Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 64 / 5 / iv / Solution

Apply the stated passive coordinate change to the background metric. Its spatial variation isCompare this with the spatial scalar decomposition $\delta g_{ij}=-2a^2\Psi {\delta }_{ij}+2a^2{\partial }_i{\partial }_j\chi$. It gives $\tilde{\Psi }=\Psi +\mathcal{H}T$ and $\tilde{\chi }=\chi -L$, where $\mathcal{H}=a^{\prime }/a$. The scalar's background depends only on time, soThe 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$Q$ is the field fluctuation on flat gauge in cosmology slices: choosing $T=-\Psi /\mathcal{H}$ sets the transformed curvature potential to zero and leaves $\widetilde{\delta \varphi }=Q$. The comoving curvature perturbation $\mathcal{R}$ is the spatial curvature potential on uniform-field slices: choose $T=\delta \varphi /{\bar{\varphi }}^{\prime }$ 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 ${\bar{\varphi }}^{\prime }\ne 0$, and the flat-slice expression requires $\mathcal{H}\ne 0$. The polynomial combination ${\bar{\varphi }}^{\prime }\Psi +\mathcal{H}\delta \varphi$ remains invariant without division; at a stationary field background $\delta \varphi$ itself is invariant under the time shift, but that field cannot serve as a comoving clock.