# Tag Info

4

Imagine a QFT with some particle content. Some of these fields will be massless and some massive. For simplicity, consider a massles scalar field and a massive scalar field with mass $M$. If we are working at some energy $E\ll M$, we won't see the massive field (as happened with the Higgs before LHC, for example). This is the IR CFT. Why IR? Because we ...

4

This is a good question. I can only answer half of it: The eigenvalues of $M^{l}$ are some times complex. Take a look at the RG flow of Einstein gravity close to its UV fixed point. The two relevant couplings (newton's constant and cosmological constant) spiral around the fixed point as they move away from it. The real part of the critical exponents ...

2

Critical exponents are properties of the RG fixed point that drives the phase transition. They are computed by linearising the RG flow equations close to the fixed point. The exponents are the derivatives of the beta functions evaluated at the fixed point. They know nothing of the way you approach the fixed point. In particular if you are flowing slightly ...

1

Based on the exact result for a supersymmetric Yang-Mills $$\beta(\alpha_s)=-\frac{\alpha_s^2}{4\pi}\frac{3N}{1-\frac{N\alpha_s}{2\pi}}$$ these guys arXiv:0711.3745 postulated the following exact result for a non-supersymmetric Yang-Mills $$\beta(\alpha_s)=-\alpha_s^2\frac{11N}{12\pi}\frac{1}{1-\frac{17N}{11}\frac{\alpha_s}{2\pi}}$$ Here the ...

1

I think that this nomenclature has nothing to do with the regulators. "UV CFT" and "IR CFT" actually refers to the end point of the RG flow which is triggered by a relevant operator that perturbs an UV fixed point, and it ends (barring certain non-unitary QFT) into a fixed point at lower energy scales, hence the name IR.

Only top voted, non community-wiki answers of a minimum length are eligible