A Proofs
In this appendix, we present the proofs which were omitted in the above sections. Every proof of a proposition or lemma from above will refer back to the number of the statement it is proving. We also introduce a new lemma here which is useful for proving some of the above statements. We will present all of the proofs here in the order with which their corresponding statements are given above.
Proof
(Proof of Proposition 1) The Hawking mass of a metric sphere is defined as
$$\begin{aligned} m_{H}(\Sigma _{t,r}) = \sqrt{\frac{\left| \Sigma _{t,r} \right| }{16\pi }}\left( 1 - \frac{1}{16\pi }\int \limits _{\Sigma _{t,r}} g\left( \mathbf {{H}},\mathbf {{H}} \right) \, dA \right) \end{aligned}$$
(67)
where \(\left| \Sigma _{t,r} \right| \) is the surface area of the metric sphere, \(\mathbf {{H}}\) is the mean curvature vector of the sphere in the spacetime, and \(dA\) is the volume element on the sphere [16, 25]. By the definition of our radial coordinate \(r\), we have that \(\left| \Sigma _{t,r} \right| =4\pi r^{2}\), but it is also easily computed in this metric since
$$\begin{aligned} dA = \sqrt{\left| d\sigma ^{2} \right| }\, d\theta \, d\varphi = \sqrt{r^{4} \sin ^{2}\theta }\, d\theta \, d\varphi = r^{2}\sin \theta \, d\theta \, d\varphi , \end{aligned}$$
(68)
which yields that
$$\begin{aligned} \left| \Sigma _{t,r} \right| = \int \limits _{\Sigma _{t,r}} dA = \int \limits _{0}^{2\pi }\int \limits _{0}^{\pi } r^{2}\sin \theta \, d\theta \, d\varphi = \int \limits _{0}^{2\pi } 2r^{2}\, d\varphi = 4\pi r^{2}. \end{aligned}$$
(69)
To compute the mean curvature vector, \(\mathbf {{H}}\), we have that
$$\begin{aligned} \mathbf {{H}}= \gamma ^{jk}\mathrm{\text {II} }(\partial _{j},\partial _{k}) \end{aligned}$$
(70)
where \(j,k \in \{\theta ,\varphi \}\), \(\gamma \) is the metric on \(\Sigma _{t,r}\) (note that we have reused the variable \(\gamma \) here to refer to the metric on the sphere and not to the metric on the \(t=constant\) hypersurfaces as we did in the section describing the framework of numerical relativity), and \(\mathrm{\text {II} }\) is the second fundamental form tensor, which sends a pair of vectors tangent to the sphere to a vector normal to the sphere. It is defined as follows for all \(X,Y \in T\Sigma _{t,r}\)
$$\begin{aligned} \mathrm{\text {II} }(X,Y) = \nabla _{X}Y -\, ^{2}\nabla _{X}Y \end{aligned}$$
(71)
where \(\nabla \) is the covariant derivative operator on \(N\) and \(^{2}\nabla \) is the induced covariant derivative operator on \(\Sigma _{t,r}\). Since \(\gamma \) is diagonal, we only need to concern ourselves with the diagonal components of this tensor in order to compute \(\mathbf {{H}}\). To perform these computations, first note that the Christoffel symbols on the sphere in these coordinates have the following property (note that a superscript of 2 will always denote that that quantity corresponds to the sphere, also for the following computations, roman letters will denote subscripts in \(\{\theta ,\varphi \}\), while Greek letters will denote subscripts ranging over all four coordinates). Since both the radial and time directions are normal to the sphere,
$$\begin{aligned} \Gamma _{jk}^{\ \ \ell }&= \frac{1}{2}g^{\ell \eta }(g_{j\eta ,k} + g_{k\eta ,j} - g_{jk,\eta }) = \frac{1}{2}g^{\ell m}(g_{jm,k} + g_{km,j} - g_{jk,m}) \nonumber \\&= \frac{1}{2}\gamma ^{\ell m}(g_{jm,k} + g_{km,j} - g_{jk,m}) = \,^{2}\Gamma _{jk}^{\ \ \ell }. \end{aligned}$$
(72)
Thus we need not distinguish between the Christoffel symbols for the metric on the sphere and those on the entire manifold. We have then that
$$\begin{aligned} \mathrm{\text {II} }(\partial _{j},\partial _{j}) = \nabla _{j}\partial _{j} -\, ^{2}\nabla _{j}\partial _{j} = \Gamma _{jj}^{\ \ \ \eta }\, \partial _{\eta } -\, ^{2}\Gamma _{jj}^{\ \ \ k}\, \partial _{k} = \Gamma _{jj}^{\ \ \ t}\, \partial _{t} + \Gamma _{jj}^{\ \ \ r}\, \partial _{r}. \end{aligned}$$
(73)
Then we need to compute the above Christoffel symbols. We have that
$$\begin{aligned} \Gamma _{\theta \theta }^{\ \ \ t}&= \frac{1}{2}g^{t\eta }(g_{\theta \eta ,\theta } + g_{\theta \eta ,\theta } - g_{\theta \theta ,\eta }) = \frac{1}{2}g^{tt}(- g_{\theta \theta ,t}) = -\frac{1}{2 \mathrm {e}^{2V}}\left( -\partial _{t}(r^{2})\right) = 0 \end{aligned}$$
(74a)
$$\begin{aligned} \Gamma _{\theta \theta }^{\ \ \ r}&= \frac{1}{2}g^{r\eta }(g_{\theta \eta ,\theta } + g_{\theta \eta ,\theta } - g_{\theta \theta ,\eta }) = \frac{1}{2}g^{rr}(- g_{\theta \theta ,r}) \nonumber \\&= \frac{1}{2}\left( 1-\frac{2M}{r} \right) \left( -\partial _{r}(r^{2})\right) = -r\left( 1 - \frac{2M}{r} \right) \end{aligned}$$
(74b)
$$\begin{aligned} \Gamma _{\varphi \varphi }^{\ \ \ t}&= \frac{1}{2}g^{t\eta }(g_{\varphi \eta ,\varphi } + g_{\varphi \eta ,\varphi } - g_{\varphi \varphi ,\eta }) = \frac{1}{2}g^{tt}(- g_{\varphi \varphi ,t}) = -\frac{1}{2 \mathrm {e}^{2V}}\left( -\partial _{t}(r^{2}\sin ^{2}\theta )\right) = 0 \end{aligned}$$
(74c)
$$\begin{aligned} \Gamma _{\varphi \varphi }^{\ \ \ r}&= \frac{1}{2}g^{r\eta }(g_{\varphi \eta ,\varphi } + g_{\varphi \eta ,\varphi } - g_{\varphi \varphi ,\eta }) = \frac{1}{2}g^{rr}(- g_{\varphi \varphi ,r}) \nonumber \\&= \frac{1}{2}\left( 1-\frac{2M}{r} \right) \left( -\partial _{r}(r^{2}\sin ^{2}\theta )\right) = -r\left( 1 - \frac{2M}{r} \right) \sin ^{2}\theta . \end{aligned}$$
(74d)
Then we have that
$$\begin{aligned} \mathrm{\text {II} }(\partial _{\theta },\partial _{\theta })&= \Gamma _{\theta \theta }^{\ \ \ t}\, \partial _{t} + \Gamma _{\theta \theta }^{\ \ \ r}\, \partial _{r} = -r\left( 1 - \frac{2M}{r} \right) \, \partial _{r} \end{aligned}$$
(75)
$$\begin{aligned} \mathrm{\text {II} }(\partial _{\varphi },\partial _{\varphi })&= \Gamma _{\varphi \varphi }^{\ \ \ t}\, \partial _{t} + \Gamma _{\varphi \varphi }^{\ \ \ r}\, \partial _{r} = -r\left( 1 - \frac{2M}{r} \right) \sin ^{2}\theta \, \partial _{r}. \end{aligned}$$
(76)
This makes (70) become
$$\begin{aligned} \mathbf {{H}}= \gamma ^{\theta \theta }\mathrm{\text {II} }(\partial _{\theta },\partial _{\theta }) + \gamma ^{\varphi \varphi }\mathrm{\text {II} }(\partial _{\varphi },\partial _{\varphi }) = -\frac{2}{r}\left( 1 - \frac{2M}{r} \right) \, \partial _{r}. \end{aligned}$$
(77)
Finally, we can compute the Hawking mass as follows.
$$\begin{aligned} m_{H}(\Sigma _{t,r})&= \sqrt{\frac{\left| \Sigma _{t,r} \right| }{16\pi }}\left( 1 - \frac{1}{16\pi }\int \limits _{\Sigma _{t,r}} g\left( \mathbf {{H}},\mathbf {{H}} \right) \, dA \right) \nonumber \\&= \sqrt{\frac{4\pi r^{2}}{16\pi }}\left( 1 - \frac{1}{16\pi }\int \limits _{\Sigma _{t,r}} g\left( -\frac{2}{r}\left( 1 - \frac{2M}{r} \right) \, \partial _{r},\, -\frac{2}{r}\left( 1 - \frac{2M}{r} \right) \, \partial _{r} \right) \, dA \right) \nonumber \\&= \frac{r}{2}\left( 1 - \frac{1}{16\pi }\int \limits _{0}^{2\pi }\int \limits _{0}^{\pi } \frac{4}{r^{2}}\left( 1 - \frac{2M}{r} \right) ^{2}g(\partial _{r},\partial _{r}) r^{2}\sin \theta \, d\theta \, d\varphi \right) \nonumber \\&= \frac{r}{2}\left( 1 - \frac{1}{16\pi }\int \limits _{0}^{2\pi }\int \limits _{0}^{\pi } 4\left( 1 - \frac{2M}{r} \right) \sin \theta \, d\theta \, d\varphi \right) \nonumber \\ m_{H}(\Sigma _{t,r})&= M \end{aligned}$$
(78)
\(\square \)
To prove Lemma 2, we will utilize the following additional lemma.
Lemma 14
For all \(\eta \in \{t,r,\theta ,\varphi \}\), let \(\nu _{\eta }\) be as defined by Eq. (31). Then the following are true.
-
1.
For all \(\eta \in \{t,r,\theta ,\varphi \}\), \(\mathrm{Ric }(\nu _{\eta },\nu _{\eta }) = g(\nu _{\eta },\nu _{\eta })g^{\eta \eta }\mathrm{Ric }_{\eta \eta }\).
-
2.
\(\displaystyle R = \sum \nolimits _{\eta } g(\nu _{\eta },\nu _{\eta })\mathrm{Ric }(\nu _{\eta },\nu _{\eta })\)
Proof
Since \(\{\nu _{\eta }\}\) is a frame field, we have that
$$\begin{aligned} \mathrm{Ric }(\nu _{t},\nu _{t}) = \mathrm {e}^{-2V}\mathrm{Ric }(\partial _{t},\partial _{t}) = -g^{tt}\mathrm{Ric }_{tt} = g(\nu _{t},\nu _{t})g^{tt}\mathrm{Ric }_{tt}. \end{aligned}$$
(79)
By similar arguments, we have that for all \(\eta \in \{t,r,\theta ,\varphi \}\),
$$\begin{aligned} \mathrm{Ric }(\nu _{\eta },\nu _{\eta }) = g(\nu _{\eta },\nu _{\eta })g^{\eta \eta }\mathrm{Ric }_{\eta \eta } \end{aligned}$$
(80)
which proves 1.
To prove 2, we use the first result and find that since \(g\) is diagonal and \(g(\nu _{\eta },\nu _{\eta })=\pm 1\), we have that
$$\begin{aligned} R = \sum _{\eta \omega } g^{\eta \omega }\mathrm{Ric }_{\eta \omega } = \sum _{\eta } g^{\eta \eta }\mathrm{Ric }_{\eta \eta } = \sum _{\eta } \frac{\mathrm{Ric }(\nu _{\eta },\nu _{\eta })}{g(\nu _{\eta },\nu _{\eta })} = \sum _{\eta } g(\nu _{\eta },\nu _{\eta })\mathrm{Ric }(\nu _{\eta },\nu _{\eta }). \end{aligned}$$
(81)
Fact 2 also follows from a well understood more general property of frame fields and traces of tensors. \(\square \)
With this, we prove Lemma 2.
Proof
(Proof of Lemma 2) In this proof, all indices range over the coordinates \(\{t,r,\theta ,\varphi \}\). By Lemma 14 and the well known formula for the components of the Riemann curvature tensor, \(\mathbf {R}\), in terms of the Christoffel symbols
$$\begin{aligned} \mathbf {R}_{jk\ell }^{\ \ \ \ m} = \Gamma _{j\ell \ \ ,k}^{\ \ m} - \Gamma _{k\ell \ \ ,j}^{\ \ m} + \Gamma _{j\ell }^{\ \ s}\Gamma _{ks}^{\ \ m} - \Gamma _{k\ell }^{\ \ s}\Gamma _{js}^{\ \ m}. \end{aligned}$$
(82)
Since \(\mathrm{Ric }_{jk} = \mathbf {R}_{j\ell k}^{\ \ \ \ \ell }\), we have that
$$\begin{aligned} \mathrm{Ric }(\nu _{t},\nu _{t})&= g(\nu _{t},\nu _{t})g^{tt}\mathrm{Ric }_{tt} \nonumber \\&= \mathrm {e}^{-2V}\left( \Gamma _{tt\ ,k}^{\ \ k} - \Gamma _{kt\ ,t}^{\ \ k} + \Gamma _{tt}^{\ \ s}\Gamma _{ks}^{\ \ k} - \Gamma _{kt}^{\ \ s}\Gamma _{ts}^{\ \ k} \right) \end{aligned}$$
(83a)
$$\begin{aligned} \mathrm{Ric }(\nu _{t},\nu _{r})&= \mathrm {e}^{-V}\sqrt{1 - \frac{2M}{r}}\, \mathrm{Ric }_{tr} \nonumber \\&= \mathrm {e}^{-V}\sqrt{1 - \frac{2M}{r}} \left( \Gamma _{tr\ ,k}^{\ \ k} - \Gamma _{kr\ ,t}^{\ \ k} + \Gamma _{tr}^{\ \ s}\Gamma _{ks}^{\ \ k} - \Gamma _{kr}^{\ \ s}\Gamma _{ts}^{\ \ k} \right) \end{aligned}$$
(83b)
$$\begin{aligned} \mathrm{Ric }(\nu _{r},\nu _{r})&= g(\nu _{r},\nu _{r})g^{rr}\mathrm{Ric }_{rr} = \left( 1 - \frac{2M}{r} \right) \mathrm{Ric }_{rr} \nonumber \\&= \left( 1 - \frac{2M}{r} \right) \left( \Gamma _{rr\ ,k}^{\ \ k} - \Gamma _{kr\ ,r}^{\ \ k} + \Gamma _{rr}^{\ \ s}\Gamma _{ks}^{\ \ k} - \Gamma _{kr}^{\ \ s}\Gamma _{rs}^{\ \ k} \right) \end{aligned}$$
(83c)
$$\begin{aligned} \mathrm{Ric }(\nu _{\theta },\nu _{\theta })&= g(\nu _{\theta },\nu _{\theta })g^{\theta \theta }\mathrm{Ric }_{\theta \theta } = \frac{1}{r^{2}}\mathrm{Ric }_{\theta \theta } \nonumber \\&= \frac{1}{r^{2}} \left( \Gamma _{\theta \theta \ ,k}^{\ \ k} - \Gamma _{k\theta \ ,\theta }^{\ \ k} + \Gamma _{\theta \theta }^{\ \ s}\Gamma _{ks}^{\ \ k} - \Gamma _{k\theta }^{\ \ s}\Gamma _{\theta s}^{\ \ k} \right) \end{aligned}$$
(83d)
$$\begin{aligned} \mathrm{Ric }(\nu _{\varphi },\nu _{\varphi })&= g(\nu _{\varphi },\nu _{\varphi })g^{\varphi \varphi }\mathrm{Ric }_{\varphi \varphi } = \frac{1}{r^{2}\sin ^{2} \theta }\mathrm{Ric }_{\varphi \varphi } \nonumber \\&= \frac{1}{r^{2}\sin ^{2}\theta } \left( \Gamma _{\varphi \varphi \ ,k}^{\ \ \ k} - \Gamma _{k\varphi \ ,\varphi }^{\ \ \ k} + \Gamma _{\varphi \varphi }^{\ \ \ s}\Gamma _{ks}^{\ \ k} - \Gamma _{k\varphi }^{\ \ \ s}\Gamma _{\varphi s}^{\ \ \ k} \right) . \end{aligned}$$
(83e)
Then to compute these, we need the Christoffel symbols. Recall that the formula for the Christoffel symbols in terms of the metric components is the following
$$\begin{aligned} \Gamma _{jk}^{\ \ \ell } = \frac{1}{2}g^{\ell m} (g_{jm,k} + g_{km,j} - g_{jk,m}) \end{aligned}$$
(84)
Using (84) and the fact that \(g\) is diagonal, we have that
$$\begin{aligned} \Gamma _{jk}^{\ \ t}&= \frac{1}{2}g^{t m}(g_{jm,k} + g_{km,j} - g_{jk,m}) = \frac{1}{2}g^{tt}(g_{jt,k} + g_{kt,j} - g_{jk,t}) \nonumber \\&= -\frac{1}{2 \mathrm {e}^{2V}}(g_{jt,k} + g_{kt,j} - g_{jk,t})\end{aligned}$$
(85a)
$$\begin{aligned} \Gamma _{jk}^{\ \ r}&= \frac{1}{2}g^{r m}(g_{jm,k} + g_{km,j} - g_{jk,m}) = \frac{1}{2}g^{rr}(g_{jr,k} + g_{kr,j} - g_{jk,r})\nonumber \\&= \frac{1}{2}\left( 1 - \frac{2M}{r} \right) (g_{jr,k} + g_{kr,j} - g_{jk,r})\end{aligned}$$
(85b)
$$\begin{aligned} \Gamma _{jk}^{\ \ \theta }&= \frac{1}{2}g^{\theta m}(g_{jm,k} + g_{km,j} - g_{jk,m}) = \frac{1}{2}g^{\theta \theta }(g_{j\theta ,k} + g_{k\theta ,j} - g_{jk,\theta })\nonumber \\&= \frac{1}{2r^{2}} (g_{j\theta ,k} + g_{k\theta ,j} - g_{jk,\theta })\end{aligned}$$
(85c)
$$\begin{aligned} \Gamma _{jk}^{\ \ \varphi }&= \frac{1}{2}g^{\varphi m}(g_{jm,k} + g_{km,j} - g_{jk,m}) = \frac{1}{2}g^{\varphi \varphi }(g_{j\varphi ,k} + g_{k\varphi ,j} - g_{jk,\varphi })\nonumber \\&= \frac{1}{2r^{2}\sin ^{2}\theta } (g_{j\varphi ,k} + g_{k\varphi ,j} - g_{jk,\varphi }) = \frac{1}{2r^{2}\sin ^{2}\theta } (g_{j\varphi ,k} + g_{k\varphi ,j}). \end{aligned}$$
(85d)
The last line is due to the fact that none of the metric components depend on \(\varphi \). To help compute these quantities, it would be useful to note the following.
$$\begin{aligned} g_{tt,t}&= \partial _{t}(- \mathrm {e}^{2V}) = -2V_{t} \mathrm {e}^{2V} \quad g_{tt,\theta } = \partial _{\theta }(- \mathrm {e}^{2V}) = 0 \nonumber \\ g_{tt,r}&= \partial _{r}(- \mathrm {e}^{2V}) = -2V_{r} \mathrm {e}^{2V} \quad g_{tt,\varphi } = \partial _{\varphi }(- \mathrm {e}^{2V}) = 0\end{aligned}$$
(86a)
$$\begin{aligned} g_{rr,t}&= \partial _{t}\left( 1 - \frac{2M}{r} \right) ^{-1} \quad g_{rr,\theta } = \partial _{\theta }\left( 1 - \frac{2M}{r} \right) ^{-1} = 0 \nonumber \\&= \left( 1 - \frac{2M}{r} \right) ^{-2}\left( \frac{2M_{t}}{r} \right) \nonumber \\ g_{rr,r}&= \partial _{r}\left( 1 - \frac{2M}{r} \right) ^{-1} \quad g_{rr,\varphi } = \partial _{\varphi }\left( 1 - \frac{2M}{r} \right) ^{-1} = 0\nonumber \\&= \left( 1 - \frac{2M}{r} \right) ^{-2}\left( \frac{2M_{r}}{r} - \frac{2M}{r^{2}} \right) \end{aligned}$$
(86b)
$$\begin{aligned} g_{\theta \theta ,t}&= \partial _{t}(r^{2}) = 0 \quad g_{\theta \theta ,\theta } = \partial _{\theta }(r^{2}) = 0 \nonumber \\ g_{\theta \theta ,r}&= \partial _{r}(r^{2}) = 2r \quad g_{\theta \theta ,\varphi } = \partial _{\varphi }(r^{2}) = 0 \end{aligned}$$
(86c)
$$\begin{aligned} g_{\varphi \varphi ,t}&= \partial _{t}(r^{2}\sin ^{2}\theta ) = 0 \quad g_{\varphi \varphi ,\theta } = \partial _{\theta }(r^{2}\sin ^{2}\theta ) = 2r^{2}\sin \theta \cos \theta \nonumber \\ g_{\varphi \varphi ,r}&= \partial _{r}(r^{2}\sin ^{2}\theta ) = 2r\sin ^{2}\theta \quad g_{\varphi \varphi ,\varphi } = \partial _{\varphi }{r^{2}\sin ^{2}\theta } = 0. \end{aligned}$$
(86d)
All the other metric components are 0. Next, if
$$\begin{aligned} \Phi = \left( 1 - \frac{2M}{r} \right) , \end{aligned}$$
(87)
then, using (85) and (86), it is a series of straightforward algebra computations to obtain the following values of the Christoffel symbols. For brevity, we only include the nonzero Christoffel symbols
$$\begin{aligned} \Gamma _{tt}^{\ \ t}&= V_{t} \quad \Gamma _{tr}^{\ \ t} = \Gamma _{rt}^{\ \ t} = V_{r} \quad \Gamma _{rr}^{\ \ t} = \frac{M_{t}}{r\mathrm {e}^{2V}}\Phi ^{-2}\end{aligned}$$
(88a)
$$\begin{aligned} \Gamma _{tt}^{\ \ r}&= V_{r}\mathrm {e}^{2V}\Phi \quad \Gamma _{tr}^{\ \ r} = \Gamma _{rt}^{\ \ r} = \frac{M_{t}}{r}\Phi ^{-1} \quad \Gamma _{rr}^{\ \ r} = \Phi ^{-1}\left( \dfrac{M_{r}}{r} - \dfrac{M}{r^{2}} \right) \nonumber \\ \Gamma _{\theta \theta }^{\ \ \, r}&= -r\Phi \quad \Gamma _{\varphi \varphi }^{\ \ \ r} = -r \Phi \sin ^{2}\theta \end{aligned}$$
(88b)
$$\begin{aligned} \Gamma _{r\theta }^{\ \ \, \theta }&= \Gamma _{\theta r}^{\ \ \, \theta } = \frac{1}{r} \quad \Gamma _{\varphi \varphi }^{ \ \ \ \theta } = -\sin \theta \cos \theta \end{aligned}$$
(88c)
$$\begin{aligned} \Gamma _{r\varphi }^{\ \ \ \varphi }&= \Gamma _{\varphi r}^{\ \ \ \varphi } = \frac{1}{r} \quad \Gamma _{\theta \varphi }^{\ \ \ \varphi } = \Gamma _{\varphi \theta }^{\ \ \ \varphi } = \dfrac{\cos \theta }{\sin \theta }. \end{aligned}$$
(88d)
Using (88), we can now continue the computations started in (83).
$$\begin{aligned} \mathrm{Ric }(\nu _{t},\nu _{t})&= \mathrm {e}^{-2V}\left( \Gamma _{tt\ ,k}^{\ \ k} - \Gamma _{kt\ ,t}^{\ \ k} + \Gamma _{tt}^{\ \ s}\Gamma _{ks}^{\ \ k} - \Gamma _{kt}^{\ \ s}\Gamma _{ts}^{\ \ k} \right) \nonumber \\&= \mathrm {e}^{-2V}\Bigg \{\partial _{t}V_{t} + \partial _{r}\left( V_{r} \mathrm {e}^{2V}\left( 1 - \frac{2M}{r} \right) \right) - \partial _{t}V_{t} - \partial _{t}\left( \frac{M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1} \right) \nonumber \\&\quad + V_{t}\left( V_{t} + \frac{M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1} \right) \nonumber \\&\quad + V_{r} \mathrm {e}^{2V}\left( 1 - \frac{2M}{r} \right) \left( V_{r} + \left( 1 - \frac{2M}{r} \right) ^{-1}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) + \frac{2}{r} \right) \nonumber \\&\quad - V_{t}^{2} - 2 V_{r}^{2} \mathrm {e}^{2V}\left( 1 - \frac{2M}{r} \right) - \frac{M_{t}^{2}}{r^{2}}\left( 1 - \frac{2M}{r} \right) ^{-2} \Bigg \} \nonumber \\&= \mathrm {e}^{-2V}\Bigg \{V_{rr} \mathrm {e}^{2V}\left( 1 - \frac{2M}{r} \right) + 2V_{r}^{2} \mathrm {e}^{2V}\left( 1 - \frac{2M}{r} \right) + V_{r} \mathrm {e}^{2V}\left( \frac{2M}{r^{2}} - \frac{2M_{r}}{r} \right) \nonumber \\&\quad - \frac{M_{tt}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1} - \frac{2M_{t}^{2}}{r^{2}}\left( 1 - \frac{2M}{r} \right) ^{-2} + \frac{V_{t}M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1} \nonumber \\&\quad + V_{r}^{2} \mathrm {e}^{2V}\left( 1 - \frac{2M}{r} \right) + V_{r} \mathrm {e}^{2V}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) + \frac{2V_{r} \mathrm {e}^{2V}}{r}\left( 1 - \frac{2M}{r} \right) \nonumber \\&\quad - 2 V_{r}^{2} \mathrm {e}^{2V}\left( 1 - \frac{2M}{r} \right) - \frac{M_{t}^{2}}{r^{2}}\left( 1 - \frac{2M}{r} \right) ^{-2} \Bigg \} \nonumber \\ \mathrm{Ric }(\nu _{t},\nu _{t})&= \left( V_{rr} + V_{r}^{2} + \frac{2V_{r}}{r} \right) \left( 1-\frac{2M}{r} \right) + \frac{V_{r}}{r}\left( \frac{M}{r} - M_{r} \right) \nonumber \\&\quad + \frac{V_{t}M_{t} - M_{tt}}{r \mathrm {e}^{2V}}\left( 1-\frac{2M}{r} \right) ^{-1} - \frac{3M_{t}^{2}}{r^{2} \mathrm {e}^{2V}}\left( 1-\frac{2M}{r} \right) ^{-2} \end{aligned}$$
(89)
$$\begin{aligned} \mathrm{Ric }(\nu _{t},\nu _{r})&= \mathrm {e}^{-V}\sqrt{1 - \frac{2M}{r}} \left( \Gamma _{tr\ ,k}^{\ \ k} - \Gamma _{kr\ ,t}^{\ \ k} + \Gamma _{tr}^{\ \ s}\Gamma _{ks}^{\ \ k} - \Gamma _{kr}^{\ \ s}\Gamma _{ts}^{\ \ k} \right) \nonumber \\&= \mathrm {e}^{-V}\sqrt{1 - \frac{2M}{r}} \Bigg \{\partial _{t}(V_{r}) + \partial _{r}\left( \frac{M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1} \right) - \partial _{t}(V_{r})\nonumber \\&\quad - \partial _{t}\left( \left( 1 - \frac{2M}{r} \right) ^{-1}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) \right) + V_{r}\left( V_{t} + \frac{M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1} \right) \nonumber \\&\quad - 2\partial _{t}\left( \frac{1}{r} \right) + \frac{M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1}\left( V_{r} + \left( 1 - \frac{2M}{r} \right) ^{-1}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) + \frac{2}{r} \right) \nonumber \\&\quad - V_{r}V_{t} - \frac{2M_{t}V_{r}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1} - \frac{M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-2}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) \Bigg \} \nonumber \\&= \mathrm {e}^{-V}\sqrt{1 - \frac{2M}{r}} \Bigg \{\partial _{r}\partial _{t}\left( -\frac{1}{2}\ln \left( 1 - \frac{2M}{r} \right) \right) - \partial _{t}\partial _{r}\left( -\frac{1}{2}\ln \left( 1 - \frac{2M}{r} \right) \right) \nonumber \\&\quad + \frac{2M_{t}}{r^{2}}\left( 1 - \frac{2M}{r} \right) ^{-1} \Bigg \}\nonumber \\ \mathrm{Ric }(\nu _{t},\nu _{r})&= \frac{2M_{t}}{r^{2} \mathrm {e}^{V}}\left( 1 - \frac{2M}{r} \right) ^{-1/2} \end{aligned}$$
(90)
$$\begin{aligned} \mathrm{Ric }(\nu _{r},\nu _{r})&= \left( 1 - \frac{2M}{r} \right) \left( \Gamma _{rr\ ,k}^{\ \ k} - \Gamma _{kr\ ,r}^{\ \ k} + \Gamma _{rr}^{\ \ s}\Gamma _{ks}^{\ \ k} - \Gamma _{kr}^{\ \ s}\Gamma _{rs}^{\ \ k} \right) \nonumber \\&= \left( 1 - \frac{2M}{r} \right) \Bigg \{\partial _{t}\left( \frac{M_{t}}{r \mathrm {e}^{2V}}\left( 1 - \frac{2M}{r} \right) ^{-2} \right) \nonumber \\&\quad + \partial _{r}\left( \left( 1 - \frac{2M}{r} \right) ^{-1}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) \right) \nonumber \\&\quad - \partial _{r}V_{r} -\partial _{r}\left( \left( 1 - \frac{2M}{r} \right) ^{-1}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) \right) \nonumber \\&\quad - 2\partial _{r}\left( \frac{1}{r} \right) + \frac{V_{t}M_{t}}{r \mathrm {e}^{2V}}\left( 1 - \frac{2M}{r} \right) ^{-2} + \frac{M_{t}^{2}}{r^{2} \mathrm {e}^{2V}}\left( 1 - \frac{2M}{r} \right) ^{-3}\nonumber \\&\quad + \left( V_{r} + \frac{2}{r} \right) \left( 1 - \frac{2M}{r} \right) ^{-1}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) + \left( 1 - \frac{2M}{r} \right) ^{-2}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) ^{2}\nonumber \\&\quad - V_{r}^{2} - \frac{2M_{t}^{2}}{r^{2} \mathrm {e}^{2V}}\left( 1 - \frac{2M}{r} \right) ^{-3} - \frac{2}{r^{2}} - \left( 1 - \frac{2M}{r} \right) ^{-2}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) ^{2} \Bigg \} \nonumber \\&= \left( 1 - \frac{2M}{r} \right) \Bigg \{\frac{M_{tt}}{r \mathrm {e}^{2V}}\left( 1 - \frac{2M}{r} \right) ^{-2} - \frac{2V_{t}M_{t}}{r \mathrm {e}^{2V}}\left( 1 - \frac{2M}{r} \right) ^{-2}\nonumber \\&\quad + \frac{4M_{t}^{2}}{r^{2} \mathrm {e}^{2V}}\left( 1 - \frac{2M}{r} \right) ^{-3} - V_{rr} + \frac{2}{r^{2}} + \frac{V_{t}M_{t}}{r \mathrm {e}^{2V}}\left( 1 - \frac{2M}{r} \right) ^{-2}\nonumber \\&\quad - \frac{M_{t}^{2}}{r^{2} \mathrm {e}^{2V}}\left( 1 - \frac{2M}{r} \right) ^{-3} - V_{r}^{2} - \frac{2}{r^{2}}\nonumber \\&\quad + \left( V_{r} + \frac{2}{r} \right) \left( 1 - \frac{2M}{r} \right) ^{-1}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) \Bigg \} \nonumber \\ \mathrm{Ric }(\nu _{r},\nu _{r})&= -(V_{rr} + V_{r}^2)\left( 1 - \frac{2M}{r} \right) - \left( \frac{2}{r^{2}} + \frac{V_{r}}{r} \right) \left( \frac{M}{r} - M_{r} \right) \nonumber \\&\quad - \frac{V_{t}M_{t} - M_{tt}}{r \mathrm {e}^{2V}}\left( 1-\frac{2M}{r} \right) ^{-1} + \frac{3M_{t}^{2}}{r^2 \mathrm {e}^{2V}}\left( 1-\frac{2M}{r} \right) ^{-2} \end{aligned}$$
(91)
$$\begin{aligned} \mathrm{Ric }(\nu _{\theta },\nu _{\theta })&= \frac{1}{r^{2}} \left( \Gamma _{\theta \theta \ ,k}^{\ \ k} - \Gamma _{k\theta \ ,\theta }^{\ \ k} + \Gamma _{\theta \theta }^{\ \ s}\Gamma _{ks}^{\ \ k} - \Gamma _{k\theta }^{\ \ s}\Gamma _{\theta s}^{\ \ k} \right) \nonumber \\&= \frac{1}{r^{2}}\Bigg \{\partial _{r}\left( -r\left( 1 - \frac{2M}{r} \right) \right) - \partial _{\theta }\left( \frac{\cos \theta }{\sin \theta } \right) - r\left( 1 - \frac{2M}{r} \right) \left( V_{r} + \frac{2}{r} \right) \nonumber \\&\quad - r\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) + 2\left( 1 - \frac{2M}{r} \right) - \frac{\cos ^{2}\theta }{\sin ^{2}\theta }\Bigg \} \nonumber \\&= \frac{1}{r^{2}}\Bigg \{-\left( 1 - \frac{2M}{r} \right) - r\left( \frac{2M}{r^{2}} - \frac{2M_{r}}{r} \right) + \frac{1}{\sin ^{2}\theta } - r\left( 1 - \frac{2M}{r} \right) \left( V_{r} + \frac{2}{r} \right) \nonumber \\&\quad + r\left( \frac{M}{r^{2}} - \frac{M_{r}}{r} \right) + 2\left( 1 - \frac{2M}{r} \right) - \frac{\cos ^{2}\theta }{\sin ^{2}\theta }\Bigg \} \nonumber \\&= \frac{1}{r^{2}}\Bigg \{-\frac{\sin ^{2}\theta }{\sin ^{2}\theta } + \frac{2M}{r} - \frac{M}{r} + M_{r} + \frac{1}{\sin ^{2}\theta } - rV_{r}\left( 1 - \frac{2M}{r} \right) \nonumber \\&\quad - 2\left( 1 - \frac{2M}{r} \right) + 2 \left( 1 - \frac{2M}{r} \right) - \frac{\cos ^{2}\theta }{\sin ^{2}\theta } \Bigg \} \nonumber \\ \mathrm{Ric }(\nu _{\theta },\nu _{\theta })&= \frac{1}{r^{2}}\left( \frac{M}{r} + M_{r} \right) - \frac{V_{r}}{r}\left( 1 - \frac{2M}{r} \right) \end{aligned}$$
(92)
$$\begin{aligned} \mathrm{Ric }(\nu _{\varphi },\nu _{\varphi })&= \frac{1}{r^{2}\sin ^{2}\theta } \left( \Gamma _{\varphi \varphi \ ,k}^{\ \ \ k} - \Gamma _{k\varphi \ ,\varphi }^{\ \ \ k} + \Gamma _{\varphi \varphi }^{\ \ \ s}\Gamma _{ks}^{\ \ k} - \Gamma _{k\varphi }^{\ \ \ s}\Gamma _{\varphi s}^{\ \ \ k} \right) \nonumber \\&= \frac{1}{r^{2}\sin ^{2}\theta } \Bigg \{\partial _{r}\left( -r\sin ^{2}\theta \left( 1 - \frac{2M}{r} \right) \right) + \partial _{\theta }\left( -\sin \theta \cos \theta \right) \nonumber \\&\quad - r\sin ^{2}\theta \left( 1 - \frac{2M}{r} \right) \left( V_{r} + \frac{2}{r} \right) - r\sin ^{2}\theta \left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) - \cos ^{2}\theta \nonumber \\&\quad + 2\sin ^{2}\theta \left( 1 - \frac{2M}{r} \right) + 2\cos ^{2}\theta \Bigg \} \nonumber \\&= \frac{1}{r^{2}\sin ^{2}\theta } \Bigg \{-\sin ^{2}\theta \left( 1 - \frac{2M}{r} \right) - r\sin ^{2}\theta \left( \frac{2M}{r^{2}} - \frac{2M_{r}}{r} \right) - \cos ^{2}\theta + \sin ^{2}\theta \nonumber \\&\quad - r\sin ^{2}\theta \left( 1 - \frac{2M}{r} \right) \left( V_{r} + \frac{2}{r} \right) + r\sin ^{2}\theta \left( \frac{M}{r^{2}} - \frac{M_{r}}{r} \right) - \cos ^{2}\theta \nonumber \\&\quad + 2\sin ^{2}\theta \left( 1 - \frac{2M}{r} \right) + 2\cos ^{2}\theta \Bigg \}\nonumber \\&= \frac{1}{r^{2}}\left( \frac{2M}{r} - \frac{M}{r} + M_{r} - rV_{r}\left( 1 - \frac{2M}{r} \right) - 2\left( 1 - \frac{2M}{r} \right) + 2\left( 1 - \frac{2M}{r} \right) \right) \nonumber \\ \mathrm{Ric }(\nu _{\varphi },\nu _{\varphi })&= \frac{1}{r^{2}}\left( \frac{M}{r} + M_{r} \right) - \frac{V_{r}}{r}\left( 1 - \frac{2M}{r} \right) = \mathrm{Ric }(\nu _{\theta },\nu _{\theta }). \end{aligned}$$
(93)
These are the only nonzero components of the Ricci curvature tensor since
$$\begin{aligned} \mathrm{Ric }(\nu _{t},\nu _{\theta })&= \frac{1}{r\mathrm {e}^{V}}\mathrm{Ric }_{t\theta } = \frac{1}{r\mathrm {e}^{V}}\mathbf {R}_{tk\theta }^{\ \ \ \, k} \nonumber \\&= \frac{1}{r\mathrm {e}^{V}}\left( \Gamma _{t\theta \ ,k}^{\ \ \, k} - \Gamma _{k\theta \ ,t}^{\ \ \, k} + \Gamma _{t\theta }^{\ \ \, s}\Gamma _{ks}^{\ \ k} - \Gamma _{k\theta }^{\ \ \, s}\Gamma _{ts}^{\ \ k} \right) \nonumber \\&= \frac{1}{r\mathrm {e}^{V}}\left( - \Gamma _{r\theta \ ,t}^{\ \ \, r} - \Gamma _{\theta \theta \ ,t}^{\ \ \ \theta } - \Gamma _{\varphi \theta \ ,t}^{\ \ \ \varphi } - \Gamma _{k\theta }^{\ \ \, r}\Gamma _{tr}^{\ \ k} - \Gamma _{k\theta }^{\ \ \, \theta }\Gamma _{t\theta }^{\ \ \, k} - \Gamma _{k\theta }^{\ \ \, \varphi }\Gamma _{t\varphi }^{\ \ \, k} \right) \nonumber \\&= \frac{1}{r\mathrm {e}^{V}}\left( - \Gamma _{\theta \theta }^{\ \ \ r}\Gamma _{tr}^{\ \ \theta } - \Gamma _{r\theta }^{\ \ \, \theta }\Gamma _{t\theta }^{\ \ \, r} - \Gamma _{\varphi \theta }^{\ \ \ \varphi }\Gamma _{t\varphi }^{\ \ \, \varphi } \right) \nonumber \\&= 0 \end{aligned}$$
(94)
$$\begin{aligned} \mathrm{Ric }(\nu _{t},\nu _{\varphi })&= \frac{1}{r\sin \theta \mathrm {e}^{V}}\mathrm{Ric }_{t\varphi } = \frac{1}{r\sin \theta \mathrm {e}^{V}}\mathbf {R}_{tk\varphi }^{\ \ \ \, k} \nonumber \\&= \frac{1}{r\sin \theta \mathrm {e}^{V}}\left( \Gamma _{t\varphi \ ,k}^{\ \ \, k} - \Gamma _{k\varphi \ ,t}^{\ \ \, k} + \Gamma _{t\varphi }^{\ \ \, s}\Gamma _{ks}^{\ \ k} - \Gamma _{k\varphi }^{\ \ \, s}\Gamma _{ts}^{\ \ k} \right) \nonumber \\&= \frac{1}{r\sin \theta \mathrm {e}^{V}}\left( - \Gamma _{r\varphi \ ,t}^{\ \ \, r} - \Gamma _{\theta \varphi \ ,t}^{\ \ \ \theta } - \Gamma _{\varphi \varphi \ ,t}^{\ \ \ \varphi } - \Gamma _{k\varphi }^{\ \ \, r}\Gamma _{tr}^{\ \ k} - \Gamma _{k\varphi }^{\ \ \, \theta }\Gamma _{t\theta }^{\ \ \, k} - \Gamma _{k\varphi }^{\ \ \, \varphi }\Gamma _{t\varphi }^{\ \ \, k} \right) \nonumber \\&= \frac{1}{r\sin \theta \mathrm {e}^{V}}\left( - \Gamma _{\varphi \varphi }^{\ \ \ r}\Gamma _{tr}^{\ \ \varphi } - \Gamma _{\varphi \varphi }^{\ \ \ \theta }\Gamma _{t\theta }^{\ \ \, \varphi } - \Gamma _{r\varphi }^{\ \ \, \varphi }\Gamma _{t\varphi }^{\ \ \, r} - \Gamma _{\theta \varphi }^{\ \ \ \varphi }\Gamma _{t\varphi }^{\ \ \, \theta } \right) \nonumber \\&= 0 \end{aligned}$$
(95)
$$\begin{aligned} \mathrm{Ric }(\nu _{r},\nu _{\theta })&= \frac{1}{r}\sqrt{1 - \frac{2M}{r}}\mathrm{Ric }_{r\theta } = \frac{1}{r}\sqrt{1 - \frac{2M}{r}}\mathbf {R}_{rk\theta }^{\ \ \ \, k} \nonumber \\&= \frac{1}{r}\sqrt{1 - \frac{2M}{r}}\left( \Gamma _{r\theta \ ,k}^{\ \ \, k} - \Gamma _{k\theta \ ,r}^{\ \ \, k} + \Gamma _{r\theta }^{\ \ \, s}\Gamma _{ks}^{\ \ k} - \Gamma _{k\theta }^{\ \ \, s}\Gamma _{rs}^{\ \ k} \right) \nonumber \\&= \frac{1}{r}\sqrt{1 - \frac{2M}{r}}\left( \Gamma _{r\theta \ ,\theta }^{\ \ \, \theta } - \Gamma _{r\theta \ ,r}^{\ \ \, r} - \Gamma _{\theta \theta \ ,r}^{\ \ \ \theta } - \Gamma _{\varphi \theta \ ,r}^{\ \ \ \varphi } + \Gamma _{r\theta }^{\ \ \, \theta }\Gamma _{r\theta }^{\ \ \, r} \right. \nonumber \\&\quad \left. + \Gamma _{r\theta }^{\ \ \, \theta }\Gamma _{\theta \theta }^{\ \ \ \theta } + \Gamma _{r\theta }^{\ \ \, \theta }\Gamma _{\varphi \theta }^{\ \ \varphi } - \Gamma _{r\theta }^{\ \ \, s}\Gamma _{rs}^{\ \ r} - \Gamma _{\theta \theta }^{\ \ \ s}\Gamma _{rs}^{\ \ \theta } - \Gamma _{\varphi \theta }^{\ \ \ s}\Gamma _{rs}^{\ \ \varphi }\right) \nonumber \\&= \frac{1}{r}\sqrt{1 - \frac{2M}{r}}\left( \frac{\cos \theta }{r\sin \theta } - \Gamma _{r\theta }^{\ \ \, \theta }\Gamma _{r\theta }^{\ \ \, r} - \Gamma _{\theta \theta }^{\ \ \ r}\Gamma _{rr}^{\ \ \theta } - \Gamma _{\varphi \theta }^{\ \ \ \varphi }\Gamma _{r\varphi }^{\ \ \, \varphi } \right) \nonumber \\&= \frac{1}{r}\sqrt{1 - \frac{2M}{r}}\left( \frac{\cos \theta }{r\sin \theta } - \frac{\cos \theta }{r\sin \theta } \right) \nonumber \\&= 0 \end{aligned}$$
(96)
$$\begin{aligned} \mathrm{Ric }(\nu _{r},\nu _{\varphi })&= \frac{1}{r\sin \theta }\sqrt{1 - \frac{2M}{r}}\mathrm{Ric }_{r\varphi } = \frac{1}{r\sin \theta }\sqrt{1 - \frac{2M}{r}}\mathbf {R}_{rk\varphi }^{\ \ \ \, k} \nonumber \\&= \frac{1}{r\sin \theta }\sqrt{1 - \frac{2M}{r}}\left( \Gamma _{r\varphi \ ,k}^{\ \ \, k} - \Gamma _{k\varphi \ ,r}^{\ \ \, k} + \Gamma _{r\varphi }^{\ \ \, s}\Gamma _{ks}^{\ \ k} - \Gamma _{k\varphi }^{\ \ \, s}\Gamma _{rs}^{\ \ k} \right) \nonumber \\&= \frac{1}{r\sin \theta }\sqrt{1 - \frac{2M}{r}}\left( \Gamma _{r\varphi \ ,\varphi }^{\ \ \, \varphi } - \Gamma _{r\varphi \ ,r}^{\ \ \, r} - \Gamma _{\theta \varphi \ ,r}^{\ \ \ \theta } - \Gamma _{\varphi \varphi \ ,r}^{\ \ \ \varphi } + \Gamma _{r\varphi }^{\ \ \, \varphi }\Gamma _{r\varphi }^{\ \ r}\right. \nonumber \\&\quad \left. + \Gamma _{r\varphi }^{\ \ \, \varphi }\Gamma _{\theta \varphi }^{\ \ \ \theta } + \Gamma _{r\varphi }^{\ \ \, \varphi }\Gamma _{\varphi \varphi }^{\ \ \ \varphi } - \Gamma _{r\varphi }^{\ \ \, s}\Gamma _{rs}^{\ \ r} - \Gamma _{\theta \varphi }^{\ \ \ s}\Gamma _{rs}^{\ \ \theta } - \Gamma _{\varphi \varphi }^{\ \ \ s}\Gamma _{rs}^{\ \ \varphi }\right) \nonumber \\&= \frac{1}{r\sin \theta }\sqrt{1 - \frac{2M}{r}}\left( - \Gamma _{r\varphi }^{\ \ \, \varphi }\Gamma _{r\varphi }^{\ \ \, r} - \Gamma _{\theta \varphi }^{\ \ \ \varphi }\Gamma _{r\varphi }^{\ \ \, \theta } - \Gamma _{\varphi \varphi }^{\ \ \ r}\Gamma _{rr}^{\ \ \varphi } - \Gamma _{\varphi \varphi }^{\ \ \ \theta }\Gamma _{r\theta }^{\ \ \, \varphi } \right) \nonumber \\&= 0 \end{aligned}$$
(97)
$$\begin{aligned} \mathrm{Ric }(\nu _{\theta },\nu _{\varphi })&= \frac{1}{r^{2}\sin \theta }\mathrm{Ric }_{\theta \varphi } = \frac{1}{r^{2}\sin \theta }\mathbf {R}_{\theta k \varphi }^{\ \ \ \ k}\nonumber \\&= \frac{1}{r^{2}\sin \theta }\left( \Gamma _{\theta \varphi \ ,k}^{\ \ \ k} - \Gamma _{k\varphi \ ,\theta }^{\ \ \, k} + \Gamma _{\theta \varphi }^{\ \ \ s}\Gamma _{ks}^{\ \ k} - \Gamma _{k\varphi }^{\ \ \, s}\Gamma _{\theta s}^{\ \ \, k} \right) \nonumber \\&= \frac{1}{r^{2}\sin \theta }\left( \Gamma _{\theta \varphi \ ,\varphi }^{\ \ \ \varphi } - \Gamma _{r\varphi \ ,\theta }^{\ \ \, r} - \Gamma _{\theta \varphi \ ,\theta }^{\ \ \ \theta } - \Gamma _{\varphi \varphi \ ,\theta }^{\ \ \ \varphi } + \Gamma _{\theta \varphi }^{\ \ \ \varphi }\Gamma _{r\varphi }^{\ \ \, r} \right. \nonumber \\&\quad \left. + \Gamma _{\theta \varphi }^{\ \ \ \varphi }\Gamma _{\theta \varphi }^{\ \ \ \theta } + \Gamma _{\theta \varphi }^{\ \ \ \varphi }\Gamma _{\varphi \varphi }^{\ \ \ \varphi } - \Gamma _{k\varphi }^{\ \ \, r}\Gamma _{\theta r}^{\ \ \, k} - \Gamma _{k\varphi }^{\ \ \, \theta }\Gamma _{\theta \theta }^{\ \ \ k} - \Gamma _{k\varphi }^{\ \ \, \varphi }\Gamma _{\theta \varphi }^{\ \ \ k}\right) \nonumber \\&= \frac{1}{r^{2}\sin \theta }\left( - \Gamma _{\theta \varphi }^{\ \ \ r}\Gamma _{\theta r}^{\ \ \, \theta } - \Gamma _{r\varphi }^{\ \ \, \theta }\Gamma _{\theta \theta }^{\ \ \ r} - \Gamma _{\varphi \varphi }^{\ \ \ \varphi }\Gamma _{\theta \varphi }^{\ \ \ \varphi } \right) \nonumber \\&= 0. \end{aligned}$$
(98)
The other components are 0 by the symmetry of the Ricci curvature tensor. To get the last statement, we note that by Eq. (93) and Lemma 14-2, we have that
$$\begin{aligned} R = \sum _{\eta } g(\nu _{\eta },\nu _{\eta })\mathrm{Ric }(\nu _{\eta },\nu _{\eta }) = -\mathrm{Ric }(\nu _{t},\nu _{t}) + \mathrm{Ric }(\nu _{r},\nu _{r}) + 2\mathrm{Ric }(\nu _{\theta },\nu _{\theta }). \end{aligned}$$
(99)
From here, it is a matter of algebra and the statements above to get the desired equation for \(R\). \(\square \)
Next, we prove Proposition 7.
Proof
(Proof of Proposition 7) In solving Eqs. (49)–(51), we already solve the equations resulting from two of the components of the Einstein equation, namely (50) and (51), which we reprint here for convenience.
$$\begin{aligned} M_{r}&= 4\pi r^{2}\mu \end{aligned}$$
(100)
$$\begin{aligned} V_{r}&= \left( 1 - \frac{2M}{r} \right) ^{-1}\left( \frac{M}{r^{2}} + 4\pi r P \right) \end{aligned}$$
(101)
In order to show that solving these two equations coupled with \(\nabla _{g} \cdot T = 0\) is sufficient to solve the entire Einstein equation, we must show that if these equations hold, so do the other components of the Einstein equation. To do so, we first need to write down the other components of the Einstein equation. Recall that the above equations come from the \(\nu _{t},\nu _{t}\) and \(\nu _{r},\nu _{r}\) components of the Einstein equation. Thus we only have the \(\nu _{t},\nu _{r}\) and \(\nu _{\theta },\nu _{\theta }\) components left to compute (the \(\nu _{\varphi },\nu _{\varphi }\) component is identical to \(\nu _{\theta },\nu _{\theta }\) component). Then, by Corollary 3 and Eq. (48), we have the following
$$\begin{aligned} G(\nu _{t},\nu _{r})&= 8\pi T(\nu _{t},\nu _{r}) \nonumber \\ M_{t}&= 4\pi r^{2} \mathrm {e}^{V}\rho \sqrt{1 - \frac{2M}{r}} \end{aligned}$$
(102)
and
$$\begin{aligned} G(\nu _{\theta },\nu _{\theta })&= 8\pi T(\nu _{\theta },\nu _{\theta }) \nonumber \\ 8\pi Q&= \left( V_{rr} + V_{r}^{2} + \frac{V_{r}}{r} \right) \left( 1-\frac{2M}{r} \right) + \left( \frac{M}{r} - M_{r} \right) \left( \frac{1}{r^{2}} + \frac{V_{r}}{r} \right) \nonumber \\&\quad + \frac{V_{t}M_{t} - M_{tt}}{r \mathrm {e}^{2V}}\left( 1 - \frac{2M}{r} \right) ^{-1} - \frac{3M_{t}^{2}}{r^{2} \mathrm {e}^{2V}}\left( 1-\frac{2M}{r} \right) ^{-2}. \end{aligned}$$
(103)
Next we compute what \(\nabla _{g} \cdot T = 0\) means in terms of the functions \(\mu \), \(P\), \(\rho \), and \(Q\). In what follows, the summing indices run through the set \(\{t,r,\theta ,\varphi \}\) and we will use the Einstein summation convention wherever it applies, while specifically denoting any other summations of a different form. First, we define
$$\begin{aligned} \varepsilon _{k} = g(\nu _{k},\nu _{k}) \end{aligned}$$
(104)
for all \(k \in \{t,r,\theta ,\varphi \}\). Then we can write the divergence of \(T\) as follows
$$\begin{aligned} \nabla _{g} \cdot T&= \sum _{k} \varepsilon _{k}(\nabla _{\nu _{k}}T)(\nu _{k},\partial _{j})\, dx^{j} \nonumber \\&= -(\nabla _{\nu _{t}}T)(\nu _{t},\partial _{j})\, dx^{j} + (\nabla _{\nu _{r}}T)(\nu _{r},\partial _{j})\, dx^{j} \nonumber \\&\quad + (\nabla _{\nu _{\theta }}T)(\nu _{\theta },\partial _{j})\, dx^{j} + (\nabla _{\nu _{\varphi }}T)(\nu _{\varphi },\partial _{j})\, dx^{j}. \end{aligned}$$
(105)
We will simplify each of these four terms individually. By Eqs. (88), (31), and (48), we have that
$$\begin{aligned} (\nabla _{\nu _{t}}T)(\nu _{t},\partial _{k})\, dx^{k}&= \mathrm {e}^{-2V}(\nabla _{t}T)(\partial _{t},\partial _{k})\, dx^{k} \nonumber \\&= \mathrm {e}^{-2V}\Big [\partial _{t}(T(\partial _{t},\partial _{k})) - \Gamma _{tt}^{\ \ m}T(\partial _{m},\partial _{k}) - \Gamma _{tk}^{\ \ m}T(\partial _{t},\partial _{m})\Big ]\, dx^{k} \nonumber \\&= \mathrm {e}^{-2V}\Big [\partial _{t}(T(\partial _{t},\partial _{t}))\, dt + \partial _{t}(T(\partial _{t},\partial _{r}))\, dr \nonumber \\&\quad - 2\Gamma _{tt}^{\ \ t}T(\partial _{t},\partial _{t})\, dt - 2\Gamma _{tt}^{\ \ r}T(\partial _{t},\partial _{r})\, dt \nonumber \\&\quad - (\Gamma _{tt}^{\ \ t} + \Gamma _{tr}^{\ \ r})T(\partial _{t},\partial _{r})\, dr - \Gamma _{tr}^{\ \ t}T(\partial _{t},\partial _{t})\, dr - \Gamma _{tt}^{\ \ r}T(\partial _{r},\partial _{r})\, dr\Big ] \nonumber \\&= \mathrm {e}^{-2V}\Bigg [\left( 2V_{t} \mathrm {e}^{2V}\mu + \mathrm {e}^{2V}\mu _{t} - 2V_{t} \mathrm {e}^{2V}\mu - 2\rho V_{r} \mathrm {e}^{3V}\sqrt{1 - \frac{2M}{r}} \right) \, dt \nonumber \\&\quad + \Bigg (V_{t} \mathrm {e}^{V}\rho \left( 1 - \frac{2M}{r} \right) ^{-1/2} + \mathrm {e}^{V}\rho _{t}\left( 1 - \frac{2M}{r} \right) ^{-1/2} \nonumber \\&\quad + \frac{ \mathrm {e}^{V}\rho M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-3/2} \nonumber \\&\quad - \left( V_{t} + \frac{M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1} \right) \mathrm {e}^{V}\rho \left( 1 - \frac{2M}{r} \right) ^{-1/2} \nonumber \\&\quad - V_{r} \mathrm {e}^{2V}\mu - V_{r} \mathrm {e}^{2V}P\Bigg )\, dr \Bigg ] \nonumber \\ (\nabla _{\nu _{t}}T)(\nu _{t},\partial _{k})\, dx^{k}&= \left( \mu _{t} - 2\rho V_{r} \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) \, dt \nonumber \\&\quad + \left( \rho _{t} \mathrm {e}^{-V}\left( 1 - \frac{2M}{r} \right) ^{-1/2} - V_{r}(\mu + P) \right) \, dr \end{aligned}$$
(106)
Next, we have
$$\begin{aligned} (\nabla _{\nu _{r}}T)(\nu _{r},\partial _{k})\, dx^{k}&= \left( 1 - \frac{2M}{r} \right) (\nabla _{r}T)(\partial _{r},\partial _{k})\, dx^{k} \nonumber \\&= \left( 1 - \frac{2M}{r} \right) \Big [\partial _{r}(T(\partial _{r},\partial _{k})) - \Gamma _{rr}^{\ \ m}T(\partial _{m},\partial _{k}) - \Gamma _{rk}^{\ \ m}T(\partial _{r},\partial _{m})\Big ]\, dx^{k} \nonumber \\&= \left( 1 - \frac{2M}{r} \right) \Big [\partial _{r}(T(\partial _{r},\partial _{t}))\, dt + \partial _{r}(T(\partial _{r},\partial _{r}))\, dr \nonumber \\&\quad - \left( \Gamma _{rr}^{\ \ r} + \Gamma _{rt}^{\ \ t} \right) T(\partial _{t},\partial _{r})\, dt - 2\Gamma _{rr}^{\ \ t}T(\partial _{t},\partial _{r})\, dr \nonumber \\&\quad - 2\Gamma _{rr}^{\ \ r}T(\partial _{r},\partial _{r})\, dr - \Gamma _{rr}^{\ \ t}T(\partial _{t},\partial _{t})\, dt - \Gamma _{rt}^{\ \ r}T(\partial _{r},\partial _{r})\, dt \Big ]\nonumber \\&= \left( 1 - \frac{2M}{r} \right) \left[ \left( \rho \mathrm {e}^{V}\left( 1 - \frac{2M}{r} \right) ^{-3/2}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) \right. \right. \nonumber \\&\quad \left. \left. + \rho V_{r} \mathrm {e}^{V}\left( 1 - \frac{2M}{r} \right) ^{-1/2} + \rho _{r} \mathrm {e}^{V}\left( 1 - \frac{2M}{r} \right) ^{-1/2} \right. \right. \nonumber \\&\quad \left. \left. - \frac{\mu M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-2} - \frac{P M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-2} \right. \right. \nonumber \\&\quad \left. \left. -\rho \mathrm {e}^{V}\left( 1 - \frac{2M}{r} \right) ^{-1/2}\left[ \left( 1 - \frac{2M}{r} \right) ^{-1}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) + V_{r} \right] \right) \, dt \right. \nonumber \\&\quad \left. + \Bigg (P_{r}\left( 1 - \frac{2M}{r} \right) ^{-1} + 2P\left( 1 - \frac{2M}{r} \right) ^{-2}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) \right. \nonumber \\&\quad \left. - \frac{2\rho M_{t}}{r \mathrm {e}^{V}}\left( 1 - \frac{2M}{r} \right) ^{-5/2} - 2P\left( 1 - \frac{2M}{r} \right) ^{-2}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) \Bigg )\, dr \right] \nonumber \\ (\nabla _{\nu _{r}}T)(\nu _{r},\partial _{k})\, dx^{k}&= \left( \rho _{r} \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} - \frac{M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1}(\mu + P) \right) \, dt \nonumber \\&\quad + \left( P_{r} - \frac{2\rho M_{t}}{r \mathrm {e}^{V}}\left( 1 - \frac{2M}{r} \right) ^{-3/2} \right) \, dr. \end{aligned}$$
(107)
Thirdly,
$$\begin{aligned} (\nabla _{\nu _{\theta }}T)(\nu _{\theta },\partial _{k})\, dx^{k}&= \frac{1}{r^{2}}(\nabla _{\theta }T)(\partial _{\theta },\partial _{k})\, dx^{k} \nonumber \\&= \frac{1}{r^{2}} \Big [\partial _{\theta }(T(\partial _{\theta },\partial _{k})) - \Gamma _{\theta \theta }^{\ \ m}T(\partial _{m},\partial _{k}) - \Gamma _{\theta k}^{\ \ m}T(\partial _{\theta },\partial _{m})\Big ]\, dx^{k} \nonumber \\&= \frac{1}{r^{2}}\Big [-\Gamma _{\theta \theta }^{\ \ r}T(\partial _{r},\partial _{t})\, dt - \Gamma _{\theta \theta }^{\ \ r}T(\partial _{r},\partial _{r})\, dr - \Gamma _{\theta r}^{\ \ \theta }T(\partial _{\theta },\partial _{\theta })\, dr \Big ]\nonumber \\&= \frac{1}{r^{2}}\left[ r\rho \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}}\, dt + r P \, dr - r Q\, dr \right] \nonumber \\ (\nabla _{\nu _{\theta }}T)(\nu _{\theta },\partial _{k})\, dx^{k}&= \frac{\rho \mathrm {e}^{V}}{r}\sqrt{1 - \frac{2M}{r}}\, dt + \frac{P - Q}{r}\, dr. \end{aligned}$$
(108)
Lastly, we have that
$$\begin{aligned} (\nabla _{\nu _{\varphi }}T)(\nu _{\varphi },\partial _{k})\, dx^{k}&= \frac{1}{r^{2}\sin ^{2}\theta }(\nabla _{\varphi }T)(\partial _{\varphi },\partial _{k})\, dx^{k} \nonumber \\&= \frac{1}{r^{2}\sin ^{2}\theta }\Big [ \partial _{\varphi }(T(\partial _{\varphi },\partial _{k})) - \Gamma _{\varphi \varphi }^{\ \ \, m}T(\partial _{m},\partial _{k}) - \Gamma _{\varphi k}^{\ \ m}T(\partial _{\varphi },\partial _{m}) \Big ]\, dx^{k}\nonumber \\&= \frac{1}{r^{2}\sin ^{2}\theta } \Big [ - \Gamma _{\varphi \varphi }^{\ \ \, r}T(\partial _{r},\partial _{t})\, dt - \Gamma _{\varphi \varphi }^{\ \ \, r}T(\partial _{r},\partial _{r})\, dr - \Gamma _{\varphi \varphi }^{\ \ \, \theta }T(\partial _{\theta },\partial _{\theta })\, d\theta \nonumber \\&\quad - \Gamma _{\varphi r}^{\ \ \, \varphi }T(\partial _{\varphi },\partial _{\varphi })\, dr - \Gamma _{\varphi \theta }^{\ \ \, \varphi }T(\partial _{\varphi },\partial _{\varphi })\, d\theta \Big ] \nonumber \\&= \frac{1}{r^{2}\sin ^{2}\theta } \Big [ r\rho \mathrm {e}^{V} \sin ^{2}\theta \sqrt{1 - \frac{2M}{r}}\, dt + r \sin ^{2}\theta P\, dr \nonumber \\&\quad + r^{2}Q\sin \theta \cos \theta \, d\theta - r Q \sin ^{2}\theta \, dr - r^{2}Q \cos \theta \sin \theta \, d\theta \Big ] \nonumber \\&= \frac{\rho \mathrm {e}^{V}}{r}\sqrt{1 - \frac{2M}{r}}\, dt + \frac{P - Q}{r}\, dr \nonumber \\ (\nabla _{\nu _{\varphi }}T)(\nu _{\varphi },\partial _{k})\, dx^{k}&= (\nabla _{\nu _{\theta }}T)(\nu _{\theta },\partial _{k})\, dx^{k}. \end{aligned}$$
(109)
Then Eq. (105) becomes
$$\begin{aligned} \nabla _{g} \cdot T&= -(\nabla _{\nu _{t}}T)(\nu _{t},\partial _{j})\, dx^{j} + (\nabla _{\nu _{r}}T)(\nu _{r},\partial _{j})\, dx^{j} \nonumber \\&\quad + (\nabla _{\nu _{\theta }}T)(\nu _{\theta },\partial _{j})\, dx^{j} + (\nabla _{\nu _{\varphi }}T)(\nu _{\varphi },\partial _{j})\, dx^{j} \nonumber \\&= \Bigg (-\mu _{t} + 2\rho \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}}\left( V_{r} + \frac{1}{r} \right) + \rho _{r} \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \nonumber \\&\quad - \frac{M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1}(\mu + P)\Bigg )\, dt + \Bigg (-\rho _{t} \mathrm {e}^{-V}\left( 1 - \frac{2M}{r} \right) ^{-1/2} \nonumber \\&\quad + V_{r}(\mu + P) + P_{r} - \frac{2\rho M_{t}}{r \mathrm {e}^{V}}\left( 1 - \frac{2M}{r} \right) ^{-3/2} + \frac{2(P - Q)}{r}\Bigg )\, dr. \end{aligned}$$
(110)
Then \(\nabla _{g} \cdot T = 0\) yields the following two equations
$$\begin{aligned} \mu _{t}&= 2\rho \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}}\left( V_{r} + \frac{1}{r} \right) + \rho _{r} \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} - \frac{M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1}(\mu + P)\end{aligned}$$
(111a)
$$\begin{aligned} \rho _{t}&= \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}}\left( V_{r}(\mu + P) + P_{r} + \frac{2(P - Q)}{r} \right) - \frac{2\rho M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1}. \end{aligned}$$
(111b)
We are now ready to show that solving \(\nabla _{g} \cdot T = 0\) along with solving Eqs. (100) and (101) on each \(t=constant\) slice also solves the remaining two unique nonzero components of the Einstein equation (102) and (103). We will show first that (100), (101), and (111) implies that (102) holds as well. To do this, we first need to show that (100) and (102) are compatible, that is, given these equations, \(M_{rt} = M_{tr}\). Differentiating (100) with respect to \(t\) yields,
$$\begin{aligned} \partial _{t}M_{r}&= \partial _{t}\left( 4\pi r^{2}\mu \right) \nonumber \\ M_{rt}&= 4\pi r^{2}\mu _{t} \end{aligned}$$
(112)
while differentiating (102) with respect to \(r\) yields,
$$\begin{aligned} \partial _{r}M_{t}&= \partial _{r}\left( 4\pi r^{2} \mathrm {e}^{V}\rho \sqrt{1 - \frac{2M}{r}} \right) \nonumber \\ M_{tr}&= 4\pi \Bigg [ 2r \mathrm {e}^{V} \rho \sqrt{1 - \frac{2M}{r}} + r^{2}V_{r} \mathrm {e}^{V}\rho \sqrt{1 - \frac{2M}{r}} \nonumber \\&\quad + r^{2} \mathrm {e}^{V}\rho _{r}\sqrt{1 - \frac{2M}{r}} - r^{2} \mathrm {e}^{V}\rho \left( 1 - \frac{2M}{r} \right) ^{-1/2}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) \Bigg ] \nonumber \\&= 4\pi r^{2}\Bigg [\mu _{t} + \frac{M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1}(\mu + P) - 4\pi r \rho \mathrm {e}^{V}\left( 1 - \frac{2M}{r} \right) ^{-1/2}(\mu + P) \Bigg ] \nonumber \\&= 4\pi r^{2}\mu _{t} \end{aligned}$$
(113)
where in the next to last line we made substitutions using Eqs. (100), (101), and (111a) and the last line used (102). Thus if \(\nabla _{g} \cdot T=0\), (100) and (102) are compatible, that is, \(M_{tr} = M_{rt}\) everywhere the equations are defined, namely, wherever \(r \ne 0\). However, by using L’Hôpital’s rule on the equations above coupled with the fact from Eq. (40) that at \(r = 0\), \(M = M_{r} = M_{rr} = 0\) for all \(t\), the equation \(M_{tr} = M_{rt}\) still holds at the central value \(r=0\). This implies that there exists a function \(M(t,r)\) which satisfies both (100) and (101) everywhere, which of course is the metric function we seek.
As per our hypothesis, if for all values of \(t\), we have solved (100) with the initial condition \(M = 0\) at \(r=0\), we will obtain some function \(M^{*}(t,r)\) which satisfies (100) everywhere. Then we have that
$$\begin{aligned} M_{r}^{*} = M_{r} \end{aligned}$$
(114)
everywhere, where \(M\) is the function that satisfies both (100) and (101). This implies that
$$\begin{aligned} M^{*}(t,r) = M(t,r) + f(t) \end{aligned}$$
(115)
for some smooth function \(f(t)\). However, we also have that \(M^{*}(t,0) = M(t,0) = 0\) for all \(t\), which implies that \(f(t) \equiv 0\). Hence \(M^{*}(t,r) = M(t,r)\) and the function we obtain by integrating (100) with compatible initial conditions necessarily also satisfies (102).
Finally, we use Eqs. (100), (101), (111), and (102) [since the first three imply Eq. (102)] to show that (103) is automatically satisfied. In order to do this, we will have to compute \(V_{rr}\) and \(M_{tt}\). They are
$$\begin{aligned} \partial _{t} M_{t}&= \partial _{t}\left( 4\pi r^{2} \mathrm {e}^{V}\rho \sqrt{1 - \frac{2M}{r}} \right) \nonumber \\ M_{tt}&= 4\pi r^{2}\left[ V_{t} \mathrm {e}^{V}\rho \sqrt{1 - \frac{2M}{r}} + \mathrm {e}^{V}\rho _{t}\sqrt{1 - \frac{2M}{r}} - \frac{ \mathrm {e}^{V} \rho M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1/2} \right] \nonumber \\&= V_{t}M_{t} + 4\pi r^{2}\mathrm {e}^{V}\left[ \rho _{t}\sqrt{1 - \frac{2M}{r}} - \frac{\rho M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-1/2} \right] \end{aligned}$$
(116)
and
$$\begin{aligned} \partial _{r}V_{r}&= \partial _{r}\left[ \left( 1 - \frac{2M}{r} \right) ^{-1}\left( \frac{M}{r^{2}} + 4\pi r P \right) \right] \nonumber \\ V_{rr}&= 2\left( 1 - \frac{2M}{r} \right) ^{-2}\left( \frac{M_{r}}{r} - \frac{M}{r^{2}} \right) \left( \frac{M}{r^{2}} + 4\pi r P \right) \nonumber \\&\quad + \left( 1 - \frac{2M}{r} \right) ^{-1}\left( \frac{M_{r}}{r^{2}} - \frac{2M}{r^{3}} + 4\pi P + 4\pi r P_{r} \right) \nonumber \\&= \left( 2\frac{V_{r}}{r} + \frac{1}{r^{2}} \right) \left( 1 - \frac{2M}{r} \right) ^{-1}\left( M_{r} - \frac{M}{r} \right) - \frac{V_{r}}{r} + 4\pi \left( 1 - \frac{2M}{r} \right) ^{-1}(2P + rP_{r}). \end{aligned}$$
(117)
Making these substitutions into (103) yields
$$\begin{aligned} 8\pi Q&= 4\pi (2P + rP_{r}) + V_{r}^{2}\left( 1 - \frac{2M}{r} \right) + \frac{V_{r}}{r}\left( M_{r} - \frac{M}{r} \right) \nonumber \\&\quad -\frac{4\pi r}{\mathrm {e}^{V}}\left[ \rho _{t}\left( 1 - \frac{2M}{r} \right) ^{-1/2} - \frac{\rho M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-3/2} \right] \nonumber \\&\quad - \frac{3M_{t}^{2}}{r^{2} \mathrm {e}^{2V}}\left( 1-\frac{2M}{r} \right) ^{-2}\nonumber \\&= 4\pi (2P + rP_{r}) + V_{r}\left( \frac{M}{r^{2}} + 4\pi r P \right) + \frac{V_{r}}{r}\left( M_{r} - \frac{M}{r} \right) \nonumber \\&\quad -\frac{4\pi r}{\mathrm {e}^{V}}\left[ \rho _{t}\left( 1 - \frac{2M}{r} \right) ^{-1/2} - \frac{\rho M_{t}}{r}\left( 1 - \frac{2M}{r} \right) ^{-3/2} \right] \nonumber \\&\quad - \frac{12\pi \rho M_{t}}{\mathrm {e}^{V}}\left( 1-\frac{2M}{r} \right) ^{-3/2} \nonumber \\&= 4\pi (2P + rP_{r}) + 4\pi r V_{r}(P + \mu ) - \frac{8\pi \rho M_{t}}{\mathrm {e}^{V}}\left( 1-\frac{2M}{r} \right) ^{-3/2} \nonumber \\&\quad - 4\pi r \left( V_{r}(\mu + P) + P_{r} + \frac{2(P - Q)}{r} \right) + \frac{8\pi \rho M_{t}}{\mathrm {e}^{V}}\left( 1 - \frac{2M}{r} \right) ^{-3/2} \nonumber \\&= 8\pi Q \end{aligned}$$
(118)
where the last two lines follow from using (111b). Thus so long as Eq. (111) holds, Eqs. (100), (101), and (102) imply Eq. (103). Since Eqs. (111), (100), and (101) imply Eqs. (102) and (111) follows from \(\nabla _{g} \cdot T = 0\), we have that solving \(\nabla _{g} \cdot T = 0\) and solving Eqs. (100) and (101) at each time \(t\) also solves the entire Einstein equation, which was the desired result. \(\square \)
The proof of Lemma 8 follows.
Proof
(Proof of Lemma 8) Recall Eq. (54). Then to compute the components of \(T\), we first have that
$$\begin{aligned} df&= f_{t}\, dt + f_{r}\, dr = p\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}}\, dt + f_{r}\, dr \end{aligned}$$
(119)
$$\begin{aligned} d\bar{f}&= \bar{f}_{t}\, dt + \bar{f}_{r}\, dr = \bar{p}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}}\, dt + \bar{f}_{r}\, dr. \end{aligned}$$
(120)
Then, since \(g\) is diagonal, this implies that
$$\begin{aligned} \left| df \right| ^{2}&= g(df,d\bar{f}) = g\left( p\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}}\, dt + f_{r}\, dr,\, \bar{p}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}}\, dt + \bar{f}_{r}\, dr \right) \nonumber \\&= \mathrm {e}^{2V}\left( 1 - \frac{2M}{r} \right) p\bar{p}\, g(dt,dt) + f_{r} \bar{f}_{r}\, g(dr,dr) \nonumber \\&= \left( 1 - \frac{2M}{r} \right) \left( \left| f_{r} \right| ^{2} - \left| p \right| ^{2} \right) . \end{aligned}$$
(121)
With these facts, we now compute the quantities in question and do so in the same order as they were presented. Thus we have the following
$$\begin{aligned} T(\nu _{t},\nu _{t})&= \mu _{0}\left( \frac{2}{\Upsilon ^{2}} df(\nu _{t})d\bar{f}(\nu _{t}) - \left[ \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} - \left| p \right| ^{2}}{\Upsilon ^2} + \left| f \right| ^{2} \right] g(\nu _{t},\nu _{t}) \right) \nonumber \\&= \mu _{0}\left( \left| f \right| ^{2} + \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} + \left| p \right| ^{2}}{\Upsilon ^{2}} \right) . \end{aligned}$$
(122)
Next, because \(\nu _{\eta }\) is an everywhere orthonormal basis,
$$\begin{aligned} T(\nu _{t},\nu _{r})&= \mu _{0}\Bigg (\frac{1}{\Upsilon ^{2}} \left[ df(\nu _{t})d\bar{f}(\nu _{r}) + d\bar{f}(\nu _{t})df(\nu _{r}) \right] \nonumber \\&\qquad - \left[ \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} - \left| p \right| ^{2}}{\Upsilon ^2} + \left| f \right| ^{2} \right] g(\nu _{t},\nu _{r})\Bigg ) \nonumber \\&= \frac{2\mu _{0}}{\Upsilon ^{2}}\left( 1 - \frac{2M}{r} \right) \mathfrak {R}(f_{r}\bar{p}). \end{aligned}$$
(123)
$$\begin{aligned} T(\nu _{r},\nu _{r})&= \mu _{0}\left( \frac{2}{\Upsilon ^{2}} df(\nu _{r})d\bar{f}(\nu _{r}) - \left[ \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} - \left| p \right| ^{2}}{\Upsilon ^2} + \left| f \right| ^{2} \right] g(\nu _{r},\nu _{r}) \right) \nonumber \\&= \mu _{0}\left( -\left| f \right| ^{2} + \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} + \left| p \right| ^{2}}{\Upsilon ^{2}} \right) . \end{aligned}$$
(124)
$$\begin{aligned} T(\nu _{\theta },\nu _{\theta })&= \mu _{0}\left( \frac{2}{\Upsilon ^{2}} df(\nu _{\theta })d\bar{f}(\nu _{\theta }) - \left[ \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} - \left| p \right| ^{2}}{\Upsilon ^2} + \left| f \right| ^{2} \right] g(\nu _{\theta },\nu _{\theta }) \right) \nonumber \\&= -\mu _{0}\left( \left| f \right| ^{2} + \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} - \left| p \right| ^{2}}{\Upsilon ^{2}} \right) . \end{aligned}$$
(125)
Finally,
$$\begin{aligned} T(\nu _{\varphi },\nu _{\varphi })&= \mu _{0}\left( \frac{2}{\Upsilon ^{2}} df(\nu _{\varphi })d\bar{f}(\nu _{\varphi }) - \left[ \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} - \left| p \right| ^{2}}{\Upsilon ^2} + \left| f \right| ^{2} \right] g(\nu _{\varphi },\nu _{\varphi }) \right) \nonumber \\&= -\mu _{0}\left( \left| f \right| ^{2} + \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} - \left| p \right| ^{2}}{\Upsilon ^{2}} \right) \end{aligned}$$
(126)
so that \(T(\nu _{\varphi },\nu _{\varphi }) = T(\nu _{\theta },\nu _{\theta })\). \(\square \)
Finally, we prove Lemma 11.
Proof
(Proof of Lemma 11) By Eq. (48) and Lemma 8, we have that
$$\begin{aligned} \mu&= \mu _{0}\left( \left| f \right| ^{2} + \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} + \left| p \right| ^{2}}{\Upsilon ^{2}} \right) \end{aligned}$$
(127a)
$$\begin{aligned} \rho&= \frac{2\mu _{0}}{\Upsilon ^2}\left( 1 - \frac{2M}{r} \right) \mathfrak {R}(f_{r}\bar{p})\end{aligned}$$
(127b)
$$\begin{aligned} P&= \mu _{0}\left( -\left| f \right| ^{2} + \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} + \left| p \right| ^{2}}{\Upsilon ^{2}} \right) \end{aligned}$$
(127c)
$$\begin{aligned} Q&= -\mu _{0}\left( \left| f \right| ^{2} + \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} - \left| p \right| ^{2}}{\Upsilon ^{2}} \right) . \end{aligned}$$
(127d)
We know from a previous proof that \(\nabla _{g} \cdot T=0\) is equivalent to Eq. (111). Moreover, by (55) and Lemma 10, we have that (53) is equivalent to (61) and (65). Thus it suffices to show that given (127), Eqs. (61) and (65) imply Eq. (111).
We will start with (111a). On one hand, the left hand side is equal to differentiating (127a) with respect to \(t\).
$$\begin{aligned} \mu _{t}&= \partial _{t}\left[ \mu _{0}\left( \left| f \right| ^{2} + \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} + \left| p \right| ^{2}}{\Upsilon ^{2}} \right) \right] \nonumber \\&= \mu _{0}\left[ f_{t}\bar{f}+ f \bar{f}_{t} - \frac{2M_{t}}{r}\left( \frac{\left| f_{r} \right| ^{2} + \left| p \right| ^{2}}{\Upsilon ^{2}} \right) \right. \nonumber \\&\quad \left. + \left( 1 - \frac{2M}{r} \right) \frac{f_{rt}\bar{f}_{r} + f_{r}\bar{f}_{rt} + p_{t}\bar{p}+ p\bar{p}_{t}}{\Upsilon ^{2}} \right] \nonumber \\ \mu _{t}&= \mu _{0}\Bigg [\bar{p}\left( f \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} + \frac{p_{t}}{\Upsilon ^{2}}\left( 1 - \frac{2M}{r} \right) \right) \nonumber \\&\quad + p \left( \bar{f}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} + \frac{\bar{p}_{t}}{\Upsilon ^{2}}\left( 1 - \frac{2M}{r} \right) \right) - \frac{2M_{t}}{r}\left( \frac{\left| f_{r} \right| ^{2} + \left| p \right| ^{2}}{\Upsilon ^{2}} \right) \nonumber \\&\quad + \left( 1 - \frac{2M}{r} \right) \frac{\bar{f}_{r}\partial _{r}\left( p \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) + f_{r}\partial _{r}\left( \bar{p}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) }{\Upsilon ^{2}} \Bigg ]. \end{aligned}$$
(128a)
On the other hand, we substitute (127) into (111a).
$$\begin{aligned} \mu _{t}&=2\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}}\left( \frac{2\mu _{0}}{\Upsilon ^2}\left( 1 - \frac{2M}{r} \right) \mathfrak {R}(f_{r}\bar{p}) \right) \left( V_{r} + \frac{1}{r} \right) \nonumber \\&\quad + \partial _{r}\left[ \frac{2\mu _{0}}{\Upsilon ^2}\left( 1 - \frac{2M}{r} \right) \mathfrak {R}(f_{r}\bar{p}) \right] \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} - \frac{2M_{t}\mu _{0}}{r}\left( \frac{\left| f_{r} \right| ^{2} + \left| p \right| ^{2}}{\Upsilon ^{2}} \right) \nonumber \\&= \bar{p}\frac{\mu _{0}}{\Upsilon ^{2}}\left( 1 - \frac{2M}{r} \right) \left[ f_{r}V_{r}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} + \frac{2f_{r}\mathrm {e}^{V}}{r}\sqrt{1 - \frac{2M}{r}} \right] \nonumber \\&\quad + p \frac{\mu _{0}}{\Upsilon ^{2}}\left( 1 - \frac{2M}{r} \right) \left[ \bar{f}_{r}V_{r}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} + \frac{2\bar{f}_{r}\mathrm {e}^{V}}{r}\sqrt{1 - \frac{2M}{r}} \right] \nonumber \\&\quad + \frac{\mu _{0}V_{r}\mathrm {e}^{V}}{\Upsilon ^{2}}\left( 1 - \frac{2M}{r} \right) ^{3/2}\left( f_{r}\bar{p}+ p \bar{f}_{r} \right) - \frac{2M_{t}\mu _{0}}{r}\left( \frac{\left| f_{r} \right| ^{2} + \left| p \right| ^{2}}{\Upsilon ^{2}} \right) \nonumber \\&\quad + \frac{\mu _{0}\mathrm {e}^{V}}{\Upsilon ^2}\sqrt{1 - \frac{2M}{r}}\Bigg [(f_{r}\bar{p}+ p \bar{f}_{r})\partial _{r}\left( 1 - \frac{2M}{r} \right) \nonumber \\&\quad + \left( 1 - \frac{2M}{r} \right) (f_{rr}\bar{p}+ p \bar{f}_{rr} + f_{r}\bar{p}_{r} + p_{r} \bar{f}_{r})\Bigg ] \nonumber \\ \mu _{t}&= \bar{p}\frac{\mu _{0}}{\Upsilon ^{2}}\left( 1 - \frac{2M}{r} \right) \left[ \partial _{r}\left( f_{r}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) + \frac{2f_{r}\mathrm {e}^{V}}{r}\sqrt{1 - \frac{2M}{r}} \right] \nonumber \\&\quad + p \frac{\mu _{0}}{\Upsilon ^{2}}\left( 1 - \frac{2M}{r} \right) \left[ \partial _{r}\left( \bar{f}_{r}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) + \frac{2\bar{f}_{r}\mathrm {e}^{V}}{r}\sqrt{1 - \frac{2M}{r}} \right] \nonumber \\&\quad - \frac{2M_{t}\mu _{0}}{r}\left( \frac{\left| f_{r} \right| ^{2} + \left| p \right| ^{2}}{\Upsilon ^{2}} \right) \nonumber \\&\quad + \frac{\mu _{0}}{\Upsilon ^2}\left( 1 - \frac{2M}{r} \right) \left[ \bar{f}_{r}\partial _{r}\left( p\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) + f_{r}\partial _{r}\left( \bar{p}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) \right] . \end{aligned}$$
(128b)
Setting (128a) equal to (128b), we obtain
$$\begin{aligned} \bar{p}&\left( \Upsilon ^{2} f \mathrm {e}^{V}\left( 1 - \frac{2M}{r} \right) ^{-1/2} + p_{t} \right) + p \left( \Upsilon ^{2}\bar{f}\mathrm {e}^{V}\left( 1 - \frac{2M}{r} \right) ^{-1/2} + \bar{p}_{t} \right) \nonumber \\&\quad = \bar{p}\left[ \partial _{r}\left( f_{r}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) + \frac{2f_{r}\mathrm {e}^{V}}{r}\sqrt{1 - \frac{2M}{r}} \right] \nonumber \\&\quad \quad + p \left[ \partial _{r}\left( \bar{f}_{r}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) + \frac{2\bar{f}_{r}\mathrm {e}^{V}}{r}\sqrt{1 - \frac{2M}{r}} \right] \nonumber \\ 0&= \bar{p}\left[ - p_{t} + \mathrm {e}^{V}\left( -\Upsilon ^{2} f \left( 1 - \frac{2M}{r} \right) ^{-1/2} + \frac{2f_{r}}{r}\sqrt{1 - \frac{2M}{r}} \right) + \partial _{r}\left( f_{r}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) \right] \nonumber \\&\quad + p\left[ \!- \bar{p}_{t} + \mathrm {e}^{V}\left( \!-\Upsilon ^{2} \bar{f}\left( 1 - \frac{2M}{r} \right) ^{-1/2} + \frac{2\bar{f}_{r}}{r}\sqrt{1 - \frac{2M}{r}} \right) + \partial _{r}\left( \bar{f}_{r}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) \right] . \end{aligned}$$
(128c)
Note that, since \(V\) and \(M\) are real valued, the second term above is the complex conjugate of the first term. Then by (61) Eq. (128c) holds.
Next, we consider (111b). We first differentiate (127b) with respect to \(t\). Using Eq. (65), this yields
$$\begin{aligned} \rho _{t}&= \partial _{t}\left[ \frac{2\mu _{0}}{\Upsilon ^2}\left( 1 - \frac{2M}{r} \right) \mathfrak {R}(f_{r}\bar{p}) \right] \nonumber \\&= \frac{\mu _{0}}{\Upsilon ^{2}}\left[ -\frac{4M_{t}}{r}\mathfrak {R}(f_{r}\bar{p}) + \left( 1 - \frac{2M}{r} \right) \left( f_{rt}\bar{p}+ \bar{f}_{rt}p + f_{r}\bar{p}_{t} + \bar{f}_{r}p_{t} \right) \right] \nonumber \\&= \frac{\mu _{0}}{\Upsilon ^{2}}\Bigg \{-\frac{4M_{t}}{r}\mathfrak {R}(f_{r}\bar{p}) + \left( 1 - \frac{2M}{r} \right) \Bigg [\partial _{r}\left( \left| p \right| ^{2}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) \nonumber \\&\quad + \left| p \right| ^{2}\partial _{r}\left( \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) + f_{r}\bar{p}_{t} + \bar{f}_{r}p_{t}\Bigg ]\Bigg \}. \end{aligned}$$
(129a)
On the other hand, we substitute (127) into (111b), to obtain
$$\begin{aligned} \rho _{t}&=\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}}\Bigg [\frac{2\mu _{0}V_{r}}{\Upsilon ^{2}}\left( 1 - \frac{2M}{r} \right) (\left| f_{r} \right| ^{2} + \left| p \right| ^{2}) \nonumber \\&\quad + \mu _{0}\partial _{r}\left( -\left| f \right| ^{2} + \left( 1 - \frac{2M}{r} \right) \frac{\left| f_{r} \right| ^{2} + \left| p \right| ^{2}}{\Upsilon ^{2}} \right) + \frac{4\mu _{0}\left| f_{r} \right| ^{2}}{r\Upsilon ^{2}}\left( 1 - \frac{2M}{r} \right) \Bigg ] - \frac{4\mu _{0}M_{t}}{r\Upsilon ^{2}}\mathfrak {R}(f_{r}\bar{p})\nonumber \\&= \frac{\mu _{0}}{\Upsilon ^{2}}\left( 1 - \frac{2M}{r} \right) \Bigg \{\bar{f}_{r}\Bigg [\partial _{r}\left( f_{r}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) \nonumber \\&\quad +\mathrm {e}^{V} \left( - \Upsilon ^{2}f\left( 1 - \frac{2M}{r} \right) ^{-1/2} + \frac{2 f_{r}}{r}\sqrt{1 - \frac{2M}{r}} \right) \Bigg ]\nonumber \\&\quad + f_{r}\left[ \partial _{r}\left( \bar{f}_{r}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) +\mathrm {e}^{V}\left( -\Upsilon ^{2}\bar{f}\left( 1 - \frac{2M}{r} \right) ^{-1/2} + \frac{2 \bar{f}_{r}}{r}\sqrt{1 - \frac{2M}{r}} \right) \right] \nonumber \\&\quad + \left| p \right| ^{2}\partial _{r}\left( \mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) + \partial _{r}\left( \left| p \right| ^{2}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) \Bigg \} - \frac{4\mu _{0}M_{t}}{r\Upsilon ^{2}}\mathfrak {R}(f_{r}\bar{p}). \end{aligned}$$
(129b)
Equating (129a) and (129b) yields
$$\begin{aligned} 0&= \bar{f}_{r}\Bigg [- p_{t} + \mathrm {e}^{V}\left( -\Upsilon ^{2} f \left( 1 - \frac{2M}{r} \right) ^{-1/2} + \frac{2f_{r}}{r}\sqrt{1 - \frac{2M}{r}} \right) \nonumber \\&\quad + \partial _{r}\left( f_{r}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) \Bigg ]\nonumber \\&\quad + f_{r}\Bigg [- \bar{p}_{t} + \mathrm {e}^{V}\left( -\Upsilon ^{2} \bar{f}\left( 1 - \frac{2M}{r} \right) ^{-1/2} + \frac{2\bar{f}_{r}}{r}\sqrt{1 - \frac{2M}{r}} \right) \nonumber \\&\quad + \partial _{r}\left( \bar{f}_{r}\mathrm {e}^{V}\sqrt{1 - \frac{2M}{r}} \right) \Bigg ] \end{aligned}$$
(129c)
As before, since \(M\) and \(V\) are real valued, the second term above is the complex conjugate of the first term. Then by Eqs. (61), (129c) holds. Then Eqs. (61) and (65) imply Eq. (111) holds, which completes the proof. \(\square \)
