# Calculating the maximum total neutrino mass by using cosmological bound

In an article its written,

$$\Omega_{\nu} = \frac{\rho_{\nu}}{\rho_{crit}}=\frac{\sum m_{i,\nu}n_{i,\nu}}{\rho_{crit}} = \frac{\sum m_{\nu}}{93.14h^2eV}$$

Now I am trying to derive this for myself but I could not. Can someone help me ?

So the values are,

$$\rho_{crit} = 1.053 75 \times 10^{-5}h^2 GeV/c^2~~cm^{-3}$$

Total neutrino average number density today : $$n_{\nu} = 339.5~cm^{-3}$$

I tried to write it like,

$$\frac{n_{\nu}\sum m_{\nu}}{\rho_{crit}} = \frac{\sum m_{\nu}}{93.14h^2eV}$$

$$\frac{n_{\nu}}{\rho_{cric}} = \frac{339.5cm^{-3}}{1.05375 \times 10^{-5}h^2 GeV/c^2~~cm^{-3}} = \frac{1}{3.103h^2 \times 10^{-8} GeV} = \frac{1}{31.0382916 h^2eV}$$

I am missing additional $$1/3$$. Where is it coming from ?

I guess its a simple question but I couldnt see the answer.

Equation 25.2

The text makes it clear that the summation is over three neutrino species. In which case when you take the number density out of the summation, it is the average density per species that you use, which is one third of 339 cm$$^{-3}$$.