Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 73 / 1 / v

The combined kinematic and cosmological redshift is most cleanly found by passing through the emitter's local comoving frame. The distance estimate then requires an explicit approximation to the Hubble law.

Let ${\nu }_c$ be the photon frequency in the local comoving frame just after emission. The kinematic redshift gives ${\nu }_e/{\nu }_c=1+z_{\mathrm{kin}}$, and propagation in the FRW metric gives ${\nu }_c/{\nu }_0=1+z_{\mathrm{cosm}}$. Multiplication givesEquivalently, $z_{\mathrm{tot}}=z_{\mathrm{cosm}}+z_{\mathrm{kin}}+z_{\mathrm{cosm}}{z}_{\mathrm{kin}}$. Both redshifts being small is what permits their approximate addition. Here the observer's own peculiar velocity is assumed already corrected to the Hubble flow frame.

In the local, nonrelativistic Hubble law approximation, $c{z}_{\mathrm{tot}}\simeq H_{0}r+v_p$. Ignoring the peculiar velocity gives an inferred distance $r_{\mathrm{est}}=r+v_p/H_0$, soThe requested accuracy therefore requiresAt exactly $120\mathrm{Mpc}$ the worst-case error is $5\%$, so “better than” requires a strictly larger distance. This is the conventional leading-order estimate. If one retains the exact Doppler factor and the product term while treating $r=c{z}_{\mathrm{cosm}}/H_0$ as exact, the positive-velocity case gives $|\Delta r|/r=(1+z_{\mathrm{cosm}})q/z_{\mathrm{cosm}}$, with $q=\sqrt{(1+600/c)/(1-600/c)}-1$. It requires $r>(c/H_0)q/(0.05-q)\simeq 125\mathrm{Mpc}$. That small refinement is conditional on the distance-redshift prescription; the simple $120\mathrm{Mpc}$ answer uses the local approximation consistently.