# Weyl Tensor invariant under conformal transformations

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1. Mar 16, 2016

### Al X

1. The problem statement, all variables and given/known data
As the title says, I need to show this. A conformal transformation is made by changing the metric:
$g_{\mu\nu}\mapsto\omega(x)^{2}g_{\mu\nu}=\tilde{g}_{\mu\nu}$

2. Relevant equations
The Weyl tensor is given in four dimensions as:
$C_{\rho\sigma\mu\nu}=R_{\rho\sigma\mu\nu}-\left(g_{\rho\left[\mu\right.}R_{\left.v\right]\sigma}-g_{\sigma\left[\mu\right.}R_{\left.v\right]\rho}\right)+\frac{1}{3}g_{\rho\left[\mu\right.}g_{\left.\nu\right]\sigma}R$

where $R_{\mu\nu}$ is the Ricci tensor, $R$ is the Ricci scalar, and $R_{\rho\sigma\mu\nu}$ is the Riemann tensor
3. The attempt at a solution
$\begin{eqnarray*} \tilde{g}_{\rho\left[\mu\right.}R_{\left.v\right]\sigma}&=&\frac{1}{2}\left(\tilde{g}_{\rho\mu}R_{\nu\sigma}-\tilde{g}_{\rho\nu}R_{\mu\sigma}\right)=\frac{1}{2}\omega(x)^{2}\left(g_{\rho\mu}R_{\nu\sigma}-g_{\rho\nu}R_{\mu\sigma}\right)\\\tilde{g}_{\sigma\left[\mu\right.}R_{\left.v\right]\rho}&=&\frac{1}{2}\left(\tilde{g}_{\sigma\mu}R_{\nu\rho}-\tilde{g}_{\sigma\nu}R_{\mu\rho}\right)=\frac{1}{2}\omega(x)^{2}\left(g_{\sigma\mu}R_{\nu\rho}-g_{\sigma\nu}R_{\mu\rho}\right)\\\tilde{g}_{\rho\left[\mu\right.}\tilde{g}_{\left.\nu\right]\sigma}&=&\frac{1}{2}\left(\tilde{g}_{\rho\mu}\tilde{g}_{\nu\sigma}-\tilde{g}_{\rho\nu}\tilde{g}_{\mu\sigma}\right)=\frac{1}{2}\omega(x)^{4}\left(g_{\rho\mu}g_{\nu\sigma}-g_{\rho\nu}g_{\mu\sigma}\right) \end{eqnarray*}$

From here, I am lost. How do I make the $\omega(x)$ vanish?

2. Mar 16, 2016

### Twigg

If $g_{\mu\nu} \to \omega (x)^{2}g_{\mu\nu}$, what about $\Gamma_{\mu\nu}^{\sigma}$? Is $\tilde{R}_{\rho\sigma\mu\nu}$ the same as $R_{\rho\sigma\mu\nu}$?

3. Mar 17, 2016

### Al X

Thanks! That gave me the push in the right direction! Managed to solve it now.