# What is a “reggized gluon”?

I'm reading a paper in which the author used these words many times assuming that the reader knows what he is talking about. Can someone please explain what it is? What is the difference between a regular gluon and a reggized gluon?

• seems to me it has to do with Regge rajectories :en.wikipedia.org/wiki/Regge_theory . This link seems to be using the concept indiana.edu/~ntceic/Talks/Szczepaniak.pdf . – anna v Sep 29 '13 at 5:04
• Assuming you're talking about the Caron-Huot paper from 2 days ago, it's actually quite comprehensive, even for people with no serious background in the BKFL business. Reggeization is explained in section 2.2. You do need some QCD background, no what Wilson lines are etc. – Vibert Sep 29 '13 at 9:56
• Supplementing the paper itself referred to by Vibert: a good set of slides on this topic is available here. For an introduction to Regge theory in general The Analytic S Matrix is a readable source. – twistor59 Sep 29 '13 at 11:20

After doing some research, I think this figure, from a new book by Kovchegov and Levin, illustrates it well: Fig. 3.10. Reggeized gluon (bold corkscrew line) represented as the sum of all leading-$\ln s$ corrections to the single-gluon exchange amplitude for $qq\to qq$ scattering.

So essentially, a reggeized gluon is an "effective particle" that is actually a quantum superposition of one gluon, two gluons, three gluons, etc. but only counting the terms in the amplitude that are leading logarithmic - that is, terms proportional to $\alpha_s(\alpha_s\ln s)^n$, not $\alpha_s^2(\alpha_s\ln s)^n$ or so on.

Mathematically, here's what's going on. The amplitude for single-gluon exchange looks like this:

$$(\text{stuff})\frac{s}{k_\perp^2}$$

The amplitude for two-gluon exchange, taking into account both the straight and crossed diagrams, looks like this:

$$-(\text{stuff})\frac{s}{k_{\perp}^2}\frac{\alpha_s N_c}{4\pi^2}\ln s\int\mathrm{d}^2\mathbf{k}_{2\perp}\frac{k_{\perp}^2}{k_{1\perp}^2 k_{2\perp}^2} \equiv (\text{stuff})\frac{s}{k_{\perp}^2}\omega_G(k_{\perp})\ln s$$

where $\mathbf{k}$ (total momentum transfer) is $\mathbf{k}_1$ (momentum of one gluon) plus $\mathbf{k}_2$ (momentum of other gluon), and $(\text{stuff})$ is the same constant in both cases. If you go on to calculate the amplitude for $n$-gluon exchange for higher $n$, keeping only the leading logarithmic terms, you find a series of the form

$$\begin{multline} (\text{stuff})\frac{s}{k_{\perp}^2} + (\text{stuff})\frac{s}{k_{\perp}^2}\omega_G(k_{\perp})\ln s + \frac{1}{2}(\text{stuff})\frac{s}{k_{\perp}^2}\biggl(\omega_G(k_{\perp})\ln s\biggr)^2 + \cdots \\ = (\text{stuff})\frac{s}{k_{\perp}^2}e^{\omega_G(k_{\perp})\ln s} \end{multline}$$

(at least, this is what the book says; I haven't done the calculation myself). So this entire series of terms describing $n$-gluon exchanges condenses to a single term $(\text{stuff})s^{\omega_G(k_{\perp})+1}/k_\perp^2$. You can interpret this as representing the exchange of a single "effective particle" which has a propagator $ig_{\mu\nu}s^{\omega_G(k_{\perp})+1}/k_\perp^2$ (compare to a single gluon which has a propagator of just $ig_{\mu\nu}/k_\perp^2$). That effective particle is the reggeized gluon.

• +1 for helping me learn something. In the total interaction the reggeized gluon is only a part , so is the assumption that it is the dominant one, or the only one? as in the old Regge theory? – anna v Sep 30 '13 at 3:27
• @anna I don't think it's necessarily assumed that reggeized gluon exchange is always the dominant part of the amplitude, but for the kinds of processes that motivated the development of the concept (i.e. high energy hadron scattering), it is dominant. Of course it's not the only contribution but it's assumed that others can be neglected at the relevant order in perturbation theory. As far as I can tell, anyway. – David Z Sep 30 '13 at 4:16