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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.

Preserving synchronous gauge means keeping both $\delta g_{00}$ and $\delta g_{0i}$ zero. Applying the metric transformation to the background givesThe first equation requires $(a{\xi }^0{)}^{\prime }=0$, so ${\xi }^0=C(\mathbf{x})/a$. For the scalar displacement ${\xi }^i={\partial }^i\lambda$, the second requires ${\lambda }^{\prime }={\xi }^0$, up to a spatially homogeneous coordinate displacement that does not affect scalar perturbations. Integrating givesThis is the residual gauge freedom in synchronous gauge. The lapse and shift conditions do not uniquely choose a slicing and spatial labeling.

A background density is a scalar, so under this remaining time displacementFor a fluid with constant equation of state, background conservation gives ${\bar{\rho }}^{\prime }/\bar{\rho }=-3(1+w)\mathcal{H}$, henceThe synchronous density gauge mode scales as ${\tau }^{-2}$ in radiation domination and as ${\tau }^{-3}$ in matter domination. It can mimic a changing density perturbation without changing physical observables. A density time dependence alone therefore does not identify a physical growing or decaying mode. Choosing coordinates comoving with cold matter fixes the residual time freedom, and a remaining spatial relabeling fixes the initial metric-trace integration convention. Gauge-invariant curvature perturbations provide another way to specify physical initial conditions.

For consistency, linear metric theory requires all background-orthonormal components of $h_{ij}$ to be small. The PDF's small-determinant condition by itself is insufficient: $h_{ij}=\mathrm{diag}(L,0,0)$ has zero determinant even for large $L$. This is small determinant does not imply a small metric perturbation.