Weyl transformation of connection and curvature tensors

  • #1

Main Question or Discussion Point

Given a Weyl transformation of the metric ##g_{\mu\nu} \rightarrow g'_{\mu\nu} = e^{\Omega(x)} g_{\mu\nu}##, I'm trying to find the corresponding connection ##\Gamma'^{\lambda}_{\mu\nu}##, and from that ##-## via the Riemann tensor ##R'^{\lambda}_{\mu\nu\kappa}## ##-## the Ricci tensor ##R'_{\mu\kappa}##. For the connection, I end up with
$$\Gamma'^{\lambda}_{\mu\nu} = \Gamma^{\lambda}_{\mu\nu} + \frac{1}{2}\{\delta^\lambda_\nu \partial_\mu\Omega + \delta^\lambda_\mu \partial_\nu\Omega - g^{\sigma\lambda}g_{\nu\mu}\partial_\sigma\Omega \}$$
The Riemann tensor is then defined in terms of the connection as
$$R'^{\lambda}_{\mu\nu\kappa} = \partial_\kappa\Gamma'^{\lambda}_{\mu\nu} - \partial_\nu\Gamma'^{\lambda}_{\mu\kappa} + \Gamma'^{\rho}_{\mu\nu} \Gamma'^{\lambda}_{\kappa\rho} - \Gamma'^{\sigma}_{\mu\kappa} \Gamma'^{\lambda}_{\nu\sigma}$$
which gives us the Ricci tensor as
$$R'_{\mu\kappa} = R'^{\lambda}_{\mu\lambda\kappa} = \partial_\kappa\Gamma'^{\lambda}_{\mu\lambda} - \partial_\lambda\Gamma'^{\lambda}_{\mu\kappa} + \Gamma'^{\rho}_{\mu\lambda} \Gamma'^{\lambda}_{\kappa\rho} - \Gamma'^{\sigma}_{\mu\kappa} \Gamma'^{\lambda}_{\lambda\sigma}$$
which is reasonable to study term by term. For the first term, we get
$$\partial_\kappa\Gamma'^{\lambda}_{\mu\lambda} = \partial_\kappa\Gamma^{\lambda}_{\mu\lambda} + \frac{1}{2}\{\delta^\lambda_\lambda \partial_\kappa\partial_\mu\Omega + \delta^\lambda_\mu \partial_\kappa\partial_\lambda\Omega - g^{\sigma\lambda}g_{\lambda\mu}\partial_\kappa\partial_\sigma\Omega \} = \partial_\kappa\Gamma^{\lambda}_{\mu\lambda} + \frac{1}{2} D \partial_\kappa\partial_\mu\Omega$$
where ##D = \delta^\lambda_\lambda## is the dimensionality. For the second term of the Ricci tensor, we get
$$\partial_\lambda\Gamma'^{\lambda}_{\mu\kappa} = \partial_\lambda\Gamma^{\lambda}_{\mu\kappa} + \frac{1}{2}\{\delta^\lambda_\kappa \partial_\lambda\partial_\mu\Omega + \delta^\lambda_\mu \partial_\lambda\partial_\kappa\Omega - \partial_\lambda (g^{\sigma\lambda}g_{\kappa\mu}\partial_\sigma\Omega) \} = \\
= \partial_\lambda\Gamma^{\lambda}_{\mu\kappa} + \frac{1}{2}\{2\partial_\kappa\partial_\mu\Omega - \partial_\lambda (g^{\sigma\lambda}g_{\kappa\mu})\partial_\sigma\Omega - g^{\sigma\lambda}g_{\kappa\mu} \partial_\lambda\partial_\sigma\Omega \}$$
For the third term of the Ricci tensor, we get
$$\Gamma'^{\rho}_{\mu\lambda} \Gamma'^{\lambda}_{\kappa\rho} = \Gamma^{\rho}_{\mu\lambda} \Gamma^{\lambda}_{\kappa\rho} + \frac{1}{4} \{ (D + 2) - 2 g^{\sigma\rho} g_{\kappa\mu}\partial_\sigma\Omega \partial_\rho\Omega \} + \frac{1}{4}g^{\sigma\lambda} (\partial_\mu g_{\lambda\sigma} \partial_\kappa\Omega + \partial_\kappa g_{\lambda\sigma} \partial_\mu\Omega) + \frac{1}{2}g^{\tau\rho}(\partial_\mu g_{\kappa\tau} + \partial_\kappa g_{\mu\tau})\partial_\rho\Omega - g^{\tau\rho}\partial_\tau g_{\kappa\mu} \partial_\rho\Omega $$
And for the fourth and last term, we get
$$\Gamma'^{\sigma}_{\mu\kappa} \Gamma'^{\lambda}_{\lambda\sigma} = \Gamma^{\sigma}_{\mu\kappa} \Gamma^{\lambda}_{\lambda\sigma} + \frac{1}{4} \{ 2D \partial_\mu\Omega \partial_\kappa\Omega + g^{\rho\sigma} g_{\kappa\mu} \partial_\rho\Omega \partial_\sigma\Omega \} + \frac{D}{4} \{ g^{\rho\sigma} \partial_\mu g_{\kappa\rho} \partial_\sigma\Omega + g^{\rho\sigma} \partial_\kappa g_{\mu\rho} \partial_\sigma\Omega - g^{\rho\sigma} \partial_\rho g_{\kappa\mu} \partial_\sigma\Omega \} + \frac{1}{4} \{ \partial_\mu\Omega g^{\tau\lambda} \partial_\kappa g_{\lambda\tau} + \partial_\kappa\Omega g^{\tau\lambda} \partial_\mu g_{\lambda\tau} - g^{\rho\sigma}g_{\kappa\mu}\partial_\rho\Omega g^{\tau\lambda}\partial_\sigma g_{\lambda\tau} \}$$
Putting it all together, I get the total Ricci tensor as
$$R'_{\mu\kappa} = R_{\mu\kappa} + \frac{1}{2}(D+2)\nabla_\mu \partial_\kappa \Omega - \frac{3}{2}\partial^\sigma g_{\kappa\mu} \partial_\sigma\Omega - \frac{1}{2}g_{\kappa\mu}\square\Omega + \frac{1}{4}(D+2) + \frac{1}{4}g_{\kappa\mu}\partial_\sigma\Omega\partial^\sigma\Omega + \frac{1}{2}g^{\sigma\lambda}\partial_\mu g_{\lambda\sigma}\partial_\kappa\Omega + \frac{1}{2}g^{\sigma\lambda}\partial_\kappa g_{\lambda\sigma}\partial_\mu\Omega - \frac{1}{4}g_{\kappa\mu}\partial^\sigma\Omega g^{\tau\lambda}\partial_\sigma g_{\lambda\tau} + \frac{1}{2}D\partial_\mu\Omega\partial_\kappa\Omega $$
where ##\nabla_\mu## is the covariant derivative, and ##\square = \partial^\sigma\partial_\sigma## is the d'Alembert operator. This seems to be wrong; as far as I can tell, from the formula for the Ricci tensor on https://en.wikipedia.org/wiki/Weyl_transformation, it should rather be something like
$$R'_{\mu\kappa} = R_{\mu\kappa} + \frac{1}{2}(2-D)\nabla_\mu \partial_\kappa \Omega - \frac{1}{2}g_{\kappa\mu}\square\Omega + \frac{1}{4} (D-2) \partial_\mu\Omega\partial_\kappa\Omega - \frac{1}{4} (D-2) g_{\mu\kappa} \partial_\sigma\Omega\partial^\sigma\Omega $$
with the metric in question.

Any suggestions on where I might have gone wrong in the above ##-## I feel a bit stuck for the moment. Thanks a lot in advance.
 

Answers and Replies

  • #2
34
18
Your Christoffel symbols do not account for the fact that ##g_{ab} \to e^{\Omega} g_{ab}## implies ##g^{ab} \to e^{-\Omega} g^{ab}##, so the Christoffel symbols (compare to those in the wiki link), and especially the derivatives of the Christoffel symbols, are off. Taking the conformal factor to be ##\Omega^2## makes things look simpler (or as ##e^{2\Omega}## even simpler), and the results in this form are in Zee's Gravity Ch. VI.I.
 
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Likes Rabindranath and dextercioby
  • #3
Your Christoffel symbols do not account for the fact that ##g_{ab} \to e^{\Omega} g_{ab}## implies ##g^{ab} \to e^{-\Omega} g^{ab}##
Thanks for your reply. I do believe, however, that I do account for the fact (correct me if I'm wrong). From the definition of the Christoffel symbol ##\Gamma^\lambda_{\mu\nu} = \frac{1}{2} g^{\sigma\lambda} \{ \partial_\mu g_{\nu\sigma} + \partial_\nu g_{\mu\sigma} - \partial_\sigma g_{\nu\mu} \}##, we get, with the transformation ##g_{\mu\nu} \to e^{\Omega} g_{\mu\nu}## ,
$$\Gamma^\lambda_{\mu\nu} \rightarrow \Gamma'^\lambda_{\mu\nu} = \frac{1}{2} e^{-\Omega} g^{\sigma\lambda} \{ \partial_\mu (e^\Omega g_{\nu\sigma}) + \partial_\nu (e^\Omega g_{\mu\sigma}) - \partial_\sigma (e^\Omega g_{\nu\mu}) \} \\ = \frac{1}{2} e^{-\Omega} g^{\sigma\lambda} \{ e^\Omega \partial_\mu g_{\nu\sigma} + e^\Omega \partial_\nu g_{\mu\sigma} - e^\Omega \partial_\sigma g_{\nu\mu} \} + \frac{1}{2} e^{-\Omega} g^{\sigma\lambda} \{ e^\Omega g_{\nu\sigma} \partial_\mu \Omega + e^\Omega g_{\mu\sigma}\partial_\nu \Omega - e^\Omega g_{\nu\mu}\partial_\sigma \Omega \} \\ = \Gamma^\lambda_{\mu\nu} + \frac{1}{2} \{ \delta^\lambda_\nu \partial_\mu \Omega + \delta^\lambda_\mu \partial_\nu \Omega - g_{\nu\mu}\partial^\lambda \Omega \} $$
which is the equation in the top of my original post, and indeed equivalent to the one at https://en.wikipedia.org/wiki/Weyl_transformation (since, in my case, ##f=f'=e^{\Omega(x)}##). Do you agree?

As for the derivatives of the Christoffel symbols, we then get (for the first term in the Riemann and Ricci tensors, respectively)
$$ \partial_\kappa \Gamma'^\lambda_{\mu\nu} = \partial_\kappa \Gamma^\lambda_{\mu\nu} + \frac{1}{2} \{ \delta^\lambda_\nu \partial_\kappa\partial_\mu \Omega + \delta^\lambda_\mu \partial_\kappa\partial_\nu \Omega - \partial_\kappa (g_{\nu\mu}\partial^\lambda \Omega) \} \text{ (Riemann)} \\ \Rightarrow \partial_\kappa \Gamma'^\lambda_{\mu\lambda} = \partial_\kappa \Gamma^\lambda_{\mu\lambda} + \frac{1}{2} \{ \delta^\lambda_\lambda \partial_\kappa\partial_\mu \Omega + \delta^\lambda_\mu \partial_\kappa\partial_\lambda \Omega - \partial_\kappa (g_{\lambda\mu}\partial^\lambda \Omega) \} \\ = \partial_\kappa \Gamma^\lambda_{\mu\lambda} + \frac{1}{2} D \partial_\kappa\partial_\mu \Omega \text{ (Ricci)} $$
Do I miss something here?
 
  • #4
34
18
I believe you are right, in exponential form those factors from the Christoffels cancel so unfortunately I'm not sure where you're going wrong without basically re-doing it in this notation.

I have a write-up of the conformal transformation of the Ricci scalar in Zee's notation attached in a pdf with all the gory details and it arrives at the correct result as given in Zee - the Ricci tensor is in most of the terms until I contract ##g^{bd}## so it would be very easy to read off most of the calculation and simplify the Ricci tensor to give the result as written in Zee's notation (included), and then compare to the notation of the wiki to figure out what's going wrong.
 

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  • #5
34
18
Peeling ##\Omega^{-2} g^{bd}## off ##I## and ##V## easily gave me the ##\Omega^{-2}## terms in Zee's Ricci scalar, the other three should give the ##\Omega^{-1}## part which I can check another day but there should be no issue.
 

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