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

[WannabeNewton]

Homework Statement

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

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!

 

[George Jones]

Hi kevinferreira. Welcome to Physics Forums!

To make LateX work, enclose mathematics by the tags

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

 

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

 

[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 $1/2\xi^\nu\nabla_\nu R = \xi^\nu\nabla^\mu R_{\mu\nu}$ ? I'm stuck exactly here. Thanks!

 

[shichao116]

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!

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!

 

[sjabbo]

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!

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).

 

[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 $\xi^{d}\nabla_{d}R = 0$.