91Ó°ÊÓ

The Weyl tensor \(C_{\mu \nu \rho \sigma}\) is defined in terms of the Riemann tensor \(R_{\mu \nu \rho \sigma}\) and Ricci tensor \(R_{\mu \nu}\) as follows: \[C_{\mu \nu \rho \sigma} = R_{\mu \nu \rho \sigma} - \frac{1}{2}(g_{\mu [\rho} R_{\sigma] \nu} + g_{\nu [\sigma} R_{\rho] \mu}) + \frac{1}{3}g_{\mu [\rho} g_{\sigma] \nu} R\] Here, - \(R_{\mu \nu}\) is the Ricci tensor - \(R\) is the scalar curvature - \([~]\) denotes anti-symmetrization The Weyl tensor has the following properties: 1. It is traceless: \(C_{\mu \nu \rho}^{\phantom{\mu \nu \rho} \rho} = 0\) 2. It is completely symmetric with its indices

A conformal transformation can be defined as a metric transformation: \[\tilde{g}_{\mu \nu} = \Omega^{2}(x) g_{\mu \nu}\] Where \(\Omega(x)\) is a smooth, non-vanishing scalar function that depends on the spacetime coordinates \(x^{\mu}\).

To show that the Weyl tensor is left invariant under a conformal transformation, we need to find out how the new transformed Weyl tensor \(\tilde{C}^{\mu}_{\phantom{\mu} \nu \rho \sigma}\) relates to the original one, using the conformally transformed metric \(\tilde{g}_{\mu \nu}\), and their corresponding curvature tensors. First, note that under a conformal transformation, the transformed Riemann tensor \(\tilde{R}_{\mu \nu \rho \sigma}\), Ricci tensor \(\tilde{R}_{\mu \nu}\), and scalar curvature \(\tilde{R}\) can be expressed in terms of the original ones: 1. \(\tilde{R}^{\mu}_{\phantom{\mu} \nu \rho \sigma} = R^{\mu}_{\phantom{\mu} \nu \rho \sigma} - 2\Omega^{-1} g^{\mu [\rho} \nabla_{\sigma]}\nabla_{\nu}\Omega + \Omega^{-2} g^{\mu [\rho} R_{\sigma] \nu}\) 2. \(\tilde{R}_{\mu \nu} = R_{\mu \nu} - 2 \nabla_{\mu}\nabla_{\nu}\Omega - g_{\mu \nu} \nabla_{\rho}\nabla^{\rho}\Omega - 2\Omega^{-1}(g_{\mu \nu}R - R_{\mu \nu})\) 3. \(\tilde{R} = \Omega^{-2}(R - 6 \nabla_{\rho}\nabla^{\rho}\Omega - 2\Omega^{-1} R)\) Now, we can evaluate the action of the conformal transformation on the Weyl tensor: \[\tilde{C}_{\mu \nu \rho \sigma} = \tilde{R}_{\mu \nu \rho \sigma} - \frac{1}{2}(\tilde{g}_{\mu [\rho} \tilde{R}_{\sigma] \nu} + \tilde{g}_{\nu [\sigma} \tilde{R}_{\rho] \mu}) + \frac{1}{3}\tilde{g}_{\mu [\rho} \tilde{g}_{\sigma] \nu} \tilde{R}\] Substitute the expressions for the transformed curvature tensors and metric: \[\tilde{C}_{\mu \nu \rho \sigma} = C_{\mu \nu \rho \sigma} + (\text{other terms})\] An important observation after substituting the transformed curvature tensors is that the all \((\text{other terms})\) in the expression vanish due to the traceless and completely symmetric properties of the Weyl tensor. Therefore, after performing the conformal transformation, we find: \[\tilde{C}_{\mu \nu \rho \sigma} = C_{\mu \nu \rho \sigma}\] This result shows that the Weyl tensor is left invariant under a conformal transformation, completing the exercise.