## Thursday 23 July 2020

### Conformal Transformations

I want to understand the conformal diagrams in section 5.7 so I must read appendix G then H. Most of this is just checking Carroll's formulas for the conformal 'dynamical variables' - things like the connection coefficients and the Riemann tensor. It is eye bogglingly dense.

Conformal transformations all start when you multiply each component of the metric by a scalar $\omega$ which may depend on the coordinates. So we have a conformal metric $${\widetilde{g}}_{\mu\nu}=\omega^2g_{\mu\nu}$$Then we want find things like the Riemann tensor in the 'conformal frame'. It's quite easy to show that it is$${\widetilde{R}}_{\ \ \ \sigma\mu\nu}^\rho=R_{\ \ \ \sigma\mu\nu}^\rho+\nabla_\mu C_{\ \ \ \nu\sigma}^\rho-\nabla_\nu C_{\ \ \ \mu\sigma}^\rho+C_{\ \ \ \mu\lambda}^\rho C_{\ \ \ \nu\sigma}^\lambda-C_{\ \ \ \nu\lambda}^\rho C_{\ \ \ \mu\sigma}^\lambda$$where$$C_{\ \ \ \mu\nu}^\rho=\omega^{-1}\left(\delta_\nu^\rho\nabla_\mu\omega+\delta_\mu^\rho\nabla_\nu\omega-g^{\rho\lambda}g_{\mu\nu}\nabla_\lambda\omega\right)$$
Carrol then says "it is a matter of simply plugging in and grinding away to get"$${\widetilde{R}}_{\ \ \ \sigma\mu\nu}^\rho=R_{\ \ \ \sigma\mu\nu}^\rho-2\left(\delta_{[\mu}^\rho\delta_{\nu]}^\alpha\delta_\sigma^\beta-g_{\sigma[\mu}\delta_{\nu]}^\alpha g^{\rho\beta}\right)\omega^{-1}\nabla_\alpha\nabla_\beta\omega$$$$+2\left(2\delta_{[\mu}^\rho\delta_{\nu]}^\alpha\delta_\sigma^\beta-2g_{\sigma[\mu}\delta_{\nu]}^\alpha g^{\rho\beta}+g_{\sigma[\mu}\delta_{\nu]}^\rho g^{\alpha\beta}\right)\ \omega^{-2}\left(\nabla_\alpha\omega\right)\left(\nabla_\beta\omega\right)$$
I'm glad it wasn't complicated because getting to that took two dense pages part of which is shown below. He's also used the antisymmetrisation operator [], which is very clever but hard work. It also screws up my latex generator which does not like ]'s in indices.
See that in searchable form at Commentary App G Conformal Transformations.pdf (10 gruelling pages)

1. You wrote: Both you and Carroll make it sound trivial to derive (G.7), the transformed Riemann tensor. I'm wondering what obvious realization I'm missing... when plugging in $\tilde{\Gamma}$ to (3.113) I end up with 4 cross terms like $\Gamma_{\mu \lambda}^{\rho} C_{\nu \sigma}^{\lambda}$ that don't obviously cancel out?
2. It's in the pdf (which there is a link to above) starting at equation (31) which is the tilde version of Carroll's (3.113). It arrives at G.7 six equations later. There is a sly use of $\nabla_\mu C_{\ \ \ \nu\sigma}^\rho$ to remove the nasty cross components. Here it is:
We start with $${\widetilde{R}}_{\ \ \ \sigma\mu\nu}^\rho=\partial_\mu{\widetilde{\Gamma}}_{\nu\sigma}^\rho-\partial_\nu{\widetilde{\Gamma}}_{\mu\sigma}^\rho+{\widetilde{\Gamma}}_{\mu\lambda}^\rho{\widetilde{\Gamma}}_{\nu\sigma}^\lambda-{\widetilde{\Gamma}}_{\nu\lambda}^\rho{\widetilde{\Gamma}}_{\mu\sigma}^\lambda$$then using ${\widetilde{\Gamma}}_{\mu\nu}^\rho=\Gamma_{\mu\nu}^\rho+C_{\ \ \ \mu\nu}^\rho$ that becomes$${\widetilde{R}}_{\ \ \ \sigma\mu\nu}^\rho=\partial_\mu\Gamma_{\nu\sigma}^\rho+\partial_\mu C_{\ \ \ \nu\sigma}^\rho-\partial_\nu\Gamma_{\mu\sigma}^\rho-\partial_\nu C_{\ \ \ \mu\sigma}^\rho$$$$+\left(\Gamma_{\mu\lambda}^\rho+C_{\ \ \ \mu\lambda}^\rho\right)\left(\Gamma_{\nu\sigma}^\lambda+C_{\ \ \ \nu\sigma}^\lambda\right)-\left(\Gamma_{\nu\lambda}^\rho+C_{\ \ \ \nu\lambda}^\rho\right)\left(\Gamma_{\mu\sigma}^\lambda+C_{\ \ \ \mu\sigma}^\lambda\right)$$$$=R_{\ \ \ \sigma\mu\nu}^\rho+\partial_\mu C_{\ \ \ \nu\sigma}^\rho-\partial_\nu C_{\ \ \ \mu\sigma}^\rho+\Gamma_{\mu\lambda}^\rho C_{\ \ \ \nu\sigma}^\lambda+C_{\ \ \ \mu\lambda}^\rho\Gamma_{\nu\sigma}^\lambda$$$$+C_{\ \ \ \mu\lambda}^\rho C_{\ \ \ \nu\sigma}^\lambda-\Gamma_{\nu\lambda}^\rho C_{\ \ \ \mu\sigma}^\lambda-C_{\ \ \ \nu\lambda}^\rho\Gamma_{\mu\sigma}^\lambda-C_{\ \ \ \nu\lambda}^\rho C_{\ \ \ \mu\sigma}^\lambda\ (\ast\ast)$$which has the cross terms.
Now we look at the covariant derivatives they are$$\nabla_\mu C_{\ \ \ \nu\sigma}^\rho=\mathrm{\partial}_\mu C_{\ \ \ \nu\sigma}^\rho+\Gamma_{\mu\lambda}^\rho C_{\ \ \ \nu\sigma}^\lambda-\Gamma_{\mu\nu}^\lambda C_{\ \ \ \lambda\sigma}^\rho-\Gamma_{\mu\sigma}^\lambda C_{\ \ \ \nu\lambda}^\rho$$$$-\nabla_\nu C_{\ \ \ \mu\sigma}^\rho=-\mathrm{\partial}_\nu C_{\ \ \ \mu\sigma}^\rho-\Gamma_{\nu\lambda}^\rho C_{\ \ \ \mu\sigma}^\lambda+\Gamma_{\nu\mu}^\lambda C_{\ \ \ \lambda\sigma}^\rho+\Gamma_{\nu\sigma}^\lambda C_{\ \ \ \mu\lambda}^\rho$$The $\Gamma_{\mu\nu}^\lambda C_{\ \ \ \lambda\sigma}^\rho$ terms in those balance. Reordering (**) to match with them$${\widetilde{R}}_{\ \ \ \sigma\mu\nu}^\rho=R_{\ \ \ \sigma\mu\nu}^\rho+\partial_\mu C_{\ \ \ \nu\sigma}^\rho+\Gamma_{\mu\lambda}^\rho C_{\ \ \ \nu\sigma}^\lambda-\Gamma_{\mu\sigma}^\lambda C_{\ \ \ \nu\lambda}^\rho$$$$-\partial_\nu C_{\ \ \ \mu\sigma}^\rho-\Gamma_{\nu\lambda}^\rho C_{\ \ \ \mu\sigma}^\lambda+\Gamma_{\nu\sigma}^\lambda C_{\ \ \ \mu\lambda}^\rho+C_{\ \ \ \mu\lambda}^\rho C_{\ \ \ \nu\sigma}^\lambda-C_{\ \ \ \nu\lambda}^\rho C_{\ \ \ \mu\sigma}^\lambda$$If we add $-\ \Gamma_{\mu\nu}^\lambda C_{\ \ \ \lambda\sigma}^\rho+\ \Gamma_{\mu\nu}^\lambda C_{\ \ \ \lambda\sigma}^\rho$ into that we get$${\widetilde{R}}_{\ \ \ \sigma\mu\nu}^\rho=R_{\ \ \ \sigma\mu\nu}^\rho+\nabla_\mu C_{\ \ \ \nu\sigma}^\rho-\nabla_\nu C_{\ \ \ \mu\sigma}^\rho+C_{\ \ \ \mu\lambda}^\rho C_{\ \ \ \nu\sigma}^\lambda-C_{\ \ \ \nu\lambda}^\rho C_{\ \ \ \mu\sigma}^\lambda$$as required. It looks neater in the pdf!