# Deep inelastic and Bjorken limit

In DIS the Bjorken limit is given by the conditions: $$Q^2 \rightarrow \infty$$, $$\nu \rightarrow \infty$$ and $$x=Q^2/(2M\nu)$$ is finite, where $$Q^2$$ is the opposite of the transferred momentum, $$M$$ is the proton mass and $$\nu$$ is the transferred energy. In this limit, the structure functions of the nucleons are functions of $$x$$ only and don't depend on $$Q^2$$.

The deep inelastic limit instead is the condition in which the invariant mass $$W$$ of the hadronic system that is produced in the electron-nucleon interaction is much bigger than the proton mass $$M$$. Please correct me if I'm wrong.

But, are the Bjorken limit and the deep inelastic limit the same condition?

I am not sure about you definition of "deep inelastic limit", but $$Q \rightarrow \infty$$ while $$x$$ fixed is not equivalent to $$W \rightarrow \infty$$. Consider $$W^2 = (P+q)^2 = Q^2 (1/x - 1) + M^2.$$ From $$Q \rightarrow \infty$$ at fixed $$x = Q^2/(2P\cdot q)$$ it is clear that $$W \rightarrow \infty$$. But $$W \rightarrow \infty$$ can also mean that $$x \rightarrow 0$$ while $$Q$$ is fixed (the mass scale $$M$$ is always considered small and fixed). This is the so-called small-x region, where there is another large scale $$1/x$$, while in the usual treatment of DIS one considers the region where $$(1/x-1) \sim 1$$, so the only large scale is $$Q/M$$. This leads to substantial differences for the two kinematic regions. There also regions where $$(1/x-1)$$ is small ($$x \rightarrow 1$$ is elastic scattering), e.g. of order $$M/Q$$, where the standard DIS treatment may also fail.