The quantity $\partial_\mu \epsilon_\nu +\partial_\nu\epsilon_\mu$ is just the variation of the metric $g_{\mu\nu}$ under the (infinitesimal) diffeomorphism you wrote, $x^\mu\to x^\mu+\epsilon^\mu$. Your equation just says
$$\partial_\mu \epsilon_\nu +\partial_\nu\epsilon_\mu = C \delta_{\mu\nu}$$
for some $C$ which means the condition that the variation of the metric under the diffeomorphism is proportional to the flat background metric (in conformal gauge) itself so it can be compensated by a Weyl scaling of the metric by some $\Omega(x)$. The equation above is indeed equivalent to yours because one may calculate $C$ (related to $\Omega-1$ which is infinitesimally small, just like $\epsilon$). Just trace my equation above over $\mu=\nu$ and you get $2\partial\cdot \epsilon$ on the left hand side and $Cd$ on the right hand side which implies $C=2(\partial\cdot\epsilon)/d$, just like your equation says.

The transformation rule for the metric, $\delta g_{\mu\nu} = \partial_\mu \epsilon_\nu +\partial_\nu\epsilon_\mu$, may be computed from your $g\to g'$ rule. Just Taylor-expand your formula $g'=()()g$ with respect to $\epsilon$ up to the linear terms in $\epsilon$. Use the Leibniz rule – which will produce two terms in the variation, one from the first $()$ and one from the second, and the fact that that $\partial x^{\prime\mu}/\partial x^\nu = \delta^\mu_\nu + \partial_\nu \epsilon^\mu $ which is just the derivative of $x'=x+\epsilon$ and which is how you get each term in the symmetrized sum.

This post imported from StackExchange Physics at 2014-05-01 12:16 (UCT), posted by SE-user Luboš Motl