 Solution

Assume an infinite, stationary, uniform thin wire, and no added external homogeneous fields. Its four-current is $j^{\mu }=(\sigma ,0,0,J)\delta (x)\delta (y)$. A boost along the wire leaves the transverse delta functions unchanged and transforms the time and longitudinal components, givingThe densities are measured per unit simultaneous length in their respective frames; treating charge alone as invariant per unchanged coordinate length would miss the current contribution.

For a coaxial cylinder, Gauss's law gives $E_r(2\pi rL)={\mu }_0\sigma L$ in the stated $c=1$ units. For a circular loop, Ampère's law gives $B_{\phi }2\pi r={\mu }_{0}J$. Thus $E_r={\mu }_0\sigma /(2\pi r)$ and $B_{\phi }={\mu }_{0}J/(2\pi r)$. The same reasoning in the primed frame gives the same expressions with primed densities.

Since $\hat{z}\times \hat{\phi }=-\hat{r}$ and $\hat{z}\times \hat{r}=\hat{\phi }$, the Lorentz transformation of electromagnetic fields gives $E_r^{\prime }=\gamma (E_r-v{B}_{\phi })$ and $B_{\phi }^{\prime }=\gamma (B_{\phi }-v{E}_r)$. Substituting the laboratory fields reproduces the primed-density formulas exactly; parallel components vanish in both frames.