Past exam of the optics and electrodynamics course of the University of Cambridge / 2012 / Paper 1 / 4 / c / Solution

In $S$, adding the stationary opposite line cancels charge density and electric field but contributes no current. HenceThe neutral current filament in a boosted frame is not neutral. Transform its line-integrated four-current:The same result follows by adding the original resting line $({\lambda }^{\prime },0)$ to the new negative line, whose primed density and current are $(-\gamma \lambda ,+\gamma \lambda u)$. In particular ${\lambda }^{\prime }-\gamma \lambda =-\gamma \lambda u^2/c^2$ because $\lambda =\gamma {\lambda }^{\prime }$.

Transforming the total fields givesFor positive $\lambda$, $\mathbf{u}\times \mathbf{B}$ points inward, explaining the electric-field sign. The primed Gauss and Ampère fields calculated from ${\lambda }_{\mathrm{tot}}^{\prime }$ and $I_{\mathrm{tot}}^{\prime }$ reproduce these expressions. Both a moving charge distribution and a moving current must be transformed; boosting just one constituent would give an inconsistent total.