Classical electrovac dual rotation

Since $\beta$ is constant, $d{\mathcal{F}}_{\beta }=\cos \beta d\mathcal{F}+\sin \beta d({\star }_g\mathcal{F})=0$ by [F1].

By [F2], ${\star }_g{\mathcal{F}}_{\beta }=\cos \beta {\star }_g\mathcal{F}+\sin \beta {\star }_g^2\mathcal{F}=\cos \beta {\star }_g\mathcal{F}-\sin \beta \mathcal{F}$. Applying $d$ and using [F1] proves $d({\star }_g{\mathcal{F}}_{\beta })=0$.

In a positive oriented orthonormal frame let $E_i=-{\mathcal{F}}_{0i}$ and $B_i=\frac{1}{2}{ϵ}_{ijk}{\mathcal{F}}_{jk}$. The component formula [F2] gives $E(\star \mathcal{F})=-B$ and $B(\star \mathcal{F})=E$, hence $E_{\beta }=E\cos \beta -B\sin \beta$ and $B_{\beta }=B\cos \beta +E\sin \beta$. Direct expansion gives $E_{\beta }^2+B_{\beta }^2=E^2+B^2$, $E_{\beta }\times B_{\beta }=E\times B$, and $E_{\beta i}{E}_{\beta j}+B_{\beta i}{B}_{\beta j}=E_i{E}_j+B_i{B}_j$: the mixed symmetric terms cancel and $B\times E=-E\times B$.

The observer-frame components of the quadratic tensor $Q_{ab}={\mathcal{F}}_{ac}{\mathcal{F}}_b{}^c-\frac{1}{4}{g}_{ab}{\mathcal{F}}^2$ are $Q_{00}=\frac{1}{2}(E^2+B^2)$, $Q_{0i}=-(E\times B{)}_i$, and $Q_{ij}=-E_i{E}_j-B_i{B}_j+\frac{1}{2}{\delta }_{ij}(E^2+B^2)$, by direct raising with $\eta =\mathrm{diag}(-1,1,1,1)$. Each is unchanged by step 1.3, so tensoriality gives $Q({\mathcal{F}}_{\beta })=Q(\mathcal{F})$ globally. This proof uses only the displayed geometric quadratic tensor and does not assert an SI normalization.

By steps 1.1-1.2 the rotated field is closed with closed dual, and by step 2.1 its quadratic tensor is unchanged. Substitution in [F1] therefore gives the same Ricci equation for ${\mathcal{F}}_{\beta }$, proving the Statement.