4
$\begingroup$

Comparing the plots for the total (inelastic) cross sections as a function of the centre of mass energy for $pp$ and $e^+e^-$ collisions:

proton-proton cross section

electron-positron cross section

one notes that the trend at high energy is opposite: the $pp$ cross section increases while the $e^+e^-$ decreases. Is there a (simple) explanation for this?

$\endgroup$
4
  • 1
    $\begingroup$ Note that the electron cross-section, away from resonances, seems to be proportional to $1/\sqrt E$ (if center-of-mass energy $E$ is what's on that unlabeled horizontal axis). Also note that the e-e plot would fit only in the left half of the hadron plot, so maybe stuff starts happening again at higher energy. I don't know, though. $\endgroup$
    – rob
    Jan 29, 2016 at 0:01
  • $\begingroup$ @rob Yes, that's obviously energy [GeV], as easily inferred from the masses of the resonances. Regarding the stuff happening at higher energy: one expects resonances at the Higgs and at the Top masses, but there are no hints that the general trend should be altered. $\endgroup$
    – DarioP
    Jan 29, 2016 at 7:51
  • $\begingroup$ There's probably a simple interpretation for the $1/\sqrt E$ cross section in e-e. Thermal (milli-eV) neutrons have the same shape, and the usual explanation is that the cross section is proportional to inverse of the speed, or to the "dwell time" near a nucleus (but of course that particular argument doesn't hold for relativistic electrons). If you can predict the $1/\sqrt E$ cross section for electrons you should find yourself making an assumption that is broken for protons. $\endgroup$
    – rob
    Jan 29, 2016 at 19:13
  • $\begingroup$ Can someone give me a peer-reviewed reference of where the $e^+e^-$ cross section graph comes from? $\endgroup$
    – juacala
    Mar 5 at 13:58

1 Answer 1

1
$\begingroup$

There are two different answers: the growth of the PDF's, or the Regge trajectory of the Pomeron.

Roughly, the reason why the proton-proton cross section grows with Mandelstam $s$ is because the parton density functions (PDF's) of a proton, in particular it's gluon PDF, grows faster than $s$, thus outcompeting the natural fall off rate of $\frac{1}{s}$ of a two particle cross section. The growth of the electron PDF's on the other hand do not outcompete the $\frac{1}{s}$ falloff rate.

A completely different answer comes from a different framework for QCD called Regge kinematics. The high energy cross section of QCD is controlled by a reggeon called the pomeron. Because the Regge trajectory has a positive slope, the cross section rises. In a sense, this is just repeating the statement that the cross section rises, but one can calculate the slope of the Regge trajectory of the Pomeron (see the BFKL equation) and find that it is indeed positive.

If you don't know what a parton density function is: The parton density function of a hadron is a measure of how many particles it contains. You might complain that a proton only has three particles, two up quarks and a down, but this description is only accurate at low energies and is mainly useful for hadron spectroscopy (classifying the hadrons). When calculating a cross section in QCD, the framework is such that you presume that there is some likelihood of finding any particle inside of a proton, for example $P_{\gamma}(E_{\gamma})$ may denote the likelihood of finding a photon of energy $E$ inside of the proton. Without going into too many details, the intuition is obvious, the more particles inside of a hadron, the greater the cross section.

$\endgroup$
1
  • $\begingroup$ Interesting, maybe it would be useful to elaborate a bit more on how/why PDFs depend on $s$? $\endgroup$
    – DarioP
    Jan 4, 2021 at 20:55

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.