Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 64 / 3 / i / Solution

The comoving observer measures $E=a{p}^0$, so $q=aE=a^2{p}^0$. Write $\mathcal{H}=a^{\prime }/a$. The null geodesic condition impliesThe time component of the geodesic equation, using $d\tau /d\lambda =p^0$, is thereforeDifferentiate $q=a^2{p}^0$. The homogeneous expansion terms cancel:At zeroth order $p^i/p^0=n^i$, with ${\delta }_{ij}{n}^i{n}^j=1$. The difference between coordinate velocity and physical unit direction is already first order and multiplies $h^{\prime }$, so it can be discarded in this expression. We obtain the synchronous photon momentum redshiftIn the unperturbed universe $q$ is constant and physical photon energy redshifts as $a^{-1}$; the displayed term is the additional anisotropic first-order redshift.

To check the direction, let $v^i=p^i/p^0$. Combining the spatial and temporal geodesic equations cancels their homogeneous Hubble terms and givesEvery remaining connection or explicit metric-derivative term is first order. A local orthonormal spatial frame defines the physical unit vector by $n^i=v^i+\frac{1}{2}{h}^i{}_j{v}^j+O(h^2)$. Its derivative adds only first-order terms, soConsequently direction deflection multiplied by an already first-order temperature anisotropy is second order. The first-order brightness calculation may use a straight unperturbed ray without neglecting the first-order gravitational frequency change.