 Solution

Use Minkowski spacetime coordinates $x^{\mu }=(ct,x,y,z)$ and the Minkowski metric of signature $(+,-,-,-)$. For a massive particle with three-velocity $\mathbf{v}=d\mathbf{x}/dt$ and $|\mathbf{v}|<c$, proper time satisfiesThus the four-velocity components areHere upper indices indicate components in the chosen coordinates; lowering with the Minkowski metric changes the signs of the spatial components. Its invariant product defined by the Minkowski metric is

The four-momentum is $p^{\mu }=(E/c,\mathbf{p})=m{U}^{\mu }$, so $E=\gamma m{c}^2$ and $\mathbf{p}=\gamma m\mathbf{v}$. The Taylor expansion of the Lorentz factor at small $|\mathbf{v}|/c$ isConsequently $\mathbf{p}=m\mathbf{v}+O(m|\mathbf{v}{|}^3/c^2)$, recovering ordinary momentum. At rest the four-momentum has energy $E_0=m{c}^2$, so subtracting this rest energy gives the kinetic energyThus the nonrelativistic limit preserves the rest-energy offset as well as the usual kinetic energy.