# Parton distribution function interpretation

I wished to clarify the interpretation of parton density plot1.

Are the following interpretation correct?

1. First fix $Q^2$ for an experiment.
2. Then the area under each curve $\int_0^1 xf(x)dx$ describes fraction of the total momentum of proton or nucleon carried by a particular parton. (And I guess this probability or fraction is valid only for that specific value of $Q^2$)

Now, what I am confused with is that since we fix $Q^2$ for a particular plot, doesn't this also fix the fraction of momentum carried by a parton? I mean to say that at fixed $Q^2$, if gluon carries 50% of the proton momentum , its x value should be 0.5 only. Then why does we get an entire range of values of x?

Also, in the plot, at around x = 0.1 or 0.2 in the plot, we see that number of up quarks are almost twice of down quarks(assuming that they refer to valence quarks). So should I interpret this x as the value where most of the experiments predict the actual composition of proton for that particular Q^2?

Thank you!

• You have apparently misunderstood the parton kinematics, and in particular the variation of x with ν. Sep 7, 2017 at 14:29

First and foremost, you should never forget that the parton density functions (PDFs) do not mean much by themselves: they are just one part of a cross section and it is always a good idea to write that cross section to clearly see the meaning of $x$ and $Q^2$. Let's take the example which is the order of the day, a cross section $pp \to X$ where $X$ stands for however complex a final state you wish:
$$\sigma_{pp\to X} = \sum_\text{partons a, b}\ \int_0^1 dx_a f_a(x_a, \mu_F^2) \int_0^1 dx_b f_b(x_b, \mu_F^2) \sigma_{ab\to X}\left(-x_a p, x_b p, \{P_\nu\}, \frac{Q^2}{\mu_F^2}\right).$$
I wrote it in the center-of-mass frame where the protons has momenta $p$ and $-p$ along the common direction of the beams. $\{P_\nu\}$ stands for the 4-momenta of the final particles. $\sqrt{Q^2}$ is the energy of $pp$ in the center-of-mass. $\mu_F$ is an arbitrary scale which had to be introduced so as to factorise into the PDFs some divergences appearing in Feynman diagrams [*]. In your question, you assumed that one always chooses $\mu_F^2=Q^2$. A popular choice it is but by no mean always the right one. In any case, as you see, $x_a$ and $x_b$ run through the segment $[0,1]$. Well, in fact, momentum conservation puts a lower bound on the $x$'s, which depends on the final particle energies and momenta along the beam direction.
As for your last question, at small $x$ the PDFs for $u$ and $d$ are dominated by sea quarks whereas at $x\approx 1$, the constraint that the sum of all $x_a$'s is 1 tend to spoil the ratio u/d. Thus there is indeed a sweet region where valence quarks still dominate enough while being far enough from 1 to avoid skewing the ratio: it depends on $\mu_F$ somewhat but details are a bit foggy in my memory.
[*] Specifically, if a quark or gluon $a$ emits another quark or gluon $b$ and $\vec{p}_b$ tends toward being parallel to $\vec{p}_a$, then a divergence appear, called a colinear divergence for this reason.