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- Homework Statement
- “Quarks carry only about half of the nucleon’s momentum.” Why do you integrate only over ## \int F_2^N(x), dx ## to get this result ?

- Relevant Equations
- Integral over structure function: ## \int F_2^N(x), dx ##, Callan-Gross relation: ## 2xF_1(x) = F_2(x)##

I study particle physics with “Particles and Nuclei” / Povh et al. and “Modern particle physics” / Mark Thomson and I am currently at “Deep-Inelastic scattering”. After introducing several scattering equations, such as Rosenbluth, that all include terms for electric AND magnetic scattering, i.e. momentum transfer, for comparing the contribution of quarks to the total momentum of the nucleon they integrate only the electric part, ##\int F_2^{pe}(x), dx ≈ 0.18## which scaled with factor 18/5 originating from the partial electric charges of the quarks gives ≈ 0.55, i.e. “about half of the nucleon’s momentum”.

Why can you omit F1 / magnetic contributions?

If I’d use the Callan-Gross relation ## 2xF_1(x) = F_2(x)## to calculate the ##F_1(x)## integral, since x <= 1 the contribution would exceed that of ##F_2(x)##.

Why can you omit F1 / magnetic contributions?

If I’d use the Callan-Gross relation ## 2xF_1(x) = F_2(x)## to calculate the ##F_1(x)## integral, since x <= 1 the contribution would exceed that of ##F_2(x)##.