Quark charges - parton-distribution function - factor 5/18

Question

I got the following from Thomson’s Modern Particle Physics and some other sources.

The factor 5/18 distinguishing electron-DIS and neutrino scattering experiments is a basic proof for quarks and their charges. As I understand, you use isospin symmetry to set $u_n(x) = d_p(x)$, which should be only an approximation since their mass is not identical, and you neglect contributions from s- and sea quarks. Then you take the average of the structure function for the deuteron: $F_2^d = (F_2^p + F_2^n)/2$. The 5/18 results from (4/9 + 1/9)/2, i.e. essentially the sum of z^2, the square of the charges of u + d divided by 2.

If you’d average over all p and n quarks the sum for $z_i^2$ of p + n would be: (4/9+4/9+1/9 + 1/9+1/9+4/9) = 15/9 which in turn would give as average over 6 quarks: 15/(9*6) = 5/18, i.e. the same value. Is this coincidence? Wouldn’t that be the easier way with fewer assumptions?

Answer 1

It is not a coincidence that the two methods give the same answer, since assuming a deuteron has 3 $u$ and 3 $d$ quarks is just a specific case of assuming equal numbers of $u$ and $d$ quarks. Any equal number of $u$ and $d$ quarks will give the 5/18 factor.

Only considering valence quarks might give the same 5/18 factor as assuming isospin invariance, but only by ignoring much of the quark content of the deuteron. Both your lines of reasoning ignore $s$ and $\bar{s}$ sea quarks, but assuming a deuteron specifically has 3 up and 3 down quarks has an additional assumption. Either there is no contribution from any sea quarks, including $u$, $\bar{u}$, $d$ and $\bar{d}$ sea quarks, which is wrong, or any such contributions are equal, which means you can't avoid assuming isospin invariance.