QCD: Incoming Particle Momenta, Factorization & Renormalization Scales

[petergreat]

I encountered a paper in which the authors presented parton-level cross sections as a function of these variables: incoming particle momenta, factorization scale, renormalization scale, and strong coupling constant at the renormalization scale. I used to think that QCD factorization scale should always be set to equal the renormalization scale.

When they are not equal, how does such a calculation proceed?

(Of course, the result is integrated with the parton distribution function at the factorization scale to give the final answer.)

 

[petergreat]

I suppose the renormalization scale will serve as UV cutoff, while the factorization scale will serve as IR cutoff for the Feynman diagrams?

 

[humanino]

Factorization scale is not the same as renormalization scale.

Factorization scale is one and the same, unique for a given process. It is the scale at which the lower twist dominates over the higher twists. When I say "dominates" it of course depends on some conventions, which are in fact related to the precision of the experiment. If one has 2% errors bars, one can say that the lower twist dominates at 2%. But it may not be valid anymore when one gets more statistics for instance. So when I say "unique" I mean to say that people usually agree on percent level being the threshold of precision physics.

A given process can be probed at different experimental scales. This sets the values one uses for the renormalization scales.

Take the simplest of unpolarized deep inelastic lepton scattering on a proton target. The lower twist starts dominating at the percent level (or so) around 3 GeV. This is the factorization scales. Nothing prevents one from making measurements at higher energies. As the energy increases, the quark starts radiating gluons and are depleted from the high xB, while the gluon distribution grows. These scaling violations are part of the renormalization, or DGLAP evolution, or dependence of the distribution on the probing scale.

 

[petergreat]

Let me make my question more precise. I encountered a formula of this form
$$\sigma(Q^2)=\sum_{a,b} \int_0^1 \frac {dx_1}{x_1} \frac {dx_2}{x_2} f_a (x_1,\mu_F^2) f_b(x_2,\mu_F^2) G_{ab}(x_1 x_2; \alpha_S(\mu_R^2), Q^2/\mu_R^2; Q^2/\mu_F^2),$$
where $Q^2$ is the experimental energy scale, $\mu_F$ is the factorization scale, $\mu_R$ is the renormalization scale.
Later in the paper the hard scattering function $G_{ab}$ is presented as an analytical expression of the 4 variables (in the bracket at the end of the formula) to NLO accuracy.
I've never previously encountered something like this. A single scale $\mu$ was always used for both the parton distribution function and strong coupling constant.
How does one perform QCD NLO calculation with the two scales set to be different?

 

[humanino]

Can you provide a reference for the paper ? The formula indeed does not look consistent with what I was saying above.

 

[petergreat]

Can you provide a reference for the paper ? The formula indeed does not look consistent with what I was saying above.

It's this paper about Higgs cross section. http://arxiv.org/abs/hep-ph/0306211 The formula is Eqn. (1) on Page 2. (My posting above has changed it into a generic form, but the spirit is the same.) On Page 4 the analytical NLO expressions are given.

 

[humanino]

I still did not have time to come back to this discussion and properly inform myself on the topic of your paper. I can only re-state my understanding and hopefully somebody more knowledgeable will step in.

The renormalization scale is as usual fixed by the actual flow of momentum in the perturbative (hard) part of the graphs. The factorization scale is related to the definition of "small" when we perform the operator product expansion : we want that the next-to-leading twist be "small" compared to the dominant twist. Since the dominance sets in at a given scale once we have picked our definition of "small", this fixes the factorization scale. We can keep doing higher energy experiments with the same definition of small NLO corrections, and to relate the different experiments we still need evolution with the renormalization scale. So those two scales have no reason to be equal in general. They deal with different singularities.

There are most certainly infrared divergencies in the structure functions which are taken care of with the factorization scale. There are also other collinear singularities in the G_{ab} above beyond LO.

 

[petergreat]

I think I more or less understand what the authors are doing now.

The renormalization scale $\mu_R$ is used for dimensional regularization of UV divergence, and produces the factor $(4\pi \mu_R^2/Q^2)^\epsilon$.
On the other hand, the factorization scale $\mu_F$ (which is probably used in a different sense than as you explained) comes into play when we regulate the IR collinear singularity. We subtract $\frac 1 \epsilon P_{ab} (4\pi \mu_F^2/Q^2)^\epsilon\times (subleading\ cross\ section)$ from the cross section calculated from massless partons, where $P_{ab}$ is the Altarelli-Parisi kernel, to obtain the IR finite parton cross section.

For the above procedure to be valid, we need to use strong coupling constant at $\mu_R$, and use parton distribution table for the scale $\mu_F$.

The two scales are often set equal to simplify the result, but they don't have to.

Does what I wrote above sound legitimate?

 

[blechman]

My understanding of the factorization scale vs renormalization scale is:

1. Renormalization scale: this is the usual scale you get in DimReg. This scale can be used to derive an RG evolution equation and thus can be used to resum large logarithms.

2. Factorization scale: this is the scale at which you no longer trust perturbation theory. You know you have terrible soft and collinear divergences, and these divergences can be absorbed into IR-unsafe quantities such as parton distrib'n functions. $\mu_F$ is the scale where you declare these soft divergences to take over. Above that scale you are perturbative and can rely on Feynman Diagram calculations, while below that scale you have an ugly matrix element of a (nonlocal) operator, which you cannot calculate but is "universal" - that is, the same for many different hard processes (such as a pdf).

The renormalization scale is arbitrary, but you want to choose it wisely! You always choose this scale so that large logarithms vanish (more precisely, are resummed into running coupling constants, etc). That is why you evaluate $\alpha_s$ at the renormalization scale.

The factorization scale is NOT arbitrary! That is set by the kinematics of your problem. As humanino said: it's the scale at which higher twist terms matter. In other words, it's the place where you don't know how to calculate anymore!

There's more to choosing $\mu_R=\mu_F$ than just "simplifying the result": you are avoiding dangerous logarithms of the ratio of these scales, which might destroy Perturbation theory. As long as there are no other scales, there is no problem. The tricky part is when there are SEVERAL scales in your problems (collinear divergences). Then it is by no means obvious what to chose for $\mu_R$. This is what "Soft Collinear Effective Theory" is all about.

Hope that helps!