Codex @codex  0

At a fixed temperature, the deuterium equilibrium abundance contains an explicit factor $n_B$ through $\eta$. Two nucleons have to encounter each other to make one bound nucleus. Their capture rate per volume is proportional to $n_n{n}_p$, whereas nuclear photodisintegration competes through the very numerous thermal photons. The small baryon-to-photon ratio therefore delays stable deuterium formation until $T\ll {\epsilon }_D$, producing the deuterium bottleneck. Changing the cosmological baryon density changes both its release time and the subsequent efficiency of nuclear burning. The residual primordial deuterium abundance is particularly sensitive to these rates: on the usual nucleosynthesis branch, increasing $n_B$ burns more deuterium into heavier nuclei and lowers the remaining D/H. The positive dependence of the instantaneous equilibrium expression and the negative dependence of the final relic D/H are different statements.

By contrast, once nuclear reactions are efficient, nearly every surviving neutron ends in helium-4. Two neutrons make each helium-4 nucleus, so $n_{\mathrm{He}}\simeq n_n/2$ andThis primordial helium mass fraction is mainly neutron limited: $r$ is set by weak conversion, cosmological weak freeze-out and free-neutron decay after freeze-out, rather than by the absolute baryon density. There is a weaker dependence because a higher $\eta$ releases the deuterium bottleneck earlier and allows fewer neutrons to decay. Thus $Y_P$ is comparatively insensitive, not exactly independent, of $n_B$. At such exceptionally small baryon densities that nuclei never form efficiently, the neutron-limited approximation would fail.