- #1

- 4

- 0

## Main Question or Discussion Point

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!

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!