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Momentum density is momentum per unit physical volume. In a local orthonormal tetrad of the measuring observer it is $q^{\hat{i}}=T^{\hat{0}\hat{i}}$. A perfect fluid in general relativity moving at a small velocity relative to that observer has $q^{\hat{i}}=(\rho +P)v^{\hat{i}}$ to first order. For isotropic background photons, $P=\rho /3$, so the factor is $4\rho /3$ rather than $\rho$. Integrating the kinetic stress-energy tensor gives $q^{\hat{i}}=\int f{p}^{\hat{i}}{d}^{3}p$.

Use a local orthonormal tetrad and units in which the speed of light is one. A particle has four-momentum $p^{\hat{a}}=E(1,\mathbf{e})$ and moves with velocity $\mathbf{e}$. Thus its energy density contributes $E$, its momentum density contributes $E{e}^i$, and its momentum flux contributes $E{e}^i{e}^j$. Integrating the phase-space distribution function gives the kinetic stress-energy tensor. The measure $d^{3}p/E$ is the future mass-shell Lorentz-invariant phase-space measure, so the same expression transforms as a spacetime stress-energy tensor; fixed state-counting or polarization factors are understood to be included in $f$.

Define $⟨F{⟩}_{\Omega }=(4\pi {)}^{-1}\int Fd\Omega$. The photon momentum measure is $d^{3}p=E^{2}dEd\Omega$, so every component of the kinetic stress-energy tensor has a radial factor $E^{3}dE=a^{-4}{ϵ}^{3}dϵ$. An isotropic background has $⟨e^i{⟩}_{\Omega }=0$ and $⟨e^i{e}^j{⟩}_{\Omega }={\delta }^{ij}/3$. HenceThis radiation pressure equation of state requires isotropy and masslessness, not a thermal spectrum.

The distribution perturbation is $\delta f=-ϵ{\bar{f}}^{\prime }(ϵ)\Theta$. Assume finite energy density and endpoint behavior ${ϵ}^4\bar{f}(ϵ)\to 0$ at zero and infinity, as for the Planck photon distribution. Integration by parts then givesConsequently the perturbed energy density and momentum density areSince $q^i=(\bar{\rho }+\bar{P})v^i=(4\bar{\rho }/3)v^i$, the photon angular temperature moments give $\delta =4⟨\Theta {⟩}_{\Omega }$ and $v^i=3⟨\Theta e^i{⟩}_{\Omega }$.

The sign of anisotropic stress must be specified. Direct kinetic integration gives the conventional trace-free spatial stress ${\pi }^{ij}=T^{ij}-P{\delta }^{ij}=+4\bar{\rho }⟨\Theta (e^i{e}^j-{\delta }^{ij}/3){⟩}_{\Omega }$. The PDF uses the opposite sign, ${\Pi }^{ij}=-{\pi }^{ij}$, equivalently $T^{ij}=P{\delta }^{ij}-{\Pi }^{ij}$. With that convention, all requested moments areTaking the spatial trace also givesThe angular temperature description assumes $\Theta$ is independent of photon energy; independent spectral distortions would require additional energy-dependent moments.

Let $⟨F⟩=(4\pi {)}^{-1}\int Fd\Omega$ and $f=\bar{f}(ϵ)-ϵ{\bar{f}}^{\prime }(ϵ)\Theta (\mathbf{e})$ with comoving energy $ϵ=aE$. If ${ϵ}^4\bar{f}$ vanishes at both integration endpoints, integration by parts gives $-\int {ϵ}^4{\bar{f}}^{\prime }dϵ=4\int {ϵ}^3\bar{f}dϵ$. The kinetic stress-energy tensor then gives ${\delta }_{\gamma }=4⟨\Theta ⟩$, $v_{\gamma }^i=3⟨\Theta e^i⟩$, $\delta P=\bar{\rho }{\delta }_{\gamma }/3$, and ${\pi }^{ij}=4\bar{\rho }⟨\Theta (e^i{e}^j-{\delta }^{ij}/3)⟩$. If $\Pi =-\pi$ is used, the last expression changes sign. Frequency independence of $\Theta$ is essential to this common temperature description.

Relativistic kinetic theory describes particles by a scalar phase-space distribution function on the future mass shell. Its Boltzmann equation uses geodesic streaming and a collision operator. Integrating its momentum moments supplies a kinetic stress-energy tensor and particle-number currents. An orthonormal tetrad makes the local momentum integrals identical in form to those of special relativity.