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A Can geodesic deviation be zero while curvature tensor is not

  1. Mar 13, 2016 #1
    I understand(or assume understand) that geodesic deviation describes how much parallel geodesics diverge/converge on manifolds while moving along these geodesic. But is not it a definition for intrinsic curvature? If it is same as Riemann curvature tensor in terms of describing curvature, why there is two distinct notion to describe it?

    Last edited: Mar 13, 2016
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  3. Mar 14, 2016 #2


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    Geodesic deviation is a special case of applying the Riemann curvature tensor.

    The informal meaning of the curvature tensor is this:

    Make a "rectangle" with sides given by vectors [itex]U^\mu[/itex] and [itex]V^\mu[/itex]. Take a third vector [itex]W^\mu[/itex], and parallel-transport it around the rectangle to get back to the start. Then the change in [itex]W^\mu[/itex] resulting from the parallel-transport is:

    [itex]\delta W^\mu = R^\mu_{\nu \lambda \sigma} W^\nu U^\lambda V^\sigma[/itex]

    Geodesic deviation only gives you information about the special case in which [itex]W = V[/itex]; the vector that you are parallel-transporting around the rectangle is the separation vector between the sides of the rectangle.
  4. Mar 16, 2016 #3


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    I am not sure about Lorentzian manifolds but for Riemannian manifolds the entire Riemann curvature tensor is determined by the sectional curvatures of tangential 2 planes.

    These are the normalized curvatures ,##<R(X,Y)Y,X>/<X,X><Y,Y> - <X,Y>^2##.

    The geodesic deviation equation ##D^2J/dt^2 = - R(J(t),c'(t))c'(t)## almost gives you the sectional curvature of the 2 plane spanned by ##J(t)## and ##c'(t)##. One only needs to take the inner product with ##J## and normalize. So for a Riemannian manifold one ought to be able to determine the entire curvature tensor from geodesic deviations. ( Note that sectional curvature is not defined for a degenerate 2 plane on a Lorentzian manifold.)

    The variation vector field of a variation through geodesics is always a Jacobi field. That is: it always satisfies the geodesic deviation equation. One can always find a variation through geodesics that has an arbitrary Jacobi field as its variation vector field. This means that one can always compute ##R(J(t),c'(t))c'(t)## at any point along the geodesic in terms of some variation through geodesics. So for a Riemannian manifold the answer to your question is yes.

    For a Lorentzian manifold I am not sure since there are degenerate 2 planes. Also the proof that every Jacobi field is the variation vector field of a variation through geodesics depends on the existence of a neighborhood of each point where each point is connected by a unique minimal geodesic. I tried to find a reference on line but without luck. I am sure one of the Relativity experts here would know the answer. In any case, the geodesic deviations ought to tell you a lot about the curvature tensor if not the whole thing. I would be interested to know the answer.
    Last edited: Mar 20, 2016
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