= Solution
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. Thus
$$
0=\frac{df}{d\eta}=\bar f'(\epsilon)\frac{d\epsilon}{d\eta}-\epsilon\bar f'(\epsilon)\left(\frac{\partial\Theta}{\partial\eta}+\mathbf e\cdot\nabla\Theta\right).
$$
For a nontrivial spectrum this identifies the linear <Free-streaming photon Boltzmann equation>:
$$
\boxed{\frac{\partial\Theta}{\partial\eta}+\mathbf e\cdot\nabla\Theta-\frac{d\ln\epsilon}{d\eta}=0.}
$$
In <Newtonian gauge in cosmology>, substitute the specified scalar gravitational-redshift source to write
$$
\dot\Theta+e^j\partial_j\Theta=\dot\phi-e^j\partial_j\psi.
$$
Multiply its <angular average> by four. Since $\langle e^j\rangle_\Omega=0$, the <photon angular temperature moments> imply the <photon continuity equation>
$$
\boxed{\dot\delta+\frac43\nabla\cdot\mathbf v-4\dot\phi=0.}
$$
For the <photon Euler equation>, multiply by $3e^i$ and average. The source is $-3\langle e^ie^j\rangle_\Omega\partial_j\psi=-\partial^i\psi$, while the second angular moment is
$$
\langle\Theta e^ie^j\rangle_\Omega=\frac{\delta\,\delta^{ij}}{12}-\frac{\Pi^{ij}}{4\bar\rho}.
$$
It follows that
$$
\boxed{\dot v^i+\frac14\delta^{ij}\partial_j\delta-\frac3{4\bar\rho}\partial_j\Pi^{ij}+\delta^{ij}\partial_j\psi=0.}
$$
The 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.
Back to article page
