Can a Killing Vector Field Prove v^\mu \nabla_\alpha R=0?

[loops496]

Homework Statement

Suppose $v^\mu$ is a Killing Vector field, the prove that:
$$v^\mu \nabla_\alpha R=0$$

Homework Equations

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

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.

 

[fzero]

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.

 

[loops496]

You're totally rigth fzero, it is $v^\mu \nabla_\mu R=0$ and I mistyped the sign killing equation (ooops sorry) shame on me :/.