This section presents a relatively simple and quantitative description of the spin-orbit interaction for an electron bound to a hydrogen-like atom, up to first order in perturbation theory, using some semiclassical electrodynamics and non-relativistic quantum mechanics. This gives results that agree reasonably well with observations.

The energy of a magnetic moment in a magnetic field is given by $$\Delta {\rm H}_{}=-{\boldsymbol \mu }\cdot \mathbf {B}$$ where $\boldsymbol \mu$ is the magnetic moment of the particle, and $\bf B$ is the magnetic field it experiences.

We shall deal with the magnetic field first. Although in the rest frame of the nucleus, there is no magnetic field acting on the electron, there is one in the rest frame of the electron. Ignoring for now that this frame is not inertial, in SI units we end up with the equation $${ \mathbf {B} =-{\frac {\mathbf {v} \times \mathbf {E} }{c^{2}}}}$$

Now, my professor is used to work with the $\bf H$-field instead of $\bf B$, and in CGS units. So the previous formula becomes: $${\bf H} = \frac 1 2 \frac {\boldsymbol {\cal E} \ \times \bf v}{c}$$
The problem is that $\frac 1 2 $ factor. He said it's the famous Thomas factor... can someone relate?


1 Answer 1


The 1/2 is an effect due to Thomas precession. The Wiki linked article has a good exposition. See also here for a pedagogical account.


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