# Derivation of Spin-orbit Coupling

I was reading Griffiths's book on quantum mechanics.

In chapter 6, he tried to derive the spin-orbit coupling using a classical approach.
$$H=-\vec{\mu}\cdot \vec{B}$$

1. Finding the relation between $\vec{μ}$ and $\vec{S}$
He consider a spinning charged ring with mass $m$, radius $r$, total charge $e$ and period $T$.
Magnetic Dipole Moment: $\vec{\mu}=i\vec{A}=\frac{e\pi r^2}{T}\hat{n}$
(Spin) Angular Momentum: $\vec{S}=\tilde{I}\vec{\omega}=\frac{2m\pi r^2}{T}\hat{n}$
$$\vec{\mu}=\frac{e}{2m}\vec{S}$$
Dirac: $\frac{e}{2m}\rightarrow\frac{e}{m}$

2. Find the magnetic field $\vec{b}$
He consider the rest frame of electron.
$$\vec{B}=\frac{\mu_0 i}{2r}\hat{n}=\frac{1}{c^2\epsilon_0}\frac{e/T}{2r}\hat{n}=\frac{1}{4\pi\epsilon_0}\frac{e}{mc^2r^3}\vec{L}$$

3. Combining both terms
$$H=-\frac{1}{4\pi\epsilon_0}\frac{e^2}{m^2c^2r^3}\vec{L}\cdot\vec{S}$$

Then he point out that when we choose the electron's rest frame, there is one problem as it is a non-inertial frame. Therefore, we need to consider the Thomas's precession, which gives a factor of 1/2. Finally, it cancels with the 2 we multiplied to the classically derived gyromagnetic ratio.

My problem is when we choose an accelerating frame, obviously we need some modifications. But, how should we modify? and how can we get the factor of 1/2 from Thomas's precession?