# Why does the weak force distinguish left and right handedness?

I'm wondering why the weak interaction only affects left-handed particles (and right-handed antiparticles).

Before someone says "because thats just the way nature is" :-), let me explain what I find needs an explanation:

In the limit of massless fermions, chirality (handedness) becomes helicity $(\vec S \cdot \hat p)$. Now, helicity is a property of the state of motion of an object in space. It is pretty unobvious to me how the internal symmetry $SU(2) \times U(1)$ would "know" about it, and be able to distinguish the two different helicity states of motion.

On a more technical level, IIRC, left and right handed spinors are distinguished by their transformation properties under certain space-time transformations, and are defined independent of any internal symmetry. If we want to get the observed V-A / parity violating behavior, we have to plug in a factor of $(1 - \gamma^5)$ explicitly into the Lagrangian.

Is there any reason this has to be like this? Why is there no force coupling only to right handed particles? Why is there no $(1 + \gamma^5)$ term? Maybe it exists at a more fundamental level, but this symmetry is broken?

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Er..."why" is sometimes tough to answer when you are asking about a plain observational fact. At least until you have a more fundamental theory. –  dmckee Jul 5 '11 at 18:29
Well, of course :-). There are some observations that one just takes as facts of nature, but sometimes one can try to seek deeper understanding by trying to explain them. The fact that all planets are (almost) in the same plane can be taken for granted, or it can tell us something about solar system formation. Or, we can just note that elementary particles have masses and put them in our equations (and live with an inconsistent theory, but nature doesn't necessary care about mathematical elegance), or we can reason about EWSB and the Higgs mechanism. –  jdm Jul 6 '11 at 1:20
@jdm: Currently there is no deeper understanding of the breakdown of parity conservation in weak interaction. If you find out, you may well receive a Nobel Prize. –  Siyuan Ren Jul 6 '11 at 3:25
@Karsus Ren: Easier said than done :-) –  jdm Jul 6 '11 at 7:54

I think you are sort of reversing the logic of chirality and helicity in the massless limit. Chirality defines which representation of the lorentz group your Weyl spinors transform in. It doesn't 'become' helicity, helicity 'becomes' chirality in the massless limit. That is, chirality is what it is, and it defines a representation of a group and that can't change. This other thing we have defined called helicity just happens to be the same thing in a particular limit.

Now once you take the massless limit the Weyl fermions are traveling at the speed of light you can no longer boost to a frame that switches the helicity. I think its best to think of a fermion mass term as an interaction in this case and remember that the massive term of a Dirac fermion is a bunch of left and right- handed Weyl guys bumping up into one another along the way. Conversely if you want to talk about a full massive Dirac fermion that travels less than c and you can boost to change the helicity, but that full Dirac fermion isn't the thing carrying weak charge, only a 'piece' of it is.

See this blog post on helicity and chirality.

As far as the left-right symmetry being broken people have certainly built models along these lines but I don't think they have worked out.

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We'll, I know chirality is an instrinsic property and doesn't change, while helicity is frame-dependent. I could have said they become the same, or there is a 1:1 correspondence between them etc.. –  jdm Jul 7 '11 at 5:05
Thanks for the links. The first one is very nice. And actually your last paragraph about left-right-symmetric theories was what I was looking for. –  jdm Jul 7 '11 at 5:10

The explanation is simple--- all particles we can see are chiral, they have only one handedness, because if they had both handedness, they could get a mass, and generically, that mass would be of order of magnitude the Planck mass. We live at energy scales which are teeny compared to the Planck mass, so we can only see massless stuff, so we only see chiral fermions (and gauge bosons).

The right question then is the other way around, if everything is chiral, why then do the electromagnetic and strong interactions not violate parity?

This is because the Higgs mechanism partners up the chiralities into massive Dirac particles at lower energies, and only the W,Z bosons know that they were chiral to begin with. At low energies, you get parity and charge-conjugation as accidental symmetries, because these are symmetries of the low-energy Dirac Lagrangian coupled to the remaining photon and gluons.

As for the neutrinos, a chiral neutrino can have a Majorana mass while only having one chirality, and this is certainly what is going on in nature, since this scheme predicts the mass correctly, and this mass is absurdly small.

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Ron,for example if both left and right fermions live in the same representation of SU(2), why would they have mass of order of Planck scale? –  Newman Jul 18 '12 at 16:18
@Newman: Because there would be a parameter in the Lagrangian with dimension mass giving them mass, and why should it be tuned to something very small? –  Ron Maimon Jul 18 '12 at 16:23
but wouldnt there be chiral symmetry in the massles limit, so that naturalness condition would be satisfied_ –  Newman Jul 19 '12 at 14:26
@Newman: I don't agree that this is the right statement of the naturalness condition. "Chiral symmetry" is not the right condition. "Gauge invariance forbidding mass terms" is the right condition. When you artificially write the 2-spinors as 2 components of the 4-spinors, this looks like chiral symmetry, because this is what decouples the other two components, but this is a notational trick. The things are 2-spinors really. Why should one impose a global chiral symmetry? I agree that if you do impose it, you can make a natural m=0 tuning, meaning that higher order terms will not break it. –  Ron Maimon Jul 19 '12 at 16:21

There's nothing a priori that says it has to be that way and (I believe) neutrino oscillations have now shown that they have a mass which implies both right and left handed neutrinos exist. That said, the lagrangian is still seriously skewed towards the left handed interaction. So it's still very left-handed in a sense but there's nothing that says a coupling to right handed particles is out.

We just don't see a strong coupling experimentally.

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Look up the (as yet unsettled) question of Dirac versus Majorana nature of the neutrinos. Non-zero mass is necessary but not sufficient to require that both handedness are present. –  dmckee Jul 6 '11 at 4:35
@dmckee Good point. It's been a while since I've active in particle phyiscs. I gotta do a refresh on that. –  unclejamil Jul 6 '11 at 16:48
@unclejamil: It is (theoretically) almost certain that the neutrinos do not have a right handed component, and get their mass from mixing with the antiparticle. This is open experimentally, but one would take any bet theoretically against Dirac masses. Although dmckee corrected you, what you wrote is the opposite of what is most likely--- it is most likely that the neutrino is a massive particle with two helicities which is its own antiparticle at nonrelativistic velocities. –  Ron Maimon Jul 19 '12 at 16:24