Apologies if I have this completely wrong (and for the general long-windedness). I've searched online but can't find anything helpful/relevant.
I'm trying to use the geodesic equation
D2ξμDλ2+Rμμβαγξαdxβdλdxγdλ=0
I start with the Riemann tensor components: Rθθϕθϕ=sin2θ, Rθθϕϕθ=−sin2θ, Rϕθθθϕ=−1, Rϕθθϕθ=1.
Let uσ≡dxσdλ. Then expand out the Riemann components to get: (Rμμθθθuθuθ+Rμμθθϕuθuϕ+Rμμϕθθuϕuθ+Rμμϕθϕuϕuϕ)ξθ+(Rμμθϕθuθuθ+Rμμθϕϕuθuϕ+Rμμϕϕθuϕuθ+Rμμϕϕϕuϕuϕ)ξϕ.
Set μ=θ to give D2ξθDλ2+(Rθμθθθuθuθ+Rθμθθϕuθuϕ+Rθμϕθθuϕuθ+Rθμϕθϕuϕuϕ)ξθ+(Rθμθϕθuθuθ+Rθμθϕϕuθuϕ+Rθμϕϕθuϕuθ+Rθμϕϕϕuϕuϕ)ξϕ=0
D2ξθDλ2+(sin2θ)(uϕuϕ)ξθ−(sin2θ)(uϕuθ)ξϕ=0
Set μ=ϕ to give D2ξϕDλ2+(Rϕμθθθuθuθ+Rϕμθθϕuθuϕ+Rϕμϕθθuϕuθ+Rϕμϕθϕuϕuϕ)ξθ+(Rϕμθϕθuθuθ+Rϕμθϕϕuθuϕ+Rϕμϕϕθuϕuθ+Rϕμϕϕϕuϕuϕ)ξϕ=0
D2ξϕDλ2+(−1)(uθuϕ)ξθ+(1)(uθuθ)ξϕ=0
D2ξϕDλ2−ξθ(uθuϕ)+ξϕ(uθuθ)=0
My reason for asking is that Misner et al give the geodesic deviation equation (the one I use in my question) as D2ξμDλ2+Rμμβαγξαdxβdλdxγdλ=0,
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