# Index Raising in Linearized General Relavitiy

1. Apr 17, 2012

### alex3

I'm reading a few textbooks (Straumann, Schutz, Hartle) on GR and am a little confused working through a small part of each on linearized GR.

1. Relevant equations

Using Straumann, the Ricci tensor is given by

$$R_{\mu\nu} = \partial_{\lambda} \Gamma^{\lambda}_{\phantom{k}\nu\mu} - \partial_{\nu} \Gamma^{\lambda}_{\phantom{k}\lambda\mu}$$

with the Christoffel symbols given by

$$\Gamma^{\alpha}_{\phantom{k}\mu\nu} = \frac{1}{2}\eta^{\alpha\beta} ( h_{\mu\beta,\nu} + h_{\beta\nu,\mu} - h_{\mu\nu,\beta} )$$

2. The problem

My problem is that the book is confusing me on the next equality. This what I expected when applying the flat metric:

$$\Gamma^{\alpha}_{\phantom{k}\mu\nu} = \frac{1}{2} ( \eta^{\alpha\beta}h_{\mu\beta,\nu} + \eta^{\alpha\beta}h_{\beta\nu,\mu} - \eta^{\alpha\beta}h_{\mu\nu,\beta} ) \\ \Gamma^{\alpha}_{\phantom{k}\mu\nu} = \frac{1}{2} ( h_{\mu\phantom{\alpha},\nu}^{\phantom{k}\alpha} + h^{\alpha}_{\phantom{\alpha}\nu,\mu} - h_{\mu\nu}^{\phantom{\mu\nu},\alpha} )$$

i.e. the flat metric raises all $\beta$'s to $\alpha$'s.

However, the book gets this

$$\Gamma^{\alpha}_{\phantom{k}\mu\nu} = \frac{1}{2} ( h^{\alpha}_{\phantom{\alpha}\mu,\nu} + h^{\alpha}_{\phantom{\alpha}\nu,\mu} - h_{\mu\nu}^{\phantom{\mu\nu},\alpha} )$$

So, the problem is in the first term: how come the book is able to swap the $\alpha$ and $\mu$ like that?

2. Apr 17, 2012

### George Jones

Staff Emeritus
Because $h_{\mu \beta , \nu}$ is symmetric in $\mu$ and $\beta$, $h_{\mu\phantom{\alpha},\nu}^{\phantom{k}\alpha} = h^{\alpha}_{\phantom{\alpha}\mu,\nu}$.

$$\eta^{\alpha\beta} h_{\mu\beta,\nu} = \eta^{\alpha \beta} h_{\beta \mu , \nu}$$

3. Apr 17, 2012

### alex3

How do we know that $h_{\alpha\beta}$ is symmetric? I can't see it mentioned anywhere. The only condition I see is $\lvert h_{\alpha\beta}\rvert \ll 1$.

4. Apr 17, 2012

### George Jones

Staff Emeritus
$h_{\alpha\beta} = g_{\alpha\beta} - \eta_{\alpha\beta}$, and $g$ and $\eta$ are both symmetric.

5. Apr 17, 2012

### alex3

Why do we assume $g_{\alpha\beta}$ is symmetric then? Is that a property we assume of all metrics? I didn't think we did. Do we assume symmetry of $g_{\alpha\beta}$ as it deviates only slightly from the Minkowski metric?

Last edited: Apr 17, 2012
6. Apr 17, 2012

### George Jones

Staff Emeritus
In standard general relativity, yes.

No, a symmetric $g$ can differ substantially from the Minkowski metric.

If the metric weren't symmetric, then it would not always have a tangent space isomorphic to Minkowski spacetime. If a metric tensor field is not symmetric, then there exists at least one point (event) at which the metric tensor for the tangent space is not symmetric.

7. Apr 17, 2012

### alex3

Got it now, thank you very much!