SW Engr here educating myself on particle physics, apologies in advance for any phrasing errors. I tried the google machine to no avail.

Q: If only left handed matter fermions experience the weak interaction, does this mean that all nucleons must have at least one of their doublet quarks be left handed, a down quark in a neutron or an up quark in a proton? Ditto for the inverse as applies to right handed anti-fermions.

Is there a weak interaction-friendly chirality constraint like this for quarks in nucleons?

If so, did the chirality constraint get sorted out when quarks were first confined to nucleons and are there thoughts on a mechanism?

Or is there no constraint and there exist nucleons that do not have quarks of the proper chirality for weak interaction that are thus immune and will never decay?

Thanks in advance,



1 Answer 1


OK, here is a schematic mathless picture that should fix your visualization.

To start with, at very short distances, far shorter than a fermi, the quarks have small masses of the order of 2.5 MeV for the u and 5 MeV for the d, the up and down quarks of the nucleons. Masses convert left-chiral quarks to right-handed quarks and vice versa. Both types exist inside the nucleons and flip-flop into each other incessantly, and so you can always catch a quark left-handed and couple it to the weak interaction mediators and have a weak decay/interaction, such as neutron decay.

In an imaginary, notional world, you might have zero quark masses, in which case the right- and left-chiral quarks would not couple to each other (yet); they'd only couple to gluons, the mediators of the strong interactions, that "glue" them together into hadrons, here nucleons. The gluons also produce a googzillion of virtual quark-antiquark pairs that give their nucleons their characteristic features, and, more importantly, they "confine" the quarks to a "large" area, a blob of diameter about a fermi. The quarks cannot escape this confinement radius.

Equally importantly, the gluons (near the confinement radius) generate a "huge" mass for such quarks, of the order of 330 MeV, about a third of the mass of the nucleon, in a subtle phenomenon called "chiral symmetry breaking". This is a strictly strong interaction phenomenon, the one that really gives the nucleons their mass of about 1000 MeV, thrice the mass of these suddenly fat quarks (called "constituent" ones), and, intriguingly, does not unfold very differently whether the original ("current") quarks started out massless or with the small, few MeV, masses mentioned.

  • $\begingroup$ Thanks for taking the time to explain, this is exactly what I was looking for. $\endgroup$
    – IknoweD
    Mar 6, 2020 at 4:31

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