User:L33th4x0r/lorentz

The Lorentz transform is given as:

$$\begin{align} t' &= \gamma \left( t - \frac{vx}{c^2} \right)  \\ x' &= \gamma \left( x - v t \right)\\ y' &= y \\ z' &= z \end{align}$$

since y' and z' do not change, focusing on just the t' and x':

$$\begin{align} t' &= \gamma \left( t - \frac{vx}{c^2} \right)  \\ x' &= \gamma \left( x - v t \right) \end{align}$$

My question is, if we already know the time parameter t will transform as:

$$\begin{align} t' &= \gamma \left( t - \frac{vx}{c^2} \right) \end{align}$$

then why can we not utilize the already transformed t' parameter and take a shortcut directly in the target reference frame by simply multiplying the transformed time t' by -v (the velocity has a negative sign because, to the primed/target reference frame, the unprimed/original reference frame is moving in the opposite direction with same speed v, hence velocity is -v) to get the transformed x' in the target reference frame, as follows?:

$$\begin{align} x' &= -vt' \\ &= -v\gamma \left( t - \frac{vx}{c^2} \right) \\ &= \gamma \left( -vt + v \frac{vx}{c^2} \right) \\ &= \gamma \left( -vt + \frac{v^2}{c^2}x \right) \\ &= \gamma \left( \frac{v^2}{c^2}x - vt \right) \end{align}$$

However, doing so results in a wrong transformation for x'. But I am wondering what the reason for this is, is it because of a failure to take into consideration length contraction aspect in addition to the time dilation when dealing with the spatial parameter x'?

Many thanks for clarification.