Are Killing Vectors the Key to Solving Complex Equations?

[TimeFall]

See below. I screwed up the edit and the use of tex.

 

[TimeFall]

EDIT: Used proper tex (hopefully!)

Hello! I'm working through Weinberg's book Gravitation and Cosmology, and I'm currently in chapter 13, symmetric spaces. I'm trying to follow his derivation of the Killing condition, and I simply cannot, for the life of me, get from equation 13.1.2 to equation 13.1.4. I plugged 13.1.3 into 13.1.2 as he says to, but what I get is very different.
13.1.2: $$g_{\mu\nu} (x) = \frac{\partial x'^\rho}{\partial x^\mu} \frac{\partial x'^\sigma}{\partial x^\nu}g_{\rho\sigma} (x')$$
And 13.1.3: $$x'^\mu = x^\mu + \epsilon \zeta^\mu (x)$$

Then, only keep the result of the substitution to first order in epsilon. When I do this, I get:
$$g_{\mu\nu} (x) = \frac{\partial x^\rho}{\partial x^\mu} \frac{\partial x^\sigma}{\partial x^\nu} g_{\rho\sigma} (x') + \epsilon \left [ \frac{\partial \zeta^\sigma (x) }{\partial x^\nu } \frac{\partial x^\rho }{\partial x^\mu } g_{\rho\sigma} (x') + \frac{\partial \zeta^\rho (x)}{\partial x^\mu } \frac{\partial x^\sigma }{\partial x^\nu } g_{\rho\sigma} (x') \right ]$$.

It's supposed to be 13.1.4: $$0 = \frac{\partial \zeta^\mu (x)}{\partial x^\rho} g_{\mu\sigma}(x) + \frac{\partial \zeta^\nu (x)}{\partial x^\sigma} g_{\rho\nu} (x) + \zeta^\mu (x) \frac{\partial g_{\rho\sigma} (x)}{\partial x^\mu}$$

All of his metrics are functions of x, not x', and he has no epsilon in the equation. That makes it seem to me that the first term on the right hand side of the equation I got has to equal the left hand side, so that they cancel and equal 0. Then the epsilon can divide out. The problem is that then there are only two terms left, as opposed to the three that he has. I'm guessing it has something to do with switching from g(x) to g(x'), but I don't see it. Any help would be greatly, greatly appreciated! Thank you very much!

 

[WannabeNewton]

Your expression reduces to $g_{\mu\nu}(x) = g_{\mu\nu}(x) + \epsilon (\zeta^{\rho}\partial_{\rho}g_{\mu\nu}(x) + g_{\mu\sigma}(x)\partial_{\nu}\zeta^{\sigma} + g_{\rho \nu}\partial_{\mu}\zeta^{\rho})$ after using the fact that $g_{\rho\sigma}(x') = g_{\rho\sigma}(x) + \epsilon \zeta^{\gamma}\partial_{\gamma}g_{\rho\sigma}(x) + O(\epsilon^2)$ hence $\zeta^{\mu}\partial_{\mu}g_{\rho\sigma}(x) + g_{\rho\nu}(x)\partial_{\sigma}\zeta^{\nu} + g_{\mu\sigma}\partial_{\rho}\zeta^{\mu} = 0$ after appropriately relabeling the indices. Don't forget that $\partial_{\nu}x^{\mu} = \delta^{\mu}_{\nu}$.

 

[TimeFall]

Thank you very much! I totally forgot about expanding the metric, as well as the delta condition.