- #1

- 3,836

- 1,418

## Main Question or Discussion Point

__Draft re Ricci vs Riemann tensors__This one is really just the beginning of a musing. I can't even remember if I came to any conclusion or just forgot about it. I started a thread in Jan 2014, a couple of months after this blog post, on the related issue of what the physical significance of the Ricci tensor is, and got some answers. I can't remember now whether they would have also covered off the questions raised in what is below.

So, despite its incompleteness and uncertainty, I'll save it here so I have the option of reactivating it.

---------------------------------------------------------------------------

I have been thinking about the fact that Einstein's field equation uses the Ricci rather than the Riemann tensor. The Ricci tensor has ten degrees of freedom and the Riemann tensor has 24, so some information has been lost in the process of contracting the Riemann tensor to get the Ricci. It seems to follow that that information has no impact on gravitation. I've been wondering what sort of information that is, and whether it can be described in some physical, intuitive way.

One way to think of it is as follows: Let ##TR=_{def}T_P^*M\otimes(T_PM)^3## be the space of (1 3) tensors at a point P in spacetime M, and let ##RiccContract(R)## denote the contraction of a (1 3) tensor on the 1st and 3rd indices.

Then, for a given Ricci tensor ##R^{Ricci}## at P, is there some pattern of the following set of vectors, which share a common Ricci contraction?

$$S(R^{Ricci},i,j,k) = \{R(\vec{e}_i, \vec{e}_j, \vec{e}_k) : R\in TR, RiccContract(R)=R^{Ricci}\}$$

The set has two degrees of freedom and so might trace out a two-dimensional surface in the 4-dimensional space-time. If that surface were a [yes, it really did stop in the middle of a sentence!]