Feynman diagram for ##\mu^+\mu^-## production in ##p\bar{p}## reaction

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In the discussion on Feynman diagrams for muon-antimuon production in proton-antiproton collisions, the main focus is on the interaction of quarks during the reaction. It is clarified that only two quarks from the proton and antiproton combine to produce a virtual photon, which then creates the muon pair. The fate of the remaining quarks is questioned, with the suggestion that they may form mesons, specifically whether they would result in two neutral pions or a charged and a neutral pion. The response indicates that both outcomes are likely, emphasizing the probabilistic nature of particle interactions. Overall, the conversation highlights the complexities of quark interactions in high-energy physics.
Nirmal Padwal
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Homework Statement
Remembering that helicity is conserved at high energies,
a) Draw a typical diagram for ##\mu^+ \mu^-##-pair production, with an invariant mass around 30 GeV, in unpolarised ##p\bar{p}## collisions.
b) Derive an expression for the angular distribution (with respect to the ##\bar{p}## direction) of ##\mu^+## in the ##\mu^+\mu^-##, centre-of-mass system.
(Hint: You will need to look up the appropriate ##d^j_{m'm}## rotation matrix elements)
Relevant Equations
1) ##d^1_{11} = \frac{1}{2}(1+\cos\theta)##
2) ##d^1_{-11} = \frac{1}{2}(1-\cos\theta)##
I was able to solve b) but I am confused for a). I understand that in the proton-antiproton collision, only two quarks (one from proton and other from anti-proton) can be combined to get a virtual photon that in turn creates muon and anti-muon. I don't understand what would happen to the other quarks? Since single quarks cannot exist independently, I think maybe they combine to form mesons. Is that correct? But which meson? If I take ##u## and ##\bar{u}## from ##p## and ##\bar{p}## respectively (please check the feynman diagram below), I am still left with ##u,d,\bar{u},\bar{d}##. Do they combine to give two ##\pi^0##s or ##\pi^+\pi^-##?
pp collision feynman diag.jpeg
 
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Nirmal Padwal said:
Homework Statement:: Remembering that helicity is conserved at high energies,
a) Draw a typical diagram for ##\mu^+ \mu^-##-pair production, with an invariant mass around 30 GeV, in unpolarised ##p\bar{p}## collisions.
b) Derive an expression for the angular distribution (with respect to the ##\bar{p}## direction) of ##\mu^+## in the ##\mu^+\mu^-##, centre-of-mass system.
(Hint: You will need to look up the appropriate ##d^j_{m'm}## rotation matrix elements)
Relevant Equations:: 1) ##d^1_{11} = \frac{1}{2}(1+\cos\theta)##
2) ##d^1_{-11} = \frac{1}{2}(1-\cos\theta)##

I was able to solve b) but I am confused for a). I understand that in the proton-antiproton collision, only two quarks (one from proton and other from anti-proton) can be combined to get a virtual photon that in turn creates muon and anti-muon. I don't understand what would happen to the other quarks? Since single quarks cannot exist independently, I think maybe they combine to form mesons. Is that correct? But which meson? If I take ##u## and ##\bar{u}## from ##p## and ##\bar{p}## respectively (please check the feynman diagram below), I am still left with ##u,d,\bar{u},\bar{d}##. Do they combine to give two ##\pi^0##s or ##\pi^+\pi^-##?
View attachment 323288
It's not so much a matter of which happens, it's a matter of which is more likely. I haven't checked the tables but I would suspect that both versions are about equally probable

-Dan
 
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...