Now I can understand its unlikely to see a $W$ or $Z$ boson decay into a pair of quarks that involve a top or anti-top quark simply because of the amount of energy it would require, but find it well within the realm of a theoretical possibility unbound by practical limitations considering that any particle can have enough velocity to obtain any energy level.

Wikipedia states that the branching ratios can be derived theoretically from coupling constants, so it appears that somehow the $W$ and $Z$ boson energy levels are limited. Im thinking maybe there are some assumptions that are taken to derive the branching ratios that include assuming the energy level of these bosons thereby excluding the top quark from the branching ratios.

Can someone confirm that it is or is not possible for a top or anti-top quark from being involved in a $W$ or $Z$ boson decay by stating what theoretical mechanism allows or prevents this?

Note: It sounds like since the reference frame of the boson and the decay products are the same then the total mass-energy of the products is limited to the rest mass of the Boson.

  • $\begingroup$ The Z boson does have quark antiquark decay modes listed right on the Particle data group information sheet. Also it has more then enough mass to decay to at least the 5 lighter quarks as pairs of particles and antiparticles $\endgroup$
    – Triatticus
    Mar 17, 2021 at 0:51
  • $\begingroup$ @Triatticus, I think the W boson decay would be something like a top and bottom (bottom or strange or down quark paired with a top) $\endgroup$
    – Jason
    Mar 17, 2021 at 0:54

1 Answer 1


The top quark mass, $m_\text{top} = 173\,\mathrm{GeV}/c^2$, is much heavier than the masses of either the charged ($m_W = 80\,\mathrm{GeV}/c^2$) or neutral ($m_Z = 91\,\mathrm{GeV}/c^2$) weak boson. So a weak boson in its rest frame does not have enough energy to decay to a final state involving top quarks.

Any particle can have enough velocity to obtain any energy

Give it any velocity you want. I'll run alongside the weak boson and observe the decay in its rest frame. (I am very fast.) There is not enough energy for the decay to happen in my reference frame, so you don't observe it either.

It appears that somehow the Z and W boson energy levels are limited.

You seem to be thinking that perhaps the $Z$ and $W$ have internal degrees of freedom and can store energy in an excited state. That's a good idea: we see that atoms and nuclei can store energy internally. Even nucleons will vibrate if you hit them hard enough, giving rise the the baryon spectrum. However, the $Z$ and $W$ (along with the quarks and the leptons) are structureless, fundamental particles. So far as we know, they don't have any internal degrees of freedom to excite.

  • $\begingroup$ not for the OP, but for general readers maybe the possibility that in future high energy colliders a virtual Z to t t_bar is possible ? $\endgroup$
    – anna v
    Mar 17, 2021 at 4:58
  • $\begingroup$ Choosing a rest frame makes no sense to me. Can you rephase or rewrite that to better explain it? $\endgroup$
    – Jason
    Mar 17, 2021 at 17:33
  • $\begingroup$ @Jason Consider a particle with four-momentum $(E,\vec p)$ and mass $m^2 = E^2 - p^2$. You can always boost into a reference frame where the four-momentum is $(m, \vec 0)$. If your theory is invariant under Lorentz transformations (as special relativity requires), you have to have the same physics in both reference frames. $\endgroup$
    – rob
    Mar 17, 2021 at 17:55
  • $\begingroup$ An electron flying around a particle accelerator can have more energy than its rest mass - assuming the m in (m, 0) is rest mass. $\endgroup$
    – Jason
    Mar 17, 2021 at 18:52
  • 1
    $\begingroup$ @Jason Only in the lab frame. The electron believes it is at rest, buffeted by interactions with a particle accelerator crashing relativistically around it. Related, related. $\endgroup$
    – rob
    Mar 17, 2021 at 19:31

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.