When a particle spins in the same direction as its momentum, it has right helicity, and left helicity otherwise. Neutrinos, however, have some kind of inherent helicity called chirality. But they can have either helicity. How is chirality different from helicity?
Quantum Field Theory – Understanding the Difference Between Helicity and Chirality
chiralityhelicityparticle-physicsquantum-field-theory
Related Solutions
You are correct that for a massive spinor, helicity is not Lorentz invariant. For a massless spinor, helicity is Lorentz invariant and coincides with chirality. Chirality is always Lorentz invariant.
Helicity defined $$ \hat h = \vec\Sigma \cdot \hat p, $$ commutes with the Hamiltonian, $$ [\hat h, H] = 0, $$ but is clearly not Lorentz invariant, because it contains a dot product of a three-momentum.
Chirality defined $$ \gamma_5 = i\gamma_0 \ldots \gamma_3, $$ is Lorentz invariant, but does not commute with the Hamiltonian, $$ [\gamma_5, H] \propto m $$ because a mass term mixes chirality, $m\bar\psi_L\psi_R$. If $m=0$, you can show from the massless Dirac equation that $\gamma_5 = \hat h$ when acting on a spinor.
Your second answer is closest to the truth:
The weak interaction couples only with left chiral spinors and is not frame/observer dependent.
A left chiral spinor can be written $$ \psi_L = \frac12 (1+\gamma_5) \psi. $$ If $m=0$, the left and right chiral parts of a spinor are independent. They obey separate Dirac equations.
If $m\neq0$, the mass states $\psi$, $$ m(\bar\psi_R \psi_L + \bar\psi_L \psi_R) = m\bar\psi\psi\\ \psi = \psi_L + \psi_R $$ are not equal to the interaction states, $\psi_L$ and $\psi_R$. There is a single Dirac equation for $\psi$ that is not separable into two equations of motion (one for $\psi_R$ and one for $\psi_L$).
If an electron, say, is propagating freely, it is a mass eigenstate, with both left and right chiral parts propagating.
In theory, it is possible to measure chirality using the exact method that you mentioned. If you have an electron and you are unsure its chirality, you can send it through a dense bucket of $W$ bosons and see if it interacts with them. If it did, then you know that you had a left handed electron.
There is, however, an important issue with this idea (and no its not the issue of figuring out how to produce a bucket of $W$ bosons). The problem is that the electron can change chirality and back again so even if you measured its chirality one moment, it doesn't mean that it had a same chirality a little time earlier.
One way to avoid this problem, is to accelerate the electrons. In this case the time it takes an electron to flip its chirality is $\sim 1/E $. This is easily seen using ordinary QM: \begin{equation} P(e_L \rightarrow e_R) = \left| \left\langle e _L \right| \exp \left( - i H t \right) \left| e _R \right\rangle \right| ^2 = \sin^2(- E t ) \end{equation} where $ H = \gamma M = \gamma \left( \begin{array}{cc} 0 & m _e \\ m _e & 0 \end{array} \right) $, $\left| e _L \right\rangle = ( 1 , 0 ) ^T , \left| e _R \right\rangle = ( 0, 1 ) ^T$, and $ E = \gamma m _e $.
If $1/E\gg $ the time it takes the electrons to pass through the apparatus then you can confidently say that the electrons you started with are the ones you measured and so you can measure its chirality.
Notice that this idea relies on going to the high energy limit, where chirality=helicity, so it may not be exactly what you were looking for. But while the two quantum numbers are seemingly completely different, our ability to measure chirality is closely linked to helicity.
While the above idea was quite hypothetical, you could in principle produce high energy chiral eigenstate electrons and collide them with positrons. Then you could measure the products of the collisions and infer if the initial states were indeed left or right chiral.
Lastly, the property of neutrino masses does not change whether or not (hypothetical) right handed neutrinos can interact weakly. Due to the necessary quantum numbers of the right handed neutrinos they would have to be standard model singlets. For this reason, they cannot interact with the weak force.
Best Answer
At first glance, chirality and helicity seem to have no relationship to each other. Helicity, as you said, is whether the spin is aligned or anti aligned with the momentum. Chirality is like your left hand versus your right hand. Its just a property that makes them different than each other, but in a way that is reversed through a mirror imaging - your left hand looks just like your right hand if you look at it in a mirror and vice-versa. If you do out the math though, you find out that they are linked. Helicity is not an inherent property of a particle because of relativity. Suppose you have some massive particle with spin. In one frame the momentum could be aligned with the spin, but you could just boost to a frame where the momentum was pointing the other direction (boost meaning looking from a frame moving with respect to the original frame). But if the particle is massless, it will travel at the speed of light, and so you can't boost past it. So you can't flip its helicity by changing frames. In this case, if it is "chiral right-handed", it will have right-handed helicity. If it is "chiral left-handed", it will have left-handed helicity. So chirality in the end has something to do with the natural helicity in the massless limit.
Note that chirality is not just a property of neutrinos. It is important for neutrinos because it is not known whether both chiralities exist. It is possible that only left-handed neutrinos (and only right-handed antineutrinos) exist.