Integration of structure function F2 to calculate quark momentum

[qatch]

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 $F_2(x)$.

 

[member 553083]

The reason why the authors of your textbooks omit the magnetic contributions when calculating the momentum of the nucleon is because they are assuming a simple quark model of the nucleon, in which the quarks have only electric charge. In this scenario, the magnetic contributions from the quarks would be zero, so integrating only the electric part (F2) will give an accurate result for the total momentum of the nucleon.However, if you were to use the Callan-Gross relation to calculate F1, then you would need to take into account the magnetic contributions from the quarks. This would result in a greater contribution from the quarks to the total momentum of the nucleon than what was calculated using only F2. Therefore, it is important to keep in mind that the assumptions made about the quark model can have a significant impact on the results obtained.