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The normalized angular average is over the unit sphere, with solid angle measure $d\Omega$. Rotational symmetry gives $⟨e^i{⟩}_{\Omega }=0$, $⟨e^i{e}^j{⟩}_{\Omega }={\delta }^{ij}/3$, and $⟨e^i{e}^j{e}^k{e}^{\ell }{⟩}_{\Omega }=({\delta }^{ij}{\delta }^{k\ell }+{\delta }^{ik}{\delta }^{j\ell }+{\delta }^{i\ell }{\delta }^{jk})/15$. The constants follow by contracting indices and using $e^i{e}_i=1$. These identities extract photon angular temperature moments. For normalized spherical harmonics, $\int Y_{\ell m}^{\ast }{Y}_{{\ell }^{\prime }{m}^{\prime }}d\Omega ={\delta }_{\ell {\ell }^{\prime }}{\delta }_{m{m}^{\prime }}$; their angular average instead has an extra factor $1/(4\pi )$.

A small direction-dependent temperature change $T=\bar{T}(1+\Theta )$ shifts a thermal spectrum according to $\bar{f}(ϵ/(1+\Theta ))=\bar{f}(ϵ)-\Theta ϵ{\bar{f}}^{\prime }(ϵ)$ at first order. Hence $\boxed{\Theta =\delta T/\bar{T}}$ is the angular photon temperature perturbation. For a general spectrum the same ansatz describes an energy-independent brightness dilation; a literal thermodynamic temperature interpretation additionally assumes a thermal shape without spectral distortions.

Use $E=ϵ/a$ in the tetrad stress-energy tensor. The background photon energy density isIntegration by parts gives $-\int dϵ{ϵ}^4{\bar{f}}^{\prime }=4\int dϵ{ϵ}^3\bar{f}$, assuming the boundary term vanishes. Thus each directional energy perturbation is four times its temperature perturbation. Writing $⟨\cdot {⟩}_{\Omega }=(4\pi {)}^{-1}\int d\Omega$, we haveIn the paper's convention for photon angular temperature moments, there is no $(2\ell +1)$ factor multiplying ${\Theta }_{\ell }$. Legendre polynomial orthogonality gives $⟨\Theta {⟩}_{\Omega }={\Theta }_0$ and $⟨e^i\Theta {⟩}_{\Omega }=-i{\hat{k}}^i{\Theta }_1/3$. Comparing with the stated density and velocity conventions yieldsFor completeness, $⟨e^i{e}^j{P}_2(\hat{\mathbf{k}}\cdot \mathbf{e}){⟩}_{\Omega }=({\hat{k}}^i{\hat{k}}^j-{\delta }^{ij}/3)/5$. The $(-i{)}^2$ quadrupole phase then gives $\boxed{{\Pi }_{\gamma }=-3{\Theta }_2/5}$. These moment relations are photon angular temperature moments and keep the anisotropic stress convention consistent with the subsequent hierarchy.

The collisionless Boltzmann equation is conservation of $f$ along the photon trajectory. To linear order, use the unperturbed trajectory $d\mathbf{x}/d\eta =\mathbf{e}$ when differentiating $\Theta$. Deflection of $\mathbf{e}$ is first order and multiplies the first-order angular dependence of $\delta f$, while the background has no angular dependence; it therefore contributes only at second order. Similarly, the energy derivative of $\delta f$ times the perturbed energy change is second order. ThusFor a nontrivial spectrum this identifies the linear Free-streaming photon Boltzmann equation:In Newtonian gauge in cosmology, substitute the specified scalar gravitational-redshift source to writeMultiply its angular average by four. Since $⟨e^j{⟩}_{\Omega }=0$, the photon angular temperature moments imply the photon continuity equationFor the photon Euler equation, multiply by $3e^i$ and average. The source is $-3⟨e^i{e}^j{⟩}_{\Omega }{\partial }_j\psi =-{\partial }^i\psi$, while the second angular moment isIt follows thatThe background energy density is spatially homogeneous, so no spatial derivative acts on $\bar{\rho }$. The minus sign of its anisotropic stress term follows from the PDF's $\Pi =-\pi$ convention. Switching to the conventional trace-free stress $\pi$ changes that term to a plus sign.

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.

In Newtonian gauge in cosmology, the Free-streaming photon Boltzmann equation is $\dot{\Theta }+\mathbf{e}\cdot \nabla \Theta =\dot{\varphi }-\mathbf{e}\cdot \nabla \psi$. Angular averaging and the photon angular temperature moments give the displayed continuity equation. Its dipole yields the photon Euler equation, with the sign of the anisotropic stress term determined by the stress convention.