# A question about variation of Christoffel connection

1. Jan 2, 2012

### shichao116

Hi all,

I'm reading Sean Carroll's Space Time and Geometry and haven't figure out how equation 4.64 is derived, where he is in the process of deriving Einstein's equation from Hilbert action.

Given there is a variation of the metric,
$g_{\mu\nu} \rightarrow g_{\mu\nu} + \delta g_{\mu\nu}$,

The corresponding variation of Christoffel connection is:

$\delta\Gamma^{\sigma}_{\mu\nu} = -1/2[g_{\lambda\mu}\nabla_{\nu}(\delta g^{\lambda\sigma})+g_{\lambda\nu}\nabla_{\mu}( \delta g^{\lambda\sigma}) - g_{\mu\alpha}g_{\nu\beta}\nabla^{\sigma}(\delta g^{\alpha\beta})]$

The first thing I don't understand is where the covariant derivatives come from. Because the Christoffel connection is defined through partial derivative of metric.

Can anyone tell me how to derive this equation explicitly? Thanks very much.

2. Jan 2, 2012

### aichi

Derivation:
lemma:

$g_{\mu \nu;\lambda} = 0$.
--
$g_{\mu \nu; \lambda} = g_{\mu \nu, \lambda} - g_{\alpha \nu} \Gamma^\alpha_{\mu \lambda} - g_{\mu \alpha} \Gamma^\alpha_{\nu \lambda}$.
Take differential on both sides, we reach:

$0 = \delta g_{\mu \nu, \lambda} - \delta g_{\alpha \nu} \Gamma^\alpha_{\mu \lambda} - \delta g_{\mu \alpha} \Gamma^\alpha_{\nu \lambda} - g_{\alpha \nu} \delta \Gamma^\alpha_{\mu \lambda} - g_{\mu \alpha} \delta \Gamma^\alpha_{\nu \lambda}$​

Easy to recognize, the first 3 terms are merely the covariant derivatives. Thus,

$0 = (\delta g)_{\mu \nu; \lambda} - g_{\alpha \nu} \delta \Gamma^\alpha_{\mu \lambda} - g_{\mu \alpha} \delta \Gamma^\alpha_{\nu \lambda}$​

There are three of these identities by cycling indices:
$(\delta g)_{\mu \nu; \lambda} = g_{\alpha \nu} \delta \Gamma^\alpha_{\mu \lambda} + g_{\mu \alpha} \delta \Gamma^\alpha_{\nu \lambda}$
$(\delta g)_{\lambda \nu; \mu} = g_{\alpha \nu} \delta \Gamma^\alpha_{ \lambda \mu} + g_{\lambda \alpha} \delta \Gamma^\alpha_{\nu \mu}$
$(\delta g)_{\lambda \mu; \nu} = g_{\alpha \mu} \delta \Gamma^\alpha_{ \lambda \nu} + g_{\lambda \alpha} \delta \Gamma^\alpha_{\mu \nu}$​

Add the first two together, subtracted by the third one, you obtain the result. Signs may differ, due to different conventions.

The reason it "appears" that the partial derivatives are simply replaced by covariant derivatives are, the chain rule (Leibniz principle) gives one more term than the partial derivative thing. Use that term and the identity in the lemma, you can also get this result.

Last edited: Jan 2, 2012
3. Jan 2, 2012

### Bill_K

Or, just realize that δΓσμν is a genuine tensor, and so when you replace gμν,σ everywhere by ∇σgμν + Γ terms, that the Γ terms must cancel.