Lie derivative and Riemann tensor

lark

Suppose you have a spacetime with an observer at rest at the origin, and the surface at t = 0 going through the origin, and passing through the surface there are geodesics along increasing time. Then as you get a small ways away from the surface, the geodesics start to deviate from each other. Within a small region around the origin, the geodesic deviation is $$\tau x^b {R_{0b0}}^d$$, where $$\tau$$ is the proper time as measured by the observer, $$x^b$$ is the position vector on the surface at t = 0, and $${R_{0b0}}^d$$ is the Riemann tensor at the origin. So the 4-velocity of points starting from rest at t = 0 is $$\tau x^b {R_{0b0}}^d + (1,0,0,0)$$.
So can you get the Riemann tensor, at least the $${R_{0b0}}^d$$ components, as some kind of Lie derivative - what happens to the position vector of a point as it's carried along by the time flow vector field? I think it'd be a second Lie derivative, the points start out at rest so the first Lie derivative would be 0.
Laura

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Orodruin

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The curvature tensor is related to the affine connection on the manifold, which has to do with how nearby tangent spaces relate to each other. The Lie derivative on the other hand does not depend on any connection, only on the pullback using the flow of a local vector field. You therefore should generally not expect the two concepts to be related. While both the connection and the Lie derivative have several properties that you would typically associate with a derivative, only the connection is a natural directional derivative as $\nabla_{fX} Y = f \nabla_X Y$, where $f$ is a scalar function. This does not hold for the Lie derivative.

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