# Conformal transformation equation

+ 6 like - 0 dislike
397 views

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

+ 6 like - 0 dislike

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 (10,248 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

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar\varnothing$sicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.