Neutrinos are produced in beta decays with a given helicity. My question is wether this helicity is a constant of this movement or is it variable?
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2$\begingroup$ Helicity is not frame-independent, if that's what you're asking. If you run faster than the neutrino, then its helicity changes. $\endgroup$– probably_someoneCommented May 3, 2018 at 4:10
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$\begingroup$ I think that the person means is it always a set value irrespective of direction $\endgroup$– TriatticusCommented May 3, 2018 at 4:11
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$\begingroup$ I mean that observed from a given rest frame, if the neutrino is produced with a given helicity, wether this helicity is constant regardless the direction to which the neutrino is heading $\endgroup$– JuanjoCommented May 3, 2018 at 4:24
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2$\begingroup$ Yes. All neutrinos have left helicity while all antineutrinos have right helicity. As mentioned above, helicity can be reversed by outrunning a neutrino (although chirality would still remain the same). However, verifying this experimentally is not possible at this time. $\endgroup$– safesphereCommented May 3, 2018 at 6:37
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$\begingroup$ I guess this is the answer, helicity is frame-independent if neutrinos have mass and therefore there is a frame where you can run faster than them. If they did not have mass, such a frame would not exist, right? I still do not get why helicity changes when you are running faster than the neutrino (can someone give a more formal answer?). Thanks $\endgroup$– JuanjoCommented May 3, 2018 at 9:50
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Your starting point that,
Neutrinos are produced in beta decays with a given helicity.
is plain unsound. (To begin with, it is an open question whether each and every neutrino is massive, each and every neutrino is massless, or whether some neutrinos are massive while others are massive.) In any case, there is a consensus that all neutrinos are produced with given left chirality. For massive neutrinos---like any massive particle---chirality is different than helicity .
By way of example, consider a positron emission resulting from $\beta$-decay. This emission effectively results from the disintegration of a virtual $W^+$. In this scenario, a right-chiral positron is produced in the nucleus' frame, in association with a left-chiral neutrino.
As the comments indicate, this mix of helicity components is frame-dependent. The consequence is that if you scooted after the neutrino and you overtook the neutrino, then the neutrino would appear to be receding from you in the opposite direction; further, the neutrino would now be mostly of positive helicity.
As a final note, I feel it important to remind you that there is no such a thing as left or right helicity: This term seems to have been invented to confuse people. Helicity can be positive or negative. Thus, I encourage you to use the term helicity here at your own risk.
answered May 4, 2018 at 13:55