A very simple calculation in relativistic dynamics show an impressive aspect that differentiates the momentum change due to a force in the direction of movement of the body and the transverse ones. By using the definition of relativistic momentum $\mathbf{p}=\gamma m \mathbf{v}$, we can write $$ \frac{d\mathbf{p}}{dt} = \frac{d\gamma}{dt} m \mathbf{v} + \gamma m \frac{d\mathbf{v}}{dt}, $$ by defining the longitudinal and transverse acceleration, with respect to the vector $\mathbf{v}$ $$ \frac{d\mathbf{v}}{dt} = \mathbf{a}_L +\mathbf{a}_T, $$ it is simple to show that the first term in the first equation is only longitudinal $$ \frac{d\gamma}{dt} m \mathbf{v} = \frac{v^2}{c^2} \gamma^3 m \mathbf{a}_L , $$ and the second term is composed of longitudinal and transverse parts $$ \gamma m \frac{d\mathbf{v}}{dt}=\gamma m (\mathbf{a}_L+\mathbf{a}_T). $$ Summing the contributions, we can separate the Newton equation with respect to transverse and longitudinal forces, obtaining $$ \frac{d\mathbf{p}}{dt}=\gamma^3m \mathbf{a}_L+\gamma m \mathbf{a}_T. $$