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I'm trying to prove identity $$\nabla^{\rho}C_{\rho\sigma\mu\nu}=\nabla_{[\mu}R_{\nu]\sigma}+\frac{1}{6}g_{\sigma[\mu}\nabla_{\nu]}R$$ for $4D$ spacetime.

The task seems simple, requiring second Bianchi identity for tensor of curvature, but I seem to be missing out something obvious. Carroll (Spacetime and Geometry pg. 147) defines Weyl tensor in $n$ dimensions as $$C_{\rho\sigma\mu\nu}=R_{\rho\sigma\mu\nu}-\frac{2}{n-2}(g_{\rho[\mu}R_{\nu]\sigma}-g_{\sigma[\mu}R_{\nu]\rho})+\frac{2}{(n-1)(n-2)}g_{\rho[\mu}g_{\nu]\sigma}R.$$

The first term cancels out the part of the second term after I act with covariant derivative. The consequence of Bianchi identity that I use is: $$\nabla^{\rho}R_{\rho\sigma\mu\nu}=\nabla_{[\mu}R_{\nu]\sigma}.$$

I have also found different definition of Weyl tensor(1):

$$C_{\rho\sigma\mu\nu}=R_{\rho\sigma\mu\nu}-\frac{1}{n-2}(g_{\rho[\mu}R_{\nu]\sigma}-g_{\sigma[\mu}R_{\nu]\rho})+\frac{1}{(n-1)(n-2)}g_{\rho[\mu}g_{\nu]\sigma}R$$

and in that case I get the right-hand of the first equation multiplied by $\frac{1}{2}$.

My guess is that I'm missing out something in the definitions/notation or my derivation of second Bianchi identity is wrong.

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    $\begingroup$ Please, post the source for the second definition, the one without the 2s $\endgroup$ – DanielC Jan 22 '18 at 18:05
  • $\begingroup$ I have edited the post. $\endgroup$ – Caneholder123 Jan 22 '18 at 18:12
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The different factors come from the different conventions for the antisymmetrisation symbol.

Some texts define $$ A_{[\mu\nu]} = A_{\mu\nu} - A_{\nu\mu}\,, $$ whereas others define $$ A_{[\mu\nu]} = \frac{1}{2} \left( A_{\mu\nu} - A_{\nu\mu} \right)\,. $$ I personally like the second convention much more!

The way you can check which convention is being used is to note that the Weyl tensor is defined as the tracefree part of the Riemann tensor. In other words, it is required to satisfy $$ C^\mu{}_{\rho\mu\sigma} = 0 \, . $$

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  • $\begingroup$ ...who in their right mind uses the first one? $\endgroup$ – Ryan Unger Jan 23 '18 at 5:20
  • $\begingroup$ @0celo7 - I know, right! $\endgroup$ – Prahar Jan 24 '18 at 13:48

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