Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook