The first vacuum <Maxwell equations> equation is preserved for every vector field because part a gives
$$
d\widetilde F=d\mathcal L_XF=\mathcal L_X(dF)=0.
$$
On two-forms in four dimensions, the mixed volume tensor $T_{ab}{}^{cd}$ represents twice the <Hodge star operator>. Part b therefore says that a conformal Killing field commutes with the Hodge star:
$$
\mathcal L_X(\star F)=\star\mathcal L_XF.
$$
It follows that
$$
d(\star\widetilde F)
=d\mathcal L_X(\star F)
=\mathcal L_Xd(\star F)=0.
$$
Hence $\widetilde F=\mathcal L_XF$ satisfies both vacuum Maxwell equations whenever $F$ does.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 309 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-309/4/c/solution