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.