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Assuming stress-energy tensor ${\rm{T}}_{\mu}^{\nu}\equiv {\rm{diag}}~\{\varepsilon,-p,-p,-p\}$ and taking trace from both sides of Einstein field equations (EFE) one obtains the relation \begin{equation} -R=\kappa~(\varepsilon-3 p). \tag{1} \end{equation} If you know your metric component $g_{rr}$, and $g_{00}$ is constant, you can easily calculate the energy density and pressure from equations \begin{equation}\label{pressure} \kappa~ p=-\frac{1-{A}^{-1}}{r^2}~,\tag{2} \end{equation} \begin{equation} \label{density} \kappa~\varepsilon=\frac{1-{A}^{-1}}{r^2}-\frac{1}{r}~\frac{{\rm d}{A}^{-1}}{{\rm d}r}~,\tag{3} \end{equation} and insert them into equation (1). For more detailed answer, see https://physics.stackexchange.com/a/679431/281096 (be aware of other notation there).

However, the assumption that $g_{00}$ is constant implies (Einstein field equations) that \begin{equation} \label{em2l} A(r)=\frac{1}{1-b r^{2}},\tag{4} \end{equation} where $b$ is some constant of dimension $L^{-2}$. As the result scalar Ricci curvature is constant, too ($R=-6~b$). Your metric describes homogeneous, static and isotropic solution, the so-called Einstein's universe, see https://mpra.ub.uni-muenchen.de/83001/ .

Assuming stress-energy tensor ${\rm{T}}_{\mu}^{\nu}\equiv {\rm{diag}}~\{\varepsilon,-p,-p,-p\}$ and taking trace from both sides of Einstein field equations (EFE) one obtains the relation \begin{equation} -R=\kappa~(\varepsilon-3 p). \tag{1} \end{equation} If you know your metric component $g_{rr}$, and $g_{00}$ is constant, you can easily calculate the energy density and pressure from equations \begin{equation}\label{pressure} \kappa~ p=-\frac{1-{A}^{-1}}{r^2}~,\tag{2} \end{equation} \begin{equation} \label{density} \kappa~\varepsilon=\frac{1-{A}^{-1}}{r^2}-\frac{1}{r}~\frac{{\rm d}{A}^{-1}}{{\rm d}r}~,\tag{3} \end{equation} and insert them into equation (1). For more detailed answer, see https://physics.stackexchange.com/a/679431/281096 (be aware of other notation there).

However, the assumption that $g_{00}$ is constant implies (Einstein field equations) that \begin{equation} \label{em2l} A(r)=\frac{1}{1-b r^{2}},\tag{4} \end{equation} where $b$ is some constant of dimension $L^{-2}$. As result of the equations (1),(2) and (3), the scalar Ricci curvature, pressure and energy density are constant too: \begin{equation} \varepsilon=3~b,~~~p=-b,~~~R=-6~b.\tag{5} \end{equation} Your metric describes homogeneous, static and isotropic solution derived by Einstein in 1917 called later Einstein's universe, see https://mpra.ub.uni-muenchen.de/83001/ .

Assuming stress-energy tensor ${\rm{T}}_{\mu}^{\nu}\equiv {\rm{diag}}~\{\varepsilon,-p,-p,-p\}$ and taking trace from both sides of Einstein field equations (EFE) one obtains the relation \begin{equation} -R=\kappa~(\varepsilon-3 p). \tag{1} \end{equation} If you know your metric component $g_{rr}$, and $g_{00}$ is constant, you can easily calculate the energy density and pressure from equations \begin{equation}\label{pressure} \kappa~ p=-\frac{1-{A}^{-1}}{r^2}~,\tag{2} \end{equation} \begin{equation} \label{density} \kappa~\varepsilon=\frac{1-{A}^{-1}}{r^2}-\frac{1}{r}~\frac{{\rm d}{A}^{-1}}{{\rm d}r}~,\tag{3} \end{equation} and insert them into equation (1). For more detailed answer, see https://physics.stackexchange.com/a/679431/281096 (be aware of other notation there).

However, the assumption that $g_{00}$ is constant implies that \begin{equation} \label{em2l} A=\frac{1}{1-b r^{2}},\tag{4} \end{equation} where $b$ is some constant of dimension $L^{-2}$. As result scalar Ricci curvature is constant, too ($R=-6~b$). Your metric describes homogeneous, static and isotropic solution, the so-called Einstein's universe, see https://mpra.ub.uni-muenchen.de/83001/ .

Assuming stress-energy tensor ${\rm{T}}_{\mu}^{\nu}\equiv {\rm{diag}}~\{\varepsilon,-p,-p,-p\}$ and taking trace from both sides of Einstein field equations (EFE) one obtains the relation \begin{equation} -R=\kappa~(\varepsilon-3 p). \tag{1} \end{equation} If you know your metric component $g_{rr}$, and $g_{00}$ is constant, you can easily calculate the energy density and pressure from equations \begin{equation}\label{pressure} \kappa~ p=-\frac{1-{A}^{-1}}{r^2}~,\tag{2} \end{equation} \begin{equation} \label{density} \kappa~\varepsilon=\frac{1-{A}^{-1}}{r^2}-\frac{1}{r}~\frac{{\rm d}{A}^{-1}}{{\rm d}r}~,\tag{3} \end{equation} and insert them into equation (1). For more detailed answer, see https://physics.stackexchange.com/a/679431/281096 (be aware of other notation there).

However, the assumption that $g_{00}$ is constant implies (Einstein field equations) that \begin{equation} \label{em2l} A(r)=\frac{1}{1-b r^{2}},\tag{4} \end{equation} where $b$ is some constant of dimension $L^{-2}$. As the result scalar Ricci curvature is constant, too ($R=-6~b$). Your metric describes homogeneous, static and isotropic solution, the so-called Einstein's universe, see https://mpra.ub.uni-muenchen.de/83001/ .

Assuming stress-energy tensor ${\rm{T}}_{\mu}^{\nu}\equiv {\rm{diag}}~\{\varepsilon,-p,-p,-p\}$ and taking trace from both sides of Einstein field equations (EFE) one obtains the relation \begin{equation} -R=\kappa~(\varepsilon-3 p). \tag{1} \end{equation} If you know your metric component $g_{rr}$, and $g_{00}$ is constant, you can easily calculate the energy density and pressure from equations \begin{equation}\label{pressure} \kappa~ p=-\frac{1-{A}^{-1}}{r^2}~,\tag{2} \end{equation} \begin{equation} \label{density} \kappa~\varepsilon=\frac{1-{A}^{-1}}{r^2}-\frac{1}{r}~\frac{{\rm d}{A}^{-1}}{{\rm d}r}~,\tag{3} \end{equation} and insert them into equation (1). For more detailed answer, see https://physics.stackexchange.com/a/679431/281096 (be aware of other notation there).

However, the assumption that $g_{00}$ is constant implies that \begin{equation} \label{em2l} A=\frac{1}{1-b r^{2}},\tag{4} \end{equation} with $b$ some constant of dimension $L^{-2}$. Es result the Ricci scalar $R$ is constant, too.

Assuming stress-energy tensor ${\rm{T}}_{\mu}^{\nu}\equiv {\rm{diag}}~\{\varepsilon,-p,-p,-p\}$ and taking trace from both sides of Einstein field equations (EFE) one obtains the relation \begin{equation} -R=\kappa~(\varepsilon-3 p). \tag{1} \end{equation} If you know your metric component $g_{rr}$, and $g_{00}$ is constant, you can easily calculate the energy density and pressure from equations \begin{equation}\label{pressure} \kappa~ p=-\frac{1-{A}^{-1}}{r^2}~,\tag{2} \end{equation} \begin{equation} \label{density} \kappa~\varepsilon=\frac{1-{A}^{-1}}{r^2}-\frac{1}{r}~\frac{{\rm d}{A}^{-1}}{{\rm d}r}~,\tag{3} \end{equation} and insert them into equation (1). For more detailed answer, see https://physics.stackexchange.com/a/679431/281096 (be aware of other notation there).

However, the assumption that $g_{00}$ is constant implies that \begin{equation} \label{em2l} A=\frac{1}{1-b r^{2}},\tag{4} \end{equation} where $b$ is some constant of dimension $L^{-2}$. As result scalar Ricci curvature is constant, too ($R=-6~b$). Your metric describes homogeneous, static and isotropic solution, the so-called Einstein's universe, see https://mpra.ub.uni-muenchen.de/83001/ .