Complete spherical Einstein and conservation reduction

In a regular spherical chart away from the angular coordinate poles, the metric in [F1] has the following nonzero Christoffel symbols, with lower-index symmetry understood: ${\mathrm{\Gamma }}^0{}_{0r}={\varphi }^{\mathrm{\prime }},{\mathrm{\Gamma }}^r{}_{00}=b{e}^{2\varphi }{\varphi }^{\mathrm{\prime }},{\mathrm{\Gamma }}^r{}_{rr}=-\frac{b^{\mathrm{\prime }}}{2b},{\mathrm{\Gamma }}^r{}_{\theta \theta }=-rb,{\mathrm{\Gamma }}^r{}_{\phi \phi }=-rb{\sin }^2\theta ,{\mathrm{\Gamma }}^{\theta }{}_{r\theta }={\mathrm{\Gamma }}^{\phi }{}_{r\phi }=\frac{1}{r},{\mathrm{\Gamma }}^{\theta }{}_{\phi \phi }=-\sin \theta \cos \theta ,{\mathrm{\Gamma }}^{\phi }{}_{\theta \phi }=\cot \theta \mathrm{.}$ They follow by inserting the diagonal metric entries $(-e^{2\varphi },b^{-1},r^2,r^2{\sin }^2\theta )$ and their derivatives into the Christoffel formula in [F2]. The identities obtained below depend only on $r$ and are tensor-component identities, so their validity extends across the two angular coordinate poles by smoothness.

Substituting the symbols of step 1.1 in the curvature formula of [F2], contracting as in [F2], and simplifying gives ${\mathrm{Ric}}^0{}_0=-b({\varphi }^{\mathrm{\prime }\mathrm{\prime }}+{\varphi }^{\mathrm{\prime }2}+\frac{2{\varphi }^{\mathrm{\prime }}}{r}+\frac{b^{\mathrm{\prime }}{\varphi }^{\mathrm{\prime }}}{2b}),{\mathrm{Ric}}^r{}_r=-b({\varphi }^{\mathrm{\prime }\mathrm{\prime }}+{\varphi }^{\mathrm{\prime }2}+\frac{b^{\mathrm{\prime }}{\varphi }^{\mathrm{\prime }}}{2b})-\frac{b^{\mathrm{\prime }}}{r},{\mathrm{Ric}}^{\theta }{}_{\theta }={\mathrm{Ric}}^{\phi }{}_{\phi }=\frac{1-b}{r^2}-\frac{b^{\mathrm{\prime }}}{2r}-\frac{b{\varphi }^{\mathrm{\prime }}}{r}\mathrm{.}$ Hence $G^0{}_0=-(1-b)\mathrm{/}{r}^2+b^{\mathrm{\prime }}\mathrm{/}r$, $G^r{}_r=-(1-b)\mathrm{/}{r}^2+2b{\varphi }^{\mathrm{\prime }}\mathrm{/}r$, and $G^{\theta }{}_{\theta }=G^{\phi }{}_{\phi }=b({\varphi }^{\mathrm{\prime }\mathrm{\prime }}+{\varphi }^{\mathrm{\prime }2}+\frac{{\varphi }^{\mathrm{\prime }}}{r})+\frac{b^{\mathrm{\prime }}}{2}({\varphi }^{\mathrm{\prime }}+\frac{1}{r})\mathrm{.}$ These equalities use only the displayed finite component formula, the listed Christoffels, and the contractions defining $G$; no spherical field equation is cited.

For the diagonal tensor $T^a{}_b$ of [F1], its radial divergence is ${\mathrm{\nabla }}_a{T}^a{}_r=p^{\mathrm{\prime }}+{\varphi }^{\mathrm{\prime }}(ϵ+p)\mathrm{.}$ Indeed, in ${\mathrm{\partial }}_a{T}^a{}_r+{\mathrm{\Gamma }}^a{}_{ac}{T}^c{}_r-{\mathrm{\Gamma }}^c{}_{ar}{T}^a{}_c$, the derivative term is $p^{\mathrm{\prime }}$, the trace term contributes $p({\varphi }^{\mathrm{\prime }}-b^{\mathrm{\prime }}\mathrm{/}(2b)+2\mathrm{/}r)$, and the last term contributes $ϵ{\varphi }^{\mathrm{\prime }}+p{b}^{\mathrm{\prime }}\mathrm{/}(2b)-2p\mathrm{/}r$. Adding the three contributions gives $p^{\mathrm{\prime }}+(ϵ+p){\varphi }^{\mathrm{\prime }}$. The other three components of ${\mathrm{\nabla }}_a{T}^a{}_b$ vanish: for $b=0$ there is no time dependence and every mixed time-radial Christoffel is zero; for $b=\theta ,\phi$, angular derivatives vanish and the only polar trace term $\cot \theta p$ cancels the matching diagonal connection term, while the $\phi$ component has no nonzero trace term.

The $00$ equation in [A1] is $-(1-b)\mathrm{/}{r}^2+b^{\mathrm{\prime }}\mathrm{/}r=-\kappa ϵ$. Multiplying by $r^2$ yields $1-b-r{b}^{\mathrm{\prime }}=\kappa ϵr^2$. Differentiating $M=c^{2}r(1-b)\mathrm{/}(2G)$ gives $M^{\mathrm{\prime }}=c^2(1-b-r{b}^{\mathrm{\prime }})\mathrm{/}(2G)=4\pi r^2ϵ\mathrm{/}{c}^2$.

The $rr$ equation in [A1] is $-(1-b)\mathrm{/}{r}^2+2b{\varphi }^{\mathrm{\prime }}\mathrm{/}r=\kappa p$. Solving for ${\varphi }^{\mathrm{\prime }}$ and substituting $1-b=2GM\mathrm{/}(c^{2}r)$ and $\kappa =8\pi G\mathrm{/}{c}^4$ gives ${\varphi }^{\mathrm{\prime }}=\frac{(1-b)+\kappa p{r}^2}{2br}=\frac{GM\mathrm{/}{c}^2+4\pi G{r}^{3}p\mathrm{/}{c}^4}{r^{2}b}\mathrm{.}$

By [F3] and [A1], ${\mathrm{\nabla }}_a{T}^a{}_r={\kappa }^{-1}{\mathrm{\nabla }}_a{G}^a{}_r=0$. Step 2.2 therefore gives $p^{\mathrm{\prime }}=-(ϵ+p){\varphi }^{\mathrm{\prime }}$. Substitution of step 3.2 yields the displayed TOV pressure equation.

Conversely, the mass and lapse equations give the $00$ and $rr$ equations by reversing steps 3.1–2.2. Define $D^a{}_b=G^a{}_b-\kappa T^a{}_b$. Then $D^0{}_0=D^r{}_r=0$, and its only possible nonzero entries are the equal angular diagonal entries $D^{\theta }{}_{\theta }=D^{\phi }{}_{\phi }$, by step 2.1. The pressure equation and step 2.2 give ${\mathrm{\nabla }}_a{T}^a{}_b=0$ for every $b$, so [F3] gives ${\mathrm{\nabla }}_a{D}^a{}_b=0$. In particular, the radial component is ${\mathrm{\nabla }}_a{D}^a{}_r=-{\mathrm{\Gamma }}^{\theta }{}_{\theta r}{D}^{\theta }{}_{\theta }-{\mathrm{\Gamma }}^{\phi }{}_{\phi r}{D}^{\phi }{}_{\phi }=-2D^{\theta }{}_{\theta }\mathrm{/}r$. Since $r>0$, it follows that both angular entries vanish. Thus $D=0$ and the full mixed field equation holds.