[MUSIC] In this video we will give you a few hints on how to construct a Feynman diagram for the two-body process a+b -> c+d. We first have to decide wether this can be an annihilation process in which the two initial state particles annihilate into a gauge boson. In that case the momentum transfer will be equal to the total energy-momentum. If they cannot annihilate, they have to exchange a gauge boson in a scattering process. In this case the q^2 is equal to the difference between the final particle momentum and the initial particle momentum. Now we must decide, which gauge boson can be exchanged. First of all we have to remember that at each vertex quantum numbers are strictly conserved. First of all baryon and lepton numbers must be conserved, and, if we are not talking about weak interactions, also the particle flavor. Then we must consider the charge at each vertex. If the electric charge is not 0 and it does not change at the vertex, we can exchange a photon. If the particle’s weak isospin is not 0 and the charge does not change, we can exchange a Z. If the charge does change by ±1 unit, we can exchange a W±. Finally, if the color is not 0 and it changes at the vertex, we can exchange a gluon. So here is a first example, the process e+e- -> u ubar. We clearly cannot couple the electron directly to a u quark so this is an annihilation process in which the e+ e- annihilate into a u bar pair. This can proceed via the exchange of a photon or a Z, since both the electron and the u quark carry both electric and weak isospin charge. It cannot go via the exchange of a gluon or a W because the e+ e- does not carry color and the charge does not change at the vertex. A second example is the process e+ e- -> nu unbar. Unless the neutrinos are electron-type neutrinos, this is again an annihilation process. It proceeds via the exchange of a Z boson. It cannot proceed via photon exchange, because the neutrino does not carry electric charge. Neither the electron nor the neutrino are colored, so gluons cannot be exchanged. The electric charge is not changing at the vertex, so W± exchange is also excluded. The third example is the process e- u -> nu_e d. In this case it must be a scattering process, since the e- cannot directly couple to the u quark. They must in fact exchange a W boson since the charge changes by 1 unit at each vertex. This is possible because both the e- and the u quark carry weak isospin. They cannot exchange a photon since the charge changes neither can they exchange a Z. And since the e- and the nu do not carry color, they cannot exchange a gluon either. The next example is the decay of a d quark into a u e- and an electron-antineutrino. To conserve baryon and lepton number we must connect the d quark to the u quark and the electron to the antineutrino. This means that at each vertex, the charge changes by 1 unit and we can exchange a W boson. We cannot exchange a photon because the charge changes, likewise we cannot exchange a Z. And since the e- and the nu don’t carry color we cannot exchange a gluon. The last example is a red d quark scattering off a blue d quark, giving a blue u quark and a red d quark. This is clearly a strong process since it involves color exchange. Since strong interactions conserve flavor, we must connect the two u quarks and also the two d quarks. That means that the two vertices are connected by a gluon of red-antiblue color such that the color is conserved at each vertex. We cannot exchange a photon since the color changes, neither a Z or a W bosons since all of these cannot change the color of a particle. [MUSIC]