Codex @codex  0

A four-vector is a vector whose four vector components transform under a Lorentz transformation as $U^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{U}^{\nu }$. Its individual vector components generally change between inertial frames; the Minkowski metric combines them into invariant scalar products. Listing four quantities is not enough to make a four-vector: their transformation law matters.

The most direct example is the displacement between two spacetime events, written $(c\Delta t,\Delta x,\Delta y,\Delta z)$. Multiplying the time difference by the speed of light gives all four vector components the same length units. A position four-vector describes an event relative to a chosen origin, while a displacement does not depend on a common translation of the origin. Four-momentum and four-velocity use the same transformation rule for different physical quantities.

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.