# Conformal transformation equation

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I am currently reading Kiritsis's string theory book, and something bugs in the CFT (fourth) chapter. He derives the equation that should satisfy an infinitesimal conformal transformation $x^{\mu} \rightarrow x^{\mu} + \epsilon^{\mu}(x)$ which is $\partial_{\mu}\epsilon_{\nu}+\partial_{\nu}\epsilon_{\mu} = \frac{2}{d}(\partial . \epsilon)\delta_{\mu\nu}$ from $g_{\mu\nu}(x)\rightarrow g'^{\mu\nu}(x')=\Omega(x)g_{\mu\nu}(x)=\frac{\partial x^{\alpha}}{\partial x'^{\mu}}\frac{\partial x^{\beta}}{\partial x'^{\nu}}g_{\alpha\beta}(x)$. I have been trying to do that with no success, is there an additional hidden hypothesis I'm missing, or am I just bad with math?

This post imported from StackExchange Physics at 2014-05-01 12:16 (UCT), posted by SE-user toot
What have you tried? It's hard to know what you might be missing without seeing at least an outline of your work.

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

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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
answered Jan 26, 2012 by (10,278 points)
Ha ha, this will probably help me too somewhere along the way (if I`m able to mend some math weak spots in a finite amount of time such that I can return to this stuff) :-)

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

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