Positive harmonic conformastatic electrovac data — Examples

The function in [A1] is smooth and strictly positive. Its derivatives are $H_i=2\lambda x^i$, $q_H=4{\lambda }^2\mathrm{\mid }x{\mathrm{\mid }}^2$, and ${\mathrm{\Delta }}_{\delta }H=6\lambda >0$. The potential ${\mathcal{A}}_{\sigma }=(H^{-1}-1)d{x}^0$ is smooth on all of $D$, and its field is $\mathcal{F}=d{\mathcal{A}}_{\sigma }$, with ${\mathcal{F}}_{i0}=-H_i\mathrm{/}{H}^2$. Since $d^2=0$, it satisfies $d\mathcal{F}=0$. The metric is Lorentzian by [F2].

In the orthonormal frame [F2], the field components satisfy $\mathcal{F}(n,e_i)=H\cdot H^{-1}{\mathcal{F}}_{0i}=H_i\mathrm{/}{H}^2$, so ${\mathcal{F}}_{ab}{\mathcal{F}}^{ab}=-2q_H\mathrm{/}{H}^4$ and $Q(n,n)=q_H\mathrm{/}(2H^4)$. Also ${\mathcal{F}}^{i0}=H_i\mathrm{/}{H}^2$ and $\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}=H^2$ by [F1, F2], hence ${\mathrm{\partial }}_i(\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}{\mathcal{F}}^{i0})={\mathrm{\partial }}_i{H}_i={\mathrm{\Delta }}_{\delta }H=6\lambda$. Using ${\mathrm{\Gamma }}^{\alpha }{}_{\alpha i}=2H_i\mathrm{/}H$ and ${\mathrm{\Gamma }}^{\alpha }{}_{\alpha 0}=0$ from [F3], together with symmetry of ${\mathrm{\Gamma }}^0{}_{ac}$ and antisymmetry of ${\mathcal{F}}^{ac}$, gives ${\mathrm{\nabla }}_a{\mathcal{F}}^{a0}={\mathrm{\partial }}_i(H_i\mathrm{/}{H}^2)+(2H_i\mathrm{/}H)(H_i\mathrm{/}{H}^2)={\mathrm{\Delta }}_{\delta }H\mathrm{/}{H}^2=6\lambda \mathrm{/}{H}^2\mathrm{\ne }0$. Define the smooth vector field $v$ by $v^b:={\mathrm{\nabla }}_a{\mathcal{F}}^{ab}$. Fix auxiliary constants ${\mu }_0>0$ and $c>0$, using $t=x^0\mathrm{/}c$, and define the smooth vector field $j:=-{\mu }_0^{-1}v$. Then ${\mathrm{\nabla }}_a{\mathcal{F}}^{ab}=-{\mu }_0{j}^b$ by construction, and $d\mathcal{F}=0$ by step 1.1, so all hypotheses of [F5] hold on the smooth, boundaryless, oriented and time-oriented chart manifold from [F1, F2]. The auxiliary $j$ is used only to apply this conditional mathematical identity. By [F5], $d({\star }_g\mathcal{F})=-{\iota }_v{\mathrm{v}\mathrm{o}\mathrm{l}}_g$. Contraction with the nowhere-zero volume form is injective, and the computed $v^0=6\lambda \mathrm{/}{H}^2$ is nonzero. Hence $d({\star }_g\mathcal{F})\mathrm{\ne }0$, violating the source-free Maxwell equation in [F4]. If this residual were interpreted as a matter current, that would require an additional sourced model; none is assumed here. The Ricci table [F3] gives $\mathrm{Ric}(n,n)=H^2{\mathrm{Ric}}_{00}=q_H\mathrm{/}{H}^4-{\mathrm{\Delta }}_{\delta }H\mathrm{/}{H}^3$. Therefore $\mathrm{Ric}(n,n)-2Q(n,n)=-{\mathrm{\Delta }}_{\delta }H\mathrm{/}{H}^3=-6\lambda \mathrm{/}{H}^3\mathrm{\ne }0$. This also violates the necessary Ricci equation [F4], so these smooth conformastatic data are not geometric electrovac. This counterexample uses the arbitrary-positive-$H$ supplier directly; the harmonic-only theorem is not applied outside its hypotheses.

Parametrize the unit sphere away from its seam and poles by $\mathrm{\Phi }(\theta ,\varphi )=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$. The tangent vectors ${\mathrm{\Phi }}_{\theta }$ and ${\mathrm{\Phi }}_{\varphi }$ are orthogonal, with squared lengths $1$ and ${\sin }^2\theta$, so the induced metric is $d{\theta }^2+{\sin }^2\theta d{\varphi }^2$ and its outward area form is ${\omega }_{S^2}=\sin \theta d\theta \wedge d\varphi$. The omitted seam and poles are chart-boundary sets of two-dimensional measure zero. Hence ${\int }_{S^2}{\omega }_{S^2}={\int }_0^{2\pi }{\int }_0^{\pi }\sin \theta d\theta d\varphi =4\pi$. The radius-$L$ sphere factor has area form $L^2{\omega }_{S^2}$ and area $4\pi L^2$. Thus its area-radius function is $R_A=\sqrt{(4\pi L^2)\mathrm{/}(4\pi )}=L$ at every $(\tau ,z)$.

The function $R_A$ is constant, so $d{R}_A=0$ everywhere. If it were a coordinate function in a chart $\phi :U\to V\subset {\mathbb{R}}^4$ at $p$, let $e_j$ be the coordinate vector for its component. Since $V$ is open, $\phi (p)+t{e}_j\in V$ for all sufficiently small $t$, and the chart's smooth coordinate parametrization gives a smooth curve $\gamma (t)={\phi }^{-1}(\phi (p)+t{e}_j)$. Then $R_A(\gamma (t))={\phi }^j(p)+t$, whose derivative at $t=0$ is $1$; thus $d{R}_A{\mathrm{\mid }}_p({\gamma }^{\mathrm{\prime }}(0))=1$, contradicting $d{R}_A{\mathrm{\mid }}_p=0$. Therefore $R_A$ cannot be a local coordinate.

By [F2], the example is an actual source-free electrovac product datum, so the failure is not caused by a missing field-equation condition. It refutes the universal claim in the Statement.

The choices in [A1] satisfy the smoothness and transverse-harmonic hypotheses in [F1]. They give $P=-a_0^2(x^2+y^2)$ and $\mathcal{F}=a_{0}du\wedge dx$, which is nonzero. Under the definition in [F2], [F3] proves this is a smooth source-free geometric electrovac datum.

Applying [F4] gives ${\mathcal{F}}_{ab}{\mathcal{F}}^{ab}=S={\mathrm{Ric}}_{ab}{\mathrm{Ric}}^{ab}=R_{abcd}{R}^{abcd}=0$. For the constant profile in [A1], [F3] also gives ${\mathrm{Ric}}_{uu}=2a_0^2\mathrm{\ne }0$. But $P_{xx}=-2a_0^2$, so $R_{uxux}=-P_{xx}\mathrm{/}2=a_0^2\mathrm{\ne }0$. Thus the field and Ricci tensor are nonzero and, by [F5], the metric is not flat although all four displayed scalar invariants vanish.

The Euclidean distances $\mathrm{\mid }x-a_1\mathrm{\mid }$ and $\mathrm{\mid }x-a_2\mathrm{\mid }$ are positive and smooth on $D$. The formula is the $N=2$ instance of the function in [F2], with positive coefficients and $\sigma =+1$, so it is positive and harmonic on $D$ and defines the metric-field data of [F1].

Apply [F2] with $N=2$, $a_1\mathrm{\ne }{a}_2$, $m_1,m_2>0$, and $\sigma =+1$. For every displayed $0<r<d$, the closed radius-$r$ ball about either center excludes the other center. Thus [F2] applies to the time-slice embeddings specified in the Example, giving the geometric electrovac equations on the full punctured domain and $q_e=m_A$, $q_m=0$ on every such sphere; these values are independent of $s$. The result is an exact mathematical example; none of the excluded centers is included as a regular point.