Freeze Out of Non-Relativistic Species

We can define the quantity ${\Delta }_Y\equiv Y-Y_{eq}$.

Now, if we solve the Boltzmann equation acquired earlier,

For early times, $1<x\ll x_f$, the yield closely follows its equilibrium value, $Y\approx Y_{eq}$, and we can assume that $d{\Delta }_Y/dx=0$. Then,

${\Delta }_Y=-\frac{\frac{d{Y}_{eq}}{dx}}{Y_{eq}}\frac{x^2}{2\lambda ⟨\sigma v⟩}$.

Thus, at freeze-out we obtain,

${\Delta }_{Y_f}\approx \frac{x_f^2}{2\lambda ⟨\sigma v⟩}$

For late times, $x\gg x_f$, we can assume that $Y\gg Y_{eq}$ and thus ${\Delta }_{Y_{\infty }}\approx Y_{\infty }$. Then,

$\frac{d{\Delta }_Y}{dx}\approx -\frac{\lambda ⟨\sigma v⟩}{x^2}{\Delta }_Y^2$.

Separating variables and integrating from freeze-out time up to nowadays,

${\int }_{{\Delta }_{Y_f}}^{{\Delta }_{Y_{\infty }}}\frac{d{\Delta }_Y}{{\Delta }_Y^2}=-{\int }_{x_f}^{x_{\infty }}\frac{\lambda ⟨\sigma v⟩}{x^2}dx$.

Since $x_{\infty }\gg x_f$,

$\frac{1}{{\Delta }_{Y_{\infty }}}=\frac{1}{{\Delta }_{Y_f}}+\frac{\lambda }{x_f}(a+\frac{b}{2x_f})$.

Note that, here, we have expanded the thermally averaged annihilation cross-section in powers of $x^{-1}$ as $⟨\sigma v⟩=a+\frac{b}{x}$.

The first term of the RHS is generally ignored. This leads to

${\Delta }_{Y_{\infty }}=Y_{\infty }=\frac{x_f}{\lambda (a+\frac{b}{2x_f})}$.

The relic density can be expressed as

$\Omega h^2=\frac{m_{\chi }{Y}_{\infty }{s}_0{h}^2}{{\rho }_c}\approx \frac{10^{-10}Ge{V}^{-2}}{a+\frac{b}{40}}\approx \frac{3\times 10^{-27}c{m}^3{s}^{-1}}{a+\frac{b}{40}}$.

${\sum }_a{t}_{ki}^a{t}_{lj}^a{t}_{k^{\prime }{i}^{\prime }}^a{t}_{l^{\prime }{j}^{\prime }}^{a\ast }\to \frac{1}{4}{\delta }_{kj}{\delta }_{li}{\delta }_{k^{\prime }{j}^{\prime }}{\delta }_{l^{\prime }{i}^{\prime }}\to \text{traced over gives }\frac{C_F^2}{N_c^2}\text{ where }{C}_F=\frac{N_c^2-1}{2N_c}$