Undergrad Hadronization in electron-positron collisions

[kelly0303]

Hello! I am looking at the plot showing the ratio of cross sections of $e^-e^+$ to hadron, to $e^-e^+$ to $\mu^+\mu^-$. Doing a first order approximation the data is in pretty good agreement (an error of about 10%). However when the first order correction to the QCD is added, coming from $e^-e^+ \to q\bar{q}g$, the agreement is almost perfect. I am a bit confused about why, experimentally, one would use the $e^-e^+ \to q\bar{q}g$ for this kind of plot. Isn't the final product, in this case, made of 3 jets, compared to the case in which no gluon is produced in the final state? So can't one only use the events with only 2 jets in the final state and in that case the prediction from first order approximation will already be perfect compared to the experiment? (I don't know if one method is better than the other I am just wondering if that approach would be possible). Thank you!

 

[Reggid]

I am looking at the plot showing the ratio of cross sections of e−e+e^-e^+ to hadron, to e−e+e^-e^+ to μ+μ−\mu^+\mu^-.

I guess you are talking about the ratio of the total cross sections \sigma(e^+e^-\to \text{hadrons})/\sigma(e^+e^-\to \mu^+\mu^-) as a function of center of mass energy (the R ratio).

But I don't really understand your question.

First of all, just because you are considering the contributions from $q\bar{q}g$, this does not mean you have a three jet event. What you call a "jet" depends highly on how you define "jet", i.e. which algorithm you are using to cluster particles into jets. In general you will only get a third jet when you radiate a hard gluon. In case the emitted gluon is almost collinear to one of the quarks or very soft, it will usually not be clustered into a separate jet.

But this is completely irrelevant to your question (if I understand it correctly), because you are looking only at the total cross sections, which means you are insensitive to the different momentum configurations of the final state particles.

So including one higher order in perturbation theory (which means including $q\bar{q}g$ events as well as virtual corrections to $q\bar{q}$ in your calculation) will give you a higher precision in the cross section $\sigma(e^+e^-\to \text{hadrons})$. Why do you think you should not be able to compare this to the cross section $\sigma(e^+e^-\to \mu^+\mu^-)$ just as before? Or are we talking about different plots?

I also don't really understand the title of your question, because nowhere in the rest of it are you talking about hadronization effects.

 

[Dr.AbeNikIanEdL]

I am a bit confused about why, experimentally, one would use the e−e+→q¯qge−e+→qq¯ge^-e^+ \to q\bar{q}g for this kind of plot.

Experimentally you are not using any quarks or gluons. You essentially just count how often you see hadrons in your collisions. From the point of view of the (perturbative) theory calculation this corresponds to the production of an arbitrary number of quarks and gluons (which then somehow form the hadrons you actually observe). However, every additional gluon comes with a coupling factor, so if you restrict yourself to the lowest order your calculation is only #e^+e^- \rightarrow q\bar{q}#, and then the next order has an additional gluon and so on.

You might of course try to construct observables that are less sensitive to higher order corrections, but these will need their specific theory calculations that might have their own issues. As @Reggid pointed out, also a (somehow defined) two jet cross section will receive higher order corrections.

 

[vanhees71]

It's one of the very basic plots in physics. Among other things it admits you to pretty easily read off that quarks carry 3 color. You find the plot, as nearly anything on high-energy-particle physics, at the particle-data-group website:

http://pdg.lbl.gov/

For some explanations, see my slides from lectures I gave to graduate students in heavy-ion collisions:

https://th.physik.uni-frankfurt.de/~hees/hqm-lectweek14/index.html

(particularly Theory Lecture I).