How do particle scattering cross sections scale with energy in colliders? Particularly photons, electrons, protons, and gold or lead nucleii? (If necessary, break this into four separate questions.)

It is stated that due to the Heisenberg uncertainty principle, it takes more energy to measure a smaller distance and the inverse wavelength energy-wavelength relationship for photons is well known.
Therefore, I would conclude that the (total scattering) cross section for photons at least would vary as the inverse square of the energy.

On the other hand, general relativity says the the longitudinal dimension shrinks with increasing velocity, but the transverse dimensions do not.
Therefore, I would conclude that the cross section of say a gold atom in RHIC or a lead atom in the LHC would basically remain constant as the energy increased.

But how about the electron, supposedly a point particle?
Due to PEP and LEP, we should have good data on this. Looking at PDG figure 41.6,
I see lots of big spikes for resonant particles, but the underlying bacground is clearly a falling powerlaw that I think looks like it could be an inverse square law.

The proton cross sections in the next few figures, e.g. 41.7 and 41.11, show a falling elastic scattering cross section, but they also show an approximately constant total cross section.

What little i've found for RHIC supports the approximately constant cross section.

Therefore I conclude that the cross sections of photons and electrons (neglecting resonances) scale as the inverse square of the energy, but the cross sections of protons and nucleii are roughly constant with energy.

Comments or corrections?

So is it true that the photons and electrons are different from the protons and nucleii?

And why don't the protons scatter as three or six point particles (quarks and gluons) instead of one big blob?

  • $\begingroup$ Have a look at the measured crossections : pdg.lbl.gov/2010/reviews/rpp2010-rev-cross-section-plots.pdf . Crossections rise with energy because as dmckee said the more the cm energy the more particles offer a target, gluons and all $\endgroup$
    – anna v
    Jul 10, 2012 at 19:34
  • $\begingroup$ This is true--- this is the pomeron. Strong interaction cross sections increase slowly with energy, remaining roughly constant, while all electromagnetic cross sections fall off as a power of the energy. The analogy with GR is apropos, it's the AdS/CFT picture of the pomeron. $\endgroup$
    – Ron Maimon
    Jul 10, 2012 at 20:08

2 Answers 2


And why don't the protons scatter as three or six point particles (quarks and gluons) instead of one big blob?

They do. Many of the contributions to the inelastic process should be understood as parton scattering with subsequent hardonization, but we can't make parton beams or parton targets, we make nucleon and electron beams and targets so it makes sense to record the nucleon cross-sections.

  • $\begingroup$ So why does one point particle (the electron) have a falling cross section, and three point particles (the quarks in a proton) have a rising cross section? TIA for any reasonable explanation. $\endgroup$
    – Jim Graber
    Jun 11, 2012 at 11:26
  • $\begingroup$ @JimGraber It is a mistake to think of a nucleon as composed of three (or, indeed, any fixed number of) partons. The number you see is a function of the momentum transfer at which you probe the system. $\endgroup$ Jun 11, 2012 at 13:22

Your question covers numerous topics, but I will try to answer several of them here.

For a proton (I am most familiar with this case), the proton asymptotically becomes a black disk Ref. The disk part is because it is Lorentz contracted at the collision point. The same process happens for nuclei as well (see the bottom left of here for example). The "black" part has to do with absorption which is a different phenomenon.

Next, we note that the radius of the disk actually increases with energy. In particular, the Froissart bound says that the total cross section of the proton grows at most like $\sigma\propto\log^2s$ where $s$ is the total center of mass energy squared (the "squared" only changes a factor of $1/2$ in the coefficient). Since the proton is "black", this means that the cross section is (basically) $\pi r^2$ so the radius goes like $\log s$. Googling "arxiv" and "black disk" or "Froissart bound" will give you several papers on the subject. Current evidence supports $\sigma_{pp}\propto\log^2s$. The derivation of the Froissart bound is a bit tricky and there are numerous approaches out there (mostly containing a partial wave expansion approach). A good reference in Block and Cahn 1985 (behind paywall I believe) but there are many others. Check the pdg to see that it is well described by $\sigma_{pp}\propto\log^2s$ and that faster growth terms are disfavored and that slower growth without $\log^2s$ doesn't work either. Martin Block has written extensively on this as well (all over on the arxiv).

As for, say, the electron, the situation is a bit different. First, we note from the same pdg reference that $\sigma_{ee}\to\text{hadrons}$ and the ratio of that quantity to the cross section to muons is plotted. The cross section to muons is easy to calculate from Feynman diagrams. We note though, that there is no "total ee cross section" information like for protons. Let's consider a simple description: Rutherford scattering which considers electric interactions only. If we integrate the expression for $d\sigma/d\Omega$ over the solid angle we get $\infty$ - which of course makes sense: any two electrons passing each other will be deflected no matter how far away they are. The electric force is a long range force. That is why only $ee\to$ hadrons is considered because then the impact parameter needs to be small enough to create $W,Z$ bosons to make hadrons (or muons).

For $\gamma\gamma$ cross sections, see the same pdg reference above (towards the bottom).


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