This paper from 2005 claims that the mass of the lightest neutrino is unconstrained. (see p9)

Oscillations are only able to constrain the differences in squares as far as I know, but perhaps constraints could come from cosmology or beta decay experiments.

Is there still no constraints today or are there lower limits on the lightest neutrino mass?


It is true that neutrino oscillations can only be used to put limits on the differences in mass between the various mass eigenstates. To determine the absolute mass of neutrinos, one must study different types of observables, such as cosmological limits or beta decay experiments. There are not many other ways to measure the mass of these particles!

There are several problems, however:

  • the cosmological limits on the sum of the neutrino masses are the strongest constraints that we have today (from Cosmic Microwave Background measurements + Baryon Acoustic Oscillations, mainly), but they are obtained from complicate analyses of data. These analyses require the assumption of a cosmological model, so any constraints that we obtain are only valid inside that model. If it is wrong, the results are wrong as well. Nowadays, the upper limit is around 0.1-0.2 eV [1] (depending on the dataset one considers) for the sum of the three neutrino masses.

  • Beta decay experiments can obtain a constraint on an effective mass of the electron neutrino, that is a combination of the masses of the three mass eigenstates, weighted by the mixing matrix elements (see Eq. 14.26 and 14.28 of [2]). This is one of the best ways to put limits, since there is no model dependence and the mass matrix is rather well known. It may sound strange, but the upper limit nowadays is around 2 eV. It is much larger than the cosmological limit because such a measure requires an extreme precision of some eV or less in determining the energy of the electron released in the beta decay. This energy, for the beta-decay of tritium atoms, is around 18.5 keV: the precision must be at least 4 orders of magnitude smaller. The experiment KATRIN is expected to reduce this upper limit to something like 0.2 eV.

To resume: we just have upper limits, no detection. If the cosmological model and consequently the cosmological limits are correct, we will have to wait several more years before measuring the absolute neutrino masses.


  1. Planck 2015 results. XIII. Cosmological parameters Planck Collaboration (P.A.R. Ade (Cardiff U.) et al.). Feb 5, 2015. 63 pp. Published in Astron.Astrophys. 594 (2016) A13
  2. C. Giunti, C. W. Kim, "Fundamentals of Neutrino Physics and Astrophysics", New York, Oxford University Press, 2007, p. 369, ISBN 978-0-19-850871-7

there are many experiments that put different contraints on the neutrino masses. Here is a good collection from the particle data group.

To summarize: There are lots of experiments that put upper bounds on the neutron mass. The PDG groups estimate is that $\nu_e < 2eV$, $\nu_{\mu} < 0.19eV$, $\nu_{\tau} < 18MeV$. All with a confidence level of about $90\%$, that means that with a probability of $90\%$ will a measured value be below the given bound.

  • $\begingroup$ I'm not sure I understand their document - are they saying that the upper bound on "electron neutrino mass" is 2eV? Surely this can't be right with limits on the sum much lower than that (0.23eV from Planck 2013) $\endgroup$ Jul 23 '15 at 22:20
  • $\begingroup$ Yes this seems to be the upper bound, estimated by the group. This is 2014 so there should be new data coming in soon. $\endgroup$
    – john
    Jul 23 '15 at 22:28
  • $\begingroup$ I don't know about Planck, but this is the source our professor recommend, and he is working as a particle physicist at Cern. $\endgroup$
    – john
    Jul 23 '15 at 22:34
  • $\begingroup$ @mihapriimek A lot of data gets logged in the PDG. If you look at the entire document, you see bounds on sum of masses such as 0.42, 0.39, 0.29, and 0.24. (eV) I suggest reading thoroughly through it to see exactly what they say, but some bounds are (significantly) lower than others. $\endgroup$
    – Omry
    Jul 24 '15 at 12:21
  • $\begingroup$ Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. $\endgroup$
    – Kyle Kanos
    Aug 4 '15 at 13:36

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