A Is the "massless quark" hypothesis ever useful?

dextercioby

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It all started with a debate on another site. Someone told me that "massless quarks" are useful (in theory I presume).
So here it goes. All known theorized quarks are found from scattering experiments to be massive, i.e. have a rest mass. Someone with more knowledge than me told me that assuming (against logic, from my point of view) quarks to be massless, i.e. just like a gas of photons, is theoretically useful. It is true? If so, why can't one simply assume that quarks are the way they are, massive, and get better theoretical results??

P.S. This is not about the Higgs mechanism.
 
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Vanadium 50

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Are you saying "why is this often a good approximation?": Same reason we talk about frictionless planes, massless pulleys, etc.
 

dextercioby

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No, those are idealizations for sake of teaching students. A different concept. Here is simply: assume mass of whatever quark equals zero. Do whatever calculations and get a physical relevant result (which you can even measure). Why assume mass equal zero in the first place?
 

Vanadium 50

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Because it makes the calculation easier. Why not use GR to solve incline plane problems?
 

vanhees71

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That's a very complicated issue. The light-quark (u, d, in some sense also s) masses are nowadays defined through lattice QCD:


The reason of the difficulty to define quark masses is of course that they are, together with gluons, "confined" in hadrons, i.e., there are no asymptotic free quarks, which could be defined as definite mass eigenstates and their mass then measured in experiments.

Now, however, the massless-quark limit, is of great importance to understand hadrons and to build effective quantum-field theoretical models, describing them. From first principles we can calculate their masses quite accurately using lattice QCD, but that's more or less it. To describe dynamical hadrons, we need effective field theories as an approximation of QCD in the low-energy regime, where perturbation theory is inapplicable, because the strong coupling constant becomes large in this regime ("asymptotic freedom" of QCD).

Now nature comes to rescue! QCD does not only have its defining fundamental local color-gauge symmetry, but in the light-quark sector and approximate chiral symmetry. In the massless-light-quark limit this symmetry becomes exact, and one can build effective hadronic QFTs by assuming chiral symmetry as an approximation. Now chiral symmetry would on the first glance imply that for each hadron of definite parity (and parity is a good quantum number as long as we can neglect weak interactions, which break parity symmetry, but dealing only with the strong interaction, parity is conserved and thus space reflection a good symmetry) there must be a "chiral partner" hadron with the opposite parity and the same mass. This is obviously not true! E.g., there's no chiral partner with the same mass of the protons and neutrons.

On the other hand, as the determination of the "current-quark masses" as explained in the above quoted summary article of the particle-date group indicates that the light u and d quarks have really small masses (a few MeV) on the typical hadronic mass scale (of about a GeV).

The solution of this apparent paradox has been found in the early 60ies, i.e., well before QCD has been known as the underlying fundamental theory of the strong interaction: The chiral symmetry is spontaneously broken, i.e., the ground state (vacuum state) is not chirally symmetric. Being a global symmetry then chiral symmetry implies that there must be some massless scalar or pseudoscalar bosons (the Nambu-Goldstone modes of broken chiral symmetry). Indeed, there are the pseudoscalar pions, which have a quite small mass of around 140 MeV. That they are not strictly massless is due to the explicit breaking of chiral symmetry.

Today the breaking of chiral symmetry is understood as the formation of a socalled quark condensate in the vacuum state of full interacting QCD, i.e., the order parameter of this symmetry ##\langle \bar{\psi} \psi \rangle \neq 0## (with ##\psi## the light-quark flavor doublet). This is a scalar field, and the usual association of the chiral partner of the pions is the ##\sigma## meson, which however is a very broad "two-pion resonance". Nevertheless it's in accord with chiral symmetry.

The explicit chiral-symmetry breaking by the quark masses can treated as a perturbation, and this works amazingly well. An entire "industry" in theoretical hadron physics has been built, known as "chiral perturbation theory", and it is used successfully on many subjects related with hadron physics.
 

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