The Maxwell field tensor is given by

$$F^{a b}=\left(\begin{array}{cccc} 0 & -E_{x} & -E_{y} & -E_{z} \\ E_{x} & 0 & -B_{z} & B_{y} \\ E_{y} & B_{z} & 0 & -B_{x} \\ E_{z} & -B_{y} & B_{x} & 0 \end{array}\right)$$

A general 4-velocity is written as $U^{a}=\gamma(1, \mathbf{v})$, where $\gamma=\left(1-|\mathbf{v}|^{2}\right)^{-1 / 2}$, and $c=1$. A general 4-current density is written as $J^{a}=(\rho, \mathbf{j})$, where $\rho$ is the charge density and $\mathbf{j}$ is the 3 -current density. Show that

$$F^{a b} U_{b}=\gamma(\mathbf{E} \cdot \mathbf{v}, \mathbf{E}+\mathbf{v} \times \mathbf{B})$$

In the rest frame of a conducting medium, Ohm's law states that $\mathbf{j}=\sigma \mathbf{E}$ where $\sigma$ is the conductivity. Show that the relativistic generalization to a frame in which the medium moves with uniform velocity $\mathbf{v}$ is

$$J^{a}-\left(J^{b} U_{b}\right) U^{a}=\sigma F^{a b} U_{b}$$

Show that this implies

$$\mathbf{j}=\rho \mathbf{v}+\sigma \gamma(\mathbf{E}+\mathbf{v} \times \mathbf{B}-(\mathbf{v} \cdot \mathbf{E}) \mathbf{v})$$

Simplify this formula, given that the charge density vanishes in the rest frame of the medium