Positive harmonic conformastatic electrovac data

Since ${\mathcal{F}}_{\sigma }=d{\mathcal{A}}_{\sigma }$, the identity $d^2=0$ in [F5] gives $d{\mathcal{F}}_{\sigma }=0$. Raising the two indices of ${\mathcal{F}}_{i0}=-\sigma H_i\mathrm{/}{H}^2$ with [F2] gives ${\mathcal{F}}^{i0}=\sigma H_i\mathrm{/}{H}^2$ and ${\mathcal{F}}^{0i}=-\sigma H_i\mathrm{/}{H}^2$. Contracting the antisymmetric pair yields ${\mathcal{F}}_{ab}{\mathcal{F}}^{ab}=2{\sum }_i{\mathcal{F}}_{i0}{\mathcal{F}}^{i0}=-2q_H\mathrm{/}{H}^4$.

The last Christoffel term in ${\mathrm{\nabla }}_a{\mathcal{F}}^{ab}={\mathrm{\partial }}_a{\mathcal{F}}^{ab}+{\mathrm{\Gamma }}^a{}_{ac}{\mathcal{F}}^{cb}+{\mathrm{\Gamma }}^b{}_{ac}{\mathcal{F}}^{ac}$ vanishes because ${\mathrm{\Gamma }}^b{}_{ac}$ is symmetric in $a,c$ and ${\mathcal{F}}^{ac}$ is antisymmetric. For $b=0$, the remaining terms are ${\mathrm{\partial }}_i(\sigma H_i\mathrm{/}{H}^2)+(2H_i\mathrm{/}H)(\sigma H_i\mathrm{/}{H}^2)=\sigma {\mathrm{\Delta }}_{\delta }H\mathrm{/}{H}^2=0$, after the product rule cancels the two $q_H\mathrm{/}{H}^3$ terms. For $b=i$, ${\mathrm{\partial }}_0{\mathcal{F}}^{0i}=0$, all ${\mathcal{F}}^{ji}=0$, and the only possible trace term is ${\mathrm{\Gamma }}^{\alpha }{}_{\alpha 0}{\mathcal{F}}^{0i}=0$. Thus $v^b={\mathrm{\nabla }}_a{\mathcal{F}}^{ab}=0$ for every $b$, and [F4] gives $d({\star }_g{\mathcal{F}}_{\sigma })=0$.

The only nonzero contraction in ${\mathcal{F}}_{0c}{\mathcal{F}}_0{}^c$ is ${\sum }_i{F}_{0i}{F}_0{}^i=q_H\mathrm{/}{H}^6$, while ${\mathcal{F}}_{ic}{\mathcal{F}}_j{}^c=F_{i0}{g}^{00}{F}_{j0}=-H_i{H}_j\mathrm{/}{H}^2$. Combining these with ${\mathcal{F}}_{ab}{\mathcal{F}}^{ab}=-2q_H\mathrm{/}{H}^4$ and $g_{00}=-H^{-2}$, $g_{ij}=H^2{\delta }_{ij}$ gives $Q_{00}=q_H\mathrm{/}(2H^6)$, $Q_{0i}=0$, and $Q_{ij}=({\delta }_{ij}{q}_H-2H_i{H}_j)\mathrm{/}(2H^2)$.

Substituting ${\mathrm{\Delta }}_{\delta }H=0$ into the arbitrary-positive-$H$ table [F3] gives ${\mathrm{Ric}}_{00}=q_H\mathrm{/}{H}^6$, ${\mathrm{Ric}}_{0i}=0$, and ${\mathrm{Ric}}_{ij}=({\delta }_{ij}{q}_H-2H_i{H}_j)\mathrm{/}{H}^2$. These are exactly twice the entries in step 2.2; together with the closedness in step 1.1 and source-free divergence in step 2.1 they are all three equations in Geometric electrovac equations and flux normalization.

On the two-dimensional base, the inverse metric components are $g^{\tau \tau }=-1\mathrm{/}(L^{2}f)$ and $g^{zz}=f\mathrm{/}{L}^2$. Applying the Christoffel formula [F4] gives, apart from lower-index symmetry, ${\mathrm{\Gamma }}^{\tau }{}_{\tau z}=f^{\mathrm{\prime }}\mathrm{/}(2f)$, ${\mathrm{\Gamma }}^z{}_{\tau \tau }=f{f}^{\mathrm{\prime }}\mathrm{/}2$, and ${\mathrm{\Gamma }}^z{}_{zz}=-f^{\mathrm{\prime }}\mathrm{/}(2f)$. Their traces ${\mathrm{\Gamma }}^a{}_{\tau a}$ and ${\mathrm{\Gamma }}^a{}_{za}$ are zero. Substitution in the coordinate curvature formula [F6] and the Ricci contraction [F5], with convention [F2], gives ${\mathrm{Ric}}_{\tau \tau }={\mathrm{\partial }}_z(f{f}^{\mathrm{\prime }}\mathrm{/}2)-2(f^{\mathrm{\prime }}\mathrm{/}2f)(f{f}^{\mathrm{\prime }}\mathrm{/}2)=f{f}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}2$; similarly ${\mathrm{Ric}}_{zz}={\mathrm{\partial }}_z(-f^{\mathrm{\prime }}\mathrm{/}2f)-\{(f^{\mathrm{\prime }}\mathrm{/}2f{)}^2+(-f^{\mathrm{\prime }}\mathrm{/}2f{)}^2\}=-f^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}(2f)$, and ${\mathrm{Ric}}_{\tau z}=0$. Since $f^{\mathrm{\prime }\mathrm{\prime }}=2$, these are $f$ and $-1\mathrm{/}f$, respectively, which equal $-L^{-2}$ times the corresponding components of the base metric.

On a regular polar chart from [A2], the unit sphere metric $d{\theta }^2+{\sin }^2\theta d{\varphi }^2$ has, by [F4], the only nonzero symbols ${\mathrm{\Gamma }}^{\theta }{}_{\varphi \varphi }=-\sin \theta \cos \theta$ and ${\mathrm{\Gamma }}^{\varphi }{}_{\theta \varphi }={\mathrm{\Gamma }}^{\varphi }{}_{\varphi \theta }=\cot \theta$. The coordinate curvature formula [F6] and contraction [F5] give ${\mathrm{Ric}}_{\theta \theta }=-{\mathrm{\partial }}_{\theta }\cot \theta -{\cot }^2\theta =1$ and ${\mathrm{Ric}}_{\varphi \varphi }={\mathrm{\partial }}_{\theta }(-\sin \theta \cos \theta )-{\cos }^2\theta +2{\cos }^2\theta ={\sin }^2\theta$; the mixed component is zero. Scaling the sphere metric by the constant $L^2$ leaves its connection unchanged, so in the orthonormal coframe of the Statement its Ricci entries are $L^{-2},L^{-2}$. The product metric has no mixed Christoffel symbols because each factor coefficient depends only on its own factor coordinates; hence its mixed Ricci entries vanish.

The field is constant in $(\tau ,z)$ and has only ${\mathcal{F}}_{\tau z}=-L$ and ${\mathcal{F}}_{z\tau }=L$. Thus $d\mathcal{F}=0$. The determinant satisfies $\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}=L^4\sin \theta$, and raising indices gives ${\mathcal{F}}^{\tau z}=L^{-3}$ and ${\mathcal{F}}^{z\tau }=-L^{-3}$. In ${\mathrm{\nabla }}_a{\mathcal{F}}^{ab}={\mathrm{\partial }}_a{\mathcal{F}}^{ab}+{\mathrm{\Gamma }}^a{}_{ac}{\mathcal{F}}^{cb}+{\mathrm{\Gamma }}^b{}_{ac}{\mathcal{F}}^{ac}$, the last term vanishes by symmetry of the connection and antisymmetry of the field. For $b=\tau ,z$, the only derivative terms are ${\mathrm{\partial }}_z{\mathcal{F}}^{z\tau }$ and ${\mathrm{\partial }}_{\tau }{\mathcal{F}}^{\tau z}$, both zero; their trace terms vanish because ${\mathrm{\Gamma }}^a{}_{a\tau }={\mathrm{\Gamma }}^a{}_{az}=0$ from the base Christoffels and the product has no mixed symbols. For $b=\theta ,\varphi$, all ${\mathcal{F}}^{ab}$ vanish. Hence ${\mathrm{\nabla }}_a{\mathcal{F}}^{ab}=0$; [F3] gives $d({\star }_g\mathcal{F})=0$.

In the orthonormal coframe, ${\mathcal{F}}_L=-L^{-1}{e}^0\wedge e^1$, so ${\mathcal{F}}_{AB}{\mathcal{F}}^{AB}=-2\mathrm{/}{L}^2$. Direct contraction in signature $(-,+,+,+)$ gives $Q_{AB}=\frac{1}{2L^2}\mathrm{diag}(1,-1,1,1{)}_{AB}$: the electric pair contributes $+1\mathrm{/}(2L^2)$ to the time entry and $-1\mathrm{/}(2L^2)$ to the radial entry, while the trace term contributes $+1\mathrm{/}(2L^2)$ to each sphere entry. Steps 1.1 and 2.1 give $\mathrm{Ric}=2Q$ on the regular polar locus. The metric and two-form are smooth on the full product by [F1], so $\mathrm{Ric}(g)$ and $Q({\mathcal{F}}_L)$ are smooth tensor fields there; by density in [A2], their equality extends by continuity to both sphere poles. The displayed coframe components are local to those polar charts. Together with the Maxwell equations in step 3.1 this proves the full definition of a geometric electrovac datum on the stated domain.

In the ordered coordinates $(u,v,x,y)$ the metric matrix has $(u,v)$ block $\left(\begin{array}{cc}P& -1\\ -1& 0\end{array}\right)$ and transverse block $I_2$. Its inverse has $g^{uv}=g^{vu}=-1$, $g^{vv}=-P$, $g^{uu}=0$, $g^{xx}=g^{yy}=1$; the determinant is $-1$. The field has only ${\mathcal{F}}_{ux}=a$ and ${\mathcal{F}}_{uy}=b$, so direct raising gives ${\mathcal{F}}^{vx}=-a$, ${\mathcal{F}}^{vy}=-b$, with ${\mathcal{F}}^{ux}={\mathcal{F}}^{uy}=0$. Direct differentiation of $\mathcal{F}=a(u)du\wedge dx+b(u)du\wedge dy$ gives $d\mathcal{F}=0$, since each derivative term contains $du\wedge du$.

The nonzero Christoffel symbols, apart from lower-index symmetry, are ${\mathrm{\Gamma }}^v{}_{uu}=-P_u\mathrm{/}2$, ${\mathrm{\Gamma }}^v{}_{ui}=-P_i\mathrm{/}2$, and ${\mathrm{\Gamma }}^i{}_{uu}=-P_i\mathrm{/}2$ for $i\in \{x,y\}$. Substitution of the metric and inverse into [F4] gives these entries; the remaining metric derivatives vanish or lie in a constant block. Each trace ${\mathrm{\Gamma }}^a{}_{ac}$ is zero. Hence ${\mathrm{\nabla }}_a{\mathcal{F}}^{ab}={\mathrm{\partial }}_a{\mathcal{F}}^{ab}$, because ${\mathrm{\Gamma }}^b{}_{ac}{\mathcal{F}}^{ac}=0$ by symmetry. The only nonzero raised field components have upper pair $vx$ or $vy$, are independent of $v$, and have no component with upper $u$; therefore every ${\mathrm{\partial }}_a{\mathcal{F}}^{ab}=0$. By [F3], $d({\star }_g\mathcal{F})=0$.

The coordinate curvature and Ricci formulas [F5, F6] with the Christoffel list in step 2.1 have zero trace-derivative and trace-product terms because ${\mathrm{\Gamma }}^a{}_{ac}=0$. In the last product ${\mathrm{\Gamma }}^a{}_{bc}{\mathrm{\Gamma }}^c{}_{ad}$, $c=u$ gives a zero second factor (no Christoffel has upper $u$), $c=v$ gives a zero first factor (no Christoffel has lower $v$), and for transverse $c=i$ a nonzero first factor requires $a=v$, making the second factor zero because it then has lower $v$. Hence that product vanishes. In the derivative term ${\mathrm{\partial }}_a{\mathrm{\Gamma }}^a{}_{\nu \beta }$, the only nonzero derivatives are ${\mathrm{\partial }}_x{\mathrm{\Gamma }}^x{}_{uu}=-P_{xx}\mathrm{/}2$ and ${\mathrm{\partial }}_y{\mathrm{\Gamma }}^y{}_{uu}=-P_{yy}\mathrm{/}2$. Thus ${\mathrm{Ric}}_{uu}=-(P_{xx}+P_{yy})\mathrm{/}2$ and every other Ricci component is zero. From [F1], $P_{xx}+P_{yy}=-4(a^2+b^2)+h_{xx}+h_{yy}=-4(a^2+b^2)$, giving the stated Ricci component.

Raising the field indices from step 1.1 gives ${\mathcal{F}}_{ab}{\mathcal{F}}^{ab}=0$ and ${\mathcal{F}}_{uc}{\mathcal{F}}_u{}^c=a^2+b^2$; all other ${\mathcal{F}}_{ac}{\mathcal{F}}_b{}^c$ entries vanish. Hence $Q_{uu}=a^2+b^2$ and all other $Q$ entries vanish. Comparing with step 3.1 proves $\mathrm{Ric}=2Q$. Together with the closedness in step 1.1, the source-free divergence in step 2.1, and [F2], this proves the geometric electrovac equations.

For $z\mathrm{\ne }0$, direct differentiation gives ${\mathrm{\partial }}_i\mathrm{\mid }z{\mathrm{\mid }}^{-1}=-z_i\mathrm{\mid }z{\mathrm{\mid }}^{-3}$ and ${\mathrm{\partial }}_i^2\mathrm{\mid }z{\mathrm{\mid }}^{-1}=-\mathrm{\mid }z{\mathrm{\mid }}^{-3}+3z_i^2\mathrm{\mid }z{\mathrm{\mid }}^{-5}$. Summing over $i$ yields ${\mathrm{\Delta }}_{\delta }\mathrm{\mid }z{\mathrm{\mid }}^{-1}=-3\mathrm{\mid }z{\mathrm{\mid }}^{-3}+3\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{\mid }z{\mathrm{\mid }}^{-5}=0$. Every term in $H$ is smooth on $D$, and every term is positive there, so $H>0$ and ${\mathrm{\Delta }}_{\delta }H=0$ on $D$.

Inverting the diagonal metric in [F1] gives $g^{00}=-H^2$ and $g^{ij}=H^{-2}{\delta }^{ij}$, so raising ${\mathcal{F}}_{i0}=-\sigma H_i\mathrm{/}{H}^2$ gives ${\mathcal{F}}^{i0}=\sigma H_i\mathrm{/}{H}^2$. The metric determinant is $(-H^{-2})(H^2{)}^3=-H^4$, so the oriented volume form is ${\mathrm{v}\mathrm{o}\mathrm{l}}_g=H^{2}d{x}^0\wedge d{x}^1\wedge d{x}^2\wedge d{x}^3$. The Hodge formula [F4], with ${ϵ}_{123}=1$ and ${ϵ}_{ijk}$ alternating, therefore gives $({\star }_g{\mathcal{F}}_{\sigma }{)}_{ij}=-\sigma {ϵ}_{ijk}{H}_k$. For each sphere $S\subset D$ and fixed $s$, the map $i_{s,S}$ in the Statement is smooth and injective, its differential is $v\mapsto (0,v)$ of rank two, and projection onto $D$ restricts to its continuous inverse on its image. Thus it is a smooth embedding as required by [F3]. The sphere is compact and disjoint from all centers, so both forms are smooth on a neighborhood of its image and their pullbacks are admissible for [F3]. Pulling back to the outward-oriented Euclidean sphere gives ${\int }_S{i}_{s,S}^{\ast }({\star }_g{\mathcal{F}}_{\sigma })=-\sigma {\int }_S{\mathrm{\partial }}_{\nu }Hd{S}_{\delta }$, because ${\mathrm{\partial }}_{\nu }H=n^k{H}_k$ and the induced oriented area form has components ${ϵ}_{ijk}{n}^{k}d{S}_{\delta }$. The pullback of ${\mathcal{F}}_{\sigma }$ to each spatial sphere is zero because every term in [F1] contains $d{x}^0$, which vanishes on its tangent vectors; hence $q_m(S)=0$. All pulled-back coefficients are independent of $s$, so both fluxes are independent of the chosen time slice.

Let $S_r(a)$ be a Euclidean sphere with center $a$ and radius $r$, and let $m\mathrm{/}\mathrm{\mid }x-b\mathrm{\mid }$ be one term in $H$, where $d:=\mathrm{\mid }b-a\mathrm{\mid }\mathrm{\ne }r$. At $x=a+rn$ with $n\in S^2$, its outward normal derivative is $-m(r-n\cdot (b-a))\mathrm{/}\mathrm{\mid }rn-(b-a){\mathrm{\mid }}^3$. The sphere area element is $r^{2}d\mathrm{\Omega }$, where $d\mathrm{\Omega }=\sin \theta d\theta d\varphi$ and ${\int }_{S^2}d\mathrm{\Omega }=2\pi {\int }_0^{\pi }\sin \theta d\theta =4\pi$. If $d=0$, direct integration gives flux $-4\pi m$. If $d>0$, choose the polar axis along $b-a$; this orthogonal change preserves the induced sphere area. With $u=\cos \theta$ and $n\cdot (b-a)=du$, the flux is $-2\pi m{r}^{2}J$, where $J:={\int }_{-1}^1\frac{r-du}{(r^2+d^2-2rdu{)}^{3\mathrm{/}2}}du\mathrm{.}$ Set $w=r^2+d^2-2rdu$. Then $du=-dw\mathrm{/}(2rd)$, the endpoint square roots for $w$ are $r+d$ and $\mathrm{\mid }r-d\mathrm{\mid }$, and $r-du=(r^2-d^2+w)\mathrm{/}(2r)$. Therefore $J=\frac{1}{4r^{2}d}((r^2-d^2){\int }_{(r-d{)}^2}^{(r+d{)}^2}{w}^{-3\mathrm{/}2}dw+{\int }_{(r-d{)}^2}^{(r+d{)}^2}{w}^{-1\mathrm{/}2}dw)\mathrm{.}$ Using antiderivatives $-2w^{-1\mathrm{/}2}$ and $2w^{1\mathrm{/}2}$ gives $J=\frac{(r^2-d^2)(\mathrm{\mid }r-d{\mathrm{\mid }}^{-1}-(r+d{)}^{-1})+(r+d-\mathrm{\mid }r-d\mathrm{\mid })}{2r^{2}d}\mathrm{.}$ If $d<r$, the numerator is $4d$, so $J=2\mathrm{/}{r}^2$; if $d>r$, the numerator is $0$, so $J=0$. Thus a Coulomb term contributes $-4\pi m$ to the outward flux exactly when its center lies inside the sphere, and zero when its center lies outside.

The conditions in [F1] now hold, so the defined data exist on the entire punctured domain. Applying [F2] proves the source-free field equations there; no equation is asserted at a removed center.

For an isolated sphere $S_A(r)$, its own center has $d=0$, while each other center has distance $d_{AB}>r$ by [A1]. Applying step 1.3 term by term gives ${\int }_{S_A(r)}{\mathrm{\partial }}_{\nu }Hd{S}_{\delta }=-4\pi m_A$. Step 1.2 yields $q_e(S_A(r))=\sigma m_A$, and that step also gives $q_m(S_A(r))=0$.

On an outer sphere that strictly encloses every center, each term has $d=\mathrm{\mid }{a}_A\mathrm{\mid }<R$. Step 1.3 gives ${\int }_{S_R}{\mathrm{\partial }}_{\nu }Hd{S}_{\delta }=-4\pi {\sum }_A{m}_A$. Step 1.2 therefore yields $q_e(S_R)=\sigma {\sum }_A{m}_A$ and $q_m(S_R)=0$.