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When reading about the masses of neutrinos in the HDM (Hot Dark Matter) model of the universe, I came across the following equation:

$$\tag{1} \sum_i m_{\nu_i} = 92 \Omega_\nu h^2 \text{eV}$$

My problem arises with the statement that follows: "where the sum runs over all neutrino species with $M_{\nu_i} \leq 1$ MeV."

This is written differently, solely as $m_\nu = 92 \Omega_\nu h^2 \text{eV}$ in Sterile Neutrinos as Dark Matter.

The last equation makes it look like the mass of each individual neutrino must be $92 \Omega_\nu h^2 \text{eV} $(which sounds wrong to me as neutrinos of different flavours must have different masses) and the equation $(1)$ makes it look like that value is the mass of all neutrinos in the universe. Is this last interpretation the correct one?

If so, does the condition below it mean that the mass of each neutrino must be smaller than $1$ MeV or has it got some different meaning I am not understanding?

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    $\begingroup$ Just to avoid confusion: This is about the individual mass of the various //types// of neutrinos, not about the summed weight of all neutrinos in the universe put together. $\endgroup$
    – rfl
    Commented Sep 28, 2020 at 9:17

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It is the sum of the masses of neutrinos that should be equal to the right side of the formula you quote, qualified that it is the low mass neutrinos that are considered, i.e. the standard model neutrinos which are all of very small mass.

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  • $\begingroup$ Then, does that condition, of $M \leq 1 \text{MeV}$ represent a limit of the mass of each individual neutrino? $\endgroup$
    – user7077252
    Commented Sep 1, 2020 at 20:45
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    $\begingroup$ Thats right.each individual standard model neutrino . High mass neutrinos are proposed in different models beyond the standard model, and they are not in the sum $\endgroup$
    – anna v
    Commented Sep 2, 2020 at 6:38

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