How does meson exchange work within large nuclei?

Question

I read the Wikipedia page on Mesons and it mentioned that there both charged and uncharged Mesons that decay into neutrino/electrons and photons, respectively. Unfortunately it didn't elaborate on exactly why that is the case, unless it was hidden in the mathematics further down the page. If there are two types of Mesons differentiated by their electric charge, does that mean that neutrons will release neutral Mesons and protons will release charged Mesons? If there is charge being carried by a Meson, how is the charge being stored if a Meson is simply two quarks and a gluon?

It makes sense to imagine nucleons as exchanging mesons at the same rate as the photons are transmitting the electromagnetic repulsion of the protons, therefore they remain clumped together. But if there are many numerous nucleons, it isn't as simple as imagining a single trade off between two nucleons. There would be a probability field of Mesons, and therefore would randomly allow for some protons to be pushed out of the nucleus via electric repulsion. But this is the job that is handled by the Weak force, transmitted by the W and Z Bosons. Where am I going wrong in imagining the inner workings of the big nuclei?

Answer 1

Your question is founded on misunderstandings. The theory of what goes on inside the nucleus is neither simple nor intuitive.

1. "Meson" is simply a name for any particle that is a bound state of a quark and an anti-quark, just like "baryon" is a name for a bound state (like a proton or a neutron) of three (anti-)quarks. It does not denote a particular particle species - like "proton", "neutron", "electron", "pion", "kaon", etc. - but is a more general category. Since quarks have either charge $+\frac{2}{3}$ or $-\frac{1}{3}$ and anti-quarks have correspondingly either charge $+\frac{1}{3}$ or $-\frac{2}{3}$, mesons as a combination of one quark and one anti-quark can have either charge $-1$, $0$, or $1$ (All charges in terms of multiples of the electron charge).

   Mesons are not "two quarks and a gluon", they are rather to be thought of as 'two quarks and an indeterminate number of sea quarks and gluons'. This is the parton model, which can be applied to mesons as well as baryons. On average, this parton soup contains the quark and anti-quark of which the meson is the bound state.

2. Quantum field theory (QFT), which forms the basis of quantum chromodynamics (QCD) - the theory of quarks, gluons and their composite states - is not amenable to be talked about merely in terms of particles exchanging other particles. Nucleons are not "exchanging mesons", just like electrically charged particles are not "exchanging photons". In electrostatic attraction there are no photons whatsoever we could ever detect - yet we derive the electrostatic Coulomb force "from the idea of photon exchange". The point is that drawing a "particle exchange" in a Feynman diagram is not synonymous with claiming an exchange of real particles happens as a detectable process. There is no "rate" at which photons, gluons, mesons or any other force-carrying particles are exchanged.

   There is no "probability field of mesons", there is just the quantum-field theoretic fact that the possibility of exchanging a meson of mass $m$ leads to a Yukawa force $\propto \frac{\mathrm{e}^{-mr}}{r^2}$ (note that for $m=0$ this is the Coulomb force of the massless photon).

   One of the mesons with the largest contribution to the nuclear force (also called the residual strong force) is the pion because it has a comparatively low mass and would be massless if isospin symmetry were exact.

3. The weak force does not "randomly push out protons from the nucleus". In fact, it does not meaningfully transmit a classical "force" at all, the presence of the W and Z bosons in the theory simply allows certain interactions to happen that we collectively group under the moniker "radioactive decay". The weak interaction therein changes the kind of the hadrons and leptons involved (turning a neutron into a proton, an electron and a neutrino in $\beta$-decay, for instance) and is distinct from the nuclear force mediated by the pions that just keeps the nucleus together.