Dyonic rotating model and gauge obstruction

By [F1] and the seed's zero magnetic flux, its geometric electric parameter is $q_0={\gamma }_e{Q}_0$ and its geometric magnetic parameter is zero. The exact seed statement [F3] supplies the geometric electrovac hypotheses for [L1]; the conditional physical-model statement [F2] supplies the source-free SI solution on its stated pole-free and full-sphere exterior domains. Each cited carrier is used at its current draft interface only.

Let $F_0={\mathcal{F}}_0\mathrm{/}\alpha$ be the SI seed field. By [L1], the geometric dual-rotated field is closed with closed Hodge dual and has the same quadratic stress as ${\mathcal{F}}_0$; since $\alpha >0$ is constant, $d{F}_{\beta }=d({\star }_g{F}_{\beta })=0$ for $F_{\beta }={\mathcal{F}}_{\beta }\mathrm{/}\alpha$. Put $V^b:={\mathrm{\nabla }}_a{F}_{\beta }^{ab}$. In an oriented coordinate chart write ${\mathrm{v}\mathrm{o}\mathrm{l}}_g=wd{x}^0\wedge d{x}^1\wedge d{x}^2\wedge d{x}^3$ with $w=\sqrt{\mathrm{\mid }\det (g_{ab})\mathrm{\mid }}$; this coefficient follows from the defining value of the metric volume form on a positively oriented orthonormal basis and the determinant change-of-basis formula. Differentiating the determinant expansion gives ${\mathrm{\partial }}_{c}w=\frac{1}{2}w{g}^{ab}{\mathrm{\partial }}_c{g}_{ab}$; the Christoffel formula and symmetry of $g$ give ${\mathrm{\Gamma }}^a{}_{ca}=\frac{1}{2}{g}^{ab}{\mathrm{\partial }}_c{g}_{ab}$. Hence $({\mathrm{\nabla }}_c{\mathrm{v}\mathrm{o}\mathrm{l}}_g{)}_{0123}={\mathrm{\partial }}_{c}w-{\mathrm{\Gamma }}^a{}_{ca}w=0$, so $\mathrm{\nabla }{\mathrm{v}\mathrm{o}\mathrm{l}}_g=0$. At each point $p$, choose a positively oriented orthonormal frame with metric matrix $\eta =\mathrm{diag}(-1,1,1,1)$. Torsion-freeness of the metric-selected connection [F4] gives $(d{\star }_g{F}_{\beta }{)}_{abc}=3{\mathrm{\nabla }}_{[a}({\star }_g{F}_{\beta }{)}_{bc]}$. The Hodge component formula [F7] has $({\star }_g{F}_{\beta }{)}_{ij}=-{ϵ}_{ijk}(F_{\beta }{)}_{0k}$ and $({\star }_g{F}_{\beta }{)}_{0i}=\frac{1}{2}{ϵ}_{ijk}(F_{\beta }{)}_{jk}$, where ${ϵ}_{ijk}$ is alternating with ${ϵ}_{123}=+1$. In this frame $V^0={\sum }_i{\mathrm{\nabla }}_i(F_{\beta }{)}_{0i}$ and $V^k=-{\mathrm{\nabla }}_0(F_{\beta }{)}_{0k}+{\sum }_i{\mathrm{\nabla }}_i(F_{\beta }{)}_{ik}$. Substituting the displayed Hodge components into $3{\mathrm{\nabla }}_{[a}({\star }_g{F}_{\beta }{)}_{bc]}$ gives $(d{\star }_g{F}_{\beta }{)}_{123}=-V^0,(d{\star }_g{F}_{\beta }{)}_{0ij}={ϵ}_{ijk}{V}^k\mathrm{.}$ In the same oriented frame, $({\iota }_V{\mathrm{v}\mathrm{o}\mathrm{l}}_g{)}_{123}=V^0$ and $({\iota }_V{\mathrm{v}\mathrm{o}\mathrm{l}}_g{)}_{0ij}=-{ϵ}_{ijk}{V}^k$, so these components establish $d({\star }_g{F}_{\beta }){\mathrm{\mid }}_p=-{\iota }_{V(p)}{\mathrm{v}\mathrm{o}\mathrm{l}}_g{\mathrm{\mid }}_p$. Its left side is zero. Contraction with the nonzero volume form is an isomorphism $T_{p}M\to {\mathrm{\Lambda }}^3{T}_p^{\ast }M$, so $V(p)=0$; hence ${\mathrm{\nabla }}_a{F}_{\beta }^{ab}=0$ throughout every stated domain. Thus the source-free SI Maxwell equations hold without using the equivalence clause of [F4]. Its electromagnetic stress is unchanged because $Q(F_{\beta })={\alpha }^{-2}Q({\mathcal{F}}_{\beta })={\alpha }^{-2}Q({\mathcal{F}}_0)=Q(F_0)$. The Einstein tensor and metric are unchanged, so the same zero-source SI equation-based model equations hold on every base domain of [F2].

On the full-sphere exterior [F8], let $y=\cos \theta$, $\mathrm{\Sigma }=r^2+a^2{y}^2$, and $R=r^2+a^2$. Differentiating the exact seed potential in [F2] gives $({\mathcal{F}}_0{)}_{r0}=q_0(r^2-a^2{y}^2)\mathrm{/}{\mathrm{\Sigma }}^2$ and $({\mathcal{F}}_0{)}_{r\varphi }=-a{\sin }^2\theta ({\mathcal{F}}_0{)}_{r0}$. Raising indices with the KN inverse metric and applying [F7] yields $({\star }_g{\mathcal{F}}_0{)}_{\theta \varphi }=q_{0}R(r^2-a^2{y}^2)\sin \theta \mathrm{/}{\mathrm{\Sigma }}^2$. Since $\frac{d}{dy}(y\mathrm{/}\mathrm{\Sigma })=(r^2-a^2{y}^2)\mathrm{/}{\mathrm{\Sigma }}^2$, integration over the outward sphere gives $\frac{1}{4\pi }{\int }_{S_r}{\star }_g{\mathcal{F}}_0=\frac{q_{0}R}{2}{\int }_{-1}^1\frac{r^2-a^2{y}^2}{(r^2+a^2{y}^2{)}^2}dy=\frac{q_{0}R}{2}{\left[\frac{y}{\mathrm{\Sigma }}\right]}_{-1}^1=q_0\mathrm{.}$ The global potential in [F8] and Stokes' theorem [F5] give ${\int }_{S_r}{\mathcal{F}}_0={\int }_{S_r}d{\mathcal{A}}_0=0$. By [F7], ${\star }_g^2=-1$ on two-forms, so integrating ${\mathcal{F}}_{\beta }=\cos \beta {\mathcal{F}}_0+\sin \beta {\star }_g{\mathcal{F}}_0$ and its Hodge dual directly yields $q_e=q_0\cos \beta$ and $q_m=q_0\sin \beta$. Therefore $q_e^2+q_m^2=q_0^2({\cos }^2\beta +{\sin }^2\beta )=q_0^2$, and every KN metric coefficient depending on charge through this sum is unchanged. This local integral calculation proves the stated flux values.

The electric conversion in [F1] gives $Q_e^{\mathrm{o}\mathrm{u}\mathrm{t}}=q_e\mathrm{/}{\gamma }_e=Q_0\cos \beta$, using the flux parameter from step 2.2. For the magnetic flux, [F1] gives ${\mathrm{\Phi }}_B^{\mathrm{o}\mathrm{u}\mathrm{t}}=4\pi q_m\mathrm{/}\alpha$, with $q_m$ from step 2.2. From ${\epsilon }_0{\mu }_0{c}^2=1$, ${\left(\frac{\alpha {\mu }_{0}c}{4\pi }\right)}^2=\frac{G_N{\mu }_0}{4\pi c^2}={\gamma }_e^2\mathrm{.}$ Both $\alpha ,{\mu }_0,c,{\gamma }_e$ are positive, hence $\alpha {\mu }_{0}c\mathrm{/}(4\pi )={\gamma }_e$. Substituting $q_m={\gamma }_e{Q}_0\sin \beta$ gives ${\mathrm{\Phi }}_B^{\mathrm{o}\mathrm{u}\mathrm{t}}={\mu }_{0}c{Q}_0\sin \beta$. These conversions preserve the stated SI flux convention and introduce no magnetic-charge relabeling.

Assume a global smooth real one-form ${\mathcal{A}}_{\beta }$ on a sphere-containing exterior with $d{\mathcal{A}}_{\beta }={\mathcal{F}}_{\beta }$ and $q_m\mathrm{\ne }0$. Represent the fixed sphere $S_r$ by radially projecting the eight oriented triangular faces of the octahedron onto the unit sphere and taking their signed sum. Each face is disjoint from the origin, so its radial projection is smooth; common edges cancel with opposite orientations. This is a finite smooth singular $2$-cycle, so $\mathrm{\partial }{S}_r=0$. By [F5] and step 2.2, $4\pi q_m={\int }_{S_r}{\mathcal{F}}_{\beta }={\int }_{S_r}d{\mathcal{A}}_{\beta }={\int }_{\mathrm{\partial }{S}_r}{\mathcal{A}}_{\beta }=0,$ a contradiction. Thus no such global potential exists on the rotating sphere-containing exterior when its magnetic flux is nonzero.

If $a=0$, the KN potential in [F6] reduces to $-q_0{r}^{-1}d{x}^0$, so its exterior derivative is ${\mathcal{F}}_0=-q_0{r}^{-2}d{x}^0\wedge dr$. On the sphere-containing exterior $f>0$, the positively oriented orthonormal coframe $e^0=\sqrt{f}d{x}^0$, $e^1=f^{-1\mathrm{/}2}dr$, $e^2=rd\theta$, $e^3=r\sin \theta d\varphi$ gives $d{x}^0\wedge dr=e^0\wedge e^1$ and ${\star }_g(e^0\wedge e^1)=-e^2\wedge e^3$ by the Hodge component convention [F7]. Hence ${\star }_g{\mathcal{F}}_0=q_0\omega$. Therefore the dual rotation is ${\mathcal{F}}_{\beta }=-q_0\cos \beta r^{-2}d{x}^0\wedge dr+q_0\sin \beta \omega ,$ which is exactly the dyonic RN field of [F6] with the flux parameters in step 2.2. The exact current interface [L3] supplies the two patch potentials and the criterion for a global real potential in this nonrotating case.

Steps 1.1-3.3 establish the conditional dual-rotated model, full-sphere flux identifications and global-potential obstruction. The conventions use only the inherited ${\mathrm{A}\mathrm{C}}_{\omega }$ in [A2]; they impose no full AC, flux quantization, or empirical claim.