# The spin-orbit interaction for a classical magnetic dipole moving in an electric field

Spin-orbit coupling is one component of the fine structure of atoms, which is explicitly concerned with the interaction of the electrons' spin with their orbital angular momentum. It can be explicitly derived from the Dirac equation by taking the nonrelativistic limit to subleading order in $1/c$, and it produces a term in the hamiltonian of the form $H\sim \mathbf L\cdot\mathbf S$, where $\mathbf L$ and $\mathbf S$ are the orbital and spin angular momenta of the relevant electron.

In addition to (or, mostly, instead of) this rigorous approach, however, the form of the spin-orbit interaction potential is often justified via a heuristic argument, which roughly looks as follows:

Consider a hydrogen atom, with an electron whizzing around the nucleus, and transform to a frame that is fixed on the electron. Then, as the electron's trajectory orbits around the nucleus, the electron not only sees the electrostatic field of the proton but it also sees an effective current as the nucleus orbits around it. This current then produces a magnetic field, which interacts with the intrinsic magnetic dipole of the electron.

This argument is obviously problematic, because there is no such thing as an electron trajectory in quantum mechanics, and there is no such thing as an "electron-mounted frame"; the heuristic calculation gives (a reasonable interpretation for the) correct results, but it cannot be expanded into an argument that makes any sense at all.

As such, it is really appealing to try and work out a similar heuristic argument that produces an interaction of the form $H\sim \mathbf L\cdot\mathbf S$ from an analysis that takes place on the rest frame of the proton, and which is not subject to those disadvantages. This is explored in a previous question, Spin-orbit coupling from the rest frame of the proton?, and probably in plenty of earlier resources, but I have yet to see such an analysis give satisfactory results.

My goal here is much more modest, in the hope that it will yield a better shot at a clearer physical picture of spin-orbit coupling.

Consider, therefore, a classical magnetic dipole $\mathbf m=IA\hat{\mathbf n}$, which consists of a rigid loop of area $A$ with unit normal $\hat{\mathbf n}$ carrying a current $I$, much smaller than the spatial dependence of any electric or magnetic fields in the problem. The shape of the circuit loop should be irrelevant to the results.

This dipole is centered at $\mathbf r_c=\mathbf r_c(t)$, which follows a trajectory that has been pre-specified ahead of time (you can think of it being mounted on a rail and pushed by a little motor, though later we might solve for $\mathbf r_c(t)$ responding to some equation of motion), and it traverses a region with a pre-specified static electric field $\mathbf E(\mathbf r)$. As an archetypal example, think of an elliptical orbit around the center of a Coulomb electric field.

The degree of freedom of interest is the orientation of the dipole, $\hat{\mathbf n}=\hat{\mathbf n}(t)$, which follows a trajectory that's in principle specified ahead of time, though again we might later make this follow some equation of motion.

On to my question, then:

• How can one derive, through considerations exclusively on the inertial frame of the initial electric field, the energy associated with the orientation of our classical magnetic dipole? In principle this should produce a spin-orbit coupling of the form $\mathbf L\cdot \mathbf m$, where $\mathbf L$ is the orbital angular momentum of the dipole's trajectory $\mathbf r_c(t)$, but how does this come out of the interactions of the moving charges inside the circuit with the static electric field?

And, finally, just to be clear: yes, this is a terrible model for an electron, because the magnetic dipole of electrons is very much not due to the circulation of electric charge. Nevertheless, I think this method of analysis can provide a separate analogy which can hopefully be clearer and less problematic than the usual "let's pretend there's an (inertial?) reference frame mounted on the electron" heuristics, i.e. by providing a model which is also (ultimately) wrong, but which is at least internally consistent.

• Have you studied how the E-M field transforms in Special Relativity? May 18, 2017 at 16:22
• @md2perpe Yes, I have, but the point is precisely to avoid doing that transformation, which is deeply problematic when applied to the quantum mechanics of the hydrogen atom. The analysis of electromagnetic situations is completely symmetric with respect to changes of frame of reference (citation), and any analysis that derives that energy in the instantaneous rest frame of the point dipole needs to have an analog analysis in the inertial frame where $\mathbf B=0$. The question asks for that analysis. May 18, 2017 at 16:34
• iopscience.iop.org/article/10.1088/0143-0807/12/6/004/pdf Maybe this is a start? May 21, 2017 at 13:04
• @Anonjohn That does seem relevant, though it seems that they let $\hat{\mathbf n}(t)$ precess freely ($\boldsymbol{\mu}(t)$ in their notation), instead of fixing its trajectory and asking for the energy required for that trajectory, as in the question. It's still a good startpoint, though. May 21, 2017 at 13:12
• There's some good stuff in that citation of yours three comments up. Someone should think about that stuff some more, I bet they could come up with some cool physics results May 23, 2017 at 10:22

This question is studied in the paper http://aapt.scitation.org/doi/abs/10.1119/1.1976708. The key point is that a moving magnetic dipole acquires an electric dipole moment proportional to its velocity. (See Fig. 2 of http://aapt.scitation.org/doi/pdf/10.1119/1.14820 for a nice nonrelativistic visual intuition of why this is the case.) Therefore, the electron's spin magnetic moment and its (semiclassical) motion around the nucleus combine to give it an effective electric dipole moment ${\bf p}$ proportional to the cross product of its velocity and its spin. This electric dipole moment couples to the electric field from the proton via the usual equation $H = -{\bf p} \cdot {\bf E}$.
The Hamiltonian is proportional to the triple product ${\bf E} \cdot ({\bf v} \times {\bf S})$, where ${\bf v}$ and ${\bf S}$ are the electron's semiclassical velocity and spin respectively. This ordering of the triple product is the natural one in the inertial proton's frame. But we can reorder the triple product as ${\bf S} \cdot ({\bf E} \times {\bf v})$. This ordering is the natural one to use in the electron's frame, because the electron sees a magnetic field proportional to ${\bf v} \times {\bf E}$ because of how electric fields transform under Lorentz boosts, so the triple product just looks like the usual coupling between a magnetic dipole and a magnetic field. In both frames, the coupling looks like a dipole-field coupling, but the dipole and the field are both electric in the inertial frame and both magnetic in the electron's frame. This rewriting of the triple product gives a nice heuristic argument for why the inertial-frame formula $H = -{\bf m} \cdot {\bf B}$ is still qualitatively correct (up to constants) in the noninertial electron's frame.