CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): Riemann tensor irreducible part Eiklm = 1 2 (gilSkm+gkmSil −gimSkl − gklSim) constructed from metric tensor gik and traceless part of Ricci tensor

Oct 10, 2005 · The first piece, the scalar part, is so called because it is built out of the curvature scalar and the metric. The second piece, the semi-traceless piece, is built out of the metric and the traceless Ricci tensor (hence the name semi-traceless). The third piece is what is left over and is called the Weyl tensor. tensor into irreducible components (Singer & Thorpe I 969):-e (trA, B, A -.ktrA, C-.3tr C), where trA =trC = scalar curvature, B is the traceless Ricci tensor and the last two components, which we denote W+ and W11, together give the conformally invariant Weyl tensor, W = W+ + W . Note that the metric is Einstein iff B = 0, conformally flat free part of A is the self-dual part W+ of the Weyl tensor, and the trace-free part of C is the anti-self-dual part W-. The matrix B gives the traceless Ricci tensor. If a manifold is conformally flat with positive scalar curvature, then A and C are the same positive multiple of the identity matrix, and Dilatation–Distortion Decomposition of the Ricci Tensor Pierre A. Millette E-mail: PierreAMillette@alumni.uottawa.ca, Ottawa, Canada We apply a natural decomposition of tensor ﬁelds, in terms of dilatations and distor-tions, to the Ricci tensor. We show that this results in a separation of the ﬁeld equations Apr 29, 2017 · Since the Weyl tensor and the traceless part of the Ricci tensor, $$\eta \ell _a\ell _b$$, as well as $$\varvec{E}$$ are 1-balanced, their arbitrary derivative is also 1-balanced and so is any covariant derivative of the Riemann tensor, i.e. $$abla ^{(k)} \varvec{R}$$ is 1-balanced for any $$k \in \mathbb {N}$$. $$\square$$

## Proof that a traceless strain tensor is pure shear deformation. Ask Question Asked 5 years, 3 months ago. Active 4 years, 7 months ago. Viewed 455 times

Nov 03, 2012 · It is not so much that they can be constructed but the vanishing of the Ricci tensor in less than 4-dim is trivial in the sense that the Riemann tensor reduces to the Ricci tensor (or to the Ricci scalar in two dim.) and making the Riemann tensor zero gets you a flat solution, wich is not GR anymore. The Petrov-Penrose types of Pleba\'nski spinors associated with the traceless Ricci tensor are given. Finally, the classification is compared with a similar classification in the complex case

### The Petrov-Penrose types of Plebański spinors associated with the traceless Ricci tensor are given. Finally, the classification is compared with a similar classification in the complex case. Now on home page

Dilatation–Distortion Decomposition of the Ricci Tensor Pierre A. Millette E-mail: PierreAMillette@alumni.uottawa.ca, Ottawa, Canada We apply a natural decomposition of tensor ﬁelds, in terms of dilatations and distor-tions, to the Ricci tensor. We show that this results in a separation of the ﬁeld equations Apr 29, 2017 · Since the Weyl tensor and the traceless part of the Ricci tensor, $$\eta \ell _a\ell _b$$, as well as $$\varvec{E}$$ are 1-balanced, their arbitrary derivative is also 1-balanced and so is any covariant derivative of the Riemann tensor, i.e. $$abla ^{(k)} \varvec{R}$$ is 1-balanced for any $$k \in \mathbb {N}$$. $$\square$$ CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): Riemann tensor irreducible part Eiklm = 1 2 (gilSkm+gkmSil −gimSkl − gklSim) constructed from metric tensor gik and traceless part of Ricci tensor Imagine that you are in a Euclidean space. You make a tiny ball around a point in that space. We know that this will be a sphere of some small but definite volume. It will be a perfect sphere around the point.