# The Bjorken-x depicitive

1. Nov 8, 2008

### blue2script

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:

$$x = \frac{Q^2}{2M\nu} = \frac{Q^2}{2P\cdot q}$$

where $$Q = k - k'$$ (k is the momentum of the electron), M the mass of the proton and $$\nu = \epsilon - \epsilon'$$ 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

$$0 = p'^2 = \left(p+q\right)^2 = 2p\cdot q + q^2 = 2p\cdot q - Q^2$$

thus

$$1 = \frac{Q^2}{2p\cdot q} = \frac{Q^2}{2\xi P\cdot q} = \frac{x}{\xi}$$

$$\Rightarrow x = \xi$$

where p is the momentum of a single quark, P the momentum of the nucleous and $$\xi$$ is the momentum fraction of the quark, meaning $$p = \xi P$$. 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?