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Color of gluon mediating quark-antiquark process?

  1. Nov 27, 2011 #1
    I have the following process:
    [itex]q(R) + \bar{q}(\bar{B}) \rightarrow q(R)+\bar{q}(\bar{B})[/itex]

    In words: a quark with red color-charge and an antiquark with an anti-blue color-charge are incoming, and a red quark and anti-blue antiquark are emerging. Since I am not sure how else to draw that, I try to do some ASCII art here:
    Code (Text):

    q(R)          q(R)
       1\           3/
         \          /
          \        /
           \      /
              ~
              ~ G (?,?)
              ~
              ~
            /   \
           /     \
          /       \
        2/        4\
    q-(B-)      q-(B-)

    ----------------------> t
     
    q are the quarks, q- the antiquarks, R is red, B- is antiblue, time runs from left to right.


    The question is: which gluon(s) can participate in this exchange?

    I have read Griffiths "Introduction to Elementary Particle Physics" on that topic and on p.290 Example 8.1 he more or less states exactly that problem. He talks about a "typical octet state" [itex]R\bar{B}[/itex]. What is that supposed to mean? I thought the octet states are always a superposition of two states, in this case:
    [tex]|1\rangle = (R\bar{B} + B\bar{R})/\sqrt{2}[/tex]

    I was thinking along the lines of color conservation. The only thing that made sense to me is that the mediating gluon has to be [itex]B\bar{B}[/itex]: the incoming red quark sends out a [itex]B\bar{B}[/itex] pair. The incoming anti-blue antiquark combines with the blue charge, and what remains is an anti-blue charge... But I somewhat feel that this logic is flawed. :(

    What's the right answer here?
     
  2. jcsd
  3. Nov 28, 2011 #2

    Bill_K

    User Avatar
    Science Advisor

    Isn't it just q(R)+qˉ(Bˉ) → g(RBˉ) → q(R)+qˉ(Bˉ)?
     
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