Recognitions:
Gold Member

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

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. $\triangledown _{\xi }R = \xi ^{\rho }\triangledown _{\rho }R = 0$.

3. The attempt at a solution
From previous parts of the problem I had shown that $\triangledown _{\mu }\triangledown _{\nu }\xi ^{\mu } = R_{\nu \rho }\xi ^{\rho }$ and we have, from the Bianchi identity, that $\triangledown ^{v}R_{v\rho } = \frac{1}{2}\triangledown _{\rho }R$ so combining the two we see that $\triangledown ^{\nu }\triangledown _{\mu }\triangledown _{\nu }\xi ^{\mu } = R_{\nu \rho }\triangledown ^{\nu }\xi ^{\rho } + \frac{1}{2}\xi ^{\rho }\triangledown _{\rho }R$. Since $\triangledown ^{\nu }\xi ^{\rho }$ is anti - symmetric and $R_{\nu \rho }$ is symmetric, their contraction vanishes so we are left with $\triangledown ^{\nu }\triangledown _{\mu }\triangledown _{\nu }\xi ^{\mu } = \frac{1}{2}\xi ^{\rho }\triangledown _{\rho }R$. 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!
 ....
 Mentor Hi kevinferreira. Welcome to Physics Forums! To make LateX work, enclose mathematics by the tags Code: $and$ (for in-line mathematics,), or Code: $$and$$ (for stand-alone mathematics).

Recognitions:
Gold Member

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

Thanks guys!

Recognitions:
Gold Member
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 $\triangledown ^{\nu }\triangledown _{\mu }\triangledown _{\nu }\xi ^{\mu } = \triangledown ^{\nu }R^{\nu }_{\mu \nu \sigma }\xi ^{\sigma }$ 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.
 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): $$\triangledown^{\mu}R_{\mu\nu}-\frac{1}{2}\triangledown_{\nu}R=0$$ and contract it with your vectorfield: $$\frac{1}{2}\xi^{\nu}\triangledown_{\nu}R=\xi^{\nu}\triangledown^{\mu}R_ {\mu\nu}.$$ You recognise what you want on the left side.
 Hi WannabeNewton, I now have the same problem as you did in this thread, can you show how you "solve it quickly" from $1/2\xi^\nu\nabla_\nu R = \xi^\nu\nabla^\mu R_{\mu\nu}$ ? I'm stuck exactly here. Thanks!
Hi WannabeNewton, I now have the same problem as you did in this thread, can you show how you "solve it quickly" from $1/2\xi^\nu\nabla_\nu R = \xi^\nu\nabla^\mu R_{\mu\nu}$ ? I'm stuck exactly here. Thanks!