# Why are sea quarks dominant only at low value of the Bjorken $x$ variable?

In electron proton deep inelastic scatterings, the sea quarks or gluon PDFs are only dominant at low value of $$x$$. Thomson explained this in his book Modern Particle Physics (see page 194) by arguing

...in reality the proton is a dynamic system where the strongly interacting quarks are constantly exchanging virtual gluons that can fluctuate into virtual $$q\bar{q}$$ pairs through [strong] processes...Because gluons with large momenta are suppressed by the $$1/q^2$$ gluon propagator, this sea of virtual quarks and antiquarks tend to be produced at low values of x.

I am not able to see what this has to do with the value of $$x$$. Does this mean when $$x$$ decreases, the four momentum $$q$$ of the gluon also decreases? If so, why is it the case?

So the Parton distribution functions are "essentially" the probability distributions (very suspicious how both are called pdf) of a probe particle to interact with a component inside your hadron. They depend both on the momentum at which we probe $$Q^2$$ and the fraction of momentum carried by said parton, the Bjorken x. They are mostly known through experiment as we have a poor understanding of non-perturbative QCD and the DGLAP equation governs the evolution ( and I'm sure more recent development I am not aware of).

So let's look at the first graph, this is the PDF of (you guess it) a proton. We see this because the up quarks peaks at around $$\sim 0.2$$ momentum fraction and has twice as much weight as the down quark. On the other hand, the gluon distribution kinda blows-up as you get close to $$0$$ momentum fraction (the graph has the gluon pdf divided by 10 near in mind!).
This actually makes sense because they are massless particles and it is much more easy to put a massless particle on shell than a massive particle. That's why massless particles tend to create particle showers in scattering experiments (EM showers and jets). Another way to think about this is that this is an IR (or soft) process and there exists a whole technology to resum these.
Same story for the sea quarks, they are quantum fluctuations away from "a hadron has 3 more quarks than anti-quarks", so their pdf is suppressed as well.

I hope I've given enough of an answer here, but really, you just need to be able to read these graphs and it will all make sense eventually. We could also discuss the relationship with $$Q^2$$ in the comments, there is also a nice physical interpretation there.

• Could you explain what you mean "put a massless particle on shell"? I am trying to understand why QCD processes like quark antiquark pair production are surpassed at high value of $x$. Jul 12 '20 at 9:07
• Right so maybe i should highlight that pdfs are the probability to interact with one parton inside a hadron. So if x is close to 1, you are asking a parton to carry most of the momentum of the hadron. Let's consider the first graph again which is relatively low hadron momentum. Then "inside" the hadron, at high x, we should only resolve the uud content. At low x though, we start to see the non-pertubative soup of gluons and sea quarks. Jul 12 '20 at 9:19
• I am making the analogy with the well-known behavior of soft photons and soft gluons, which are low momentum emissions of massless ptcle in scattering process (nothing to do with pdfs here). These soft emissions lead to IR divergences that need to be regularised. And I using this picture to justify the prevalence of low x gluons in the pdf. Now this analogy is limited but i think it gets the point across. Jul 12 '20 at 9:25
• Ultimately the details are super messy (and to be frank i don't know much beyond dglap) and any interpretation in terms of perturbative diagram is wrong. Well maybe not wrong, but incomplete at least. I have just a grad student's knowledge on this topic so maybe some else can jump in. Jul 12 '20 at 9:27
• Unfortunately, at my level I can't understand more involved examples such as soft photons or IR divergence. But I did find a simple explanation by Thomson himself in his slides. See if you can spot any logical mistakes in my answer. Jul 12 '20 at 11:23

My QCD knowledge is very limited - I haven't learnt it anywhere else other than in some qualitative discussion in Thomson's particle physics textbook. So I was only seeking a qualitative answer to aid my understanding in PDFs.

Thanks to @anna_v 's answer I realised I hadn't gone through Thomson's own slides of the chapter and the answer lies there all along.

If we consider the following quark-antiquark pair production strong process in an electron proton deep inelastic scattering experiment

Since the gluon is suppressed by the $$1/q^2$$ gluon propagator:

$$-i \frac{g_{\mu\nu}}{q^2} \delta^{ab}$$

the pair production is more likely to happen at low value of $$q$$. Now it is easy to see that the momentum of the pair produced sea quarks is just the momentum of the virtual gluon, so a low $$q$$ indeed indicates a low momentum fraction $$x$$ of the interacting sea quarks.

The slide can be found here under chapter 8 or in the first link in @anna_v 's answer.

• Yes this is what Thomson must have meant, I agree. And actually, here lies what I meant by "soft radiation", you need an arbitrarily small amount of momentum to produce a gluon (or any massless particle). What is interesting about qcd is that colour confinement forces them to split and create jets of particles. What I don't like about this interpretation is that, given a probing scale Q, if you take x to be low enough, you reach the scale at which perturbation theory breaks down (because qcd is asymptotically free). And this is the whole reason why we need PDFs in the first place! Jul 12 '20 at 12:27
• too many "q"s in total Jul 12 '20 at 13:25

I copy these definitions of the scattering variables"

I am not able to see what this has to do with the value of x. Does this mean when x decreases, the four momentum q of the gluon also decreases? If so, why is it the case?

The definition of x is dependent on the four momentum carried by the propagator, in the diagram small $$q$$ vector , which is connected with $$Q$$ . This should clear for you that the larger $$Q$$ by the four vector dot product of $$q$$ with itself. From the definition, large $$x$$ mean large $$Q$$ So from the definition the larger the momentum transfers to the gluon are suppressed.

This explains the gluon propagator statement you quote, but you should read the link to see how the parton distributions are found:

Ultimately the parton distribution functions are obtained from a fit to all experimental data including neutrino scattering

In this link , parton distribution functions are discussed, but as the other answer states, it is not a simple mathematically project.

• I don't think the $q$ in my question is the same momentum here in the slide. In the slide, $q$ is the momentum of the virtual photon, but in the passage I quoted from Thomson, $q$ is the momentum of the virtual gluon of a quark-antiquark pair production. Jul 12 '20 at 9:09
• The definition of the x variable plotted in the dpf splots and Q2 is fixed independent of the exchanged particle, it is the four momentum of the interaction . one would need the exact quote of where the statement is made. . Jul 12 '20 at 9:14
• I have updated the question including a longer quote and where you can find it. Jul 12 '20 at 9:19
• There are two issues here: 1)the parton plots , where low x means there was small momentum transfer to the proton, and high x where also the momentum transfer is large., i.e. there is a lot of momentum and energy to be distributed among the partons. 2) why the sea partons, gluons, quarks antoquarks , have a smaller probability of being at high x. (see plot in other answer) . The valence quarks get high probability.It seems to me the statement is a misstatement, but lets see whether there will be an answer that will defend the statement. Jul 12 '20 at 9:51
• Yes I agree with @anna, this is a strong statement to make. In my comments I make the analogy with soft radiation and I think that is what the author means by suppression by $1/q^2$. But I also think this is a flawed interpretation. Jul 12 '20 at 10:53