Ivancevic_Applied-Diff-Geom

140 Applied Differential Geometry: A Modern Introductionp, and let ɛu be the vector from p to q. Since the two particles start outat rest relative to one other, the velocity of the particle at q is obtained byparallel transporting v along ɛu.Now let us wait a short while. Both particles trace out geodesics as timepasses, and at time ɛ they will be at new points, say p ′ and q ′ . The pointp ′ is displaced from p by an amount ɛv, so we get a little parallelogram,exactly as in the definition of the Riemann curvature:Next let us calculate the new relative velocity of the two particles. Tocompare vectors we must carry one to another using parallel transport. Letv 1 be the vector we get by taking the velocity vector of the particle at p ′and parallel transporting it to q ′ along the top edge of our parallelogram.Let v 2 be the velocity of the particle at q ′ . The difference v 2 − v 1 is thenew relative velocity. It follows that over this passage of time, the averagerelative acceleration of the two particles is a = (v 2 − v 1 )/ɛ. By equation(3.3),v 2 − v 1alimɛ→0 ɛ 2 = R(u, v)v, therefore lim = R(u, v)v.ɛ→0 ɛThis is the simplified form of the geodesic deviation equation. From the definitionof the Riemann curvature it is easy to see that R(u, v)w = −R(v, u)w,so we can also write this equation asa αlimɛ−→0 ɛ = −Rα βγδv β u γ v δ . (3.4)Using geodesic deviation equation (3.4) we can work out the secondtime derivative of the volume V (t) of a small ball of test particles thatstart out at rest relative to each other. For this we must let u range over anorthonormal basis of tangent vectors, and sum the ‘outwards’ componentof acceleration for each one of these. By equation (3.4) this gives¨Vlim ∣ = −R αV −→0 V t=0βαδv β v δ .In terms of the so–called Ricci tensor, which is a contracted Riemann tensor,R βδ = R α βαδ ,we may write the above expression as¨Vlim ∣ = −R βδ v β v δ .V −→0 V t=0