Are Colour Degrees of Freedom Relevant in e+e- Annihilation Cross Section?

[genloz]

Homework Statement

Describe how the colour degree of freedom of the quarks can be used to explain the fact that, at electron energies of about 20GeV the cross section for e+e- annihilation into hadrons is of the order 4 times the cross section for e+e- -> mu+mu-.

Homework Equations

None given

The Attempt at a Solution

But I thought that colour degrees of freedom related to hadrons, not leptons, so how does it fit into an electron positron annihilation.

 

[malawi_glenn]

in e+ e- annihilation you get high energy photon, that photon can pair produce any kind of particle that interact via EM force. For each pair of quarks you have 3 colours, so you expect to get more quarks then leptons at that energy. Also this is strenghten becase quakrs also do not carry one unit elementary charge, but 1/3 or 2/3.

 

[genloz]

Thanks for that... So because the energy from the photon is high, you expect to get quarks rather than leptons? I think I'm missing the logical step from your first to your second sentence. Wouldn't it be harder to create a hadron because it involves 3 appropriate quarks combined? Rather than a single particle, a lepton?

I guess that this possible relates to something I read somewhere that I also didn't quite understand... that even though proton/antiproton collisions have the same energy available for new particles as proton/proton collisions, the particles in the annihilation have much more energy than the particles in the proton/proton collision... I'm a bit unclear on the why of that as well...

 

[nrqed]

Thanks for that... So because the energy from the photon is high, you expect to get quarks rather than leptons? I think I'm missing the logical step from your first to your second sentence. Wouldn't it be harder to create a hadron because it involves 3 appropriate quarks combined? Rather than a single particle, a lepton?

he first step of the reaction is that a photon is converted into a single quark-antiquark pair. Then there is a complicated hadronization process leading finally to the hadrons observed.

Yes, it is harder to create hadrons than a single lepton-antilepton pair (there is less phase space available0 BUT fi the energy ismuch larger than the rest mass of the things produced, the calculation becomes a matter of counting degrees of freedom. This si why they mention such a large energy (

compared to the rest mass of electrons, pions, etc)

I guess that this possible relates to something I read somewhere that I also didn't quite understand... that even though proton/antiproton collisions have the same energy available for new particles as proton/proton collisions, the particles in the annihilation have much more energy than the particles in the proton/proton collision... I'm a bit unclear on the why of that as well...

In the p pbar annihilation, all the energy is available to create new particles.

 

[malawi_glenn]

No I just said that you have high energy, so you can produce particles heavier than electrons, muons and quarks for example.

And as nrgred said, at these very high energies, cross sections are much more related to degrees of freedom. A quark can have 3 different colours, I am not talking about hadrons now, just that a quark can have 3 more degrees of freedom than a lepton. Therefore the cross section for making quarks are more probable. Imagine this: you have four options, and three of them are related to another option. You can get a cat, a shark a baracuda or a pike, and all are equal in probability. But since shark, baracuda and pike are all fishes, it is more probable that you get a fish, than a cat.

The process of hadronizations comes afterwards..as nrged said.

 

[genloz]

That's clear and makes a lot of sense... thanks... are there any formulas show that it is so in a numerical manner?

 

[malawi_glenn]

yes there are formulas, but I can't write them down for you: See "particles and nuclei" by Povh, page 123 and forward. Or books at your library, or wait til someone with more available time will answer.