= Solution
Use the <position four-vector> $X^a=(ct,\mathbf x)$ and Minkowski signature $(+,-,-,-)$. Along a massive particle's timelike path,
$$
c^2d\tau^2=c^2dt^2-|d\mathbf x|^2,
\qquad d\tau=dt\sqrt{1-|\mathbf v|^2/c^2}=dt/\gamma.
$$
Thus its <four-velocity> is $U^a=dX^a/d\tau=\gamma(c,\mathbf v)$, and multiplying by the invariant <rest mass> gives
$$
\boxed{p^a=MU^a=(M\gamma c,M\gamma\mathbf v).}
$$
Its temporal component is $E/c$, and the invariant relation is $(E/c)^2-|\mathbf p|^2=M^2c^2$.
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, $p^ap_a=0$ implies $E/c=|\mathbf p|$. Hence
$$
\boxed{p^a=(|\mathbf p|,\mathbf p),\qquad E=c|\mathbf p|.}
$$
The original PDF retains the spatial vector in this expression; the TeX conversion incorrectly replaced it by its magnitude.
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