# Infinitesimal coordinate transformation of the metric

## Homework Statement:

The metric transforms as ##g'_{\mu\nu}(x') = \frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^\nu} g_{\alpha\beta}(x)##. Show that under the infinitesimal transformation ##x^\alpha \rightarrow x'^\alpha = x^\alpha + \epsilon^\alpha## (##\epsilon^\alpha## is small), ##g'_{\mu\nu}(x^\mu + \epsilon^\mu) = g_{\mu\nu}(x) + (\partial_{\mu} \epsilon^\mu + \partial_{\nu} \epsilon^\nu) g_{\mu\nu}(x) = g_{\mu\nu}(x) + \partial_{\mu} \epsilon_\nu + \partial_{\nu} \epsilon_\mu##.

## Relevant Equations:

I kinda know how to do this problem, it is just that I hit a sign problem. If I take the partial derivative of the coordinate transformation with respect to ##x'^\mu##, I get

writing it first in the inverse form, ##x^\alpha = x'^\alpha - \epsilon^\alpha##

##\frac{\partial x^\alpha}{\partial x'^\mu} = \delta^\alpha_\mu - \partial'_\mu \epsilon^\alpha = \delta^\alpha_\mu - \frac{\partial x^\alpha}{\partial x'^\mu} \partial_\alpha \epsilon^\alpha##

transposing terms,

##\frac{\partial x^\alpha}{\partial x'^\mu} (1 + \partial_\alpha \epsilon^\alpha) = \delta^\alpha_\mu##

##\frac{\partial x^\alpha}{\partial x'^\mu} \approx \delta^\alpha_\mu - \delta^\alpha_\mu \partial_\alpha \epsilon^\alpha \quad## (to first order)

Now this is going to give me a minus sign instead of a plus sign in the right hand side of the equation to be shown. Did I do something wrong?