When nucleons bind to form a nucleus, does each nucleon "retain ownership" over its quarks (is each nucleon a truly unique entity)? Or is it possible for quarks to swap from one nucleon to another? Or does it not make sense to talk so exactly about them?

Submitted by ChrisGnam t3_xwae1i in askscience

I was reading about the strong force and color confinement recently, but everything I read seemed to only talk about an isolated nucleon.

A brief reading of the Nuclear Force (the binding force between nucleons) which can be viewed as a "residual strong force" indicates it is mediated by virtual mesons. To my completely uneducated mind, that reads as though the nuclear force is almost like nucleons swapping quarks between one another (as a meson is a quark-antiquark pair).

I'm assuming I'm reading too much into it and letting my uninformed imagination run a bit too wild. Because my loose understanding of virtual particles is that they're just a byproduct of random fluctuations in the quantum fields. But then that makes me think that maybe it doesn't make sense to say that a nucleon has "ownership" over its quarks at all, if its quarks can't really be distinguished. Which only furthers my question of if each nucleon is clearly a separate entity while in a nucleus or not.

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Ast0815 t1_ir682my wrote

In short: Yes nucleons can exchange quarks via mesons, but because of confinement the baryons will always have three valence quarks.

But all, e.g., up quarks are indistinguishable. So it makes no sense to say that a particular quarks left the proton and travelled to another (as part of a pion for example). At the same time a new up-quark must have been created and remained in the original proton.

And then there is the issue of energy scale. The quarks are still bound inside the nucleons, meaning the energy required to get one out is larger than the random energy fluctuations in the system (= temperature). That is why you only exchange virtual mesons rather than real particles. Mathematically this makes it convenient to describe the nucleons as their own things that than interact with one another, rather than looking at all the quarks and gluons and whatnot separately.

You can create states of matter where the quarks are no longer bound, i.e. it makes sense to model things on the quark level. It is called "quark gluon plasma", and it exists at much higher energies/temperatures. E.g. it is created briefly in particle colliders, when you smash nuclei into each other.

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Amadis001 t1_ir6r4aa wrote

The density of a nucleus is about 1/6 of the density of its individual nucleons. That means that even within a nucleus, it is still mostly empty space. Still, being bound within that nucleus does have some effects on the nucleons, including some weakening of quark confinement. Hence the radius of individual protons and neutrons within the nucleus is slightly larger than for free protons and neutrons. They swell because the bag pressure is reduced. Further, while the primary mechanism for nuclear binding is virtual pion exchange, it is also true to some very small extent that individual quarks are free to wander from nucleon to nucleon. At extreme pressures, i.e., inside a neutron star or in high energy nuclear collisions, the quarks become free-er and free-er to wander, until in the limit of a quark gluon plasma they can wander across the entire extended volume of the nuclear matter. At normal nuclear temperature and pressure, this still happens, but only to a very tiny, practically immeasurably small, degree.

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lordvadr t1_ir9wmdi wrote

Replying to Amadis001 (#38,602)

Is this quark-meandering essentially what triggers nuclear decay? I.e. as they move around, do they eventually get into a configuration that makes firing off the particle energetically beneficial?

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Zelrak t1_ir9ws1e wrote

Replying to Ast0815 (#37,605)

Q from someone with high energy theory background but who didn't do much nuclear physics: Given that the nucleus is in a strong coupling regime, what does it mean to talk about a perturbative picture of nuclei exchanging mesons? I always thought of the nucleus as a bag of stuff we don't understand and all we can say is that the whole bag has some given colour, baryon, etc charges, where you can measure form factors and such other phenomenological parameters

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Amadis001 t1_irbm20v wrote

Replying to lordvadr (#43,754)

Hmm. You've asked a classical question about a quantum-mechanical system, which makes it a bit hard to answer precisely. But the answer to the spirit of your question is "yes".

It's more precise not to think about subatomic particles in bound systems as "meandering" so much as having non-zero probability of being found in different places. And the probability of any scattering interaction, including nuclear decay, will depend (amongst other things) on the overlap of the initial and final state wave functions. So yes, it matters where the particles "meander."

A great example of this is K-capture, or inverse beta decay. This only happens almost entirely with S-shell electrons, since P and higher angular-momentum electron wave functions go to zero at the origin. Only an S electron "spends enough time near the nucleus" to be captured with any significant probability.

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