A Geometrical interpretation of Ricci and Riemann tensors?

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1. Jul 15, 2016

Victor Alencar

I do not get the conceptual difference between Riemann and Ricci tensors. It's obvious for me that Riemann have more information that Ricci, but what information?
Riemann tensor appears when you compare the change of the sabe vector(or other tensor) when it takes two different paths. You can see it comutanting two differents cov. derivatives of a vector ou computing the parallel displacement.
Studying general relativity I saw : "If Riemann tensor is zero, the space is flat; if the Ricci tensor is zero the space is empty". Someone knows some mathematical proof of this affirmation?
And what the Ricci scalar say to us? It's always directly proporcional to curvature?

2. Jul 16, 2016

Markus Hanke

Ricci can be taken as the trace of the Riemann tensor, hence it is of lower rank, and has fewer components. If you have a small geodesic ball in free fall, then ( ignoring shear and vorticity ) the Ricci tensor tells you the rate at which the volume of that ball begins to change, whereas the Riemann tensor contains information not only about its volume, but also about its shape.

If the Riemann tensor is zero, then the equation of geodesic deviation reduces to the equation of a straight line, meaning that the separation vector between geodesics is constant. Hence, initially parallel lines will remain parallel everywhere - you are dealing with a flat manifold.
As for empty space, this is just a consequence of the Einstein equations. If you write them in trace-reversed form, and set T=0 ( empty space ), you get a vanishing Ricci tensor. Hence, empty space implies Ricci flatness.

The Ricci scalar is the trace of the Ricci tensor, and it is a measure of scalar curvature. It can be taken as a way to quantify how the volume of a small geodesic ball ( or alternatively its surface area ) is different from that of a reference ball in flat space.

Perhaps you might find this helpful :

http://arxiv.org/pdf/gr-qc/0401099.pdf