 Solution

Use $\mathcal{H}=a^{\prime }/a$ and the Fourier convention $\nabla \to i\mathbf{k}$. A valid linearization requires all components of $h_{ij}$ to be small: the printed small-determinant condition alone is insufficient, as the matrix $\mathrm{diag}(1,1,0)$ has zero determinant without being small. This is small determinant does not imply a small metric perturbation.

Define the density contrast $\delta =(\rho -\bar{\rho })/\bar{\rho }$ and the scalar velocity potential by $v^i={\partial }^i\theta$, or $v_{\mathbf{k}}^i=i{k}^i{\theta }_{\mathbf{k}}$. Hence $\nabla \cdot \mathbf{v}=-k^2\theta$. To first order, $u^0=a^{-1}$, $u^i=a^{-1}{v}^i$ and $g^{ij}=a^{-2}({\delta }^{ij}-h^{ij})$. The product $u^i{u}^j$ is second order. Substitution into the perfect fluid stress-energy tensor givesHere the equation of state is used locally, so the pressure perturbation is $w\bar{\rho }\delta$.

The contraction in ${\nabla }_{\mu }{T}^{0\mu }=0$ isUsing the displayed connections gives, through first order,In particular ${\Gamma }_{\mu 0}^{\mu }=4\mathcal{H}+h^{\prime }/2$; the spatial connection contraction adds $3\mathcal{H}P+\bar{P}{h}^{\prime }/2$, while differentiating $a^{-2}\rho$ supplies $-2\mathcal{H}\rho$. Subtract the background relation ${\bar{\rho }}^{\prime }+3\mathcal{H}(1+w)\bar{\rho }=0$. The background terms multiplying $\delta$ cancel, leavingFor comoving cold dark matter, $w=0$ and $\theta =0$, so ${\delta }_c^{\prime }=-h^{\prime }/2$ and ${\delta }_c+h/2$ is constant at each spatial point. The proper-volume element of a comoving cell is $a^3\sqrt{\det (I+h)}{d}^{3}x=a^3(1+h/2)d^{3}x$. Conserved mass in the cell then gives the density change opposite to the perturbation of its volume. In comoving synchronous coordinates the linear CDM density change is a volume change, not transport across cell boundaries.