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?

  • 2
    $\begingroup$ You just have to brute force calculate $G_{\mu\nu}\equiv R_{\mu\nu}+\frac12 Rg_{\mu\nu}$ $\endgroup$ Mar 13, 2021 at 8:39
  • $\begingroup$ Yes maybe, but I am very confused..I can't understand how...can you give me even only an hint on how starting? $\endgroup$
    – Nik
    Mar 13, 2021 at 8:42
  • $\begingroup$ Above all how I can pass then form $G_{\mu\nu}$ to $G_{\mu}^{\nu}$? $\endgroup$
    – Nik
    Mar 13, 2021 at 8:42
  • 1
    $\begingroup$ Compute the unique Christoffel symbols, then the Riemann curvature tensor, then the Ricci tensor and the Ricci scalar, with which you construct $G_{\mu\nu}$ (I don't think there's a simpler way). Once you have $G_{\mu\nu}$, you can raise the index by doing standard matrix multiplication with the inverse metric $\endgroup$ Mar 13, 2021 at 8:44
  • $\begingroup$ @Nik: If you don't know how to raise and lower an index by use of the metric you should go back and revise this before going on to tackle more complex questions as these operations are used without explanation everywhere. Basically $G^{\alpha}_{\beta} = g^{\alpha\gamma}.G_{\gamma\beta}$ here $g$ is the metric. There are two forms, $g^{\alpha\gamma}$ and $g_{\alpha\gamma}$ amd they are inverses of each other. $\endgroup$ Mar 13, 2021 at 10:32

2 Answers 2


You have your metric ansatz:


where $λ=λ(t,r), ν=ν(t,r)$.

You want to compute the Einstein tensor which is defined as:

$$G_{αβ} = R_{αβ} - \cfrac{1}{2}g_{αβ}R$$

In order to compute the Einstein tensor you have to compute the Ricci tensor which is defined as:

$$R_{αβ}=Γ^{ρ}_{βα,ρ} - Γ^{ρ}_{ρα,β} +Γ^{ρ}_{ρλ}Γ^{λ}_{βα} -Γ^{ρ}_{βλ}Γ^{λ}_{ρα}$$

where the Christoffel symbols are given by:

$$Γ^{δ}_{βγ}= \frac{1}{2}g^{αδ}(\frac{\partial g_{αβ}}{\partial x^{γ}} + \frac{\partial g_{αγ}}{\partial x^{β}} - \frac{\partial g_{βγ}}{\partial x^{α}})$$

So you have at first to compute the Christoffel symbols from which you can get the expression for the Ricci tensor. Then you need to compute the Ricci scalar which is defined as:

$$R = g^{αβ}R_{αβ} = g^{tt}R_{tt} +g^{rr}R_{rr}+ g^{θθ}R_{θθ} + g^{\phi\phi}R_{\phi\phi}$$

After careful calculations you can obtain the Einstein tensor. If you want to derive the Schwarzchild solution, since the Einstein equation reads:


you can trace this equation to see that the Ricci scalar vanishes so Einstein equation becomes $R_{αβ}=0$ and you can avoid unnecessary calculations.

EDIT 1: Calculation of one of the Christoffel symbols

The equation is: $$Γ^{δ}_{βγ}= \frac{1}{2}g^{αδ}(\frac{\partial g_{αβ}}{\partial x^{γ}} + \frac{\partial g_{αγ}}{\partial x^{β}} - \frac{\partial g_{βγ}}{\partial x^{α}})$$ For $α=δ=r$ we have: $$Γ^{r}_{βγ}= \frac{1}{2}g^{rr}(\frac{\partial g_{rβ}}{\partial x^{γ}} + \frac{\partial g_{rγ}}{\partial x^{β}} - \frac{\partial g_{βγ}}{\partial r})$$

For $β=γ=t:$ $$Γ^{r}_{tt}= \frac{1}{2}g^{rr}(\frac{\partial g_{rt}}{\partial t} + \frac{\partial g_{rt}}{\partial t} - \frac{\partial g_{tt}}{\partial r})$$

Since metric is diagonal this becomes:

$$Γ^{r}_{tt}= \frac{1}{2}g^{rr}(- \frac{\partial g_{tt}}{\partial r}) = \cfrac{1}{2}e^{-2λ(t,r)}2ν'(t,r)e^{-2ν} =ν'(t,r)e^{-2(λ+ν)}$$

You have to do the same for all possible combinations of $α,β,γ,δ$. Remember that the factor in front of the Christoffels is $g^{αδ}$ which is zero for any $α \neq δ$ so you need to check only the cases where $α=δ$. Hope this hepls!

  • $\begingroup$ Thanks now I will prove with your indications! another thing if I can: I have not a lot of knowledge in differential geometry, I am learning by myselpf, and surely it is the reason why I have difficulties in my question. Can give me how compute the expression even of one cristofell symbols, e.g. $\Gamma_{\beta\alpha, \rho}^{\rho}$? $\endgroup$
    – Nik
    Mar 13, 2021 at 9:32
  • $\begingroup$ I added the calculation of $Γ^{r}_{tt}$. $\endgroup$
    – Noone
    Mar 13, 2021 at 9:54

From the metric coefficients, you can obtain the Christoffel symbols \begin{align} \Gamma^{\alpha}_{\;\;\mu \nu}=\frac{1}{2} g^{\alpha \lambda} \left(g_{\mu \lambda,\nu}+ g_{\lambda \nu,\mu}-g_{\mu \nu, \lambda}\right) \;, \end{align} which you will use to determine the components of the Ricci tensor (it's a bit tedious, but straightforward) \begin{align} R_{\mu \nu} &= \Gamma^{\rho}_{\;\;\mu\nu,\rho}-\Gamma^{\rho}_{\;\;\mu\rho,\nu} + \Gamma^{\rho}_{\;\;\lambda\rho}\Gamma^{\lambda}_{\;\;\mu\nu} - \Gamma^{\rho}_{\;\;\lambda\mu}\Gamma^{\lambda}_{\;\;\rho\nu} \;, \end{align} then contract the Ricci tensor with the metric to obtain the Ricci scalar \begin{align} R = g^{\mu \nu} R_{\mu \nu} \end{align} and finally plug in everything in \begin{align} G_{\mu \nu} = R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu} \end{align} to obtain the Einstein tensor. In vacuum, $G_{\mu \nu}=0$, but it is not in presence of matter, where we should have \begin{align} G_{\mu \nu} = \frac{8\pi G}{c^4} T_{\mu \nu} \;, \end{align} with $T_{\mu \nu}$ the matter content of the Universe.


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