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QCD Jets

  1. Feb 20, 2010 #1
    In e+e- annihilation, are the jets that are produced quarks and gluons, or hadrons?

    What is the basic idea behind a QCD calculation for such a process? For example, do I take the two initial states to be e+e-, and the two final states to be q+q- (quark and antiquark), but also add the possibility of emission of low-energy gluons and quarks (in much the same way one treats infrared divergences of soft photons in QED)?

    Also, if the particles in jets are hadrons and not a stream of quarks and gluons, how does one represent them in QCD calculations? I'm only aware of representing hadrons in the context of effective field theories below the QCD scale .2 GeV, but these experiments are at much higher energies.

    Also, what about cross-section for e+e- to go into a lepton+antilepton- pair rather than a quark+antiquark- pair? Does this become negligible at these energies?
     
    Last edited: Feb 20, 2010
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  3. Feb 20, 2010 #2

    blechman

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    electron and positron annihilate into a photon or a Z boson (or a higgs, or some other particle yet to be discovered) and out comes a q-qbar pair. At high enough energies (roughly around 10 GeV or higher) these quarks will separate and form jets of hadrons. At higher energies they can also emit hard gluons and you can get a "3 jet event", where the gluon fragments to hadrons as well.

    that's the idea. Once you have the q qbar pair, you can start exchanging gluons or bremming off gluons. In the end, you resort to things like "parton-hadron duality" and allow each hard parton (quark/antiquark/gluon) to be matched onto your jet. In practice you can use MC like Pythia to simulate the fragmentation process.

    at jet-scale energies, it doesn't have to be "low energy gluons" - you can brem off a hard gluon as well to make a third jet.

    You don't. You do what I said above -- match them onto the jet and let the computer fragment them for you. There are some models that do this, but you have to take them with an appropriate grain of salt (they're not QCD).

    No, they're still there. But two muons don't look anything like two jets in the detector, so they're easy to cut out of the jet sample.
     
  4. Feb 20, 2010 #3

    blechman

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    WARNING: Technical!
    Let me add something to this: when doing a jet physics calculation, there are many energy scales in the problem. That means that you must resort to Renormalization Group techniques to properly resum all the potentially large logarithms you get in such calculations. One of these scales is the energy of the interaction, and another is that lower 1 GeV energy scale where the jets are formed (as you said in the above quote). So what you do is you use RG to resum all the high energy effects into the Wilson Coefficients and then you are left with matrix elements that must be evaluated at the low scale. THIS is where you match to the jet algorithms.
     
  5. Feb 21, 2010 #4
    When you say "some other particle yet to be discovered", would this particle have to be greater than 200 GeV in mass, since LEP reached those energies? I know you're talking about a virtual particle, but even if it's 300 GeV in mass, you will see signs of it through the effects on the cross-section, even though you can't produce it in a final state. Also, how can you match QCD-experiments with theoretical calculations when you don't even know all the particles that participate virtually?

    According to a chart I have, at only 130 MeV, a phase diagram for heavy-ion collisions jumps from 3 degrees of freedom (the 3 pions) to 52 (quarks, antiquarks, gluons, and all their colors and spins and polarizations). So I assume that means above 130 MeV quarks and gluons do not hadronize? Does this verify that [tex]\Lambda_{QCD} [/tex] is around .2 GeV which is kind of close to 130 MeV? What is the difference between heavy ion collisions and e+e- annihilation - they seem to operate at the same energy of 100 GeV (LEP and RHIC).

    But asymptotic freedom makes the probability of third and more jets unlikely.

    So the energy of the interaction is around 100 GeV, and 1GeV is hadronization. Does this mean the quark/antiquark pair leave at 100 GeV each and start bremming gluons and other stuff until each particle in the jet is around 1GeV at which point they begin to hadronize?

    So in principle, if you could sustain a temperature greater than 1GeV, you could actually see quarks and gluons because they won't hadronize until the temperature gets lower than 1GeV?
     
  6. Feb 21, 2010 #5

    blechman

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    That's why you do experiments! You calculate the cross section based on your model, and then compare to what you see. If you are missing a particle, then the results of the experiment would not agree with theory calculation.

    It's a bit more subtle than you're suggesting, because even though the particles you created might have been produced at LEP, if their couplings are small enough you would not have seen them. The classic example is the Higgs: even if the Higgs was 114 GeV, like we think it is, because it has such a tiny coupling to electrons (since the electron is so light), it plays no significant role in jet production at LEP.

    Density. LEP ran electron on positron, while RHIC is running 100+ nucleons (protons+neutrons) on 100+ nucleons. That's the difference.

    On the contrary, asymptotic freedom predicts the 2 jet probability vanishes exponentially at high energies! This is the so-called "Sudakov Double-Log suppression".

    That is more or less correct. If the temperature of the universe were greater than several GeV, you would not see hadrons. That's what RHIC is testing.
     
  7. Feb 21, 2010 #6
    I'm learning in statistical mechanics that temperature is a phenomenon that requires lots of particles at equilibrium to define (or something like that: I was a bit unclear what the teacher was talking about): [tex]T=\frac{\partial E}{\partial S}[/tex] at constant particle number and volume .

    I've also heard that temperature is really energy, and that Boltzmann's constant was only invented because a long time ago people didn't realize temperature was really energy, so they needed a conversion factor.

    In Planck units: [tex]\hbar=c=k=1 [/tex], so all dimensions are powers of mass.

    So the RHIC is testing collisions of lots of particles at once: is this the reason a temperature can be defined, because you need more than one particle to define a temperature according to statistical mechanics? However, statistical mechanics seems to require 10^(23) particles, and not 400 or so particles that the RHIC is colliding at once. 400 particles (2 gold nuclei) is better than just colliding 2 protons like the LHC, but 400 is still not close to 10^23 particles required by statistical mechanics.

    So if the LHC operates at 1 TeV, can one convert that into a temperature via Boltzmann's constant (1/k)*(1 TeV) and say that the LHC is exploring those temperatures, in spite of the fact the LHC is just colliding one particle with one particle?

    I did an exercise where I showed that adding thermal corrections to the Higgs potential results in restoration of symmetry at some temperature: that is, the mass-squared in front of the quadratic term [tex]-\mu^2 \phi^2 [/tex] of the Lagrangian becomes positive instead of negative (to be more correct I should say that the Green's function terms become positive, i.e., the "dressed" two-point vertex involving no momenta [mass-type terms]). I was wondering if symmetry restoration can be achieved with high energy of just a few particles instead of high temperature of many particles, i.e., if I wanted to observe symmetry restoration, would I need to collide heavy ions, or just two protons?

    It seems weird that with the RHIC one talks about phase diagrams, but with the LHC one doesn't talk about phase diagrams, but if energy were freely converted to temperature, then why not both?
     
    Last edited: Feb 21, 2010
  8. Feb 21, 2010 #7

    blechman

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    All correct.

    Deep inside the nuclear collision, the density is high enough to talk about a system that reaches some sort of thermal equilibrium. This means that you can apply stat mech there. This is a nontrivial statement and requires some pretty intense calculations to verify, but it seems to be the case. Since the LHC has protons colliding on their own in an effective vacuum, there is nothing for the collision products to come into equilibrium with, and the system is not properly described by stat mech.


    Again, see my above comment. It doesn't really work to pretend that there is a temperature at the LHC - it is effectively T=0. Although when the universe was at temperatures comparable to the LHC energy, then the processes that are occurring at the LHC were in thermal equilibrium. It is in that sense that we say the LHC is "studying the behavior and conditions of the early universe."

    That is correct, and at energies at the LHC the EW symmetry is restored in some sense (the Z boson starts to behave as though it were massless). This is true even at the Tevatron. But this is not a finite-temperature effect! This is true even at zero-temperature but high collision energies. Then the Higgs vacuum expectation value effectively decouples.


    Again, you cannot apply finite-temperature field theory to the LHC, since the density isn't high enough, the system is not in a "thermal equilibrium", etc. So it makes no sense to talk about phase diagrams, etc.
     
    Last edited: Feb 21, 2010
  9. Feb 21, 2010 #8

    blechman

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    To understand the use of words like "temperature", I remind you of the Zeroth Law of Thermodynamics:

    Two bodies of the same temperature are in thermal equilibrium.

    The idea is that in order to be able to talk about temperature at all, you need to be in thermal equilibrium! Without that condition, you cannot really talk about "temperature."

    Just because it has the same units as energy doesn't mean that it can be interchanged with any other form of energy in any problem! Temperature represents the energy involved in a thermal equilibrium.

    Without that, you cannot apply the techniques of statistical mechanics.

    Aside: I should say that there is a thing called "nonequilibrium stat mech," but this is something quite different...
     
  10. Feb 21, 2010 #9
    Thanks! That was very helpful.

    My main area is more towards neutrinos, but I feel to be well-rounded at physics, I need to know some things about QCD so I can follow what's going on in the accelerators because that's the big news in physics, the restart of the LHC.
     
  11. Feb 23, 2010 #10

    jal

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    This seems like a good thread to ask my question.

    My understanding is that Jets are produced because of the change of pressure/density of the perfect liquid of quarks gluons. I guess that would be because it is expanding.

    Is anyone coming up with possible ways of controlling the expansion and controlling the production of jets?
    Are there papers that I could read?
    jal
     
  12. Feb 23, 2010 #11

    blechman

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    jets form because when a quark (or antiquark or gluon) flies away with sufficiently large energies, the QCD force grows to the point that the quark starts radiating large amounts of gluons. It's exactly the same phenomenon as hadronization (where a b quark becomes a B meson, for example), except there's too much energy for hadrons to form so you just get a spray of particles.

    It's not really an issue of pressure or density. This is more or less single-particle dynamics.
     
  13. Feb 23, 2010 #12
    I just looked at the Wikipedia page for the RHIC and it says temperatures of 7 trillion Kelvin were reached. Multiplying by Boltzmann's constant I get half a GeV.

    This seems to be a little too close to [tex]\Lambda_{QCD}=.2 \mbox{ }GeV[/tex].

    Does hadronization occur exactly below .2 GeV, so that if it's .2001, then it won't hadronize?

    Because the way that [tex]\Lambda_{QCD} [/tex] is calculated, it's just the value where perturbation theory predicts the strong coupling to go to infinity. Of course, as soon as the coupling becomes large, perturbation theory no longer becomes reliable, so [tex]\Lambda_{QCD} [/tex] is just arbitrary and has nothing to do with anything physical like a phase transition. Has the hadronization temperature been predicted from theory?
     
  14. Feb 23, 2010 #13

    blechman

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    No, hadronization does not have such a crisp and well-defined energy scale. Physics never works like that!! Hadronization sets in at energies OF ORDER [itex]\Lambda_{\rm QCD}[/itex]. So RHIC is able to get the temperature of the quark-gluon bath up to several hundred GeV, as you say, right at or slightly above the phase transition temperature. ALICE (at the LHC) will get even higher.

    Your last paragraph is not really correct, either. [itex]\Lambda_{\rm QCD}[/itex] is uniquely defined in terms of the strong coupling constant: it is a physical quantity (up to ambiguity in the renormalization algorithm) -- we can measure it! By dimensional analysis and the belief (assumption, but a very believable one!!) that hadronization is actually described by nonperturbative QCD, this is the ONLY energy scale you can use to define a transition.
     
  15. Feb 23, 2010 #14
    I don't think we can measure [tex]\Lambda_{QCD}[/tex]. For example, on page 6 of a layperson's introduction to QCD:

    http://www.frankwilczek.com/Wilczek_Easy_Pieces/298_QCD_Made_Simple.pdf

    there is a nice figure of QCD experiments and their measured value for the strong coupling. At the lowest energies is an experiment called "Sum Rules", but at 100 GeV there is e+e-, and at 8 GeV there is deep inelastic scattering. Anyways, at the experiment "Sum Rules", the strong coupling is already at .3 to .4. There are no experiments at an energy less than 1 GeV (where the couplings heads off to infinity). The theoretical curve is derived using an experiment at the Z-mass which is around 80 GeV, serving as the boundary condition of the RG equation (I don't know how accurate is the RG equation they're using, if it's more than 1-loop order or not).

    So as I understand it, [tex]\Lambda_{QCD}[/tex] is the value of the energy at which the coupling goes to infinity, assuming that the RG equation is still valid in the regime where the coupling is greater than 1. They say that the Landau pole of QED is an artifact of perturbation theory: QED really doesn't have an infinite coupling constant at the Landau pole, because the Landau pole assumes that perturbation theory works and that the RG-equation is always valid even at high energies.

    Anyways, that's why I say that [tex]\Lambda_{QCD}[/tex] is not physical and can't correspond to a phase transition. I know [tex]\Lambda_{QCD}[/tex] changes slightly with renormalization scheme and that the value is usually quoted for MS-bar. Any parameter depends on renormalization scheme, so in my view, a value of a parameter can't be physical. But a phase transition is physical.
     
  16. Feb 23, 2010 #15

    blechman

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    Not quite. [itex]\Lambda_{\rm QCD}[/itex] and [itex]\alpha_s[/itex] are completely interchangable - you can use one or the other! This is the famous phenomenon of "dimensional transmutation".

    In other words: If you tell me the value of [itex]\alpha_s(M_Z)[/itex], I will tell you the precise value of [itex]\Lambda_{\rm QCD}[/itex]. You can then express all your calculations in terms of [itex]\alpha_s[/itex] OR [itex]\Lambda_{QCD}[/itex].

    That is how you "measure" the strong coupling scale - it is done at the Z mass. [itex]\Lambda_{\rm QCD}[/itex] is just as physical as [itex]\alpha_s[/itex].

    It is a little misleading to think of the QCD scale as "the scale where the strong coupling is infinity" -- such a definition makes no sense, as you are rightfully pointing out. But that's not how we actually define the scale! We have a much more quantitative and well-defined definition for it, using the RG equations.
     
  17. Feb 23, 2010 #16

    blechman

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    This is a dangerous argument, where we get lost in the horrors of semantics!

    If you REALLY want to be picky, there are only two things in all of particle physics that are "physical": cross sections and branching fractions. Everything else is "unphysical". But that doesn't mean you cannot talk unambiguously about the mass of a particle, or the fine structure constant, once you set your definitions.
     
  18. Feb 23, 2010 #17
    IMO hadronization occurs when there isn't sufficient energy to continue spawning new particles any more. You try to pull two colored quarks away from each other, as long as their energies are sufficient to break the gluon string that connects them by bringing in two new quarks, that's what's going to happen.
     
  19. Feb 23, 2010 #18

    Haelfix

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    Its worthwhile to keep in mind hadronization is really a statistical, many body phenomenon, and isn't super sharp. We know order of magnitude estimates, but the exact details will depend on environmental details, modeling assumptions and physics that isn't entirely well understood.

    In principle, hadronization can occur at higher energy scales as well, its just suppressed.
     
  20. Feb 23, 2010 #19
    lol, I agree it's just semantics. When you measure a parameter, it's physical in the sense that you just measured a real cross-section to get that parameter. So it's a physical result.

    I was probably sloppy with terminology.
     
  21. Feb 23, 2010 #20

    blechman

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    Right. Sorry if I sound a little obnoxious, but the word "physical" brings out the worst in me! :wink:

    But the point of [itex]\Lambda_{\rm QCD}[/itex] being defined in terms of [itex]\alpha_s(M_Z)[/itex] is very important, so I just want to make sure I made that point. We never need to do a nonperturbative QCD calculation to define this scale!
     
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