 Solution

For $|\mathbf{v}|<c$, define the Lorentz factor $\gamma =(1-|\mathbf{v}{|}^2/c^2{)}^{-1/2}$. A particle of rest mass $m$ hasThese expressions give the relativistic energy-momentum relation directly:so $\boxed{E^2=c^2\mathbf{p}\cdot \mathbf{p}+m^2{c}^4}$.

The four-momentum iswhere $U^{\mu }$ is the four-velocity, the proper-time derivative of $(ct,\mathbf{x})$. It is a four-vector, so its Minkowski metric norm is invariant under Lorentz transformations. Choose metric signature $(+,-,-,-)$; in the particle's rest frame, $P^{\mu }=(mc,0)$ and $P_{\mu }{P}^{\mu }=m^2{c}^2$. In any frame the same norm is $E^2/c^2-|\mathbf{p}{|}^2$. Equating the two yields the same relativistic energy-momentum relation. With the opposite metric convention both norm expressions change sign, leaving the energy relation unchanged.