# A Can geodesic deviation be zero while curvature tensor is not

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1. Mar 13, 2016

### Elnur Hajiyev

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?

Thanks.

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

### stevendaryl

Staff Emeritus
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 $U^\mu$ and $V^\mu$. Take a third vector $W^\mu$, and parallel-transport it around the rectangle to get back to the start. Then the change in $W^\mu$ resulting from the parallel-transport is:

$\delta W^\mu = R^\mu_{\nu \lambda \sigma} W^\nu U^\lambda V^\sigma$

Geodesic deviation only gives you information about the special case in which $W = V$; the vector that you are parallel-transporting around the rectangle is the separation vector between the sides of the rectangle.

3. Mar 16, 2016

### lavinia

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