 Solution

Use the position four-vector $X^a=(ct,\mathbf{x})$ and Minkowski signature $(+,-,-,-)$. Along a massive particle's timelike path,Thus its four-velocity is $U^a=d{X}^a/d\tau =\gamma (c,\mathbf{v})$, and multiplying by the invariant rest mass givesIts temporal component is $E/c$, and the invariant relation is $(E/c{)}^2-|\mathbf{p}{|}^2=M^2{c}^2$.

For a massless particle, use the null energy–momentum relation rather than substituting $M=0$ into $M\gamma$: a massless particle has no proper-time parametrization of a timelike rest frame. With positive energy, $p^a{p}_a=0$ implies $E/c=|\mathbf{p}|$. HenceThe original PDF retains the spatial vector in this expression; the TeX conversion incorrectly replaced it by its magnitude.