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I know this question has been asked a few times before here in various ways, but I haven't found answers which helped me a lot. For one, the class I am in is not using any of the underlying math, so all the answers about Dirac spinors or Weyl spinors were a bit beyond me. I found this post here, but the top answer still leaves me a bit confused. I have lots of follow-up questions about it, so I thought it might warrant a new post.

When the $W$ boson decays into the fermion and the antineutrino, the antineutrino must be right handed, and it was my understanding that the $W$ boson can only interact with a left-handed muon. But the muon must also be right-handed to satisfy conservation of spin. The resolution to this seems to be that since the muon is not massless, it must have both left-handed and right-handed components, but this doesn't make much sense to me.

To start, given my understanding of helicity, the muon will simply have some helicity, left- or right-handed, which can change based on your frame of reference (since you can boost to change the muons observed motion); and the phrase "left-handed component" doesn't make sense to me. What does it really mean for the $W$ boson to only interact with a left-handed muon, if that property is frame-dependent? What does it mean for a muon to have a left-handed "component", and since we originally decided that the muon must be right handed to conserve spin, how does any left-handed component not still violate spin conservation?

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There are literally dozens of questions on this site dealing with chirality (Lorentz-invariant, but violated by masses) and helicity (conserved, but not Lorentz invariant). They are very hard to contrast and relate without math and discussion of properties of the Dirac equation. Probably impossible, but let me give you a trail map of the math to study and understand.

  • In the decay frame of the spinless pion, the two decay products come out back-to-back. By conservation of angular momentum, the spins of them have to be opposite, so their helicities the same: both negative or both positive.

  • The $W^-$ intermediating the decay decays to a left-chiral muon and a right-chiral antineutrino.

  • Helicity and chirality are associated, but not identical. Left-chirality is associated with negative helicity (-1/2) and right chirality with positive 1/2 helicity, mostly. By comparing the corresponding spinor components, you must see that this association is violated/reversed by small components proportional to m/E. This is the heart of you question, really, and it is all math. The role of the mass term in the Dirac equation is to mix left and right chirality states, so, then, convert one into the other.

  • The above association is strongest for the smallest mass, in this case the antineutrino's, so the right chiral antineutrino must have positive helicity. Thus, by above, the muon must also have positive helicity, despite its left chirality, the disfavored one from above, enhanced by the muon mass. This is why the decay channel is $\mu \bar \nu$ and not $e\bar\nu$: because the the muon's high mass violates the decay-suppressing association more than the electron's low mass. (If you wanted a mental-fantasy mnemonic of this, you might dream of the muon born left-chiral off the W, but converted to a right-chiral one through the mass, which is then predominantly of positive helicity.)

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  • $\begingroup$ So helicity is essentially a function of both chirality and mass/energy (and I assume potentially some other stuff), hence higher mass gives a larger disconnect between chirality and helicity? I previously understood helicity to simply be the projection of spin onto momentum - how does this change sign from a mass change, or is this definition of helicity simplified/incomplete? $\endgroup$
    – Vedvart1
    Apr 19, 2022 at 0:52
  • $\begingroup$ Yes, your definition is right, and also the higher the mass, the bigger the disconnect: the tilt between helicity and chirality. I personally can only understand this by looking at the spinor solutions of the Dirac equation. $\endgroup$ Apr 19, 2022 at 0:55
  • $\begingroup$ Wait, so, with helicity being the projection of spin onto momentum, we have $h = s \cdot \hat{p}$. To my understanding, larger mass does not change either $s$ or $\hat{p}$; so how does larger mass change $h$? $\endgroup$
    – Vedvart1
    Apr 19, 2022 at 1:05
  • $\begingroup$ Larger mass does not change the helicity. It misaligns the chirality off the associated helicity! $\endgroup$ Apr 19, 2022 at 2:28
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It sounds like you're at about the same stage in your education as me, so this is how I was taught to think about it:

Unlike chirality, helicity is not Lorentz invariant for particles with mass. Helicity is based upon whether or not the particle's intrinsic spin aligns with it's linear momentum vector (direction of travel); because of this, for a particle with mass, there exists a frame which moves faster than the particle, accessible via an allowed Lorentz boost, as the particle cannot travel at the speed of light. In this frame, the helicity flips; you now see the particle move backwards relative to you, yet it's spin still points towards you. This is why for a massless particle LH helicity is equivalent to LH chirality; you cannot boost to a frame that swaps it's helicity as it is travelling at the speed of light.

This is also why more massive decay products experience less helicity suppression - for two body decay from a stationary parent particle, the energy in the decay process is fixed, thus the momenta of the decay products are also fixed; however the velocity is equal to the momentum divided by the mass so heavier products will inherit less velocity, and there will be a greater space of accessible frames in which the "wrong" helicity becomes the "correct" helicity. This is why the decay width of X to an electron and anti neutrino is far more suppressed than X to a muon and anti neutrino, which in turn is more suppressed than X to a Taon and anti neutrino. The more massive the decay product, the less helicity suppression it encounters.

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