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Rob Woodside
Dec24-04, 11:08 PM
P: 90
Riemann curvature tensor derivation

Quote Quote by weio

So far this is how I understand it, though I know I could be very wrong. If you have two geodesics parallel to each other, with tangents [tex] V [/tex] and [tex] V' [/tex] , in which the coordinate [tex] x^\alpha [/tex] point along both geodesics. There is some connecting vector [tex] w^\alpha [/tex] between them. Let the affine parameter on the geodesics be [tex] \lambda [/tex]

Riemman tensor calculates the acceleration between these two geodesics. so you calculate the acceleration at some point A, A' on each geodesic , and subtract them. this gives you an expression telling how the components of [tex] w^\alpha [/tex] change.

[tex] \frac{d^2w^\alpha} {d\lambda^2} = \frac{d^2x^\alpha} {d\lambda^2} | A' - \frac {d^2x^\alpha} {d\lambda^2} | A = - \Gamma^\alpha_0_0_\beta w^\beta [/tex]

After that you calculate the full 2nd covariant derivative along V, ie , you get something like
[tex] \bigtriangledown v \bigtriangledown v w^\alpha = (\Gamma^\alpha_\beta_0_0 - \Gamma^\alpha_0_0,\beta) w^\beta [/tex]
[tex] = R^a_0_0\beta w^\beta [/tex]
[tex] = R^a_u_v_\beta V^u V^v w^\beta [/tex]

That's where the tensor arises. so basically it's a difference in acceleration as geodesics don't maintain their seperation in curved space.
Yes it arises there and in many other places, including the one you asked about and that I told you about.