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Let $A=\bar{\rho }+\bar{P}$, and assume $A\ne 0$ so that the density-defined curvature is meaningful. Under the cosmological gauge transformation in part ii,Therefore the two time-shift terms cancel:This is the uniform-density curvature perturbation with the paper's overall sign. On a uniform-density slice it is the spatial curvature potential $\Psi$. Its opposite overall sign is also a common convention, so the sign must be kept consistent rather than imported from another metric convention.

To establish time conservation, write $D_t={\bar{N}}^{-1}{\partial }_t$ and $c_a^2=\dot{\bar{P}}/\dot{\bar{\rho }}$. Background stress-energy conservation impliesFor constant $w$ in the given barotropic equation of state, $c_a^2=w$. Differentiate $\zeta =\Psi -\delta \rho /(3A)$:Insert both perturbed evolution equations from the hint, keeping the spatial Laplacian $\Delta ={\nabla }^2/a^2$:Thus the lapse perturbation $\Phi$ and the trace perturbation $\kappa$ cancel exactly. The remaining pressure term is the non-adiabatic pressure perturbation $\delta P_{\mathrm{nad}}=\delta P-c_a^2\delta \rho$. In the assumed barotropic fluid, the perturbed equation of state gives $\delta P=w\delta \rho$ and $c_a^2=w$, so it vanishes. For a Fourier mode,For regular long-wavelength perturbations, the velocity and shear potentials in this combination do not grow as $k^{-2}$, so the right-hand side is gradient-suppressed when $k\ll aH$. Therefore $\zeta$ is conserved to leading superhorizon order, which is the superhorizon conservation of uniform-density curvature requested.

The statement is not exact for arbitrary finite $k$: the displayed gradient term gives its correction. It also requires adiabatic cosmological perturbations; entropy perturbations in a multicomponent system or a nonbarotropic fluid can give $\delta P_{\mathrm{nad}}\ne 0$ and evolve $\zeta$ even at zero gradient order. At exact $w=-1$, $A=0$ and the question's variable is undefined, rather than a well-defined conserved density-slicing perturbation.

Use the spatial-potential convention of this paper and assume $\bar{\rho }+\bar{P}\ne 0$. A passive time displacement changes the spatial scalar by $\tilde{\Psi }-\Psi =\bar{N}H{\xi }^0$ and changes the energy density byThe two contributions to the uniform-density curvature perturbation therefore cancel:A spatial scalar displacement changes $E$ but not this combination. Hence $\zeta$ is gauge invariant at linear order. Pure cosmological-constant matter, for which $\bar{\rho }+\bar{P}=0$, does not supply a energy density clock and this formula is not defined for that component alone.

Now take the definite constant-$w$ barotropic equation of state and use physical-time differentiation $\mathcal{D}={\bar{N}}^{-1}{\partial }_t$. Put $A=\bar{\rho }+\bar{P}$. Background stress-energy conservation gives $\mathcal{D}A=-3H(1+w)A$. Differentiating $\zeta =-\Psi +\delta \rho /(3A)$, then substituting the two supplied perturbation equations, givesThus the lapse and expansion perturbations cancel without imposing a gauge. In terms of the non-adiabatic pressure perturbation $\delta P_{\mathrm{nad}}=\delta P-w\delta \rho$,For the stipulated barotropic equation of state, $\delta P=w\delta \rho$, so the first term vanishes. For a Fourier mode, $\Delta =-k^2/a^2$. For regular long-wavelength solutions the remaining gradients are suppressed when $k\ll aH$. Consequently $\dot{\zeta }\simeq 0$ on superhorizon scales, with exact conservation in the leading zero-gradient limit. It is not an exact finite-$k$ identity. Entropy perturbations or a singular gradient expansion can invalidate that conclusion; the supplied adiabatic assumption is what removes the unsuppressed pressure source.

This superhorizon conservation of uniform-density curvature is useful because inflationary fluctuations leave the horizon during cosmic inflation and may re-enter only much later. In an adiabatic single-clock evolution, the conserved uniform-density curvature perturbation transports their initial amplitude through reheating and the later radiation in cosmology and matter eras without solving every short-distance process. It then determines the initial conditions for large-scale structure and Cosmic microwave background anisotropy. Extra fluctuating fields or fluids can introduce isocurvature perturbations and convert them into curvature, so conservation must be checked rather than assumed during such transitions.

For $W=\bar{\rho }+\bar{P}\ne 0$, the conserved uniform-density curvature in the spatial-metric convention $1-2\Phi$ isThe source's plus sign in front of both terms is not gauge invariant and cannot yield the claimed conservation. Choosing the opposite overall sign, $\Phi -\delta \rho /(3W)$, is equally valid, but changing only one term is necessary. A passive time displacement $t\mapsto t+T$ gives $\tilde{\Phi }=\Phi +\bar{N}HT$ and $\widetilde{\delta \rho }=\delta \rho +3\bar{N}HWT$, so their combination above is unchanged. This is the uniform-density curvature sign test.

The lapse perturbation in the perturbed energy-conservation source is $\Psi$, not the spatial-curvature perturbation $\Phi$. Indeed the perturbation of the proper-time density derivative is $\mathcal{D}\delta \rho -\Psi \mathcal{D}\bar{\rho }$, while the normal expansion changes by $-\kappa$. The consistent equation isFor a constant barotropic equation of state, $\delta P=w\delta \rho$ and $\mathcal{D}W=-3H(1+w)W$. Differentiate ${\zeta }_{\mathrm{ud}}$ and cancel the density terms:More generally, $\delta P_{\mathrm{nad}}=\delta P-(\mathcal{D}\bar{P}/\mathcal{D}\bar{\rho })\delta \rho$ givesRegular scalar modes have gradient terms suppressed by $(k/aH{)}^2$. Thus adiabatic uniform-density curvature is conserved to leading order on superhorizon scales, ${\dot{\zeta }}_{\mathrm{ud}}\simeq 0$. This is a long-wavelength limit, not exact finite-wavenumber conservation; it assumes a regular gradient expansion and excludes $W=0$, where this variable is undefined. In contrast the literal printed variable obeys $\mathcal{D}{\zeta }_{\mathrm{printed}}=2\mathcal{D}\Phi +O(k^2)$ even after repairing the energy equation, and generally evolves across changes of the background equation of state.

There are three practical benefits. First, uniform-density curvature is gauge invariant, so matching it does not depend on coordinate slicing. Second, superhorizon conservation of uniform-density curvature transports primordial amplitudes through changes in the expansion history without solving each short-scale dynamical equation. Third, when a mode re-enters the horizon it supplies initial conditions for density growth and Cosmic microwave background anisotropy; non-adiabatic pressure is also an explicit diagnostic of when that simple transport fails.

Because density is a scalar field, $\tilde{\rho }(\tilde{x})=\rho (x)$. Write both sides as background plus perturbation and expand the background time shift:The coordinate shift acting on an already first-order perturbation contributes only at second order. Thus the density perturbation gauge transformation isThe uniform-density curvature perturbation in the paper's sign convention isThe derivative on ${\bar{\rho }}^{\prime }$ is essential and is present in the PDF; the TeX transcription drops it. The scalar-curvature combination transforms aswhile the density term changes by $-\mathcal{H}T$. Hence the two time-slicing changes cancel:On a uniform-density slice, $\delta \rho =0$ and this variable is the signed spatial-curvature perturbation. The construction assumes ${\bar{\rho }}^{\prime }\ne 0$; a pure constant-density cosmological constant does not define such a time slicing.

In Newtonian gauge in cosmology, $E=B=0$ and $C=-\Phi$. The background cosmological perfect-fluid continuity equation gives ${\bar{\rho }}^{\prime }=-3\mathcal{H}(\bar{\rho }+\bar{P})$, so, writing $D=\bar{\rho }+\bar{P}$,Set the cosmological adiabatic sound speed $c_a^2={\bar{P}}^{\prime }/{\bar{\rho }}^{\prime }$. Then $D^{\prime }=-3\mathcal{H}D(1+c_a^2)$. Differentiate the previous expression and insert the perturbed energy-conservation equation:Consequently the curvature evolution equation in this sign convention isThe last equality defines the total energy-frame velocity by $\mathbf{q}=D\mathbf{v}$. For adiabatic cosmological perturbations, the non-adiabatic pressure perturbation vanishes. On superhorizon scales, the stated suppression of the velocity divergence makes the remaining gradient term negligible. Thus $\zeta$ is conserved to leading order in $k/\mathcal{H}$. The positive sign of the pressure source follows from the paper's definition of $\zeta$ and must not be replaced by a formula using the opposite curvature convention.

Superhorizon conservation of uniform-density curvature lets one carry a primordial perturbation from its inflationary generation through otherwise complicated eras and predict the initial conditions for later Cosmic microwave background anisotropy and structure growth. It relies on adiabaticity: an isocurvature perturbation can source curvature evolution, so the same conclusion is not automatic in an arbitrary multifield model.

For spatial metric $a^2(1-2\Phi ){\delta }_{ij}$, the combination $\zeta =-\Phi +\delta \rho /[3(\bar{\rho }+\bar{P})]$ is invariant under a time displacement: $\Phi$ shifts by $\bar{N}HT$ and $\delta \rho$ by $3\bar{N}H(\bar{\rho }+\bar{P})T$. The opposite overall sign is also valid. The sum $\Phi +\delta \rho /[3(\bar{\rho }+\bar{P})]$ is not invariant. This is a useful convention check before applying superhorizon conservation of uniform-density curvature.