Direction derivative of Ricci scalar w.r.t. killing field

WannabeNewton

1. The problem statement, all variables and given/known data
I didn't really know if this belonged here or in the math section but it is from a physics book so what the heck =D. I have to show that the directional derivative of the ricci scalar along a killing vector field vanishes i.e. [itex]\triangledown _{\xi }R = \xi ^{\rho }\triangledown _{\rho }R = 0[/itex].

3. The attempt at a solution
From previous parts of the problem I had shown that [itex]\triangledown _{\mu }\triangledown _{\nu }\xi ^{\mu } = R_{\nu \rho }\xi ^{\rho }[/itex] and we have, from the Bianchi identity, that [itex]\triangledown ^{v}R_{v\rho } = \frac{1}{2}\triangledown _{\rho }R[/itex] so combining the two we see that [itex]\triangledown ^{\nu }\triangledown _{\mu }\triangledown _{\nu }\xi ^{\mu } = R_{\nu \rho }\triangledown ^{\nu }\xi ^{\rho } + \frac{1}{2}\xi ^{\rho }\triangledown _{\rho }R[/itex]. Since [itex]\triangledown ^{\nu }\xi ^{\rho }[/itex] is anti - symmetric and [itex]R_{\nu \rho }[/itex] is symmetric, their contraction vanishes so we are left with [itex]\triangledown ^{\nu }\triangledown _{\mu }\triangledown _{\nu }\xi ^{\mu } = \frac{1}{2}\xi ^{\rho }\triangledown _{\rho }R[/itex]. Here's where I'm stuck. I tried playing around with the left side, by using the definition of a killing field, to see if I can show that the left side must vanish (possibly by anti - symmetry and\or dummy index relabeling tricks) but I can't seem to simplify it further. Any help is much appreciated thanks!

kevinferreira

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George Jones

Hi kevinferreira. Welcome to Physics Forums!

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WannabeNewton

Thanks guys!

WannabeNewton

Actually Kevin, I'm not sure where your calculations went but I'm not sure about your last line with the riemann tensor. You wrote that [itex]\triangledown ^{\nu }\triangledown _{\mu }\triangledown _{\nu }\xi ^{\mu } = \triangledown ^{\nu }R^{\nu }_{\mu \nu \sigma }\xi ^{\sigma }[/itex] but I don't think is even allowed since you have two repeated indices on the top and one on the bottom which doesn't make sense (two nu's on the top and one nu on the bottom) so I don't know if that expression is valid in accordance with the summation convention.

kevinferreira

Actually, I screwed up the whole thing, it was wrong. so I just wanted to keep the least trace of it possible! =D

Anyway, I've been working on this, and here's what I have.
Start with the twice contracted Bianchi identity (that you implicitly derived in your previous calculations):
[tex] \triangledown^{\mu}R_{\mu\nu}-\frac{1}{2}\triangledown_{\nu}R=0[/tex]
and contract it with your vectorfield:
[tex] \frac{1}{2}\xi^{\nu}\triangledown_{\nu}R=\xi^{\nu}\triangledown^{\mu}R_ {\mu\nu}.[/tex]
You recognise what you want on the left side.

WannabeNewton

Ah yes that solves it quite quickly. I solved it as well in the interim between your responses but my calculations had a few extra steps so I like yours better in the end. Thanks mate, cheers!

shichao116

Hi WannabeNewton, I now have the same problem as you did in this thread, can you show how you "solve it quickly" from [itex]1/2\xi^\nu\nabla_\nu R = \xi^\nu\nabla^\mu R_{\mu\nu}[/itex] ? I'm stuck exactly here. Thanks!

shichao116

Hi WannabeNewton, I now have the same problem as you did in this thread, can you show how you "solve it quickly" from [itex]1/2\xi^\nu\nabla_\nu R = \xi^\nu\nabla^\mu R_{\mu\nu}[/itex] ? I'm stuck exactly here. Thanks!