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Use the photon phase-space coordinates $(\mathbf{x},q,\mathbf{n})$, with $q=ap$ and $\mathbf{n}$ its propagation direction. Divide the Collisionless Boltzmann equation by $d\tau /d\lambda$ and apply the chain rule. The homogeneous blackbody distribution is time independent at fixed comoving momentum because its physical temperature scales as $a^{-1}$. ThereforeOn the unperturbed ray $d{x}^i/d\tau =n^i$; its correction is first order and multiplies the first-order spatial gradient of $f_1$. Similarly $q^{\prime }$ and $(n^i{)}^{\prime }$ are first order, so their products with derivatives of $f_1$ are second order. The zeroth-order distribution has no directional dependence. Keeping only linear terms and using $q^{\prime }=-q{h}_{ij}^{\prime }{n}^i{n}^j/2$ givesAfter Fourier transforming with $e^{i\mathbf{k}\cdot \mathbf{x}}$ and $\mu =\hat{\mathbf{k}}\cdot \mathbf{n}$, the Free-streaming photon Boltzmann equation in synchronous variables isDirection deflection is a real first-order geodesic effect, but its effect on the already perturbed distribution starts at second order here.

Let $I_0={\int }_0^{\infty }{q}^3{f}_0(q)dq$, so in the phase-space normalization being used $a^4{\bar{\rho }}_{\gamma }=4\pi I_0$. The photon brightness perturbation is $\Delta =I_0^{-1}\int q^3{f}_{1}dq$. This agrees with the stated density normalization, and $I_0$ is independent of conformal time. Integrating the distribution equation givesFor a Planck distribution, the endpoint term $[q^4{f}_0{]}_0^{\infty }$ vanishes. Integration by parts yields $\int q^4{f}_0^{\prime }dq=-4I_0$. Thus the synchronous photon brightness equation isFor a small directional blackbody temperature shift $\Theta =\Delta T/T$, expansion gives $f_1=-q{f}_0^{\prime }(q)\Theta$. The same integral then gives $\Delta =4\Theta$, also following from blackbody energy density proportional to $T^4$. This temperature interpretation presumes a perturbed blackbody shape; frequency-integrated brightness can still be defined for more general distributions.

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.

Write the total derivative of the phase-space distribution function asSince $a\bar{T}$ is constant for the background photon gas, expand the Bose-Einstein distribution asAt first order the four terms are respectivelyThe angular-deflection velocity is already first order and multiplies the first-order angular dependence of $f$, so its contribution is second order. Thus the collisionless left-hand side of the Free-streaming photon Boltzmann equation is

Insert the photon temperature multipole expansion into the Fourier-space transport equation. The Legendre polynomial recurrence relation turns multiplication by $\mu$ into nearest-neighbour multipole couplings, while Orthogonality of Legendre polynomials projects onto a fixed $\ell$. The monopole projection has no collision term because Thomson scattering conserves photon number:The dipole projection receives the electron-velocity source,For every $\ell \ge 2$, the collision term damps the anisotropic multipole and the photon Boltzmann hierarchy isThus the Free-streaming photon Boltzmann equation moves angular structure between neighbouring multipoles, whereas Thomson scattering suppresses all multipoles above the dipole in the tight-coupling approximation.

Let the photon phase-space distribution be a Planck distribution whose local temperature is $T(\eta )[1+\Theta (\eta ,\mathbf{x},\hat{\mathbf{p}})]$. In the absence of collisions, Liouville theorem says that the distribution is constant along a photon null geodesic. Linearizing $df/d\eta =0$ about the homogeneous distribution and using $d\mathbf{x}/d\eta =\hat{\mathbf{p}}$ givesA spatial Fourier transform sends $\hat{\mathbf{p}}\cdot \nabla$ to $i\hat{\mathbf{p}}\cdot \mathbf{k}$, soThis is the Free-streaming photon Boltzmann equation: the second term transports angular structure, while the last term is the gravitational redshift source.

The collisionless equation states that $f$ is constant along a photon geodesic:At linear order $d{x}^i/d\eta ={\hat{p}}^i$. The last term is second order because the background distribution is isotropic. Inserting the stated temperature parametrization into this Free-streaming photon Boltzmann equation and using the energy equation givesFor a Fourier mode, with $\mu =\hat{\mathbf{k}}\cdot \hat{\mathbf{p}}$, this is

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.