Thus we assume the Ricci scalar to be constant which leads to a substantial simplification of the field equations. We prove that a vacuum solution to quadratic gravity with traceless Ricci tensor of type N and aligned Weyl tensor of any Petrov type is necessarily a Kundt spacetime.
More generally, we show that the field equations of such theories reduce to an equation linear in the Ricci tensor for Kerr-Schild spacetimes having type-N Weyl and type-N traceless Ricci tensors is constructed algebraically using the metric tensor and the traceless part of the Ricci tensor where g a b is the metric tensor . The Weyl tensor or conformal curvature tensor is completely traceless, in the sense that taking the trace, or contraction , over any pair of indices gives zero. Ricci curvature tensor plays an important role in general relativity, where it is the key term in the Einstein field equations. It is known, the Ricci tensor defined by the Riemannian curvature The Ricci curvature, or trace component of the Riemann tensor contains precisely the information about how volumes change in the presence of tidal forces, so the Weyl tensor is the traceless component of the Riemann tensor. A NOTE ON TRACELESS METRIC TENSOR UDC 514.763.5(045)=20 Dragi Radojević Mathematical Institute SANU, 11 000 Belgrade, Knez Mihailova 35 E-mail: dragir@turing.mi.sanu.ac.yu Abstract. We present the coordinate transformations which transform the diagonal Minkowski metric tensor in a metric tensor with all zero diagonal components. Some of stable metric g , if the L2 norm of the traceless Ricci tensor Tg is small relative to suitable geometric quantities, then one can deform g to an Einstein metric through the Ricci flow. The concept "stability" is defined as follows. (Hence-forth we omit the subscript g in notations for geometric quantities associated with g.) 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
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.
The Schur lemma for the Ricci tensor. Suppose (M, g) is a smooth Riemannian manifold with dimension n. Recall that this defines for each element p of M: the sectional curvature, which assigns to every 2-dimensional linear subspace V of T p M a real number sec p (V) the Riemann curvature tensor, which is a multilinear map Rm p : T p M × T p M However, the "proof" that traceless stress tensor implies conformal symmetry in that book doesn't seem to make sense to me because it omitted the essential transformation of fields. Playing with conformal scalar field theory (e.g. page 38 Di Francesco), we can see the traditional stress tensor is only traceless on shell while the generalized 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.
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
Quite literally, a traceless tensor T is one such that Tr(T)=0. The trace of a tensor (in index notation) can be thought of as contracting one of a tensor’s indices with another: i.e. in general relativity, the Ricci curvature scalar is given by t