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Weyl Tensor invariant under conformal transformations

  1. Mar 16, 2016 #1
    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. jcsd
  3. Mar 16, 2016 #2

    Twigg

    User Avatar
    Gold Member

    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}##?
     
  4. Mar 17, 2016 #3
    Thanks! That gave me the push in the right direction! Managed to solve it now.
     
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