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Variation of metric

  1. Aug 4, 2014 #1
    if we know

    K^{a b}= (∇^a*ζ^b -∇^b*ζ^a)/2,

    ζ is a killing vector,

    under the variation of metric g_{a b}→g_{a b}+δ(g_{a b}) which preserves the Killing vector δ(ζ^a)=0,

    h_{a b} = δ(g_{a b}) = ∇^a*ζ^b +∇^b*ζ^a,

    how to prove

    δ(K^{a b})= ζ_c*∇^a*h^{b c} - h^{c a}*∇_c*ζ^b - ( ζ_c*∇^b*h^{a c} - h^{c b}*∇_c*ζ^a )

    thanks!
     
  2. jcsd
  3. Aug 5, 2014 #2

    WannabeNewton

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    ##h_{ab} = 2\nabla_{(a}\zeta_{b)} = 0## for a Killing vector so what you've written down is a completely trivial statement. Any variation in the metric along the flow generated by ##\zeta^a## vanishes by definition of a Killing vector i.e. ##\mathcal{L}_{\zeta}g_{ab} = 0##. And indeed we have ##\mathcal{L}_{\zeta}K_{ab} = \mathcal{L}_{\zeta}\nabla_{a}\zeta_b = \zeta^c \nabla_c \nabla_a \zeta_b + \nabla_c \zeta_b \nabla_a \zeta^c + \nabla_a \zeta_c \nabla_b \zeta^c = \zeta^c \zeta^d R_{dcab} +\nabla_c \zeta_b \nabla_a \zeta^c - \nabla_a \zeta^c \nabla_c \zeta_b = 0##.
     
  4. Aug 5, 2014 #3
    Thanks for your reply!

    There is something wrong with your reply, maybe the reason is I don't explain it clear. h_{a b}≠0. If you want to know more, please refer to arXiv:1306.2138[hep-th]. the 1st equation in the last page.
     

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    Last edited: Aug 5, 2014
  5. Aug 8, 2014 #4

    WannabeNewton

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    Feiye, is this the paper: http://arxiv.org/pdf/1306.2138.pdf to which you refer?

    In your original post you said that ##\zeta^a## was a Killing field and that ##\delta g_{ab} = 2\nabla_{(a}\zeta_{b)}## but this would mean that the infinitesimal transformation of ##g_{ab}## occurs through the flow of ##\zeta^a##. If ##\zeta^a## is a Killing field as you said then ##\delta g_{ab} = 0## by definition. So I don't think ##\zeta^a## is a Killing field in the paper. Rather I think we choose a diffeomorphism generated by the flow of a vector field ##\zeta^a## such that if ##\xi^a## is a Killing field then ##\delta \xi^a = 0## when ##g_{ab} \rightarrow g_{ab} + \delta g_{ab} = g_{ab} + 2\nabla_{(a}\zeta_{b)}##. Is that more in agreement with what the paper is doing?

    And as an aside the arxiv paper I linked seems to be different from the paper you're reading based on the difference in notation and equation numberings between the arxiv and your attachments. If it's possible could you link the paper you're reading instead? It's ok if it's behind a paywall I can access it.
     
  6. Aug 9, 2014 #5

    Thanks very much for your reply. You are so kind!

    The arxiv "1306.2138"(you linked) is the original paper I read, and the attachment is an PPT which is just a work related with 1306.2138. So the equation numberings is different. But the problem is in the 1306.2138.

    we will start from the equation(15), we get Neother potential ##K^{\mu \nu}##, and in the Einstein Gravity, ##K^{\mu \nu} = 2*\nabla^{[\mu}*\xi^{\nu]}##. Then we will get the ADT potential(equation(16)). The first term of (16) is ##\delta K^{\mu \nu}##. Substituting ##K^{\mu \nu} = 2*\nabla^{[\mu}*\xi^{\nu]}## into it, the result is the sum of the 1st and 2nd terms of the 1st equation in page 4.

    The case of Einstein gravity is just an simple example to calculate the conserve charge, but I can't get it.

    You are right. ##\zeta^{^a}## is a diffeomorphism. And under the diffeomorphism we get the conserved current(equation (3)). And as the paper says "In order to see the relation of the off-shell current for the diffeomorphism to the linearized conserved current for a Killing vector , it is useful to consider the change in the Noether potential under the variation of the metric which preserves the Killing vector ##\delta \xi^a = 0##".

    The core of the problem is variation of ##K^{\mu \nu}## with respect metric with assuming ##\delta \xi^a = 0##.

    The attachment is my derivation process. the 1st is with assuming ##\delta \xi_a = 0##. The 2nd is with assuming ##\delta \xi^a = 0##. Both of them I can't get the result, so I think maybe I make some mistakes. But I can't find where I am wrong. Please help me, thanks!
     

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  7. Aug 10, 2014 #6
    I have got the right answer from my senior fellow apprentice. It is in the attachment. Whatever thank you very much!
     

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  8. Aug 11, 2014 #7

    WannabeNewton

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    Cool, and thanks for providing the final calculation as well! I'm sorry I wasn't able to reply to you sooner. I was actually calculating ##\delta K^{ab} = \delta (\nabla^a \xi^b) = \delta (\nabla^{[a}\xi^{b]})## under ##\delta g_{ab} = 2\nabla_{(a}\zeta_{b)}## in the exact same way you did using ##\delta \xi^a =0## and got stuck in the same place that you did as well. I didn't even think about expanding ##\nabla^a \xi^b## out in terms of ##\Gamma^a_{bc}## and considering ##\delta \Gamma^a _{bc}## haha.
     
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