# Why small $x$ corresponds to high energy process?

I just start to study Quantum Chromodynamics, I read in some references that small $$x$$ corresponds to process of high energy, but I cannot find a straightforward explanation.

In the case of electron deep inelastic collision, $$x$$ is defined as the fration of the momentum of the hadron (eg. proton) carried by the parton which interacts with the incident electron. The probablility of find such a parton is determined by the parton distribution function. Also, for gluons, the distribution function becomes large at small $$x$$, which, as I understand, indicates a degree of saturation. But why small $$x$$ is related to high energy process?

Thank you for your explanation.

As I understand, when one talks about high energy in the context of QCD, there are, among others, two relevent quantities. The first quantity is $$Q^2$$, and the second one is $$x$$.

Considering the collision between an electron and a proton shown in the above plot,

$$e(k)+p(p)\rightarrow e(k')+X(p_X) .$$

In particular, the process is relevant to the present discussion when the wave length of the virtual photon $$\gamma$$ is much smaller than the size of the proton $$r_p$$, namely,

$$\lambda \ll r_p .$$

Here, the first quantity,

$$Q^2\equiv -q^2=-(k-k')^2 ,$$

being large is equivalent to the above condition. Physically, this is when the photon is able to probe the substructure of the proton. Subsequently, the parton model, as well as perturbative QCD, provides a reasonable description of process. As referred to be high energy above.

The second quantity is

$$x\equiv \frac{Q^2}{2p\cdot q}=\frac{Q^2}{Q^2+W^2-m_p^2} ,$$

where

$$W^2=(p+q)^2=m_p^2+\frac{Q^2}{x}-Q^2\simeq \frac{Q^2}{x}(1-x).$$

Now, as $$x\to 0$$, $$W$$ becomes more significant. Therefore, in this context, this is also frequently referred to as high energy in literature. As it is known that the gluon distribution function (PDF) becomes more significant as $$x\to 0$$, the small $$x$$ region corresponds to where the gluon saturation takes place. As $$Q^2$$ and $$x$$ are two independent variables, the region of small $$x$$ does not necessarily implies large $$Q^2$$. Although, for perturbative QCD to work, $$Q^2$$ still need to be moderately large. In this case, the physics is closely associated with that of Color Glass Condensate. I hope this answers your question.

• thx for the detailed answer! – pincolino May 10 at 1:47