Big Bang Nucleosynthesis

Standard big bang nucleosynthesis only involves particles which are part of the standard model - ie. excludes dark matter.

Is this justified? Well, dark matter is non-interacting (or so weakly interacting that we can't detect it), therefore it does not interact strongly with baryons and photons during the epoch of nucleon formation.

However there certainly is theoretical work that investigates non-standard big bang models that do include additional degrees of freedom due to new neutrino species or that include residual annihilation of weak scale massive dark matter particles. Apparently, these pieces of new physics have a significant effect on the nucleosynthesis of the light elements - deuterium, tritium, lithium. A reasonable review (I have not read it) appears to be given by Jedamzik & Pospelov (2009).

Some of these ideas have been used to suggest a solution to the problem that standard big bang nucleosynthesis appears to give too much Li (e.g. Bailly 2011).

This answer is based on David H. Weinberg from Ohio State University:

When the universe is about one second old, the particle species present are photons, neutrinos, and (at an abundance smaller by a factor ∼ $10^9$ ) protons, neutrons, and electrons. Presumably there are also dark matter particles, but they do not matter for BBN. The universe is still dense and hot enough that weak interactions involving neutrinos can convert neutrons to protons and vice versa. Since neutrons are more massive than protons, they are less abundant — conversion of a neutron to a proton is less probable than conversion of a proton to a neutron. In thermal equilibrium

$\frac{n_n}{n_p} = e^{-Q/kT} , Q \equiv (m_n - m_p) c^2 = 1.2934 MeV $

The conversion reactions become slow compared to the age of the universe at t ∼ 3 seconds, kT ∼ 0.7 MeV. Since the interaction rate is dropping quickly as the density and temperature of the universe decline, there are no subsequent conversions. The neutron-to-proton ratio “freezes in” at

$\frac{n_n}{n_p} = e ^{-1.2934/0.7} = 1/6$