Warped two-sphere curvature and charged mass conservation

Let $A,B,C\in \{0,1\}$ be base indices and $i,j,k\in \{2,3\}$ sphere indices. In base coordinates, write $r_A:={\mathrm{\partial }}_{A}r$ and $r^A:=q^{AB}{r}_B$, the components of the vector field $q^{AB}{\mathrm{\partial }}_{B}r$ (the $q$-dual of $dr$). The inverse metric is $g^{AB}=q^{AB}$ and $g^{ij}=r^{-2}{\gamma }^{ij}$. Substituting the block metric $g_{AB}=q_{AB}$, $g_{ij}=r^2{\gamma }_{ij}$, $g_{Ai}=0$ into [F4] gives the mixed Christoffels ${\mathrm{\Gamma }}^A{}_{ij}=-\frac{1}{2}{q}^{AB}{\mathrm{\partial }}_B(r^2{\gamma }_{ij})=-r{r}^A{\gamma }_{ij}$ and ${\mathrm{\Gamma }}^i{}_{Aj}=\frac{1}{2}{r}^{-2}{\gamma }^{ik}{\mathrm{\partial }}_A(r^2{\gamma }_{kj})=r^{-1}{r}_A{\delta }_j^i$. All other mixed Christoffels vanish because the inverse metric is block diagonal, $q_{AB}$ is independent of sphere coordinates, and ${\gamma }_{ij}$ is independent of base coordinates. The unit round sphere has ${\mathrm{\Gamma }}^{\theta }{}_{\varphi \varphi }=-\sin \theta \cos \theta$ and ${\mathrm{\Gamma }}^{\varphi }{}_{\theta \varphi }={\mathrm{\Gamma }}^{\varphi }{}_{\varphi \theta }=\cot \theta$ on the polar chart of [F6]. Substitution in [F5] and contraction by [F3] give $\mathrm{Ric}(\gamma {)}_{\theta \theta }=-{\mathrm{\partial }}_{\theta }\cot \theta -{\cot }^2\theta =1$ and $\mathrm{Ric}(\gamma {)}_{\varphi \varphi }={\mathrm{\partial }}_{\theta }(-\sin \theta \cos \theta )-{\cos }^2\theta +2{\cos }^2\theta ={\sin }^2\theta$, with mixed component zero. Thus $\mathrm{Ric}(\gamma )=\gamma$ on that chart and everywhere by smoothness.

In an oriented orthonormal coframe for the two-dimensional Lorentzian base and the round sphere, ${ϵ}_{AB}{ϵ}^{AB}=-2$ and the sphere area form has squared norm $2\mathrm{/}{r}^4$. The two summands of $\mathcal{F}$ occupy orthogonal blocks, so ${\mathcal{F}}_{ab}{\mathcal{F}}^{ab}=2(q_m^2-q_e^2)\mathrm{/}{r}^4$. The same two-dimensional contractions give ${\mathcal{F}}_{AC}{\mathcal{F}}_B{}^C=-q_e^2{r}^{-4}{q}_{AB}$ and ${\mathcal{F}}_{ik}{\mathcal{F}}_j{}^k=q_m^2{r}^{-4}{g}_{ij}$; mixed contractions vanish. Substituting these three identities into [F1]'s definition of $Q$ yields the stated base and sphere components.

In the coordinate Ricci formula, the terms in $\mathrm{Ric}(g{)}_{AB}$ with only base indices give $\mathrm{Ric}(q{)}_{AB}$. The remaining sphere-index terms are $-{\mathrm{\partial }}_B(2r_A\mathrm{/}r)+(2r_C\mathrm{/}r){\mathrm{\Gamma }}^C{}_{AB}-2r_A{r}_B\mathrm{/}{r}^2=-2r^{-1}({\mathrm{\partial }}_B{r}_A-{\mathrm{\Gamma }}^C{}_{AB}{r}_C),$ which is $-2r^{-1}({\mathrm{\nabla }}_A^{q}dr{)}_B$. For $\mathrm{Ric}(g{)}_{ij}$, the pure-sphere terms give $\mathrm{Ric}(\gamma {)}_{ij}={\gamma }_{ij}$. The base derivative terms are $-[{\mathrm{\partial }}_A(r{r}^A)+{\mathrm{\Gamma }}^C{}_{CA}r{r}^A]{\gamma }_{ij}=-(r{\mathrm{\Delta }}_{q}r+z){\gamma }_{ij}$. The mixed trace ${\mathrm{\Gamma }}^k{}_{kA}{\mathrm{\Gamma }}^A{}_{ij}=-2z{\gamma }_{ij}$ cancels the two mixed products $-{\mathrm{\Gamma }}^k{}_{jA}{\mathrm{\Gamma }}^A{}_{ik}-{\mathrm{\Gamma }}^A{}_{jk}{\mathrm{\Gamma }}^k{}_{iA}=2z{\gamma }_{ij}$. For $\mathrm{Ric}(g{)}_{Ai}$, the derivative ${\mathrm{\partial }}_i{\mathrm{\Gamma }}^{\alpha }{}_{\alpha A}$ vanishes because ${\mathrm{\Gamma }}^{\alpha }{}_{\alpha A}={\mathrm{\Gamma }}^B{}_{BA}+2r_A\mathrm{/}r$ depends only on the base coordinates, and all derivatives ${\mathrm{\partial }}_{\alpha }{\mathrm{\Gamma }}^{\alpha }{}_{iA}$ vanish because the only nonzero such coefficients are ${\mathrm{\Gamma }}^j{}_{iA}=(r_A\mathrm{/}r){\delta }_i^j$. In the remaining products, the term ${\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{\mathrm{\ell }k}{\mathrm{\Gamma }}^k{}_{iA}=(r_A\mathrm{/}r){\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{\mathrm{\ell }i}$ is canceled by $-{\mathrm{\Gamma }}^j{}_{ik}{\mathrm{\Gamma }}^k{}_{jA}=-(r_A\mathrm{/}r){\mathrm{\Gamma }}^j{}_{ij}=-(r_A\mathrm{/}r){\mathrm{\Gamma }}^j{}_{ji}$; every other product contains a zero mixed Christoffel. Thus $\mathrm{Ric}(g{)}_{Ai}=0$. These identities give the three stated warped Ricci formulas.

The base Einstein equation, steps 2.1 and 1.2, and [F2] yield $\frac{1}{2}{S}_q{q}_{AB}-2r^{-1}({\mathrm{\nabla }}_A^{q}dr{)}_B=(\mathrm{\Lambda }-Q^2\mathrm{/}{r}^4)q_{AB}\mathrm{.}$ Its right-hand side and the first term are scalar multiples of $q$, so ${\mathrm{\nabla }}^{q}dr$ is a scalar multiple of $q$. Taking its trace identifies that multiple as $\frac{1}{2}{\mathrm{\Delta }}_{q}r$. The sphere Einstein equation gives $1-z-r{\mathrm{\Delta }}_{q}r=\mathrm{\Lambda }{r}^2+Q^2\mathrm{/}{r}^2$, which rearranges to the stated formula for ${\mathrm{\Delta }}_{q}r$.

Differentiate $z=q^{AB}{r}_A{r}_B$. Metric compatibility gives $dz=2({\mathrm{\nabla }}^{q}dr)({\mathrm{\nabla }}^{q}r,\cdot )={\mathrm{\Delta }}_{q}rdr$ by step 3.1. Therefore $d\varpi =\frac{1}{2}(1-z)dr-\frac{r}{2}dz-\frac{Q^2}{2r^2}dr-\frac{\mathrm{\Lambda }{r}^2}{2}dr\mathrm{.}$ Substituting ${\mathrm{\Delta }}_{q}r=(1-z)\mathrm{/}r-\mathrm{\Lambda }r-Q^2\mathrm{/}{r}^3$ from step 3.1 makes the coefficient of $dr$ zero term by term. Thus $d\varpi =0$, completing the conditional calculation.