- #1
Dazed&Confused
- 190
- 3
This is from a general relativity book but I think this is the appropriate location.
The proof that <br /> \nabla_{[a} {R_{bc]d}}^e=0
is as follows:
Choose coordinates such that \Gamma^a_{bc}=0 at an event. We have <br /> \nabla_a {R_{bcd}}^e = \partial_a \partial_b \Gamma^e_{cd} - \partial_a \partial_c \Gamma^e_{bd} + \textrm{ terms in } \Gamma \partial \Gamma \textrm{ and } \Gamma \Gamma \Gamma.
Because the first term on the right-hand side is symmetric in ab and the second in ac, and because the other terms vanish at the event, we have <br /> \nabla_{[a}{R_{bc]d}}^e=0
at the event in this coordinate system. However, this is a tensor equation, so it is valid in every coordinate system.
The bit I'm unsure of is how ''the other terms'' vanish at the event, so help would be appreciated.
The proof that <br /> \nabla_{[a} {R_{bc]d}}^e=0
is as follows:
Choose coordinates such that \Gamma^a_{bc}=0 at an event. We have <br /> \nabla_a {R_{bcd}}^e = \partial_a \partial_b \Gamma^e_{cd} - \partial_a \partial_c \Gamma^e_{bd} + \textrm{ terms in } \Gamma \partial \Gamma \textrm{ and } \Gamma \Gamma \Gamma.
Because the first term on the right-hand side is symmetric in ab and the second in ac, and because the other terms vanish at the event, we have <br /> \nabla_{[a}{R_{bc]d}}^e=0
at the event in this coordinate system. However, this is a tensor equation, so it is valid in every coordinate system.
The bit I'm unsure of is how ''the other terms'' vanish at the event, so help would be appreciated.