Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 58 / 2

Write $\mathcal{H}=a^{\prime }/a$, where primes denote conformal time. In the passive convention for a cosmological gauge transformation, retain terms linear in $h$ or $\xi$ and discard their products. The linear metric gauge-transformation law then givesFor a single nonzero Fourier mode, suppress the common factor $e^{i\mathbf{k}\cdot \mathbf{x}}$. Its derivatives areUsing ${\eta }_{00}=-1$ and ${\eta }_{ij}={\delta }_{ij}$ therefore yieldsThe term proportional to $\mathcal{H}T$ comes from the time dependence of the background scale factor; omitting it would transform only the flat reference metric.

For the synchronous-to-Newtonian scalar Fourier transformation, the coefficient of ${\hat{k}}_i{\hat{k}}_j$ in the transformed spatial metric perturbation must vanish. Thus $h_s+2kL=0$. Vanishing shift then gives $L^{\prime }=kT$. For $k\ne 0$ these conditions fixThe lapse in conformal Newtonian gauge is $h_{00}^{\mathrm{N}}=-2\Phi$, soThe remaining coefficient of ${\delta }_{ij}$ isgivingThe homogeneous $k=0$ perturbation requires separate background-coordinate choices; the inverse powers of $k$ do not define a transformation for it.

Preserving synchronous gauge means $T^{\prime }+\mathcal{H}T=0$ and $L^{\prime }=kT$. The first equation is $(aT{)}^{\prime }=0$, and integration of the second gives the complete residual scalar transformation for each nonzero Fourier mode:Changing the lower limit only changes $g$. Taking the trace and trace-free longitudinal part of $\Delta h_{ij}$ givesConsequently $\Delta h_s^{\prime }=2k^{2}T$ and $\Delta h_s^{\prime \prime }=2k^2{T}^{\prime }$. The cosmological Newtonian potential changes bywhile its spatial counterpart changes byThus both potentials are invariant under residual synchronous-gauge freedom, although $h$ and $h_s$ individually are not.