Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 55 / 3

Let ${\gamma }_{ij}$ denote the spatial induced metric and $\gamma =\det {\gamma }_{ij}$. Its conformally rescaled version ${\tilde{\gamma }}_{ij}={\gamma }^{-1/3}{\gamma }_{ij}$ has determinant one. The Jacobi determinant derivative formula consequently givesThus the determinant measures the volume-changing part of the spatial metric, while the unit-determinant part contains shape changes. Metric compatibility for the spatial covariant derivative also gives ${\gamma }^{ij}{D}_i{N}_j=D_i{N}^i$.

There is a sign inconsistency in the PDF. Contracting its defining expression for $K_{ij}$, with its threading convention $d{x}^i-N^{i}dt$ and its future normal $n^{\mu }=N^{-1}(1,N^i)$, givesnot the displayed trace with a minus sign before the divergence. This is the trace of extrinsic curvature with a negative shift. For a concrete check, take ${\gamma }_{ij}={\delta }_{ij}$, $N=1$ and $N^i=b{x}^i$. The defining formula gives $K_{ij}=-b{\delta }_{ij}$ and $K=-3b$, whereas the printed trace would give $+3b$. Reversing the shift convention can produce the other trace formula, but also changes the shift terms in $K_{ij}$ and the normal; the two conventions cannot be combined.

For the requested expansion interpretation, set the shift vector to zero. The sign discrepancy then disappears. A small coordinate volume has proper volume proportional to $\sqrt{\gamma }$, and proper time along its normal is $ds=Ndt$. Its expansion rate isThereforeThis is the local volume Hubble parameter, the mean of the three local directional expansion rates. In a homogeneous isotropic background $\gamma =a^6$, it reduces to $\dot{a}/(Na)$. The interpretation does not require the local expansion to be isotropic: the trace-free part of the extrinsic curvature can still describe shear.

Use $T={\xi }^0$ and ${\xi }^i={\partial }^i\lambda$, and define the physical background Hubble parameter by $H=\dot{a}/(\bar{N}a)$. At first order, the spacetime metric components are $\delta g_{00}=-2{\bar{N}}^2\Phi$, $\delta g_{0i}=a^2{B}_{,i}$ and $\delta g_{ij}=-2a^2(\Psi {\delta }_{ij}+E_{,ij})$. Apply the given passive cosmological gauge transformation component by component. The temporal component changes by $2\bar{N}\dot{\bar{N}}T+2{\bar{N}}^2\dot{T}$; the mixed component changes by ${\bar{N}}^2{T}_{,i}-a^2{\dot{\lambda }}_{,i}$; and the spatial component changes by $-2a\dot{a}T{\delta }_{ij}-2a^2{\lambda }_{,ij}$. HenceThese are the scalar gauge transformations with a background lapse in the paper's sign convention.

Starting in synchronous gauge, preserving both $\Phi =0$ and $B=0$ requiresThe first equation integrates to $\bar{N}T=C(\mathbf{x})$; substituting in the second and integrating givesThe arbitrary lower integration limit is absorbed into $D$. Thus specifying synchronous lapse and shift does not fix the origins of the freely falling clocks or all spatial coordinate labels. This is the residual gauge freedom in synchronous gauge.

During cosmic inflation, choosing these clocks remains arbitrary. For example, the $C$ mode shifts the spatial potential by $HC$ and the density perturbation by $3HC(\bar{\rho }+\bar{P})$. In the de Sitter approximation, $H$ varies little, so a nearly constant metric perturbation can contain a pure coordinate contribution. An inflaton fluctuation or density contrast in synchronous gauge therefore cannot be declared physical solely from its time behavior. Use a uniform-density curvature perturbation or an appropriate field-based gauge-invariant variable, or fix a physical initial slicing. The exact de Sitter case has no evolving homogeneous density clock, which makes density-defined slicing degenerate.

During the standard hot Big Bang, a cold-matter rest frame supplies a convenient additional condition: imposing zero cold-matter velocity removes the nontrivial spatially varying time-shift mode. Before that condition is imposed, the pure-gauge density contrast for constant $w$ is $3H(1+w)C$, which decays as $t^{-1}$ in either radiation or matter domination. In conformal time this is proportional to ${\tau }^{-2}$ during radiation domination and ${\tau }^{-3}$ during matter domination. It can be confused with an independent decaying physical mode unless gauge freedom is fixed. The $D$ mode is a time-independent spatial relabeling and must likewise be fixed by a coordinate or initial-metric convention.

Finally, a scalar density transforms by evaluating the homogeneous scalar at the shifted time:To reach Newtonian gauge from synchronous quantities, impose $\tilde{E}=\tilde{B}=0$. For the nonhomogeneous scalar modes, chooseUsing background stress-energy conservation, $\dot{\bar{\rho }}=-3\bar{N}H(\bar{\rho }+\bar{P})$, gives the synchronous-to-Newtonian density transformation with a lapse:The expression includes the lapse factors because dots refer to the arbitrary time coordinate, not automatically to proper time.

Let $A=\bar{\rho }+\bar{P}$, and assume $A\ne 0$ so that the density-defined curvature is meaningful. Under the cosmological gauge transformation in part ii,Therefore the two time-shift terms cancel:This is the uniform-density curvature perturbation with the paper's overall sign. On a uniform-density slice it is the spatial curvature potential $\Psi$. Its opposite overall sign is also a common convention, so the sign must be kept consistent rather than imported from another metric convention.

To establish time conservation, write $D_t={\bar{N}}^{-1}{\partial }_t$ and $c_a^2=\dot{\bar{P}}/\dot{\bar{\rho }}$. Background stress-energy conservation impliesFor constant $w$ in the given barotropic equation of state, $c_a^2=w$. Differentiate $\zeta =\Psi -\delta \rho /(3A)$:Insert both perturbed evolution equations from the hint, keeping the spatial Laplacian $\Delta ={\nabla }^2/a^2$:Thus the lapse perturbation $\Phi$ and the trace perturbation $\kappa$ cancel exactly. The remaining pressure term is the non-adiabatic pressure perturbation $\delta P_{\mathrm{nad}}=\delta P-c_a^2\delta \rho$. In the assumed barotropic fluid, the perturbed equation of state gives $\delta P=w\delta \rho$ and $c_a^2=w$, so it vanishes. For a Fourier mode,For regular long-wavelength perturbations, the velocity and shear potentials in this combination do not grow as $k^{-2}$, so the right-hand side is gradient-suppressed when $k\ll aH$. Therefore $\zeta$ is conserved to leading superhorizon order, which is the superhorizon conservation of uniform-density curvature requested.

The statement is not exact for arbitrary finite $k$: the displayed gradient term gives its correction. It also requires adiabatic cosmological perturbations; entropy perturbations in a multicomponent system or a nonbarotropic fluid can give $\delta P_{\mathrm{nad}}\ne 0$ and evolve $\zeta$ even at zero gradient order. At exact $w=-1$, $A=0$ and the question's variable is undefined, rather than a well-defined conserved density-slicing perturbation.