= Solution
For a boost along positive $x$, let $\gamma=(1-v_x^2/c^2)^{-1/2}$. Parallel field components are unchanged, while transverse components mix. The <Lorentz transformation of electromagnetic fields> gives
$$
\boxed{\begin{aligned}
E'_x&=E_x,&E'_y&=\gamma(E_y-v_xB_z),&E'_z&=\gamma(E_z+v_xB_y),\\
B'_x&=B_x,&B'_y&=\gamma(B_y+v_xE_z/c^2),&B'_z&=\gamma(B_z-v_xE_y/c^2).
\end{aligned}}
$$
These signs correspond to $S'$ moving with $+v_x$ relative to $S$, not the inverse boost. They can be obtained from $F'=\Lambda F\Lambda^{\mathsf T}$ for the <electromagnetic field tensor>.
Back to article page
