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: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

- 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 $S_{′}$

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:

$LL_{′} =x_{2}=x_{2} −x_{1}−x_{1}t_{2}=t_{1} $

By plugging those values into the Lorentz transformation, we can eliminate $t_{2}andt_{1}$, and conclude that for any $t_{2}=t_{1}$, the length contraction relation holds:

$L_{′}=γL $

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.

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- Charged particle moving at the same speed of electrons thought experiment | 192
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- Transversal time dilation | 221