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

1. Dec 24, 2012

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

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!

2. Dec 25, 2012

### George Jones

Staff Emeritus
Hi kevinferreira. Welcome to Physics Forums!

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3. Dec 25, 2012

### WannabeNewton

Thanks guys!

4. Dec 25, 2012

### 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 $\triangledown ^{\nu }\triangledown _{\mu }\triangledown _{\nu }\xi ^{\mu } = \triangledown ^{\nu }R^{\nu }_{\mu \nu \sigma }\xi ^{\sigma }$ 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.

5. Dec 25, 2012

### 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.
$$\triangledown^{\mu}R_{\mu\nu}-\frac{1}{2}\triangledown_{\nu}R=0$$
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.

6. Dec 25, 2012

### 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!

7. Feb 28, 2013

### shichao116

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!

8. Feb 28, 2013

### shichao116

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!

9. Jun 18, 2013

### sjabbo

I hope that by now your question is answered, but for those who are still looking for one, the reason has been mentioned above: $\triangledown ^{\nu }\xi ^{\rho }$ is anti - symmetric and $R_{\nu \rho }$ is symmetric so $\xi^\nu\nabla^\mu R_{\mu\nu}$ vanishes (with the use of the product rule of course).

10. Jul 4, 2013

### WannabeNewton

I didn't notice someone responded to this thread after all this time (well granted it wasn't that long a time lmfao). $\xi^{a}\nabla^{b}R_{ab}$ does not vanish because of an antisymmetric and symmetric contraction. There is no contraction of an antisymmetric tensor and symmetric tensor in the above expression; $\nabla^{a}\xi^{b} = \nabla^{[a}\xi^{b]}$ yes but that is not what you have in the above expression. I don't know in what way exactly you are alluding to the product rule but for the above that just gives $\xi^{a}\nabla^{b}R_{ab}=\nabla^{b}(\xi^{a}R_{ab})$ and you must do further calculations in order to show that this vanishes identically; it is not immediate.

For anyone still interested, here is one way to solve it very easily (building upon the last sentence above): first note that $\nabla^{a}R_{ab} = \frac{1}{2}\nabla_{b}R$ from the second Bianchi identity. We also have that $\nabla_{a}\nabla_{b}\xi^{a} = R_{bd}\xi^{d}$ hence $\nabla^{b}\nabla_{a}\nabla_{b}\xi^{a} = \nabla^{b}(R_{bd}\xi^{d}) = \xi^{d}\nabla^{b}R_{bd} = \frac{1}{2}\xi^{d}\nabla_{d}R$. Now $\nabla^{b}\nabla_{a}(\nabla_{b}\xi^{a}) - \nabla_{a}\nabla^{b}(\nabla_{b}\xi^{a}) = R_{ae}\nabla^{e}\xi^{a} - R_{be}\nabla^{b}\xi^{e} = 0$ so $\frac{1}{2}\xi^{d}\nabla_{d}R = \nabla^{a}\nabla^{b}\nabla_{b}\xi_{a} = -\xi^{d}\nabla^{a}R_{ad} = -\frac{1}{2}\xi^{d}\nabla_{d}R$ hence $\xi^{d}\nabla_{d}R = 0$.