I am studying the Schwarzschild solution of the Einstein field equation in vacuum and I have encountered a problem in obtaining the expression for Einstein tensor for the following metric $g$ considered, i.e the spherically symmetric metric in Schwarzschild coordinates: $$g=-e^{2\nu}dt^2+e^{2\lambda}dr^2+r^2d\Omega^2$$ with $d\Omega^2=d\theta^2+\sin{\theta}^2d\phi^2$.
So this is associated to the metric $g_{\alpha \beta}=\begin{pmatrix} -e^{2\nu} & 0 & 0 & 0\\ 0 & e^{2\lambda} & 0 & 0\\ 0 & 0 & r^2& 0\\ 0 & 0 & 0 & r^2\sin{\theta}^2\\ \end{pmatrix}$
Now in order to write down the Einstein tensor, to solve the Einstein field equation, I have to consider that $G_{\mu\nu}=0\iff G_{\beta}^{\alpha}=g^{\alpha\mu}G_{\mu\beta}=0$ since $g$ is no degenerate.
But now my problem is to determine the components $G_{\beta}^{\alpha}$!
I have read I should obtain: $$G_0^0=\frac{-1}{r^2}+e^{-2\lambda}(\frac{-1}{r^2}-\frac{2 \lambda'}{r})$$ $$G_0^1=\frac{2\dot \lambda}{r}+e^{-\lambda-\nu}$$ $$G_1^1=\frac{-1}{r^2}+e^{-2\lambda}(\frac{1}{r^2}+\frac{2 \nu'}{r})$$ $$G_2^2=G_3^3=e^{-2\lambda}(\nu'^2-\nu'\lambda'+\nu''+\frac{\nu'-\lambda'}{r})+e^{-2\nu}(-\dot \lambda^2+\dot\lambda\dot\nu-\ddot \lambda)$$ $$G_{\nu}^{\mu}=0 \text{ elsewhere}$$ with $'=\frac{\partial}{\partial r}$ and $\dot{}=\frac{\partial}{\partial t}$.
Sorry I know maybe it is a trivial question for this iste, but I don't know how starting in order to find out the expression above...can you help me please? Thanks in advance
