Length Contraction And Simultaneity

Question

I wanted to derive length contraction from the Lorentz transformation, but I keep getting stuck on problems with simultaneity in my derivation. At some point, I came across Wikipedia's article on Length Contraction where, in the section labeled "Derivation", it states

In an inertial reference frame $S^\prime$, $x_1^{\prime}$ and $x_2^{\prime}$ shall denote the endpoints for an object of length $L_0^\prime$ at rest in this system. The coordinates in $S^\prime$ are connected to those in $S$ by the Lorentz transformations as follows:

$ x_1^{\prime} = \frac{x_1-vt_1}{\sqrt{1-\frac{v^2}{c^2}}} $ and $ x_2^{\prime} = \frac{x_2-vt_2}{\sqrt{1-\frac{v^2}{c^2}}} $

As this object is moving in $S$, its length $L$ has to be measured according to the above convention by determining the simultaneous positions of its endpoints, so we have to put $t_1=t_2$. Because $L=x_2-x_1$ and $L_0^\prime=x_2^\prime-x_1^\prime$, we obtain

$ L_0^{\prime} = \frac{L}{\sqrt{1-\frac{v^2}{c^2}}}. $

I understand everything except for the line which reads $L_0^\prime=x_2^\prime-x_1^\prime.$ After setting $t_1=t_2$, it is possible to write $L=x_2-x_1,$ since events $1$ and $2$ are simultaneous. However, in the primed inertial reference frame, which is moving with a nonzero velocity with respect to the laboratory frame, the events are not simultaneous. Therefore, how can it be that $ L_0^\prime=x_2^\prime-x_1^\prime, $ since $ t_1^\prime \neq t_2^\prime? $

Answer 1

You just confused the frames. In the derivation, $S'$ is the rest frame of the object of length $L'_0=x'_2-x'_1$. In that rest frame, the endpoints are simply at coordinates $x'_1$ and $x'_2$, regardless of $t'$. The length $L'_0$ in that rest frame is measured as the distance between two simultaneous events in that frame, i.e. events with $t'_1=t'_2$, and the result is, once again, $L'_0=x'_2-x'_1$. The world lines of both end points are vertical in that frame. Simple.

We want to know how this object looks like in another frame, $S$, which is moving relatively to $S'$. To measure the length $L_0$ in $S$, we have to see both endpoints simultaneously in $S$ i.e. impose $t_1=t_2$. That's exactly what they did and derived that in the frame $S$ where the object is moving, the length is measured to be $L_0 = L'_0 \cdot \sqrt{1-v^2/c^2}$. It is Lorentz-contracted.

Their derivation may have unnaturally exchanged various things, $S$ with $S'$, and they wrote the final relationship in the inverted way so that it looks like a "length dilatation", but the actual content of the derivation is right given their conventions.

Answer 2

But $ t_1^\prime = t_2^\prime? $ since in $ S^\prime $ the ends of the rod are at rest in that system and therefore can be measured simultaneously in that system.

Answer 3

You can't leave the two time endpoints the same. At time $t_{1}$, you take $L = x_{2}-x_{1}$. Now, take some time $t_{1}^{\prime}$. At this time, $x_{2}^{\prime}$ and $x_{1}^{\prime}$ are well-defined, and you have $L^{\prime} = x_{2}^{\prime} - x_{1}^{\prime}$. But, if $(t_{1},x_{1})$ is the same event as $(t_{1}^{\prime},x_{1}^{\prime})$, this will, as you right intuit, mean that $t_{2}^{\prime},x_{2}^{\prime}$ is a different event than $(t_{2},x_{2})$, because the time coordinates will label different times.

This issue is very closely related to the pole in the barn paradox.