- #1

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## Main Question or Discussion Point

I am trying to follow a proof that given a Kiling vector ##V^{u}##, the quantity ##V_{u}U^{u} ## is conserved along a geodesic.

I am given the Killiing Equation: ## \bigtriangledown_{(v}U_{u)}=0 ## [1]

Below ## U^{u} ## is tangent vector ## U^{u} = \frac{dx^{u}}{d\lambda} ##

The proof considers ## U^{u}\bigtriangledown_{v}(V_{u}U^{u}) =U^{v}U^{u}\bigtriangledown_{v}V_{u}+V_{u}U^{v}\bigtriangledown_{v}U^{u} ##

It then says that the first term is zero by the Killing equation. I can't see this. I see from [1] that ## \bigtriangledown_{v}U_{u} + \bigtriangledown_{u}U_{v} =0 ##. So unless this somehow implies each term must indivually be zero in here, I don't follow.

It also says the second term is ##0## as ##x^{u}(\lambda) ## is a geodesic. I'm unsure here too, so the covariant derivative of a tangent vector along a geodesic is zero? Could someone point me to some notes on this?

Finally, I see the expression has considered ## U^{u}\bigtriangledown_{v}(V_{u}U^{u}) ## as a pose to ## \bigtriangledown_{v}(V_{u}U^{u}) ## is this just because we know things about the former but the latter would be harder to proof?

And if we proove that the former is zero we are done as ## U^{u} ## is non-zero?

/

I am given the Killiing Equation: ## \bigtriangledown_{(v}U_{u)}=0 ## [1]

Below ## U^{u} ## is tangent vector ## U^{u} = \frac{dx^{u}}{d\lambda} ##

The proof considers ## U^{u}\bigtriangledown_{v}(V_{u}U^{u}) =U^{v}U^{u}\bigtriangledown_{v}V_{u}+V_{u}U^{v}\bigtriangledown_{v}U^{u} ##

It then says that the first term is zero by the Killing equation. I can't see this. I see from [1] that ## \bigtriangledown_{v}U_{u} + \bigtriangledown_{u}U_{v} =0 ##. So unless this somehow implies each term must indivually be zero in here, I don't follow.

It also says the second term is ##0## as ##x^{u}(\lambda) ## is a geodesic. I'm unsure here too, so the covariant derivative of a tangent vector along a geodesic is zero? Could someone point me to some notes on this?

Finally, I see the expression has considered ## U^{u}\bigtriangledown_{v}(V_{u}U^{u}) ## as a pose to ## \bigtriangledown_{v}(V_{u}U^{u}) ## is this just because we know things about the former but the latter would be harder to proof?

And if we proove that the former is zero we are done as ## U^{u} ## is non-zero?

/

**Thanks very much in advance.**