Weak field general relativity

1. Mar 7, 2006

Pietjuh

I was asked to do a problem about the gauge freedom one has in the weak field limit in general relativity. I am given a coordinate transformation $$x^{\mu '} = x^{\mu} - \zeta^{\mu}$$. Now I have to show how the perturbation of the metric transforms under this coordinate transformation.

The result should be:
$$h_{\mu\nu}^{new} = h_{\mu\nu}^{old} - \zeta_{\mu ,\nu} - \zeta_{\nu ,\mu}$$

The most simple way to do this, i thought, was to just use the transformational law for tensors.

so: $$h_{\mu ' \nu '} = \partial_{\mu '}x^{\mu} \partial_{\nu '}x^{\nu} h_{\mu \nu}$$

Using the fact that we can neglect the product of the derivatives of the small perturbation, I find that:

$$h_{\mu ' \nu '} = h_{\mu\nu} - (\partial_{\mu '}\zeta^{\mu} + \partial_{\nu '}\zeta^{\nu})h_{\mu\nu}$$

From this point I don't know how to get rid of the metric factor in the second term, because if I can lose it, i think i would be able to come to the correct form that is wanted.

Can someone give me some hints? :)

Last edited: Mar 7, 2006
2. Mar 8, 2006

dextercioby

It's more rigurous if you start with the Pauli-Fierz action and do the canonical analysis and determine the irreducible abelian infinitesimal gauge transformations for the P-F fields $h_{\mu\nu}$.

Just a thought.

Daniel.

3. Mar 8, 2006

Galileo

If you start from $g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$,
then the tranformation law gives:

$$g'_{\mu \nu}=\partial_{\mu} x^{\alpha}\partial_{\nu} x^{\beta}g_{\alpha \beta}$$
The left hand sides gives $g'_{\mu \nu}(x')\approx \eta_{\mu \nu}+h_{\mu \nu}'(x)$ and you can work out the right hand side explicitly.

Last edited: Mar 8, 2006