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

In summary, to prove that v^\mu is a Killing Vector field, it must satisfy the equation v^\mu \nabla_\mu R=0. This can be done by using normal coordinates and the Christoffel symbols, along with the fact that the Riemann tensor appears when two covariant derivatives act on a vector field. The 2nd Bianchi identity and the Killing equation, which has a positive sign, will also come into play and make certain expressions vanish due to symmetry of indices.
  • #1
loops496
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Homework Statement


Suppose [itex]v^\mu[/itex] is a Killing Vector field, the prove that:
[tex]v^\mu \nabla_\alpha R=0 [/tex]

Homework 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]

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.
 
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  • #2
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.
 
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  • #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 :/.
 

1. What is a Killing Vector field?

A Killing Vector field is a vector field on a manifold that preserves the metric of the manifold. In other words, it represents a set of infinitesimal transformations that leave the geometry of the manifold unchanged.

2. How is a Killing Vector field related to the concept of symmetry?

A Killing Vector field is related to the concept of symmetry because it represents the symmetry of a manifold. It describes the infinitesimal transformations that leave the manifold unchanged, which can be thought of as symmetries of the manifold.

3. What is the significance of Killing Vector fields in physics?

Killing Vector fields are significant in physics because they provide a mathematical tool for understanding the symmetries of physical systems. They are particularly useful in general relativity, where they are used to study the symmetries of spacetime.

4. How are Killing Vector fields used to solve differential equations?

Killing Vector fields can be used to solve differential equations by providing a set of symmetries that simplify the equations. This is known as the method of symmetry reduction and is commonly used in physics and engineering to solve complex differential equations.

5. Are there any limitations to using Killing Vector fields?

Yes, there are limitations to using Killing Vector fields. They can only be applied to manifolds with a well-defined metric, such as Riemannian manifolds. Additionally, they may not exist for all manifolds, and even when they do exist, they may not provide a complete solution to a problem.

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