Magnetic field generated by an electron in an electric field While studying the corrections of fine structure in the hydrogen atom my book tries to justify the spin-orbit term with some arguments about magnetic energy an spin. I'm fine with the calculations but I don't understand where his starting point comes from. It says that the magnetic field generated by an electron
moving in an eletric field $\vec{E}$ with velocity $\vec{v}$ is $$\vec{B}=-\frac{1}{c^2}\vec{E}\times\vec{v}$$
I've studied electromagnetism and a bit of special relativity, but I don't understand where this formula comes from.
 A: Here's how I would show it very briefly using Special Relativity:
Let's consider two frames $S$ and $S'$, with $S'$ moving with respect to $S$ at a velocity $v$. There is an electric field $\mathbf{E}$ in $S$, and the charge is at rest in $S'$. According to an observer in $S$, therefore, we have a charge moving in the electric field $\mathbf{E}$.
Now let's look at it from the charge's point of view: the charge only sees a static electric field in the universe, and describes it using a scalar potential. The four-potential the charge sees is therefore:
$$A'^\mu = \begin{pmatrix}\frac{\phi'}{c}\\0\\0\\0\end{pmatrix}$$
(note that primed quantities represent quantities measured in the $S'$ frame). How can we relate this to the four-potential in the $S$ (or "lab") frame? Well, using a Lorentz Transformation it's easy to see that (if the $S'$ is moving along the positive $\hat{\mathbf{x}}$ direction),
$$A^\mu = \begin{pmatrix}\gamma \frac{\phi'}{c}\\\gamma \frac{v}{c^2}\phi'\\0\\0\end{pmatrix},$$
which means that the new electric and magnetic potentials are
\begin{aligned}\phi &= \gamma \phi'\\ A_x &= \frac{v}{c^2}\gamma\phi',\end{aligned}
and substituting for $\phi'$, you can show that $$A_x = \frac{v}{c^2}\phi.$$
Of course, what we have done so far assumes the velocity to be only along $\mathbf{\hat{x}}$ which is just to simplify our work. It's possible (but for more tedious) to show that for an arbitrary velocity $\mathbf{v}$,
$$\mathbf{A} = \frac{\mathbf{v}}{c^2} \phi.$$
From here, it's a very short step to calculate the magnetic field, since
$$\mathbf{B} = \mathbf{\nabla \times A} = \frac{1}{c^2}\mathbf{\nabla \times }(\mathbf{v}\phi).$$
Since $\mathbf{v}$ is a constant, we can use a standard relation from vector calculus, that for a scalar function $\psi$ and a constant vector $\mathbf{C}$:
$$\mathbf{\nabla \times}(\psi \mathbf{C}) = \psi (\mathbf{\nabla \times C}) + (\nabla\psi)\times\mathbf{C}.$$
Using this relation, it can be shown very simply (since $\mathbf{v}$ is constant) that
$$\mathbf{B} = \frac{1}{c^2} (\nabla \phi) \times \mathbf{v},$$
and the definition of the electric field tells us that $$\mathbf{E} = - \nabla\phi,$$
and so $$\mathbf{B} = -\frac{1}{c^2} \left(\mathbf{E} \times \mathbf{v}\right)$$
