Traceless ricci tensor

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

Nov 30, 2016 · Acceleration of an area spanned by two vectors connecting three geodesics is proportional to the Ricci tensor. In this video I give a proof of this. which is conformal to g. Letting Edenote the traceless Ricci tensor, we recall the transformation formula: if g= ˚ 2^g, then E g= E ^g + (n 2)˚ 1 r2˚ ( ˚=n)^g; where nis the dimension, and the covariant derivatives are taken with respect to ^g. Since gis Einstein, we have E ^g = (2 n)˚ 1 r2˚ 1 n ( ˚)g: constant), Sis the Ricci tensor and ris the scalar curvature of g. They are ob- the Einstein tensor S R 2 g, 2. ˆ= 1 n, the traceless Ricci tensor S R n g, 3. ˆ traceless components of the metric perturbation. This analysis helps to clarify which degrees of freedom in general relativity are radiative and which are not, a useful exercise for understanding spacetime dynamics. Section 3 analyses the interaction of GWs with detectors whose sizes are small compared to the wavelength of the GWs. (12.45) is the difference of two four-vectors, the relation is a valid tensor equation, which holds in any curvilinear coordinate system. In addition, the fourth(!) rank tensor in Eq. (12.45) R σ μ ν σ, the Riemann curvature tensor, is independent of the vector A ρ used in the construction. The above can be decomposed into the trace and a traceless tensor, ij, as h ij(t;~x) = h ij+ ij; (24) where ij is ij= 2G c6r d2 dy02 J ij(y0) y0=ct r; (25) and J ij is the reduced quadrupole-moment tensor of the source distribution J ij= I ij 1 3 ijI k k: (26) 1.4 Polarization states and e ect on free particles

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-

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 where h = h.As before, we can raise and lower indices using and , since the corrections would be of higher order in the perturbation.In fact, we can think of the linearized version of general relativity (where effects of higher than first order in h are neglected) as describing a theory of a symmetric tensor field h propagating on a flat background spacetime. Jun 01, 1970 · A tensor is symmetric if its components are unaltered by an interchange of any pair of their indices. A traceless symmetric tensor of order m has 2w+ 1 indepen- dent components, and corresponds to a (2m + 1)-dimensional irreducible representa- tion of the proper orthogonal group in three dimensions. 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

With this convention, the Ricci tensor is a (0,2)-tensor field defined by R jk =g il R ijkl and the scalar curvature is defined by R=g jk R jk. Define the traceless Ricci tensor = −, and then define three (0,4)-tensor fields S, E, and W by

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. tensor, R is the Ricci scalar and β and α are dimensionless parameters. The critical condition in a dS or AdS background is β = 6α. This leads to critical gravity where the massive spin two physical ghost becomes a massless spin two graviton. In contrast to the original work on critical gravity, no Einstein gravity with a cosmological