2
$\begingroup$

So my understanding of the nuclear force so far is this (please correct anything I have wrong):

Being a residual of the strong force, the nuclear force is mediated (in part) by the emission of virtual gluons by quarks, not completely dissimilar to the exchange of virtual photons between electrically charged particles that mediates the electromagnetic force. However, unlike the strength of the electric field between two electrically charged particles, the strength of the gluon field between two color charged particles is constant as a function of the distance between them. This is due to the so-called "flux tubes" that are formed. My first attempt at wrapping my head around these has left me scarred and deformed, but I think the concept is that they effectively behave as strings and exhibit a tension force.

As a result of the energy requirement of maintaining the flux tube increasing linearly with distance ( ΔE = W = F ⋅ d ), it would be more energetically favorable for a gluon emitted by a quark confined within one nucleon to a quark confined within another nucleon to decay into a meson. [Perhaps this has something to do with the principle of least action?]

One of the more common (I think?) decay products of gluons involved in the nuclear force is the neutral pion π0. Neutral pions usually decay into gamma rays, which I'm assuming would be observable to some degree. This leads me to believe that virtual mesons never decay. Is this true? If so, then how so? Something something uncertainty principle?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ A virtual meson is a straight or dotted line in a Feynman diagram, which itself describes one term in a perturbation series that yields the rate of some process. If you're drawing Feynman diagrams for nucleon-nucleon scattering, you wouldn't draw a diagram where a nucleon emits a meson which decays, because that would be a different process. So the answer is no, by definition. $\endgroup$
    – knzhou
    Jul 20, 2019 at 13:45
  • $\begingroup$ A completely separate question is whether it is possible for a hadron to emit a meson which subsequently decays. The answer is yes, and this is described by a different set of Feynman diagrams. In these diagrams, you have dotted lines which branch off into other dotted or squiggly lines. $\endgroup$
    – knzhou
    Jul 20, 2019 at 14:14
  • $\begingroup$ @knzhouThat's an answer, not a comment. Could you convert it? $\endgroup$
    – rob ♦
    Jul 20, 2019 at 17:14

2 Answers 2

Reset to default
1
$\begingroup$

for a gluon emitted by a quark confined within one nucleon to a quark confined within another nucleon to decay into a meson.

The strong nuclear force, i.e. what keeps the nucleons within a nucleus, is modeled with pion and rho exchanges between nucleons, color neutral quark antiquark pairs. So a gluon is never emitted by a quark into another nucleon because of color neutrality at the distance of fermi that enclosed the nucleons in a nucleus.

If one wants in a nuclear model to use gluons, one has to take care of color neutrality.

The whole interactions that keep the nucleus together are modeled with the pions and rhos and ... off mass shell, i.e. virtual particles. Virtual particles have the quantum numbers of their name, but are off mass shell, i.e. their mass varies under the integration which will give the final crossection. Thus the neutral mesons exchanged cannot decay, because they are not real.

The gluons do not decay, they are part of the elementary point particles of the standard model.

Improve this answer
$\endgroup$
1
$\begingroup$

There is a disconnect between the title and detailed text of your question because you have switched from meson exchange language to gluon exchange language. I am going to ignore the title and address the issues posed in the text.

You are correct in suspecting that the flux tubes supposedly formed between colored objects are energetically unstable. (I say supposedly because this appealing picture may be dead wrong. Electric flux tubes are forbidden in QED by one of Maxwell’s equations, and nobody has ever painted the picture in QCD convincingly, say by describing the quasi-static glue field configuration in terms of vector potentials in some gauge.)
Jets are thought to arise from the breakdown of flux tubes via the creation of additional $q\bar{q}$ pairs, a process humorously called pionization. Consider inelastic $\gamma p$ scattering, in which the photon kicks one quark to kingdom come: $qq.......q$ turns into $qq(q\bar{q})(q\bar{q})q$ and thence into $(qqq)(\bar{q}q)(\bar{q}q)$.

How do the additional $q\bar{q}$ pairs get created? A virtual gluon could decay into such a pair. (Is that what you meant by the title of the question? You said mesons, but that term is usually applied to $q\bar{q}$ objects, not to gluons.) The virtual gluon might be emitted by the quark that got the hard kick, and there are further possibilities.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I intended to be referring to meson exchange. It was my understanding that mesons exchange always begins as gluon exchange but the greater distance causes the gluon to decay into a meson mid-flight. Is this wrong? $\endgroup$
    – QuaternionsRock
    Jul 28, 2019 at 0:09
  • 1
    $\begingroup$ Gluons couple to $q\bar{q}$, but not every $q\bar{q}$ combination is a meson. There are no mesons with the same quantum numbers as a gluon. Mesons are color-singlets. Two gluons might do the trick, but not one. $\endgroup$
    – Bert Barrois
    Jul 28, 2019 at 10:32

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.