- #1

JD_PM

- 1,131

- 158

- Homework Statement
- Show that the Weyl tensor ##C^{\mu}_{ \ \nu \sigma \rho}## is left invariant under a conformal transformation of the metric i.e.

\begin{equation*}

g_{\mu \nu}(x) \to \tilde g_{\mu \nu}(x) = \Omega^2(x) g_{\mu \nu}(x)

\end{equation*}

- Relevant Equations
- N/A

The Weyl tensor is given by (Carroll's EQ 3.147)

\begin{align*}

C_{\rho \sigma \mu \nu} &= R_{\rho \sigma \mu \nu} - \frac{2}{n-2}\left(g_{\rho [\mu}R_{\nu]\sigma} - g_{\sigma [\mu}R_{\nu]\rho}\right) \\

&+ \frac{2}{(n-1)(n-2)}g_{\rho [\mu}g_{\nu]\sigma}R

\end{align*}

Where ##n## are the number of dimensions

Rising the first index (i.e. ##C^{\mu}_{ \ \nu \rho \sigma}=g^{\mu \lambda}C_{\lambda \nu \sigma \rho}##) yields

\begin{align*}

C^{\rho}_{ \ \sigma \mu \nu} &= R^{\rho}_{ \ \sigma \mu \nu} - \frac{2}{n-2}\left(\delta^{\rho}_{ [\mu}R_{\nu]\sigma} - g_{\sigma [\mu}R_{\nu]}^{\rho}\right) \\

&+ \frac{2}{(n-1)(n-2)}\delta^{\rho}_{ [\mu}g_{\nu]\sigma}R

\end{align*}

Let us work out the individual terms that explicitly have the ##g_{\mu \nu}## term

\begin{equation*}

\delta^{\rho}_{[\mu}g_{\nu] \sigma} = \frac{1}{2} \left(\delta^{\rho}_{\mu} \tilde g_{\nu \sigma} - \delta^{\rho}_{\nu} \tilde g_{\mu \sigma} \right) = \frac{1}{2} \Omega^2(x) \left(\delta^{\rho}_{\mu} g_{\nu \sigma} - \delta^{\rho}_{\nu} g_{\mu \sigma} \right)

\end{equation*}

\begin{equation*}

\tilde g_{\sigma[\mu} R_{\nu]}^{ \ \ \rho} = \frac{1}{2}\left( \tilde g_{\sigma \mu} R_{\nu}^{ \ \ \rho} - \tilde g_{\sigma \nu} R_{\mu}^{ \ \ \rho} \right) = \frac{1}{2} \Omega^2 (x) \left( g_{\sigma \mu}R_{\nu}^{ \ \ \rho} -g_{\sigma \nu}R_{\mu}^{ \ \ \rho} \right)

\end{equation*}

Now the question are

1) How to work out the Riemann tensor ##R^{\rho}_{ \ \sigma \mu \nu}##? I am quite sure that we should not go for the brute force method i.e. apply the definition of the Riemann tensor straightaway. I have been thinking that using Riemann normal coordinates could be a good idea, as the ##\Gamma \Gamma## terms would vanish. Is this the right approach?

2) What to do with ##\delta^{\rho}_{ [\mu}R_{\nu]\sigma}## term?

I have been trying to perform the transformation but I do not get rid of the terms with ##\Omega(x)##

Thank you!

PS: The exact same question was asked here but I happen to have the same main doubt: how do the ##\Omega(x)## terms cancel each other out.

PS2: I have been thinking a lot about this one. Any little help will be much appreciated.

\begin{align*}

C_{\rho \sigma \mu \nu} &= R_{\rho \sigma \mu \nu} - \frac{2}{n-2}\left(g_{\rho [\mu}R_{\nu]\sigma} - g_{\sigma [\mu}R_{\nu]\rho}\right) \\

&+ \frac{2}{(n-1)(n-2)}g_{\rho [\mu}g_{\nu]\sigma}R

\end{align*}

Where ##n## are the number of dimensions

Rising the first index (i.e. ##C^{\mu}_{ \ \nu \rho \sigma}=g^{\mu \lambda}C_{\lambda \nu \sigma \rho}##) yields

\begin{align*}

C^{\rho}_{ \ \sigma \mu \nu} &= R^{\rho}_{ \ \sigma \mu \nu} - \frac{2}{n-2}\left(\delta^{\rho}_{ [\mu}R_{\nu]\sigma} - g_{\sigma [\mu}R_{\nu]}^{\rho}\right) \\

&+ \frac{2}{(n-1)(n-2)}\delta^{\rho}_{ [\mu}g_{\nu]\sigma}R

\end{align*}

Let us work out the individual terms that explicitly have the ##g_{\mu \nu}## term

\begin{equation*}

\delta^{\rho}_{[\mu}g_{\nu] \sigma} = \frac{1}{2} \left(\delta^{\rho}_{\mu} \tilde g_{\nu \sigma} - \delta^{\rho}_{\nu} \tilde g_{\mu \sigma} \right) = \frac{1}{2} \Omega^2(x) \left(\delta^{\rho}_{\mu} g_{\nu \sigma} - \delta^{\rho}_{\nu} g_{\mu \sigma} \right)

\end{equation*}

\begin{equation*}

\tilde g_{\sigma[\mu} R_{\nu]}^{ \ \ \rho} = \frac{1}{2}\left( \tilde g_{\sigma \mu} R_{\nu}^{ \ \ \rho} - \tilde g_{\sigma \nu} R_{\mu}^{ \ \ \rho} \right) = \frac{1}{2} \Omega^2 (x) \left( g_{\sigma \mu}R_{\nu}^{ \ \ \rho} -g_{\sigma \nu}R_{\mu}^{ \ \ \rho} \right)

\end{equation*}

Now the question are

1) How to work out the Riemann tensor ##R^{\rho}_{ \ \sigma \mu \nu}##? I am quite sure that we should not go for the brute force method i.e. apply the definition of the Riemann tensor straightaway. I have been thinking that using Riemann normal coordinates could be a good idea, as the ##\Gamma \Gamma## terms would vanish. Is this the right approach?

2) What to do with ##\delta^{\rho}_{ [\mu}R_{\nu]\sigma}## term?

I have been trying to perform the transformation but I do not get rid of the terms with ##\Omega(x)##

Thank you!

PS: The exact same question was asked here but I happen to have the same main doubt: how do the ##\Omega(x)## terms cancel each other out.

PS2: I have been thinking a lot about this one. Any little help will be much appreciated.

Last edited: