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

Write $\mathcal{L}=-\frac{1}{2}{g}^{\alpha \beta }{\partial }_{\alpha }\varphi {\partial }_{\beta }\varphi -V$. Vary the covariant metric while holding the scalar fixed. The supplied determinant and inverse-metric variations giveBy the definition $\delta S=\frac{1}{2}\int \sqrt{-g}{T}^{\mu \nu }\delta g_{\mu \nu }$, lowering its indices yields the stress-energy tensor of a canonical scalar field:The sign in the determinant variation is positive for varying $g_{\mu \nu }$ and would be negative for varying the inverse metric; matching that convention is essential.

For a homogeneous scalar in conformal time, $(\partial \bar{\varphi }{)}^2=-{\bar{\varphi }}^{\prime 2}/a^2$. The time component of the stress-energy tensor is $T_{00}={\bar{\varphi }}^{\prime 2}/2+a^{2}V$, while $T_{ij}={\delta }_{ij}[{\bar{\varphi }}^{\prime 2}/2-a^{2}V]$. Raising one index and applying the prescribed definitions givesThese are the scalar kinetic energy plus/minus potential energy. In proper cosmological time $d{t}_{\mathrm{proper}}=adt$, the same kinetic term is ${\dot{\bar{\varphi }}}^2/2$. Potential domination gives negative pressure, while kinetic domination gives $P\simeq \rho$.

The mixed stress tensor has the convenient exact formSince the background scalar has no spatial gradient, retaining first-order terms gives ${\partial }^0\varphi =-{\bar{\varphi }}^{\prime }/a^2$ and ${\partial }_i\varphi ={\partial }_i\delta \varphi$ in the product. ThereforeIt is a gradient momentum density, hence a scalar velocity perturbation. With a rolling background, uniform-field slices also have vanishing scalar momentum density and provide the field's comoving slicing.

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.