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High Energy, Nuclear, Particle Physics
Cross section from subgraph in electron-proton DIS
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[QUOTE="Reggid, post: 6258998, member: 654378"] The leptonic part is not thrown out, and the amplitude is not squared in the way you write it out here. You look at the process ##e+P\to e+X##. Let's denote the momentum of the electron before and after the scattering by ##k## and ##k^\prime##, the proton momentum by ##P## and the momentum of the hadronic final state as ##P_X##, and define the invariants [tex]Q^2=-q^2[/tex] [tex]x_B=\frac{-q^2}{2P\cdot q}[/tex] [tex]s=(P+k)^2[/tex] with ##q=k-k^\prime##. Then the differential cross section is [tex]\frac{\mathrm{d}\sigma}{\mathrm{d}x_B\mathrm{d}Q^2}=\frac{1}{(4\pi)^2}\frac{Q^2}{s^2x_B^2}\sum_X\mathrm{d}\Pi_X(2\pi)^4\delta^{(4)}(P+q-P_X)|\mathcal{M}|^2[/tex] The amplitude ##\mathcal{M}## to leading order in QED is [tex]\mathcal{M}(eP\to eX)=\bar{u}(k^\prime)(-ie\gamma_\mu)\frac{-i}{q^2}(ie)\langle X|J^{\mu}(0)|P\rangle[/tex] with the electromagnetic current of the quark fields ##q_i## of all possible flavors ##i## with charge quantum number ##\mathcal{Q}_i## [tex]J^{\mu}(x)=\sum_i\mathcal{Q}_i\bar{q}_i(x)\gamma^{\mu}q_i(x)[/tex] Then the spin-averaged square of the amplitude is [tex]|\mathcal{M}|^2_{\rm{spin-av.}}=\frac{32\pi^2\alpha^2}{Q^4}L_{\mu\nu}\langle P|J^{\mu}(0)|X\rangle\langle X|J^{\nu}(0)|P\rangle[/tex] where the leptonic tensor is defined as [tex]L_{\mu\nu}=k_\mu k^\prime_{\nu}+k_{\nu}k^{\prime}_{\mu}-k\cdot k^{\prime}g_{\mu\nu}[/tex] You see that the leptonic side is of course not thrown away. When you define the hadronic tensor ##W^{\mu\nu}## as [tex]W^{\mu\nu}=\frac{1}{4\pi}\sum_X\mathrm{d}\Pi_X(2\pi)^4\delta^{(4)}(P+q-P_X)\langle P|J^{\mu}(0)|X\rangle\langle X|J^{\nu}(0)|P\rangle[/tex] then the cross section can be written in the simple form [tex]\frac{\mathrm{d}\sigma}{\mathrm{d}x_B\mathrm{d}Q^2}=\frac{4\pi\alpha^2}{Q^2s^2x_B^2}L_{\mu\nu}W^{\mu\nu}[/tex] The hadronic tensor can be also written in the form [tex]W^{\mu\nu}=\frac{1}{4\pi}\int\mathrm{d}^4x\mathrm{e}^{ixq}\langle P|J^{\mu}(x)J^{\nu}(0)|P\rangle[/tex] (which is equivalent to the definition above) and that is the object for which you can then calculate QCD corrections and which you can factorized into the PDF and the hard Wilson coefficient. [/QUOTE]
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High Energy, Nuclear, Particle Physics
Cross section from subgraph in electron-proton DIS
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