Parton distribution function ($q_f(x),{\bar{q}}_f(x)$)

A density of partons with a given flavour and momentum fraction at a specified parton factorization scale. It is a number density, not generally a distribution normalized to one.

A nuclear parton distribution describes parton momentum fractions in a bound nuclear target. Its normalization must specify whether it is per nucleus or per nucleon and how $x$ is defined. Nuclear binding, shadowing and other nuclear modifications mean it need not equal a simple sum of free-nucleon distributions.

In the collinear approximation, a parton carries a fraction $0<x<1$ of its parent hadron's longitudinal momentum. A parton distribution function $f_i(x,Q^2)$ gives its distribution at the specified factorization scale. For production of one on-shell particle with no transverse recoil, $x_{+}{x}_{-}s=M^2$ and $x_{\pm }=(M/\sqrt{s})e^{\pm y}$, where $y$ is its rapidity.

For charged current production $u\bar{d}\to W^{+}$, charge conjugation of the antiproton distributions gives $u(x_{+})d(x_{-})+\bar{d}(x_{+})\bar{u}(x_{-})$. With a common sea quark distribution $S$, the valence quark terms are $u_v{d}_v$, the mixed terms are $u_{v}S+S{d}_v$, and the two sea assignments give $2S(x_{+})S(x_{-})$. Omitting the second beam assignment undercounts the pure sea contribution by two.

The approximate equality of proton and neutron parton distribution functions after interchanging up quarks and down quarks, including their antiquarks. It follows from the corresponding isospin symmetry when electromagnetic and light-quark mass-difference effects are neglected. Isospin-singlet sea distributions are unchanged.

The parton distribution function is a number density. Integrating the quark distribution minus the antiquark distribution counts the net flavour charge of the hadron, since every sea quark-antiquark pair contributes zero. The corresponding flavour-vector conserved current counts particles minus antiparticles. For a proton, the net up/down numbers are two and one; for a neutron they are one and two. The rule fixes the difference of distributions, not their sum, and does not imply absence of a sea.

With only up/down valence quarks and no antiquarks, the quark number sum rule and leading parton model expression give the charge-squared number moment $\int F_2(x)dx/x={\sum }_f{Q}_f^2{N}_f$. The squared charges are $4/9$ for up and $1/9$ for down, so the proton has $8/9+1/9=1$ and the neutron has $4/9+2/9=2/3$. Restricting the flavour list to up/down does not itself remove antiquarks; the no-sea assumption is essential for these individual moments.

The scale at which collinear contributions are assigned to parton distribution functions rather than the hard scattering coefficient. The separate factors depend on this scale, while a consistently calculated physical differential scattering cross-section does not.