= Solution
Apply the stated passive coordinate change to the background metric. Its spatial variation is
$$
\widetilde g_{ij}-g_{ij}=-2a^2\partial_i\partial_jL-2aa'T\delta_{ij}.
$$
Compare this with the spatial scalar decomposition $\delta g_{ij}=-2a^2\Psi\delta_{ij}+2a^2\partial_i\partial_j\chi$. It gives $\widetilde\Psi=\Psi+\mathcal HT$ and $\widetilde\chi=\chi-L$, where $\mathcal H=a'/a$. The scalar's background depends only on time, so
$$
\boxed{\widetilde\Psi=\Psi+\mathcal HT,\qquad
\widetilde{\delta\phi}=\delta\phi-\bar\phi'T.}
$$
The spatial displacement $L$ changes the shear potential but not these two scalar amplitudes in this decomposition. Direct cancellation of $T$ gives either of the useful invariant combinations
$$
\boxed{Q=\delta\phi+\frac{\bar\phi'}{\mathcal H}\Psi,\qquad
\mathcal R=\Psi+\frac{\mathcal H}{\bar\phi'}\delta\phi,\qquad
Q=\frac{\bar\phi'}{\mathcal H}\mathcal R.}
$$
$Q$ is the field fluctuation on <flat gauge in cosmology> slices: choosing $T=-\Psi/\mathcal H$ sets the transformed curvature potential to zero and leaves $\widetilde{\delta\phi}=Q$. The <comoving curvature perturbation> $\mathcal R$ is the spatial curvature potential on uniform-field slices: choose $T=\delta\phi/\bar\phi'$ and use (iii) to see that these slices also have zero scalar momentum density. For this spatial sign convention the intrinsic curvature perturbation is $\delta R^{(3)}=4a^{-2}\nabla^2\mathcal R$ in that gauge; some conventions define the curvature perturbation with the opposite sign. The canonically useful <Mukhanov-Sasaki variable> is $v=aQ$.
The interpretation of $\mathcal R$ requires $\bar\phi'\ne0$, and the flat-slice expression requires $\mathcal H\ne0$. The polynomial combination $\bar\phi'\Psi+\mathcal H\delta\phi$ remains invariant without division; at a stationary field background $\delta\phi$ itself is invariant under the time shift, but that field cannot serve as a comoving clock.
Back to article page
