1
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.