Exact rational Kerr-Newman component derivation

The coordinate change $y=\cos \theta$ is a diffeomorphism on $0<\theta <\pi$ because $dy\mathrm{/}d\theta =-\sin \theta \mathrm{\ne }0$. Expanding the metric from [F1] gives the five nonzero entries $g_{00}=-(\mathrm{\Delta }-a^{2}b)\mathrm{/}\mathrm{\Sigma },g_{03}=-ab(R-\mathrm{\Delta })\mathrm{/}\mathrm{\Sigma },g_{33}=b(R^2-a^2\mathrm{\Delta }b)\mathrm{/}\mathrm{\Sigma },g_{11}=\mathrm{\Sigma }\mathrm{/}\mathrm{\Delta },g_{22}=\mathrm{\Sigma }\mathrm{/}b\mathrm{.}$ Their $0,3$ block determinant is $-\mathrm{\Delta }b$ and the full determinant is $-{\mathrm{\Sigma }}^2$. Multiplying the stated inverse entries in Kerr-Newman metric and potential on the exact Boyer-Lindquist domain by these blocks gives $g{g}^{-1}=I_4$; the $1,1$ and $2,2$ entries are reciprocal diagonal factors. Since $\mathrm{\Sigma }>0$, the volume polynomial is $w=\mathrm{\Sigma }$. The input and inverse denominators are products of $\mathrm{\Sigma }$, $\mathrm{\Delta }$ and $b$; each is nonzero under [A1], so all divisions in this exact calculation are admissible.

The potential components from [F1] are ${\mathcal{A}}_0=-qr\mathrm{/}\mathrm{\Sigma }$, ${\mathcal{A}}_3=qrab\mathrm{/}\mathrm{\Sigma }$, and ${\mathcal{A}}_1={\mathcal{A}}_2=0$. Applying ${\mathcal{F}}_{ij}={\delta }_i{\mathcal{A}}_j-{\delta }_j{\mathcal{A}}_i$ with ${\delta }_1={\mathrm{\partial }}_r$, ${\delta }_2={\mathrm{\partial }}_y$, using [F4] and [F6], gives the four independent field entries in the Statement; antisymmetry gives the remaining entries.

Put $C:=r^2(3y^2-1)+a^2{y}^2(3-y^2)$ and $A:=r^2-a^2{y}^2$. Raising the field components in the Statement with the inverse in [F1] gives the nonzero radial/latitude density components $\mathrm{\Sigma }{\mathcal{F}}^{10}=-\frac{qRA}{{\mathrm{\Sigma }}^2},\mathrm{\Sigma }{\mathcal{F}}^{20}=-\frac{2qr{a}^{2}by}{{\mathrm{\Sigma }}^2},\mathrm{\Sigma }{\mathcal{F}}^{13}=-\frac{qaA}{{\mathrm{\Sigma }}^2},\mathrm{\Sigma }{\mathcal{F}}^{23}=-\frac{2qray}{{\mathrm{\Sigma }}^2}\mathrm{.}$ Direct differentiation gives ${\mathrm{\partial }}_r(\mathrm{\Sigma }{\mathcal{F}}^{10})=-\frac{2qr{a}^{2}C}{{\mathrm{\Sigma }}^3},{\mathrm{\partial }}_y(\mathrm{\Sigma }{\mathcal{F}}^{20})=\frac{2qr{a}^{2}C}{{\mathrm{\Sigma }}^3},{\mathrm{\partial }}_r(\mathrm{\Sigma }{\mathcal{F}}^{13})=\frac{2qar(r^2-3a^2{y}^2)}{{\mathrm{\Sigma }}^3},{\mathrm{\partial }}_y(\mathrm{\Sigma }{\mathcal{F}}^{23})=-\frac{2qar(r^2-3a^2{y}^2)}{{\mathrm{\Sigma }}^3}\mathrm{.}$ The residuals for $j=0,3$ cancel pairwise; those for $j=1,2$ vanish because their potentially differentiated radial/latitude components are zero. Thus all four Maxwell-density residual numerators are exactly zero.

Substitution of the metric entries and inverse from [F1] into the algorithm's full Christoffel and Ricci schedules [F2], using curvature convention [F4], gives the exact sixteen-entry Ricci output table $\left({\mathrm{Ric}}_{ij}\right)=\left(\begin{array}{cccc}\frac{q^2(\mathrm{\Delta }+a^{2}b)}{{\mathrm{\Sigma }}^3}& 0& 0& -\frac{q^{2}ab(\mathrm{\Delta }+R)}{{\mathrm{\Sigma }}^3}\\ 0& -\frac{q^2}{\mathrm{\Sigma }\mathrm{\Delta }}& 0& 0\\ 0& 0& \frac{q^2}{\mathrm{\Sigma }b}& 0\\ -\frac{q^{2}ab(\mathrm{\Delta }+R)}{{\mathrm{\Sigma }}^3}& 0& 0& \frac{q^2(a^2\mathrm{\Delta }{b}^2+b{R}^2)}{{\mathrm{\Sigma }}^3}\end{array}\right)\mathrm{.}$ All omitted off-diagonal entries in this full matrix are zero. The complete rational certificate linked below supplies all 64 Christoffel entries as well as all 16 ordered Ricci entries, including zeros. For each Christoffel entry, substitute its certified numerator/denominator into the Christoffel schedule; for each ordered Ricci entry substitute those certified Christoffel entries and their quotient-rule derivatives into the Ricci schedule. Polynomial cross-multiplication gives the displayed table; the certificate checker performs these 64 and 16 checks separately, without inferring an uncomputed component by symmetry.

Direct contraction of the field entries from step 1.2 with the same inverse gives ${\mathcal{F}}_{ij}{\mathcal{F}}^{ij}=-\frac{2q^2(r^4-6a^2{r}^2{y}^2+a^4{y}^4)}{{\mathrm{\Sigma }}^4},(2Q_{ij}(\mathcal{F}))=\left({\mathrm{Ric}}_{ij}\right),g^{ij}{\mathrm{Ric}}_{ij}=0,{\mathrm{Ric}}_{ij}{\mathrm{Ric}}^{ij}=\frac{4q^4}{{\mathrm{\Sigma }}^4}\mathrm{.}$ Thus the full $4\times 4$ Einstein-residual matrix is zero entry by entry, and the Maxwell-density table is zero by step 2.1. The certificate supplies the full $4\times 4$ table of $2Q$, all raised field entries, and all 16 zero Einstein residual pairs; each $2Q$ entry is checked against $2{\sum }_{k,l}{\mathcal{F}}_{ik}{g}^{kl}{\mathcal{F}}_{jl}-g_{ij}{\mathcal{F}}^2\mathrm{/}2$. Its checker expands these sums and both scalar contractions by polynomial cross-multiplication. Thus the equality to the Ricci matrix is checked componentwise, including every zero, rather than inferred from the correctness lemma. No floating-point evaluation or parameter sample is used.

By [A1], $\mathrm{\Sigma }$, $\mathrm{\Delta }$ and $b$ are nonzero, so all recorded denominator factors remain nonzero after real parameter specialization. The certificate uses reduced representatives in the same fraction field, whereas [F2] may retain unreduced pairs. By fraction equality in [F3], reduction preserves every scheduled operation and its numerator-zero flag. For the inverse, the determinant is $-{\mathrm{\Sigma }}^2$; all other divisions are by rational constants or the chart factors and their products. Even an uncancelled determinant numerator is $-{\mathrm{\Sigma }}^2$ times its nonzero denominator, so it introduces no new excluded locus. Induction through the finite schedule identifies its output with these checked representatives. Correctness and termination are [F3]; hence steps 2.1, 2.2 and 3.1 prove the stated smooth coordinate identities on every admitted chart component.