Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This is in continuation of my previous question, IR divergence and renormalization scale in dimensional regularization.

Lubos gave a nice answer there but I want to get to a very specific example which is not found in usual QFT - (in QFT all examples I have seen are such that the number of momenta powers are the same in the denominator and the numerator).

Consider the 3 dimensional integrals in A.5 (page 19) of arXiv:1301.7182. I guess one can say that these are UV divergent when 3>2(ν1+ν2) but the gap can be arbitrarily large depending on the values of ν1 and ν2 - right? Then how do the authors justify using d=3+ϵ expansion get the UV divergence?

  • Also I guess these are IR divergent if 3<2ν1..right?

    How does one define the $\epsilon$ to get this IR divergence in the $\epsilon$-expansion?

  • Also by looking at this A.5 is it clear that $d=3$ is somehow the "critical dimension"? (...the terminology is from Lubos's comments in the previous question...)

share|improve this question
    
They already do calculus with $d=3$ in reference [6]. So the formulas $A5$ and others are only formulae with $d$ as variable .$v_1$ and $v_2$ are also variables in the general mathematical formula $A5$ (The pratical values are given in the details of A3 and A4 formulae). UV divergence is $d > 2(v_1 +v_2)$ and IR divergence is $d < 2(v_1 +v_2)$ –  Trimok Jun 22 '13 at 10:20
    
@Trimok Thanks for the help. Aren't they saying that the UV divergence of A.5 is seen by a d=3+epsilon expansion? Or to put it otherwise how do they know that 3+epsilon is the correct expansion to do to see the UV divergence in 3 dimensions? –  user6818 Jun 23 '13 at 21:11
    
@Trimok Also when the integrating momentum goes near 0, I would think that $\vert \vec{q} - \vec{k} \vert$ can be thought as just k and hence it doesn't affect the IR properties. So I would think that IR divergence happens when $3 < 2\nu_1$ - anything wrong with this argument? –  user6818 Jun 23 '13 at 21:13
    
Aargh! Of course, you are right for the IR divergence (with the exception $\vec k = \vec 0$). At the beginning of appendix $A$, it is said that "For the n's leading to a log-divergence in three-dimensions, we need to know how the loop integrals depend on d explicitely, in order to be able to take the limit $\epsilon \rightarrow 0$ where $\epsilon = d-3$ and and extract the 􀀀$\epsilon^{-1}1$ pole as well as the finite term $\epsilon^0$. So, a technical problem leads them to use a variable $d$, then replacing $d$ by $3$, at the end $A.9, A.10$ –  Trimok Jun 24 '13 at 7:21
    
The dimension 3 seems to come from the definition of $\Delta$ quantities from the $P$ quantities (formula $2.19,2.20$) (There is a $k^3$ in the formula). Then it seems that a explitit calculus of $\Delta$ (formula $3.1$) gives the rules for divergences : table $1$ page $10$. But I suppose, that one have to read and understand the whole paper, which seems not easy... –  Trimok Jun 24 '13 at 7:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.