Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 310 2 b Solution Created 2026-10-03 Updated 2026-10-06
The same invariant interval can be written as . Applying the chain rule to the differentials proves the metric transformation lawFor first-order scalar cosmological perturbations and , compare perturbations at the same background coordinate label. Expanding the Jacobians and the shifted background gives the linear metric gauge-transformation lawFor the background , , the Lie derivative components areHere primes denote conformal time and is the conformal Hubble parameter. These expressions establish all four alternatives, including any chosen pair.
The component is , immediately giving the lapse function perturbation transformation. The component is , giving the shift vector scalar-potential transformation. The trace of the spatial perturbation is , so tracing the spatial Lie derivative gives the transformation of . Its trace-free scalar part is , giving the transformation of . 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 or . 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.