The books are correct. The statement is a definite relation that is being imposed between the 'old' metric structure and the transformed one, for the transformation to be conformal.

The equation you're unhappy about,
$$
g_{\mu \nu}'(x') = \Omega(x) g_{\mu \nu}(x)
$$
states that the transformed metric $g_{\mu \nu}'$ at the transformed point $x'$ can be obtained from the old metric $g_{\mu \nu}$ evaluated at the untransformed point $x$ by simply multiplying it, as a matrix by a scalar quantity $\Omega$ which is a function of the spacetime point in question.

This is distinct from the general change-of-coordinates transformation of the metric, which involves multiplying from left and right with the transformation jacobian: the inner structure of the metric is not being meddled with, which keeps the angles constant.

The old metric $g_{\mu \nu}$ is naturally a function of the old position, $x$. Because $x'$ is a function of $x$ and viceversa, writing $g_{\mu \nu}(x')$ is not formally incorrect, but *it is misleading and should be avoided*.

The scalar factor $\Omega$ is a function of the spacetime point in question and can be written as an explicit function of either the old or the transformed coordinates, as long as one is consistent. One would usually, though, keep the $x$s to the right of the equal sign, and the $x'$s to the left.

The point is that $x'$ is a function of $x$, and vice versa, and of course only one of the two is ever independent. Thus, you could write the equation as $g_{\mu \nu}'(x') = \Omega(x(x')) g_{\mu \nu}(x(x')) $, where $x'$ is the only independent variable, or $g_{\mu \nu}'(x'(x)) = \Omega(x) g_{\mu \nu}(x) $, where $x$ is independent. Both variants are quite ugly. Thus we usually write in a simpler notation with primed quantities on one side related to unprimed ones on the other, in the understanding that only one set of the two is ever independent but that they are, of course, completely equivalent. To a symmetric situation we respond, where possible, with symmetric notation.

Regarding your edit, I can't really say where you're getting tripped up, particularly as your definition of the Lie derivative is rather at odds with the more standard, more abstract ones. For an infinitesimal transformation that's "anchored" at the old point $x$, I would strongly urge you to *not* work in the transformed plane. Bring transformed quantities back to the untransformed plane and work with them there, instead of the other way. If you're careful, both ways should be equivalent, but it's a lot easier to mess up in the transformed plane, because it's "moving".

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Emilio Pisanty