Hubble expansion rate and reaction rates In terms of early universe cosmology I often stumble over sentences like:

We assume that, at sufficiently early times, reaction rates for
particle interactions  are much faster than the expansion rate, so
that the cosmic fluid is in thermal  equilibrium.

The quote is from the book "The Cosmic Microwave Background" by Ruth Durrer, page 14, ( https://www.cambridge.org/core/books/cosmic-microwave-background/10D066B56BBBA899F3B89A29E0B3B78B )
How can I understand this better? Are there some equations where I can explicitly see that the production of a specific particle species goes drastically down when the Hubble rate is larger than the reaction rate? And with reaction rate one means the rate which an observer in the lab would see?
 A: Early universe is very dense and hot, in this condition protons can't capture electrons to form atoms. The mean time of interraction between 2 particles is negligeable than the expansion rate give approximatively by $1/H$.
For example interaction rate for neutrino with the $l$ species ($l=$ electron for example) is
$$\Gamma_\nu = <n_l\sigma v>$$
where $n_l$ is the density of the species l, $\sigma$ the cross-section of neutrino (=probability of interaction) and $v$ the relative velocity between the two particles. Neutrino interact with weak interaction with coupling constant $G_F\sim 10^{-5}$ (Fermi's constant). You can verify that $<\sigma v> \sim G_F^2 T^2$ and $n_l \sim T^3 $ (with fermi-dirac statistic: https://en.wikipedia.org/wiki/Fermi%E2%80%93Dirac_statistics) so $\Gamma_\nu \sim G_F^2 T^5$.
At early times Universe is dominated by radiation and the radiation density is such as $\rho_r \propto T^4$. Friedman equation give you $H \propto \rho_r^{1/2} \propto T^2$
So the time between two interaction of neutrino is $\tau \propto \frac{1}{\Gamma_\nu}\propto T^{-5}$ and expansion time is $t\sim\frac{1}{H}\propto T^{-2}$. Comparing the two you can see that for high temperature $\tau \ll t$.
Note that with temperature decreasing $\tau$ finally become larger than $t$, it's the decoupling of neutrino.
$T^{-2}$ vs $T^{-5}$" />
I hope this clarify your question !
A: This is a question that puzzled me once. This is simply a misleading statement. For now, a more satisfying one would be:
Once the interaction rate drop to $\Gamma\approx H$, the neutrino, for instance, is expected to have less than one scattering until the end of time($t\rightarrow\infty$). Namely, calculate this:
$$N=\int_{t}^{\infty}\Gamma(t')dt'=\int_{0}^{T}\frac{\Gamma(T')}{H(T')}\frac{dT'}{T'}$$
Noting that $H(T)\propto T^{2}$,assuming $\Gamma(T)\propto T^{n}$,you obtain:
$$N=\frac{1}{n-2}\frac{\Gamma(T)}{H(T)}$$
Now you can see it! Then, if there is less than one scattering, your intuition can tell that there won't be thermal equilibrium. For quantitative description of this statement, I refer you to Boltzmann equation dealing with out of equilibrium situations. It is treated in every standard cosmology book. And from it you can better understand why $\Gamma\ll H$ makes it impossible for one species to reach equilibrium.
