User:L33th4x0r/lorentz

The Lorentz transform is given as:

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

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

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

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

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

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?:

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

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.