Relativistic force ($F=dp/dt$)

For three-momentum $p=\gamma mv$, force and acceleration satisfy

The associated four-force uses proper time instead of coordinate time and is a different, four-dimensional object.

The three-force measured in a particle's instantaneous rest frame. For a pure force, ${\mathbf{f}}_0=m_0{\mathbf{a}}_0$, where ${\mathbf{a}}_0$ is the rest-frame acceleration. In that frame the four-force has zero temporal component and spatial part ${\mathbf{f}}_0$; the four-force itself is a different object.

Under a constant force $F$ with $p(0)=0$, momentum is $p=Ft$ andThe speed approaches $c$ without reaching it at finite time.