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?


1 Answer 1


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.


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