Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 312 / 1 / a

Write $h_{ij}$ for the spatial metric tensor and lower the shift vector with $h_{ij}$. For the negative-shift convention, the four-dimensional metric and its inverse have componentsThe normal has covariant components $n_{\mu }=(-N,0,0,0)$, so $n_{\mu }{n}^{\mu }=-1$ and it annihilates every tangent vector to the spatial slice. Since $n_i=0$, its covariant derivative gives $n_{i;j}=N{\Gamma }^0{}_{ij}$ and therefore $K_{ij}=-N{\Gamma }^0{}_{ij}$.

Compute the required Christoffel symbol using the symmetric connection formula. Its time-index term and spatial-index term combine to giveThe spatial expression in parentheses is ${\dot{h}}_{ij}+D_i{N}_j+D_j{N}_i$, where $D$ is the covariant derivative of $h$. Thus the extrinsic curvature with a negative shift isThis derivation fixes both signs from the actual 3+1 decomposition of spacetime, rather than transferring a formula using the opposite shift vector convention. The connection formula used here has $g_{\lambda \kappa ,\nu }$ as its second differentiated term; the printed repeated-index version is a typographical error.

Set $h=\det h_{ij}$. Contract the extrinsic curvature with a negative shift with $h^{ij}$. The Jacobi determinant derivative formula and metric compatibility giveConsequentlyThe local volume element is $\sqrt{h}{d}^{3}x$, so for zero shift vector the proper-time derivative along the normal is $d/ds=N^{-1}{\partial }_t$ andHere $\theta$ is the expansion scalar. Writing $a_{\mathrm{loc}}=h^{1/6}$ defines a local linear scale from this volume, whence the local volume Hubble parameter isThe factor of three converts volume expansion into linear expansion, and $N$ converts coordinate time into proper time. This agrees with the usual Hubble parameter in a homogeneous FLRW metric.

For a volume-shape decomposition the unimodular spatial metric is ${\gamma }_{ij}=a_{\mathrm{loc}}^{-2}{h}_{ij}$, with determinant one. If instead the printed positive power $a_{\mathrm{loc}}^2{h}_{ij}$ is taken literally, its determinant is $a_{\mathrm{loc}}^{12}$; it is a conformal rescaling but not the unimodular shape metric. The trace calculation does not need that rescaling.

Decompose the mixed extrinsic curvature of a spatial hypersurface into $K^i{}_j=\frac{1}{3}K{\delta }^i{}_j+{\tilde{K}}^i{}_j$. In the long-wavelength approximation in cosmology, the inflationary shear damping law makes ${\tilde{K}}^i{}_j$ fall as the inverse local volume. During cosmic inflation, many e-folds therefore suppress the anisotropic expansion, while $K$ still measures the local isotropic expansion through the local volume Hubble parameter.

Neglecting spatial gradients at leading order gives approximately independent expanding patches. Restrict also to scalar cosmological perturbations, and neglect the independent frozen tensor cosmological perturbation. The remaining spatial scale can be written as a background scale times a local fluctuation:This is the nonlinear scalar metric in a cosmological gradient expansion. The exponential keeps the metric positive and permits finite $\zeta$; no expansion in its amplitude is needed. Scalar lapse and shift perturbations describe the remaining freedom in time slicing and spatial threading. Choose $N=1+\Psi$ and $N_i=-a_{\mathrm{bg}}^2{\partial }_{i}B$. The negative shift vector convention then produces the positive mixed term $2a_{\mathrm{bg}}^2{\partial }_{i}Bdtd{x}^i$.

The exact completed-square form hasThe quadratic shift expression must be understood as this norm. If the raised derivative in the displayed ansatz is intended as a flat derivative, the corresponding factors of the spatial scale are needed. In either convention the shift-square term is of second gradient order, and the leading long-wavelength approximation in cosmology is unaffected.

Damping of shear motivates locally isotropic scalar expansion, not the disappearance of every tensor perturbation. A time-independent anisotropic shape can survive; the conformally flat scalar ansatz additionally neglects that tensor sector.