Can we calculate L-S coupling without Dirac equation? It is known that there exists an orbital and spin angular momentum coupling for an electron moving in the atom. And the Hamiltonian can be directly derived using Dirac equation. I want to use a classical picture. That is, for an electron moving with velocity $v$ in an electric field, we can obtain a magnetic field by Lorentz transformation:$$\vec{B}=\vec{E}\times\vec{v}/c$$ Where $v\ll c$. And $\vec{E}=-\triangledown \phi$, $\vec{v}=\vec{p}/m$. We can know the magnetic field, and couple with spin $\vec{S}$, we will have a $L-S$ coupling term. The problem is that this term differs from the correct result obtained by Dirac equation only on a factor $\frac{1}{2}$. I think it may be the circular motion of the electron that accounts for the problem. If I take the accelerate motion of the electron, I may get the same result. But I don not know how to do with this. Should I use GR or the derivation is totally wrong?
 A: You are right, it happen due to circular motion. This effect is called Thomas precession. 
In 1925 Thomas relativistically recomputed the precessional frequency of the doublet separation in the fine structure of the atom. He thus found the missing factor 1/2, which came to be known as the Thomas half.
There are several method to derive it. For example one of them is written in Moller`s book The theory of relativity and Itzykson Zuber quantum field theory.
But I prefer another way.
For description of spin motion in special theory of relativity one can use spin four-vector $a^{\mu}$. This four-vector must be satisfy the following identity. 
In the co-moving frame $a^{0}=0$ and $a^{i}=s^i$(This four-vector looks like usual spin in this frame)
Because of this one can write another identity
$$v a=0$$ where $v$ is four-vector of velocity $v^\mu=\frac{p^\mu}{m}$
Clearly that if velocity change the spin will be change.
The spin precession of an electron in an external electromagnetic field is described by the Bargmann–Michel–Telegdi equation.
$$\dot{a}^\mu=\frac{g e}{2m}F^{\mu\nu}a_{\nu}-\frac{e}{m}(\frac{g}{2}-1)u^\mu F^{\nu\tau}u_\nu a_\tau $$
where $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$ and g is a gyromagnetic ratio.
For derivation Thomas precession it is necessary to make non-relativistic expansion. 
I  will not write intermediate calculations because all of it is written in following good book: V. B. Berestetskii, E. M. Lifshitz, and L. P. Pitaevskii, Quantum Electrodynamics (Elsevier,
Oxford, 1982).  paragraph 41 Motion of spin in an external field.
But I will write the answer
$$\dot{\mathbf{\xi}}=\frac{\mu}{s}\left[\xi\times \mathbf{H}\right]+\left(\frac{\mu}{s}-\frac{e}{2m}\right)\left[\xi\times \left[\mathbf{E}\times \mathbf{v}\right]\right]$$
In our case $\mathbf H=0$ and for electron $\left(\frac{\mu}{s}-\frac{e}{2m}\right)=\left(\frac{d e}{2 m}-\frac{e}{2m}\right)=\frac{\mu}{2s}$ in this identity I use that $g=2$ 
And we obtain 
$$\dot{\mathbf{\xi}}=\frac{\mu}{2s}\left[\xi\times \left[\mathbf{E}\times \mathbf{v}\right]\right]$$
Thus we obtain the Thomas half
