Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 64 / 1

Write $h_{ij}={}^{(3)}{g}_{ij}$ for the induced metric and lower the shift vector with it, $N_i=h_{ij}{N}^j$. In the stated negative-shift convention the four-dimensional components and inverse components areMultiplying the given future unit normal by this metric tensor gives $n_{\mu }=(-N,0,0,0)$. In particular the tangential normal components vanish, but their covariant derivatives do not:The relevant Christoffel symbol isFor the second equality, the last parenthesis is $2h_{k\ell }{}^{(3)}{\Gamma }^{\ell }{}_{ij}$, so its contraction subtracts the two spatial connection terms in the symmetrized derivative of $N_i$. ThereforeThis is extrinsic curvature with a negative shift. Both the future-normal orientation and the definition $K=-\nabla n$ matter: replacing the threading by $d{x}^i+N^{i}dt$ reverses the shift terms. The spatial induced metric is generally a function of time as well as position; restricting it to time-independent data would simply make the displayed time derivative vanish.

Retain the homogeneous lapse function $\bar{N}(t)$, and define the physical Hubble parameter $H=\dot{a}/(\bar{N}a)$. At first order the scalar metric perturbations give $\delta g_{00}=-2{\bar{N}}^2\Phi$, $\delta g_{0i}=a^2{B}_{,i}$, and $\delta g_{ij}=-2a^2(\Psi {\delta }_{ij}+E_{,ij})$. Applying the passive coordinate displacement component by component gives the scalar gauge transformations with a background lapseFor example the $0i$ transformation is $\delta {\tilde{g}}_{0i}=a^2{\partial }_{i}B-a^2{\partial }_i\dot{\lambda }+{\bar{N}}^2{\partial }_i{\xi }^0$. The spatial isotropic part acquires $-2a\dot{a}{\xi }^0{\delta }_{ij}$, explaining the positive sign in the transformation of $\Psi$.

Preserving synchronous gauge requires both transformed $\Phi$ and $B$ to remain zero. The first condition is ${\partial }_t(\bar{N}{\xi }^0)=0$, and the second is $\dot{\lambda }={\bar{N}}^2{\xi }^0/a^2$. Integrating them givesThus fixing the lapse function and shift vector perturbations does not exhaust the scalar coordinate freedom. The functions $C,D$ describe residual synchronous-gauge freedom; an integration-origin change in the integral is absorbed into $D$.

To move from synchronous gauge to Newtonian gauge, choose $\lambda =-E_S$ to set $\tilde{E}=0$. Requiring $\tilde{B}=0$ then givesA scalar energy density transforms as $\delta {\rho }_N=\delta {\rho }_S-\dot{\bar{\rho }}{\xi }^0$. Dividing by the background energy density, which changes the contrast only at second order if perturbed in that denominator, yields the synchronous-to-Newtonian density transformation with a lapseThe last form uses background stress-energy conservation, $\dot{\bar{\rho }}=-3\bar{N}H(\bar{\rho }+\bar{P})$. With cosmic time $\bar{N}=1$ this is ${\delta }_N={\delta }_S-3(1+w)a^{2}H{\dot{E}}_S$; with conformal time $\bar{N}=a$ it is ${\delta }_N={\delta }_S-3(1+w)\mathcal{H}{E}_S^{\prime }$, where $\mathcal{H}=a^{\prime }/a$. Keeping these clock conventions distinct is essential to the factors of $a$.

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.