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-
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. This energy momentum tensor agrees with the symmetric and gauge{invariant electromagnetic energy{momentum tensor obtained by \improving" the canonical one. Note that it is traceless: g T = 0. Since a gas of photons is made up of electromagnetic eld, its energy-momentum tensor must be traceless too, which implies that w= 1=3, as stated above. It is a simple computation to check that $$ \Psi \circ T = \mathrm{Id}, T \circ \Psi = \mathrm{Id} $$ Since the space of algebraic Riemann tensor and that of Ricci tensors have the same number of dimension (with the manifold dimension = 3), this means that every Riemann tensor is uniquely determined (rank-nullity theorem) by its Ricci tensor Interestingly, can be transformed to the Ricci tensor-Ricci scalar equation which indicates that the traceless Ricci tensor is coupled to the Ricci scalar. At this stage, we observe that ( 25 ) may become a massless propagating tensor equation for as which suggests a way of defining a massless spin-2 in gravity.
This energy momentum tensor agrees with the symmetric and gauge{invariant electromagnetic energy{momentum tensor obtained by \improving" the canonical one. Note that it is traceless: g T = 0. Since a gas of photons is made up of electromagnetic eld, its energy-momentum tensor must be traceless too, which implies that w= 1=3, as stated above.
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
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
Trace-free Ricci tensor. In Riemannian geometry and pseudo-Riemannian geometry, the trace-free Ricci tensor (also called traceless Ricci tensor) of a Riemannian or pseudo-Riemannian n-manifold (M,g) is the tensor defined by = −, 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