- #1
binbagsss
- 1,259
- 11
Homework Statement
How to show that lie deriviaitve of metric vanish ##(L_v g)_{uv}=0## <=> metric is independent of this coordinate, for example if ##v=\partial_z## then ##g_{uv} ## is independent of ##z## (and vice versa)
2. Relevant equation
I am wanting to show this for the levi-civita symbol as the connection i.e. metric compatability were we have ## \nabla_{a}g^{uv} =0 ## so the first term of ##(L_v g)_{uv}=0## vanishes trivially.
The Attempt at a Solution
[/B]
##(L_ug)_{uv} = U^{\alpha}\nabla_{\alpha}g_{uv}+g_{u\alpha}\nabla_vU^{\alpha}+g_{\alpha v}\nabla_u U^{\alpha}## (1)
If the metric is independent of some coordinate, e.g ##z## then we have ## \partial_z g_{ab} =0## but , assuming metric compatibility ##\nabla_{\alpha}g_{uv}=0## anyway, so there's no need to split it up into the partial and connection term, which seemed to me the most obvious way and only way I can see to substitute the information of metric independence of (say) ##z## into (1)
So we have ##(L_ug)_{uv} = g_{u\alpha}\nabla_vU^{\alpha}+g_{\alpha v}\nabla_u U^{\alpha}##, where there are no derivatives on the metric in these two terms, so I'm pretty stuck..
Many thanks