# Estimate of neutrino masses from the values of $\Delta m_{21}^2$ and $\Delta m_{13}^2$ only!

The blog here says that

The measured mass differences between the eigenstates are $$\Delta m_{21}^2=7.5\times 10^{-5} {\rm eV}^2$$ and $$\Delta m_{13}^2=2.5\times 10^{-3} {\rm eV}^2$$, suggesting that all Standard Model neutrinos have masses below 0.1 eV. That is well in line with cosmological observations which find that the radiation budget of the early universe is consistent with the existence of exactly 3 neutrinos with the sum of the masses less than 0.2 eV.

This suggests that assuming there are only three light neutrinos and without using the cosmological bound on the sum of the three light neutrinos masses is it possible to make a judicious estimate of the individual neutrino masses from the experimental knowledge that $$\Delta m_{21}^2=7.5\times 10^{-5} {\rm eV}^2$$ and $$\Delta m_{13}^2=2.5\times 10^{-3} {\rm eV}^2$$. The estimate tells that the neutrinos masses must be below $$0.1 {\rm eV}$$. It can then be used to cross-check whether it is consistent with the cosmological bound from Planck.

Can someone explain how can we make this estimate? The blog seems to suggest that independent of informations other than the mass-squared differences one can say that individual masses are below $$0.1$$ eV. Given $$\Delta m^2_{ij}$$ values only, why can't neutrinos masses, for example, greater than 1 eV? Or even 10 eV?

• Keep in mind that there are also upper bounds from null attempts to measure $\left< m_{\nu_e} \right>$ from the energy end-point. These are not at all stringent, but they have shaped the way people think about neutrinos for a long time. KATRIN of course, expects to put the whole business to bed. – dmckee Dec 16 '18 at 18:58
• @dmckee Yes. I know. But the blog seems to suggest that independent of informations other than the mass-squared differences one can say that individual masses are below $0.1$ eV. Given $\Delta m^2_{ij}$ values why can't neutrinos masses, for example, greater than 1 eV? Or even 10 eV? – SRS Dec 17 '18 at 14:12
• I would say that you can't logically infer that that all neutrinos have masses below 1 eV just from the mass differences between two pairs of them. The author of the blog says that it is a suggestion. I think it could suggest that if you have more background information, but not just those 2 mass differences. – anonymous Dec 17 '18 at 14:22
• @anonymous I thought if one doesn't allow much fine-tuning, one could estimate individual masses. Otherwise, this is an insufficient information. – SRS Dec 17 '18 at 14:28
• I guess that looking at the mass hierarchy for the charged leptons and quarks, if you expect something similar for neutrinos, that could suggest, together with the mass differences, something about the total masses. – anonymous Dec 17 '18 at 14:32

$$\Delta m_{21}^2 \equiv m_2^2 - m_1^2$$ $$\Delta m_{32}^2 \equiv m_3^2 - m_2^2$$
The quantities on the left are what is measured by neutrino experiments. In fact, what we currently know (2018) is only the value of $$\Delta m_{21}^2$$ and the absolute value of $$\Delta m_{32}^2$$. It is currently not known whether $$m_3$$ is heavier or lighter than $$m_2$$. This is known as the neutrino mass hierarchy problem. Currently, data from the experiments NOVA and T2K suggest that it is most likely that $$m_3 > m_2$$ (called the "normal" hierarchy). From these you can see that you are free to choose any mass for any one of the neutrinos and still end up with the particular values we have measured for the difference of the masses squared.
I can only speculate on why the author said that. Perhaps he or she was imagining a "naturalness" argument that the mass squared differences should be the same order of magnitude as the masses themselves, i.e. it would be weird if the mass differences squared were of order $$10^{-5}~\mathrm{eV}^2$$ while the masses themselves were of order $$10~\mathrm{eV}$$. However since we don't have any model which explains why the neutrinos have mass, this is not good reasoning.