# Why is the square of the neutrino mass negative?

Why is the square of the neutrino mass negative?

In arXiv:hep-ph/0009291 this is explained by giving the example of:

$$m^2_{\nu_e}= -2.5 \pm 3.3 \text{eV}^2 \tag{1}$$

"Thee negative value of the neutrino mass-square simply means:" $$E^2/c^2 -p^2=m^2_{\nu_e} c^2 < 0$$ "The right-hand side in Eq. (3) can be rewritten as ($$-m_s^2 c^2$$), then $$m_s$$ has a positive value."

What is the meaning of $$m_s$$? This isn't explained in this article, and it makes no sense to me that the value for the square of a mass could be negative.

A similar question has been posted in Negative Neutrino Mass squared and in Negative Mass Square I do not understand the answers given.

• Rather than posting a new question, I recommend adding comments to the linked questions & the answers there. Try to ask for specific clarifications, as "I do not understand the answers" leaves us floundering as how to help you. Commented Sep 9, 2020 at 12:19
• Does this answer your question? Negative Mass Square Commented Sep 9, 2020 at 12:19
• The only reason I didn't do that is because when I have done that before, I got little to no feedback, whilst when posting it again, I tend to get answers from a different POV that help me understand. That link was posted in my question and I did not understand the explanation. Commented Sep 9, 2020 at 12:29
• You are referring to >20 year old data. The difficult-to-interpret parameter fits have not disappeared to date but they are much milder. Of course that fit parameter is not supposed to make sense as a fundamental physics quantity. Are you familiar with the fine mechanics of measurement? Commented Sep 9, 2020 at 15:42

The mass squared of the electron neutrino obviously cannot be negative as then the neutrino would have an imaginary mass, so the obvious conclusion is that there is some unknown (probably systematic) error in the experiment. The latest measurements I am aware of are the results from the KATRIN experiment, and while they do still give a negative value $$m^2 = -1.0_{-1.1}^{+0.9}~\text{eV}^2$$ this is not significantly different from $$0$$.