Quantifying the baryon asymmetry requires normalizing the baryon number density w.r.t comoving entropy density or photon density-Why?

Question

This question has been edited a little to clarify the confusion I have.

The matter-antimatter asymmetry of the Universe quantified by the baryon asymmetry as $$Y_B=\frac{n_B-n_{\bar{B}}}{s}=\frac{n_B}{s}$$ i.e., the difference in the number densities of baryons $n_B$ and antibaryons $n_{\bar{B}}$ normalized w.r.t the comoving entropy density $s$. Sometimes the baryon assymetry is also expressed by normalizin the baryon number density w.r.t the comoving photon density $n_\gamma$.

What is the significance of normalizing the baryon asymmetry w.r.t $s$? Does it have anything to do with the fact that $sa^3$ is conserved (where $a$ is the scale factor)?

Addendum Why is it that people don't quote simply $(n_B-n_{\bar{B}})$ i.e., the number density of baryons (more precisely, that of baryons minus antibaryons) as the baryon asymmetry? Why do we have to normalize it w.r.t either the photon density $n_\gamma$ or the entropy density $s$?

I understand that as the Universe expands, this number density dilutes. Does it mean that the normalized quantity $Y_B$ don't dilute? If yes, is this the reason for normalizing it?

Answer 1

You have to normalize it to something because $n_B - n_\bar B$ is just the number of baryons in the universe. I more frequently see the baryon asymmetry normalized by the number of CMB photons (e.g. this PDG table.

The Particle Data Group's review of Big-Bang cosmology says on page 12

For photons [...] $$ d(sR^3)/dt = 0. $$ For radiation, this corresponds to the relationship between expansion and cooling, $T \propto R^{-1}$ in an adiabatically expanding universe. Note also that both $s$ and $n_\gamma$ scale as $T^3$.

I interpret this to mean that your entropy-scaled baryon asymmetry and my photon-scaled baryon asymmetry are simply proportional to one another, for just the reason that you guessed in your question.