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Sep 15, 2014 at 21:13 comment added Andrew McAddams @Orbifold : and please check my comment about $n$ factor.
Sep 15, 2014 at 21:13 comment added Andrew McAddams @Orbifold : here is the set of actions: $$ \tilde {\gamma}_{\Gamma}(\tilde{g}) - \gamma_{\Gamma} (g) - q(g)\beta (g)\partial_{g} \frac{1}{q(g)} = 0 \Rightarrow \tilde {\gamma}_{\Gamma}(\tilde{g}) = \gamma_{\Gamma}(g) + q(g)\beta (g)\partial_{g} \frac{1}{q(g)} = $$ $$ = \gamma_{\Gamma}(g) - \beta(g)q(g)\frac{q'(g)}{q^{2}(q)} = \gamma_{\Gamma}(g) - \beta (g)\partial_{g}ln (q(g)). $$
Sep 15, 2014 at 21:08 comment added Orbifold regarding the minus sign: thats fine! but when you evaluate for $\tilde{\gamma}$ will is not become $+$ when going on the other side of the equation?
Sep 15, 2014 at 21:08 comment added Andrew McAddams @Orbifold : as for the first one, it's only the redefinition of the definition of $Z$-constant (as you can see from OP definition, there is $Z^{-1}$, not $Z^{-n}$ at n-point function.
Sep 15, 2014 at 21:03 comment added Andrew McAddams @Orbifold : as for the second, the relative $-$ sign have arised from derivation of $\frac{1}{q(g)}$ function: $$ q(g)\partial_{g}\frac{1}{q(g)} = -q(g)\frac{q(g)'}{q^{2}(q)} = -\partial_{g}ln(q(g)). $$ As for the first, the OP definition
Sep 15, 2014 at 20:46 comment added Orbifold Also could you explain the relative '$-$' sign in $\gamma_{\Gamma}(g)-\beta(g)\partial_gln(q(g))$
Sep 15, 2014 at 20:41 comment added Orbifold The CS equation for an n-point function $\Gamma$ contains a factor of $n$ along with the anomalous dimension. How did you neglect that? It would result in a $1/n$ factor in the final answer!
Sep 15, 2014 at 19:41 vote accept CommunityBot
Sep 15, 2014 at 18:23 history edited Andrew McAddams CC BY-SA 3.0
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Sep 15, 2014 at 14:05 history answered Andrew McAddams CC BY-SA 3.0