0
$\begingroup$

The positive-pion $\pi^+$ is comprised of 1 valence up-quark and 1 valence anti-down-quark. Assuming isospin symmetry, the parton distribution functions (PDFs) for these two valence quarks should be identical. The valence-PDF is defined as the PDF of the valence quark minus its sea component, e.g. $u^v(x)=u(x)-\bar{u}(x)$. The fact that the $\pi^+$ has one valence $u$-quark and one valence $\bar d$-quark is encoded in the standard sum-rule:

$$\int_0^1 dx\left(u(x)-\bar{u}(x)\right)=\int_0^1 dx\, u^v(x)=1$$

Of course, the PDF depends on the factorization scale $\mu_F$, used in the renormalization of collinear divergences. The convolution of the PDFs with partonic cross sections gives physical quantities which do not depend on $\mu_F$. We commonly set $\mu_F=\mu_{UV}=Q$ where $Q$ is the momentum-transfer in the presumed scattering process, in order to cancel logarithmic corrections like $\log \left(\dfrac{\mu_F}{Q} \right)$ and $\log \left(\dfrac{\mu_{UV}}{Q} \right)$.

There are certain properties of the valence PDF that I would expect to hold for all values of $Q$. First, I would expect the relevant sum-rules to hold (e.g. the aforementioned equation)).

The second is that, I would expect the distribution to be symmetric about $x=1/2$, or at least peak there. The reason is, definitely the valence $u$-quark in the $\pi^+$ is most likely to carry half the hadron's momentum. DGLAP evolution resolves quark and gluon loops (i.e. the sea) so I don't see why it would change salient features of the valence PDF. (I must be wrong about this)

However when I consult some recent calculations for the valence PDF for the pion [source], the result is not even close to being symmetric around $x=1/2$.

enter image description here

Should the valence PDF of the pion be symmetric about $x=1/2$, or at least peak there, for any value of $Q$? Is there any way to understand how DGLAP is "physically" affecting the valence PDF? I just want to understand the above PDF a little better.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Linked. Red alert : you are quoting the wrong graph! Take the right-hand column graphs of Fig. 16 of your source! They, of $xf^\pi_v(x)$, instead, quantify the fraction of the pion's momentum carried by the valance quarks, and peak at about 1/2, as opposed to 1/3 for baryons, etc... Like all parton distributions, they fuzz up and vary with energy. To grasp their meaning, you might start with baryons whose pdf's are better researched, understood, and interpreted.... $\endgroup$
    – Cosmas Zachos
    Commented Dec 29, 2021 at 17:27
  • $\begingroup$ @CosmasZachos I meant to quote the valence distribution itself, rather than the momentum fraction distribution, as the physical interpretation of it seems clearer to me. Regardless, how can you possibly say that the plots on the right of Fig. 16 (plots of $xf^{\pi}_v(x)$) peak at about 1/2? They definitely do not. The peak is at about $x\approx 0.35$, and the average is $\langle x \rangle \approx 0.22$. What I suspect is that for small Q we would get a symmetric valence distribution, but at large Q it somehow shifts to lower-x (despite the fact that the sea is factored out). $\endgroup$
    – Arturo don Juan
    Commented Dec 29, 2021 at 18:02
  • $\begingroup$ @CosmasZachos I agree with your point about first trying to understand the nucleon valence PDFs, that sounds like a great starting point. You mentioned that they fuzz up and vary with energy, and I would like to understand this a little better, especially how it applies to the valence distributions. A separate question that I have is how much the Q-dependence is affected by different renormalization schemes, and how this mismatch between schemes changes by including more terms in the perturbative expansion (e.g. of the corresponding DGLAP kernel). $\endgroup$
    – Arturo don Juan
    Commented Dec 29, 2021 at 18:09
  • 1
    $\begingroup$ The pion is the absolutely worst starting point for the parton model, as pions are pseudogldstone freak modes. The peakings are still close to 1/2 in this metaphorical picture of them: they are so broad! Remember, you have no "right" to the conventional-wisdom intuition: it is an informal privilege on a parameterization of an infinite-component quantum collectivity, helpful in orienting yourself. Maybe you could ask a question on it, in general, in the parton picture. $\endgroup$
    – Cosmas Zachos
    Commented Dec 29, 2021 at 18:52
  • 1
    $\begingroup$ Might start here. $\endgroup$
    – Cosmas Zachos
    Commented Jan 25, 2022 at 17:13

0

Reset to default