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1

Since you found the critical point via numerical simulations, you probably have little analytical insight into its properties. This makes it hard to extract the central charge, because it often appears in expressions combined with speed of sound or other quantities (e.g. in the free energy for a 2d CFT). So you need to find a universal quantity, easily ...

-2

Well, theoretical suggestions are as many;  but lab techniques suggest for scaled electroscopes to do as such. There also are technical issues Measuring the q charge - contacting method 1) Sure one has to contact the "point" charge with the top head of the electroscope. 2) The scale has to agree with the physical systems of measurements, SI or else. ...

2

There are a few different ingredients going into this: Firstly, the (holomorphic) current generating the infinitesimal conformal transformation $\delta z=\epsilon v(z)$ is $j(z)=i v(z) T(z)$ (there's a similar antiholomorphic one too). The general Ward identity (at $z=0$) for this current gives $$\frac{1}{i\epsilon}\delta\mathcal A(0,0)=Res_{z\rightarrow ... 1 1) First, looking at (2.3.4), you see that j^a is the coefficient of \partial_a\rho. An application of this (2.3.12), (2.3.13). To make connection with this formalism, it is preferable to choose the variations : X^\mu\rightarrow X^\mu-\epsilon \rho(\sigma) v^c\partial_c X^\mu From this, we get : \partial_a X^\mu\rightarrow ... 1 The action is$$ S = \frac{1}{2\pi \alpha'} \int d^2 \sigma \sqrt{\gamma} \gamma^{ab} \partial_a X^\mu \partial_b X_\mu $$The definition of the stress tensor from GR is$$ T_{ab} = \lambda\frac{4\pi}{\sqrt{\gamma}} \frac{ \delta S}{ \delta \gamma^{ab}}  Usually $\lambda = 1$, but different books use different conventions. I do not remember what convention ...

2

Note that it's the residue of the combination $j(z)\mathcal A(z_0,\bar{z_0})$ that we need: the poles come from the OPE where the operators come together. Now it could be that $j(z)\mathcal A(z_0,\bar{z_0})$ is more singular than $\frac{1}{z-z_0}$ at $z=z_0$. Only the simple poles contribute by the residue theorem. And if there is no simple pole (maybe ...

2

First of all, note that the radial operator ordering ${\cal R}$ is implicitly implied in many textbooks of CFT (e.g. Ref. 1). For instance, eq. (2.2.7) on p. 39 in Ref. 1 is discussing Wick's theorem between two operator ordering prescriptions. In this case between normal ordering $:~:$ and radial ordering ${\cal R}$. See also e.g. this Phys.SE post. The ...

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