# Massive-ness of Neutrinos as a Consequence of Special Relativistic Kinematics?

As I understand it is still unclear if all neutrinos in the standard model are massive. Neutrino oscillations show that certainly not all neutrinos can be massless, but it might still be the case that one neutrino species is massless.

I found a very simple argument why all neutrinos should be massless, based only on energy-momentum conservation of special relativity and one particular interaction in the standard model. I'm wondering if it is a valid argument, if it is used anywhere, and if it is considered of any value.

Consider the standard model interaction vertex of two neutrinos going into a Z-boson, $\nu + \nu\to Z$. Here $\nu$ can be any neutrino. Suppose that this neutrino were massless. Then, denoting the two neutrino four-momenta by $p^\mu$ and $q^\mu$, and using $E^2 = p^2+m^2$ with $m=0$, we will have $p^0=\pm p^1$. Similarly, $q^0 = \pm q^1$. For simplicity we consider the case where both signs are a $+$, so that we have $p^0 = p^1$ and $q^0 = q^1$. Now, denoting the $Z$-boson four-momentum by $k^\mu$, energy-momentum conservation in the vertex requires that $k^0=p^0 + q^0$, as well as $k^1 = p^1+q^1 = p^0 + q^0$. We see immediately that this implies that $k^1 = k^0$, meaning that the $Z$-boson must be massless.

Of course we know that the $Z$-boson has a mass, so we have arrived at a contradiction. Hence our initial assumption, that the neutrino $\nu$ was massless, must be false. Therefore all neutrinos must be massive.

EDIT: As Anna points out in the comments, the process should not be $\nu + \nu\to Z$ but $\nu + \bar\nu\to Z$, i.e., incorporating an antineutrino. It does not change the argument, of course.

EDIT: I see I totally forgot to mention that I'm working in $1+1$ dimensions. I also now see, thanks to the answer by Cosmas Zachos, that choosing both $\pm$ signs as a $+$ is questionable, and may not be possible. Perhaps the $\nu + \bar\nu\to Z$ vertex only occurs when the two $\pm$ signs are opposite, in which case my argument above does not work. Or it could be that, as some other people say, the process only occurs in cases where for instance the $Z$ is virtual, so that it can be off-shell.

• Actually the incorrect assumption is that the process you describe is possible by itself. It's kinematically forbidden, just like an electron-positron annihilation always produces at least two photons. – Javier Feb 27 '18 at 13:27
• standard model neutrinos cannot be in this interaction, lepton number violation. Z carries no lepton number. @Javier Z can decay to a neutrino antineutrino pair. in the question Z could be virtual, decaying again to a particle antipaticle pair. – anna v Feb 27 '18 at 14:02
• @anna But in that decay the energy of the neutrinos would be fixed, right? You can see that in the Z rest frame. So neutrino+antineutrino into Z is forbidden unless they have exactly the right energy. And virtual particles need not be on shell, so the argument wouldn't apply. Is this correct? – Javier Feb 27 '18 at 14:58
• As soon as you say "at a vertex", you admit the possibility of off-shell terms. Interior lines in Feynman diagrams are not required to have the mass of the particle they correspond to. – dmckee Feb 27 '18 at 15:21

You may appreciate this is a bogus argument, if you remind yourself that a massive $\pi^0\to \gamma \gamma$, two massless particles, alright.
First go to the center of momentum frame of the neutrinos, so $p^1+q^1=0$, and hence $M_Z= p^0+q^0=2p^0$, just as the two photons would do.
With oscillation experiments, you can only probe mass differences between two mass eigenstates, and this imply that just one of the eigenstates can be massless. We don't even know which could be the lightest, since we don't know the hierarchy of the mass spectrum: normal hierarchy or inverted hierarchy? Unfortunately, it seems we can get an answer only from neutrinoless double $\beta$-decay experiments, which should help us to decide whether neutrino is a Dirac or Majorana particle (or both, in generalized theories beyond the SM).