 Solution

Use rationalized electromagnetic units with $c=1$ and electrostatic constant $1/(4\pi )$. The four-potential $A^a=(\phi ,A_x,A_y,A_z)$ transforms by the same matrix as the coordinates:Write $r=\sqrt{y^2+z^2}$ and ${\mathbf{e}}_r=(0,y,z)/r$. Gauss's law for the line at rest gives ${\mathbf{E}}^{\prime }=\sigma {\mathbf{e}}_r/(2\pi r)$, ${\mathbf{B}}^{\prime }=0$. A suitable potential is ${\phi }^{\prime }=-\sigma \log (r/r_{\ast })/(2\pi )$, ${\mathbf{A}}^{\prime }=0$. The inverse Lorentz transformation gives $\phi =\gamma {\phi }^{\prime }$, $A_x=\gamma v{\phi }^{\prime }$. The fields from $\mathbf{E}=-\nabla \phi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$ are thereforeThe sign agrees with a line charge moving in the positive $x$ direction: its density is $\gamma \sigma$ and its current $\gamma v\sigma$. In SI units insert $1/{ϵ}_0$ in the electric expressions and replace $\mathbf{B}=\mathbf{v}\times \mathbf{E}$ by $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$.

The two invariants check explicitly:For $|v|\ll 1$, the electric field differs from the static field only at order $v^2$, while the magnetic field is of order $v$ and is the field of the moving line current. These fields are evaluated off the charged axis.