Boltzmann equation (Number density) I'm trying to understand the Boltzmann equations use in the early Universe. The derivation is somewhat tedious, but in the end I end up with:
$$a^{-3}\frac{d}{dt}\left(n_1a^3\right) = n_1^{(0)}n_2^{(0)}\langle\sigma v\rangle\left[\frac{n_3 n_4}{n_3^{(0)}n_4^{(0)}} - \frac{n_1 n_2}{n_1^{(0)}n_2^{(0)}}\right]$$
So the evolution of species $n_1$ is given by the production and elimination of the reaction between 1 with 2, and 3 with 4 - in short. So the $n_i$ is the number density of each species, but my question is (Which I haven't been able to figure out): What does the supscript $0$ means (i.e. $n_i^{(0)}$) ?
What I have been able to find is, that it is supposed to be the equilibrium number density, but I'm not quite sure I understand what that actually means. So I was hoping someone could tell me :)
Thanks in advance.
 A: The above is the Boltzmann equation for annihilation/collisions in an expanding universe.
Short answer
The superscript in this case stands for equilibrium number density, as you pointed out. This means the $n_i^{(0)}$ stands for particle species $i$ number density in number density equilibrium, meaning equilibrium in which each particle species $i$ is produced at the same rate as they it is annihilated. This implies that $\mu_i=0$
The number densities:
\begin{align}
n_i&=g_i e^{\mu_i/T} \int \mathrm{d}^3p\space e^{-E_i/T}/(2\pi )^3\\
n_i^{(0)}& = g_i \int \mathrm{d}^3p\space e^{-E_i/T}/(2\pi )^3
\end{align}
Long answer
Let $1+2\leftrightarrow 3+4$ describe particles 1 and 2 annihilating to particles 3 and 4.
Assume reversibility; the Boltzmann equation is:
\begin{align}a^{-3}\frac{d(n_1a^3)}{dt}&=\int \frac{d^3p_1}{(2\pi)^3 2E_1} \int \frac{d^3p_2}{(2\pi)^3 2E_2} \int \frac{d^3p_3}{(2\pi)^3 2E_3} \int\frac{d^3p_4}{(2\pi)^3 2E_4} \\
&\times (2\pi )^4 \delta^3(p_1+p_2-p_3-p_4)\delta(E_1+E_2-E_3-E_4) |M|^2\\
&\times\{f_3f_4[1\pm f_1][1\pm f_2] - f_1 f_2[1\pm f_3][1\pm f_4] \}\end{align}
Where $a$ is the scale factor, $p_i$ is momentum,$E_i$ is energy, $M$ is the invariant amplitude, $f_i$ is the phase space.
Now, rates of annihilation are proportional to:
\begin{align}
rate(1+2\rightarrow 2+3)&\propto f_1 f_2(1\mp f_3)(1\mp f_4)\\
rate(1+2\leftarrow 2+3)&\propto f_3 f_4(1\mp f_1)(1\mp f_2)
\end{align}
And hence in annihilation equilibrium rates balance:
$$rate(1+2\rightarrow 2+3) = rate(1+2\leftarrow 2+3)$$
$$\Rightarrow f_1^{eq} f_2^{eq}(1\mp f_3^{eq})(1\mp f_4^{eq}) = f_3^{eq} f_4^{eq}(1\mp f_1^{eq})(1\mp f_2^{eq})$$
So what does this mean for the number density?
Assume kinetic equilibrium (scattering/annihilation extremely rapid, so we can assume Bose-Einstein or Fermi-Dirac forms)
$$f_i = \frac{1}{e^{(E_i-\mu_i)/T_i}\mp 1}$$
Assume $T<<E-\mu$
$$f_i \approx \frac{1}{e^{(E_i-\mu_i)/T_i}}$$
Hence $rate(1+2\rightarrow 3+4)-rate(3+4\rightarrow 1+2)$ becomes:
$f_3 f_4(1\mp f_1)(1\mp f_2)-f_1 f_2(1\mp f_3)(1\mp f_4) = f_1 f_2(1\mp f_3)(1\mp f_4)[-1 + \exp((E_1-\mu_1)/T_1 + (E_2-\mu_2)/T_2 - (E_3-\mu_3)/T_3 - (E_4-\mu_4)/T_4)]$
Assume $T_i=T_j$ and energy conservation $E_1+E_2=E_3+E_4$
$$e^{-(E_1+E_2)/T}[e^{(\mu_3+\mu_4)/T} - e^{(\mu_1+\mu_2)/T}]$$
Now number density $n$ is:
$$n_i=g_i e^{\mu_i/T} \int \mathrm{d}^3p\space e^{-E_i/T}/(2\pi )^3$$
And in the case of constant number density, $\mu_i=0$:
$$n_i=n_i^{(0)} = g_i \int \mathrm{d}^3p\space e^{-E_i/T}/(2\pi )^3$$
Now we simply re-write the equation with $n_i$ and $n_i^{(0)}$:
$$e^{-(E_1+E_2)/T}\left[\frac{n_3 n_4}{n_3^{(0)} n_4^{(0)}} - \frac{n_1 n_2}{n_1^{(0)} n_2^{(0)}}\right]$$
Note that while $n_i=n_i^{(0)}$ forces rates of annihilation to balance out, they are also balanced out if $\mu_1+\mu_2=\mu_3 + \mu_4$.
Sources:

*

*Scott Dodelson's Modern Cosmology

*Andrew J. S. Hamilton's General Relativity, Black Holes, and Cosmology
