## 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. 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?

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?

I made that look nicer by putting two # characters before and after each bit of latex.

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!