What is the relation between energy density ($\Omega$) and the number density of neutrinos ($n$)?

Question

What is the numerical relation between energy density ( $\Omega_\text{s}, \Omega_\alpha$), and the number density of neutrinos (sterile - $n_s$ , active - $n_\alpha$)?

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Background info that might help better understand my question:

This is based on an equation derived through the relationship of sterile neutrinos distribution function and the active neutrinos distribution function:

$$f_s = k \sin^2 2\theta f_\alpha \tag{1}$$

integrating in terms of momentum gives:

$$\int \frac{d^3 p}{(2\pi)^3} f_s= \int \frac{d^3 p}{(2\pi)^3} k \sin^2 2\theta f_\alpha \tag{2}$$

where $k$ is a numerical constant, $f_i$ is the distribution function of $i$ neutrinos and $\theta$ is the collision angle of neutrinos.

Based on the definition of number density, $n_i= \int \frac{d^3 p}{(2\pi)^3} f_i$ , we get:

$$\frac{n_s}{n_\alpha}= \gamma \tag{3}$$

where $\gamma = \frac{1}{n_\alpha}\int \frac{d^3 p}{(2\pi)^3} \sin^2 2\theta f_\alpha$

**I'm trying to connect this last equation to $\Omega$ but I need to understand the relation in my question first. **

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I've seen in the following article the following statement, based on $(4)$ below:

$$f_s = k f_\alpha \tag{4}$$

$f_s$ has the same functional form as $f_\alpha$ and therefore $\Omega_s/ \Omega_\alpha = (m_s/m_\alpha) (f_s/f_\alpha)$.

I still don't understand where this equation was deduced from.

Answer 1

If I'm not mistaken, the quantities $\Omega$ represent the energy density of the corresponding particle relative to the critical density $\rho_\text{crit} = 3 H_0^2 m_\text{pl}^2$. Thus, to understand the claim, let's write down the number density $$n_i = \int\frac{{\rm d}^3p}{(2\pi)^3}f_i\,.$$ Once whatever process is generating the particles shuts off, the number of species $i$ per co-moving volume is fixed as long as it doesn't decay. When the universe cools below the mass of the species, it becomes non-relativistic, and its energy denisty scales like $m_i n_i$. Therefore, $$\Omega_i = \frac{m_i n_i}{\rho_\text{crit}}\,.$$ Now let's take the ratio of species for which we know $f_a\propto f_s$. Suppose that $\alpha \equiv f_a/f_s$ is the constant of proportionality. Then $$\frac{n_a}{n_s} = \frac{\int\frac{{\rm d}^3 p}{(2\pi)^3}f_a}{\int\frac{{\rm d}^3 p}{(2\pi)^3}f_s} = \frac{\alpha\int\frac{{\rm d}^3 p}{(2\pi)^3}f_s}{\int\frac{{\rm d}^3 p}{(2\pi)^3}f_s} = \alpha\,.$$ Hence $$\frac{\Omega_a}{\Omega_s} = \frac{m_a n_a}{m_s n_s} = \frac{m_a}{m_s}\alpha = \frac{m_a f_a}{m_s f_s}\,.$$