By how much protons dipole moment inside a nucleus attenuate the culomb force between them? As up quarks repel more than down quarks the protons should be oriented with the positive side looking away from the centre of the nucleus. In that case the strong force have more job to do with the dipole moment that wants to break apart the particle instead of the residual strong force that have dipole moment as a 'friendly' force?
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Do you mean the proton’s magnetic dipole moment? Or do you mean its intrinsic electric-dipole moment (a CP-violating observable whose value is consistent with zero)? Or do you mean the induced electric dipole moment due to the proton’s electric polarizability? – rob Jan 26 '22 at 21:18
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@rob The third one. Because an electric field should make the nucleons act as a dielectric material... – Krešimir Bradvica Jan 27 '22 at 15:06
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The electric polarizabilities for the proton and neutron, catalogued for example by the Particle Data Group, are about a thousand times smaller than you would expect from doing dimensional analysis. In a hand-waving way, this is because strong interaction makes the "medium" inside of a proton "stiffer" than the "medium" within a hydrogen atom (where the dimensional-analysis result is okay).
In low-mass nuclei, the statement that "strong isospin is a good symmetry" is basically equivalent to "you have permission to neglect nucleon charge." If you can effectively predict the excitation spectrum for a nucleus without considering electric charge at all, it's probably also reasonable to neglect the small electromagnetic correction due to polarizability.
If you wanted to compute this, I'd use a mean-field approach. Choose a nucleus of interest and model it as a uniform-density sphere of charge, whose electric field is zero at the origin, linear in radius to the edge of the nucleus, then $1/r^2$ to infinity. Then convert your nucleon polarizability into an electric susceptibility for nuclear matter. The energy density of the electric field will shrink a little within the nucleus, as you move from $\vec E$ to $\vec D$. The difference in the integrated electric field energy is an order-of-magnitude estimate for the polarizability correction to the energies of nuclear states.
rob
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