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For a massive particle of rest mass $m$, its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$, where the four-velocity is $U^{\mu }=\gamma (c,\mathbf{v})$, $E=\gamma m{c}^2$ and $\mathbf{p}=\gamma m\mathbf{v}$. With metric signature $(-,+,+,+)$, $P\cdot P=-m^2{c}^2$. A photon has four-momentum $(E/c,(E/c)\hat{\mathbf{n}})$, a future-pointing null vector. Four-momentum conservation states that the total incoming and outgoing four-momenta agree for an isolated interaction; this includes both energy and momentum conservation in every inertial frame.

Take the incoming photon direction as $x$. Its energy is $\hslash \omega =\mu m{c}^2$, so the total initial four-momentum isThe invariant mass available after photon absorption is therefore $M=m\sqrt{1+2\mu }$. In the centre-of-momentum frame, the two equal daughters have opposite momenta. Their total energy is at least their combined rest energy $2\alpha m{c}^2$, with equality precisely when both daughters have zero momentum in that frame. Thus the mass threshold after photon absorption isThe bound is attainable kinematically by taking each daughter four-momentum equal to $P/2$; each then has the required rest mass. It immediately gives ${\alpha }_{\max }\to 1/2$ as $\mu \to 0$.

At this threshold both daughters move with the centre-of-momentum frame. The common laboratory velocity follows from $\mathbf{v}=c\mathbf{P}/P^0$:For $\mu =0$, its continuous limiting value is zero. At smaller allowed daughter rest masses, some of the centre-of-momentum energy appears as their relative kinetic energy, so their four-momenta need not be $P/2$.

In the local orthonormal tetrad, the components of the photon number current areThey are respectively the photon number density and the photon number flux. Thus $J^{\mu }$ is the photon number four-current, not an energy current or an electric charge current. Photons carry no electric charge. For the isotropic background,Integrating the covariant Boltzmann equation over the invariant mass-shell measure, with vanishing momentum-boundary flux, giveswhere $C[f]$ is the collision operator in the covariant streaming equation. ThereforeCollisionless photon propagation satisfies this condition, as do Thomson scattering and ordinary Compton scattering when all participating photons are counted: each collision has one incoming and one outgoing photon. Photon emission, photon absorption, bremsstrahlung, double Compton scattering and electron-positron annihilation in cosmology or its inverse generally give a nonzero number source. A cancellation of net creation and destruction can nevertheless make the integrated source vanish even when individual number-changing reactions occur. Cosmological redshift itself changes photon energy, not photon number; in an expanding homogeneous universe a conserved photon number current gives ${\bar{n}}_{\gamma }\propto a^{-3}$.

In an orthonormal tetrad, $J^{\hat{0}}=\int f{d}^{3}p$ is the photon number density and $J^{\hat{i}}=\int f{e}^i{d}^{3}p$ is its number flux. Integrating the Boltzmann equation gives ${\nabla }_{\mu }{J}^{\mu }=\int C[f]d^{3}p/E$ when momentum-boundary terms vanish. Collisionless propagation and photon-number-conserving scattering give zero divergence. Photon emission, photon absorption, electron-positron annihilation in cosmology, bremsstrahlung and double Compton scattering generally do not. This is a number current, not an electric charge current: photons are neutral.