 Solution

Define the momentum density $q_i=(\bar{\rho }+\bar{P})v_{\perp i}$, lowering its spatial label with ${\delta }_{ij}$, so the mixed stress-energy tensor obeys $T^i{}_0=q_i$ and $T^0{}_i=-q_i$. All products of $J_{ij}$ or its derivatives with $q_i$ are second order and must be dropped in linear cosmological perturbation theory.

For $\nu =0$, the three terms in stress-energy conservation areHere the trace $J_{ii}^{\prime }$ vanishes, ${\Gamma }_{00}^i={\Gamma }_{i0}^0=0$, and ${\partial }_i{q}_i=(\bar{\rho }+\bar{P}){\partial }_i{v}_{\perp }^i=0$. ThusFor $\nu =i$, spatial pressure gradients vanish because this is a purely vector cosmological perturbation. The first two terms areIn the remaining contraction, the two background-connection times momentum density terms cancel:The other terms are ${\Gamma }_{0i}^0\bar{\rho }=0$ and $-\bar{P}{\Gamma }_{ji}^j=0$. ConsequentlyExpanding this conservation law and using the background cosmological continuity equation gives the equivalent peculiar velocity equation, when $\bar{\rho }+\bar{P}\ne 0$,