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In nuclear physics, nuclear force, also known as the residual strong force, is mediated by pions exchanged between protons and neutrons. It doesn't seem like this should be limited to protons and neutrons, though, since the mechanism by which pions are exchanged comes up from the fact that protons and neutrons are made of quarks. Would hadrons with quarks other than up and down quarks and antiquarks exchange more exotic mesons?

Follow-up question: If all hadrons experience the nuclear force, do different hadrons experience it differently, and if so, how?

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    $\begingroup$ An impossibly open question. All hadrons, including those with s,c,b,t valence quarks are largely composed of a soup of gluons and soft light quark partons, so of course they couple to pions. They decay weakly if they cannot decay strongly. Asking for the complete nuclear physics of highly unstable particles might be too much, no? $\endgroup$ Oct 19 '21 at 22:23
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the residual strong force […] is mediated by pions exchanged between protons and neutrons.

This is a pretty big oversimplification of the strong nuclear force. The pion, as the lightest member of the meson spectrum, can be associated with the longest-range part of the residual strong force. But if you’re also interested in the phenomenon that nuclear matter has constant density, you already have to go further up in the meson spectrum than the pion to find a repulsive interaction. In many-body systems or in high-energy interactions, the meson-exchange picture rapidly stops being a useful way to make quantitative predictions.

Your question about “all hadrons” suggests you are also curious about long-range interactions between mesons. That’s basically impossible to measure directly, because all mesonic hadrons are short-lived. It’s one thing to build an accelerator that makes a beam of pions or kaons; it’s a different thing altogether to make two such accelerators and point the beams at each other. Beam-beam interaction experiments, like the Large Hadron Collider, make use of stored beams of stable particles. (A possible exception was a neutron-neutron scattering length measurement which used the simultaneous detonation of two nuclear bombs in the same underground cavity. Its results were never published; I have heard variously that a blast door failed and the DAQ was destroyed, that the experiment was proposed but never attempted, and that the whole thing is an extremely niche urban legend.)

To the extent there are residual meson-meson interactions, virtual mesons will participate in them as well, and those meson-meson interactions show up as modifications to your model of the meson-mediated nuclear force. That recursive relationship gets messy fast. For a taste of how complicated it is, look for literature about whether the “fictitious $\sigma$” (renamed $f_0(500)$ in the current PDG) is a “real” meson or a two-pion bound state.

Long-lived baryons absolutely participate in the nuclear force, illustrated most clearly by hypernuclei made of protons, neutrons, and exotic baryons. (In practice there’s just one hyperon, for the same reason as the absence of meson-meson colliders.)

For short-lived baryons, it’s not clear whether the meson-mediated approximation would be useful under any circumstances. The effective range of the pion-mediated force is related to the pion’s mass:

$$ r_\pi = \frac{\hbar c}{m_\pi c^2} \approx 1.3\rm\,fm \approx r_\text{nucleon} $$

An unstable state with lifetime $\tau$ has an intrinsic uncertainty $\Gamma$, or width, in its total energy,

$$ \Gamma \tau \approx \hbar $$

which you can think of as a kind of energy-time version of the uncertainty principle. If we use $\hbar = c = 1$ units so that energy, mass, (inverse) time, and (inverse) distance are all measured in the same units, you might say that a particle whose decay width is larger than the pion mass, $\Gamma > m_\pi$, will probably have already decayed in the time it takes for light to cross a nucleon. It’s not clear (to me) what it would mean to “interact” with a particle which has already decayed before you finish touching it.

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  • $\begingroup$ This is a wonderful answer, but I have a few questions. First, what did you mean by "virtual mesons will participate in the as well?" I think this was a typing error. Second, can you define "lifetime" more precisely? Do you mean the mean lifetime? $\endgroup$
    – zucculent
    Oct 20 '21 at 22:23
  • $\begingroup$ Typo fixed. Between “mean lifetime” and “half-life,” and between the Breit-Wigner width and the uncertainty principle, there are some factors of two or log two or the like which are totally irrelevant at the level of this answer. I use “$\approx$” instead of “$=$” to sweep them under the rug; if you want to get them right, start from a textbook. $\endgroup$
    – rob
    Oct 21 '21 at 1:46
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The strong nuclear force is a residual force from the Quantum Chromo Dynamics (QCD) force, one of the three basic forces in the Standard Model, SM, of particle physics.

In the context of atomic nuclei, the same strong interaction force (that binds quarks within a nucleon) also binds protons and neutrons together to form a nucleus. In this capacity it is called the nuclear force (or residual strong force). So the residuum from the strong interaction within protons and neutrons also binds nuclei together.

The strong residual force is analogous to the Wan der Waals, WW, forces in the solid state of atoms. In quantum mechanical terms:

The force (WW) results from a transient shift in electron density(the orbitals of the electrons about the nucleus). Specifically, the electron density may temporarily shift more greatly to one side of the nucleus. This generates a transient charge to which a nearby atom can be either attracted or repelled.

Do all hadrons experience the strong nuclear force?

To try to explain fundamental quark interactions in terms of the pion exchange model used in nuclear physics is futile. See what the nuclear force is in feynman diagrams of the quark content of the nuclear interaction of a proton with a neutron in the nucleus in terms of quarks and gluons.

proton neutron interaction

The pion exchange model comes from a different simpler Quantum Field Theory, QFT for nuclei, example., which ignores the quarks and concentrates on protons , neutrons and various mesons for the exchanged particles in the Feynman diagrams.

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  • $\begingroup$ The nuclear force is not analogous to the van der Waals force in molecular/atomic interactions. Due to the running of the coupling constant of the strong force, the effect of the strong force on length scales where confinement does not forbid two-gluon exchange is too weak in order to explain the nuclear force. $\endgroup$
    – Samuel
    Oct 20 '21 at 16:53
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    $\begingroup$ @Samuel While the analogy between the van der Waals force and the long-range nuclear force may not be good enough to be useful quantitatively, it’s still useful as a pedagogical tool. Atoms are electrically neutral, but have an electromagnetic “contact” interaction described by van der Waals. Nucleons are color-neutral, but have a QCD-associated contact interaction. $\endgroup$
    – rob
    Oct 20 '21 at 17:13
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All hadrons experience the strong force: that's more or less what defines them. In particular, it's the residual kind. The coupling constants to specific gauge mesons might vary by species (I'm not a QCD expert), but qualitatively they behave the same.

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  • $\begingroup$ So is that a yes to my question? $\endgroup$
    – zucculent
    Oct 19 '21 at 22:19
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    $\begingroup$ @zucculent The exchanged virtual mesons won't become more exotic based on the hadrons' valence quarks, because the sea quarks aren't any different. $\endgroup$
    – J.G.
    Oct 19 '21 at 22:23
  • $\begingroup$ @zucculent As well as pions, there are also rho & omega mesons involved. But bear in mind that these exchange mesons are all virtual, not real particles. $\endgroup$
    – PM 2Ring
    Oct 20 '21 at 6:05
  • $\begingroup$ @PM2Ring Isn't one of the quarks in the meson real though, since it came from inside a very real hadron? $\endgroup$
    – zucculent
    Oct 20 '21 at 14:22
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    $\begingroup$ @zucculent This. $\endgroup$
    – J.G.
    Oct 20 '21 at 16:31

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