The Lorentz transform for the 4 vector $( ct , x)$ is :

$L(\beta) = \gamma\begin {pmatrix} 1& \beta \\ \beta & 1 \end{pmatrix}$

the Lorentz transform can be consider as a rotation in hyperbolic space.

$L(\chi) = \begin{pmatrix} cosh(\chi) & sinh(\chi) \\ -sinh(\chi) & cosh(\chi) \end{pmatrix}$

such that

$tanh(\chi) = \beta$

$cosh(\chi)=\gamma$

thus, for the energy mass equation, it is just a rotation on the hyperbolic space from the rest frame of the particle.

$P=(E,p_x) = m ( cosh(\chi ), sinh( \chi ) )$

This bring a great simplification on changing frame. when it is very low speed,

$sinh(\chi ) \rightarrow 0$

$cosh(\chi ) \rightarrow 1$