${g_a}_b=\stackrel{a\downarrow b\to }{\left[\begin{array}{cccc}B& 0& 0& 0\\ 0& A& 0& 0\\ 0& 0& r^2& 0\\ 0& 0& 0& {\left(r\sin \left(\theta \right)\right)}^2\end{array}\right]}$ ${g^a}^b=\stackrel{a\downarrow b\to }{\left[\begin{array}{cccc}\frac{1}{B}& 0& 0& 0\\ 0& \frac{1}{A}& 0& 0\\ 0& 0& r^{-2}& 0\\ 0& 0& 0& {\left(r\sin \left(\theta \right)\right)}^{-2}\end{array}\right]}$ ${G_a}_b=\stackrel{a\downarrow b\to }{\left[\begin{array}{cccc}\frac{B\left(A-A^2-r\frac{\partial A}{\partial r}\right)}{A^2{r}^2}& 0& 0& 0\\ 0& \frac{B-AB+r\frac{\partial B}{\partial r}}{B{r}^2}& 0& 0\\ 0& 0& \frac{-A{r}^2{\frac{\partial B}{\partial r}}^2-2r{B}^2\frac{\partial A}{\partial r}-B{r}^2\frac{\partial B}{\partial r}\frac{\partial A}{\partial r}+2ABr\frac{\partial B}{\partial r}+2AB{r}^2\frac{{\partial }^{2}B}{\partial r^2}}{4A^2{B}^2}& 0\\ 0& 0& 0& \frac{-A{r}^2{\frac{\partial B}{\partial r}}^2+A{r}^2{\frac{\partial B}{\partial r}}^2{\cos \left(\theta \right)}^2-2r{B}^2\frac{\partial A}{\partial r}+2r{B}^2{\cos \left(\theta \right)}^2\frac{\partial A}{\partial r}-B{r}^2\frac{\partial B}{\partial r}\frac{\partial A}{\partial r}+B{r}^2{\cos \left(\theta \right)}^2\frac{\partial A}{\partial r}\frac{\partial B}{\partial r}+2ABr\frac{\partial B}{\partial r}+2AB{r}^2\frac{{\partial }^{2}B}{\partial r^2}-2AB{r}^2{\cos \left(\theta \right)}^2\frac{{\partial }^{2}B}{\partial r^2}-2ABr{\cos \left(\theta \right)}^2\frac{\partial B}{\partial r}}{4A^2{B}^2}\end{array}\right]}$
${T_a}_b=\stackrel{a\downarrow b\to }{\left[\begin{array}{cccc}P+\rho +BP& 0& 0& 0\\ 0& AP& 0& 0\\ 0& 0& P{r}^2& 0\\ 0& 0& 0& P\left(r^2-r^2{\cos \left(\theta \right)}^2\right)\end{array}\right]}$
$\frac{B\left(A-A^2-r\frac{\partial A}{\partial r}\right)}{A^2{r}^2}=8\pi \cdot \left(P+\rho +BP\right)$
$\frac{B-AB+r\frac{\partial B}{\partial r}}{B{r}^2}=8\pi \cdot AP$
$\frac{-A{r}^2{\frac{\partial B}{\partial r}}^2-2r{B}^2\frac{\partial A}{\partial r}-B{r}^2\frac{\partial B}{\partial r}\frac{\partial A}{\partial r}+2ABr\frac{\partial B}{\partial r}+2AB{r}^2\frac{{\partial }^{2}B}{\partial r^2}}{4A^2{B}^2}=8\pi \cdot P{r}^2$
$\frac{-A{r}^2{\frac{\partial B}{\partial r}}^2+A{r}^2{\frac{\partial B}{\partial r}}^2{\cos \left(\theta \right)}^2-2r{B}^2\frac{\partial A}{\partial r}+2r{B}^2{\cos \left(\theta \right)}^2\frac{\partial A}{\partial r}-B{r}^2\frac{\partial B}{\partial r}\frac{\partial A}{\partial r}+B{r}^2{\cos \left(\theta \right)}^2\frac{\partial A}{\partial r}\frac{\partial B}{\partial r}+2ABr\frac{\partial B}{\partial r}+2AB{r}^2\frac{{\partial }^{2}B}{\partial r^2}-2AB{r}^2{\cos \left(\theta \right)}^2\frac{{\partial }^{2}B}{\partial r^2}-2ABr{\cos \left(\theta \right)}^2\frac{\partial B}{\partial r}}{4A^2{B}^2}=8\pi \cdot P\left(r^2-r^2{\cos \left(\theta \right)}^2\right)$
difference between 1st and 2nd...
$A\frac{\partial B}{\partial r}+B\frac{\partial A}{\partial r}=-8r\pi \cdot A^2\left(P+\rho \right)$
$\frac{\partial }{\partial r}\left(AB\right)=-8\pi \cdot r{A}^2\left(\rho +P\right)$
difference between 2nd and 3rd...
$A{r}^2{\frac{\partial B}{\partial r}}^2-4A^2{B}^2+4A{B}^2+2r{B}^2\frac{\partial A}{\partial r}+B{r}^2\frac{\partial B}{\partial r}\frac{\partial A}{\partial r}+2ABr\frac{\partial B}{\partial r}-2AB{r}^2\frac{{\partial }^{2}B}{\partial r^2}=0$
differences between 3rd and 4th...
$0=0$