Finite positive multicenter MP exterior and fluxes

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 $q_m(S_R)=0$.