5
$\begingroup$

The question says it all. Is there the theoretical possibility that one of the neutrino mass eigenstates have zero mass, whereas the other two have not? Is there a theory that allows this option?

I seem to remember that this would mean that the zero mass neutrino state has always only one helicity, and his antiparticle always only the opposite one.

Is such a combination of massless and massive neutrinos possible? If it is possible, can one say whether the massive states are Majorana or Dirac?

$\endgroup$
  • $\begingroup$ Duplicate?. Whether there is, or is not, conceptually, a lowest-mass right-handed neutrino is strictly "academic", as, in the absence of a mass term $m_1=0$, it would never couple to its left mate and thence to our weakly interacting world; except through gravitons: so it would be really, really invisible. A Majorana mass term coupling it to itself would also be undetectable, except possibly through cosmological means (pfllllttt!). It would be the ultimate invisible particle... $\endgroup$ – Cosmas Zachos Feb 24 at 18:00
7
$\begingroup$

Yes, it is possible that the rest mass of the lightest neutrino is zero. It seems implausible, but is allowed by theory and not ruled out (yet!) by experiment. That's independent of their Dirac/Majorana nature.

$\endgroup$
9
$\begingroup$

Experiments so far have pinned down the two mass differences of the three neutrino masses. The lowest one, $m_1$, may conceivably be zero, even though in physics massless states need some type of theoretical "protection" (gauge invariance, etc...), which the lightest species could not plausibly have, its other two brothers lacking it! Implausible is not impossible.

Now, if it were the case that $m_1=0$, you could, indeed, get by without a right-chiral neutrino.

But, even if you had one such, could you tell? No. A strictly "academic" issue, then: in the absence of a mass term, it would never couple to its left-chiral (active) mate and thence to our entire weakly interacting world, our only handle in detecting neutrinos directly; it would be perfectly sterile, so it would be really, truly invisible.

Of course, such extra particles would have gravitational interactions, and dodgy cosmological arguments could account for them, but how do you propose to contrast them to "conventional" dark matter?

Lastly, this extraneous sterile species, cut off from our WI world by its lack of a Dirac mass coupling it to the left-chiral active species, could still have a Majorana mass term coupling it to itself. But this would also be undetectable, except possibly through cosmological means (Hah!). It would be the ultimate invisible particle. An invisible massive sterile neutrino alongside a massless active one! Calling it "neutrino" would be a stretch, then.

The other two neutrino mass eigenstates could violate lepton number (via extra Majorana masses) or not, completely imperviously to it. It is really cut off from their world, and cannot affect it in any way.

(Our entire discussion above avoided the "convenient fakes", the flavor states, and the corresponding PMNS mixings defining them.)

$\endgroup$
0
$\begingroup$

"Of course, such extra particles would have gravitational interactions, and dodgy cosmological arguments could account for them, but how do you propose to contrast them to "conventional" dark matter?" --- On the contrary, all right-chiral Weyl spinors without any SM charges are dark matter and conversely, all dark matter right-chiral Weyl spinors are sterile neutrinos. And we know from dark matter distributions that dark matter particles do have interactions with each other besides gravity. Hopefully, they also have interactions with SM particles, however (very) weakly. And they are not 'extra' at all; we know they exist.

$\endgroup$
  • $\begingroup$ Welcome to Physics.StackExchange! I think you meant to comment Cosmas Zachos' answer. In that case you should click the "add a comment" link below his answer, and insert your comment there. Thanks! $\endgroup$ – Timo Kärkkäinen Mar 10 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.