Why can't I equate time components of Lorentz-transformed spacetime points to find the contracted distance between them?

Question

I define a reference frame $S$ and a reference frame $S^\prime$ moving with velocity $v$ in the positive $x$-direction relative to $S$. There is a meter stick at rest in $S$ with left and right ends $L^\mu$ and $R^\mu$ respectively, which can be represented as spacetime points:

$L^\mu=(t,0,0,0)$

$R^\mu=(t,1,0,0)$.

With $c=1$, the Lorentz transformation from $S$ to $S^\prime$ in matrix form is

$ \Lambda^{\mu^\prime}_{\;\;\mu} = \begin{pmatrix} \gamma & -\gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} . $

Using this transformation on a general spacetime point in $S$, $x^\mu=(t,x,y,z)$, produces the coordinate relationships:

$t^\prime=\gamma t -\gamma v x$

$x^\prime=\gamma x -\gamma v t$.

Using these relationships, $L$ and $R$ can be written as points in $S^\prime$:

$L^{\mu^\prime}=(\gamma t -\gamma v x_L, \gamma x_L -\gamma v t, 0, 0)$

$R^{\mu^\prime}=(\gamma t -\gamma v x_R, \gamma x_R -\gamma v t, 0, 0)$.

(I believe my logical flaw is here) If our goal is to measure the distance of the meter stick with respect to $S^\prime$, we need to find the difference in the $x^\prime$ components of $L^{\mu^\prime}$ and $R^{\mu^\prime}$ when their $t^\prime$ components are equal, so we equate their time components:

$\gamma t -\gamma v x_L = \gamma t -\gamma v x_R$.

With $x_L=0$ and $x_R=1$, we get

$0=\gamma v x_R$,

which is only true when $v=0$ (which it isn't) or $x_R=0$ (which it isn't).

My question is: What is the logical flaw in equating the time components of the left and right spacetime points of the meter stick in order to find their $x^\prime$ components at a synchronized time $t^\prime$ so that the meter stick's contracted length can be measured as $\Delta x^\prime=R^{x^\prime}-L^{x^\prime}$?

edit: I want to clarify why I thought to equate their time components in the first place. My understanding is that the length of an object in a reference frame $F$ is the difference in their spatial coordinates when measured at the same time within that reference frame.

Answer 1

Using the Lorentz transformations the position of the left end of the rod at some time $t_L$ in $S^\prime$ is:

$$ (t_L, 0) \rightarrow (\gamma t_L, -\gamma vt_L) \tag{1} $$

Likewise the position of the right end of the rod at some time $t_R$ in $S^\prime$ is (taking the rod length to be $\ell$:

$$ (t_R, \ell) \rightarrow \left( \gamma(t_R-v\ell), \gamma(\ell-vt_R) \right) \tag{2} $$

We want to compare the $x^\prime$ position of the ends at the same $t^\prime$ but this will be at different values of $t$ i.e. $t_L \ne t_R$. If we require that $t^\prime$ be the same for the left and right ends then we get:

$$ \gamma t_L = \gamma(t_R-v\ell) $$

giving us:

$$ t_R = t_L + v\ell $$

Now we take the equation for $x^\prime_R$ from equation (2) above:

$$ x^\prime_R = \gamma(\ell-vt_R) $$

and substitute $t_R = t_L + v\ell$ to get:

$$\begin{align} x^\prime_R &= \gamma(\ell-v(t_L + v\ell)) \\ &= \gamma\ell(1 - v^2) -\gamma vt_L \\ &= \frac{\ell}{\gamma} + x^\prime_L \end{align}$$

which is the correct result.

Looking at your approach I think you are assuming that $t^\prime_L = t^\prime_R$ and $t_L = t_R$ and this can't be correct because both times can't be equal in both frames.