Let’s think of the double-charmed omega baryon as a kind of second-generation proton, and the single-charmed omega as a second-generation neutron. I made up the names “puotron” and “muotron” for these particles.

The proton and neutron are made up of first-generation quarks (up-up-down and up-down-down respectively) and a process of electron capture creates a neutron and an electron neutrino from a proton and an electron

proton + electron -> neutron + electron neutrino

So I was thinking about whether this process also happens in the second-generation baryons. Specifically, if a

  • double charmed omega baryon (charm-charm-strange / second generation proton / puotron) and muon

can produce a

  • charmed omega baryon (charm-strange-strange / second generation neutron /muotron) and a muon neutrino.

more simply: puotron + muon =? muotron + muon neutrino

Thanks in advance for any feedback!

  • $\begingroup$ Note p + en + ν is only necessary because the proton is "accidentally" lighter than the neutron, but this does not obtain for the unobserved doubly charmed Ω. So your "puotron" may decay to the singly charmed "muotron" + a positron (or a μ+) and a neutrino, in principle. Of course, your "muotron" lifetime is just $2.7 ~~10^{-13}$s, !, so freakishly hard to detect with a missing neutrino... $\endgroup$ Nov 13, 2021 at 17:52

1 Answer 1


The proton and the neutron behave like two isospin projections of the same particle, a “nucleon.” The evidence for isospin symmetry comes from nuclear physics, where total isospin appears to be (approximately) a good quantum number. The clearest places to see this are in the excited states of mirror nuclei, where the angular momenta and energies of the excited states are similar even though the nuclear charge is different.

Isospin symmetry is a useful approximation when

  1. the strong interaction is more important than electrical force, so we can ignore the proton’s charge

  2. the nucleon binding energies are large compared to the neutron-proton mass difference.

We have no reason to suspect a similar isospin-like symmetry between the charm and strange quarks, because the charm quark is substantially heavier than the strange quark. The two charge states of the nucleon (that is, the proton and neutron) differ in mass by only about one part per thousand, about 1 MeV out of 1 GeV. The neutral lambda baryon, with quark content $\Lambda^0 = \rm uds$, is only about 10% heavier than the nucleon. But the charmed lambda is twice as massive than the neutral lambda, and the single-charmed omega (your “muotron”) is likewise a full GeV heavier than its triple-strange counterpart.

For particles made only from the up and down quarks, and to some extent for particles including strange quarks, the lion’s share of the mass comes from the virtual quark-gluon sea rather than from the valence quarks. But for charm and beyond, the bare quark mass is unavoidably meaningful.

You sometimes read papers where the basic idea of isospin (an $SU(2)$ symmetry) is extended to $SU(3)$ to include the strange quark. As far as I understand, this approach is useful in a narrow range of energies where you can treat the strange quark as “massless” but the charm quark as “massive.” But if you want your flavor symmetry to include transformations involving the 1 GeV charm quark, you will be remiss to neglect the 5 GeV bottom quark.

The specific interaction you ask about,

$$ \Omega_{cc} + \mu \to \Omega_c + \nu_\mu $$

is allowed, but impossible to realize. The only double-charmed baryon to have been observed at all thus far is the $\Xi^{++}_{cc} = \rm ucc$, and the status of its decays is “seen.” I wouldn’t hold my breath on the detection of any interaction between an $\Omega_{cc}$ and another unstable particle like a muon; we’ll be limited to observing its decay products.

But furthermore, there is no special connection between the second generation of quarks and the second generation of leptons. The strange $K^+$ meson is more likely to decay to a muon than to an electron, but so is the first-generation $\pi^+$ meson: the reason for that has to do with the weak interaction’s preference for left-handed particles and right-handed antiparticles. With a third particle to share the angular momentum, the two decays

\begin{align} K^+ &\to \pi^0 e^+ \nu_e & 5.1\% &\text{ branching ratio}\\ K^+ &\to \pi^0 \mu^+ \nu_\mu & 3.4\% &\text{ b.r.} \end{align}

are consistent with lepton universality.

  • $\begingroup$ thank you very much! this explains it perfectly. $\endgroup$
    – Nazuid
    Nov 13, 2021 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.