Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 55 / 3 / a / Solution

In Newtonian gauge, the coordinate-comoving observer has normalized four-velocity $u^{\mu }=a^{-1}(1-\psi ,0)$, hence $u_0=a(1+\psi )$. In the $(+---)$ convention the measured photon energy is $E=u_{\mu }{p}^{\mu }$. Thus to first order

Use the time component of the geodesic equation and divide by $p^0=d\eta /d\lambda$:Let $\mathcal{H}=a^{\prime }/a$ and take the photon path direction $\mathbf{e}$ at zeroth order. The left side divided by $ϵ/a^2$ is ${ϵ}^{\prime }/ϵ-2\mathcal{H}+2\mathcal{H}\psi -d\psi /d\eta$. The contraction on the right side, using the provided connection coefficients, isThe homogeneous expansion cancels. Therefore the scalar photon energy-redshift law isThe total derivative is $d/d\eta ={\partial }_{\eta }+\mathbf{e}\cdot \nabla$ on the unperturbed ray. Perturbing that path inside a first-order potential would contribute only at second order. The homogeneous $a^{-1}$ energy redshift is already removed by the comoving momentum variable $ϵ$.