[![enter image description here][1]][1]


The 3+1-Lorentz transformation is
\begin{align}                 
    \mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=}  \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot}  \mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,t
\tag{01a}\label{01a}\\
 c\,t^{\boldsymbol{\prime}} & \boldsymbol{=}   \gamma_{\mathrm u}\left(c\,t\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathbf{x}}{c}\right)
\tag{01b}\label{01b}\\
\gamma_{\mathrm u} & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\frac12}
\tag{01c}\label{01c} 
\end{align}
and in differential form
\begin{align}                 
    \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=}  \mathrm d\mathbf{x}\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \left(\mathbf{u}\boldsymbol{\cdot}  \mathrm d\mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,\mathrm dt
\tag{02a}\label{02a}\\
c\, \mathrm dt^{\boldsymbol{\prime}} & \boldsymbol{=}   \gamma_{\mathrm u}\left(c\,\mathrm dt\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c}\right)
\tag{02b}\label{02b}      
\end{align}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

**For time-like vectors**

Could you check in \eqref{02a} what would be the quantity $\mathrm d\mathbf{x}^{\boldsymbol{\prime}}$ if in this same equation you replace
\begin{equation}
\mathbf{u}\boldsymbol{\longrightarrow}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}
\tag{03}\label{03}      
\end{equation}
under the assumption
\begin{equation}
\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert <c
\tag{04}\label{04}      
\end{equation}

--------------------------------

**For space-like vectors**

Could you check in \eqref{02b} what would be the quantity $\mathrm d t^{\boldsymbol{\prime}}$ if in this same equation you replace 
\begin{equation}
\mathbf{u}\boldsymbol{\longrightarrow}\dfrac{c^2}{\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert^2}\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}
\tag{05}\label{05}      
\end{equation}
under the assumption
\begin{equation}
\left\Vert\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}\right\Vert >c
\tag{06}\label{06}      
\end{equation}




  [1]: https://i.sstatic.net/ylPza.png
  [2]: https://i.sstatic.net/ssoru.png