Integrating the <Collisionless dark-matter Vlasov equation> over momentum makes the force term a vanishing momentum-space boundary term. Since $\rho=m a^{-3}\int f\,d^3p$, the zeroth moment is
$$
\boxed{\rho'+3\mathcal H\rho+\partial_i(\rho v^i)=0}.
$$
Multiplying by $p^i/(am)$ before integrating gives the first moment. Decompose the second velocity moment as
$$
\langle u^iu^j\rangle=v^iv^j+\sigma^{ij}
$$
using the <velocity-dispersion tensor of collisionless matter>. Combining the result with the continuity equation gives
$$
\boxed{v^{i\prime}+\mathcal Hv^i+v^j\partial_jv^i
=-\partial^i\phi-\frac1\rho\partial_j(\rho\sigma^{ij})}.
$$
The single-stream pressureless-fluid equations follow when $\sigma^{ij}=0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 312 / 3 / ii / Solution

{title2=$\sigma^{ij}$}

The velocity-dispersion tensor is the second central moment of a phase-space distribution,
$$
\sigma^{ij}=\langle u^iu^j\rangle-v^iv^j.
$$
It contributes the stress-divergence term $\rho^{-1}\partial_j(\rho\sigma^{ij})$ to the Euler equation and vanishes in the ideal single-stream limit.


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