Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 310 / 2

A cosmological gauge transformation changes the coordinates used to identify a perturbed spacetime with its homogeneous background. It can produce apparent metric perturbations even in a physically unperturbed FRW metric. For ${\tilde{x}}^i=x^i+{\xi }^i(\tau ,\mathbf{x})$ and unchanged time, the inverse differentials are, to first order,Inserting these into the background line element gives the transformed componentsThe spatial components can now depend on position, although no physical density inhomogeneity or spacetime curvature perturbation has been introduced. For a scalar displacement ${\xi }^i={\partial }^{i}L$, these are pure coordinate-generated scalar cosmological perturbations: $E=-L$, $C=-{\nabla }^{2}L/3$, $B=-L^{\prime }$ and $A=0$ in the transformed coordinates.

The gauge problem is that individual perturbation variables mix this coordinate freedom with physical fluctuations. A nonzero coordinate-dependent field therefore need not represent a physical effect. One must either choose a definite gauge, such as Newtonian gauge in cosmology, or construct invariant combinations, while ensuring that observables do not depend on that choice.

The same invariant interval can be written as $g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }={\tilde{g}}_{\alpha \beta }d{\tilde{x}}^{\alpha }d{\tilde{x}}^{\beta }$. Applying the chain rule to the differentials proves the metric transformation lawFor first-order scalar cosmological perturbations and ${\xi }^{\mu }=(T,{\partial }^{i}L)$, compare perturbations at the same background coordinate label. Expanding the Jacobians and the shifted background gives the linear metric gauge-transformation lawFor the background ${\bar{g}}_{00}=-a^2$, ${\bar{g}}_{ij}=a^2{\delta }_{ij}$, the Lie derivative components areHere primes denote conformal time and $\mathcal{H}$ is the conformal Hubble parameter. These expressions establish all four alternatives, including any chosen pair.

The $00$ component is $\delta g_{00}=-2a^{2}A$, immediately giving the lapse function perturbation transformation. The $0i$ component is $a^2{\partial }_{i}B$, giving the shift vector scalar-potential transformation. The trace of the spatial perturbation is $6a^{2}C$, so tracing the spatial Lie derivative gives the transformation of $C$. Its trace-free scalar part is $2a^2({\partial }_i{\partial }_j-{\delta }_{ij}{\nabla }^2/3)E$, giving the transformation of $E$. ThusAs usual, identifying scalar potentials from their derivatives uses the standard boundary conditions, or nonzero Fourier modes, to remove homogeneous ambiguities.

At first order, tensor cosmological perturbations are spatial transverse-traceless tensors. The coordinate-generated spatial perturbation consists of a trace term and symmetrized derivatives of the displacement. In Fourier space, the latter terms carry a factor $k_i$ or $k_j$. The transverse-traceless projector removes these longitudinal terms and the trace, including the derivative of a transverse vector displacement. Therefore the tensor perturbation is gauge invariant at linear order around the homogeneous background. This is a first-order statement, not a claim of automatic invariance at arbitrary perturbative order.

Because density is a scalar field, $\tilde{\rho }(\tilde{x})=\rho (x)$. Write both sides as background plus perturbation and expand the background time shift:The coordinate shift acting on an already first-order perturbation contributes only at second order. Thus the density perturbation gauge transformation isThe uniform-density curvature perturbation in the paper's sign convention isThe derivative on ${\bar{\rho }}^{\prime }$ is essential and is present in the PDF; the TeX transcription drops it. The scalar-curvature combination transforms aswhile the density term changes by $-\mathcal{H}T$. Hence the two time-slicing changes cancel:On a uniform-density slice, $\delta \rho =0$ and this variable is the signed spatial-curvature perturbation. The construction assumes ${\bar{\rho }}^{\prime }\ne 0$; a pure constant-density cosmological constant does not define such a time slicing.

In Newtonian gauge in cosmology, $E=B=0$ and $C=-\Phi$. The background cosmological perfect-fluid continuity equation gives ${\bar{\rho }}^{\prime }=-3\mathcal{H}(\bar{\rho }+\bar{P})$, so, writing $D=\bar{\rho }+\bar{P}$,Set the cosmological adiabatic sound speed $c_a^2={\bar{P}}^{\prime }/{\bar{\rho }}^{\prime }$. Then $D^{\prime }=-3\mathcal{H}D(1+c_a^2)$. Differentiate the previous expression and insert the perturbed energy-conservation equation:Consequently the curvature evolution equation in this sign convention isThe last equality defines the total energy-frame velocity by $\mathbf{q}=D\mathbf{v}$. For adiabatic cosmological perturbations, the non-adiabatic pressure perturbation vanishes. On superhorizon scales, the stated suppression of the velocity divergence makes the remaining gradient term negligible. Thus $\zeta$ is conserved to leading order in $k/\mathcal{H}$. The positive sign of the pressure source follows from the paper's definition of $\zeta$ and must not be replaced by a formula using the opposite curvature convention.

Superhorizon conservation of uniform-density curvature lets one carry a primordial perturbation from its inflationary generation through otherwise complicated eras and predict the initial conditions for later Cosmic microwave background anisotropy and structure growth. It relies on adiabaticity: an isocurvature perturbation can source curvature evolution, so the same conclusion is not automatic in an arbitrary multifield model.