On a photograph, it seems you might not even see the contraction in some cases as mentioned at https://en.wikipedia.org/wiki/Terrell_rotation

Measuring a length means to measure the $x_2 - x_1$ difference for a single point in time in your frame ($t2 = t1$).

So what we want to obtain is $x'_2 - x'_1$ for any given time $t'2 = t'1$.

In summary, we have:
$$
\begin{aligned}
L  &= x_2  &- x_1 \\
L' &= x'_2 &- x'_1
t'_2 = t'_1
\end{aligned}
$$

By plugging those values into the <Lorentz transformation>, we can eliminate $t_2 and t_1$, and conclude that for any $t'_2 = t'_1$, the length contraction relation holds:
$$
L' = \frac{L}{\gamma}
$$

The key question that needs intuitive clarification then is: but how can this be symmetric? How can both observers see each other's rulers shrink?

And the key answer is: because to the second observer, the measurements made by the first observer are not simultaneous. Notably, the two measurement events are obviously <spacelike-separated events> by looking at the <light cone>, and therefore can be measured even in different orders by different observers.


Length contraction