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High Energy, Nuclear, Particle Physics
Cross section from subgraph in electron-proton DIS
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[QUOTE="Reggid, post: 6261271, member: 654378"] Well, for the cross section you just start from the normal ##1+2\to 3+4+...## cross section that you find in every textbook, and just separate the phase space integration of the outgoing electron with momentum ##\vec{k}^\prime## from the hadronic final state particles (and set the masses of the proton and the electron to zero). [tex]\mathrm{d}\sigma=\frac{1}{2s}\frac{\mathrm{d}^3k^\prime}{(2\pi)^3}\frac{1}{2|\vec{k}^\prime|}\sum_X\mathrm{d}\Pi_X(2\pi)^4\delta^{(4)}(P+q-P_X)|\mathcal{M}|^2[/tex] Then you can derive the relation [tex]\frac{\mathrm{d}^3k^\prime}{(2\pi)^3}\frac{1}{2|\vec{k}^\prime|}=\frac{\mathrm{d}x_B\mathrm{d}Q^2}{(4\pi)^2}\frac{Q^2}{x_B^2s}[/tex] and arrive at the cross section given in my first post. For the hadronic tensor you can use [tex](2\pi)^4\delta^{(4)}(P+q-P_X)=\int\mathrm{d}^4x\,\mathrm{e}^{i(P+q-P_X)x}[/tex] the fact that ##|X\rangle## and ##|P\rangle## are eigenstates of the momentum operator ##\hat{P}##, the fact that the momentum operator is the generator of spatial translation, i.e. [tex]\mathrm{e}^{i\hat{P}x}J^{\mu}(0)\mathrm{e}^{-i\hat{P}x}=J^{\mu}(x)[/tex] and the completeness relation [tex]\sum_X\mathrm{d}\Pi_X\,|X\rangle\langle X|=\mathbf{1}[/tex] [/QUOTE]
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High Energy, Nuclear, Particle Physics
Cross section from subgraph in electron-proton DIS
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