
#1
Dec2412, 07:57 AM

C. Spirit
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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! 



#2
Dec2512, 02:27 PM

P: 123

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#3
Dec2512, 02:34 PM

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P: 6,038

Hi kevinferreira. Welcome to Physics Forums!
To make LateX work, enclose mathematics by the tags




#4
Dec2512, 04:15 PM

C. Spirit
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Direction derivative of Ricci scalar w.r.t. killing field
Thanks guys!




#5
Dec2512, 04:38 PM

C. Spirit
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#6
Dec2512, 06:46 PM

P: 123

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. 



#7
Dec2512, 06:54 PM

C. Spirit
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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!




#8
Feb2813, 11:02 PM

P: 13

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!




#9
Feb2813, 11:03 PM

P: 13





#10
Jun1813, 06:53 AM

P: 1





#11
Jul413, 12:48 AM

C. Spirit
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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##. 


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