# How (why!?) does one introduce an UV cut-off in dimensional regularization?

This question is in reference to the confusing equation 3.7 (page 14) of this paper.

• One sees the 1-loop answers in their theory as given in their A.7 and A.8 on page 20. Each of the terms is a product of Gamma functions multiplied by $k^{2n+d}$. But how does this become their 3.7? (where an UV cut-off of $k_{NL}$ has been introduced)

• Now such a combination of Gamma functions as in 3.7 can easily produce double or higher order poles. Then in what sense does their 3.9 follow from 3.7?

(..also what is the physics behind writing $(\frac{k}{k_{NL}})^{\epsilon}$ as $\sim 1 + \epsilon ln (\frac{k}{k_{NL}} )$?..)

And even if the Gamma function combination in 3.7 had a simple pole and the above rewriting were to be done then still a $1/\epsilon$ term should have been there in 3.9. Why is that not there?

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For your $\epsilon$ problem, it' simply mathematics, when $\epsilon \rightarrow 0$ : $(\frac{k}{k_{NL}})^{\epsilon} = e^{\epsilon ln(\frac{k}{k_{NL}})} \sim 1 + \epsilon ln (\frac{k}{k_{NL}})$ – Trimok May 25 '13 at 10:36
@Trimok That is not my question! I asked for the physics of it - as in what is the point of doing this Taylor series expression? (by doing this one sees that the coeffieint of the 1/epsilon from the Gamma function part actually contributes to the finite part - and the coefficient of all the poles change after doing this..) – user6818 May 25 '13 at 23:15

The quantity $k_{NL}$ in their paper isn't a "cutoff scale"; it is a scale at which the nonlinearities (therefore NL) in some quantities become substantial.

There is no cutoff scale $\Lambda$ in dimensional regularization; it's one of the main features and virtues of the dimensional regularization. Instead, $\Lambda\to \infty$ is replaced by the $\epsilon = d-4\to 0$ limit (well, it's $d-3$ in this paper). These two limits ultimately play a very similar role, with the map $1/\epsilon\leftrightarrow \ln(\Lambda/\mu)$ where $\mu$ is an additional scale. The dimensional regularization requires to introduce a scale $\mu$ as well, for dimensional reasons, but this is not cutoff.

The amplitude (3.7) doesn't produce higher-order singularities. When several gamma functions in the numerator coincide, they're also canceled by some singularities in the denominator. This statement may be verified by looking at the arguments of the gamma functions, as detailed in (A.7) and (A.8): if one subtracts the arguments of the gamma functions in the numerator, one always gets an argument from the denoninator (up to an addition of an integer with the right sign). It's a general consistency condition in physics that good theories may only produce simple poles, not higher-order singularities.

The physics behind $$\left(\frac{k}{k_{NL}}\right)^\epsilon = 1+\epsilon\ln\left(\frac{k}{k_{NL}}\right)+O(\epsilon^2)$$ is that it is a mathematical identity, the leading two terms in a Taylor expansion in $\epsilon$, and it's always allowed to use mathematical identities – facts – in physics. They're not only allowed but, in this case, useful because $\epsilon$ is small in the physically interested limit. In such situations, to expand in $\epsilon$ is not only common sense but sort of an obligation. If you do not know how to get the formula above, note that the general power $x^y$ is more intelligently rewritten as $\exp(y\ln x)$.

The reasons why the "more divergent terms" in (3.9) were dropped is explained below the equation (3.9).

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You can try $n=-3$ and do the $d=3+\epsilon$ on their (A.7 + A.8) and you will get a double pole in epsilon. Also how does one introduce the $k_{NL}$ in 3.7 starting from the expression in A.7,A.8 - that is my question! I didn't ask for the proof of the Taylor series you wrote down - I wanted to know what is the "physics" of doing it? Why would one do this? – user6818 May 25 '13 at 23:11
I am confident that I have answered and explained the physical reason of expanding in $\epsilon$ - it's because $\epsilon$ is infinitesimal in the physical limit. Also, I don't believe you have actually calculated A.7 and A.8 to get a double pole. A double pole, if any, ultimately cancels. The power of $(k/k_{NL})$ is a dimensional analysis explained in and above 2.25 and in and above B.13. – Luboš Motl May 26 '13 at 18:16
Motl I took A.7 and A.8 and put it in Mathematica with the $d$ replaced by $3+epsilon$ and $n=-3$ and when asked to expand in "Series" it put out a double pole. (...also clearly the Gamma functions in the denominator don't contribute to the pole..) – user6818 May 26 '13 at 19:02
Motl Also A.7 and A.8 comes with a factor of $k^{2n+d}$ (which has dimensions) but in 3.7 they want to think of the momentum dependence as $(k/k_{NL})^{2n+d}$ which is dimensionless - how could both of these be the same expression!? – user6818 May 26 '13 at 19:05
Motl Should mention that the order 2 pole always comes from A.8 but not from A.7 - I don't know how to prove this but experimentally so - but of course its only the sum of A.7 and A.8 that is physically relevant.. – user6818 May 28 '13 at 0:02