Statements in the chapter 18.3 of Lie Algebras in Particle Physics by Georgi

Question

In chapter 18.3 of Lie Algebras in Particle Physics by Georgi, he said (see the bottom, which seemed confusing or wrong):

If the $SU(2)\times U(1)$ symmetry remained unbroken, the quarks and electrons would have to be massless particles

1. Shouldn't the $SU(2)$ be confined thus the particles still become massive? (at least the left-handed particles)

...because the weak interactions treat their left-handed and right-handed helicity components differently, which is consistent with relativity only for massless particles.

2. What does this sentence try to emphasize or clarify? (We now know that the weak interactions treat their left-handed and right-handed helicity components differently, but both the left-handed and right-handed helicity components can be paired up to be massive not massless.)

Here is the text:

There is something peculiar going on here. The $SU(2)\times U(1)$ cannot really be a symmetry. If it were, the weak interactions would have long range, like the electromagnetic interactions. Instead, the weak interactions have very short range and their force particles, the $W$ and $Z$, are massive. Furthermore, if the $SU(2)\times U(1)$ symmetry remained unbroken, the quarks and the electron would have to be massless particles, because the weak interactions treat their left-handed and right-handed components differently, which is consistent with relativity only for massless particles. Some new physics gives mass to the quarks and leptons, and to all but one linear combination (the photon) of $R_3$ and $X$, without destroying the consistency of the theory. What is this physics?

Answer 1

Ugh... if you are cool with dismissing potential Majorana masses violating lepton number, then his playful hypothetical does make sense:

1. In a mass term you connect left-chiral fermions with right-chiral ones. Since SU(2) is chiral, it wouldn't break chiral symmetry like QCD, generating masses, and it certainly would not couple the left-chiral fermions to unconfined right chiral ones, the coupling we have in our world underlying masses. Think of neutrino physics before the detection of neutrino masses: people did not even expect a right handed sterile neutrino to exist, since, they thought, the active left-chiral neutrinos would be massless.

2. Unless you know the breathtaking answer, the hypothetical is not that meaningful. Since the weak-active fermions are left-chiral, and since they mix with the right-chiral weak-sterile ones via the mass term, you can't have a relativistically consistent mass term, and gauge invariance (which he assumes), unless you "cheat" through the Higgs mechanism, the second job of the Higgs. You manage by Higgs-saturating the left-doublet fermions into gauge-invariant singlets.

You need not overthink it: you cannot have gauge invariant and relativistically invariant masses, except through this Higgs mechanism.

Answer 2

I'll talk just about leptons since I think this is enough to discuss this issue. We build our model with one ${\rm SU}_L(2)\times{\rm U}_Y(1)$ symmetry. In this model the leptons are represented by a left chiral fermion $\ell_L$ and a right chiral fermion $\ell_R$.

The $\ell_L$ forms an ${\rm SU}_{L}(2)$ doublet with its corresponding neutrino $\nu^\ell$. We may call this doublet $$L_\ell = \begin{pmatrix}\nu^\ell \\ \ell_L\end{pmatrix}.$$

On the other hand $\ell_R$ is an ${\rm SU}_L(2)$ singlet. From this we may read off the ${\rm SU}_L(2)$ isospin quantum numbers $I_3$. Indeed, $\nu^\ell$ has $I_3$ quantum number $1/2$ while $\ell_L$ has $I_3$ quantum number $-1/2$ and $\ell_R$ has $I_3$ quantum number $0$.

Now we have the hypercharge $Y$. It is attributed with the Gell-Mann Nishijima relation $$Q=I_3+\dfrac{Y}{2}.$$

From this, since the lepton must have electric charge $Q = 1$, since $\ell_L$ has $I_3$ equal to $1/2$ it must have $Y=1$. On the other hand since $\ell_R$ has zero $I_3$ it must have $Y=2$. Finally since $\nu^\ell$ has charge zero and $I_3$ equal to $1/2$ it must have $Y = 1$.

Now a mass term for the fermion couples $\ell_L$ and $\ell_R$ in terms of the form $\overline{\ell}_L\ell_R$ and $\overline{\ell}_R\ell_L$. Now observe:

1. $\ell_L$ is one component of an ${\rm SU}_L(2)$ doublet while $\ell_R$ is an ${\rm SU}_L(2)$ singlet. Therefore the mass terms can never be ${\rm SU}_L(2)$ singlets.

2. $\ell_L$ has hyperchage $Y = 1$ and ${\overline{\ell}}_R$ has hypercharge $Y=-2$. Therefore $\overline{\ell}_R\ell_L$ has hypercharge $Y=-1$ and is also not invariant under ${\rm U}_Y(1)$. Likewise ${\overline{\ell}}_L$ has hypercharge $Y =-1$ and $\ell_R$ has hypercharge $Y=2$, therefore $\overline{\ell}_L\ell_R$ has hypercharge $Y = 1$ and is also not invariant under ${\rm U}_Y(1)$.

In summary a mass term is simply not compatible with the ${\rm SU}_L(2)\times {\rm U}_Y(1)$ symmetry in the sense that they lead to a Lagrangian which is not invaraint under this symmetry.

Answer 3

In the Wigner classification, you have four classes: the homogeneous class (which the vacuum belongs to), the bradyons, the luxons and the tachyons. For the homogeneous class, I don't think helicity can even be defined. For the other classes, is helicity not an invariant, except for a subclass of the luxons - which I've termed the helical luxons. The photon belongs to this class.

The weak force, for all intents and purposes, has left helicity as its charge for matter, right helicity for anti-matter, which means - first and foremost - that helicity is an invariant property for weak charges. Ipso facto: nothing has weak charge except helical luxons! Period. Full stop.

You can pair left and right helical states for bradyons, like you say. But for bradyons, handedness is not invariant. A bradyon that is left helical to a stationary observer is right helical to an observer who overtakes it and vice versa.

Georgi's comment is misleading in the following way: he says were but for symmetry breaking, the "massive" fundamental particles would be massless. What's misleading is that they are massless, period. The Higgs doesn't give them mass, but merely the appearance of mass by driving the back-and-forth left-right swapping that characterizes the behavior of a massive Dirac fermion. Essentially, it drives an almost Brownian-motion like zig-zagging whose average motion is that of a sub-light particle - i.e. a bradyon - for particles that are actually still light speed. So, in the Dirac part of the Lagrangian $\bar{ψ} \left(γ^μ \left(iħ\overleftrightarrow{∂_μ} + e_a A^a_μ\right) - mc\right) ψ$, the $mc$ part becomes a field $φ$ + a coupling $g$: $mc → gφ$, while the actual $m$ is zero. The left-right coupling is the action of $φ$ rather than the effect of $m$. The extra term $\bar{ψ} gφ ψ$, that replaces $\bar{ψ} mc ψ$, is a little more complicated than that - but not by much. You can arrange it to look just like that with the right kind of syntactic gynmastics.