 Solution

$\partial_{t} f = \partial_{t} f_{1}$ is first order; the zeroth-order collisionless equation has already made $\partial_{t} f_{0} = 0$ .
$\partial_{x^{i}} f$ is first order, while $d x^{i} / d t$ has a zeroth-order free-streaming velocity and first-order corrections. Their product contributes at first order only through that background velocity.
$\partial_{q} f = f_{0}^{'} (q) + \partial_{q} f_{1}$ begins at zeroth order; $d q / d t$ begins at first order. Its leading product is $f_{0}^{'} \overset{q}{˙}_{1}$ .
$\partial_{n^{i}} f = \partial_{n^{i}} f_{1}$ begins at first order, and $d n^{i} / d t$ begins at first order because background trajectories keep their direction. Their product is second order and drops out.

Thus the linear Collisionless Boltzmann equation is

\partial_{t} f_{1} + v_{0}^{i} \partial_{x^{i}} f_{1} + \overset{q}{˙}_{1} f_{0}^{'} (q) = 0.

(1)

For photons

v_{0}^{i} = n^{i} / a

. For a particle of mass

m

v_{0}^{i} = q n^{i} / [a q^{2} + a^{2} m^{2}]

. The order counting does not assume that the first-order angular deflection itself vanishes; only its multiplication by an already perturbed angular distribution makes it higher order.

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