Can strong interaction be repulsive?

Question

I know that the repulsion between nuclei is mostly caused by electrostatic repulsion and Pauli's exclusion principle. But in the sub-nucleus level, is there a condition where the strong interaction acts repulsively between elementary particles?

Answer 1

The answer as you can see from the comments differs whether you are asking about the residual strong force (nuclear force) or the strong force between quarks inside neutrons and protons.

1. residual strong force

You can read a lot on this site about whether the nuclear force is attractive at large distances and repulsive at short distances. Though, this is more complicated.

A nucleon is a composite object made out of three quarks. The nucleon is color-neutral, so to first order, we expect that a nucleon should not interact with another nucleon at all. This is in fact approximately what we do see, since at large distances the nucleon-nucleon interaction falls off exponentially. But the cancellation is not exact, and at small distances we do get an interaction. This is called a residual interaction, and it's exactly analogous to the residual interaction between two electrically neutral atoms, which is the van der Waals force, often modeled by a Lennard-Jones potential.

How does the nature of nuclear force change between attractive or repulsive based on distance?

The residual strong force is described by the NN potential. This is different from the strong force that acts between the quarks inside the neutron and proton because the residual force is mediated by color neutral lighter mesons.

2. between quarks inside neutrons and protons, strong force

The strong force attracts quarks (contrary to popular belief this is not always true depending on the quarks type), but it acts like a string, when the quarks get closer, the attraction becomes weaker. This is caused by the asymptotic freedom.

The strong force does pull quarks together, but it also gets weaker as the quarks get closer (i.e. it acts sort of like a spring), in a phenomenon known as "asymptotic freedom." In this way, the strong force is very different than electromagnetism, where the force gets stronger if the charges are closer together. As such, there's no reason to expect that quarks which are placed close together will immediately annihilate, as there's just not a lot of force on them in the first place.

What keeps quarks separate (strong force pulls, but what repels to equal out)

Contrary to popular belief, the strong force between quarks is not always attractive.

Quarks of different color experience an attractive force between them; whereas quarks of a like color repel each other.

https://www.sciencedirect.com/topics/earth-and-planetary-sciences/quark

Answer 2

I want to address the "repulsive" of color interactions part and why there is not much in literature discussing it.

How does "repulsion" work for electromagnetic interactions? two like charges repel the closer they are and if point charges an infinite repulsion would exist in overlap, because of the $1/r{2}$ behavior of the electric field. At the mathematics of the quantum level with photon exchanges this affects the matrix elements which give the appropriate probabilistic behavior .The two frameworks merge mathematically at the overlap between micro and macro. Keep in mind that the coupling entering in each diagram is 1/137.

In QCD there is asymptotic freedom and a coupling constant of 1, used when Feynman diagrams are displayed. The reason that one does not find much about repulsion for the color force is a) because of asymptotic freedom. The smallest the distance between two quarks the less the force (as a classical analogy to electromagnetism). b)because the gluons are self interactive, and interactions between quarks will always have an attractive component due to the alternate-colored gluon exchanges possible . Together with the coupling constant of one this makes repulsion practically a force that will only affect the parton functions by its existence, in modifying the interactions of quarks antiquarks and gluons in the soup that a proton, for example, is.

In any case the perturbation image of QCD is illustrative and not useful for calculations because the series does not converge. That is why QCD on the lattice is used for calculations.