The Ricci tensor, the Einstein tensor, and the traceless Ricci tensor are symmetric 2-tensors: R j k = R k j {\displaystyle R_{jk}=R_{kj}} G j k = G k j {\displaystyle G_{jk}=G_{kj}}

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

## action contains terms that are second order in the curvature tensor, namely S= Z M d4x √ −g(c1CµνρσCµνρσ +c2R2 +c3R˜µνR˜µν), (1) where ci are dimensionless coupling constants, Cµνρσ is the Weyl tensor and R˜µν = Rµν − 1 4 gµνRis the traceless part of the Ricci tensor Rµν. Partly due to their scale in-

The non-relativisitic tidal tensor Ei j ik @ 2 @xk@xj; (5.5) determines the tidal forces, which tend to bring the particles together. This is the fundamental object for the description of gravity and not their individual accelerations g i= @ i! Exercise Assume the tidal tensor Ei j to be reduced to diagonal form, as in the example below. Show 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 on the use of the traceless stress tensor (TST). It is shown that it naturally leads to the appearance of a modiﬁed viscosity given by C. =3/ tr.˝/ where is the shear-viscosity coefﬁcient, the relaxation time and tr(˝) the trace of the extra stress tensor. This modiﬁed viscosity reaches high values near singular points, the troublesome

### 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

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 traceless Ricci tensor Ric0 to an endomorphism of Λ2M(anti-commuting with ∗), and W± are respectively the self-dual and anti-self-dual parts of the Weyl tensor W. The self-dual Weyl tensor W+ is viewed as a section of the bundle S2 0(Λ +M) of symmetric, traceless endomorphisms of Λ+M(also 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.) hand, the Ricci tensor “straddles” this gap and has elements in both subspaces. Since the self-dual part of the conformal tensor has 5 independent (complex) components, a useful way to present them is via a (complex-valued), traceless 3 matrix, which we will describe below. 5 In Riemanniangeometry, the decomposition of the curvature tensor into its irre-ducible summands under the orthogonal group is regarded as fundamental. There are three such summands, the scalar curvature, the traceless Ricci curvature, and the Weyl curvature.1 The metrics for which one or more of these irreducible tensors action contains terms that are second order in the curvature tensor, namely S= Z M d4x √ −g(c1CµνρσCµνρσ +c2R2 +c3R˜µνR˜µν), (1) where ci are dimensionless coupling constants, Cµνρσ is the Weyl tensor and R˜µν = Rµν − 1 4 gµνRis the traceless part of the Ricci tensor Rµν. Partly due to their scale in-