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The Bjorken-x depicitive

  1. Nov 8, 2008 #1
    Hello!

    Sorry to bother you with all these questions coming to my mind in preparing for my oral exam in QCD. I was just wondering about the nature of the Bjorken x. You see, it is defined as:

    [tex]x = \frac{Q^2}{2M\nu} = \frac{Q^2}{2P\cdot q}[/tex]

    where [tex]Q = k - k'[/tex] (k is the momentum of the electron), M the mass of the proton and [tex]\nu = \epsilon - \epsilon'[/tex] is the energy difference of the electron (in the rest frame of the nucleon).

    So, what we find is that we can measure the Bjorken-x solely from the initial and final state of the electron. On the other side, we know

    [tex]0 = p'^2 = \left(p+q\right)^2 = 2p\cdot q + q^2 = 2p\cdot q - Q^2[/tex]

    thus

    [tex]1 = \frac{Q^2}{2p\cdot q} = \frac{Q^2}{2\xi P\cdot q} = \frac{x}{\xi}[/tex]

    [tex]\Rightarrow x = \xi[/tex]

    where p is the momentum of a single quark, P the momentum of the nucleous and [tex]\xi[/tex] is the momentum fraction of the quark, meaning [tex]p = \xi P[/tex]. So, what we find is that the Bjorken x is the momentum fraction of the quark. However, we can calculate x and therefore the momentum fraction solely from the electron. That means there is a very deep relation between the momentum transfer through the photon and the momentum fraction of the quark.

    My question is then: Where does this deep connection comes from? Is there any depictive way to make this argument, this connection, more visual, transparent?

    Thanks for all your comments! With best regards,
    Blue2script

    PS: Another question of the same kind is: The connection above binds Q^2 to x. But then, how can we treat Q^2 and x independent in the naive parton model?
     
  2. jcsd
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