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Killing Vector field question

  1. Aug 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex]v^\mu[/itex] is a Killing Vector field, the prove that:
    [tex]v^\mu \nabla_\alpha R=0 [/tex]

    2. Relevant equations
    1) [itex]\nabla_\mu \nabla_\nu v^\beta = R{^\beta_{\mu \nu \alpha}} v^\alpha[/itex]
    2) The second Bianchi Identity.
    3) If [itex]v^\mu[/itex] is Killing the it satisfies then Killing equation, viz. [itex] \nabla_\mu v_\nu - \nabla_\nu v_\mu=0[/itex]

    3. The attempt at a solution
    I know I should use normal coordinates making my life easier with the Christoffels and use the that the Riemann tensor appears when I have two covariant derivatives acting on a vector field, but I'm stuck and can't figure out how to proceed :(. Any help will be greatly appreciated.

    M.
     
    Last edited: Aug 29, 2015
  2. jcsd
  3. Aug 30, 2015 #2

    fzero

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    The equation you wrote down to prove would require that ##\nabla_\alpha R =0##. Perhaps you meant ## v^\mu \nabla_\mu R =0##? Also the Killing equation has a + sign: i.e. ##\nabla_\mu v_\nu + \nabla_\nu v_\mu =0##.

    If so, you should be able to start with the expression ##\nabla_\nu (v^\mu {R^\nu}_\mu)##. You will need to use (1), the 2nd Bianchi identity in the form ##2 \nabla_\nu {R^\nu}_\mu = \nabla_\mu R## and the Killing equation will make various expressions vanish by symmetry of indices.
     
  4. Aug 30, 2015 #3
    You're totally rigth fzero, it is [itex]v^\mu \nabla_\mu R=0[/itex] and I mistyped the sign killing equation (ooops sorry) shame on me :/.
     
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