Deep Inelastic Scattering cross section: Smearing of the $F_2(x)$ graph and relation of form factors with Fourier Trasform of nucleon distibutions

Consider Deep Inelastic Scattering cross section

$$\frac{d^2 \sigma}{d \Omega d \nu}=(\frac{d \sigma}{d \Omega})|_{Mott} \{ W_2(Q^2,\nu)+2 W_1(Q^2,\nu) \tan^2(\theta/2) \}$$

It is said that the Bjorken variable $x$ is "the fraction of proton momentum carried by the quark in IFM (infinite momentum frame), where masses and transerse momenta can be neglected".

Nevertheless in a graph of $F_2(x)=W_2/\nu$ versus $x$, it is claimed that the momentum exchange between parton (quarks) determines a smearing of the elastic scattering peak.

So why is $x$ considered the fraction of momentum only in IMF (where, by hypotesis, partons do not "talk" with each other or exchange momentum), but at the same time it is said that the previous graph is smeared by the exchange of momentum between partons (which makes x oscillate between a central value)?

Also why in D.I.S. cross section $W_1$ and $W_2$ are interpreted as sort of Fourier trasforms of some distributions (charge or magnetic moment) that describes the "shape" of the nucleon?

I'm asking this because I see that the indipendency of $W_1$ and $W_2$ from $Q^2$ is used as a clue that nucleons are point-shaped. Nevertheless I do not see how are $W_1$ and $W_2$ related to the Fourier trasforms of such distributions, since this formula is evidently different from Rosenbluth cross section.

• – Cosmas Zachos Jun 10 '18 at 21:25

Is the IMF the same as "on the light-cone"? Let's assume it is. In the IMF the proton is Lorentz contracted and time dilated into a static disk full of patrons. Quantum mechanically, it is a superposition of many states--along comes the virtual photon, and it selects a parton from one of these state with which interact.

If that parton is a valence quark, it doesn't have exactly $1/3$ the momentum fraction, rather it is spread out by interactions with the other patrons. Note that the spreading is not occurring during the time-scale of the scattering. The momentum of the valence quark is a superposition of many momenta around $1/3$, and the scattering selects one of them. Over the course of an experiment, many events reconstruct the shape in the figure (likewise for sea quarks).

Regarding the Fourier transform, I'm not sure where you are confused. So you agree the elastic form factors $F_1(Q^2)$ and $F_2(Q^2)$ recover the known electric and magnetic properties of the proton at $Q^2=0$:

$$F_1(0) = 1$$

$$F_2(0) = \mu_p = 2.79$$

And you understand that the Sach's form factors are Fourier transform of the charge and magnetic density

$$G_E = F_1 - \frac{Q^2}{4M_p^2} F_2$$

$$G_M = F_1 + F_2$$

(if $Q^2 << M^2$)?

That can be show in the Breit frame ($\nu \equiv q_0=0$) via:

$$G_E(Q^2) =\int{d^3{\bf x}e^{i{\bf xq}}} \approx 1 + \frac 1 6 Q^2\langle r^2 \rangle + \cdots$$

That's really no different from optical scattering where a photon with wavenumber ${\bf q}$ selects the target's fourier component $\rho({\bf q})$.

But your question is about the deep inelastic form factors, which satisfy:

$$\lim_{Q^2\rightarrow \infty, {\rm fixed\ }x}{\nu W_2(Q^2, \nu)} = MF_2(x)$$

$$\lim_{Q^2\rightarrow \infty, {\rm fixed\ }x}{ W_1(Q^2, \nu)} = F_1(x)$$

The DIS cross section is:

$$\sigma = \sigma_0\big[W_2+ 2W_1\tan^2{\frac{\theta}2}\big]$$

Where $\sigma_0$ is the point scattering cross section.

What is the point of confusion? In short, there is an initial electron $|k\rangle = \exp{(ikx)}$ and a final electron $\langle k'|=\exp{(-ik'x)}$, and they are coupled by a nucleon:

$$\langle k'| N |k\rangle$$

That's pretty much a Fourier transform in:

$$q_{\mu} = k_{\mu} - k'_{\mu}\rightarrow e^{i{\bf qx}}$$

That the result is independent of $Q^2$ means that there is no size-scale present in the parton: it is a point (the FT of a delta function is flat and extends over all space).

Of course, there still is $x$ dependence--that's the parton density (charge weighted) per momentum fraction.