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?

  • 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_someone May 3 '18 at 4:10
  • $\begingroup$ I think that the person means is it always a set value irrespective of direction $\endgroup$ – Triatticus May 3 '18 at 4:11
  • $\begingroup$ 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 $\endgroup$ – Juanjo May 3 '18 at 4:24
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    $\begingroup$ 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. $\endgroup$ – safesphere May 3 '18 at 6:37
  • $\begingroup$ 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 $\endgroup$ – Juanjo May 3 '18 at 9:50

Your starting point

Neutrinos are produced in beta decays with a given helicity.

is plain unsound. Neutrinos are produced with a given chirality, which, for massive neutrinos, is distinctly different than helicity, \https://en.wikipedia.org/wiki/Chirality_(physics)#Chirality_and_helicity.

Consider a positron emission β-decay, so, effectively, the disintegration of a virtual $W^+$. A right-chiral positron is produced in the nucleus' frame, in association with a left-chiral neutrino (actually a massive "electron neutrino", which is a "fictitious" convenience linear superposition of three massive neutrinos, but let's not muddy the waters, for now).

This left-chiral neutrino will have a wave-function with a huge component of negative helicity, and a minuscule, $O(m_\nu/p)$ component of positive helicity, normally ignored, as it would vanish for a massless neutrino and underlies your confusion about "definite helicity production".

As the comments indicate, this mix of helicity components is frame-dependent, as always. So, if you scooted after the neutrino and you overtook it, it would appear to be receding from you in the opposite direction: so it would now be mostly of positive helicity.

(There is, of course, no such thing as left- or right- helicity... terms virtually invented to confuse. Use at your own risk.)

Frame dependent or not, helicity is a wonderfully useful concept, much more intuitive than chirality, and has simple accounting rules for physical systems, which is why it is routinely used.

For more formal detail, you may see this.

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