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

Quark-antiquark creation in strong interactions

  1. May 14, 2013 #1
    (please note this is similar to a homework question I have posted but it is not the same and here I am just trying to understand a concept)

    In a book I was reading it said

    [itex]
    \Delta^{+} \longrightarrow n + \pi^{+}
    [/itex]

    is a viable process via the strong force because all that is needed is a down antidown quark pair, which can be provided via the strong force. My question is why? If [itex]\Delta^{+}[/itex] is uud and [itex]\Delta^{+}\pi^{+}[/itex] is udd + [itex]u\bar{d}[/itex] we can see that the right hand side only has one antidown quark therefore there has only been one quark-antiquark pair created. Converting one quark into two only creates one extra particle effectively so I see a deficit on the left hand side.

    Please can someone explain
     
    Last edited: May 14, 2013
  2. jcsd
  3. May 14, 2013 #2

    Bill_K

    User Avatar
    Science Advisor

    Δ+ → n + π0

    :confused: What you have written does not conserve charge.
     
  4. May 14, 2013 #3
    You're right. I've changed it now it. My question really is about how a 3 quark particle can become two particles with a total of 5 quarks :smile:

    we don't do this stuff in lectures
     
  5. May 14, 2013 #4

    Hepth

    User Avatar
    Gold Member

    The process would be

    d -> d + g -> d + (q qbar)

    A gluon gets emitted that then decays. You're thinking its one d decaying to two, when its really 1 to 3
     
  6. May 14, 2013 #5
    oh I see. does the q-qbar-pair have to be the same type? If I understand it right in this case the gluon must become an up and an antidown pair.
     
  7. May 14, 2013 #6

    Hepth

    User Avatar
    Gold Member

    The q can be anything, but not mixed. A pi0 is not u dbar. It is u ubar and d dbar.
     
  8. May 14, 2013 #7
    sorry my question was wrong. it should have been pi+ :redface:

    does this mean this process is forbidden?
     
  9. May 14, 2013 #8

    Hepth

    User Avatar
    Gold Member

    I'll write ubar as u~

    You have
    uud > udd + ud~
    So quark content is fine.

    Draw it out, the gluon makes a d d~ but the d goes to the neutron, the d~ to the pion.
     
  10. May 14, 2013 #9
    I think it just went click.
    The Delta+ decays into two parts (e.g. ud and u) thereby producing a gluon which becomes dd~ pair except the d is taken up by ud to form a neutron and d~ joins with u to form a pi+.

    Is that more or less right?
     
  11. May 14, 2013 #10

    Hepth

    User Avatar
    Gold Member

    Yes that is correct.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook