All galaxies in the population share the same formation time $t_f$, so the difference between their stellar ages equals the difference between their cosmic emission times. The <redshift-time relation> gives
$$
\frac{dz}{dt}=-(1+z)H(z).
$$
For a close pair with $|\Delta z|\ll1$,
$$
\boxed{\Delta t\simeq-\frac{\Delta z}{(1+z)H(z)}},
\qquad
\boxed{H(z)\simeq-\frac1{1+z}\frac{\Delta z}{\Delta t}}.
$$
A <cosmic chronometer> measurement therefore reconstructs $H(z)$ directly from differential galaxy ages. Inserting those values into the matter-plus-dark-energy Friedmann expression constrains $w(z)$.

Supernova distances integrate $1/H(z)$, while $H(z)$ already contains an integral of $w(z)$. Cosmic chronometers avoid the distance integral, so rapid oscillations in $w(z)$ suffer one fewer smoothing operation and can leave a more visible signal.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 310 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-310/1/d/solution