- #1
PhyAmateur
- 105
- 2
I was trying to see what is the covariant derivative of a covector. I started with
$$ \nabla_\mu (U_\nu V^\nu) = \partial_\mu (U_\nu V^\nu) = (\partial_\mu U\nu) V^\nu + U_\nu (\partial_\mu V^\nu) $$ since the covariant derivative of a scalar is the partial derivative of the latter.
Then I know $$ \nabla_\mu (U_\nu V^\nu) = \nabla_\mu (U_\nu) V^\nu +U_\nu \nabla_\mu V^\nu$$ according to Leibniz rule.
Setting them equal to each other I get $$ \nabla_\mu (U_\nu) V^\nu = (\partial_\mu U\nu) V^\nu - U_\nu (\Gamma ^\nu _{\mu\lambda} V^\lambda) $$
Can this last term be equally written as $$U_\lambda(\Gamma ^\lambda _{\mu\nu} V^\nu)$$?? Because the proof was to get finally this form: $$ \nabla_\mu U_\nu = \partial_\mu U_\nu - \Gamma^\lambda_{\mu\nu}U_\lambda $$ I just got stuck at the last part. Thank you.
$$ \nabla_\mu (U_\nu V^\nu) = \partial_\mu (U_\nu V^\nu) = (\partial_\mu U\nu) V^\nu + U_\nu (\partial_\mu V^\nu) $$ since the covariant derivative of a scalar is the partial derivative of the latter.
Then I know $$ \nabla_\mu (U_\nu V^\nu) = \nabla_\mu (U_\nu) V^\nu +U_\nu \nabla_\mu V^\nu$$ according to Leibniz rule.
Setting them equal to each other I get $$ \nabla_\mu (U_\nu) V^\nu = (\partial_\mu U\nu) V^\nu - U_\nu (\Gamma ^\nu _{\mu\lambda} V^\lambda) $$
Can this last term be equally written as $$U_\lambda(\Gamma ^\lambda _{\mu\nu} V^\nu)$$?? Because the proof was to get finally this form: $$ \nabla_\mu U_\nu = \partial_\mu U_\nu - \Gamma^\lambda_{\mu\nu}U_\lambda $$ I just got stuck at the last part. Thank you.