Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was wandering around the particle date group page for meson and couldn't find a meson for top-bottom, which from symmetry you would expect.

Q1: Is this because it hasn't been found? Q2: There is an underlying reason why it can't exist?

share|cite|improve this question
I think that the production of the top quark and even bottom quark is so violent - requires so much energy - and the quarks decay to lighter ones so quickly that it's unlikely that you produce two such heavy quarks, moreover with low enough relative velocity that they will be able to form a meson before they decay. So I would guess that the bound state doesn't exist in any useful sense. The non-existence of the related "toponium" (top-antitop bound state) is explicitly stated, for the same reason I mentioned, at – Luboš Motl May 29 '12 at 5:55
up vote 2 down vote accepted

The top quark is so massive relative it's partners that it decays on time scales faster than the hadronization time scale.

Look at the tables don't find any top quark mesons. Instead the top--uniquely among the quarks--has decay modes. (PDF link)

share|cite|improve this answer
hmm. I thought all quarks "decay" by changing flavor through the weak interaction anyway. The up quark looks stable. . As Lubos says, it is the width that is different in these decays. – anna v May 29 '12 at 15:42
@annav Yes, they do. But the top is unique in so far as it's high mass means it decays (mostly to $W + b$) at least as fast as hardonization occurs, so we never observe distinguishable bound states. That is why PDG lists decay processes for the top but not for the strange, bottom or charm which are handled by the decay of the hadrons they form. – dmckee May 29 '12 at 16:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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