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.

  • $\begingroup$ Do you understand what "momentum fraction" means in this context? Do you understand that $q(x)$ is a parton distribution function and what that implies? $\endgroup$ Apr 15, 2013 at 21:14
  • $\begingroup$ Well $2M\nu - Q^2 >0 \Rightarrow x\in(0,1)$ because $x=\frac{Q^2}{2M\nu}$ that's all I know - It is the "Lorentz invariant scaling variable" I don't understand what the functions mean really. $\endgroup$
    – shilov
    Apr 15, 2013 at 21:33
  • $\begingroup$ You'll want to read about Bjorken scaling; $x$ represents the fraction of the composite target's momentum carried by the particular parton that you've interacted with in a frame where the target's total momentum is very high (called the "infinite momentum frame"). That probably sounds odd, but the whole point is that some features of the interaction physics get very simply when expressed in those terms. I'd have to bone up a little to be confident of writing a good answer, so maybe someone else will jump in. $\endgroup$ Apr 15, 2013 at 23:23


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