I'm trying to show that the connection is compatible with the metric, to do this, I must evaluate

$$ \nabla_{\sigma} g^{\mu \nu} = \partial_{\sigma}g^{\mu \nu} +\Gamma^{\mu}_{\sigma \lambda}g^{\lambda \nu}+\Gamma^{\nu}_{\sigma \lambda}g^{\mu \lambda} = 0. $$

The space-time I'm considering is the following general static spherically symmetric line element

$$ ds^{2} = -A(r)dt^{2}+B(r)^{-1}+ r^{2}(d\theta^{2}+\sin^{2}\theta \ d\phi^{2}) \,\, . $$

What is my advance until now:

I have to evaluate the christoffel symbols, so I proceed, for example,

$$ \Gamma^{0}_{00} = -\frac{1}{2}A(r)^{-1}(0+0-0) = 0 \,\, , $$

My doubt is: Do I need to evaluate all the components of the covariant derivative $\nabla_{\sigma}g^{\mu \nu}$ to the connection be compatible with the metric?

What's the relation between the free indice $\sigma$ of the covariant derivative and the free indice $\lambda$ of the Christoffel symbol? Both range from 0 to 3?

My difficult lies in the evaluation of each component this tensorial equation, once there are lots of indices I get confused, how to perform the evaluation of these components? Could show me the procedure to evaluate at least one component to use as example?

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    $\begingroup$ The expression you have is incorrect. It should be $\nabla_\sigma g^{\mu \nu} = \partial_\sigma g^{\mu \nu} + \Gamma ^{\mu} {}_{\sigma \lambda} g^{\lambda \nu} + \Gamma ^{\nu} {}_{\sigma \lambda} g^{\mu \lambda}.$ $\endgroup$ Jul 17, 2017 at 2:02
  • $\begingroup$ @MichaelSeifert Thank you, I'm gonna update the question, could you make an answer to me using the right expression? I would like to upvote if it's good answer! $\endgroup$ Jul 17, 2017 at 2:15
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    $\begingroup$ @WaynerKlën I can write up an answer for you but as is your question is incomplete. Do you have a specific metric that you want to take the covariant derivative of? Are you wanting to do this in general? If you have a specific metric, have you already worked out the Christoffel symbols for that metric? $\endgroup$
    – Mason
    Jul 17, 2017 at 3:36
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    $\begingroup$ Yep, I think I understand. I'll write up an answer now. I think you should edit your question so that you either supply the metric you are working with or just say "how do I do this given that I already have the form of the metric and Christoffel Symbols?" $\endgroup$
    – Mason
    Jul 17, 2017 at 3:52
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    $\begingroup$ @WaynerKlën Actually, from your comment I'm not unsure I understood you correctly. Do you know how to calculate the Christoffel symbols? Like if I asked you what $\Gamma^0_{00}$ was, could you calculate it? $\endgroup$
    – Mason
    Jul 17, 2017 at 3:54

1 Answer 1


So given a metric and the Christoffel symbols, we want to show that \begin{equation} \nabla_\sigma g^{\mu \nu} = \partial_\sigma g^{\mu \nu} + \Gamma^\mu_{\sigma \lambda} g^{\lambda \nu} + \Gamma^\nu_{\sigma \lambda} g^{\mu \lambda} = 0 \end{equation}

First of all, if we want, we can simply our lives here by noticing that $g^{\mu \nu} = g^{\nu \mu}$ and hence we really only need to show that \begin{equation} \partial_\sigma g^{\mu \nu} + 2 ~\Gamma^\mu_{\sigma \lambda} g^{\lambda \nu} = 0~~. \end{equation}

However, I think you probably want more instruction with the index notation so I won't make that simplification.

So let's say you want to calculate the $0,0,0$ component of this tensor, i.e. we set $\sigma = \mu = \nu = 0$ so we have

\begin{equation} \nabla_0 g^{00} = \partial_0 g^{0} + \Gamma^0_{0 \lambda} g^{\lambda 0} + \Gamma^0_{0 \lambda} g^{0 \lambda}~. \end{equation}

But now we have two sets of repeated indices so we need to sum over both of them independently so we do it like this \begin{align} \nabla_0 g^{00} &= \partial_0 g^{0} + \Gamma^0_{0 0}g^{00} + \Gamma^0_{0 1}g^{10} + \Gamma^0_{0 2}g^{02} + \Gamma^0_{0 3}g^{03} + \Gamma^0_{0 \lambda} g^{0 \lambda} \end{align} Now we just have one set of contracted indices so we expand those as well \begin{align} \nabla_0 g^{00} = \partial_0 g^{00} + \Gamma^0_{0 0}g^{00} + \Gamma^0_{0 1}g^{10} + \Gamma^0_{0 2}g^{20} + \Gamma^0_{0 3}g^{30} + \Gamma^0_{0 0}g^{00} + \Gamma^0_{0 1}g^{01} + \Gamma^0_{0 2}g^{02} + \Gamma^0_{0 3}g^{03} \end{align}

Now given all the components of $\mathbf{\Gamma}$ and $\mathbf{g}$ we can plug in the numbers and show that this evaluates (hopefully) to zero! This can now be repeated for every value of $\sigma,~ \mu$ and $\nu$.

I hope this helps, if not or if there's anything else you'd like clarification on please ask!


I just saw your edit so I'll answer those extra questions here:

Do I need to evaluate all the components of the covariant derivative [of the metric for] the connection be compatible with the metric?

Yes, I believe so.

What's the relation between the free indice σ of the covariant derivative and the free indice λ of the Christoffel symbol? Both range from 0 to 3?

The two indices are effectively unrelated. $\lambda$ is a dummy index denoting a dot product between the Christoffel symbol and the (inverse) metric.

  • $\begingroup$ Isn't it missing one more indice in the partial derivative of the metric tensor, i.e, shouldn't it be $\partial_{0}g^{00}$? $\endgroup$ Jul 17, 2017 at 4:34
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    $\begingroup$ @HerrSchrödinger Yep, you're right. Just a typo! Other than that, does this make sense? $\endgroup$
    – Mason
    Jul 17, 2017 at 4:39
  • $\begingroup$ Yes, It makes sense! By now it's just this! Thanks @Mason! $\endgroup$ Jul 17, 2017 at 4:52
  • $\begingroup$ I've updated my answer in response to your update by the way $\endgroup$
    – Mason
    Jul 17, 2017 at 5:23

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