In a real collider experiment e.g. LHC / LEP how can one reconstruct the energy and momentum of the resultant $q\bar{q}$ pair produced from the process $e^+e^-\to q\bar{q}$?

Specifically, how can we relate the kinematic variables before ($q\bar{q}$) and after hadronisation (observed jets)?


I was working on a question that asserted as a side comment that one could reconstruct the final state energy and momentum of the resultant $q\bar{q}$ pair produces from the process $e^+e^-\to q\bar{q}$ observed experimentally in a collider.

The question posed itself was easy (and not relevant here) however, the assertion of the ability to reconstruct the energy and momentum of the quark and antiquark caused me some confusion.


I have studied Thompson Modern Particle Physics and the only explanation I have is rather hand wavy and I'm not sure of its legitimacy.

I figured that we could observe the cross section of the hadron jet process and compare at specific centres of mass $\sqrt{s}$ to known experimental results (in plot below) form the ratio,

$$ R= \frac{e^+e^- \to hadrons} {e^+e^- \to \mu^+\mu^-} $$

If we were clever about picking $\sqrt{s}$ then perhaps results could peak at a meson production process that we would be aware of and then hence, we could determine the final state energy and momentum (this is the really hand wavy bit as I don't actually know how one might do that!)

enter image description here


2 Answers 2


Hadronization still doesn't break conservation of energy and momentum. So getting the total energy, or the total momentum, of the quark-antiquark pair is easy: just add up the total energy and momentum of all the reaction products.

To get the individual energy or momentum of one particle, i.e. just the quark (or antiquark), we rely on the fact that they come out of the detector in opposite directions. That means it's easy to separate the outgoing particles into two clusters: one came from the quark and one came from the antiquark. You can then add up the energies and momenta within each jet to find the momentum of the quark or antiquark that produced it.

This image, which I found on Quantum Diaries, shows the kind of separation I'm talking about. Fair warning, though: this particular image came from a proton-proton collision, which means there's a bit of extra "junk" emitted in all directions, as you can see by looking between the jets. That doesn't happen much in electron-positron collisions. Unfortunately I couldn't find a better picture to illustrate it.

CMS dijet

Another option for identifying the momentum of a progenitor quark (or antiquark, or even gluon) is to look at only one outgoing particle; usually the highest-momentum hadron. Suppose that particle has a fraction $z$ of its parent quark's momentum, and is moving in the same direction. We can apply some quantum field theory and a lot of data analysis to figure out the probability distribution for different values of $z$. This distribution (or something roughly like it) is called a fragmentation function, denoted $D(z)$ or something like $D_{\pi^+/u}(z)$ if you want to limit yourself to a specific kind of hadron and/or quark.

Fragmentation functions are more useful in proton-proton collisions or ion collisions, where the "junk" I mentioned gets in the way of cleanly grouping the outgoing particles into two jets.


To complete David's answer, here is a two jet event in LEP, an e+e- collider, the ALEPH detector. The experiment is running on the Z mass.

The central part of ALEPH consists of several different tracking detectors. The points where charged particles interact with the tracking detectors (hits) are shown as squares and the tracks fitted to the hits are drawn as thin lines. The outer part of ALEPH consists of calorimeters, i.e. detector components that measure the energy of the particles. The total energy deposited per calorimeter cell is drawn as a histogram. For penetrating particles (muons) one can see the hits in tracking chambers of the outermost calorimeter (hadron calorimeter) and muon detectors. There are also some blue hits. These originate from particles for which the trajectory does not point back to the vertex or which have low momentum.

The beam is perpendicular to the plane of the images.

two jet

A 2-jet event resulting from the Z decaying into two quarks

and a three jet event

three jet

A 3-jet event coming from a Z decaying into two quarks and one gluon

  • $\begingroup$ thanks (: perfect timing as well as I was just looking at the Z-decay width from LEP in preparation for my exam tomorrow. Some entertaining, and no doubt, confusing corrections that had to be made to get the correct level of accuracy! $\endgroup$ May 16, 2016 at 20:28
  • $\begingroup$ I do not know what you mean by "corrections". The measured energies and momenta are summed , the missing energies and momenta are fitted using appropriate models within measurement errors, and final errors calculated . one gets the accuracy that the experiment provides. $\endgroup$
    – anna v
    May 17, 2016 at 4:33
  • $\begingroup$ Ah right, corrections due to rock moving under tidal pressure, but more entertaining: The issues they faced with a charge leak form a local train track. That must have been incredibly confusing for a while! $\endgroup$ May 17, 2016 at 9:00
  • $\begingroup$ sorry, but ALEPH was in LEP , the old e+e- experiment which stopped before 2000. no train there. yes the earth-tide effects on the beam location were even then taken into account. $\endgroup$
    – anna v
    May 17, 2016 at 9:43
  • $\begingroup$ As an ex-ALEPHer myself, I can explain the "train" comment. In addition to the lunar-cycle synchronous deviations, LEP operators also noticed apparently random fluctuations occurring a few times day. They were finally traced to "ground bounce" (when the ground potential rises above zero) caused by the Geneva-Paris TGV rumbling by on the local Bellegarde line. $\endgroup$ May 24, 2018 at 11:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.