= Solution
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.
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