we got,
\(\gamma_{23}^2=\frac{1}{y}\), \(\gamma_{32}^2=-\frac{1}{y}\), \(\gamma_{22}^3=-\frac{1}{y}\), \(\gamma_{33}^3=-\frac{1}{y}\) Rest of all components in christoffels symbols are automatically zero.
Now we have to find Ricci tensor's components by tedious formulea but i am skipping this and writing ricci components,
R11=R12=R13=R21=R23=R31=R32=0, But we have some non-zer components too which is
R22=\(-\frac{1}{y^2}\), R33=\(-\frac{1}{y^2}\)
Now i am finding Reimann tensor component after huge calculation i finally reached there is only one non zero Reimann component which is
R2323=\(-\frac{1}{y^4}\)
Curvature scalar can be found by following relations :
R= g11R11+g22R22+g33R33
=\(-1\left(0\right)\)+\(y^2\left(-\frac{1}{y^2}\right)\)+\(y^2\left(-\frac{1}{y^2}\right)\) [ as \(g^{\alpha\beta}=\frac{1}{g_{\alpha\beta}}\)]
= -2
Finally we have reached last step by calculating Einstein's Tensor which is equal to
\(G_{\alpha\beta}=R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R\) [ where \(\alpha,\beta=1,2,3\)]
\(G_{11}=R_{11}-\frac{1}{2}g_{11}R\)
= \(0-\frac{1}{2}\left(-1\right)\left(-2\right)\)
=-1
\(G_{22}=R_{22}-\frac{1}{2}g_{22}R\)
=\(-\frac{1}{y^2}-\frac{1}{2}\left(\frac{1}{y^2}\right)\left(-2\right)\)
= 0
\(G_{33}=R_{33}-\frac{1}{2}g_{33}R\)
= \(-\frac{1}{y^2}-\frac{1}{2}\left(\frac{1}{y^2}\right)\left(-2\right)\)
= 0
rest of all components in Einstein's tensor is zero in matrix form we can wrirte this in
\(\left[-1\ \ \ \ 0\ \ \ \ 0\right]\)
I showed that hyperbolic geometry has constant negtaive curvature Thus, this metric is a solution for the vacuum Einstein equations. This seems strange because this is a space of constant negative curvature, which would seem to require some sort of matter/energy. \ref{257213}\ref{257213}\cite{Sharan_2009,Padmanabhan_2014,1997}\cite{Ahsan_2008}\cite{Lerner}