0
$\begingroup$

I have naive question about GR and Covariant derivative. You know \begin{align} \nabla_{\gamma} g_{\alpha \beta}=\nabla_{\gamma} g^{\alpha \beta}=0 \end{align}

And I would like to compute covariant derivative of $g_{\mu \nu}A^{\mu \nu}$, where $A^{\mu \nu}$ is a suitable tensor. \begin{align} \nabla_{\gamma} (g_{\mu \nu}A^{\mu \nu}) \end{align} Now, $g_{\mu \nu}A^{\mu \nu}$ is just scalar, it's possible to expect this answer $\partial_{\gamma} (g_{\mu \nu}A^{\mu \nu})$. This can be checked explicitly. \begin{align} \nabla_{\gamma} (g_{\mu \nu}A^{\mu \nu})&=g_{\mu \nu}\nabla_{\gamma} A^{\mu \nu}\\ &=g_{\mu \nu}(\partial_{\gamma} A^{\mu \nu}+\Gamma^{\mu}_{\ \lambda\gamma}A^{\lambda\nu}+\Gamma^{\nu}_{\ \lambda\gamma}A^{\mu\lambda})\\ &=g_{\mu \nu}\partial_{\gamma} A^{\mu \nu}+\Gamma_{\nu\lambda\gamma}A^{\lambda\nu}+\Gamma_{\mu \lambda\gamma}A^{\mu\lambda}\\ &=g_{\mu \nu}\partial_{\gamma} A^{\mu \nu}+2\Gamma_{\nu\lambda\gamma}A^{\lambda\nu}\\ &=g_{\mu \nu}\partial_{\gamma} A^{\mu \nu}+(\partial_{\gamma}g_{\nu\lambda}+\partial _{\lambda}g_{\nu \gamma}-\partial_{\nu}g_{\lambda\gamma })A^{\lambda\nu}\\ &=g_{\mu \nu}\partial_{\gamma} A^{\mu \nu}+\partial_{\gamma}g_{\nu\lambda}A^{\lambda\nu} \end{align} where I used the fact that $A^{\mu \nu}$ is a symmetric tensor because it is contracted with $g^{\mu \nu}$. However, when I tried to calculate it in another way, I could not. \begin{align} \nabla_{\gamma} (g^{\mu \nu}A_{\mu \nu})&=g^{\mu \nu}\nabla_{\gamma} A_{\mu \nu}\\ &=g^{\mu \nu}(\partial_{\gamma} A_{\mu \nu}-\Gamma^{\lambda}_{\ \mu\gamma}A_{\lambda\nu}-\Gamma^{\lambda}_{\ \nu\gamma}A_{\mu\lambda})\\ &=g^{\mu \nu}(\partial_{\gamma} A_{\mu \nu}-2\Gamma^{\lambda}_{\ \mu\gamma}A_{\lambda\nu})\\ &=g^{\mu \nu}(\partial_{\gamma} A_{\mu \nu}-g^{\lambda \tau}(\partial_{\gamma}g_{\tau \mu}+\partial_{\mu} g_{\tau \gamma}-\partial_{\tau} g_{\mu \gamma})A_{\lambda \nu})\\ &=g^{\mu \nu}\partial_{\gamma} A_{\mu \nu} -(\partial_{\gamma}g_{\tau \mu}+\partial_{\mu} g_{\tau \gamma}-\partial_{\tau} g_{\mu \gamma})A^{\tau \mu}\\ &=g^{\mu \nu}\partial_{\gamma} A_{\mu \nu}-\partial_{\gamma}g_{\tau \mu}A^{\tau \mu}\\ \end{align} However this is not $\partial_{\gamma} (g^{\mu \nu}A_{\mu \nu})$ in general. What can I do from here? Or is there a rule that the metric must have lower indexes in this case?

$\endgroup$
1
  • $\begingroup$ Note that $\partial_\gamma g_{\tau\mu} A^{\tau\mu} = - \partial_\gamma g^{\tau\mu} A_{\tau\mu}$ $\endgroup$
    – Prahar
    Apr 29, 2021 at 19:50

1 Answer 1

1
$\begingroup$

As you said, the contraction is a scalar so that $$ \begin{align} \nabla_{\gamma} (g^{\mu \nu} A_{\mu \nu}) &= \partial_{\gamma}(g^{\mu \nu}A_{\mu \nu}) \\ &= A_{\mu \nu} \, \partial_{\gamma}g^{\mu \nu} + g^{\mu \nu} \partial_{\gamma} A_{\mu \nu} \ \end{align} $$ and the same thing for $\nabla_{\gamma} (g_{\mu \nu} A^{\mu \nu})$. Your equations are correct, but the second term in your second calculation (with the minus sign) has $A^{\mu \nu} \partial_{\lambda} g_{\mu \nu}$ rather than $A_{\mu \nu} \partial_{\gamma}g^{\mu \nu}$, which is probably what you're looking for. The solution is that swapping them over gives a minus sign, i.e: $$ A_{\mu \nu} \partial_{\gamma}g^{\mu \nu} = - A^{\mu \nu} \partial_{\lambda} g_{\mu \nu} \ . $$ You can expand this out if you want to check this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.