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Energy Conservation in Quark Confinement

  1. Apr 14, 2012 #1
    Hi guys,

    I was reading about quark confinement and came up with a doubt i can't find the answer to:
    You have to use energy to try to 'pull away' one of the quarks in a meson, right? And when you give enough energy, a new quark/anti-quark pair is created and that's why you never find only one quark. But, what if i give energy to the system but not enough so that the quark/anti-quark pair is created? What happens to the energy i gave to the system? Is it reemitted or does it stay in the system as potential energy?

    I tried to think about it but can't figure this out.

    Thanks for the help in advance. :)
     
  2. jcsd
  3. Apr 14, 2012 #2
    I would bet the quarks would dance around for a while --until some relaxation time will have passed-- and then emit radiation (photon, or whatever).
     
  4. Apr 15, 2012 #3

    tom.stoer

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    Science Advisor

    You can look at an experiment to clarify the situation, e.g. deep inelastic scatterin

    [tex]p\,+\,e\,\to\,X\,+\,e[/tex]

    The energy is provided by an electron and transferred to the proton by a (virtual) photon. The final state contains again the electron plus a (quite complex) hadronic state X which is usually not discussed (we neglect non-hadronic states X like "Bremsstrahlung" due to electromagnetic interaction b/c they have nothing to do with confinement).

    Now one can look at the possible states X depending on the c.o.m. energy E.

    The lowest possible state is X=p, i.e. elastic scattering.
    The next state would be something like X=Δ+ where only orbital angular momentum is transferred to the proton, i.e. the Δ+ has the same quark content (uud) but spin 3/2.
    Another possible final state is a X=p+π° or X=n+π+ i.e. pion production. This is the first example where a quark is pulled out of a proton, where the energy is converted into a quark-antiquark pair which eventually forms a new particle (a pion) plus the original proton. In that case you can even draw a simple Feynman diagram for process.

    Now for each such process there is a so-called threshold energy which can be determined kinematically w/o any knowledge about the details of QCD dynamics. For X=p the threshold is zero, elastic scattering is always allowed. For X=Δ+ it's a nice exercise to do the calculation; it's nothing else but energy-momentum conservation. For X=p+π° or X=n+π+ it's more comoplicated b/c now there are three particles (including the electron) in the final state.

    So the answer is quite simple: if the energy is not sufficient to pull out a quark from the proton and to form a new, complex hadronic state X, it simply does not happen; the nucleon stays intact (elastic scattering) or is excited (Δ-resonance).
     
    Last edited: Apr 15, 2012
  5. Apr 15, 2012 #4
    Thanks a lot for the answer, tom.stoer, it clarified a lot of things for me.
     
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