= Solution
The $\nu=0$ result is the background <cosmological perfect-fluid continuity equation>. For a comoving region of physical <volume> proportional to $a^3$, it is precisely
$$
d(\bar\rho a^3)=-\bar P\,d(a^3),
$$
expressing local <conservation of energy> including the <pressure-volume work> done during <cosmic expansion>. A <transverse vector field> has no velocity <divergence>, and the vector spatial <metric tensor> has no trace perturbation, so neither produces a first-order compression or density change.
The spatial result is <vector-fluid momentum dilution> and integrates immediately:
$$
(a^4q_i)'=0,\qquad
\boxed{q_i(\eta)=q_i(\eta_i)\left(\frac{a_i}{a}\right)^4.}
$$
Thus, for any specified background <perfect fluid>,
$$
\boxed{v_{\perp i}(\eta)=v_{\perp i}(\eta_i)\left(\frac{a_i}{a}\right)^4\frac{\bar\rho_i+\bar P_i}{\bar\rho+\bar P}.}
$$
A statement about $\mathbf q$ alone does not fix the velocity scaling independently of the <equation of state>. For constant <equation-of-state parameter> $w\ne-1$, the <cosmological continuity equation> gives $\bar\rho\propto a^{-3(1+w)}$ and hence
$$
\boxed{\mathbf v_\perp\propto a^{3w-1}.}
$$
For <pressureless matter>, $w=0$ and $\mathbf v_\perp\propto a^{-1}$, exactly as in the Newtonian calculation in part (a). For <radiation in cosmology>, $w=1/3$ and the vector <peculiar velocity> remains constant, although its <momentum density> still falls as $a^{-4}$. More generally, for a barotropic <perfect fluid> with <cosmological adiabatic sound speed> $c_a^2=\bar P'/\bar\rho'$, the equivalent equation is $\mathbf v_\perp'+\mathcal H(1-3c_a^2)\mathbf v_\perp=0$. The Newtonian damping law describes nonrelativistic inertia; it need not apply to a relativistic <perfect fluid>. A pure <cosmological constant> has $\bar\rho+\bar P=0$ and supplies no such fluid velocity degree of freedom.
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