perfectly spherical nucleus

[pbutkiewicz5]

Yes, because the field generated by electrons is not uniform in the space occupied by the nucleus (for example, look at the electric field generated by a uniformly charged ring). At the center of the nucleus r=0 the field can have a different sign and value than at the surface of the nucleus and at every point in between. Hence there will be a different energy (lower because electrons are on average closer to the nucleus than in the case of a point nucleus) for the case where we have a point nucleus vs. with a spatial distribution. Using the gravitational analogy – we have to integrate over B(r1) where r1 is indicates over the distribution of the magnetic moments of the nucleus u_I.
I just don’t understand why we didn’t discuss a similar correction for the case of electric monopole shift. After all, for a spherical distribution of nuclear charges there is no quadrupole moment (and higher moments), so the only moment is monopole. But the correction appears only when electrons enter the nucleus (<r^2>\rho_e(0)). Why doesn’t the energy correction occur only from the fact that we have made a finite-sized (serically symmetric) nucleus from a point nucleus?

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