Say we are given the scattering cross section for neutrinos from $d$ and $\bar{u}$ quarks as $\frac{d\sigma^{d}}{dQ^2}=\frac{G_F^2}{\pi}$, $\frac{d\sigma^{\bar{u}}}{dQ^2}=\frac{G_F^2}{\pi} (1-y)^2$, and the $u,\bar{d}$ cross sections are zero, how is the average nucleon cross section determined?
The solution involves this expression: $\frac{d\sigma^p}{dQ^2}=\displaystyle \sum_q \int _0^1 dx\left(q(x) \frac{d\sigma^{q}}{dQ^2} + \bar{q}(x) \frac{d \sigma^{\bar{q}}}{d Q^2} \right).$
Why is this the cross section for the proton? What exactly is the meaning of those functions, $q(x)$?
$q(x)dx$ is the expectation value of the number of $q$ quarks in the hadron whose momentum fraction is within $[x,x+dx]$.
Why are we using $\frac{d\sigma}{dQ^2}$ and what does it represent? After this we can take $\frac{d}{dx}\frac{d\sigma^p}{dQ^2}$ do the same for the neutron and take averages. The result is:
$\frac{d^2 \sigma^N}{dQ^2 dx} = \frac{G_F^2}{2\pi}\left(u^p(x)+d^p(x)+(1-y)^2(\bar{d}^n(x) + \bar{u}^p(x))\right)$
But I don't understand the origin behind the different quantities used here and the logic behind the derivation very well.
