# The electron and neutrino spin

Can you please explain what is intended when they say that the spin of a neutrino is left-handed and equal to -1/2?

Does it mean its angular momentum is equal in direction and strength to the spin of the electron ( $h /4\pi$), even though its mass is about one million times smaller? If so, is there an explanation for that? Is there any experimental direct evidence of that magnitude?

Also, since the electron g-factor is 2, does that mean that its value is $h /2 \pi$ , equal to the orbital angular momentum in the hydrogen atom?

• Neutrinos travel at or almost at, the speed of light, so you cannot exceed their speed, look back and see a right handed spin, as you can with a massive particle such as an electron. From the picture you have used I guess you have already read hyperphysics.phy-astr.gsu.edu/hbase/particles/neutrino3.html but just in case... – user108787 Jun 4 '16 at 7:46
• Hi ally, its a while since I studied this so rather than give you a wrong answer, I would prefer to leave to someone better than me, best of luck with it though. – user108787 Jun 4 '16 at 7:55
• Of course you can exceed their speed, look back, and see a right-handed spin. It just takes a lot of acceleration to do so. – Peter Shor Jun 4 '16 at 12:27

My question is about the spin of the neutrino, and the evidence (or speculative arguing) for its magnitude which is not proportionate to its mass.

You are dealing with "things" that are supposed to be point particles, so they should not, in classical terms, have a mass in the first place. The math takes over from common sense notions on this scale, and the spin of the neutrino has to be 1/2 to conserve quantum numbers. The mass is the link between the classical world, in which we expect it to be a certain value, and the quantum world, in which the particles are not real in a way we understand.

I hopefully will not annoy you because you already know this. Because "spin" is an analogy, with absolutely nothing to do with ordinary soccer ball type spin, you could have asked the same question about the top quark, which has a huge mass but still has spin angular momentum of 1/2, because that is what we define it to have.

So comparing the mass of one point object (which classically should not have a mass) with another point object with lower mass, and expecting a different angular momentum will not make sense in ordinary terms, but this is not the ordinary world you are asking about.

• @ally Ally, please read this link below and post the above as a separate question, the g in the link is 2.00232, not just 2, because of the Lamb Shift. Also, search the site for a duplicate, its a separate question so needs a separate post and I will do my best to answer it. Thanks hyperphysics.phy-astr.gsu.edu/hbase/spin.html#c4 – user108787 Jun 5 '16 at 8:31
• @a!ly A separate post gives me time to think about it :) and you a better chance of proper (professional) answer, – user108787 Jun 5 '16 at 8:45

The spin of a fundamental fermionic particle always has the absolute value $\frac{\hbar}{2}$. This does not relate to its orbital angular momentum, it's just what the spin of a fundamental fermion is.

It now turns out that, for massless particles, there is the notion of helicity (see also What is polarisation, spin, helicity, chirality and parity?), which is the projection (i.e. the relative direction) of spin and momentum.

In your picture, someone decided to orient the axis along which one component of the spin is measured parallel to the particles momentum, so that $+1/2$ spin corresponds to the spin being parallel to the linear momentum, and $-1/2$ to them being anti-parallel. In situations where "neutrinos are massless" is a good approximation, this picture is Lorentz-invariant and indeed a good description for the otherwise rather technical notion of chirality.

The electron cannot correspond to an anti-neutrino because the latter has zero electrical charge. Spin is only one of many quantum numbers.

• But neutrinos aren't really massless. – Peter Shor Jun 4 '16 at 12:28
• @PeterShor: They aren't, and their handedness is properly explained as chirality, not helicity. But we can't draw nice pictures explaining chirality because it's a technical property based on a decomposition of the Dirac spinor into two Weyl spinors, so many people opt to explain the handedness with the helicity pictures instead, since for massless fermions the two notions coincide. – ACuriousMind Jun 4 '16 at 12:50