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The Bjorken-x depicitive

  1. Nov 8, 2008 #1

    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]


    [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,

    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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