In atomic physics, the spin-orbit is a small correction between 1/1000 and 10ppm, so fairly small. In contrast, in nuclear physics the inclusion of the spin-orbit interaction is necessary to reproduce the basic level scheme, see picture below (from Kenneth Krane, Introductory Nuclear physics). Why is that?

I imagine it has to do with the $r$-dependence: In a semi-classical calculation, the energy correction due to the spin-orbit interaction has a $1/r^3$ dependency. However, I have only seen this calculation for the atomic case and not for the nuclear one.

From Kenneth S. Krane Introductory Nuclear Physics


2 Answers 2


The nuclear force is inherently spin-dependent, which is why the interaction between nucleons is different, depending on whether spins are parallel or not:

The nuclear force is not simple, though, as it depends on the nucleon spins, has a tensor component, and may depend on the relative momentum of the nucleons.

(quoted from Wikipedia, which cites here the book by Krane, mentioned in the OP.)

Related: Why nuclear force is spin dependent?


To self-answer my question:

Yes, it is indeed because of the $1/r^3$ dependency of the spin-orbit interaction. Intuitively speaking, in the atomic case the electron orbits the nucleus on the order of $10^{-10}$m. In the nuclear case, one nucleon orbits another one with a distance on the order of $10^{-15}$m.

Another more mathematical point of view is that considering a series expansion of all the forces involved, the spin-orbit-interaction is a higher order term and thus scales with $\nabla^n F$. Since $F$ is much more close-ranged for the nucleus, $\nabla^n F$ is much higher and contributes more strongly.

  • $\begingroup$ Though you expect the size of spin orbit forces to be larger, isn't this possibly counteracted by the nuclear force being much larger than the em force? Is it obvious from your answer that the relative sizes of the two contributions are more significant in the nuclear case? $\endgroup$
    – jim
    Sep 21 at 8:11

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