How do fermions acquire mass as opposed to gauge bosons?

mmaa

Hello, if someone could enlighten me I'd be most grateful.

Also, if anybody could point me in the direction of some really good free resources that would be great too. Thanks.

bapowell

The fermions can obtain masses in a gauge-invariant way via the Higgs mechanism, just like the gauge bosons. One way to implement this is to consider a Yukawa coupling between the fermion, [itex]\psi[/itex], and the Higgs field, [itex]\phi[/itex] in the Lagrangian:

[tex]\mathcal{L}_{\rm Yukawa} \sim g\bar{\psi}\phi\psi[/tex].

When symmetry breaking occurs and the Higgs takes on its vacuum expectation value, terms of the form [itex]g \bar{\psi}v\psi[/itex] develop, where [itex]v[/itex] is the vev. These are mass terms with [itex]m\sim gv[/itex].

Not sure about specific references, but there should be plenty of information on the internet.

JustinLevy

Since the vev is just one value, the only difference in the terms for each fermion would then be is "coupling constant" with the Higgs field. Why isn't this quantized like the coupling with the electric field -> quantized electric charge or strong nuclear force -> quantized color charge?

humanino

The gauge quantum numbers such as electric charges are quantized because the gauge groups are compact. In contradistinction, the Poincare group is not compact, so the mass (as a Casimir invariant classifying the Lie representation) cannot have a discrete spectrum.

fzero

The Yukawa couplings are inputs to the theory, we can't do anything about the fact that flavor symmetries are badly broken in nature. However for each fermion, there is a global flavor charge, one for each quark and lepton (associated to the U(1) rotation on the fermion and its antiparticle). This charge is quantized naturally in all bound states.

tom.stoer

I don't think it's so easy. One must distinguish between the coupling constant e, g, ... in QED, QCD, ... which could have any value, and the charge Q, Q^a, ... as qm generators of U(1), SU(n), .... The latter one is quantized in the sense of the first Casimir Q^aQ^a. But how is this related to the coupling constant? The charge operator in QCD is something like

[tex]Q^a = \int d^3x\, g\,\bar{\psi}_i (T^a)_{ik}\psi_k[/tex]

The Casimir operator has a discrete spectrum, but still g is an arbitrary multiplicative constant which is not "quantized"

humanino

Tom, I do not see where the difficulty you are raising relates to my comment. I agree that there is a distinction between coupling constant and charge. You agree that charges in compact Lie groups are discrete, which is all I was saying. I did not say that the quantization of the charge was related to the coupling constant (vev of the Higgs).

Neither I nor the standard model offer an explanation for what the fermion masses are unfortunately. They just are what they are in the standard model, a set of ad-hoc numbers. JustinLevy was asking why those numbers are not quantized as say electric charges. The quantization of electric charges stem from the compactness of U(1).

tom.stoer

My guess was that JustinLevy didn't see the distinction between "charge e" and "charge Q". That was the reason for my comment. Sorry!

humanino

Thank you for the clarification, that is an important distinction. I now understand the difficulty you are raising. After re-reading JustinLevy's post, I realize this might indeed be the confusion.

tom.stoer

Then one should add that in the context of the Higgs phenomenon applied to fermions Q doesn't play a role at all, only different g's.

mmaa

Wow, thanks everybody. This forum is great!

hch71

Actually, as far as i know there is no reason in U(1) for the charge to be quantized. If you postulate magnetic monopoles, you get the Dirac quantization condition, but if not there is no restriction.

tom.stoer

Thinking about that I can't find a good reason that compactness alone is sufficient and whether electric charge Q is necessarily quantized.

Here you are confusing e and Q. The Dirac monopole forces e to be quantized, but humanino is talking about Q being quantized (better: the spectrum of the operator Q).

As I tried to explain in my previous post one must distinguish between Q (charge of a physical state) and e (coupling constant) which is not directly observable.

Lapidus

I rembember learning that most of the mass of protons and neutrons does not come from the Higgs mechanism, but from the binding energy that holds the three quarks together.

Is that correct? And could someone explain a bit further what goes on there..

thank you

tom.stoer

Yes, this is correct.

In deep inelastic scattering one probes the "high energy sector" of QCD. Here one finds "asymptotic free" quarks. These elementary quarks only have a mass of a few MeV, which means that be taking three times their mass cannot explain nucleon mass at the GeV scale. (In the old constituent quark model before invention of QCD one used three quarks and nothing else; these constituent quarks are not fundamental degrees of freedom but something like "dressed" quarks with surrounding "virtual" quark and gluon "clouds"; please don't take this too literally).

Unfortunately I do not know a simple picture that is able to describe - based on nearly massless elementary quarks plus QCD interaction - how the nucleon masses do arise. One could try to visualize the nucleon as a lump of energy with quark and gluon chromo-electric and chromo-magnetic fields forming a nucleon.

hch71

That is correct. The mass of a quark is not a experimentally accessible quantity, but the light (up and down) quarks have a mass of ~2-5MeV. The proton on the other hand has a mass of ~940MeV. So the contribution of the individual quarks' masses is very small.

The rest of the mass of the proton is in fact the kinetic and binding energy of the quarks. Quarks interact via the strong interaction and this strong interaction has the peculiar feature that it does not allow any particle that carries a strong charge (such as a quark) to exist alone. So quarks under normal circumstances only appear in combinations that are neutral to the strong force. One such combination is the proton. This phenomenon is called quark confinement.

Inside the proton, quarks are not at rest. They strongly interact and consequently there is a lot of energy. It is this energy that - via the energy mass equivalence - we see as mass.

In fact, there is one theory, called technicolor, that asserts that the mechanism for electroweak symmetry breaking is very similar to this mechanism of the strong force.
In this scenario all particles gain mass in a similar way as the proton through the strong force. Hence the Higgs would be a composite particle such as a hadron in the strong force.

JustinLevy

Yes I was making that confusion. Is there a technical name for these, to help distinguish them? I'm realizing now in retrospect that a gradstudent and I were severely misunderstanding each other when discussing this a month ago. (He seemed to talk as if there were only the "Q" and I thought there were only the "e", and only confused each other. Neat to see that we were both missing something very important.)

I assume the same applies for the other forces as well.
So the standard model can't explain why the "e" charges for electromagnetism or color force are quantized? (Are the 'charges' for the weak force also 'quantized' in the sense that 'e' charges are? ... or does the symmetry breaking ruin this.)