Conformal constraint weights and charged time-symmetric data

In any positively oriented coordinate chart, $\det h={\psi }^{12}\det \bar{h}$, so $\sqrt{\det h}={\psi }^6\sqrt{\det \bar{h}}$. The Levi-Civita formula and the determinant derivative give ${\mathrm{\Gamma }}^i{}_{ik}={\mathrm{\partial }}_k\log \sqrt{\det h}$, hence $D_i{V}^i=(\sqrt{\det h}{)}^{-1}{\mathrm{\partial }}_i(\sqrt{\det h}{V}^i)$ for every smooth vector field $V$. Substituting $V=E={\psi }^{-6}\bar{E}$ and then $V=B={\psi }^{-6}\bar{B}$ gives $D_i{E}^i={\psi }^{-6}{\bar{D}}_i{\bar{E}}^i$ and $D_i{B}^i={\psi }^{-6}{\bar{D}}_i{\bar{B}}^i$. Since $e={\alpha }^{-1}E$, $b={\alpha }^{-1}B$, and both factors are positive, the two SI divergence constraints are equivalent to the two barred divergence equations.

The oriented volume forms satisfy ${\mathrm{vol}}_h={\psi }^6{\mathrm{vol}}_{\bar{h}}$. In coordinates, $(E{\times }_{h}B{)}_i=({\mathrm{vol}}_h{)}_{ijk}{E}^j{B}^k={\psi }^{-6}(\bar{E}{\times }_{\bar{h}}\bar{B}{)}_i$. Also $\mathrm{\mid }E{\mathrm{\mid }}_h^2={\psi }^{-8}\mathrm{\mid }\bar{E}{\mathrm{\mid }}_{\bar{h}}^2$ and $\mathrm{\mid }B{\mathrm{\mid }}_h^2={\psi }^{-8}\mathrm{\mid }\bar{B}{\mathrm{\mid }}_{\bar{h}}^2$, by inserting $h_{ij}={\psi }^4{\bar{h}}_{ij}$ and the two vector weights. Therefore $(\kappa \mathrm{/}{\mu }_0)(\mathrm{\mid }e{\mathrm{\mid }}_h^2+\mathrm{\mid }b{\mathrm{\mid }}_h^2)=2{\psi }^{-8}(\mathrm{\mid }\bar{E}{\mathrm{\mid }}_{\bar{h}}^2+\mathrm{\mid }\bar{B}{\mathrm{\mid }}_{\bar{h}}^2)$ and $(\kappa \mathrm{/}{\mu }_0)(e{\times }_{h}b{)}_i=2{\psi }^{-6}(\bar{E}{\times }_{\bar{h}}\bar{B}{)}_i$, using [F3].

Taking the $h$-trace of $K=(\tau \mathrm{/}3)h+{\psi }^{-2}\bar{A}$ gives ${\mathrm{tr}}_{h}K=\tau +{\psi }^{-6}{\bar{h}}^{ij}{\bar{A}}_{ij}=\tau$, because $\bar{A}$ is trace-free. In the squared norm, the cross term between $(\tau \mathrm{/}3)h$ and ${\psi }^{-2}\bar{A}$ vanishes by the same trace condition; the pure trace term is ${\tau }^2\mathrm{/}3$, and the trace-free term is ${\psi }^{-12}\mathrm{\mid }\bar{A}{\mathrm{\mid }}_{\bar{h}}^2$. Thus $\mathrm{\mid }K{\mathrm{\mid }}_h^2={\tau }^2\mathrm{/}3+{\psi }^{-12}\mathrm{\mid }\bar{A}{\mathrm{\mid }}_{\bar{h}}^2$.

Put $u=2\log \psi$, so $h=e^{2u}\bar{h}$. In a chart, metric compatibility and torsion-freeness give $\mathrm{\Gamma }[h{]}^k{}_{ij}=\frac{1}{2}{h}^{k\mathrm{\ell }}({\mathrm{\partial }}_i{h}_{j\mathrm{\ell }}+{\mathrm{\partial }}_j{h}_{i\mathrm{\ell }}-{\mathrm{\partial }}_{\mathrm{\ell }}{h}_{ij})$, hence $C^k{}_{ij}:=\mathrm{\Gamma }[h{]}^k{}_{ij}-\mathrm{\Gamma }[\bar{h}{]}^k{}_{ij}={\delta }_i^k{u}_j+{\delta }_j^k{u}_i-{\bar{h}}_{ij}{u}^k$. Expanding the curvature definition and contracting gives $\mathrm{Ric}(h{)}_{ij}=\mathrm{Ric}(\bar{h}{)}_{ij}+{\bar{D}}_k{C}^k{}_{ij}-{\bar{D}}_i{C}^k{}_{kj}+C^k{}_{ij}{C}^{\mathrm{\ell }}{}_{k\mathrm{\ell }}-C^k{}_{i\mathrm{\ell }}{C}^{\mathrm{\ell }}{}_{kj}$. Here $C^k{}_{kj}=3u_j$, ${\bar{D}}_k{C}^k{}_{ij}=2{\bar{D}}_i{u}_j-{\bar{h}}_{ij}\bar{\mathrm{\Delta }}u$, and ${\bar{D}}_i{C}^k{}_{kj}=3{\bar{D}}_i{u}_j$ because $du$ is closed. Expanding the products gives $C^k{}_{ij}{C}^{\mathrm{\ell }}{}_{k\mathrm{\ell }}=6u_i{u}_j-3{\bar{h}}_{ij}\mathrm{\mid }du{\mathrm{\mid }}_{\bar{h}}^2$ and $C^k{}_{i\mathrm{\ell }}{C}^{\mathrm{\ell }}{}_{kj}=5u_i{u}_j-2{\bar{h}}_{ij}\mathrm{\mid }du{\mathrm{\mid }}_{\bar{h}}^2$; tracing with $h^{ij}={\psi }^{-4}{\bar{h}}^{ij}$ therefore yields $R(h)={\psi }^{-4}(\bar{R}-4\bar{\mathrm{\Delta }}u-2\mathrm{\mid }du{\mathrm{\mid }}_{\bar{h}}^2)$. Substitution of $u=2\log \psi$ and $\bar{\mathrm{\Delta }}\log \psi ={\psi }^{-1}{\mathrm{\Delta }}_{\bar{h}}\psi -{\psi }^{-2}\mathrm{\mid }d\psi {\mathrm{\mid }}_{\bar{h}}^2$ cancels the gradient-square terms and gives $R(h)={\psi }^{-5}(\bar{R}\psi -8{\mathrm{\Delta }}_{\bar{h}}\psi )$.

Substituting steps 1.3 and 1.4 into the left side of the SI Hamiltonian constraint [F2], and step 1.2 into its right side, gives ${\psi }^{-5}(-8{\mathrm{\Delta }}_{\bar{h}}\psi +\bar{R}\psi )-{\psi }^{-12}\mathrm{\mid }\bar{A}{\mathrm{\mid }}_{\bar{h}}^2+\frac{2}{3}{\tau }^2=2\mathrm{\Lambda }+2{\psi }^{-8}(\mathrm{\mid }\bar{E}{\mathrm{\mid }}_{\bar{h}}^2+\mathrm{\mid }\bar{B}{\mathrm{\mid }}_{\bar{h}}^2)$. Multiplication by the positive ${\psi }^5$ is reversible and yields exactly the transformed scalar equation in the Statement; step 1.1 gives the two equivalent Maxwell constraints.

By step 1.2, the right side of the SI momentum constraint in [F2] is $-2{\psi }^{-6}(\bar{E}{\times }_{\bar{h}}\bar{B}{)}_i$, which proves the displayed retained momentum equation. If $K=0$, its left side vanishes; if also $\bar{E}{\times }_{\bar{h}}\bar{B}=0$, its right side vanishes. No step has inferred this vector equation from the scalar constraint.

Put $C:=m^2-q^2$ and $w:={\psi }^2$. Since $2r+m>0$, the condition $w>0$ is equivalent to $2r+m>\mathrm{\mid }q\mathrm{\mid }$, giving exactly ${\mathcal{D}}_{m,q}$. Thus $\psi$ is smooth and positive there. In Euclidean spherical coordinates, ${\bar{D}}_i{\bar{E}}^i=r^{-2}{\mathrm{\partial }}_r(r^{2}q{r}^{-2})=0$, while ${\bar{D}}_i{\bar{B}}^i=0$.

On this domain $w:={\psi }^2=1+m\mathrm{/}r+C\mathrm{/}(4r^2)$, $w^{\mathrm{\prime }}=-m\mathrm{/}{r}^2-C\mathrm{/}(2r^3)$, and $w^{\mathrm{\prime }\mathrm{\prime }}=2m\mathrm{/}{r}^3+3C\mathrm{/}(2r^4)$. Since ${\mathrm{\Delta }}_{{\mathbb{R}}^3}\psi ={\psi }^{\mathrm{\prime }\mathrm{\prime }}+2{\psi }^{\mathrm{\prime }}\mathrm{/}r$, ${\psi }^{\mathrm{\prime }}\mathrm{/}\psi =w^{\mathrm{\prime }}\mathrm{/}(2w)$ and ${\psi }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}\psi =w^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}(2w)-(w^{\mathrm{\prime }}{)}^2\mathrm{/}(4w^2)$ give ${\mathrm{\Delta }}_{{\mathbb{R}}^3}\psi \mathrm{/}\psi =(w^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}2+w^{\mathrm{\prime }}\mathrm{/}r)\mathrm{/}w-(w^{\mathrm{\prime }}{)}^2\mathrm{/}(4w^2)=(Cw\mathrm{/}{r}^4-(w^{\mathrm{\prime }}{)}^2)\mathrm{/}(4w^2)=-q^2\mathrm{/}(4r^4{w}^2)$; here $w^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}2+w^{\mathrm{\prime }}\mathrm{/}r=C\mathrm{/}(4r^4)$ and direct expansion gives $Cw\mathrm{/}{r}^4-(w^{\mathrm{\prime }}{)}^2=-q^2\mathrm{/}{r}^4$. Thus ${\mathrm{\Delta }}_{{\mathbb{R}}^3}\psi =-q^2\mathrm{/}(4r^4{\psi }^3)$. With $\bar{R}=0$, $\bar{A}=\tau =\mathrm{\Lambda }=\bar{B}=0$, the transformed scalar equation is $-8\mathrm{\Delta }\psi -2q^2{r}^{-4}{\psi }^{-3}=0$.

The seed Gauss constraint holds by step 1.1, and step 1.2 verifies the Hamiltonian equation. The momentum right side vanishes because $\bar{B}=0$, and its left side vanishes because $K=\tau =0$. The omitted point $r=0$ is not in the manifold, so the smooth divergence calculation on ${\mathcal{D}}_{m,q}\times S^2$ makes no distributional assertion at the puncture.

Suppose $m>\mathrm{\mid }q\mathrm{\mid }$, so $C>0$ and $r_0=\sqrt{C}\mathrm{/}2$. Then $R_{\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}}=r+m+C\mathrm{/}(4r)$, $R_{\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}}^{\mathrm{\prime }}=1-C\mathrm{/}(4r^2)$, and $R_{\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}}^{\mathrm{\prime }\mathrm{\prime }}=C\mathrm{/}(2r^3)>0$. Hence $R_{\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}}$ has its unique minimum on $(0,\mathrm{\infty })$ at $r_0$, with value $m+\sqrt{C}$. The area formula follows directly from the induced metric on $S_r$, namely $r^2{\psi }^4{\gamma }_{S^2}$.

The outward unit normal to $S_r$ is $\nu ={\psi }^{-2}{\mathrm{\partial }}_r$. Its mean curvature, defined as the trace of the second fundamental form, is $H={\mathrm{div}}_h\nu =(\sqrt{\det h}{)}^{-1}{\mathrm{\partial }}_r(\sqrt{\det h}{\psi }^{-2})=2R_{\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}}^{\mathrm{\prime }}\mathrm{/}({\psi }^2{R}_{\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}})$. Thus $H(r_0)=0$, and $H$ is nonzero for every other centered coordinate sphere.

Write $r_0^2=C\mathrm{/}4$ and $R=R_{\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}}(r)$. On $r>r_0$, $R>m+2r_0$, so $f(R)>0$. Direct algebra gives $R^{2}f(R)=(R-m{)}^2-C=(r-r_0^2\mathrm{/}r{)}^2$ and $dR\mathrm{/}dr=1-r_0^2\mathrm{/}{r}^2=(r-r_0^2\mathrm{/}r)\mathrm{/}r$. Therefore $f(R{)}^{-1}(dR\mathrm{/}dr{)}^2=R^2\mathrm{/}{r}^2={\psi }^4$, which proves that $h={\psi }^{4}d{r}^2+R^2{\gamma }_{S^2}$ equals the radial and angular spatial metric in [F3] on this branch.

As $r\to \mathrm{\infty }$, ${\psi }^4=1+2m\mathrm{/}r+O(r^{-2})$; differentiating this explicit rational expression for ${\psi }^4$ gives first- and second-coordinate-derivative falloffs $O(r^{-2})$ and $O(r^{-3})$. For the inner end put $\tilde{r}=r_0^2\mathrm{/}r$. Then $\psi (r{)}^2{r}^2\mathrm{/}{r}_0^2=1+m\mathrm{/}\tilde{r}+r_0^2\mathrm{/}{\tilde{r}}^2=\psi (\tilde{r}{)}^2$ and $d{r}^2+r^2{\gamma }_{S^2}=(r^4\mathrm{/}{r}_0^4)(d{\tilde{r}}^2+{\tilde{r}}^2{\gamma }_{S^2})$. Hence the inner metric is exactly the same isotropic form in $\tilde{r}$, and it has the same falloff and derivative estimates. This proves the two stated end descriptions without extending the static areal chart through $f=0$.