# Metric tensor, infitesimal transformation

1. Jan 31, 2008

### gouranja

Hi,

I don't think this belongs in the homework section since this is a graduate course.
My question is regarding making a field theory generally covariant by including a metric tensor $$g_{\mu\nu}(x)$$in the Lagransian density, and it's transformation under infinitesimal coordinate change.

If I transform the coordinates according to:
$$x^{\mu}\rightarrow x^{\mu}'=x^{\mu}+\epsilon^{\mu}(x)$$

The metric must be transformed according to:
$$g_{\mu\nu}(x)\rightarrow g'_{\mu'\nu'}(x')=\frac{\partial x^{\alpha}}{\partial x^{\mu}'}\frac{\partial x^{\beta}}{\partial x^{\nu}'}g_{\alpha\beta}(x)$$

which I understand well. But according to the textbook I'm reading the infitesimal result is:
$$\delta g_{\mu\nu}(x)=\epsilon_{\mu;\nu}+\epsilon_{\nu;\mu}$$
(with covariant derivatives).

I have tried using:
$$\frac{\partial x^{\alpha}}{\partial x^{\mu}'}\simeq\delta_{\mu}^{\alpha}-\frac{\partial\epsilon^{\alpha}}{\partial x^{\mu}}+O(\epsilon^{2})$$

to obtain:
$$\delta g_{\mu\nu}(x)=-g_{\mu\alpha}\partial_{\nu}\epsilon^{\alpha}-g_{\nu\alpha}\partial_{\mu}\epsilon^{\alpha}$$

but I don't know how to obtain the textbook result.
Can someone clue me in on how to do it?

Thanks

2. Jan 31, 2008

### gouranja

Never mind, I solved it...