1
$\begingroup$

On page 3 of this document:

https://studentportalen.uu.se/uusp-filearea-tool/download.action?nodeId=1247106&toolAttachmentId=247022

it shows how to calculate the contracted Christoffel symbol by

\begin{align*} \Gamma^{\mu}_{\mu \lambda} &= \frac{1}{2}g^{\mu \rho}(\partial_{\mu}g_{\rho\lambda} + \partial_{\lambda}g_{\mu\rho} - \partial_{\rho}g_{\mu\lambda} ) \\ &= \frac{1}{2}(\partial^{\rho}g_{\rho\lambda} + g^{\mu \rho}\partial_{\lambda}g_{\mu\rho} - \partial^{\mu}g_{\mu\lambda} ) \\ &= \frac{1}{2}g^{\mu \rho}\partial_{\lambda}g_{\mu\rho} \end{align*}

Can someone kindly please help and explain what's happening in step $2$ and $3$? I just don't get it.

$\endgroup$
  • $\begingroup$ Nothing much is happening there. From line one to two $g^{\mu\rho}$ is pulled into the bracket and indices are raised. From step two to three the summation index in the first term is renamed to $\mu$ (or equivalently in the third to $\rho$). Then the first and last terms cancel. If that is not clear to you I would suggest reintroducing the suppressed sums explicitly. $\endgroup$ – N0va Sep 21 '17 at 11:17
0
$\begingroup$

The tensor product in step 1 is expanded and the 3 terms in step 1 are transformed into 3 terms in step 2 according to these rules: $$g^{\mu \rho}\partial_{\mu}g_{\rho\lambda} \to \partial^{\rho}g_{\rho\lambda} $$ and $$ g^{\mu \rho}\partial_{\rho}g_{\mu\lambda} \to \partial^{\mu}g_{\mu\lambda} $$

this is because you can contract the index of a derivative (only the left-most one, if there are 2 or more derivatives acting on a tensor), while the middle term (the second one) is left untouched.

The first and third term in step 2 are the same , once you see that they differ only in the dummy indices, so they cancel each other and in step 3 we are left only with the second term of step 2.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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